Genetics - Analysis of Genes and Genomes, 9th Edition

Copyright © 2017. Jones & Bartlett Learning, LLC. Todos os direitos reservados. Hartl, Daniel L. e Bruce Cochrane. Genetics: Analysis of Genes and Genomes: Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Criado por Utah em 2021-08-09 18:44:16.

NINTH EDITION

GENETIC ANALYSIS OF GENES AND GENOMES

Daniel L. Hartl, PhD Harvard University Cambridge, Massachusetts

Bruce J. Cochrane, PhD

University of Miami Oxford, Ohio

JONES & BARTLETT LEARN

Hartl, Daniel L. e Bruce Cochrane. Genetics: Analysis of Genes and Genomes: Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Erstellt von utah am 2021-08-09 18:44:47.

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Summary Unit I Definition and work with GENES

1 Genes, Genomes and Genetic Analysis 2 DNA Structure and Genetic Variation Unit II Transmission Genetics

3 Mendelian Genetics: The Principles of Segregation and Assortment

4 The chromosomal basis of inheritance 5 Genetic linkage and chromosome mapping 6 Human karyotypes and chromosomal behavior 7 The genetic basis of complex traits 8 Genetics of bacteria and their viruses

Unit III Organization and replication of chromosomes and DNA

9 Molecular organization of chromosomes and genomes

10 DNA Replication and Sequencing 11 Mutation, Repair, and Recombination Unit IV Gene Expression

12 Molecular Biology of Gene Expression 13 Molecular Mechanisms of Gene Regulation

14 Manipulation of Genes and Genomes 15 Genetic Control of Development 16 Molecular Genetics of the Cell Cycle and V Unit Variation of Cancer

17 Mitochondrial DNA and extranuclear inheritance 18 Genes in Populations 19 Molecular and Human Evolutionary Genetics

Contents Preface The Student Experience What's New in the Ninth Edition Resources Acknowledgments About the Authors

Unit I Definition and work with GENES

1 Genes, Genomes, and Genetic Analysis Learning Objectives and Science Skills 1.1 DNA as Genetic Material Experimental Evidence of the Genetic Function of DNA Genetic Role of DNA in Bacteriophage 1.2 Structure and Replication of DNA

Overview of DNA Replication 1.3 Genes and Proteins Inborn Errors of Metabolism as a Cause of Hereditary Diseases 1.4 Gene Analysis Mutant Genes and Defective Proteins Complementation Testing for Mutations in the Same Gene Complement Data Analysis Other Applications of Gene Analysis 1.5 Gene Expression: a central transcription of the dogma

Translation The Genetic Code 1.6 Mutation and Variation Variation in Populations 1.7 Genes and Environment 1.8 The Molecular Unit of Life Procarian, Archaeal, and Eucarian Genomes and Proteomes ROOTS OF THE DISCOVERY: Black Urine Disease ROOTS OF THE DISCOVERY: One Gene, One Enzyme

2 DNA Structure and Genetic Variation Learning Objectives and Science Skills 2.1 Genetic Differences Between Individuals DNA Markers as Landmarks on Chromosomes 2.2 The Terminology of Genetic Analysis

2.3 The molecular structure of the polynucleotide chains of DNA The double helix Base pairing and base stacking Antiparallel strand DNA structure in terms of function 2.4 The separation and identification of genomic DNA fragments Restriction enzymes and site-specific DNA cleavage Electrophoresis in gel Nucleic acid hybridization 2.5 Amplification of specific DNA DNA for detection and purification of DNA replication: primers and 5' to 3' strand extension

The Polymerase Chain Reaction 2.6 Types of DNA Markers Present in Genomic DNA Restriction Fragment Length Polymorphisms Single Nucleotide Polymorphisms Tandem Repeat Polymorphisms Copy Number Variation 2.7 Applications of DNA Markers Genetic Markers, Gene Mapping, and " disease genes" Other uses for DNA markers ROOTS OF DISCOVERY: The double helix THE CUTTING EDGE: High-throughput SNP genotyping

Unit II Genetic Transmission

3 Mendelian genetics: the principles of segregation and assortment Learning outcomes and scientific competences 3.1 Morphological and molecular phenotypes 3.2 Single gene segregation

Phenotypic relationships in the F2 generation The principle of segregation Verification of segregation Testing of crossbreeding and backcrossing 3.3 Segregation of two or more genes The principle of independent selection Testing of crossbreeding with independent selection of genes Three or more genes 3.4 Analysis of the human pedigree Characteristics of the dominant and recessive inheritance Most human genetic variations are not "bad" molecular markers in human pedigrees

3.5 Incomplete Dominance and Epistasis Multiple Alleles Human ABO Blood Groups Epistasis 3.6 Probability in Genetic Analysis Elementary Outcomes and Events Probability of Merging Events Probability of Overlapping Events 3.7 Conditional Probability and Pedigrees Bayes Theorem ROOTS OF DISCOVERY: What did Gregor Mendel think? He found out? ROOTS OF DISCOVERY: This land is your land

4 The Chromosomal Basis of Inheritance Learning Outcomes and Science Skills 4.1 The Stability of Chromosome Complements 4.2 Mitosis

Prophase Metaphase Anaphase Telophase 4.3 Meiosis The First Meiotic Division: Reduction The Second Meiotic Division: Equation 4.4 Sex-Linked Inheritance Chromosomal Sex Determination X-Linked Inheritance Genealogical Characteristics of X-Linked Human Inheritance Heterogametic Females

Nondisjunction as proof of the chromosomal theory of inheritance Sex determination in Drosophila 4.5 Probability in predicting distributions of offspring using the binomial distribution in genetics Importance of the binomial coefficient 4.6 Testing the goodness of fit to a genetic hypothesis Random variables and distributions The method chi-square Is Mendel's data too good to be true? DISCOVERY ROOTS: Grasshopper, grasshopper DISCOVERY ROOTS: The man with the white eyes

5 Genetic linkage and chromosome mapping Learning outcomes and science skills 5.1 Linkage and recombination of genes on a chromosome

Linkage versus rejection of synthetic alleles The chi-square test for linkage Each pair of linked genes has a characteristic frequency of recombination Recombination in females versus males 5.2 Genetic mapping Distance and frequency of recombination Crossing over Recombination between genes results from a physical exchange between chromosomes Crossover occurs at the four-strand stage of meiosis rather than multiple matings 5.3 Genetic mapping in a three-point test mating Chromosomal interference in double matings

Characteristics of genetic mapping Genetic mapping Distance and physical distance 5.4 Mapping by tetrad analysis Disordered tetrads Ordered tetrads 5.5 Human genetic mapping Derived links in pedigrees Association mapping 5.6 Peculiarities of recombination Recombination within genes Mitotic recombination ROOTS OF DISCOVERY: Genes All in one line ROOTS OF DISCOVERY: Mapping markers in the human genome

6 Human Karyotypes and Chromosomal Behavior Learning Outcomes and Scientific Skills

6.1 The human karyotype Standard karyotypes Centromere and chromosomal stability Dose compensation of X-linked genes The tabby cat Pseudoautosomal inheritance Active genes on the “inactive” X chromosome Gene content and evolution of the Y chromosome 6.2 Chromosomal abnormalities in human pregnancies Down syndrome and other viable trisomies Trisomal segregation Sex chromosome abnormalities Environmental influences on nondisjunction 6.3 Chromosomal deletions and duplications

Deletions Mapping of deletions Duplications Uneven crossing over in human red-green color blindness Copy number variation with reciprocal risks of autism and schizophrenia Gene duplication and evolution of new proteins 6.4 Genetics of chromosomal inversions Paracentric inversion (without the centromere the centromere) 6.5 Translocations Chromosomal Reciprocal translocations Pseudolinkage in heterozygotes Robertsonian translocations and trisomy 21 translocation complexes in Oenothera 6.6 Effects of genomic position on gene expression 6.7 Polyploidy in plant evolution

Sexual Versus Asexual Polyploidization Autopolyploid and Allopolyploid Monoploid Organisms ROOT DISCOVERY: Lyonization of an X Chromosome ROOT DISCOVERY: The first identified human chromosomal disorder

7 The genetic basis of complex traits Learning objectives and scientific competences 7.1 Complex traits Continuous, categorical, and threshold traits The normal distribution Correspondence of Mendelian inheritance to continuous traits

7.2 Causes of variation Genotypic variation Environmental variation Genetics and environment Genotype-environment combination interaction and association 7.3 Heritability The number of genes that affect complex traits Strict sense heritability Strict sense heritability Twin studies 7.4 Correlation between relatives Covariance and correlation The graphical meaning of the Correlation 7.5 Heritability and selection Phenotypic change with individual selection: a predictive equation Long-term artificial selection Heritability of threshold traits 7.6 Misconceptions about heritability

7.7 Identification of Genes Affecting Complex Traits Linkage Analysis in Quantitative Trait Loci Gene Mapping Genome-Wide Association Studies Candidate Genes for Complex Traits ROOT DISCOVERY: One Corn and One Grass THE TECHNOLOGY: Crowd-Sourced Genomics

8 Genetics of bacteria and their viruses Learning objectives and science skills 8.1 Plasmids and genetic exchange The F plasmid: A conjugated plasmid

8.2 Bacterial genetics Mutant phenotypes Isolation of auxotrophs Mechanisms of gene exchange 8.3 DNA-mediated transformation 8.4 Gene exchange and conjugation Cointegrated formation and Hfr cells Mapping the time of entry of F' plasmids 8.5 Transduction The phage lytic cycle Generalized transduction 8.6 Genetics of bacteriophages Formation of plaques and phage mutants Genetic recombination in the lytic cycle Fine structure of the rII gene in bacteriophage T4 8.7 Lysogenesis and specialized transduction Site-specific recombination and lysogenesis Specialized transduction ROOTS OF THE DISCOVERY: The sex life of bacteria ROOTS OF THE DISCOVERY: Artoo

THE CUTTING EDGE: Survive in a hostile world

Unit III Organization and replication of chromosomes and DNA

9 Molecular Organization of Chromosomes and Genomes Learning Outcomes and Scientific Skills 9.1 Genome Size and Evolutionary Complexity: The CV Value Paradox 9.2 DNA Supercoiling

Topoisomerase enzymes 9.3 The structure of bacterial genomes Mobile DNA in prokaryotes Mobilization of non-conjugative plasmids 9.4 The structure of eukaryotic genomes The nucleosome: the structural unit of chromatin The core particle of the nucleosome Chromosomal areas in the nucleus Chromosomal condensation Polytene chromosomes 9.5 Analysis of sequence complexity in eukaryotic genomes

Kinetics of DNA renaturation Analysis of genome size and repetitive sequences by renaturation kinetics 9.6 Unique and repetitive sequences in eukaryotic genomes Unique sequences Highly repetitive sequences Medium repeat sequences and transposable elements Mechanisms of molecular transposition Transposable elements in the human genome 9.7 Molecular structure of the human genome centromere 9.8 Molecular structure of telomeres Telomere length limited the number of cell duplicates ROOT DISCOVERY: Telomeres: the beginning of the end

10 Learning Objectives and Science Skills of DNA Replication and Sequencing

10.1 Problems with initiation, elongation, and incorporation errors 10.2 Semiconservative replication of double-stranded DNA The Meselson-Stahl experiment Theta replication of circular DNA molecules Rolling circle replication 10.3 Initiation of DNA replication Unwinding, stabilization, and stress relief Formation of DNA Primosomal complex 10.4 The process of elongation and correction Delayed discontinuous strand replication The joining of precursor fragments 10.5 Terminating replication 10.6 Replication in eukaryotes

Semi-Conservative DNA Replication in Chromosomes Multiple Origination and Bidirectional Replication in Eukaryotes Initiation Termination and the Telomere Problem 10.7 Exploring the System: DNA Sequencing Sanger Sequencing Next-Generation Sequencing ROOTS OF DISCOVERY: Replication by Moieties

11 Mutation, repair and recombination Learning outcomes and science skills 11.1 Types of mutations Germline mutations and somatic mutations Conditional mutations

Classification by function 11.2 The molecular basis of mutation Nucleotide substitutions Missense mutations: the sickle cell example Insertions, deletions, and frameshift mutations Dynamic mutation of trinucleotide repeats Cytosine methylation and gene silencing 11.3 Transposable elements as agents of mutation 11.4 Spontaneous mutation A non-adaptive nature of the mutation Estimation of mutation rates Mutation hotspots 11.5 Mutagens

Depurination Oxidation Base analogues Mutage Chemical agents that modify DNA Ultraviolet radiation Ionizing radiation Genetic effects of the Chernobyl nuclear accident Repair system 11.7 Reverse mutations and suppressor mutations Intragene suppression

Intergenic suppression The Ames test for the detection of mutagens/carcinogens 11.8 The relationship between repair and recombination Double-strand breakage and the repair model Recombination hotspots ROOTS OF THE DISCOVERY: X-Ray Daze

Unit IV Gene Expression

12 Molecular Biology of Gene Expression Learning Outcomes and Science Skills 12.1 Amino Acids, Polypeptides, and Proteins 12.2 Collinearity Between Coding Sequences and Polypeptides 12.3 Overview of Transcription 12.4 Transcription in Prokaryotes

Initiation Genetic evidence for promoters End of elongation 12.5 Transcription in eukaryotic RNA polymerases Promoters and promoter recognition Initiation End of elongation 12.6 Messenger RNA and RNA processing 5'-capping and 3'-polyadenylation Splicing of intervening sequences Characteristics of human transcripts Coupling of transcription and RNA processing mechanism of RNA splicing

Effects of intron mutations Exon shuffling on the emergence of new genes 12.7 Initiation of protein synthesis in prokaryotes Initiation in eukaryotes Elongation release Translational revision and premature termination Senseless mediated decay Polysomes 12.8 Protein folding and chaperones 12.9 The standard genetic code Genetic evidence for a triplet code How the code was broken features of the standard code of transfer RNA and aminoacyl-tRNA synthetase enzymes redundancy and wobble suppression of nonsense ROOTS OF DISCOVERY: Messenger "Light"

ROOTS OF THE DISCOVERY: Poly-U

13 Molecular Mechanisms of Gene Regulation Learning Objectives and Science Skills 13.1 Transcriptional Regulation in Prokaryotes Inducible and Repressible Systems of Negative Regulation Positive Regulation Stochastic Noise in Gene Expression 13.2 The Operon System of Gene Regulation Lac-mutants

Inducible and constitutive synthesis and repression Repressors, operators, and promoters The operon system of transcription regulation Stochastic noise in Lac expression Upregulation of the lactose operon Regulation of the tryptophan operon 13.3 Regulation by transcription termination Attenuation riboswitches 13.4 Regulation in the bacteriophage lambda 13.5 Regulation of transcription transcription in eukaryotes

Galactose metabolism in yeast transcription activating proteins Transcription enhancers and transcription silencers Deletion scanning The eukaryotic transcription complex Chromatin remodeling complexes Alternative promoters 13.6 Epigenetic mechanisms of transcription regulation Cytosine methylation Transcription methylation and inactivation Genomic imprinting in the female and male germ line 13.7 RNA regulation - Processing and decay Alternative splicing Stability of messenger RNA 13.8 Non-coding RNAs and regulatory interfering RNAs Long non-coding RNA 13.9 Translational control Regulatory RNAs that control translation

13.10 Programmed DNA Rearrangements Gene Amplification Antibody and T-Cell Receptor Variability Mating Type Interconversion Transcriptional Control of Mating Type ROOTS OF THE DISCOVERY: Operator? Operator? ROOTS OF THE DISCOVERY: Dual Trouble

14 Manipulation of genes and genomes Learning objectives and science skills 14.1 Specific DNA cleavage and cloning vectors Production of DNA fragments with defined ends Recombinant DNA molecules Plasmid, lambda, and cosmid vectors 14.2 Cloning strategies Connected DNA fragments Insertion of a specific DNA molecule in a vector The use of reverse transcriptase: cDNA and RT-PCR 14.3 Detection of recombinant molecules

Inactivation of genes in the vector Screening molecules for specific recombinants 14.4 Genomics Genomic sequencing Human genome sequencing Genome annotation Comparative genomics 14.5 Functional genomics Chromatin immunoprecipitation Two-hybrid analysis of protein-protein interactions 14.6 Transgenic organisms Germline transformation in animals

Genetic Engineering in Plants Transformational Rescue Mutagenesis and Mutations Knockout Knockdown Gene Expression with RNAi 14.7 Genetic Editing CRISPR–Cas9 in Practice 14.8 Some Applications of Genetic Engineering Giant Salmon with Artificial Growth Hormone Nutritionally Modified Rice Production of Beneficial Proteins Genetic Engineering with Animal Viruses DA ROOT DISCOVERY: Hi Dolly! THE CUTTING EDGE: Muscular Dystrophy Gene Editing

15 Genetic Control of Development Learning Objectives and Science Skills 15.1 Genetic Determinants of Development

15.2 Early embryonic development in animals Autonomic development and intercellular signaling Oocyte composition and organization Early development and activation of the zygotic genome 15.3 Genetic analysis of development in nematode cell lineage analysis Mutations affecting cell lineages Programmed cell death Loss of function and gain of function Epistasis alleles in the analysis of developmental switches 15.4 Genetic control of development in Drosophila maternal effect genes and zygotic genes

Genetic basis of pattern formation in early development Coordinate genes Gap genes Pair rule genes Segment polarity genes Interactions in the regulatory hierarchy Adult fly metamorphosis Homeotic genes Master control genes in evolution 15.5 Regulatory RNAs in development MicroRNA-based regulation in C. elegans miRNAs in the regulation of the HOX gene Long intergenic non-coding RNAs 15.6 Genetic control of development in higher plants Flower development in Arabidopsis Combinatorial determination of floral organs ROOTS OF DISCOVERY: Distinct Lineages ROOTS OF DISCOVERY: Embryogenesis

16.1 The Cell Cycle Major Cell Cycle Events Transcriptional Program of the Cell Cycle 16.2 Genetic Analysis of Cell Cycle Mutations Affecting Cell Cycle Progression 16.3 Cell Cycle Progression Cyclins and Cyclin-Dependent Protein Kinases Targets of Cyclin-CDK G1/S Complexes and G2 triggers /M transitions

Protein degradation helps regulate the cell cycle 16.4 Checkpoints in the cell cycle DNA damage checkpoint Centrosome duplication checkpoint Spindle assembly checkpoint Genetics of acute leukemias 16.8 The genomics and transcriptomics of cancer Cancer cancer: the mutational landscape Pancreatic cancer: the transcriptional landscape From genes and transcripts to therapy? ROOTS OF DISCOVERY: Cycle-Ops ROOTS OF DISCOVERY: Two hits, two misses

Variation of Unit V

Learning outcomes and scientific competences 17.1 Origin and molecular genetics of organelles Genomes of organelles RNA editing The genetic codes of organelles 17.2 Patterns of extranuclear inheritance Maternal inheritance of animal mitochondria Maternal inheritance and maternal effects Heteroplasmy Mitochondrial genetic diseases

Tracing population history by mitochondrial DNA Cytoplasmic male sterility in plants 17.3 Vegetative segregation Leaf variegation in four-hour plants Respiratory defects Mitochondrial mutants in yeast 17.4 Cytoplasmic transmission of symbionts Symbiont bacteria of aphids Killer strains of Paramecium Wolbachia in arthropods 17.5 Maternal action in snails involute BREAKTHROUGH ROOTS Gathering BREAKTHROUGH ROOTS: How Aphids Acquired PVT

18 Genes in Populations Learning Outcomes and Science Skills

18.1 Population genetics Allele frequencies and genotype frequencies Random mating and the Hardy-Weinberg principle Implications of the Hardy-Weinberg principle A test for the random mating frequency of heterozygous genotypes Multiple alleles DNA typing of X-linked genes Genetic variation and linkage 18.2 Inbreeding The inbreeding coefficient Allelic identity by ancestry Calculating the inbreeding coefficient from pedigrees Effects of inbreeding

18.3 Genetics and Evolution 18.4 Mutation and Migration Irreversible Mutation Reversible Mutation 18.5 Random Genetic Drift Loss of Genetic Variation in Endangered Species 18.6 Natural Selection Selection in a Laboratory Experiment Selection in Diploid Organisms Components of Fitness Selection-Mutation Equilibrium Heterozygous Superiority Molecular Selection Signals ROOTS OF DISCOVERY: A Christmas message from Dr. Hardy THE LATEST: CRISPR-Cas9 for Disease Control?

19 Learning Objectives and Skills in Molecular Genetics and Human Evolutionary Sciences

19.1 Molecular Evolutionary Analysis Gene Trees Bootstrapping Gene Trees and Species Trees Molecular Clock of Evolution Rates of Change in DNA Evolution Rates of Evolution in Protein-Coding Regions Origins of New Genes: Orthologues and Paralogs 19.2 Ancient DNA 19.3 Where Humans Fit in the Tree of Life Evidence that humans are more closely related to chimpanzees Similarities in genomic DNA

Analysis of differences between various genetic elements between human and chimpanzee genomes 19.4 What do the genetic differences between humans and chimpanzees mean? Molecular Adaptations Unique to Humans DNA The Neanderthal Genome The Denisovan Genome 19.7 Measuring Human Diversity

Tracing human history with genetic markers The division of variation within groups and between groups Tracing human history across the Y chromosome 19.8 Genetic adaptations unique to humans Adaptation of amylase and dietary starch to parasites and diseases Evolutionary adaptations that affect skin color Affect Western Europe ROOTS OF A DISCOVERY: Contrast of strength

Appendix A: Answers to Even Number Problems Word Roots: Prefixes, Suffixes, and Combining Forms Concise Dictionary of Genetics and Genomics

Index

Preface Students' curiosity about genetics is fueled by almost daily reports in the media about new discoveries related to genetic differences in drug response, genetic risk factors for disease, and genetic evidence for human origins and history. They are also fascinated by ethical controversies surrounding genetics: Should genetic manipulation be used on patients to treat disease? Should human fetuses be used in research? Should humans be cloned? What are the biological and ethical implications of gene editing? And there are social controversies: should there be laws regulating genetic privacy? Who should have access to genetic testing records and for what purpose? What is the role of direct-to-consumer genotyping services? The genetics teacher, in turn, faces numerous challenges: ■ Maintaining students' enthusiasm for the discipline

■ Motivate your desire to understand the principles of genetics comprehensively and rigorously. ■ Guide students to understand that genetics is not just a set of principles, but also an experimental approach to solving a variety of biological problems. ■ Helping Students Learn Think about genetic issues and the broader social and ethical issues that arise from genetics and genomics Genetics: Analyzing Genes and Genomes, the Ninth Edition addresses these challenges while emphasizing beauty and logic

Clarity and unity of subject. Our pedagogical approach is to treat transmission genetics, molecular genetics and evolutionary genetics as fully integrated subjects. This integration appeals to most modern geneticists, who recognize that distinctions between subfields are artificial. The chapters in this text are organized into five sections: a general introduction to genes and genetic analysis, transmission genetics, DNA organization and replication, gene expression and regulation, and evolutionary genetics. Recognizing that these topics can be organized in a variety of ways, we've tried to allow individual instructors to customize the order in which they're presented.

Our goal is to provide a clear, comprehensive, rigorous, and balanced introduction to genetics and genomics at the university level. We believe that a good course must maintain the right balance between two important aspects of genetics: (1) genetics as a body of knowledge about genetic transmission, function, and mutation; and (2) genetics as an experimental or "toolkit" approach to studying biological processes such as development or behavior. Every student claiming knowledge of genetics should achieve the following milestones: ■ Understand the basic processes of gene transfer, mutation, expression, and regulation. ■ Be familiar with the main experimental methods that geneticists and molecular biologists use in their studies and recognize the advantages and limitations of these approaches. ■ Thinking like a geneticist at the elementary level of the ability to formulate genetic hypotheses, work out their implications and the results to be tested against observed data

■ Explain genetic principles in your own words and put key genetic terms in context. ■ Solve problems of various types, including single-concept exercises that require the application of definitions or basic genetic principles, problems in genetic analysis that require the application of several concepts in a logical order, and problems in quantitative analysis that require some numerical calculations. the social and historical context in which genetics and genomics have evolved and continue to evolve

■ Acquire a basic familiarity with the genetic resources and information available on the Internet

The Student Experience We've included many special features to help students achieve these learning goals. The text is clear and concise, written in a relaxed, student-friendly prose style.

■ Each chapter begins with an explicit statement of learning objectives and scientific competencies that students should achieve. These guides are designed to help students identify key concepts and use them at different learning levels, including comprehension, application, analysis, and synthesis. They also serve as powerful learning tools as you review course material and exam preparation.

■ A unique feature of this text are the boxes at the discovery roots. Each chapter contains one or two of these boxes, in which textual material is linked to excerpts from classic articles reporting important experiments in genetics or raising important social, ethical, or legal issues in genetics. Each Roots of Discovery feature includes a short introduction that explains the significance of the experiment and the historical context in which it is set.

conducted, followed by brief excerpts from the original genetic literature, interspersed with commentary linking the work's findings and conclusions to the topics discussed in the chapter. Many of these boxes offer excerpts from historical articles, such as Mendel's article, but

it is by no means all “old” paper. In fact, many of these “classic” works are very recent.

Some of the pieces were originally published in French, others in German. These excerpts appear in the English translation. In articles that use obsolete or unknown terminology or archaic gene symbols, we substitute the modern equivalent because consistent terminology is used throughout the text and in the roots of

Discovery features make the material more accessible to the student. ■ In addition, several chapters contain cutting-edge boxes designed to introduce students to some of the most innovative work in the field. Topics such as the bacterial genetics of CRISPR-Cas9, genomic association studies based on personal genomics, and the potential of genetic transmission are designed to help students see progress

be done at the forefront of genetics. While the Roots of Discovery boxes are designed to highlight the experimental foundations of genetics, the Cutting Edge boxes are designed to give students a taste of where the journey is headed. Each of these features focuses on recent work that has the potential to be transformative in the future or that has introduced new methods of analysis that can be applied to questions in genetics and genomics. Each also includes references to original work that can be incorporated into class assignments and activities. We hope that the Cutting Edge feature will be useful and exciting, and that the format we are using can be applied to other new advancements as they emerge.

■ The arts program is spectacular and a learning aid in itself. Each chapter is richly illustrated with beautiful graphics that use color functionally to enhance each illustration's value as a learning aid. The illustrations are also heavily annotated with explanatory boxes that explain step by step what is happening at each level of the illustration.

These labels make art inviting and informative. They also allow the illustrations to be relatively independent of the text, allowing students to review the material without having to reread the entire chapter. The art program is used not only for its visual appeal, but also to enhance the educational value of the

Text.

Distinct colors and shapes have been used throughout the text to indicate different types of molecules - DNA, mRNA, tRNA, and so on. For example, DNA is presented in a variety of ways, depending on the level of detail required for the illustration, and each time a certain level of detail is presented, DNA is shown in the same way.

Many possible confusions are avoided by ensuring that DNA, RNA, and proteins are presented in the same way in each chapter. Numerous color photographs show molecular models in three dimensions; these provide strong visual reinforcement to the concept of macromolecules as physical entities with defined three-dimensional shapes and charge distributions that serve as the basis for interactions with other macromolecules.

The website design is clean, clear and uncluttered. This makes the text nice to look at and easy to read.

■ To help students focus on the most important details of this challenging subject, each numbered section ends with a summary that highlights key concepts. Then,

At the end of each chapter, a bulleted chapter summary summarizes key learning points for students. ■ Each chapter ends with approximately 50 problem-solving tasks that students can use to test their understanding. Problems are of several types.

Review Basic Questions Review key concepts by asking discussion questions. These questions require that genetic principles be repeated in the student's own words; some are a matter of definition or require the application of elementary principles.

The Troubleshooting Guide shows typical problems applying the principles. Problems are fully covered, showing the concepts needed to solve the problem and justifying the answer. This resource serves as a review of important concepts used in work problems. It also highlights some of the most common mistakes made by beginners and provides tips on how the student can avoid falling into these conceptual pitfalls.

A large number of analysis and application problems are provided (with solutions to even problems at the end of the text). These more traditional types of genetic problems require several concepts to be applied in a logical order and some numerical calculations may be required. The level of mathematics is that of arithmetic and elementary probability related to genetics. None of the problems use math beyond elementary algebra.

Challenge problems are similar to analysis and application problems, but they are more challenging because they usually require more extensive data analysis before the question can be answered.

■ At the end of the text are answers to even problems that fully explain the rationale for each answer. Answers are thorough, explaining the rationale for the solution and introducing the methods used. We've found that many of our students, like students everywhere, tend to skim the answer before trying to solve a problem. That's a shame. Working backwards from the answer should be a last resort, because problems are valuable learning opportunities. Problems that the student cannot solve tend to be more important than those that the student can solve, because advocates often identify areas of weakness, areas of confusion, or gaps in understanding. Therefore, we encourage our students to try to answer each question before looking at the answer. Answers to the odd-numbered questions will be made available to the instructor as part of the grading resource pack.

■ At the end of the text, we have included a short list of root words used in many genetic terms. Word Roots are designed to help students interpret and remember the meaning of key terms and master basic terminology in genetics and genomics.

■ The small lexicon of genetics at the end of the text allows students to check their understanding of key words or look up forgotten technical terms. It includes not only keywords but also genetic terms that students are likely to encounter as they explore the web or continue reading.

What's New in the Ninth Edition? This edition has been completely revised and updated. Worldwide

Chapters have been completely revised and the organization of topics as reflected in the chapter order has been revised. This reorganization was fueled in large part by developments in the field – particularly the explosion of data generated by high-throughput DNA sequencing, particularly in humans. These advances revolutionized the study of genetic variation in populations, giving rise to genome-wide association studies (GWAS) and related methods. The topic of GWAS is introduced in the context of gene mapping (Chapter 5) and extended to the genetics of complex traits (now Chapter 7). Other applications of high-throughput approaches in genetics include greatly expanded coverage of RNA-seq (Chapter 14), cancer genomics (Chapter 16), and ancient DNA analysis (Chapter 19). The Roots of Discovery boxes (formerly Connections boxes) have been rewritten, in most cases reducing the scope of direct citations from the original articles and providing readers with focused connections to the main points of the subject being covered. All Roots of Discovery boxes contain full quotes from the profiled work, including URLs, so they can be easily explored by interested students or as part of a teacher-designed activity. Tip Boxes, a new feature, also provide links to literature for easy exploration.

Chapter Organization To help students follow the main topics and not get distracted by the details, each chapter begins with an outline outlining the area ahead. This is followed by a list of learning objectives and science competencies that describe key concepts and principles. An introductory paragraph provides an overview of the chapter, illustrates the subject with some specific examples, and shows how the material relates to genetics as a whole. The text uses richly numbered bulleted lists to help students organize their learning, as well as summary statements highlighted in special font to emphasize important principles. Each numbered section ends with a bulleted list titled “Summary” that highlights the main points covered in the section and helps the student review and master the material covered. Each chapter ends with a chapter summary, discussion questions titled "Review the Basics," a troubleshooting guide, problems to be solved in a section titled "Analysis and Applications," and challenge problems.

Content and organization Today, most students learn about DNA in elementary or high school; In our class, we thought it was artificial to pretend that DNA didn't exist until the middle of the semester. A key feature, therefore, is the presence of two introductory chapters that provide a broad overview of DNA, genes, and genomes—what they are, how they work, how they change through mutation, and how they evolve over time. Introductory chapters are designed to connect the more advanced concepts students will learn with what they already know. They also serve to give each student a solid framework for integrating later material. An important role they play is to provide an adequate foundation in the

molecular foundations of genetics—DNA structure and replication, and the central dogma of molecular biology—so that molecular concepts can be integrated into chapters covering “classic” topics such as Mendelian genetics and gene mapping. In each chapter there is a balance between observation and theory, between principle and concrete example, and between challenge and motivation. Molecular, classical, and evolutionary genetics are integrated throughout the text. Reference is often made to human genetics. The text is also generously filled with applications to other animals and plants, including major model organisms used in genetics and genomics.

Below, we describe some of the highlights of the five main sections of the text.

Defining and Working with Genes ■ Genes, Genomes, and Genetic Analysis (Chapter 1) is an overview of genetics designed to bring students from different backgrounds to a common level of understanding. This chapter allows for the integration of classical, molecular, and evolutionary genetics with the rest of the text. This chapter contains the basic concepts of molecular genetics: DNA structure,

Replication, expression and mutation. A feature of this chapter is a detailed examination of the work of Beadle and Tatum that led to the one-gene, one-enzyme hypothesis and the use of complementation analysis as a tool for genetic inference. ■ DNA Structure and Genetic Variation (Chapter 2) emphasizes that the modern geneticist's most important tools are methods for the experimental manipulation of DNA. It provides a more detailed look at the structure of DNA and introduces the main methods used in DNA manipulation, including restriction enzymes, electrophoresis, DNA hybridization, Southern blotting, and polymerase chain reaction (PCR). We also introduce single nucleotide polymorphisms (SNPs) and copy number polymorphisms (CNPs) and discuss how these types of genetic markers can be probed at the genome level using oligonucleotide microarrays (DNA chips). The use of simple sequence repeat variations for genotyping is expanded.

Transmission Genetics ■ Mendelian Genetics: The Principle of Assortment and Assortment (Chapter 3) introduces the basic Mendelian concepts of random assortment and independent assortment, not only because they can be derived from crosses with experimental organisms such as peas and flies, but also by analyzing trees human genealogy. The role of probability in genetics

Analysis is also introduced. A unique feature of this chapter is the integration of molecular genetics with Mendel's experiments. We describe the molecular basis of the wrinkled mutation and show how a modern geneticist would conduct Mendel's study, examining molecular phenotypes on the one hand and morphological phenotypes on the other. This pedagogy provides a solid foundation for understanding not only Mendel's experiments as he actually performed and interpreted them, but also the use of modern molecular approaches in genetic analysis. Molecular markers are also integrated into human genetic analysis.

■ The chromosomal bases of inheritance (Chapter 4) covers the cell cycle, mitosis and meiosis, and hypothesis testing in genetics. The chi-square test approach was revised by introducing random variables and their distributions, first in terms of binomial distribution and then in terms of chi-square distribution. Computer simulated distributions are used to illustrate the logic of the significance test, after which the mechanics of performing a chi-square test are described. ■ Genetic linkage and chromosome mapping (Chapter 5) covers linkage analysis and gene mapping. The classic approach with test crosses in test organisms has been maintained, as has the simplified coverage of tetrad analysis in fungi. New to the Ninth Edition is a section on human genetic mapping, which includes expanded coverage of lod score analysis of pedigrees, as well as an all-new section on gene location through association analysis. ■ Human karyotypes and chromosomal behavior (Chapter 6) discusses the principles of chromosome mechanics with particular reference to the number and structure of human chromosomes and the types of aberrations found in human chromosomes.

The genetic implications of chromosomal abnormalities - duplications, defects, inversions and translocations - are also discussed. The evolutionary importance of gene duplication is introduced and the genetic consequences of translocation heterozygosity (pseudolinking) are described. ■ The genetic basis of complex traits (Chapter 7) covers quantitative genetics. It is a revision of Chapter 18 of the previous edition. We decided to present this material earlier

due to the growing importance of genome-wide association studies in agriculture and medicine. Crohn's disease is used as a model for this, and the strengths and weaknesses of the GWAS as a tool are highlighted. The heredity record has been revised and a new section on misconceptions about it has been added. ■ Genetics of Bacteria and Their Viruses (Chapter 8) extends the use of genetic analysis to bacteria (mainly Escherichia coli) and viruses, and serves as a transition to the more molecular topics covered in subsequent chapters. The plasmid section has been revised to focus more closely on the F plasmid; A more detailed treatment of bacterial plasmids in general has been transferred to the chapter "Molecular organization of chromosomes and genomes". Some of the classic experiments and methods, such as replica plating and the "U-tube" experiment, are highlighted. The description of the origin of the term "cistron" and the cis-trans test has been expanded, and CRISPR-Cas9 as a bacterial system is presented in an edge box.

Organization and replication of chromosomes and DNA ■ Molecular organization of chromosomes and genomes (Chapter 9) continues the transition to molecular basis

Heredity, encompassing the organization of DNA in the genomes of prokaryotes and eukaryotes. It now includes an introduction to mobile DNA in both types of organisms. The section on Cot kinetics has been shortened, acknowledging that while this methodology is historically and conceptually important, it is no longer actively used. The description of chromosomal condensation has been updated and its connection to mitosis (and meiosis) made more explicit.

■ DNA Replication and Sequencing (Chapter 10) covers the mechanics of DNA replication and sequencing methods as a starting point for an in-depth discussion of the molecular basis of genetics. DNA replication mechanisms are described first for prokaryotes and then for eukaryotes. Coverage of Sanger sequencing is followed by a description of sequencing by synthesis using the Illumina platform as a model. ■ Mutation, Repair, and Recombination (Chapter 11) covers mutation, DNA repair, and recombination. In previous editions, mutation was presented in a later chapter, but by placing it immediately after the treatment of DNA replication, we hope to highlight the importance of all these processes both in maintaining genetic continuity and in generating new variations. Added coverage of the classic Luria-Delbrück experiment showing that mutation is random in terms of adaptation. Treatment of recombination mechanisms has been included in this chapter, emphasizing the similarity between these mechanisms and post-replication repair mechanisms.

Gene Expression ■ Molecular Biology of Gene Expression (Chapter 12) covers the processes of transcription, translation and RNA

Processing. Transcription mechanisms in prokaryotes and eukaryotes are more clearly differentiated; The concept of consensus sequences is introduced more fully, along with their visualization as "sequence logos". ■ Molecular Mechanisms of Gene Regulation (Chapter 13) deals with the regulation of gene expression. Mechanisms in prokaryotes and eukaryotes were separated to better compare and contrast them. The section about

The role played by non-coding RNAs in regulation has been expanded, and their role in processes such as X chromosome inactivation, alternative mRNA splicing, and regulation of translation has been highlighted. ■ Gene and genome manipulation (Chapter 14) covers experimental methods of gene manipulation, starting with classical gene cloning methods through genome analysis, transcriptomics, and gene editing with CRISPR-Cas9. These include the use of restriction enzymes and vectors in recombinant DNA, cloning strategies, site-directed mutagenesis, production of genetically defined transgenic animals and plants, and genetic engineering applications. Coverage of whole exome sequencing has been expanded and a new section on functional genomics has been added. This material provides extended RNA-seq coverage and a description of quantitative PCR for transcription quantitation. Added a subsection on using RNAi to knock down gene expression, as well as a main section on gene editing with CRISPR-Cas9. The latter includes a cutting-edge box for applying these methods in a mouse model of muscular dystrophy. ■ Genetic control of development (Chapter 15) focuses on the genetic analysis of development in nematodes (Caenorhabditis elegans) and Drosophila and involves a thorough investigation

the genetic basis of flower development in Arabidopsis thaliana. Added a new section on the role of regulatory RNAs in controlling development, including examples from nematodes, flies, and mammalian Hox genes. ■ Molecular Genetics of the Cell Cycle and Cancer (Chapter 16) examines cancer from the point of view of the genetic control of cell division, with a focus on checkpoints that result in inhibition of cell division or programmed cell death (apoptosis) in normal cells. Cancer results from a series of sequential mutations, usually in somatic cells, that override the normal checkpoints that control cell proliferation. A new section on cancer genomics has been added, using pancreatic cancer as a model to characterize mutational and transcriptisomal changes associated with tumor progression.

Variation

■ Mitochondrial DNA and Extranuclear Inheritance (Chapter 17) discusses non-nuclear genetics, including genetic defects in human mitochondrial DNA. Added a new section describing Wolbachia transmission in arthropods. ■ Genes in Populations (Chapter 18) covers population genetics (material on molecular evolution has been moved to the Molecular and Human Evolutionary Genetics chapter). Updated DNA fingerprint section to reflect the use of short tandem repeats as markers. A new linkage disequilibrium subsection has been added, along with "heat map" illustrations of disequilibrium derived from genomic data. When discussing evolutionary processes, the order of topics has been rearranged so that genetic drift – a random process common to all finite populations – is discussed before examining the deterministic process of natural selection. A new section above

Molecular selection signals have been added, including material on lactase persistence in humans. A cutting-edge box describes the principles of the "gene drive" based on CRISPR-Cas9 and explains how it can be used to combat malaria-carrying mosquitoes. ■ Molecular and Human Evolutionary Genetics (Chapter 19) covers both molecular and human evolution, since most advances in the latter follow on from developments in the former. Other changes in this chapter include an expanded section on ancient DNA analysis and the use of Alu1 insertion sites to resolve hominid phylogeny. The record of diversity in the genus Homo was revised to include new knowledge about H. denisova and H. floresiensis. The H. sapiens genetic variation section has been revised to include data from the 1000 Genomes Project (as opposed to HapMap in previous editions) and the use of genetic mapping methods ("STRUCTURE") to characterize modern human variation. A new tip box focuses on the use of ancient DNA genotyping to characterize the Western European population.

Flexibility There is no need to start at the beginning of this text and go straight to the end. In fact, in a typical one-semester course, this borders on the impossible. In this ninth edition, each chapter has been designed as an independent unit. This feature gives instructors the ability to use any order and skip specific chapters that don't fit the course design. We've incorporated molecular and classical principles throughout the text so you can start a course with almost any of the chapters. Most instructors prefer to start with the overview of the Gene, Genome, and Genetics sections.

Analysis chapter because it brings all students to the same basic level of understanding. The chapter on DNA structure and genetic variation presents the basic experimental manipulations used in modern genetics and serves to integrate molecular and classical genetics into Mendel's discussion in the chapter Transmission Genetics: The Principle of Segregation. Some other approaches instructors can use to structure the beginning of their course are described here.

approach

chapter at the beginning

Men's initial format

Genes, Genomes, and Genetic Analysis Transmission Genetics: The Principle of Segregation DNA Structure and Genetic Variation

Initial chromosome shape

Genes, Genomes, and Genetic Analysis Chromosomes and Sex Chromosome Inheritance DNA Structure and Genetic Variation Genetic Transmission: The Principle of Segregation

Genetic linkage and chromosome mapping

Genomes-first-Format

Genes, Genomes, and Genetic Analysis DNA Structure and Genetic Variation Chromosomes and Sex Chromosome Inheritance

The writing and illustration program is designed to accommodate a wide variety of formats, and we encourage teachers to use this flexibility to suit their own needs.

Teaching Resources Jones & Bartlett Learning offers a variety of traditional and interactive multimedia supplements to help educators and students master genetics. Additional information and review copies of the following items are available from your Jones & Bartlett Learning sales representative.

■ PowerPoint Image Bank is an easy-to-use multimedia tool that provides all illustrations and photos from the text (to which Jones & Bartlett Learning owns the electronic reproduction rights) for use in classroom presentations. You can select the required images or easily create your own slideshows or print the files to create transparencies. Many images have already been included in the PowerPoint lesson outline presentations for ease of use.

■ A PowerPoint presentation accompanies the detailed outlines of each chapter in Genetics: Analysis of Genes and Genomes, Ninth Edition. This presentation which is designed

to mirror the text, it is flexibly adapted to your lecture

Organization. The sketch opens, so you can provide whatever elements you think are necessary, whether that's new text or more stock images.

■ A test database of over 1,000 questions and complete answers is available. Questions are a mix of factual, descriptive, and quantitative types. A typical chapter contains multiple-choice, fill-in-the-blank, and short-answer questions. The test bench is provided as a Microsoft Word document.

■ Instructor Guide contains chapter summaries, teaching tips, references, suggested activities, and more.

Acknowledgments We are grateful to the many colleagues whose advice, thoughtful reviews, and comments have helped in the preparation of this and previous editions. Their expert recommendations are reflected in the content, organization and presentation of the material. Laura Adamkewicz, George Mason University, Fairfax, VA Jeremy C. Ahouse, Brandeis University, Waltham, MA Mary Alleman, Duquesne University, Pittsburgh, PA Peter D. Ayling, University of Hull, Hull, UK John C. Bauer, Stratagene, Inc., La Jolla, CA Anna W. Berkovitz, Purdue University, West Lafayette, IN Mary KB Berlyn, Yale University, New Haven, CT Justin Blumenstiel, University of Kansas, Lawrence, KS Thomas A. Bobik, University of Florida, Gainesville , FL David Botstein, Princeton University, Princeton, NJ

Colin G. Brooks, The Medical School, Newcastle, Vereinigtes Königreich Pierre Carol, Université Joseph Fourier, Grenoble, Frankreich Domenico Carputo, University of Naples, Neapel, Itália Sean Carroll, University of Wisconsin, Madison, WI Chris Caton, University of Birmingham, Birmingham, AL John Celenza, Boston University, Boston, MA Estelle Chamoux, Bishop's University, Sherbrooke, Quebec, Canadá Alan C. Christensen, University of Nebraska–Lincoln, Lincoln, NE

Christoph Cremer, Heidelberg University, Heidelberg, Germany Marion Cremer, Ludwig Maximilian University, Munich, Germany Thomas Cremer, Ludwig Maximilian University, Munich, Germany Leslie Dendy, University of New Mexico, Los Alamos, NM Stephen J. DíSurney, University of Mississippi , Oxford, MS John W. Drake, National Institute of Environmental Health Sciences, Research Triangle Park, NC Kathleen Dunn, Boston College, Boston, MA Chris Easton, State University of New York, Binghamton, NY Joel C. Eissenberg, St. Louis; Louis University, St. Louis University. Louis, MO Ioannis Eleftherianos, The George Washington University, Washington, DC Wolfgang Epstein, University of Chicago, Chicago, IL Brian E. Fee, Manhattan College, Riverdale, NY Gyula Ficsor, Western Michigan University, Kalamazoo, MI Robert G. Fowler, San Jose State University, San Jose, CA David W. Francis, University of Delaware, Newark, DE

Matthew K. Fujita, University of Texas at Arlington, Arlington, TX James Fuller, University of North Carolina, Chapel Hill, NC Gail Gasparich, Towson University, Towson, MD Elliott S. Goldstein, Arizona State University, Tempe, AZ Kent G. Golic, University of Utah, Salt Lake City, UT Lesley H. Greene, Old Dominion University, Norfolk, VA Patrick Guilfoile, Bemidji State University, Bemidji, MN Jeffrey C. Hall, Brandeis University, Waltham, MA Mark L. Hammond, Campbell Universidade, Buies Creek, NC

Steven Henikoff, Fred Hutchinson Cancer Research Center, Seattle, WA Charles Hoffman, Boston College, Boston, MA Ivan Huber, Fairleigh Dickinson University, Madison, NJ Kerry Hull, Bishop's University, Quebec, Canadá Lynn A. Hunter, University of Pittsburgh, Pittsburgh , PA Richard Imberski, University of Maryland, College Park, MD Bradley Jacob Isler, Ferris State University, Big Rapids, MI Nitya P. Jacob, Oxford College of Emory University, Oxford, GA Joyce Katich, Monsanto, Inc., St. Louis , MO Jeane M. Kennedy, Monsanto, Inc., St. Louis, MO Jeffrey King, Universität Bern, Bern, Schweiz Anita Klein, University of New Hampshire, Durham, NH Tobias A. Knoch, Deutsches Krebsforschungszentrum, Heidelberg, Deutschland Yan B Linhart, University of Colorado, Boulder, CO K. Brooks Low, Yale University, New Haven, CT Elena R. Lozovsky, Harvard University, Cambridge, MA Sally A. MacKenzie, Purdue University, West Lafayette, IN Irina Makarevitch , Hamline University , Saint Paul, MN Copyright © 201 7. Jones & Bartlett Learning, LLC. Alle Rechte vorbehalten.

Gustavo Maroni, University of North Carolina, Chapel Hill, NC Lauren McIntyre, University of Florida, Gainesville, FL Eric Mendenhall, University of Alabama at Huntsville, Huntsville, AL Jeffrey Mitton, University of Colorado, Boulder, CO Robert K. Mortimer, University of California, Berkeley, CA Gisela Mosig, Vanderbilt University, Nashville, TN Erin W. Norcross, Mississippi College, Clinton, MS Steve O'Brien, National Cancer Institute, Frederick, MD

Daniel Odom, California State University, Northridge, Northridge, CA Kevin O'Hare, Imperial College, Londres, Vereinigtes Königreich Ronald L. Phillips, University of Minnesota, St. Paul, MN David Pilbeam, Harvard University, Cambridge, MA Robert Pruitt, Purdue University, West Lafayette, IN Pamela Reinagel, California Institute of Technology, Pasadena, CA Rebecca Reiss, New Mexico Tech, Socorro, NM Susanne Renner, University of Missouri, St. Louis, MO Michael Robinson, Miami University, Oxford, OH Andrew J. Roger, Dalhousie University, Halifax, Nova Scotia, Kanada Kenneth E. Rudd, National Library of Medicine, Bethesda, MD Thomas F. Savage, Oregon State University, Corvallis, OR Joseph Schlammandinger, Universität Debrecen, Ungarn Julia M. Schmitz , Piedmont College, Athens, GA Barbara B. Sears, Michigan State University, East Lansing, MI David Shepard, University of Delaware, Newark, DE Alastair G. B. Simpson, Dalhousie University, Halifax, Nova Scotia, Kanada

Navin K. Sinha, Rutgers University, Piscataway, NJ Leslie Smith, Instituto Nacional de Ciências da Saúde Ambiental, Research Triangle Park, NC Charles Staben, University of Kentucky, Lexington, KY Mark Sturtevant, Oakland University, Rochester, MI Johan H. Stuy, Florida State University, Tallahassee, FL David T. Sullivan, Syracuse University, Syracuse, NY Jeanne Sullivan, West Virginia Wesleyan College, Buckhannon, WV

Millard Susman, University of Wisconsin, Madison, WI Barbara Taylor, Oregon State University, Corvallis, OR Irwin Tessman, Purdue University, West Lafayette, IN Asli Tolun, Bogazici University, Istanbul, Türkei Yoshi Tomoyasu, Miami University, Oxford, OH Maria Tsompana , University of North Carolina, Chapel Hill, NC Micheal Tully, University of Bath, Bath, Vereinigtes Königreich David Ussery, The Technical University of Denmark, Lyngby, Dänemark George von Dassow, Friday Harbor Laboratories, Friday Harbor, WA Denise Wallack, Muhlenberg College , Allentown, PA Kenneth E. Weber, University of Southern Maine, Gorham, ME Tamara Western, Okanagan University College, Kelowna, British Columbia, Kanada Darla J. Wise, Concord University, Athens, WV Jay Zimmer, Gardner-Webb University, Boiling Federn, Carolina do Norte

We would also like to thank Barbara B. Sears of Michigan State University for her many helpful suggestions for the organelle genetics chapter in earlier editions of this book. We also want to thank the outstanding editorial and production team at Jones & Bartlett Learning who made this text possible: Matt Kane, Audrey Schwinn, Loren-Marie Durr, Nancy Hitchcock, Wes DeShano, Troy Liston, and Scott Moden. Much of the credit for the attractiveness and readability of the text must go to them. In addition, we thank Jones & Bartlett Learning for their continued commitment to quality work in book production. We are grateful to the many people mentioned in the character captions who contributed photographs, drawings, and photomicrographs from their own research and publications.

About the authors Daniel L. Hartl is the Higgins Professor of Biology at Harvard University and a member of the National Academy of Sciences and the American Academy of Arts and Sciences. He received his BS and PhD from the University of Wisconsin and did postdoctoral research at the University of California, Berkeley. His research interests include molecular genetics, genomics, molecular evolution, and population genetics.

Bruce J. Cochrane is Professor of Biology at the University of Miami. He received his BA in biology from Cornell University and his PhD in genetics from Indiana University and did postdoctoral research at the University of North Carolina. His research interests include population genetics, evolutionary biology, and bioinformatics.

UNIT 1 Defining and Working with Genes Chapter 1 Genes, Genomes and Genetic Analysis Chapter 2 DNA Structure and Genetic Variation

Medical Imaging RM/Jane Hurd.

CHAPTER 1 Genes, Genomes, and Genetic Analysis CHAPTER OVERVIEW 1.1 DNA as Genetic Material 1.2 Structure and Replication of DNA 1.3 Genes and Proteins 1.4 Genetic Analysis 1.5 Gene Expression: The Central Dogma 1.6 Mutation and Variation 1.7 Genes and Environment 1.8 The Molecular Unit of life ROOTS OF DISCOVERY: Black Urine Disease

Archibald E. Garrod (1908) Inborn errors of metabolism ROOTS OF THE DISCOVERY: One gene, one enzyme George W. Beadle and Edward L. Tatum (1941) Genetic control of biochemical reactions in Neurospora

LEARNING OBJECTIVES AND SCIENTIFIC SKILLS Once you understand and are able to apply the principles of genes, genomes, and genetic analysis discussed in this chapter, you will have acquired the following scientific skills: ■ Know the basic sequence of a transcribed chain of proteins that encode DNA , give the sequence of bases in the appropriate region of messenger RNA and the corresponding amino acid sequence in the protein. ■ For a mutation in DNA where one base is replaced by another, show how the mRNA and protein are changed.

■ Given a linear metabolic pathway for an essential nutrient, determine which intermediates restore viability in mutant strains lacking one of the enzymes in the pathway. ■ Interpret data indicating which intermediates in a linear pathway restore viability to mutants to determine the order in which enzymes and intermediates occur in the pathway. ■ Categorize mutations into groups, each corresponding to a different gene.

Each type of living organism has a unique set of inheritances.

Characteristics that distinguish them from all other species. Each species has its own developmental blueprint – often referred to as a sort of “blueprint” for building the organism – which is encoded in the DNA molecules within it.

your cells. This developmental plan determines which characteristics are inherited. Since organisms of the same species have the same developmental plan, organisms that are members of the same species generally resemble one another. for example is

Humans easily distinguished from chimpanzees or gorillas. A human being is typically erect and has long legs, relatively little body hair, a large brain, and a flat face with a prominent nose, prominent chin, pronounced lips, and small teeth. All these characteristics are inherited – they are part of our evolutionary blueprint – and help to differentiate us as Homo sapiens. But people are by no means identical. Many observable traits or traits differ from one person to another. In addition to the notable differences between men and women, there are major differences in hair color, eye color, skin color, height, weight, personality traits, and other characteristics. Some human characteristics (such as gender) are transmitted biologically, others culturally. The color of our eyes is a result of biological inheritance, but the native language we learn as children is a result of cultural inheritance. Many traits are jointly influenced by biological heredity and environmental factors. For example, weight is determined in part by heredity, but in part also by the environment: how much food we eat, its nutritional value, our exercise program, and so on. Genetics is the study of biologically inherited traits, including traits that are influenced in part by the environment. Genomics is the study of all the genes in an organism to understand their molecular understanding

Organization, function, interaction and evolutionary history. The basic concept of genetics and genomics is as follows: inherited characteristics are determined by the elements of heredity passed from parents to offspring in reproduction; these elements of heredity are called genes.

The existence of genes and the rules for their transmission from generation to generation were first articulated by Gregor Mendel in 1866 (described in the chapter entitled Mendelian Genetics: The Principles of Segregation and Assortment). Mendel's formulation of heredity referred to the abstract rules by which hereditary elements (he called them "factors") are transmitted from parents to children. His objects of study were peas with variable characteristics, such as pea color and plant height. Mendel thus introduced the study of genetics through the analysis of offspring resulting from mating. Genetic differences between species have been impossible to define because organisms from different species generally do not mate; When they mate, they usually produce hybrid offspring that die or are sterile. The approach to studying genetics by analyzing the offspring of matings is often referred to as classical genetics or organismic or morphological genetics. In contrast, molecular genetics is the study of the chemical nature of genes and their products. Advances in this field have made it possible to study the structure and function of genes, as well as the differences between species, comparing and analyzing DNA itself. However, there is no fundamental distinction between classical genetics and molecular genetics. Rather, they are different and complementary methods of studying the same thing: the function of genetic material. This text contains many examples

shows how molecular and classical genetics can be used in combination to improve the performance of genetic analysis. In the remainder of this chapter, we focus on the pioneering observations and experiments that form the basis of our current understanding of gene structure and function. Before doing so, however, we must consider the four necessary properties of genetic material:

1. Genetic material must encode information. 2. This information must be used to control the functioning of cellular processes. 3. The information contained must be transferable from cell to cell and from generation to generation. 4. There must be a potential for mutation that can give rise to the physical variation that exists between individuals and between species.

1.1 DNA as Genetic Material The establishment of genetics as a molecular science dates back to 1869, just three years after Mendel reported his experiments. That year, Friedrich Miescher discovered a new type of weak acid that is abundant in the nuclei of white blood cells. Miescher's weak acid turned out to be the chemical we now call DNA (deoxyribonucleic acid). For many years, the biological

The function of DNA was unknown and no role in heredity has been attributed to it. This section describes how DNA came to be isolated and identified as the material that makes up genes. That the nucleus plays a key role in heredity was recognized in the 1870s, when scientists observed that the nuclei of male and female reproductive cells fuse in the process of fertilization. Soon after, chromosomes were observed for the first time in the nucleus of the cell as thread-like objects that become visible under a light microscope when the cell is stained with certain dyes. Chromosomes have been found to exhibit a characteristic "division" behavior in which each daughter cell formed by cell division receives an identical set of chromosomes (see the chapter entitled "The Chromosomal Basis of Inheritance"). Further evidence of the importance of chromosomes was provided by the observation that, although the number of chromosomes in each cell can vary among biological species, the number of chromosomes in cells of a given species is almost always constant. These features of chromosomes were well known in 1900 and made it likely that chromosomes were the carriers of genes.

In the 1920s, a lot of indirect evidence pointed to a close relationship between chromosomes and DNA. Microscopic examinations using special dyes have shown that DNA is present in chromosomes. Chromosomes also contain different types of proteins, but the amount and type of chromosomal proteins vary greatly from cell type to cell type, while the amount of DNA per cell is constant. Furthermore, almost all of the DNA present in the cells of higher organisms is present in the chromosomes. However, these arguments for DNA as genetic material were unconvincing, as crude chemical analyzes suggested (wrongly, it seems) that DNA lacked the chemical diversity necessary to encode the complex set of instructions needed to build it, even in a complex way. a single cell or organism. The preferred candidate for genetic material was protein, as proteins were known to be an extremely diverse collection of molecules. Thus, proteins became widely accepted as genetic material, and DNA was believed to function only as the structural backbone of chromosomes. The experiments described in this section showed that DNA is indeed the genetic material.

Experimental evidence for the genetic function of DNA In 1928, Frederick Griffith took an important first step in elucidating the role of DNA when he showed that genetic material could be transferred from one bacterial cell to another. Griffith worked with two strains of the bacteria Streptococcus pneumoniae, identified as S and R. When a bacterial cell is grown in a solid

In the middle, it undergoes repeated cell divisions to form a visible cluster of cells called a colony. The S type of S. pneumoniae synthesizes a gelatinous capsule composed of complex carbohydrates (polysaccharides). The surrounding capsule makes each colony large and gives it a glossy or smooth appearance (S) (FIGURE 1.1). This capsule also allows the bacteria to trigger pneumonia, protecting it from the defense mechanisms of an infected animal. R strains of S. pneumoniae are unable to synthesize capsular polysaccharide; they form small colonies with a rough surface (R) (Figure 1.1). The R strain of bacteria does not cause pneumonia because, without the capsule, the bacteria is inactivated by the host's immune system. Both types of bacteria "really reproduce" in the sense that the offspring formed by cell division have the parent's capsule type, S or R.

FIGURE 1.1 Colonies of rough (R, small colonies) and smooth (S, large colonies) strains of Streptococcus pneumoniae. S colonies are larger due to the gelatinous capsule in S cells. Photo reproduced from Avery, O.T., MacLeod, C.M., & McCarty, M. (1944). Investigations into the chemical nature of the metabolizing substance of the types of pneumococci. Journal of Experimental Medicine, 79(2), 137-158. Copyright 1994 by Rockefeller University Press. Reprinted with permission from Rockefeller University Press. It hurts:

10.1084/jem.79.2.137.

Mice injected with live S cells develop pneumonia. In contrast, mice injected with live R cells or heat-killed S cells remain healthy. Griffith's main discovery was that mice injected with a mixture of live R cells and heat-destroyed S cells developed the disease and often died of pneumonia (FIGURE 1.2). Bacteria isolated from blood samples from these dead mice produce S cultures with a capsule typical of injected S cells, although the injected S cells were killed by heat. Apparently, the injected material from the dead S cells contains a substance that can be transferred to living R cells and thus allows the R cells to synthesize the Stype capsule. In other words, R bacteria can be changed - or transformed - into S bacteria.

that synthesizes the capsule is inherited by the progeny of the transformed bacterium.

FIGURE 1.2 Griffith experiment demonstrating bacterial transformation. A mouse remains healthy when injected with the nonvirulent R strain of S. pneumoniae or heat-killed cell fragments of the normally virulent S strain. The R cells are converted to the virulent S strain in the presence of heat-killed S cells, causing pneumonia in the mouse.

Transformation into Streptococcus was originally discovered in 1928, but it was not until 1944 that the chemical responsible for the conversion of R cells to S cells was identified. In a landmark experiment, Oswald Avery, Colin MacLeod, and Maclyn McCarty showed that the substance causing the conversion of R cells to S cells was DNA. When conducting their experiments, these researchers first had to develop chemical methods to isolate almost pure DNA from cells, something that had never been done before. When they added DNA isolated from S cells to growing cultures of R cells, they saw transformation – that is, some S cells were produced. Although DNA preparations contain

Trace amounts of protein and RNA (ribonucleic acid, an abundant cellular macromolecule chemically related to DNA), transforming activity was not altered by treatments that destroyed protein or RNA. On the other hand, treatments that destroyed the DNA eliminated the transforming activity (FIGURE 1.3). These experiments indicated that

The substance responsible for the genetic transformation was the cell's DNA - and therefore DNA is the genetic material.

FIGURE 1.3 A diagram of the Avery-MacLeod-McCarty experiment showing that DNA is the active material in bacterial transformation. (A) Purified DNA extracted from heat-killed S cells can convert some living R cells to S cells, but the material may still contain undetectable traces of protein and/or RNA. (B) The transforming activity is not destroyed by either protease or RNase. (C) The transforming activity is destroyed by DNase and therefore probably consists of DNA.

Genetic role of DNA in the bacteriophage Another fundamental discovery was reported in 1952 by Alfred Hershey and Martha Chase. They examined cells of the intestinal bacteria Escherichia coli after infection with the T2 virus. A virus that attacks bacterial cells is called a bacteriophage, a term often shortened to phage. Bacteriophage means "bacteria eater". The structure of a

The bacteriophage T2 particle is shown in Figure 1.4. This particle is extraordinarily small, but still has a complex structure composed of a head (which contains the phage DNA), collar, tail, and tail fibers. (The head of a human sperm cell is about 30 to 50 times larger than the T2 head in both length and width.) Hershey and Chase already knew that T2 infection occurs through binding of the tail tip of a phage particle to the bacteria. A cell wall occurs, entry of phage material into the cell, proliferation of this material to form a hundred or more progeny phages, and release of progeny phages by disruption (lysis) of the bacterial host cell. They also knew that T2 particles are composed of approximately equal amounts of DNA and protein.

FIGURE 1.4 (A) Drawing of E. coli phage T2 showing different components. DNA is confined to the inside of the head. (B) An electron micrograph of T4 phage, a closely related phage. Electron micrograph courtesy of Robert Duda, University of Pittsburgh.

Since DNA contains phosphorus but no sulfur, while most proteins contain sulfur but no phosphorus, it is possible to label DNA and proteins differently using radioactive isotopes of the two elements. Hershey and Chase produced particles containing radioactive DNA by infecting E. coli cells that were grown for several generations in medium containing 32 P (an isotope of radioactive phosphorus) and then collecting the phage progeny. Other particles containing labeled proteins were obtained in the same way using a medium containing 35 S (an isotope of radioactive sulfur). In the experiments summarized in FIGURE 1.5, non-radioactive E. coli cells were infected with phages labeled with 32 P (Part A) or 35 S (Part B), allowing DNA and proteins to be traced along their individual pathways. 🇧🇷 Infected cells were separated from unbound phage particles by centrifugation and resuspended in fresh water.

medium and then shaken vigorously in a food processor to separate adherent phage material from cell surfaces. This treatment was found to have no effect on the further course of the infection, implying that the phage genetic material must enter infected cells soon after phage attachment. The kitchen blender has proven to be a critical piece of equipment. Other methods have been tried to pluck the phage heads from the surface of the bacterial cell, but nothing has worked reliably. Hershey later explained: "We tried different grinding arrangements, with results

this was not very encouraging. When Margaret McDonald made her kitchen blender available to us, the experiment was a success.”

FIGURE 1.5 The Hershey-Chase (“shambler”) experiment showing that DNA, not protein, is responsible for directing T2 phage reproduction in infected E. coli cells. (A) Radioactive DNA is transferred to the progeny phage in substantial amounts. (B) The radioactive protein is transferred to the progeny phage in negligible amounts.

After the phage heads were removed by the mixed treatment, the infected bacteria were examined. It was found that most of the radioactivity of the 32P-labeled phage was associated with the bacteria, while only a small fraction of the radioactivity of the 35S was present in infected cells. Retention of most of the labeled DNA, as opposed to loss of most of the labeled protein, implies that a T2 phage transfers most of its DNA, but very little of its protein, to the cell it infects. The key finding (Figure 1.5) was that about 50 percent of the transferred 32P-tagged DNA, but less than 1 percent of the transferred 35S-tagged protein was inherited by the offspring.

phage particles. Hershey and Chase interpreted this result to mean that the genetic material in the T2 phage is DNA. The experiments by Avery, MacLeod, and McCarty, as well as those by Hershey and Chase, are classics for proving that genes are made of DNA. Today, the equivalent of the transformation experiment is performed daily in many research laboratories around the world, often using cultures of bacteria, yeast, or animal or plant cells. These experiments show that DNA is the genetic material in these organisms as well as in the T2 phage. While there are no known exceptions to the generalization that DNA is the genetic material in all cellular organisms and in many viruses, the genetic material in many types of viruses is RNA. Such RNA-containing viruses include the human immunodeficiency virus HIV-1, which causes AIDS (acquired immune deficiency syndrome).

SUMMARY ■ DNA was identified as part of chromosomes, but its importance as genetic material was not initially recognized. ■ Frederick Griffith showed that genetic information can be transferred in a process now known as transformation. ■ Avery, McLeod and McCarty showed that the "transforming principle" was DNA.

■ Hershey and Chase showed that DNA—but not protein—is transferred into host cells during phage infection.

1.2 Structure and replication of DNA The conclusion that DNA is the genetic material left many questions unanswered. How is the DNA in a gene duplicated when a cell divides? How does the DNA in a gene control an inherited trait? What happens to DNA when a mutation (a change in DNA) occurs in a gene?

In the early 1950s, a number of researchers began to elucidate the detailed molecular structure of DNA, hoping that structure alone would provide answers to these questions. In 1953, James Watson and Francis Crick of the University of Cambridge proposed the first essentially correct three-dimensional structure of the DNA molecule. Striking in its elegance, the framework was revolutionary in pointing out how DNA replicates, controls inherited traits and mutates. Even when his tin-and-wire model of the DNA molecule was incomplete, Crick would visit his favorite pub and exclaim, "We've discovered the secret of life." of subunits, each twisted around the other to form a double-stranded helix. The double helix is right-handed, which means that, looking down the barrel, each chain follows a clockwise path as it moves forward. You can visualize the curve on the right in part A of FIGURE 1.6 if you imagine looking at the structure from below. The dark spheres circle the "spine" of each individual strand, rotating clockwise. The subunits of each strand are nucleotides, each containing one of four chemical moieties called bases attached to a phosphorylated molecule of the five-carbon sugar deoxyribose. The four bases of DNA are:

■ Adenine (A) ■ Guanine (G) ■ Thymine (T) ■ Cytosine (C)

FIGURE 1.6 Molecular structure of the DNA double helix in the standard “B form”. (A) A space-filling model in which each atom is represented as a sphere. (B) A diagram highlighting the helical strands around the outside of the molecule and the A-T and G-C base pairs on the inside.

The chemical structures of nucleotides and bases need not concern us at this point. They are discussed in the chapter DNA Structure and Genetic Variation. An important point for our present purposes is that the bases in the double helix are paired, as shown in Figure 1.6B:

At any position on the paired strands of a DNA molecule: ■ If one strand has an A, then the partner strand has a T. ■ If one strand has a G, then the partner strand has a C. The pairing between A and T and between G and C will be complementary; the complement of A is T and the complement of G is C. Complementary pairing means that each base along one strand of DNA corresponds to a base in the opposite position on the other strand. Also: nothing restricts the sequence of bases on a single strand, so any sequence can be present along a strand.

This principle explains how just four bases in DNA can encode the vast amount of information needed to create an organism. It's the sequence of bases along the DNA that encodes genetic information, and the sequence is completely unrestricted. Complementary pairing is also known as Watson-Crick pairing. In the three-dimensional structure shown in Figure 1.6A, the base pairs are represented by the lightest spheres that fill the interior of the double helix. The base pairs lie almost flat, stacked one on top of the other perpendicular to the long axis of the double helix, like coins in a roll. When it comes to a DNA molecule, biologists usually refer to the individual strands as single-stranded DNA and the double helix as double-stranded DNA or duplex DNA.

Each DNA strand has a polarity or direction, like a chain of circus elephants connected by trunk and tail. In this analogy, each elephant corresponds to a nucleotide along the DNA strand. The polarity is determined by the direction in which the nucleotides point. The "stem" end of the tape is referred to as the 3' end of the tape and the "tail" end is referred to as the 5' end. In double-stranded DNA, the paired strands are aligned in opposite directions, with the 5' end of one strand aligned with the 3' end of the other strand. The molecular basis of polarity and the reason for the opposite orientation of strands in duplex DNA is explained in the chapter DNA Structure and Genetic Variation. In illustrating DNA molecules in this text, we use an arrow-shaped strand to represent the backbone and we use protruding tabs on the strand to represent nucleotides. The polarity of a DNA strand is indicated by the direction of the arrow-shaped band. The tail of the arrow represents the 5' end of the DNA strand, the tip to the 3' end.

Beyond the most optimistic hopes, knowledge of the structure of DNA immediately provided clues to its function: 1. The sequence of bases in DNA could be copied using each of the separate "partner" strands as a template to create a new partner strand with a similar sequence. complementary bases. 2. DNA could contain genetic information encoded in the sequence of bases, analogous to the letters printed on a strip of paper. 3. Changes in genetic information (mutations) can result from copying errors in which the base sequence of the DNA has been altered. In the remainder of this chapter, we'll discuss some of the implications of these tips.

An overview of DNA replication Watson and Crick found that the very structure of DNA suggested a mechanism for its replication. "It has not escaped our notice," they write, "that the specific base pairing we postulate points directly to a copying mechanism."

it's called replication. The replication mechanism that Watson and Crick had in mind is shown in FIGURE 1.7.

FIGURE 1.7 DNA replication. (A) Replication of a DNA duplex originally envisioned by Watson and Crick. As the parent strands separate, each parent strand serves as a template for the formation of a new daughter strand through A-T and G-C base pairing. (B) More detail showing how each parent strand serves as a template for the production of a complementary daughter strand that increases in length by the successive addition of single nucleotides to the 3' end.

As shown in panel A of Figure 1.7, the strands of the original (parent) duplex separate, and each individual strand serves as a template or template for the synthesis of a new strand (replica). The replica strands are synthesized by successively adding nucleotides so that each base in the replica is complementary (in the Watson-Crick sense) to the base on the other side of the template strand (Figure 1.7B). Although the mechanism shown in Figure 1.7 is simple in principle, it is a complex process with geometric problems and requires a lot of effort.

Variety of enzymes and other proteins. Details are explored in the chapter on DNA replication and sequencing. The end result of replication is that a single double-stranded molecule is replicated into two copies with identical sequences:

Here, the bases on the newly synthesized strands are shown in red. In the left duplex, the upper strand is the template for the starting molecule and the lower strand is newly synthesized; in the duplex on the right, the bottom strand is the template for the starting molecule and the top strand is newly synthesized. Notice in Figure 1.7B that as each new strand is synthesized, new nucleotides are added only at the 3' end of the growing strand: obligatory elongation of a DNA strand only at the 3' end is an essential feature of DNA replication. . This was a point not recognized by Watson and Crick in 1953, but which became evident when characterizing the cellular machinery of DNA replication.

SUMMARY ■ DNA consists of a double helix, with the two strands held together by complementary base pairs. ■ The two strands of the double helix have opposite polarities.

■ Each strand of a double-stranded DNA molecule serves as a template for the synthesis of a new one during DNA replication.

1.3 Genes and Proteins The structure of DNA alone suggests how two of the four necessary properties of genetic material can be explained. First, complementary base pairing provides a mechanism for replication (Figure 1.7). Second, if you think of the sequence of nucleotides along DNA as a series of letters on a piece of paper, you can think of genes as distinct.

Words that form sentences and paragraphs that make sense of the letter pattern. However, based on the structure of the DNA alone, it is not clear what it encodes. As we shall see, a simultaneous series of experiments showed that it encodes proteins, a class of macromolecules that carry out most of the biochemical activities in the cell. Cells are composed mainly of proteins. These include structural proteins that give the cell rigidity and mobility, proteins that form pores in the cell membrane to control the transport of small molecules into and out of the cell, and receptor proteins that regulate cell activities in response to molecular signals from the growth medium. . or other cells. Proteins are also responsible for most of the metabolic activities of cells. They are essential for the synthesis and breakdown of organic molecules and for generating the chemical energy necessary for cellular activities. In 1878, the term enzyme was introduced to refer to biological catalysts that accelerate biochemical reactions in cells. Around 1900, thanks largely to the work of the German biochemist Emil Fischer, it was demonstrated that enzymes were proteins. As is often the case in science, the "glitches" of nature provide clues about how things work. This was the case with linking genes to disease, because a 'fault' in a gene (a mutation) can result in a 'glitch' (malfunction) in the corresponding protein. This provided a fruitful research avenue for the study of genetics.

Inborn errors of metabolism as a cause of hereditary diseases More than a century ago, the British physician Archibald Garrod recognized that certain hereditary diseases follow the rules of transmission that Mendel had described for his peas. In 1908, Garrod gave a series of lectures in which he presented a basic hypothesis about the relationship between heredity, enzymes and disease: any hereditary disease in which cellular metabolism is abnormal results from an inherited defect in an enzyme. Such diseases became known as congenital metabolic disorders, a term still used today.

Garrod studied several inborn errors of metabolism in which patients excreted abnormal substances in their urine. One such disease was alkaptonuria. In this case, the abnormal substance excreted is homogentisic acid:

An old name for homogentisic acid was alkaptone - hence the name alkaptonuria for this disease. Although alkaptonuria is rare, with an incidence of about one in 200,000 people, it was known before Garrod studied it. The disease itself is relatively mild, but it has one prominent symptom: the patient's urine turns black due to oxidation of homogentisic acid (FIGURE 1.8).

For this reason, alkaptonuria is also known as black urine disease. One of the first cases was described in 1649: the patient was a boy, who was urinating black and, at the age of fourteen, was undergoing a drastic regimen designed to assuage the burning heat in his intestines, which caused the condition in question. become charred. and charred should cause your bile to darken. for the measures

Hemorrhages, purgatives, baths, cold and watery diet and medicines in abundance were prescribed. None of this had any apparent effect, and finally the patient, fed up with futile and unnecessary therapies, decided to let things take their natural course. None of the predicted evils came to pass. He married, raised a large family, and lived a long, healthy life, always urinating like black ink. (Narrated by Garrod, 1908.)

Figure 1.8 The urine of a person with alkaptonuria turns black because the homogentisic acid it contains is oxidized.

Courtesy of Daniel De Aguiar.

Garrod was primarily interested in the biochemistry of alkaptonuria, but he noted familial studies that suggested the disease was inherited as if it resulted from a defect in a single gene. Regarding biochemistry, he concluded that the problem with alkaptonuria was the patients' inability to break the six-carbon phenyl ring present in homogentisic acid. Where is this ring from? Most animals get it from food in their diet. Garrod proposed that homogentisic acid is formed as a breakdown product of two amino acids, phenylalanine and tyrosine, which also contain a phenyl ring. An amino acid is one of the "building blocks" that make up proteins. Phenylalanine and tyrosine are components of normal proteins. The scheme illustrating the relationship between the molecules is shown in FIGURE 1.9. Each of these sequences of biochemical reactions is called a biochemical pathway or a metabolic pathway. Every arrow in the way

represents a single step that represents the transition from the "input" or substrate molecule, shown at the arrowhead, to the "output" or product molecule, shown at the top. Biochemical pathways are usually oriented vertically with arrows pointing down, as in Figure 1.9, or horizontally, with arrows pointing left to right. Garrod did not know all the details of the reaction pathway in Figure 1.9, but he understood that the key step in the degradation of homogentisic acid is the breaking of the phenyl ring, and that the phenyl ring is formed in homogentisic acid.

of dietary phenylalanine and tyrosine.

FIGURE 1.9 Pathway for degradation of phenylalanine and tyrosine. Every step along the path illustrated

by an arrow, requires a specific enzyme to catalyze the reaction. The key step in homogentisic acid degradation is phenyl ring cleavage.

What enables each step in a biochemical pathway? Garrod correctly surmised that each step required a specific enzyme to catalyze the chemical transformation reaction. Individuals with an inborn error of metabolism, such as alkaptonuria, have a defect

in a single step of a pathway because a functional enzyme is missing for that step. When an enzyme is defective in a pathway, the pathway is said to be blocked in that step. A common consequence of a blocked pathway is that the defective enzyme substrate builds up. Garrod observed the accumulation of homogentisic acid in patients with alkaptonuria and suggested that there must be an enzyme whose function is to open the phenyl ring of homogentisic acid and that this enzyme is absent in these patients. The enzyme that opens the phenyl ring of homogentisic acid was not isolated until 50 years after Garrod's lectures. In the average person, it is found in liver cells, and just as Garrod predicted, the enzyme is defective in patients with alkaptonuria. The pathway for phenylalanine and tyrosine degradation, as understood today, is shown in FIGURE 1.10. In this figure, the emphasis is on the enzymes and not on the structures of the metabolites or small molecules on which the enzymes act. Each step along the way requires the presence of a specific enzyme that catalyzes that step.

Fig. 1.10 Inborn errors of metabolism that affect the breakdown of phenylalanine and tyrosine. An inherited disease occurs when one of the enzymes is missing or defective. Alkaptonuria results from a mutated homogentisic acid 1,2-dioxygenase; Phenylketonuria results from a mutation in phenylalanine hydroxylase.

Although Garrod only knew of alkaptonuria, in which the defective enzyme is homogentisic acid 1,2-dioxygenase, we now know the clinical consequences of defects in the other enzymes. Unlike alkaptonuria, which is a relatively benign inherited condition, the others are very serious. The condition known as phenylketonuria (PKU) results from a lack (or defect) of the enzyme phenylalanine hydroxylase (PAH). When this step in the pathway is blocked, phenylalanine builds up. Excess phenylalanine is broken down into harmful metabolites that cause

Defects in myelin formation that damage a child's developing nervous system and lead to severe intellectual disability.

ROOTS OF THE DISCOVERY Black Urine Disease Archibald E Garrod (1908) St Bartholomew's Hospital London, England

Hereditary Metabolic Disorders Although he was a respected physician, Garrod's lectures on the relationship between heredity and inborn errors of metabolism did not have an impact when they were delivered. The important concept that one gene corresponds to one enzyme (the "one gene, one enzyme hypothesis") was developed independently in the 1940s by George W. Beadle and Edward L. Tatum, who used the bread mold Neurospora crassa as a experimental organism. Finally, when Beadle learned of Garrod's hypothesis about inborn errors of metabolism, he praised it generously. This excerpt shows Garrod at his best, weaving together history, clinical medicine, heredity, and biochemistry in his account of alkaptonuria. The excerpt also illustrates how the severity of a genetic condition depends on its social context. Garrod writes as if alkaptonuria were a harmless rarity. In fact, this is especially true when life expectancy is short. With today's longer life expectancy, alkaptonuria sufferers accumulate the dark pigment in their cartilage and joints and may eventually develop arthritis.

"We can also imagine that the cleavage of the phenyl ring in normal metabolism is the work of a special enzyme and that in the

”

Congenital alkaptonuria lacks this enzyme.

Garrod's book Inborn Errors of Metabolism, published in 1903, covered various hereditary metabolic disorders, but two chapters (four and five) were devoted to alkaptonuria. In it he proposes two critical hypotheses. The first hypothesis refers to the origin of homogentisic acid in the urine of affected individuals. Two explanations are possible [for] the fact that alkaptonuric individuals excrete homogentisic acid while normal individuals do not. Either akaptonic acid is a strictly abnormal product formed by a perverse metabolism of tyrosine and phenylalanine, or it is an intermediary of normal metabolism that prevents further alterations in alkaptonuric patients... it far surpasses what can be disputed (p. 38). .

The second hypothesis is the one that proved to be the cornerstone of modern genetics, relating Mendel's laws to metabolic traits. It has been pointed out [by others] that the mode of occurrence of alkaptonuria finds a simple explanation if the anomaly is considered to have a rare recessive character in the Mendelian sense... Of the cases of alkaptonuria, a very large proportion involved the children of a first cousin. ... It is also important to note that when considering families with five or more children [with both normal parents and at least one child with alkaptonuria], the totals are computed strictly according to Mendel's law, that is, 57 [ normal children]:19 [affected children] in a ratio of 3:1 (p. 23-25). It's totally understandable why Beadle praised Garrod. In fact, the Inborn Errors of Metabolism hypotheses were essentially those tested by Beadle and Tatum in their Neurospora experiments. Remarkably, Garrod arrived

synthesizing medical and physiological observations (own and foreign) that were not made as part of the genetic analysis. In fact, in 1903, this context was still incipient. Source: Harris, H. (1963). Garrod, A. (1903) Congenital metabolic disorders. Oxford University Press.

However, if PKU is diagnosed in children shortly after birth, they may be put on a specially formulated diet that is low in phenylalanine. The child receives only the amount of phenylalanine that can be used for protein synthesis, ensuring that excess phenylalanine does not build up. The special diet is very strict. Excludes meat, poultry, fish, eggs, milk and dairy products, legumes, nuts, and baked goods made with regular flour. These foods are replaced with an expensive synthetic formula. However, with the special diet, the adverse effects of excess phenylalanine on intellectual development can be largely avoided, although in adult women with phenylketonuria who are pregnant, the fetus is at risk.

In many countries, including the United States, all newborns have their blood tested for chemical signs of phenylketonuria. Routine screening is inexpensive because PKU is relatively common. In the United States, the incidence of Caucasian births is about 1 in 8,000. The disease is less common in other ethnic groups. In the metabolic pathway shown in Fig. 1.10, defects in the degradation of tyrosine or 4-hydroxyphenylpyruvic acid lead to forms of tyrosinemia. Like PKU, these diseases have serious implications: type II is associated with skin lesions and intellectual disability, type III with severe liver impairment.

The genes for the enzymes in the biochemical pathway of Figure 1.10 were all identified and the DNA nucleotide sequence determined. In the following list and throughout the text, we use the standard typographical convention that genes are italicized, while gene products are not italicized. This convention is convenient because it means that the protein product of a gene can be represented with the same symbol as the gene itself, but while the gene symbol is italicized, the protein symbol is not. In Figure 1.10, the numbers correspond to the following genes and enzymes: 1. The PAH gene on the long arm of chromosome 12 encodes phenylalanine hydroxylase (PAH). 2. The TAT gene on the long arm of chromosome 16 encodes tyrosine aminotransferase (TAT). 3. The HPD gene on the long arm of chromosome 12 encodes 4-hydroxyphenylpyruvic acid dioxygenase (HPD). 4. The HGD gene on the long arm of chromosome 3 encodes homogentisic acid 1,2-dioxygenase (HGD).

SUMMARY ■ Archbold Garrod identified the genetic basis of 'hereditary metabolic disorders'. ■ Diseases such as alkaptonuria result from the absence of a specific enzyme in a biochemical pathway. ■ In some inborn errors of metabolism, such as phenylketonuria, dietary control can prevent the development of undesirable symptoms.

1.4 Genetic Analysis The genetic implications of Garrod's research on birth defects in

metabolism was not universally appreciated, probably because his writings focused primarily on biochemical pathways rather than heredity. And, of course, as genes and enzymes were studied in humans, the potential for classical genetic analysis was limited. The definitive link between genes and enzymes emerged from studies conducted in the 1940s by George W. Beadle and Edward L. Tatum using a filamentous fungus Neurospora crassa commonly referred to as red bread mold - an organism they chose because genetic analysis and biochemistry were possible without any problems. In these experiments, Beadle and Tatum identified new mutations, each of which caused a block in the pathway for the synthesis of a necessary nutrient, and they showed that each of these blocks corresponded to a defective enzyme necessary for a step in the pathway. His research was important not only because it cemented the link between genetics and biochemistry, but also because his experimental approach, now known as genetic analysis, has been used successfully to understand a wide range of biological processes, from the genetic control of the cell cycle and cancer on genetic influences on development and behavior. For that reason, we'll look at the Beadle-Tatum experiments in a little more detail.

ROOTS OF THE DISCOVERY One Gene, One Enzyme George W. Beadle and Edward L. Tatum (1941) Stanford University, Stanford, California

Genetic control of biochemical reactions in Neurospora How do genes control metabolic processes? The suggestion that genes control enzymes was made very early in the history of genetics, most notably by the British physician Archibald

Garrod in his 1903 book Inborn Errors of Metabolism. Despite this, the exact relationship between genes and enzymes was still uncertain. Perhaps each enzyme is controlled by more than one gene, or perhaps each gene contributes to the control of multiple enzymes. The classic experiments by Beadle and Tatum showed that the relationship is often remarkably simple: one gene codes for one enzyme. Their pioneering experiments combined genetics and biochemistry, and the "one gene, one enzyme" concept won Beadle and Tatum the 1958 Nobel Prize (Joshua Lederberg shared the prize for his contributions to microbial genetics). Now that we know that some enzymes contain polypeptide chains encoded by two (or occasionally more) different genes, a more accurate statement of principle is "one gene, one polypeptide". Beadle and Tatum's experiments also show the importance of choosing the right organism. Neurospora had been introduced as a genetic organism only a few years earlier, and Beadle and Tatum realized that they could take advantage of this organism's ability to grow in a simple medium composed of known substances.

“These preliminary results seem to suggest to us that the approach could hold promise as a method to learn more about how genes regulate development.

”

Occupation.

Beadle and Tatum's work was published in 1941 and can be found in the reference at the end of this article. In it, they point out the limitations of starting with the physiological basis of a trait (such as black urine disease) and trying to determine its genetic basis. First, these analyzes are limited to traits whose variants are non-lethal. Second, variants must have visible effects. To get around these problems, Beadle and Tatum turned the problem on its head.

[These limitations] led us to examine the general problem of the genetic control of developmental and metabolic responses by reversing the usual procedure... [beginning to] determine whether and how genes control known biochemical responses... Whether the organism should be capable of carrying out a given chemical reaction in order to survive in a given medium, a mutant unable to do so will obviously be fatal in that medium. genetically blocked was added.... So instead of starting with observed differences in characteristics between individuals, Beadle and Tatum started by generating mutations (in their case, mutations resulting from X-ray irradiation of Neurospora cells) and then identifying those that are lethal in minimal medium but in supplemented medium. with the normal product of the mutant gene. This experimental approach is one of the most important experimental tools in genetic analysis. Source: G.W. Beadle and EL Tatum, Genetic Control of Biochemical Reactions in Neurospora. Proc.Natl. Academic Sciences. USA 27 (1941): 499-506.

Mutant genes and defective proteins

N. crassa grows in the form of filaments on a variety of substrates, including laboratory media containing only inorganic salts, a sugar, and the vitamin biotin. This minimal medium contains only the nutrients necessary for the growth of the organism. N. crassa filaments consist of a mass of branching filaments divided into interconnected, multinucleated compartments that allow free exchange of nuclei and cytoplasm. Each nucleus contains a unique set of seven chromosomes. Beadle and Tatum recognized that the ability to

Growing Neurospora on minimal medium implies that the organism must be able to synthesize all metabolic components except biotin. If the biosynthetic pathways necessary for growth are controlled by genes, then a mutation in a gene responsible for synthesis of an essential nutrient would render a strain unable to grow unless the strain is supplied with the nutrient. These ideas were tested as follows. Non-mutant Neurospora spores were irradiated with X-rays or ultraviolet light to generate mutant strains with different nutritional requirements. (Why these treatments cause mutations is discussed in the Mutation, Repair, and Recombination chapter.) Isolating a set of mutants that affect a biological process—in this case, metabolism—is called mutant screening. In the first step of mutant identification, summarized in FIGURE 1.11, irradiated spores (purple) were used in crosses with an untreated strain (green). The ascospores produced by the sexual cycle in fruiting bodies were germinated individually in complete medium - that is, H. a complex medium enriched in a variety of amino acids, vitamins and other substances considered essential metabolites whose synthesis can be blocked by mutation. Even those ascospores containing a new mutation that affects the synthesis of an essential nutrient are expected to germinate and grow in complete medium.

Figure 1.11 Beadle and Tatum obtained mutants of the filamentous fungus Neurospora crassa by exposing asexual spores to X-rays or ultraviolet light. Treated spores were used to initiate the sexual cycle in fruiting bodies. After any pair of cells and their nuclei fuse, meiosis occurs almost immediately, resulting in eight sexual spores (ascospores) contained within a single ascus. These are removed individually and grown in complete medium. Ascospores carrying new feeding mutants are further identified by their inability to grow on minimal medium.

To identify which of the irradiated ascospores contained a new mutation that affected the synthesis of an essential nutrient, the conidia from each culture were transferred to minimal medium (FIGURE 1.12A). The vast majority of cultures can be grown on minimal media; these were discarded as they did not have any new mutations of the desired type. The retained cultures were the small number that failed to grow on minimal medium because they contained a new mutation that blocked the synthesis of an essential nutrient.

FIGURE 1.12 (A) Mutant spores can grow on complete medium but not on minimal medium. (B) Each new mutant is tested for growth in minimal medium supplemented with a nutrient mixture. (C) Mutants capable of growing in minimal medium supplemented with amino acids are tested with each amino acid individually. (D) Mutants that cannot grow in the absence of arginine are tested with probable arginine precursors.

Samples of a portion of each mutant culture were then transferred to a variety of media to determine whether the mutation resulted in a requirement for a vitamin, amino acid, or other substance (FIGURE 1.12B). In the example shown, the mutant strain requires one (or possibly more than one) amino acid, as a mixture of all amino acids added to the minimal medium allows for growth. As the proportion of irradiated cultures with new mutations was very small, only an insignificant number of cultures contained two or more newly emerging mutations.

simultaneously. For nutritional mutants that require amino acids, additional experiments testing each amino acid individually generally revealed that only one additional amino acid was needed in minimal medium to support growth. In FIGURE 1.12C, the mutant strain requires the amino acid arginine. Some of the possible intermediate steps in amino acid biosynthesis were identified as early as the 1940s. They were recognized by their chemical similarity to the amino acid and by their presence in small amounts in the cells of organisms. In the case of arginine, two candidates were ornithine and citrulline. All mutants requiring arginine were therefore tested in medium supplemented with ornithine alone or citrulline alone (Fig. 1.12D). One class of arginine-requiring mutants, designated class I, was able to grow on minimal medium supplemented with ornithine, citrulline, or arginine. Other mutants, designated Class II, were able to grow on minimal medium supplemented with citrulline or arginine, but not ornithine. A third class, class III, could only grow on minimal medium supplemented with arginine. The types of mutants that require arginine illustrate the logic of genetic analysis applied to metabolic pathways. The logic is easier to see in the context of the metabolic pathway depicted in FIGURE 1.13, where arginine is the end product of a linear process.

Metabolic pathway starting with some precursor metabolites and ornithine and citrulline are intermediates along the way. Mutants support this pathway structure for the following reasons: ■ Mutants that can grow in the presence of ornithine, citrulline, or arginine must have a metabolic block between the parent metabolite and the two intermediates. ■ Mutants that can only grow in the presence of arginine must have a metabolic block in the pathway between arginine and the most “downstream” intermediate.

■ Mutants that can grow in the presence of citrulline or arginine but not ornithine imply that ornithine is “upstream” of citrulline in the pathway.

FIGURE 1.13 Arginine biosynthetic pathway derived from genetic analysis of Neurospora mutants.

The structure of the pathway was further confirmed by the observations that class III mutants accumulate citrulline and class II mutants accumulate ornithine. Finally, it was demonstrated by direct biochemical experiment that the inferred enzymes were indeed present in non-mutant strains, but were absent or non-functional in mutant strains.

Complementation testing for mutations in the same gene Beadle and Tatum were lucky to study metabolic pathways in a relatively simple organism, in which one gene corresponds to another

Enzyme. In such a situation, genetic analysis of the mutants reveals much more about the metabolic pathway than just the order of the intermediates. If each mutation is classified according to the particular gene it resides in, and if all the mutations in each gene are grouped together, each set of mutations—and therefore each individual gene—corresponds to an enzymatic step in the pathway. For example, in Figure 1.13, the mutants in class I each correspond to one of three genes, implying that there are three steps in the path between the parents.

and ornithine. Likewise, class III mutants each correspond to one of two genes, implying that there are two steps in the pathway between citrulline and arginine. However, all class II mutants have mutations in the same gene, implying only one step in the pathway between ornithine and citrulline. Mutations that cause defects in the same gene are identified by a complementation test, which pools two mutations in the same cell. In most multicellular organisms and even some sexual single-celled organisms, this is usually done through mating. When two parents, each carrying one of the two mutations, are crossed, fertilization brings together the reproductive cells that contain the two mutations. As a result of normal cell division, each cell of the offspring carries one copy of each mutated gene. This method does not work in Neurospora because nuclear fusion is almost immediately followed by the formation of ascospores, each of which has only one set of chromosomes. Complementation assays are still possible in Neurospora due to the multinucleated nature of the filaments. Certain strains, including those studied by Beadle and Tatum, have the property that when the filaments of two mutant (or non-mutant) organisms come into physical contact, the filaments and the new filament fuse together.

contains several cores from each of the partners involved. This type of hybrid strand, called a heterokaryote, contains mutated forms of both genes. The word roots of the term heterokaryon means "different nuclei". (A list of the most commonly used stems in genetics is provided at the end of this text.) When a heterokaryon formed by two nutritional mutants is inoculated into minimal medium, it may or may not grow. When grown in minimal medium, the mutated genes are said to undergo complementation and this result indicates that the mutations are in different genes. On the other hand, if the heterokaryon does not grow in minimal medium, the result indicates non-complementation and it is concluded that the two mutations are in the same gene.

To illustrate this concept, consider two hypothetical examples: 1. Heterokaryons are formed between class II and class III mutants. In this case, nuclei containing the Class II mutants are unable to convert ornithine to citrulline, but produce the enzyme (or enzymes) needed to convert citrulline to arginine. In contrast, class III mutants can convert ornithine to citrulline but are unable to convert citrulline to arginine. Both steps can therefore take place in the same heterokaryon, so that growth takes place on minimal medium. These two mutations are considered complementary. 2. Heterokaryotes are formed between two independently derived class II mutations. None of these mutations can make the step from ornithine to citrulline; therefore, no growth occurs on minimal medium. These mutations are not complementary.

Let us now consider two class III mutations. Are they complementary? If the conversion of citrulline to arginine involves only a single biochemical step, then we would expect them to be non-complementary. However, Beadle and Tatum found that some pairs of mutants complemented each other, while others did not. The theory that a gene codes for an enzyme led them to conclude that more than one enzyme is needed for this step. Completion or non-complementation conclusions

result from the logic shown in FIGURE 1.14. Here the multinucleated strand is shown, and the mutated nuclei are color-coded according to which of two different genes (red or purple) is mutated. The red and purple lines represent the proteins encoded in the mutant nuclei, and an asterisk represents a defect in the protein at that position resulting from a mutation in the corresponding gene.

FIGURE 1.14 Molecular interpretation of a heterokaryotic complementation assay to determine whether two mutant strains have mutations in different genes (A) or mutations in the same gene (B). In (A) each nucleus contributes an unmutated form of one or another polypeptide chain, so that the heterokaryon is able to grow in minimal medium. In (B) both cores contribute a mutant form of the same polypeptide chain; therefore, a non-mutant form of this polypeptide cannot be synthesized and the heterokaryon is unable to grow on minimal medium.

Part A shows the situation where mutant strains have mutations in different genes. In the heterokaryote, red nuclei produce mutant forms of red protein and normal forms of purple protein, while purple nuclei produce mutant forms of protein.

the purple protein and the normal forms of the red protein. The result is that the red/purple heterokarion has normal forms of the red and purple proteins. He also has mutated forms of both proteins, but that doesn't matter. What matters is that the normal forms allow the heterokaryon to grow in minimal media because all the necessary nutrients can be synthesized. In other words, the normal violet gene in the red nucleus complements the defective violet gene in the violet nucleus, and vice versa. The logic of complement is captured in the old nursery rhyme, "Jack Sprat couldn't eat fat / His wife couldn't eat lean meat / And so between the two of them / They licked the plate clean", because each partner compensates for the deficiency in the other.

Panel B in Figure 1.14 shows a heterokarion formed between mutants with defects in the same gene—purple in this case. Both purple cores encode a normal form of red protein, but each purple core encodes a defective purple protein. When the nuclei are together, two different mutant forms of the purple protein are produced, so the biosynthetic pathway that requires the purple protein is still blocked and the heterokaryon is unable to grow on minimal medium. In other words, mutants 2 and 3 in Figure 1.14 do not complement each other, so they are assumed to have mutations in the same gene. The complementation test is based on the following principle:

Principle of complementation: A complementation test brings together two mutated genes in the same cell or organism. If that cell or organism is not a mutant, the mutations are said to be complementary to each other, and the parental lineages must have mutations in different genes. If the cell or organism mutates, the mutations cannot complement each other and the parental mutations must be in the same gene. These last mutations form a complementation group.

Supplementary data analysis

In screening for arginine-requiring Neurospora mutants, Beadle and Tatum found that mutants in different classes (class I, class II, and class III in Figure 1.13) were always complementary. This result makes sense because the genes of each class encode enzymes that act at different levels among the known intermediates. However, some of the class I mutants could not complement other class I ones, and some of the class III ones could not complement other class III ones. These results allow identifying the number of genes in each class. To illustrate this aspect of genetic analysis, consider six class III mutant strains. These strains were taken in pairs to form heterokaryons and their growth in minimal medium was evaluated. The data are shown in FIGURE 1.15. Mutant genes in the six strains are designated x1, x2, etc., and the data are presented in the form of a matrix where a plus sign (+) indicates growth on minimal medium (complementation) and a minus sign (-) indicates no growth on minimal medium (lack of complementation). The diagonal entries are all minus signs, reflecting the fact that two copies of the identical mutation cannot show complementation. The pattern of + and − signs in the matrix

indicates that the x1 and x5 mutations are not complementary; therefore, x1 and x5 are mutations in the same gene. Likewise, the x2, x3, x4, and x6 mutations do not complement each other in all combinations of pairs; therefore, x2, x3, x4, and x6 are all mutations in the same gene, but in a different gene than the one represented by x1

e x5.

FIGURE 1.15 Interpretation of complementation test results. Rows and columns are mutation specific and independently derived. Plus signs indicate pairs that complement each other; Minuses are those that do not. A complementation group is defined as all mutations that do not complement each other but complement all other mutations. In this case x1 and x4 form one group while x2, x3, x4 and x6 form a second group.

Each of the groups of non-complementary mutations is called a complementation group. As we have seen, each complementation group defines a gene, so the complementation test really provides the geneticist's working definition:

A gene is experimentally defined as a set of mutations that form a single complementation group. Each pair of mutations within a complementation group cannot complement each other. The mutations in Figure 1.15, therefore, represent two genes, and mutation of one of them results in the strain's inability to convert citrulline to arginine. Assuming that a gene encodes an enzyme, which is largely true for neurospora metabolic enzymes, the pathway from citrulline to arginine shown in Figure 1.13 must consist of two steps, with an unknown intermediate between the steps. This intermediate was later determined to be argininosuccinate. Likewise, class I mutants define three complementation groups, so there are three enzymatic steps from precursor to ornithine. These intermediaries were also soon identified. Finally, Class II mutations are not mutually complementary, and finding only one complementation group means that there is only a single enzymatic step that converts ornithine to citrulline.

Other Applications of Genetic Analysis The type of genetic analysis pioneered by Beadle and Tatum is immensely powerful for identifying the genetic control of complex biological processes. Their approach outlines a systematic way—a cookbook, if you will—for gene discovery, a way we'll see throughout this text: 1. Decide which process you want to study and find out what properties mutated organisms would exhibit a disruption in that process. process .

2. Screen mutants for mutants that exhibit these characteristics. 3. Do complementation tests to find out how many different genes you have identified. 4. Identify the products of these genes and determine what they do, how they interact with each other, and the order in which they function. Beadle and Tatum analyzed many metabolic pathways across a broad spectrum

variety of essential nutrients, but his experiments were particularly important in deciphering the pathways of amino acid biosynthesis. Their discoveries over a period of just a few years were remarkable, as they were able to deduce far more about the nature of biosynthetic pathways than they had learned in decades of biochemical research. Beadle and Tatum were awarded the 1958 Nobel Prize in Physiology or Medicine for their research, and in the following years many more Nobel Prizes in Physiology or Medicine were awarded in which genetic analysis modeled on Beadle and Tatum played a significant role. Here is a citation list of official Nobel Foundation citations: ■ 1958: George Beadle and Edward Tatum "for their discovery that genes function by regulating certain chemical processes," along with Joshua Lederberg "for their discoveries in genetic recombination and the organization of the genetic material of bacteria.” ■ 1965: François Jacob, André Lwoff and Jacques Monod “for their discoveries concerning the genetic control of enzyme and virus synthesis”. ■ 1995: Edward B. Lewis, Christiane Nüsslein-Volhard, and Eric F. Wieschaus "for their discoveries concerning the genetic control of early embryonic development."

■ 2001: Leland H. Hartwell, Tim Hunt, and Sir Paul Nurse "for their discoveries of key regulators of the cell cycle." ■ 2002: Sydney Brenner, H. Robert Horvitz, and John E. Sulston "for their discoveries in the genetic regulation of organ development and programmed cell death." ■ 2007: Mario R. Capecchi, Martin J. Evans, and Oliver Smithies "for discovering the principles for introducing specific genetic changes into mice using embryonic stem cells." ■ 2009: Elizabeth Blackburn, Carol Greider, and Jack Szostak "for discovering how chromosomes are protected by telomeres and the enzyme telomerase." ■ 2013: James Rothman, Randy Shekman, and Thomas Sudhof "for their discoveries of mechanisms that regulate vesicle trafficking, an important transport system in our cells."

■ 2015: Tomas Lindahl, Paul Modrich, and Aziz Sancar “for mechanistic studies of DNA repair”. Beadle and Tatum's experiments showed that a defective enzyme results from a mutated gene, but how? As far as they knew, genes were enzymes. That would have been a logical assumption at the time. We now know that the relationship between genes and enzymes is more indirect. With few exceptions, each enzyme is encoded in a specific sequence of nucleotides present in a region of DNA. The region of DNA that encodes the enzyme, as well as neighboring regions that regulate when and in which cells the enzyme is produced, form the "gene" that encodes the enzyme. Next, we'll look at how genes code for enzymes and other proteins.

SUMMARY ■ Genetic analysis is an experimental technique in which inferences based on the results of genetic crosses can be used to elucidate a variety of biological processes. ■ Using Neurospora crassa, Beadle and Tatum developed methods for screening mutants. ■ Complement analysis is a powerful tool for determining the number of steps in a biochemical pathway.

■ These analyzes led Beadle and Tatum to the “one gene, one enzyme” hypothesis.

1.5 Gene expression: the central dogma Watson and Crick were right when they proposed that the genetic information in DNA is contained in the sequence of bases, analogous to the letters printed on a strip of paper. In a region of DNA that controls the synthesis of a protein, the protein's genetic code is contained on a single strand and is decoded in linear order. A typical protein consists of one or more

polypeptide chains; Each polypeptide chain consists of a linear sequence of amino acids joined end to end. For example, the PAH enzyme consists of four identical polypeptide chains, each 452 amino acids long. When decoding DNA, each consecutive "code word" in the DNA specifies the next amino acid to be added to the polypeptide chain during its manufacture. The amount of DNA needed to encode the PAH polypeptide chain is therefore 452 x 31356 nucleotide pairs. The entire gene is much longer - about 90,000 nucleotide pairs. Only 1.5% of the gene is used to code for amino acids. The non-coding part contains some sequences that control gene activity, but it is not known how much of the gene is involved in regulation. There are 20 different amino acids encoded by DNA. Only four bases encode these 20 amino acids, and each "word" of the genetic code is composed of three contiguous bases. For example, the ATG base sequence specifies the amino acid methionine (Met), TCC specifies serine (Ser), ACT specifies threonine (Thr), and GCG specifies alanine (Ala). There are 64 possible combinations of three bases, but only 20 amino acids because some combinations specify the same amino acid. For example, TCT, TCC, TCA, TCG, AGT and AGC encode serine (Ser) and CTT, CTC, CTA, CTG, TTA and TTG encode leucine (Leu).

An example of the relationship between the nucleotide sequence in a DNA duplex and the amino acid sequence of the corresponding protein is shown in FIGURE 1.16. This particular DNA duplex is the human sequence encoding the first seven amino acids in the PAH polypeptide chain. The scheme sketched in Figure 1.16

points out that DNA does not encode proteins directly, but indirectly through transcription and translation processes. Thus, the indirect form of transmission of information is:

FIGURE 1.16 DNA sequence encoding the first seven amino acids in a polypeptide chain. The DNA sequence specifies the sequence of amino acids through an RNA molecule that serves as an intermediary "messenger". Although the decoding process is indirect, the end result is that each amino acid in the polypeptide chain is specified by a set of three adjacent bases in the DNA. In this example, the polypeptide chain is that of phenylalanine hydroxylase (PAH).

This relationship brings us to the central tenet of molecular genetics. The term dogma means "belief"; dates back to when the idea was first theorized. Since then, the "dogma" has been confirmed experimentally, but the term endures. The core concept of the central dogma, as stated by Francis Crick in 1958, is as follows:

Once "information" enters the proteins, it cannot get out. Regarding gene expression in most cells, the central dogma can be stated as shown in FIGURE 1.17. The "information" is what is encoded in the DNA and specifies the protein sequences. Its transition from DNA to protein does not occur directly, but through interactions with the intermediary molecule known as ribonucleic acid (RNA). The structure of RNA is similar but not identical to the structure of DNA. There is a difference in the sugars in DNA and RNA (RNA contains the sugar ribose instead of deoxyribose), RNA is usually single-stranded (no duplex), and RNA contains the base uracil (U) instead of thymine (T ). present in the DNA. The flow of information between DNA and RNA follows the rules of Watson-Crick base pairing and, as we will see in other parts of this text, in some cases it can occur in either direction (DNA → RNA or RNA → DNA). However, the flow of information from RNA to proteins requires the translation of nucleic acid sequences into polypeptide sequences. This process is irreversible. There is no known case where a protein sequence or a nucleic acid sequence (either DNA or RNA) encodes.

FIGURE 1.17 Gene expression at the molecular level. DNA encodes RNA and RNA encodes protein. The DNA → RNA step is transcription and the RNA → protein step is translation.

Three types of RNA are involved in protein synthesis: ■ A molecule, messenger RNA (mRNA), which carries genetic information from DNA and serves as a template for polypeptide synthesis. In most mRNA molecules, there is a high proportion of nucleotides that actually code for amino acids. For example, the mRNA for PAH is 2400 nucleotides long and encodes a polypeptide of 452 amino acids; in this case, more than 50% of the mRNA length encodes amino acids. ■ Four types of ribosomal RNA (rRNA), which are the main components of cellular particles called ribosomes, on which polypeptide synthesis takes place. ■ A set of approximately 45 transfer RNA (tRNA) molecules, each carrying a specific amino acid, and a

Three-base recognition region that forms base pairs with a group of three adjacent bases in the mRNA. As each tRNA participates in translation, its amino acid becomes the terminal subunit that is added to the length of the growing polypeptide chain. A tRNA that carries methionine is called tRNAMet, one that carries serine is called tRNASer, and so on. (Since there are more than 20 different tRNAs but only 20 amino acids, some amino acids correspond to more than one tRNA.) The central dogma is the fundamental principle of molecular genetics because it summarizes how the genetic information in DNA is expressed in the amino acid sequence of acids in a polypeptide chain:

The sequence of nucleotides in a gene provides the sequence of nucleotides in a messenger RNA molecule; The sequence of nucleotides in messenger RNA, in turn, indicates the sequence of amino acids in the polypeptide chain. Given a conceptually simple process like DNA-coding proteins, what might account for the added complexity of RNA mediators? One possible explanation is that an RNA intermediate provides another level of control - for example, degrading the mRNA to an unnecessary protein. Another possible reason may be historical. The structure of RNA is unique in that it has informational content in its base sequence and a complex, folded, three-dimensional structure that endows some RNA molecules with catalytic activities. Many scientists believe that RNA played roles in early forms of life, both as genetic information and in catalysis. In the course of evolution, the information function was transferred to the DNA and the

catalytic role for proteins. However, RNA was locked into its central location as an intermediary in the processes of information transfer and protein synthesis. This hypothesis implies that the involvement of RNA in protein synthesis is a relic of the earliest stages of evolution - a "molecular fossil". It is supported by a variety of observations. For example, (1) DNA replication requires an RNA molecule to begin, (2) an RNA molecule is essential for the synthesis of the ends of chromosomes, and (3) some RNA molecules catalyze key reactions in protein synthesis. .

transcription

FIGURE 1.18 shows how genetic information is transferred from DNA to RNA. The DNA opens and one of the strands is used as a template for the synthesis of a complementary strand of RNA. (How the template strand is selected is discussed in the Molecular Biology of Gene Expression chapter.) The process of making an RNA strand from a DNA template is called transcription, and the RNA molecule made is transcription. The sequence of bases in RNA is complementary (in the sense of Watson-Crick pairing) to that of the DNA template, except that U (which pairs with A) is present in RNA instead of T. The rules of base pairing between DNA and ANN are summarized in FIGURE 1.19.

Figure 1.18 Transcription is the production of a strand of RNA that is complementary in base sequence to a strand of DNA. In this example, the DNA strand below is transcribed into an RNA strand. Note that in an RNA molecule, the base U (uracil) plays the role of T (thymine) by pairing with A (adenine). Each A-U pair is marked.

Figure 1.19 Base pairing in DNA and RNA. DNA bases A, T, G, and C pair with RNA bases U, A, C, and G, respectively.

Each strand of RNA has a polarity - a 5' end and a 3' end - and, as in DNA synthesis, nucleotides are added only at the 3' end of a growing strand of RNA. Therefore, the 5' end of the RNA transcript is synthesized first, and transcription proceeds along the DNA template strand in the 3' to 5' direction. Each gene contains nucleotide sequences that initiate and terminate transcription. The RNA transcript made from each gene begins at the initiation site on the template strand, located "upstream" of the amino acid coding region, and ends at the termination site, located "downstream" of the amino acid coding region. For each gene, the length of the RNA transcript is much smaller than the length of the DNA in the chromosome. For example, the PAH gene transcript for phenylalanine hydroxylase is about 90,000 nucleotides long, but the DNA in chromosome 12 is about 130,000,000 nucleotide pairs. In this case, the length of the PAH transcript is less than 0.1% of the length of the DNA in the chromosome. A different gene on chromosome 12 would be transcribed from a different region of the DNA molecule on chromosome 12 and perhaps from the opposite strand, but again the transcribed region would be small compared to the total length of DNA on the chromosome.

Translation Synthesis of a polypeptide under the direction of an mRNA molecule is called translation. Although the sequence of bases in mRNA encodes the sequence of amino acids in a polypeptide, the molecules that actually do the "translation" are the tRNA molecules. The mRNA molecule is translated into non-overlapping groups of three bases called codons. For every codon in mRNA that specifies an amino acid, there is a tRNA molecule that contains a complementary set of three adjacent bases that can pair with the codon. The correct amino acid is added to one end of the tRNA, and when the tRNA comes in line, the amino acid to which it is attached becomes the most recent addition to the growing end of the polypeptide chain. The role of tRNA in translation is shown in FIGURE 1.20 and can be described as follows:

The mRNA is read codon by codon. Each codon that specifies an amino acid corresponds to a complementary set of three contiguous bases on a single tRNA molecule. One end of the tRNA is attached to the correct amino acid, so the correct amino acid is matched.

Figure 1.20 The role of messenger RNA in translation is to transport the information contained in a sequence of DNA bases to a ribosome, where it is translated into a polypeptide chain. Translation is mediated by transfer RNA (tRNA) molecules, each of which can base pair with a set of three contiguous bases in the mRNA. Each tRNA also carries an amino acid. As each tRNA is delivered to the ribosome in turn, the growing polypeptide chain is elongated.

The tRNA molecules used in translation do not line up along the mRNA simultaneously, as shown in Figure 1.20. The translation process takes place on a ribosome, which attaches to a single mRNA and moves step by step along it from end to end, three nucleotides at a time (codon by codon). With each new codon, the next tRNA binds to the ribosome. Then the growing end of the polypeptide chain is attached to the amino acid in the tRNA. In this way, each tRNA, in turn, serves temporarily to hold the polypeptide chain during its synthesis. As the polypeptide chain is transferred from each tRNA to the next in the series, the tRNA that previously contained the polypeptide is released from the ribosome. The polypeptide chain increases by one amino acid at each step until one of three distinct codons that specify "stop" is encountered. At this point, synthesis of the chain of amino acids occurs

acids is complete and the polypeptide chain is released from the ribosome. (This brief description of the translation glosses over many of the details presented in the chapter on the molecular biology of gene expression.)

The Genetic Code Figure 1.20 shows that the mRNA codon AUG specifies methionine (Met) in the polypeptide chain, UCC specifies Ser

(serine), ACU specifies Thr (threonine), and so on. The complete decoding table is called the genetic code and is shown in TABLE 1.1. For each codon, the left column corresponds to the first nucleotide in the codon (reading from the 5' end), the row above corresponds to the second nucleotide, and the right column corresponds to the third nucleotide. The full codon is given in the body of the table, along with the amino acid (or translation "stop") that specifies the codon. Each amino acid is identified by its full name plus a three-letter abbreviation and a one-letter abbreviation. Both types of abbreviations are used in molecular genetics.

TABLE 1.1 The standard genetic code

The code in Table 1.1 is the "standard" genetic code used in translation in the cells of almost all organisms. In the Molecular Biology of Gene Expression chapter, we examine the general features of the standard genetic code and the subtle differences found in the genetic codes of specific organisms and cell organelles. Here we are primarily interested in how the genetic code is used to translate the codons in the mRNA into amino acids in a polypeptide chain. In addition to the 61 codons that code for amino acids only, there are four codons that have specialized functions: ■ The AUG codon, which specifies Met (methionine), is also the "start" codon for polypeptide synthesis. The placement of a tRNAMet linked to AUG is one of the first steps in the initiation of polypeptide synthesis, therefore, all polypeptide chains begin with Met. (For many polypeptides, the initial Met is cleaved after translation is complete.) uses met

for translation initiation is the same tRNAMet used to specify methionine at internal positions in a polypeptide chain. ■ The UAA, UAG, and UGA codons are each a “stop” that indicates the completion of translation, leading to release of the complete polypeptide chain from the ribosome. These codons do not have tRNA molecules that recognize them, but they are recognized by protein factors that terminate translation. How the genetic code table is used to infer the amino acid sequence of a polypeptide chain can be illustrated again using PAH - specifically the DNA sequence encoding amino acids 1-7. The DNA sequence is:

This region is transcribed from left to right in RNA. As the RNA grows by adding successive nucleotides to the 3' end (Figure 1.18), the bottom strand is transcribed. The RNA nucleotide sequence is that of the top strand of DNA, except that U replaces T, so the mRNA for amino acids 1 through 7 is:

Codons are read from left to right according to the genetic code shown in Table 1.1. The AUG codon encodes Met (methionine), UCC encodes Ser (serine), and so on. In general, the amino acid sequence of this region of the polypeptide is:

or in terms of one-letter abbreviations:

The complete decoding operation for this region of the PAH gene is shown in FIGURE 1.21. In this figure, the AUG initiation codon is highlighted because some patients with PKU have a mutation in this particular codon. As would be expected from the fact that AUG is the initiation codon for polypeptide synthesis, cells in patients with this particular mutation do not produce any of the PAH polypeptides.

Then mutations and their consequences are considered.

Figure 1.21 The central dogma of typical gene expression. The DNA encoding PAH serves as a template for making a messenger RNA, and the mRNA serves to specify the sequence of amino acids in the PAH polypeptide chain through interactions with the ribosome and tRNA molecules. The total number of amino acids in the PAH polypeptide chain is 452, but only the first 7 are shown.

SUMMARY ■ Proteins are made up of polypeptides, which are chains of amino acids whose order is determined by the sequence of bases in the genes that encode them. ■ The central dogma of molecular biology states that information in the form of molecular sequences passes from nucleic acids to proteins, but not vice versa. ■ In gene expression, this process involves the transcription of DNA into messenger RNA and the translation of mRNA into protein. ■ In addition to mRNA, ribosomal RNA and transfer RNA play a crucial role in the translation process.

■ The standard genetic code consists of 64 three-letter words or codons, 61 of which code for amino acids.

1.6 Mutation and Variation We now turn to the fourth essential property of genetic material - namely, its ability to generate variation through mutation. The term mutation refers to any heritable change in a gene (or more generally in the genetic material) or in the process by which such a change occurs. One type of mutation results in a change in the sequence of bases in DNA. This type of change can be simple,

how to replace a base pair in a duplex molecule with another base pair. For example, a C-G pair in a duplex molecule can mutate to T-A, A-T, or G-C. Changing the nucleotide sequence can also be more complex, such as deleting or adding base pairs. These and other types of mutations are covered in the Mutation, Repair, and Recombination chapter.

THE WOMEN IN THE WEDDING PHOTO are sisters. Both have two copies of the same mutated PAH gene. The bride is the younger of the two. She was diagnosed just three days after birth and was put on the PKU diet soon after. Her older sister, the maid of honor, was diagnosed too late to start the diet and has mental issues. The two-year-old in the photo on the right is the couple's daughter. They planned the pregnancy: nutritional control was strict from conception to delivery to avoid the dangers of excess phenylalanine, which harms the fetus. Your daughter has passed all her developmental milestones with flying colors.

Courtesy of Charles R. Scriver, Montreal Children's Hospital Research Institute, McGill University Health Center.

Geneticists also use the term mutant, which refers to the result of a mutation. A mutation results in a mutated gene, which in turn produces a mutated mRNA, a mutated protein, and finally a mutated organism that exhibits the effects of the mutation – for example, an inherited metabolic disorder.

DNA from PKU patients from around the world was examined to determine which types of mutations are responsible for the birth defect. A wide variety of mutant types have been identified. More than 400 different mutations have been described in the PAH gene. In some cases, part of the gene is missing, leaving the genetic information to make a complete PAH enzyme missing. In other cases, the genetic defect is more subtle, but the result is a failure to produce a PAH protein or the production of an inactive PAH protein. at the

The mutation shown in FIGURE 1.22, the replacement of a G–C base pair with the normal A–T base pair at the first position in the coding sequence, changes the normal AUG (Met) codon used for translation initiation to the GUG codon , which normally specifies valine (Val) and cannot be used as a "start" codon. The result is that translation of the PAH mRNA cannot occur, so that no PAH polypeptide is produced. This mutant is designated M1V because the codon for M (methionine) at amino acid position 1 in the PAH polypeptide has been changed to a codon for V (valine). Although the M1V mutant is quite rare worldwide, it is common in some areas, such as the province of Quebec, Canada.

FIGURE 1.22 The M1V mutant in the PAH gene. The methionine codon required for initiation is mutated to a valine codon. Translation cannot be initiated and no PAH polypeptide is produced.

A very common PAH mutant is designated R408W, which means that codon 408 in the PAH polypeptide chain has been changed from an arginine (R) coder to a tryptophan (W) coder. This mutation is one of the four most common in European Caucasians with PKU. The molecular basis of the mutant is shown in FIGURE 1.23. In this case, the first base pair at codon 408 is changed from base C–G to T–A. As a result, the PAH mRNA has a mutated codon at position 408; Specifically, it has UGG instead of CGG. Translation occurs in this mutant because everything else in the mRNA is normal, but the result is that the mutant PAH carries a tryptophan (Trp) instead of an arginine (Arg) at position 408 in the polypeptide chain. The consequence of the seemingly minor change in an amino acid is very drastic. Although the R408W polypeptide is complete, the enzyme has less than 3% of the normal enzyme activity.

Figure 1.23 The R408W mutant in the PAH gene. Codon 408 for arginine (R) is mutated into a codon for tryptophan (W). The result is that position 408 in the mutant PAH polypeptide is occupied by tryptophan instead of arginine. The mutant protein has little PAH enzyme activity.

Variation in Populations Much of the classical work in genetics since Gregor Mendel has involved the genetic analysis of apparent differences in characteristics of organisms that are the result of certain mutations. Most of the variants used, such as white eyes in Drosophila, shriveled kernels in corn, or even hereditary diseases like alkaptonuria in humans, are quite rare. Therefore, early geneticists tended to view populations as composed of a few rare variants made up of large numbers of "wild-type" individuals, suggesting that the overall level of genetic variation was quite low. This picture later changed dramatically when, based on the enzyme-gene hypothesis, it became possible to quantify genetic variation—first at the protein sequence level and then at the DNA sequence level. The first works with protein variations in

Unexpected results emerged in both the fruit fly (Drosophila pseudoobscura) and humans: when examining the protein products of several genes, it was found that about 30% of them were polymorphic, meaning that there were two or more variants or alleles for them . present in the sampled individuals. Similar

Results have been obtained for almost all species studied using these methods, leading to the recognition that variation – rather than uniformity – is the norm at the genetic level in most species. In recent years, the methods we described in the chapter titled DNA Replication and Sequencing have allowed researchers to shift their focus from single genes to whole genomes. One result of these efforts has been the 1000 Genomes Project, an ongoing international research program aimed at studying common human genetic variations and identifying common human genetic variants associated with disease. To date, the project has sequenced 2,504 genomes from 26 global populations, divided into 5 'superpopulations' (FIGURE 1.24). More than 80 million polymorphisms have been identified, most of which are single nucleotide polymorphisms (SNPs). These publicly available data provide a global perspective on genetic variation in Homo sapiens and have become an invaluable tool for researchers in the fields of medical genetics, evolutionary biology, and anthropology. Most importantly, analysis of the genome of humans and other species has shown that the classical view, which posits genetic variations of shared wild-type alleles and rare variants, needs to be fundamentally revised. Much of a given species' genome is indeed immutable, but for significant parts of it, variation is the rule rather than the exception.

FIGURE 1.24 Geographic distribution of populations sampled in the 1000 Genomes Project. Each dot indicates a study population; Colors indicate overpopulations as indicated in the legend. Data from the International Genome Sample Resource. (2016). IGSR and the 1000 Genomes Project. Retrieved from http://www.internationalgenome.org.

SUMMARY ■ All new genetic variation originates from a mutation or alteration in the DNA sequence. ■ Different and independent mutations can occur in the same gene. ■ Genome sequencing has provided powerful tools for characterizing the extent and nature of genetic variation in many species, including humans.

1.7 Genes and Environment Inborn errors of metabolism illustrate the general principle that genes encode proteins and mutated genes encode mutated proteins. In diseases like PKU, the mutated proteins cause such a drastic change in metabolism that it creates a serious genetic defect. However, biology is not necessarily fate - organisms are also influenced by the environment. The PKU serves as an example for

this principle because patients who adhere to a phenylalanine-restricted diet develop their intellectual abilities within the normal range. What applies in this example applies in general: most traits are determined by the interaction of genes and environment. It is also true that most traits are influenced by multiple genes. No one knows how many genes are involved in the development and maturation of the brain and nervous system, but the number must be in the thousands. This number is in addition to the genes needed in all cells to carry out metabolism and other basic life functions. With extreme examples like PKU, where a single mutation can have such a drastic impact on intellectual development, it's easy to lose sight of gene diversity. The situation is the same as with any complex machine. An airplane can function properly as long as thousands of parts work in harmony, but just one broken part can bring down a vital system. Likewise, the development and functioning of each trait requires a large number of genes working in harmony, but in some cases a single mutated gene can have disastrous consequences. In short, the relationship between a gene and a trait is not necessarily straightforward. The biochemistry of organisms is a complex network of ramifications in which several enzymes may be involved.

substrates, deliver the same products, or respond to the same regulatory elements. The most visible characteristics of organisms are the net result of many genes acting together and in combination with environmental factors. The PKU provides examples of each of the three principles that govern these interactions:

1. A gene can affect more than one trait. Children with extreme forms of PKU often have blonde hair and reduced body pigmentation. These manifestations occur because the absence of PAH is a metabolic blockade that prevents the conversion of phenylalanine into tyrosine, which is the precursor of the pigment melanin. The relationship between severe intellectual disability and decreased pigmentation in PKU only makes sense when one understands the metabolic relationships between phenylalanine, tyrosine, and melanin. If these connections were not known, the resources would appear completely unrelated. PKU is not uncommon in this regard. Many mutated genes affect multiple traits through their secondary or indirect effects. The various, sometimes seemingly unrelated, effects of a mutated gene are called pleiotropic effects, and the phenomenon itself is known as pleiotropy. Figure 1.25 shows a cat with a white coat and blue eyes, a pigmentation pattern often associated (about 40% of the time) with deafness. Deafness can therefore be seen as a pleiotropic effect of white fur and blue eye color. The evolutionary basis of this pleiotropy is unknown. 2. Each trait can be influenced by more than one gene. We discussed this principle earlier in relation to the large number of genes required for normal brain and nervous system development and function. Among them are genes that affect the functioning of the blood-brain barrier, formed by specialized glial cells that tightly surround the capillary walls of the brain, forming a

Blocking the passage of most water-soluble molecules from the blood to the brain. The blood-brain barrier therefore affects the extent to which excess free phenylalanine in the blood can enter the brain itself. Because the effectiveness of the blood-brain barrier varies from person to person, patients with

PKU, who have very similar blood phenylalanine levels, can have dramatically different levels of cognitive development. This factor also partly explains why adherence to a controlled phenylalanine diet is critical in children but less so in adults; The blood-brain barrier is less developed in children and therefore less effective in blocking excess phenylalanine. Several genes influence even simpler metabolic traits. The degradation and excretion of phenylalanine serves as a practical example. The metabolic pathway shown in Figure 1.10 shows that four enzymes are involved in this process, but more enzymes are involved in what is known as "downstream degradation." Because differences in the activity of each of these enzymes can affect the rate at which phenylalanine is broken down and excreted, all enzymes in the signaling pathway are important in determining the amount of excess phenylalanine in the blood of patients with phenylketonuria. 3. Most traits are influenced by environmental factors and genes. Here we are back to the low phenylalanine diet. Children with PKU are not doomed to severe intellectual disability; their performance can be brought within the normal range by dietary treatment. PKU serves as an example of what motivates geneticists to discover the molecular basis of inherited diseases. The hope is that knowledge of the metabolic basis of a disease will eventually lead to methods of clinical intervention through diet, medication or other treatments that alleviate the severity of the disease.

Figure 1.25 Of cats with white coats and blue eyes, about 40% are born deaf. This is because pigment cells migrate from the neural crest to various tissues, including hair follicles, eyes and the middle ear, where their function is essential for hearing. Defective pigment cells that lead to white hair and blue eyes can therefore lead to deafness, which can be considered the pleiotropic effect of white hair and blue eyes.

SUMMARY ■ Most traits are influenced by multiple genes. ■ One gene can influence several traits. ■ Most traits are influenced by a combination of genes and environment.

1.8 The Molecular Unit of Life The degradation and excretion pathway of phenylalanine is not exclusive to humans. One of the remarkable generalizations that have emerged from molecular genetics is that very different organisms – plants and animals, for example – share many characteristics in their genetics and biochemistry. These similarities point to a basic "unit of life": all creatures on Earth share many features of the genetic apparatus, including genetic information encoded in the basic sequence of DNA, transcription into RNA, and translation into protein on ribosomes Use of transmission RNAs. All living things also share certain characteristics in their biochemistry, including many enzymes and other proteins that are similar in amino acid sequence, three-dimensional structure, and function.

Procarean, archaeal, and eukaryan organisms share a common set of similar genes and proteins because they evolved through descent from a common ancestor. The process of evolution occurs when a population of organisms gradually changes in its genetic makeup over time. Evolutionary changes in genes and proteins lead to differences in the metabolism, development and behavior of organisms, making them increasingly able to adapt to their environment. From an evolutionary perspective, the unity of basic molecular processes in organisms living today reflects inheritance from a distant common ancestor in which molecular mechanisms already existed.

Not only the unity of life, but many other features of living organisms become understandable from an evolutionary point of view. For example, inserting an RNA intermediate into the basic flow of genetic information from DNA to RNA to protein makes sense if early forms of life used RNA for both genetic information and enzyme catalysis. The importance of the evolutionary perspective in understanding aspects of biology that seem meaningless or unnecessarily complex is summed up in a famous aphorism by evolutionary biologist Theodosius Dobzhansky: "Nothing in biology makes sense except in the light of evolution."

Biologists distinguish three main domains of organisms: 1. Prokarya. Most bacteria and cyanobacteria (formerly called blue-green algae) belong to this group. The cells of these organisms lack a nucleus and membrane-bound mitochondria, are surrounded by a cell wall, and divide by binary fission. 2. Archaea. This group was originally discovered among microorganisms that produce methane gas or live in extreme environments such as hot springs or high salt concentrations. They are also widely used in more normal environments. Superficially resembling bacteria, archaeal cells exhibit important differences in the way their membrane lipids are chemically linked. The machinery for DNA replication and transcription in Archaea is similar to that of Eukarya, while its metabolism resembles that of bacteria. DNA sequence analysis shows that about half of the genes found in the Archaea kingdom are unique to that group. 3. Eucarya. This group includes all organisms whose cells contain an elaborate network of internal membranes, a membrane-enclosed nucleus, and mitochondria. your DNA is

they are in the form of linear molecules organized into true chromosomes, and cell division occurs by mitosis (discussed in The Chromosomal Basis of Inheritance). Eukaryotes include plants and animals, as well as fungi and many protozoa, such as amoebas and ciliated protozoa.

Genomes and proteomes The entirety of DNA in a cell, cell nucleus, or organelle is called the genome. When used in relation to a species of organism (for example, in terms such as "the human genome"), the term genome is defined as the DNA present in a normal reproductive cell.

Modern DNA sequencing methods (discussed in the DNA sequencing and replication chapter) are so fast and efficient that the complete DNA sequence is now known for over 1,000 different types of organisms. This includes the genomes of various representatives of all groups of organisms, including Neanderthals (an extinct species closely related to Homo sapiens, sequenced from DNA extracted from fossil bones). TABLE 1.2 shows a small selection of sequenced genomes. Genome size is given in megabases (Mb) or millions of base pairs. The genome of the bacterium Haemophilus influenzae, like that of most bacteria, is very compact because it encodes mostly proteins. A high gene density relative to the amount of DNA is also found in the budding yeast, the nematode worm, the fruit fly, and the small flowering plant Arabidopsis thaliana. In contrast, the human genome contains large amounts of non-coding DNA. The comparison with the nematode is instructive. Although the human genome is about 30 times larger than that of the worm, the number is

of genes is less than 2 times higher. This discrepancy reflects the fact that only about 1.5% of the human genome sequence encodes proteins. (Approximately 27% of the human genome is contained in genes, but much of the DNA sequence contained in genes does not code for proteins.) TABLE 1.2 Comparison of Genes and Proteins Organism

unique size,

number of

Divided

MBA

Gene

distinguishable

Protein

(about)

proteins inside

The family

Proteoma b (aproximado) Haemophilus

1.9

1700

1400

6000

4400

100

20.000

9500

} 3000

120c

16.000

8000

} 5000

2500

25.000

10.000

} 7000d

2900e

25.000

10.000

} 9900f

Influenzae (causes bacterial meningitis) Saccharomyces cerevisiae (flowering yeast) Caenorhabditis elegans (soil nematode)

Drosophila melanogaster (fruit fly) Mus musculus (laboratory mouse) Homo sapiens (human)

a Million base pairs.

b excludes “families” of proteins with similar sequences (and related functions). c Excludes 60 MB of specialized DNA (“heterochromatin”) with a very low content of

genes d Based on sequence similarity in messenger RNA (mRNA). e For simplicity, this estimate has been rounded to 3000 MB elsewhere in this text. f Based on the observation that only about 1% of mouse genes lack a similar

gene no human genome and vice-versa.

The complete set of proteins encoded in the genome is called the proteome. In less complex genomes, such as yeast, worm, and fruit fly, the number of proteins in an organism's proteome is approximately equal to the number of genes. However, as we will see in the chapter "The Molecular Biology of Gene Expression", some genes encode two or more proteins through a process called alternative splicing, in which segments of the original RNA transcript are joined together in a variety of combinations to create different messengers to make RNAs. Alternative splicing is particularly widespread in the human genome. At least one-third of human genes and possibly as many as two-thirds are alternatively spliced; Among alternatively spliced genes, the number of different mRNAs per gene ranges from 2 to 7. Thus, the human genome, with its seemingly limited repertoire of 25,000 genes, can generate approximately 60,000 to 90,000 different mRNAs. The widespread use of alternative splicing to multiply the coding capacity of genes is a source of human genetic complexity. Most eukaryotic organisms contain families of related proteins that can be grouped together by similarities in their amino acid sequence. These families exist because the development of a new gene function is typically preceded by the duplication of an existing gene, followed by changes in the nucleotide sequence in one of the genes.

Copies that give rise to the new function. The new function is generally similar to the previous one (for example, a change in the substrate specificity of a transport protein), such that the new protein retains enough amino acid sequence similarity to the original for its common ancestor to be discerned.

The molecular unit of life is reflected in the similarity of proteins in the proteome of different types of organisms. Such comparisons are shown in the right column of Table 1.2. In this table, each family of related proteins is counted only once to estimate the number of proteins in the proteome that are "distinguishable" in the sense that their sequences are different. In yeast, worms, and flies, the number of distinct proteins is about 4400, 9500, and 8000, respectively. The square brackets in Table 1.2 indicate the number of distinct proteins that share sequence similarity across species. From these comparisons, it appears that most multicellular animals share 5,000 to 10,000 proteins that are similar in sequence and function. About 3000 of them are shared with eukaryotes as distantly related as yeast and about 1000 with prokaryotes as distantly related as bacteria. What these comparisons between proteomes imply is that biological systems are based on thousands of protein components. This complexity is hard to fathom, but the challenge is far less daunting than it seemed in the past, when human cells were thought to make up to 1 million different proteins.

SUMMARY ■ DNA replication, transcription and translation processes, as well as the genetic code, are almost universal features of life. ■ Cellular organisms are divided into three domains: prokarya, eukarya, and archaea. ■ The genomes of organisms vary greatly in overall size, but less so in the number of proteins they encode. ■ An organism's proteome consists of families of related proteins, many of which are shared between organisms in different walks of life.

CHAPTER SUMMARY ■ Hereditary traits are influenced by genes. ■ Genes are made up of the chemical deoxyribonucleic acid (DNA).

■ DNA replicates to make copies of itself that are identical (except for rare mutations). ■ DNA contains a genetic code that tells what types of enzymes and other proteins are made in cells. ■ DNA occasionally mutates, and mutated forms specify altered proteins that have reduced activity or stability. ■ A mutated enzyme is an “inborn error of metabolism” that blocks a step in a biochemical pathway for the metabolism of small molecules. ■ Traits are influenced by environment and genes. ■ The molecular unit of life can be seen by comparing genomes and proteomes.

CHECK THE BASICS ■ What were the major experiments that showed that DNA is the genetic material? ■ How has understanding the molecular structure of DNA provided clues about its ability to replicate, encode proteins, and mutate? ■ Why is complementary base pairing a key feature of DNA replication? ■ What is the process of transcription and how does it differ from DNA replication? ■ What is the difference between transcription and translation? ■ What are the three types of RNA involved in protein synthesis and what role does each type of RNA play? ■ What is the “genetic code” and how is it relevant to the translation of a polypeptide chain from a messenger RNA molecule? ■ What is an inborn error of metabolism? How did this concept serve as a bridge between genetics and biochemistry?

■ How does the “core dogma” explain Garrod's discovery that dysfunctional enzymes result from mutated genes? ■ Explain why many mutant forms of phenylalanine hydroxylase have a single amino acid change, but the mutant polypeptide chains are absent or present in very small amounts. ■ What is a pleiotropic effect of a genetic mutation? Give an example ■ What are some major differences in cellular organization between prokarya, archaea and eukarya?

TROUBLESHOOTING GUIDE

PROBLEM 1 In the human gene for the huntingtin protein (so named because it is associated with Huntington's disease), the first 21 nucleotides in the amino acid coding region are:

What is the sequence of a partner tape? ANSWER The base pairing between the strands is A to T and C to G, but it is equally important that the strands in a DNA duplex have opposite polarity. The complementary strand is therefore oriented with its 5' end to the left. The base sequence of the partner strand is:

PROBLEM 2 If the DNA duplex for huntingtin in Problem 1 was transcribed from left to right, deduce the base sequence of the RNA in this coding region. ANSWER To deduce the RNA sequence we need to apply three concepts. First, the base pairing is such that A, T, C, and G in the DNA template are transcribed as U, A, G, and C, respectively, in RNA. Second, the DNA template and the RNA transcript have opposite polarity. Third (and critical to this problem), the RNA strand is always transcribed in the 5' to 3' direction. Since we are told that transcription occurs from left to right, we can conclude that the transcribed strand is the one given in Task 1. The RNA transcript therefore has the base sequence

PROBLEM 3 Given the RNA sequence deduced in Problem 2 that encodes part of human huntingtin, what is the amino acid sequence in that part of huntingtin?

ANSWER The polypeptide chain is translated into consecutive groups of three nucleotides (codons) starting at the 5' end of the coding sequence and moving in the 5' to 3' direction. The amino acid corresponding to each codon can be found in the genetic code table. The first seven amino acids in the polypeptide chain are:

TASK 4 Let's assume that a mutation in the human huntingtin gene occurs in the DNA sequence shown in Task 1. This mutation replaces the red G with a T. What is the nucleotide sequence of this region of the DNA duplex (both strands)? and messenger RNA, and what is the amino acid sequence of mutant huntingtin?

ANSWER DNA, RNA, and polypeptide chains have the following sequences, with differences from the non-mutated gene shown in red. The mutation results in a stop codon and premature termination of huntingtin synthesis.

ANALYSIS AND APPLICATIONS 1.1 Classify each of the following statements as true or false. (a) Each gene is responsible for only one visible trait.

(b) Each trait is potentially influenced by many genes. (c) The sequence of nucleotides in a gene specifies the sequence of amino acids in a protein encoded by the gene. (d) There is a one-to-one correspondence between the set of codons in the genetic code and the set of encoded amino acids.

1.2 What did Watson and Crick conclude from their study of the structure of DNA about the likely replication mechanisms, coding ability and mutation of DNA? 1.3 What does it mean to say that each strand of a duplex DNA molecule has polarity? What does it mean to say that the paired strands in a duplex molecule have opposite polarity?

1.4 What important observation about the S and R strains of Streptococcus pneumoniae led Avery, MacLeod, and McCarty to study this organism? 1.5 In the Avery, MacLeod, and McCarty transformation experiments, what was the strongest evidence that the substance responsible for the transformation was DNA rather than protein? 1.6 A chemical called phenol (carbolic acid) destroys proteins but not nucleic acids, and strong bases, such as sodium hydroxide, destroy both proteins and nucleic acids. In transformation experiments with Streptococcus pneumoniae the result would be

expect if the S strain extract had been treated with phenol? Had it been treated with a strong alkali? 1.7 What feature of the physical organization of the T2 bacteriophage made it suitable for use in Hershey-Chase experiments? 1.8 Like DNA, RNA molecules contain large amounts of phosphorus. When Hershey and Chase grew their T2 phage in bacterial cells in the presence of radioactive phosphorus, the RNA must also have incorporated the labeled phosphorus, but the experimental result was not affected. Why not? 1.9 DNA extracted from a bacteriophage contains 16% A, 16% T, 34% G, and 34% C. What can you conclude about the structure of this DNA molecule?

1.10 DNA extracted from a bacteriophage consists of 20% A, 34% T, 35% G, and 11% C. What is unusual about this DNA? What can you conclude about its structure? 1.11 A region along a transcribed strand of DNA does not contain A. Which base will be missing from the corresponding region of RNA? 1.12 If one strand of a DNA duplex has the sequence 5'ATCAG-3', what is the sequence of the complementary strand? (Write the answer with the 5' end on the left.) 1.13 Suppose a double-stranded DNA molecule is separated into its individual strands and the strands are purified using a high-speed centrifuge. At the

one of the strands, the basic composition is 25% A, 18% T, 20% G, and 37% C. What is the basic composition of the other strand? 1.14 Consider a region along one strand of a double-stranded DNA molecule composed of tandem repeats of the 5'-GTA-3' trinucleotide such that the sequence on that strand is 5'-GTAGTAGTAGT…-3'. What is the sequence on the other tape? (Write the answer with the 5' end to the left.) 1.15 If the template strand of a DNA duplex has the sequence 5'-TCAG-3', what is the sequence of the RNA transcript? (Write the answer with the 5' end to the left.)

1.16 If the non-template strand of a DNA duplex has the sequence 5'-ATCAG-3', what is the sequence of the RNA transcript in this region? (Write the answer with the 5' end to the left.) 1.17 A duplex DNA molecule contains a random sequence of four nucleotides in equal proportions. What is the average distance between consecutive occurrences of the 5'-ATGC3' sequence? Between consecutive occurrences of the sequence 5'-TACGGC-3'? 1.18 An RNA molecule folds back on itself to form a “hairpin” structure held together by a base-pairing region. A segment of the molecule in the paired region has the base sequence 5'-UAUCGUAU-3'. What is the basic sequence that this thread is paired with?

1.19 A synthetic mRNA molecule consists of the repeated sequence of bases

When this molecule is translated in vitro using ribosomes, transfer RNAs, and other necessary components of E. coli, the result is a polypeptide chain composed of the repeated amino acid Lys-Lys-Lys-.... Assuming the genetic code is a code triplet, what does this mean for the lysine (Lys) codon?

1.20 A synthetic mRNA molecule composed of the repeated sequence of bases

is terminated by the addition of a single nucleotide containing A at the right end. When translated in vitro, the resulting polypeptide consists of a repeating sequence of phenylalanines terminated by a single leucine. What does this result mean about the codon for leucine?

1.21 In in vitro translation of an RNA into a polypeptide chain, translation can begin anywhere along the RNA molecule. A synthetic RNA molecule has the sequence

How many reading frames are possible when this molecule is translated in vitro? How many reading frames are possible when this molecule is translated in vivo, starting with the AUG codon?

1.22 You have sequenced both strands of a double-stranded DNA molecule. To assess the potential of this molecule to encode amino acids, conceptually transcribe it into RNA and then conceptually translate the RNA into a polypeptide chain. How many reading frames would you need to scan? 1.23 A synthetic mRNA molecule consists of the repeated sequence of bases

When this molecule is translated in vitro, a polypeptide chain is formed composed of alternating amino acids Thr-His-Thr-His-Thr-His-.... Why do amino acids alternate? What does this result mean about the codons for threonine (Thr) and histidine (His)?

1.24 A synthetic mRNA molecule consists of the repeated sequence of bases

When this molecule is translated in vitro, the result is a mixture of three different polypeptide chains. One consists in repeating isoleucines (Ile - Ile - Ile - Ile -...), another in repeating serines (Ser - Ser - Ser - Ser -...) and the third in repeating histidines (His - His - His -His –…) . What does this result say about the way an mRNA is translated?

1.25 How is it possible that a gene with a mutation in the coding region encodes a polypeptide with the same amino acid sequence as the non-mutated gene?

1.26 A polymer has a random sequence that is 75% G and 25% U. What is the expected frequency of Trp among the amino acids in polypeptide chains resulting from in vitro translation? From Vale? Of Phé?

1.27 The results of the complementation test for the six independent recessive mutations a–f are shown in the attached matrix, where + indicates complementation and – lack of complementation. Sort the mutations into complementation groups and write the names of the mutations in each complementation group in the blanks below. (Some of the blanks can be left blank.)

Mutations in Complementation Group 1: ___ ___ ___ ___ ___ Mutations in Complementation Group 2: ___ ___ ___ ___ ___ Mutations in Complementation Group 3: ___ ___ ___ ___ ___ Mutations in Complementation Group 4: ___ ___ ___ ___ ___

1.28 A mutant screen is performed in Neurospora crassa for mutants unable to synthesize an amino acid that we will symbolize as G. Several mutants are isolated and divided into four groups according to their ability to grow (+) or not to grow (-) on supplemented minimal medium with possible

Precursors D, E and F. Data are presented in the attached table. D

class 1

2nd degree

−

Class 3

−

class 4

−

Complete the diagram shown below. Each arrow indicates one or more biochemical reactions. In each circle, write the class of mutants (1–4) whose products contribute to the reactions symbolized by the arrow; In the squares, write the name of the amino acid or precursor (D–G) at that position in the pathway.

(Video) Lecture 9 - Analyzing Genes and Genomes

1.29 The coding sequence in messenger RNA for amino acids 1 to 10 of human phenylalanine hydroxylase is:

(a) What are the first 10 amino acids? (b) What sequence would result from a mutant RNA in which red A was changed to G? (c) What sequence would result from a mutant RNA in which the red C was changed to G?

(d) What sequence would result from a mutant RNA in which the red U was changed to C? (e) What sequence would result from a mutant RNA in which the red G was changed to U?

1.30 A "frameshift" mutation is a mutation in which a specified number of base pairs, which are not a multiple of three, are inserted or removed from a coding region of DNA. The result is that, at the point of the frameshift mutation, the translational reading frame of the protein is changed relative to the non-mutated gene. To see the consequence of a frameshift mutation, consider what the coding sequence in messenger RNA is for the first 10 amino acids in human beta-hemoglobin (part of the oxygen-carrying protein in blood).

(a) What is the sequence of amino acids in this part of the polypeptide chain?

(b) What would be the consequence of a frameshift mutation that causes an RNA to lack the red U? (c) What would be the consequence of a frameshift mutation that would result in an RNA with a G inserted just before the red U?

1.31 A headline reads “Gene identified for schizophrenia”. Does this necessarily mean that schizophrenia in an individual can be diagnosed by genetic analysis alone? explain your answer 1.32 Using the data in Table 1.2, construct a graph representing genome size for different organisms

the x-axis and the number of genes on the y-axis. Now consider the case of the Mexican axolotl (Ambystoma mexicana) with an estimated genome size of 32 billion base pairs. From your graph, can you predict the number of genes that are present in this genome? explain your answer

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 For the mutant and wild-type RNA molecules described in Task 1.30, deduce the sequence of bases on both strands of the corresponding double-stranded DNA for: (a) the wild-type sequence (b ) the single base deletion (c) the single base insertion

CHALLENGE PROBLEM 2 You are performing Beadle Tatum-type experiments to analyze a metabolic pathway in Neurospora. You know that the precursor of the pathway is a molecule symbolized by P and the product is a vitamin symbolized by Z. You are sure that the path from P to Z is linear and that molecules W, X, and Y are in between. However, there may be other intermediaries that have not yet been identified. You get 10 independent mutations that cannot grow on minimal medium supplemented with P but can grow on minimal medium supplemented with Z. The 10 mutants fall into four classes that can grow (+) or not grow (-) on minimal medium supplemented with . Nutrients W, X or Y. Data are presented in the attached table.

Class I (Mutant 5)

−

Class II (Mutants 1, 3, 4, 6, 7)

−

Class III (Mutation 2)

−

Class IV (Mutants 8, 9, 10)

−

Draw a linear pathway with P on the left and Z on the right, in which the W, X, and Y intermediates are shown in the order they occur in the synthesis pathway from Z to P.

CHALLENGE PROBLEM 3 The 10 mutants in Challenge Problem 2 were tested for complementation in all paired combinations using heterokaryons. The results are presented in the matrix, where + indicates the ability of the heterokaryon to grow on minimal medium and - indicates the inability to grow on minimal medium.

Mutant 1 2 3 4 5

6 7 8 9

Assume that each complementation group defines a different gene, and assume further that each gene encodes an enzyme that catalyzes a single step in the pathway that converts a substrate molecule into a product molecule. one. (a) Redraw the metabolic pathway derived from challenge task 2. Use a right arrow to indicate each enzymatic step in the pathway, and label each arrow with the mutant number 1 through 10 that blocks the enzymatic step. In some cases, you cannot tell the order in which the enzymes occur in the pathway, so you can write them in any order. B. (b) How many unknown intermediates exist between precursor P and vitamin Z in the pathway you deduced from the data?

FOR FURTHER READING

The 1000 Genomes Consortium. (2015). A global reference for human genetic variation. Nature, 526(7571(68-74) http://doi.org/10.1038/nature15393 A review of the 1000 genomes project. Avery, O. T. (1944) Studies on the chemical nature of the substance that causes the transformation of induced pneumococcal types: induction of transformation by a deoxyribonucleic acid fraction isolated from Pneumococcus type III, Journal of Experimental Medicine, 79(2), 137-158, http://doi.org/10.1084/jem.79.2.137 Considered by most to be definitive evidence DNA is considered to be the genetic material.

Beadle, G.W. & Tatum, E.L. (1941). Genetic control of biochemical reactions in Neurospora. Proceedings of the National Academy of Sciences of the United States of America, 2(11), 499-506. http://www.ncbi.nlm.nih.gov/pubmed/16588492 The origin of the one gene and one enzyme hypothesis. Crick, F. (1970). Central dogma of molecular biology. Nature, 227 (5258), 561-563. http://doi.org/10.1038/227561a0

Crick's clearest statement of the central dogma about the flow of information from DNA to protein, but not vice versa. Griffith, F. (1928). The importance of types of pneumococci. Journal of Hygiene, 27(2), 113. http://doi.org/10.1017/S0022172400031879 The first demonstration that there is a "transformative principle". Hershey, A.D. & Chase, M. (1952). Independent roles of viral protein and nucleic acid in bacteriophage growth. Journal of General Physiology, 36(1), 39–56. http://doi.org/10.1085/jgp.36.1.39

Another confirmation that DNA is the genetic material as opposed to protein. Watson, J.D. & Crick, F.H. (1953). Molecular Structure of Nucleic Acids: A structure for the nucleic acid of deoxyribose. Nature, 171(4356), 737-738. http://www.ncbi.nlm.nih.gov/pubmed/13054692 The original publication describing the double helix structure of DNA. Watson, J.D. & Crick, F.H. (1953). Genetic implications of deoxyribonucleic acid structure. Nature. http://doi.org/10.1038/171964b0

A lesser known but still very important work, in which Watson and Crick are the first to propose a basis for the genetic code.

CHAPTER 2 DNA Structure and Genetic Variation CHAPTER OVERVIEW 2.1 Genetic Differences Between Individuals 2.2 The Terminology of Genetic Analysis 2.3 The Molecular Structure of DNA 2.4 Separation and Identification of Genomic DNA Fragments

2.5 Specific DNA amplification for detection and purification

2.6 Types of DNA markers in genomic DNA 2.7 Applications of DNA markers

ROOTS OF THE DISCOVERY: The Double Helix James D. Watson and Francis H.C. Crick (1953) A Structure for Deoxyribose Nucleic Acid THE TECHNOLOGY: High Throughput SNP Genotyping

LEARNING OBJECTIVES AND RESEARCH SKILLS Applying the principles of DNA structure and genetic variation discussed in this chapter will enable you to solve the following types of problems: ■ Explain the structure of DNA and how that structure facilitates its manipulation. ■ Construct a restriction map of a DNA fragment based on gel patterns of fragments generated by digestion with multiple restriction enzymes. ■ Use DNA marker analysis data to identify individuals and relationships.

■ For a given DNA sequence, choose primer-oligonucleotide sequences that would allow each specific fragment of the molecule to be amplified in the polymerase chain reaction. Until the mid-1970s, classical and molecular genetics, while addressing the same issues, were often treated as separate disciplines. Since then, studies in genetics have undergone a revolution based on the use of increasingly sophisticated methods to isolate and identify specific DNA fragments, essentially merging the two approaches. The culmination of these developments was large-scale genome sequencing - the ability to determine the correct sequence of the base pairs that make up the DNA throughout the genome and to identify the sequences associated with genes. because many of

Laboratory organisms used in genetic experiments have relatively small genomes, these sequences were completed first. The techniques used to sequence these simpler genomes have been extended to the sequencing of much larger genomes, including the human genome. This has greatly expanded our research capabilities.

even the most complex features at the molecular, cellular, and organismic level.

2.1 Genetic Differences Between Individuals The human genome in a reproductive cell consists of approximately 3 billion base pairs organized into 23 different chromosomes (each chromosome contains a single duplex DNA molecule). A typical chromosome may contain several hundred to several thousand genes arranged in linear order along the DNA molecule present in the chromosome. The protein coding sequences

some of these genes actually make up only about 1.3% of the entire genome. The other 98.7% of the sequences do not encode proteins. Some encode RNAs that are not mRNAs but molecules that contribute to a variety of cellular functions. Others are non-coding sequences, many of which are relatively short sequences found in hundreds of thousands of copies scattered throughout the genome. Still other non-coding sequences are the decayed remnants of genes called pseudogenes. And still others are non-coding sequences whose functions are the subject of much current investigation. As you would expect, given the vast history of non-coding DNA in the human genome, identifying protein-coding genes is a challenge in itself. Geneticists often speak of the nucleotide sequence of the "human genome" because the corresponding DNA sequences from any two individuals are identical in approximately 99.9 percent of their nucleotide locations. These shared sequences are our evolutionary heritage: they contain the genetic information that makes us human. In reality, however, there are many different human genomes. Geneticists are most interested in the 0.1% of the human DNA sequence – 3 million base pairs – that differs from genome to genome. Most of these variations are "normal", but these differences also include the mutations responsible for genetic disorders like phenylketonuria (PKU) and others.

inborn errors of metabolism, as well as mutations that increase an individual's risk of developing more complex diseases such as heart disease, breast cancer and diabetes. Fortunately, only a small fraction of all DNA sequence differences are associated with disease. Some of the others are associated with hereditary differences in height, weight, hair color, eye color, facial features, and other characteristics. Most genetic differences between people are completely harmless. Many have no detectable effects on appearance or health. Such differences can only be examined by directly examining the DNA itself. These differences are still important because they serve as genetic markers.

DNA markers as reference points in

Chromosomes In genetics, a genetic marker is any difference in DNA, however undetectable, whose pattern of transmission can be traced from generation to generation. Each individual that carries the marker also carries a stretch of the chromosome on both sides, so the marker marks a specific region of the genome. Any difference in DNA sequence between two individuals can serve as a genetic marker. Although often harmless in themselves, genetic markers make it possible to locate the positions of disease genes along chromosomes and to isolate, identify and study their DNA. Genetic markers detected by direct DNA analysis are often referred to as DNA markers. DNA markers are important in genetics because they serve as markers on long DNA molecules, like those found in chromosomes, that allow researchers to track genetic differences between individuals. In that sense they are

like signposts along a highway. Using DNA markers as reference points, geneticists can identify the locations of normal genes, mutated genes, chromosome breaks and other features important for genetic analysis. A hypothetical example is shown in FIGURE 2.1 involving the PAH gene that encodes phenylalanine hydroxylase. In this figure we see examples of two copies of the chromosome 12 segment, one encoding the normal protein and the other encoding the normal protein.

for a protein associated with PKU. These two segments also differ with respect to a non-coding nucleotide. This nucleotide could be a genetic marker, since the G residue associated with the PKU variant is not the difference that causes PKU, its presence in a given chromosome could be used as a marker to indicate the presence of that variant.

FIGURE 2.1 This schematic shows two DNA molecules containing PAH variants and a DNA marker - in this case, an A or a G at the position indicated. In this hypothetical case, the A allele is associated with the disease-associated variant and could, therefore, be used as an indicator of its presence in an individual's genome.

Of course, we face the problem of how to take all of an individual's DNA and isolate a specific gene of interest so that we can identify genetic differences between individuals. In this chapter, we'll look at some of the major ways DNA is manipulated to achieve this feat, whether or not these differences result in observable differences. An overview of the steps required

is shown in FIGURE 2.2. The use of these methods expands the scope of genetics and makes it possible to carry out genetic analyzes in any organism. As a result, detailed genetic analysis is no longer limited to humans, domesticated animals, crops and the relatively small number of model organisms favorable for genetic studies. The direct study of DNA eliminates the need for prior identification of genetic differences between individuals; it even eliminates the need for controlled crossings. The molecular analysis methods discussed in this chapter have

transformed genetics and are the main techniques used in almost all modern genetics laboratories; Furthermore, a basic understanding of them allows one to appreciate the general unity of classical and molecular genetics.

FIGURE 2.2 DNA markers serve as landmarks that identify physical positions along a DNA molecule, e.g. B. DNA from a chromosome. A DNA marker can also be used to identify bacterial cells into which a particular DNA fragment has been introduced. The DNA cloning process is not as simple as it is here; is discussed further in the chapter Manipulation of Genes and Genomes.

SUMMARY ■ Every human genome is unique, but only a small fraction of the genome differs from individual to individual. ■ Most existing variations have no obvious physical implications. Variants associated with hereditary diseases are rare. ■ Protein-coding genes make up only a small fraction of the DNA in a human genome.

■ The study of DNA markers greatly enhances the power of genetic analysis.

2.2 The terminology of genetic analysis To discuss genetic analysis at any level, we must first introduce some key terms that provide the essential vocabulary of genetics. These terms can be understood with reference to Figure 2.3. In the chapter Genes, Genomes and Genetic Analysis, we defined a gene as an element of heredity that is passed from parents to offspring in reproduction and that affects one or more genes

Trains. Chemically, a gene is a sequence of nucleotides along a DNA molecule. In a population of organisms, not all copies of a gene may have the exact same nucleotide sequence. For example, while one form of a gene may have the GCA codon at a certain position, another form of the same gene may have the GCG codon. Both codons specify alanine. Therefore, the two forms of the gene encode the same amino acid sequence, but differ in the DNA sequence. The alternative forms of a gene are called the gene's alleles. Different alleles can also code for different amino acid sequences, sometimes with drastic consequences. Recall the example of the PAH gene for phenylalanine hydroxylase in the Genes, Genomes and Genetic Analysis chapter, where a change of codon 408 from CGG (arginine) to TGG (tryptophan) results in an inactive enzyme that is considered an inborn error. of phenylketonuria metabolism.

FIGURE 2.3 Key concepts and terms used in modern genetics. Note that a single gene can have any number of alleles in the population as a whole, but no more than two alleles can be present in an individual.

Within a cell, genes are arranged in linear sequence along microscopic, threadlike bodies called chromosomes, which we examine in detail in chapters entitled The Chromosomal Basis of Inheritance and Human Karyotypes and Chromosomal Behavior. Each human reproductive cell contains a complete set of 23 chromosomes containing 3 × 109 base pairs of DNA. A typical chromosome contains several hundred to several thousand genes. In humans, the average is about 1000 genes per chromosome. Each chromosome contains a single molecule of duplex DNA complexed with proteins and coiled very tightly along its length. The DNA in the average human chromosome, when fully expanded, has relative dimensions comparable to that of a moist spaghetti noodle 40 km long; When DNA is rolled up into a chromosome, its physical compactness is comparable to that of the same noodles being rolled up and packed into an 18-foot canoe. The physical location of a gene along a chromosome is called a gene locus. In most higher organisms, including humans

In living things, all cells except the sperm or egg contain two copies of each type of chromosome—one from the mother and one from the father. Each member of such a chromosome pair is considered homologous to the other. (Sex-determining chromosomes are an important exception, which we will ignore here.) Thus, each individual carries two alleles at each locus, because one allele is present at a corresponding position on each of the maternal and paternal homologous chromosomes (Figure 2.3). . The genetic makeup of an individual is called the genotype. If, for a given gene, the two alleles are indistinguishable at the locus of an individual, the individual's genotype is said to be homozygous for the present allele. If the two alleles differ from each other at the locus, the individual's genotype is said to be heterozygous for the present alleles. Typographically, genes are indicated in italics and alleles are usually distinguished by uppercase or lowercase letters (A versus a), subscripts (A1 versus A2), superscripts (a+ versus a−), or sometimes just + and −. Use

These symbols would represent homozygous genes by one of these formulas: AA, aa, A1A1, A2A2, a+a+, a−a−, +/+ or −/−. As in the last two examples, the slash is sometimes used to separate alleles present on homologous chromosomes to avoid ambiguity. Heterozygous genes would be represented by one of the formulas Aa, A1A2, a+a- or +/-. In Figure 2.3, the Bb genotype is heterozygous because the B and b alleles are distinguishable (which is why they are given different symbols), whereas the CC genotype is homozygous. These genotypes can also be written as B/b and C/C, respectively. While the alleles present in an individual constitute its genotype, the physical or biochemical expression of the genotype

is called a phenotype. Simply put, the difference is that an individual's genotype is what's on the inside (the alleles in the DNA), while the phenotype is what's on the outside (the observable traits, including biochemical traits, behavioral traits). properties, etc.). The distinction between genotype and phenotype is crucial, as there is usually not a one-to-one correspondence between genes and traits. More complex traits – such as hair color, skin color, height, weight, behavior, lifespan and disease susceptibility – are influenced by

many genes. Most properties are also influenced to a greater or lesser extent by the environment. Thus, the same genotype can give rise to different phenotypes depending on the environment. For example, compare two people with a genetic risk for lung cancer: if one smokes and the other doesn't, the smoker gets the disease much more often. Environmental influences also imply that the same phenotype can result from more than one genotype. Smoking is another example, as most smokers who are not genetically vulnerable can also develop lung cancer. Harmful mutations represent only a small fraction of the total allelic variation in a species, and most of these alleles are very unusual. In contrast, in many cases, different alleles at a locus may not have obvious effects on phenotypes; often several alleles of such loci are found with relatively high frequency (greater than 5%). These cases are called polymorphisms; The term polymorphism literally means "multiple forms". At the sequence level, these variants are called DNA polymorphisms.

SUMMARY ■ The genome of a typical human cell contains 23 pairs of homologous chromosomes, each carrying an average of 1,000 genes. ■ Variants of certain genes are called alleles. ■ Individuals with two identical alleles of a given gene are homozygous; those with two different alleles are heterozygous. ■ The combination of alleles an individual carries is the genotype. ■ The physical manifestation of the genotype is the phenotype.

■ Many loci in the human genome are polymorphic.

2.3 The Molecular Structure of DNA Modern experimental methods for manipulating and analyzing DNA arose from a detailed understanding of its molecular structure and replication. Therefore, to understand these methods, one needs to know something about the molecular structure of DNA. As we saw in the chapter Genes, Genomes and Genetic Analysis, DNA is a helix composed of two paired and complementary strands, each composed of an ordered chain of nucleotides, each nucleotide being one of the bases A (adenine), T (thymine), G (guanine) or C (cytosine). Watson-Crick base pairing between A and T and between G and C on the complementary strands holds the strands together. Complementary strands are also critical to replication, as each strand can serve as a template for the synthesis of a new complementary strand. Let us now take a closer look at the structure of DNA and the main features of its replication.

Polynucleotide chains From a biochemical point of view, a strand of DNA is a polymer - a large molecule composed of repeating units. The units in DNA are composed of 2'-deoxyribose (a five-carbon sugar), phosphoric acid, and the four nitrogenous bases labeled A, T, G, and C. The chemical structures of the bases are shown in FIGURE 2.4. Note that two of the bases have a double-ring structure; these are called purines. The other two bases have a single ring structure; these are called pyrimidines.

FIGURE 2.4 Chemical structures of the four nitrogenous bases of DNA: adenine, thymine, guanine, and cytosine. The nitrogen atom attached to the deoxyribose sugar is indicated. Atoms shown in red are involved in hydrogen bonds between DNA base pairs.

■ The purine bases are adenine (A) and guanine (G).

■ The pyrimidine bases are thymine (T) and cytosine (C). In DNA, each base is chemically linked to a molecule of the sugar deoxyribose, forming a compound called a nucleoside. If a phosphate group is also attached to the sugar, the nucleoside becomes a nucleotide (FIGURE 2.5). Therefore, a nucleotide is a nucleoside plus a phosphate. Using the conventional numbering of carbon atoms in the sugar in Figure 2.3, the carbon atom to which the base is attached is the 1' carbon atom. (Atoms in sugar are assigned prime numbers to distinguish them from atoms in bases.) The nomenclature of nucleosides and nucleotides derived from DNA bases is summarized in TABLE 2.1. Most of these terms are not needed in this text; they are included because they are likely to be encountered in further reading.

Figure 2.5: A typical nucleotide showing the three main nucleotides

Components (phosphate, sugar, and base) that make the difference between DNA and RNA. A nucleoside consists only of sugar and base. Nucleotides are monophosphates (having a phosphate group). Nucleoside diphosphates contain two phosphate groups and nucleoside triphosphates contain three.

TABLE 2.1 Basis of DNA nomenclature

nucleosides

nucleotide

Adenine (A)

deoxyadenosine

Desoxiadenosina-5' monophosphate (dAMP) diphosphate (dADP) triphosphate (dATP)

Guanina (G)

deoxyguanosine

Desoxiguanosina-5' monophosphato (dGMP) diphosphato (dGDP) triphosphato (dGTP)

Timina (T)

deoxythymidine

Desoxitimidine-5′ monophosphato (dTMP) diphosphato (dTDP) triphosphato (dTTP)

Cytosine (C)

Deoxycytidine

Desoxicitidine-5′ monophosphato (dCMP) diphosphato (dCDP)

Triphosphate (dCTP)

In nucleic acids such as DNA and RNA, nucleotides combine to form a polynucleotide chain in which the phosphate attached to the 5' carbon atom of a sugar is attached to the hydroxyl group attached to the 3' carbon atom of the growing chain (FIGURE 2.6 ). The chemical bonds that link the sugar building blocks of neighboring nucleotides through the phosphate groups are called phosphodiester bonds. The 3'-5'-3'-5' orientation of these bonds continues along the chain, which normally consists of millions of nucleotides. Note that the end groups of each

The polynucleotide chain has a 5'-phosphate (5'-P) group at one end

and a 3'-hydroxyl (3'-OH) group at the other end. The asymmetry of the ends of a DNA strand implies that each strand has a polarity determined by which end carries the 5' phosphate and which carries the 3' hydroxyl.

Figure 2.6 Three nucleotides at the 5' end of a single polynucleotide strand. (A) The chemical structure of the sugar-phosphate bonds showing the 5' to 3' orientation of the strand (the red numbers are those assigned to the carbon atoms). (B) A common schematic way of representing a polynucleotide strand.

A few years before Watson and Crick proposed their essentially correct three-dimensional structure of DNA as a double helix, Erwin Chargaff developed a chemical technique for measuring the amount of each base present in DNA. In describing this technique, we denote the molar concentration of any base by the symbol in square brackets for the base; for example, [A] denotes the molar concentration of adenine. Chargaff used his technique to measure the [A], [T], [G] and [C] content of DNA from a variety of sources. He discovered that the basic composition of DNA, defined as the percentage G + C, varies from species to species, but is constant in all cells of an organism and within a species. Data on the basic DNA composition of a variety of organisms is provided in TABLE 2.2.

TABLE 2.2 Basic DNA Composition of Various Organisms Base (and percentage of total bases) Organism

adenine

thymine

guanine

cytosine

Base Composition (Percentage G + C)

Bacteriófago T7

26,0

24,0

48,0

36,9

36.3

14,0

12.8

26,8

30.2

29,5

21.6

18.7

40.3

Escherichia coli

24,7

23.6

26,0

25.7

51,7

luteal pregnancy

13.4

12.4

37.1

74.2

Bakterien Clostridium perfringens Streptococcus pneumoniae

Pilze Saccharomyces

31.7

32.6

18.3

17.4

35,7

23,0

22.3

27.1

27.6

54,7

Wheat

27.3

27.2

22.7

22,8*

45,5

Most

26,8

27.2

22.8

23,2*

46,0

30.8

29.4

19.6

20.2

39,8

Pig

29.4

29.6

20,5

41,0

Salmon

29,7

29.1

20.8

20.4

41.2

Human being

29,8

31.8

20.2

18.2

38.4

cerevisiae Neurospora crassa Higher plants

Animais Drosophila melanogaster

* Contains a quarter of 5-methylcytosine, a modified form of cytosine found in most plants

more complex than algae and in many animals.

Chargaff also observed certain regular relationships between the molar concentrations of different bases. These relationships are now called Chargaff's rules: ■ The amount of adenine equals that of thymine: [A] = [T]. ■ The amount of guanine corresponds to that of cytosine: [G] = [C]. ■ The amount of purine bases corresponds to that of pyrimidine bases: [A] + [G] = [T] + [C].

Although the chemical basis for these observations was not known at the time, one of the attractive features of the Watson-Crick structure of complementary pair chains was that it explained Chargaff's rules. Since A is always paired with T in double-stranded DNA, it must follow that [A] = [T]. Likewise, since G is paired with C, we know that [G] = [C]. The third rule follows by adding the other two: [A] + [T] = [G] + [C]. In the next section, we examine the molecular basis of base pairing in more detail.

the double helix

In addition to Chargaff's rules and basic nucleotide chemistry, Watson and Crick drew on structural observations by Rosalind Franklin and Maurice Wilkins. These scientists used X-ray crystallography to study the three-dimensional structure of DNA. In this method, crystals of large molecules are exposed to X-rays. The shape of crystals causes X-rays to be scattered or diffracted in a way determined by the structure of the crystal. The diffraction pattern is then recorded on photographic film, and the crystallographer can use this to draw conclusions about the structure of the crystal. Perhaps the most famous photo ever taken was Franklin's Photo 51 (FIGURE 2.7). Crystallographers were able to draw the following conclusions from this: ■ The crystal has a helical structure. ■ Filament diameter is 20 angstroms (Å). ■ The helix has two levels of repeating structure or periodicity along its length. One occurs every 3.4 Å, the other every 34 Å.

FIGURE 2.7 “Photo 51” by Rosalind Franklin, the X-ray diffraction pattern obtained from crystallized DNA. The criss-cross nature of the image indicates the helical structure of DNA. © Source of Science.

Watson and Crick used all of this data when they built their double helix model.

Base Pairing and Stacking In the three-dimensional structure of the DNA molecule proposed by Watson and Crick in 1953, the molecule consists of two polynucleotide chains twisted around each other to form a double-stranded helix in which adenine and thymine, and guanine and cytosine , are on opposite paired chains (FIGURE 2.8). In the standard structure, called the B-form of DNA, each strand makes a complete turn every 34 Å. The propeller is right-handed, which means that each chain, looking down the barrel, follows a clockwise path as it goes. The bases are 3.4 Å apart,

so there are 10 bases around the helix on each strand and 10 bases

pairs around the double helix. These correspond to the two periodicities derived from X-ray crystallographic data.

Figure 2.8 Two DNA graphs showing the three-dimensional structure of the double helix. (A) In a ribbon diagram, sugar-phosphate backbones are represented as ribbons, using horizontal lines to represent base pairs. (B) A computer model of the B shape of a DNA molecule. The stick figures are the sugar phosphate chains wriggling out of the stacked base.

Pairs forming a major sulcus and a minor sulcus. Color coding for base pairs is as follows: A, red or pink; T, dark green or light green; G, dark brown or beige; C, dark blue or light blue. Bases shown in dark colors are those attached to the blue sugar-phosphate backbone; The bases, shown in light colors, are attached to the beige backbone. (B) Courtesy of Antony M. Dean, University of Minnesota.

Strands exhibit base pairing, where each base is paired through hydrogen bonds with a complementary base on the other strand. (A hydrogen bond is a weak bond in which two participating atoms share a hydrogen atom between them.) Hydrogen bonds provide a type of force that holds wires together. In Watson-Crick base pairing, adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C). The hydrogen bonds formed at the adenine-thymine base pair and the guanine-cytosine base pair are shown in FIGURE 2.9. Note that an A-T pair (Figure 2.9A and B) has two hydrogen bonds and a G-C pair (Figure 2.9C and D) has three hydrogen bonds. This means that the hydrogen bond between G and C is stronger in the sense that it takes more energy to break; For example, the amount of heat required to separate the paired strands in a DNA duplex increases with the percentage of G + C. Since nothing constrains the sequence of bases on a single strand, any sequence can be present along a strand. This explains Chargaff's observation that the DNA of different organisms can differ in base composition. However, because the duplex DNA strands are complementary, Chargaff's rules [A] = [T] and [G] = [C] apply regardless of base composition.

FIGURE 2.9 Normal base pairs in DNA. On the left, the hydrogen bonds (dashed lines) with the connected atoms are shown in red. (A and B) A-T base pairing. (C and D) GC base pairing. For the room filling models (B and D), the colors are as follows: C, gray; N, blue; The Red; and H (shown on bases only), white. Each hydrogen bond is represented as a white disc squeezed between the atoms that share the hydrogen. The figures on the outside represent the thorns that surround the pairs of stacked bases. Source: (B,D) Space fill models courtesy of Antony M. Dean, University of Minnesota.

In the B form of DNA, the paired bases are stacked on top of each other like coins on a roll. The top and bottom of any nitrogenous base are relatively flat and nonpolar (no charge). These surfaces are called hydrophobic because they bond

bad for water molecules that are very polar. (Polarity refers to the asymmetrical distribution of charge across the V-shaped water molecule; the oxygen atom at the base of the V tends to be very negative, while the hydrogen atoms at the ends are very positive.) Due to the repulsion of water molecules, the paired nitrogenous bases tend to stack on top of each other in order to exclude as much water as possible from within the double helix. This feature of double-stranded DNA is known as base stacking. A double-stranded DNA molecule

therefore, it has a hydrophobic core composed of stacked bases, and it is the energy of base stacking that gives double-stranded DNA much of its chemical stability (base pairing alone would not be sufficient to stabilize the molecule under physiological conditions). When it comes to a DNA molecule, molecular biologists often refer to the individual strands as single strands or single stranded DNA; they refer to the double helix as double-stranded DNA or a duplex molecule. The two spiral grooves running outside the double helix are not symmetrical; One sulcus, called the major sulcus, is larger than the other, called the minor sulcus. Proteins that interact with double-stranded DNA usually have regions that make base-pair contact fitting in the major groove, the minor groove, or both (Figure 2.8B).

Antiparallel Chains Each structure in a double helix consists of alternating deoxyribose sugars with phosphate groups connecting the 3' carbon of one sugar to the 5' carbon of the next in the series (Figure 2.5). The two polynucleotide strands of the double helix have opposite polarity, in the sense that the 5' end of one strand is paired with the 3' end

the other strand. Wires with such an arrangement are called antiparallel. One implication of the presence of antiparallel strands in duplex DNA is that in each base pair, one base is attached to a sugar that is above the pairing plane and the other base is attached to a sugar that is below the pairing plane. Other

It follows that each end of the double helix has a 5'-P group (on one strand) and a 3'-OH group (on the other strand), as shown in FIGURE 2.10.

Figure 2.10 A segment of a DNA molecule showing the antiparallel orientation of the complementary strands. The shaded blue arrows indicate the 5' to 3' direction of each strand. Phosphates (P) link the 3' carbon atom of one deoxyribose to the 5' carbon atom of adjacent deoxyribose.

The DNA duplex diagram in Figure 2.8 is static and therefore somewhat misleading. DNA is actually a dynamic molecule that is constantly in motion. In some regions, the filaments may separate briefly and then come together in the same or a different conformation. The right double helix at

Figure 2.8 is the standard B-shape, but depending on conditions, DNA can actually form more than 20 slightly different variants of a right-hand helix, and some regions can even form helices with the strands twisted to the left (called Z-Shape). ). of DNA). When complementary stretches of nucleotides occur on the same strand, a single strand separated from its partner can fold back on itself like a hairpin. In DNA regions that contain suitable base sequences, even three-stranded triple helices can form.

DNA structure in relation to function

In the structure of the DNA molecule we can see how the four essential requirements of a genetic material are met. 1. All genetic material must be able to be replicated exactly so that the information it contains is replicated exactly and inherited by daughter cells. The basis for the exact duplication of a DNA molecule is the pairing of A with T and G with C in the two polynucleotide chains. Unwrapping and separating the strands, copying each free strand, creates two identical double helices. 2. The genetic material must contain encrypted information. Obviously, the order of bases in DNA provides the basis for such a code – more specifically, a genetic code in which groups of three bases specify amino acids. Because the four bases in a DNA molecule can be in any order, and because the sequence can vary from one part of the molecule to another and from organism to organism, DNA can contain many unique regions, each of which can be a distinct region. . gene 3. The genetic material must have the ability to direct the organization and metabolic activities of the cell. as we saw

In the chapter "Genes, genomes and genetic analysis", genes can direct the synthesis of a protein molecule - a polymer made of repeating units of amino acids. The sequence of amino acids in a protein determines its chemical and physical properties. A gene is expressed when its protein product is synthesized, and a requirement of the genetic material is that it determines the sequence in which amino acid units are added to the end of a growing protein molecule. 4. Genetic material must be able to undergo occasional mutations that alter the information it contains. For these mutations to be heritable, the mutant molecule must be able to replicate as faithfully as the parent molecule. This feature is needed to explain the evolution of different organisms through the slow accumulation of favorable mutations. Watson and Crick proposed that heritable mutations in DNA may be possible through rare base mismatches, resulting in the incorporation of the wrong nucleotide into a replicating strand of DNA.

SUMMARY ■ DNA is a spiral molecule composed of two antiparallel strands. ■ The two strands of DNA are complementary: The T and G pairs with C. ■ The helix is stabilized by the hydrogen bonds that form between complementary bases and by the stacking interactions that occur between adjacent pairs of bases. ■ The structure of DNA follows the four necessary properties of genetic material.

ROOTS OF THE DISCOVERY The double helix James D. Watson and Francis H.C. Crick (1953) Cavendish Laboratory Cambridge, England

A structure for deoxyribose nucleic acid This is one of the publications that marked the turning point in biology in the 20th century. Watson and Crick benefited enormously from knowing that their structure was consistent with unpublished structural studies by Maurice Wilkins and Rosalind Franklin. The same issue of Nature that contained the Watson and Crick paper also contained an article by the Wilkins group and another by the Franklin group, detailing their data and the agreement of their data with the proposed structure. Franklin was said to be only two and a half steps away from making the discovery himself. Anyway, Watson, Crick and Wilkins received the 1962 Nobel Prize for their discovery of the structure of DNA. Rosalind Franklin tragically died of cancer in 1958 at the age of 38.

“If only certain pairs of bases can be formed, it follows that if the sequence of the bases in a chain is given, then the sequence

”

on the other chain is determined automatically.

Watson and Crick's 1953 paper was concise, occupying only one printed page in Nature. Describes the basic molecular model - the helical structure, antiparallel chains and the physical positioning of bases and phosphate groups. Next, base pairing and its importance are described.

Only certain base pairs can bind. These pairs are adenine (purine) with thymine (pyrimidine) and guanine (purine) with cytosine (pyrimidine). The sequence of bases in a single strand does not appear to be restricted in any way. However… it follows that, given the sequence of bases in one chain, the sequence in the other chain is automatically determined…. It has not escaped our notice that the specific pairing we have postulated points directly to a plausible copying mechanism for the genome. That last sentence is one of the most quoted lines in all of biology. In later, more detailed work, Watson and Crick went a step further and considered how information might be encoded in DNA given the proposed molecular structure. The phosphate-sugar backbone of our model is perfectly regular, but any sequence of base pairs can fit into the structure. It follows that many different permutations are possible in a long molecule, and it seems likely that the exact sequence of bases is the code that carries the genetic information.

While Watson and Crick's first publication is considered iconic, the second extensively explores the biological and genetic significance of the double helix. Source: J.D. Watson and F.H.C. Crick, Molecular Structure of Nucleic Acids: A Structure for Deoxyribose Nucleic Acid. Nature 171 (1953): 737-738.

2.4 The separation and identification of genomic DNA fragments The following sections show how an understanding of DNA structure and replication was put into practice in the development of methods to separate and identify specific DNA fragments. These methods are primarily used to identify DNA markers or to help isolate specific DNA fragments that are of genetic interest. For example, consider a familial breast cancer family tree in which a particular piece of DNA serves as a marker for a region of the chromosome that also contains the gene whose allele is responsible for the increased risk. In this case, the ability to identify the marker genotype for each woman in the pedigree is critical to predicting breast cancer risk. As another example, suppose someone is testing the hypothesis that an allelic variant associated with an inherited disease is present in a particular DNA fragment. In this situation, it is important to be able to locate this fragment using genetic markers to isolate the fragment from affected individuals, verify that the hypothesis is true and identify the nature of the mutation.

Most methods for separating and identifying DNA fragments fall into two general categories: ■ Those that identify a specific DNA fragment present in genomic DNA, taking advantage of the fact that complementary single-stranded DNA sequences under the right conditions can form a double molecule. These methods are based on nucleic acid hybridization. ■ Those who use prior knowledge of the sequence at the ends of a DNA fragment to specifically and repeatedly replicate that genomic DNA fragment. These procedures are based on

selective DNA replication (amplification) using the polymerase chain reaction. The main difference between these approaches is that the first (based on nucleic acid hybridization) identifies fragments present in the genomic DNA itself, while the second (based on DNA amplification) identifies experimentally made copies of fragments whose original templates (but not the replicas) ) were present in genomic DNA. This difference has practical implications: ■ Hybridization methods require a larger amount of genomic DNA for experimental procedures, but relatively large fragments can be identified and no prior knowledge of the DNA sequence is required.

■ Amplification methods require extremely small amounts of genomic DNA for experimental procedures, but amplification is usually limited to relatively small fragments and some prior knowledge of DNA sequence is required. Historically, hybridization methods have been the main means of detecting specific DNA fragments. However, with the development of polymerase chain reaction and automated DNA sequencing technologies in the late 1980s, amplification-based methods came to dominate these investigations. In the following section, we briefly describe hybridization methods, after which we delve into amplification-based methods.

Restriction Enzymes and Site-Specific DNA Cleavage In methods that use nucleic acid hybridization to identify particular fragments present in genomic DNA, the first step is usually to experimentally cut the genomic DNA into specific fragments

manageable size. When genomic DNA is isolated from cells, it is usually fragmented into pieces averaging about 50,000 bases or 50 kilobases (kb) in length. It is possible to fragment them further, but the breakage is quite random. Therefore, only a small fraction of all fragments contain a specific DNA sequence, and these fragments vary in size (FIGURE 2.11). This creates a "needle in a haystack" problem: how to separate this sequence of interest from all the others?

FIGURE 2.11 Representation of random DNA fragments, some of which carry a sequence of interest (shown in red).

One of the most important discoveries in the history of molecular genetics (for which Werner Arbers, Daniel Nathans and Hamilton Smith shared the 1978 Nobel Prize) was that of restriction endonucleases, or restriction enzymes. These enzymes found in bacteria can cleave DNA molecules at positions where certain short DNA sequences (usually 4 to 6 base pairs) occur. Restriction enzymes function in nature to protect bacteria by selectively degrading the genomes of attacking bacteriophages. For example, the restriction enzyme BamH1 (known technically as a type II restriction endonuclease) recognizes the sequence

and cleaves each strand between the two G-bearing nucleotides as shown in FIGURE 2.12.

FIGURE 2.12 The mechanism of DNA cleavage by the restriction enzyme BamHI. Wherever the duplex contains a BamHI restriction site, the enzyme makes a single cut in the backbone of each strand of DNA. Each cut creates a new 3' end and a new 5' end, separating the duplex into two fragments. In the case of BamHI, the cuts are staggered cuts so that the resulting ends end in single-stranded regions, each four nucleotides long.

Figure 2.13 shows nine of the several hundred known restriction enzymes. Most restriction enzymes are named after the species in which they are found. For example, BamHI was isolated from the bacterium Bacillus amyloliquefaciens strain H and is the first (I) restriction enzyme to be isolated from this organism. As the first three letters of the name of each restriction enzyme represent the bacterial species of origin, this

The letters are in italics; the rest of the symbols in the name are not italicized.

Figure 2.13 Recognition sites for various restriction enzymes. The vertical dashed line indicates the axis of symmetry in each sequence. The red arrows indicate the interfaces. The enzymes in the first two columns create sticky ends; those in the third column produce blunt ends. The TaqI enzyme produces sticky ends that consist of two nucleotides, whereas the sticky ends produced by the other enzymes contain four nucleotides. R and Y refer to any complementary purines and pyrimidines, respectively.

Most restriction enzymes recognize only a short sequence of bases, usually four or six nucleotide pairs. At these sites, the enzyme binds to the DNA and breaks each strand of the DNA molecule, creating free 3'-OH and 5'-P groups at each position. The nucleotide sequence recognized by a restriction enzyme for cleavage is called the enzyme's restriction site. Some restriction enzymes cleave their restriction site asymmetrically (at different sites on the two strands of DNA), but other restriction enzymes cleave the site symmetrically (at the same site on both strands).

wires). The former leave sticky ends (also called sticky ends) because each end of the cleavage site has a small, single-stranded overhang that is complementary in base sequence to the other end (Figure 2.12). In contrast, enzymes with symmetrical cleavage sites produce DNA fragments with blunt ends. In virtually all cases, the restriction site of a restriction enzyme has the same reading on both strands, as long as the opposite polarity of the strands is taken into account; for example, each strand in the BamHI restriction site reads 5'-GGATCC-3'

(Figure 2.10). A DNA sequence with this type of symmetry is called a palindrome. (In common English, a palindrome is a word or phrase that reads the same backwards, like "madam".)

Restriction enzymes have the following important properties: ■ Most restriction enzymes recognize a single restriction site. ■ The restriction site is recognized regardless of the DNA source. ■ Since most restriction enzymes recognize a unique restriction site sequence, the number of cuts in the DNA of a

a given organism is determined by the number of restriction sites present. The DNA fragment created by a pair of adjacent cuts in a DNA molecule is called a restriction fragment. A large DNA molecule is usually cut into many differently sized restriction fragments. For example, an E. coli DNA molecule containing 4.6 × 10 6 base pairs is cut into several hundred to several thousand

Fragments, and mammalian genomic DNA is cut into over a million fragments. More importantly, these fragments are not randomly generated; rather, their ends are determined by the presence of restriction sites in the DNA to be digested. Therefore, in digested DNA, a sequence of interest must always appear in fragments with identical ends. This has three important implications: 1. Restriction sites can be used as genetic markers when sequence variation exists within specific loci in a population of individuals. 2. If a particular restriction fragment can be isolated, then the sequences it contains can be further characterized. 3. Two fragments of DNA, produced by digestion with an enzyme like BamH1, have complementary ends that can (at least in theory) base pair with each other. Indeed, this base pairing can be easily performed and forms the basis for DNA cloning, shown in Figure 2.4 and described in the chapter Manipulation of Genes and Genomes.

Gel Electrophoresis So now that we have the genomic DNA that's been digested, how do we separate the thousands of different fragments? They can be separated by size based on the fact that DNA is negatively charged and moves in response to an electric field. If the staples

For example, when a source of electrical energy is connected to opposite ends of a horizontal tube containing a DNA solution, the DNA molecules move toward the positive end of the tube at a rate that depends on the strength of the electric field and the shape of the tube. and tube size. molecules. The movement of charged molecules in an electric field is called electrophoresis.

The most commonly used type of electrophoresis in genetics is gel electrophoresis. An experimental setup for DNA gel electrophoresis is shown in FIGURE 2.14A. A thin plate of gel, usually agarose or acrylamide, is made and contains small slits (called wells) into which samples are placed. An electric field is applied and the negatively charged DNA molecules enter and move through the gel towards the anode (the positively charged electrode). A gel is a complex molecular network containing narrow, tortuous passages that allow smaller DNA molecules to pass more easily; therefore, the speed of movement increases as the size of the DNA fragment decreases. FIGURE 2.14B shows the result of electrophoresis of a set of double-stranded DNA molecules in an agarose gel. Each discrete region containing DNA is called a band. The bands can be visualized under ultraviolet light after immersing the gel in the ethidium bromide dye, whose molecules intercalate into the duplex DNA and make it fluorescent. In Figure 2.14B, each band on the gel results from the fact that all DNA fragments of a given size have migrated to the same position on the gel. At least about 5 × 10-9 grams of DNA are needed to generate a visible band, yielding about 109 molecules for a 3 kb fragment. The issue is that there must be a very large number of copies of a given DNA fragment to give a visible band on an electrophoresis gel.

FIGURE 2.14 DNA gel electrophoresis. (A) Liquid gel is set with an appropriately shaped mold to form slits for the samples. After electrophoresis, DNA fragments located at different positions on the gel are visualized by immersing the gel in a solution containing a reagent that binds or reacts with the DNA. (B) After staining with a fluorescent dye (ethidium bromide), the separated fragments appear as bands in a sample when the gel is exposed to ultraviolet light.

Due to the sequence specificity of the cleavage, a given restriction enzyme will generate a unique set of fragments for a given DNA molecule. A different enzyme will produce a different one

Set of fragments of the same DNA molecule. FIGURE 2.15 shows this principle for the digestion of a 10 kb long circular double-stranded DNA molecule. When digested with EcoRI restriction enzyme (Figure 2.15A), the circular molecule produces 4 kb and 6 kb bands. This pattern would result from circled EcoRI restriction sites at the relative positions shown below the gel. The circle is randomly oriented with one of the EcoRI(E) sites at the top. Likewise, digestion of the circle with the enzyme BamHI (Figure 2.15B) results in bands of 3 kb and 7

kb, implying that the circle contains BamHI sites at the positions indicated in the diagram below the gel. Again, the circle is randomly oriented, this time with one of the BamHI sites (B) at the top. A diagram that shows the cleavages of one or more restriction sites along a DNA molecule is called a restriction map.

FIGURE 2.15 Gel graphs showing the sizes of restriction fragments generated by digesting a 10 kb circular molecule of double-stranded DNA with (A) EcoRI, (B) BamHI, and (C) both enzymes together. Below each diagram is a circular DNA restriction map showing the location of the restriction sites. The restriction map in (C) takes into account the sizes of fragments generated in (A) and (B) and by digestion with both enzymes.

When EcoRI and BamHI are used together, the resulting DNA fragments show where EcoRI sites and BamHI sites are relative to each other. In this case, digestion with both enzymes produces bands of 1 kb, 2 kb, 3 kb, and 4 kb (Figure 2.15C). The restriction map shown below the gel indicates where the two types of restriction sites must be located to obtain these band sizes. This restriction map can be obtained by superimposing those of part B on those of part A and rotating the result until the distances between adjacent pairs of restriction sites are equal to 1, 2, 3 and 4 kb (not necessarily in that order). In this case you just need

Rotate the constraint map in part B 2 kb to the right. Note that in the restriction enzyme summary in Figure 2.13C, the 4 kb fragment is not the same as the 4 kb fragment seen in part A, and the 3 kb fragment is not the same as the 3 kb fragment seen in part B. This discrepancy arises because each restriction enzyme cleaves fragments produced by the other. The orientation of the constraint map in Figure 2.13C is arbitrary. It can be flipped or rotated any amount in any direction and it will still be the same constraint map. The construction of such a map facilitates two results. First, individual fragments can be isolated from the gel, inserted into self-replicating molecules such as bacteriophages, plasmids, or even small artificial chromosomes (Figure 2.4), and introduced into bacterial cells (DNA cloning, a topic we'll cover in Manipulating Genes and Genomes). . Second, and more importantly with regard to DNA markers, we have taken the first step in identifying specific positions in DNA - those where enzymes divide. This raises the question of how we can apply restriction analysis to complex genomes.

Nucleic Acid Hybridization Most genomes are large and complex enough that digestion with a restriction enzyme yields many bands of the same or similar size. Identifying a particular DNA fragment against the background of many other fragments of similar size presents a needle in a haystack problem. For example, suppose we are looking at a particular 3.0 kb BamHI fragment of the human genome that we are interested in, which serves as a marker that indicates the presence of a genetic risk factor for breast cancer in women in a given pedigree. . Based on size alone, this 3.0 kb fragment is indistinguishable from fragments about 2.9 to

3.1 KB. How many fragments in this size range are expected? When human genomic DNA is digested with BamHI, the average length of a restriction fragment is 46 = 4096 base pairs, and the expected total number of BamHI fragments is about 730,000; in the 2.9-3.1 kb size range, the expected number of fragments is around 17,000. Therefore, although we know that the fragment of interest is 3 kb long, it is only one of 17,000 fragments so similar in size that length alone cannot distinguish the fragment we are looking for from the others. In fact, as seen in FIGURE 2.16,

Digested genomic DNA, when viewed, appears as a smear rather than a series of discrete bands.

FIGURE 2.16 An example of genomic DNA from different plants digested with EcoR1 and HindIII enzymes. Reprinted from Kang, T.J., & Yang, M.S. Fast and reliable extraction of genomic DNA from various wild-type and transgenic plants. BMC Biotechnol. 2004;4:20. doi:10.1186/1472-6750-4-20. Creative Commons license available: https://creativecommons.org/licenses/by/2.0/

This identification task is actually more difficult than looking for a needle in a haystack, as haystacks are usually dry. A more accurate analogy would be looking for a needle in a haystack dropped into a pool full of water. This analogy is more relevant because, although gels contain a supporting matrix that makes them semi-solid, they are mostly water, and each DNA molecule within a gel is completely surrounded by water. Obviously, we need a method that can visualize or purify specific fragments.

At this point, we need to return to the structure of DNA (Figure 2.8), examine how the two strands in a double helix can be pulled apart to form single strands, and see how, under the right conditions, two single strands that are complementary or nearly complementary in sequence can join together. rejoin to form another double helix. The separation of strands is called denaturation, and the joining of complementary strands is called nucleic acid hybridization or renaturation. (The term hybridization is appropriate because the two strands that join to form a DNA duplex may not be exactly the same strands that were paired before denaturation.) The practical applications of nucleic acid hybridization are many: ■ A small part of a DNA fragment can be hybridized with a much larger DNA fragment. This principle is used to identify specific fragments of DNA in a complex mixture, such as B. the 3 kb BamHI marker for breast cancer that we have considered. Applications of this type include tracking genetic markers in pedigrees and isolating fragments containing a particular mutant gene. ■ A DNA fragment from one gene can be hybridized with similar fragments from other genes in the same genome; This principle is used to identify different members of gene families that

they are similar in order but not identical and have related functions. ■ A fragment of DNA from one species can hybridize with similar sequences from other species. This allows for the isolation of genes that have the same or related functions across multiple species. This method is used to study aspects of molecular evolution, such as B. how differences in sequence correlate with differences in function, and the patterns and rates of change in gene sequences during their development.

As we saw in Section 2.3, The Molecular Structure of DNA, the double-stranded helical structure of DNA is maintained by base stacking and by hydrogen bonds between complementary pairs of bases. When solutions containing DNA fragments are heated to temperatures in the range of 85°C to 100°C or to the high pH of strong alkaline solutions, the paired strands begin to separate. The unwinding of the helix takes less than a few minutes (the time depends on the length of the molecule). A common method for detecting denaturation is to measure the ability of DNA in solution to absorb ultraviolet light with a wavelength of 260 nm, since the absorbance at 260 nm (A260) of a solution of single-stranded molecules is 37 % greater than the absorbance of double-stranded molecules at the same concentration. As shown in FIGURE 2.17, the progress of denaturation can be followed by slowly heating a solution of double-stranded DNA and recording the A260 value at different temperatures. The temperature required for denaturation increases with G+C content, not only because G-C base pairs have three hydrogen bonds and A-T base pairs have two such bonds, but also because successive G-C base pairs exhibit more stacking. of bases.

FIGURE 2.17 The mechanism of thermal denaturation of DNA. The temperature at which 50 percent of the base pairs are denatured is the melting temperature, symbolized as Tm.

In order for the denatured DNA strands to be renatured, two requirements must be met: 1. The salt concentration must be high (greater than 0.25 M) to neutralize the negative charges of the phosphate groups, which would otherwise cause the complementary strands repel each other. other . 2. The temperature must be high enough to break hydrogen bonds that randomly form between short sequences of bases within the same strand, but not so high as to destroy stable base pairs between complementary strands. The initial phase of renaturation is a slow process, as the speed is limited by the possibility of a region of two complementary strands randomly joining together to form a short sequence of correct base pairs. This initial pairing step is followed by rapid pairing of the remaining complementary bases and rewinding of the

propeller. Rewinding takes place in a matter of seconds and its speed is independent of DNA concentration, as the complementary strands have already met. Importantly, when two DNA molecules from different sources contain identical sequences and can reassemble with each other, part of the renatured DNA will be heteroduplex DNA - that is, H. of molecules containing strands with base pairs originating from different sources (FIGURE 2.18). If one DNA source is genomic DNA and the other is a DNA probe consisting of part of our sequence of interest that is appropriately labeled (e.g., with radioactive 32 P), then all heteroduplexes generated will be

radioactively labeled. Geneticists say that the probe's DNA hybridizes with fragments of DNA that contain sequences that are similar and not complementary to the probe.

FIGURE 2.18 Detection of DNA by nucleic acid hybridization. The duplex genomic DNA is heat-denatured, after which it is combined with a labeled probe sequence homologous to a specific sequence in the genomic DNA. The mixture is cooled, allowing the formation of heteroduplex molecules containing a strand of genomic DNA paired with the labeled probe sequence.

This process can be used to solve the problem of viewing specific fragments in a conceptually simple way. If the denatured DNA is immobilized on a membrane (typically nitrocellulose or nylon membrane), then this membrane can be incubated in a solution containing a radioactively labeled probe consisting of copies of the sequence of interest. This process, shown in FIGURE 2.19, is called a Southern blot (named after E.M. Southern, who developed the technique). In this procedure, digested genomic DNA is transferred from an agarose gel to a

Membrane. The DNA on the membrane is then denatured, after which it is incubated with a denatured probe under conditions that allow renaturation and heteroduplex formation. The hybridization pattern can then be visualized by exposure to X-ray film.

FIGURE 2.19 Southern blot: an experimental procedure for identifying the position of a specific DNA fragment on a gel.

Returning to our original problem of testing DNA for a breast cancer risk allele marker in a pedigree, we could use this technique to analyze the DNA of any individual from whom we could obtain genomic DNA. In fact, this process is extremely sensitive: under typical conditions, a band can be seen on film containing only 5 × 10−12 grams of DNA – 1,000 times less DNA than the amount needed to produce a visible band on the gel itself.

SUMMARY ■ Restriction enzymes cleave DNA at specific recognition sites and generate specific fragments. ■ These fragments can be analyzed by gel electrophoresis. ■ Restriction sites can be mapped to DNA molecules and isolated restriction fragments for further study. ■ DNA can be denatured and annealed.

■ Restriction analysis of genomic DNA can be performed by hybridizing electrophoresed DNA with specific labeled probes.

2.5 Specific DNA amplification for detection and purification Despite its sensitivity, Southern blotting has its limitations. Requires a relatively large amount of genomic DNA, and for some applications (e.g. analysis of DNA evidence at a crime scene or from ancient remains) this type of sample is not required

accessible. Furthermore, although it allows the identification of a certain DNA fragment when it is present in a complex mixture of fragments, it does not allow the fragment to be separated from the others and purified. Obtaining the fragment in a purified form requires cloning, which is straightforward but time consuming. (Methods of cloning are discussed in the Manipulation of Genes and Genomes chapter.) However, if the fragment of interest is not very long and if the nucleotide sequence at each end is known, it becomes possible to obtain large quantities of the fragment simply by selective replication. This process is called amplification. Once this is achieved, the amplified fragment can be analyzed directly. In fact, these methods have largely replaced hybridization methods in genetic screening protocols. How would one know the nucleotide sequences at the ends of interest? Returning to our example of the 3.0 kb BamHI fragment used to mark a breast cancer risk allele in certain pedigrees. Suppose this fragment is cloned and sequenced from an affected individual and the BamHI fragment is found to be missing a 500 base pair region compared to the sequence in unaffected individuals. At this point, the sequences at the ends of the fragment are known and we can also conclude that amplification of genomic DNA from individuals with the risk factor will result in a 3.0 kb band, whereas amplification of genomic DNA from non-carriers will provide a band . be 3.5 kb. This difference makes it possible to diagnose each person in the family tree as a carrier or carrier

Non-carriers only by DNA amplification. To understand how amplification works, it is first necessary to examine some key features of DNA replication.

Limitations in DNA replication: primers and 5' to 3' strand extension

As with restriction analysis, amplification uses a biological process - in this case, DNA replication. To do this, it uses the enzyme that forms the sugar-phosphate bond (the phosphodiester bond) between adjacent deoxynucleotides in a DNA strand, called DNA polymerase. A variety of DNA polymerases have been purified and, to amplify a DNA fragment, DNA synthesis is performed in vitro by combining purified cellular components in a test tube under strictly defined conditions. (In vitro, literally “in glass,” meaning no living cells are present.) In order for DNA polymerase to catalyze the synthesis of a new DNA strand, there must be pre-existing single-stranded DNA. Any single-stranded DNA molecule present in the reaction mixture can serve as a template on which a new partner strand is generated by the DNA polymerase. For DNA replication to occur, the 5' triphosphates of the four deoxynucleoside triphosphates must also be present. This requirement is pretty obvious, as these are the precursors from which new DNA strands are made. The necessary triphosphates are the compounds identified in Table 2.1 as dATP, dGTP, dTTP and dCTP, which contain the bases adenine, guanine, thymine and cytosine, respectively. Details of the structures of dCTP and dGTP are shown in FIGURE 2.20, which indicates the phosphate groups cut during DNA synthesis. DNA synthesis requires all four nucleoside 5'-triphosphates and will not occur if any of them are omitted.

Figure 2.20 Two deoxynucleoside triphosphates used in DNA synthesis. The two outer phosphate groups are removed during synthesis.

A characteristic of all DNA polymerases is that a DNA polymerase can only lengthen one strand of DNA. It is not possible for DNA polymerase to initiate synthesis of a new strand even when a template molecule is present. An important implication of this principle is that DNA synthesis requires a preexisting segment of nucleic acid that is hydrogen bonded to the template strand; this segment is called the primer. Since the primer molecule can be very short, it is an oligonucleotide, which literally means "a few nucleotides". As we'll see in the chapter on DNA replication and sequencing, in living cells the primer is a short piece of RNA; in contrast, in in vitro DNA amplification, the primer used is usually DNA. The 3' end of the primer is essential, since DNA synthesis occurs only by adding consecutive nucleotides to the 3' end of the growing strand. In other words, chain elongation always occurs in the 5'-to-3' direction (5' → 3'). The reason

for the 5' → 3' direction of chain extension is shown in FIGURE 2.21: The reaction catalyzed by DNA polymerase is the formation of a phosphodiester bond between the free 3'-OH group of the chain to be extended and the nearest phosphorus atom. internal to the nucleoside triphosphate, which is attached to the 3' end, is incorporated. Recognition of the appropriate incoming nucleoside triphosphate after replication depends on base pairing with the opposite nucleotide on the template strand. DNA polymerase normally catalyzes the polymerization reaction that incorporates the new nucleotide into DNA.

Primer terminal only when correct base pair is present. The same DNA polymerase is used to add each of the four deoxynucleoside phosphates to the 3'-OH terminus of the growing strand.

Figure 2.21 Addition of nucleotides to the 3'-OH terminus of a growing strand. The recognition step is shown as the formation of hydrogen bonds between G and C. The chemical reaction consists of the 3'-OH group at the 3' end of the growing chain attacking the innermost phosphate group of the incoming trinucleotide.

The polymerase chain reaction

The requirement for an oligonucleotide primer and the restriction that strand elongation must always be in the 5' → 3' direction make it possible to obtain large amounts of a given DNA sequence by selective amplification in vitro. The process of selective amplification is called a polymerase chain reaction.

(PCR). Californian Kary B. Mullis received the Nobel Prize in 1993 for his invention. PCR amplification uses DNA polymerase and a pair of short synthetic oligonucleotide primers, typically 18-22 nucleotides in length, that are complementary in sequence to the ends of the DNA sequence to be amplified. FIGURE 2.22 shows an example where the primer oligonucleotides (green) are 9-mers. These are too brief for most practical purposes, but they are for illustrative purposes only. The original duplex molecule (part A) is shown in blue. This duplex is mixed with a large excess of primer molecules, DNA polymerase and all four nucleoside triphosphates. As the temperature increases, the duplex strands denature and separate. When the temperature is lowered again to allow renaturation, the primers, being present in large excess, hybridize (or anneal) with the separated template strands (Part B). Note that the sequences of the primers differ from each other, but are complementary to the sequences present on opposite strands of the original DNA duplex flanking the region to be amplified. Primers are oriented with their 3' ends toward the region to be amplified, as each DNA strand extends only at the 3' end. After the primers are annealed, each is extended by DNA polymerase using the original strand as a template, and the newly synthesized DNA strands (in red) grow toward each other as synthesis progresses (panel C). Note that a region of duplex DNA present in the original reaction mix can only be amplified by PCR if the region is flanked by primer oligonucleotides.

FIGURE 2.22 Role of primer sequences in PCR amplification. (A) Target DNA duplex (blue) showing sequences chosen as primer binding sites flanking the region to be amplified. (B) Primer (green) bound to denatured strands of target DNA. (C) First round of amplification. The newly synthesized DNA is shown in pink. Note that each primer is extended beyond the other primer site. (D) Second round of amplification (only one tape shown); in this round, the

The newly synthesized strand ends at the opposite site of the primer. (E) Third round of amplification (only one tape shown); in this round, both strands are cut at the primer locations. Primer sequences are generally at least twice as long as shown here.

To initiate a second round of PCR amplification, the temperature is raised again to denature the duplex DNA. As the temperature is lowered, the original parental strands hybridize to the primers.

and are replicated as shown in Figure 2.22B and C. The daughter strands produced in the first round of amplification are also joined with primers and are replicated as shown in part D. In this case, however, the daughter duplex molecules are identical in sequence to the original parent molecule, consisting entirely of primer oligonucleotides and non-parental DNA synthesized in the first or second round of PCR. As successive cycles of denaturation, priming, and elongation occur, the original parental strands are diluted by proliferation of new daughter strands until nearly all molecules produced in PCR have the structure shown in panel E. The power of PCR amplification derives from the fact that the number of template copies increases in exponential progression: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, etc., where it doubles with each replication cycle. Starting with a mixture containing just one molecule of the fragment of interest, repeated rounds of DNA replication exponentially increase the number of amplified molecules. For example, if you start with a single molecule, 25 rounds of DNA replication will result in 225 = 3.4 × 10 7 molecules. This number of molecules in the amplified fragment is so much greater than that of the other non-amplified molecules in the original mixture that the amplified DNA can often be used without further purification. For example, a single 3 kb file

The fragment in E. coli represents only 0.06% of that organism's DNA. However, if that single fragment were replicated in 25 rounds of replication, 99.995 percent of the resulting mixture would consist of the amplified sequence. A 3 kb fragment of human DNA represents only 0.0001 percent of the total genome size. Amplifying a 3 kb fragment of human DNA to 99.995% purity would require approximately 34 cycles of PCR. FIGURE 2.23 gives an overview of the polymerase chain

Reaction. The DNA sequence to be amplified is again shown in blue and the oligonucleotide primers in green. The oligonucleotides attach to the ends of the sequence to be amplified and become substrates for chain elongation by DNA polymerase. In the first round of PCR amplification, the DNA is denatured to separate the strands. The denaturing temperature is usually around 95°C. The temperature is then lowered to allow for annealing in the presence of a large excess of primer oligonucleotides. The annealing temperature is typically in the range of 50°C-60°C and is largely dependent on the G+C content of the oligonucleotide primers. To complete the cycle, the temperature is slightly increased to about 70°C to prolong each primer. The denaturation, renaturation and replication steps are repeated 20 to 30 times, and at each cycle the number of molecules in the amplified sequence is doubled.

FIGURE 2.23 Polymerase chain reaction (PCR) used to amplify specific DNA sequences. Oligonucleotide primers (green) complementary to the ends of the target sequence (blue) are used in repeated rounds of DNA denaturation, annealing, and replication. Newly replicated DNA is shown in pink. The number of copies of the target sequence doubles with each round of replication, eventually outnumbering any other sequences that may be present.

Performing PCR with conventional DNA polymerases is not feasible, as the polymerase itself irreversibly unfolds and becomes inactive at the high temperature required for denaturation. However, DNA polymerase isolated from certain types of organisms is heat stable, as these organisms typically live in hot springs with temperatures well above 90°C, such as those found in Yellowstone National Park. These organisms are called thermophiles. The most used thermostable DNA polymerase is called Taq polymerase because it was native

isolated from the thermophilic bacterium Thermus aquaticus. PCR amplification is very useful for generating large amounts of a specific DNA sequence. The main limitation of the technique is that the DNA sequences at the ends of the region to be amplified must be known to synthesize oligonucleotide primers. Furthermore, sequences larger than about 5,000 base pairs cannot be efficiently replicated by conventional PCR methods, although some PCR modifications allow amplification of longer fragments. On the other hand, many applications require the amplification of relatively small fragments. The main advantage of PCR amplification is that only traces of template DNA are needed. In theory, only one template molecule is needed, but in practice amplification of a single molecule can fail because the molecule can be accidentally broken or damaged. In practice, amplification is usually reliable with only 10 to 100 template molecules, making PCR amplification 10,000 to 100,000 times more sensitive than detection by nucleic acid hybridization. PCR greatly simplifies the genotyping of individuals. Rather than going through all the necessary steps in Southern blotting, a small amount of genomic DNA can be used in combination with appropriate primers to amplify the sequence of interest. Going back to our breast cancer risk marker example, FIGURE 2.24

illustrates what we would observe if we genotyped two individuals, one homozygous and one heterozygous, for the allele of interest. Furthermore, the extraordinary sensitivity of PCR amplification led to its use in DNA typing for criminal cases where a small amount of biological material was left behind.

Culprit (skin cells on a cigarette butt or hair root cells on a single hair can provide enough template DNA for amplification).

FIGURE 2.24 Determination of a PCR-based DNA marker genotype. The genotypes of two individuals are shown. Individual 1 is homozygous; Individual 2 is heterozygous for two alleles that differ in length due to a deletion in the second allele (shown in orange). PCR products can be analyzed directly without blotting, resulting in one subject 1 band and two subject 2 bands.

In research, PCR is often used to study independent mutations in a gene whose sequence is known, to identify the molecular basis of each mutation, and to examine the DNA sequence.

Variations between alternative forms of a gene that may be present in natural populations, or to study differences between genes with the same function in different species. The PCR method has also found widespread use in clinical laboratories for diagnosis. To cite just one very important example, the presence of the Human Immunodeficiency Virus (HIV), which causes Acquired Immunodeficiency Syndrome (AIDS), can be detected in trace amounts in blood banks by PCR using primers complementary to sequences in the genetic material viral is . This and other PCR applications are facilitated by the fact that the method lends itself to automation through the use of mechanical robots to set up and run reactions.

SUMMARY ■ Specific DNA fragments contained in a complex mixture, such as genomic DNA, can be selectively amplified using the polymerase chain reaction (PCR).

■ PCR requires the presence of template DNA, nucleoside triphosphates, DNA polymerase, and two oligonucleotide primers flanking the region to be amplified. ■ PCR requires repeated cycles of DNA synthesis, denaturing, and primer annealing. This is accomplished using a thermal cycler and a heat-stable DNA polymerase. ■ Using PCR greatly simplifies DNA marker genotyping and can be used for many other applications including pathogen detection and forensic DNA analysis.

2.6 Types of DNA Markers in Genomic DNA Genetic variation in the form of polymorphisms exists in most natural populations of organisms. The DNA manipulation methods examined in Sections 2.4 and 2.5 can be used in a variety of combinations to reveal differences between individuals. Anyone reading the modern genetics literature will encounter a bewildering array of acronyms that refer to the different species in which genetic polymorphisms are found. Different approaches are used because no technique is ideal for all applications, each technique has its own advantages and disadvantages, and new techniques are constantly being developed. In this section, we look at some of the most important methods used to detect DNA polymorphisms in individuals.

Constraint Fragment Length

Polymorphisms The simplest type of genetic polymorphism is the single nucleotide polymorphism (SNP) (pronounced "snip"). This is simply a position in the genome where two or more different bases are found in different genomes within a population. We now know that SNPs are ubiquitous: the 1000 Genomes Project has cataloged over 80 million of them in the human genome. However, before the development of modern high-throughput genotyping, it was difficult to detect SNPs and other forms of sequence variation. Restriction enzymes proved to be valuable tools to elucidate their presence. We have already seen that restriction enzymes can be used to map the occurrence of certain sequences (the recognition sites) in DNA molecules. If a recognition site contains a DNA polymorphism,

then digestion can be used as a means of detection. An example is shown in FIGURE 2.25, where a T-A nucleotide pair occurs in some molecules and a C-G pair in others. In this example, the polymorphic nucleotide site is contained within a cleavage site for the restriction enzyme EcoRI (5'-GAATTC-3'). The two closest flanking EcoRI sites are also shown. In such a situation, DNA molecules with T-A nucleotide pairs at both the flanking sites as well as the intermediate site are cleaved, producing two EcoRI restriction fragments. In contrast, DNA molecules will be with C-G nucleotide pairs

cleaved at both flanking sites but not at the middle site (because the presence of C-G destroys the EcoRI restriction site), giving only a larger restriction fragment. An SNP that eliminates a restriction site is known as a restriction fragment length polymorphism (RFLP) (pronounced "riflip" or spelling it).

FIGURE 2.25 A difference in the DNA sequence of two molecules can be detected when the difference eliminates a restriction site. (A) This molecule contains three restriction sites for EcoRI, including one at each end. It is divided into two fragments by the enzyme. (B) This molecule has an altered EcoRI site in the middle where 5'-GAATTC-3' becomes 5'-GAACTC-3'. The altered site cannot be cleaved by EcoRI, so treatment of this molecule with EcoRI results in a larger fragment.

In the early days of genomic DNA analysis, before the development of automated DNA sequencing (see chapter DNA Replication and Sequencing), RFLP analysis was an extremely important tool for mapping and detecting chromosomes

DNA polymorphisms, as such polymorphisms can be detected by Southern blotting. However, with the advent of PCR, the procedure can be greatly simplified. As shown in FIGURE 2.26, primers flanking the RFLP site can be used to amplify the DNA containing it, and then this can be done by digestion and simple gel electrophoresis.

used to analyze the product. Since an amplified fragment can also be sequenced, comparison of products from different alleles can then be extended to all positions in the fragment.

FIGURE 2.26 In a restriction fragment length polymorphism (RFLP), alleles can differ in the presence or absence of a DNA cleavage site. This example shows a PCR fragment containing two single base pair alleles. Primer sequences are shown in blue. The a allele lacks a restriction site present in the DNA of the A allele. The difference in fragment length can be detected by digesting PCR products containing the cleavage site. The RFLP alleles are codominant, meaning that the DNA of the heterozygous Aa genotype gives rise to each of the distinct bands seen in the DNA of the homozygous AA and aa genotypes.

Single Nucleotide Polymorphisms RFLP analysis can detect only allelic variants that affect the digestion pattern of the enzyme(s) used, which is only a small fraction of the large number of allelic differences that exist in genomic DNA. If the fragments are not digested, but their sequence is fully determined, the SNPs can be detected. An SNP is a specific nucleotide site in DNA that differs from individual to individual in terms of the identity of the nucleotide pair occupying the site. For example, some DNA molecules may have a T-A base pair at a particular nucleotide location, while other DNA molecules in the same population may have a C-G base pair at the same location. This difference forms an SNP. The SNP defines two alleles for which there can be three genotypes among individuals in the population: homozygous with T–A at the corresponding position on both homologous chromosomes, homozygous with C–G at the corresponding position on both homologous chromosomes, or heterozygous with T – A on one chromosome and C–G on the homologous chromosome. In the human genome, two randomly chosen DNA molecules are likely to differ at one SNP site

approximately every 1000 bp in non-coding DNA and at an SNP site approximately every 3000 bp in protein-coding DNA. Note that when discussing an SNP, the stipulation is that DNA molecules must differ "often" in nucleotide location. This determination excludes rare genetic variations of species that occur in less than 1% of DNA molecules in a population. Extremely rare genetic variants are generally not as useful in genetic analysis as more common variants.

SNPs are the most common form of genetic differences between people. About 10 million SNPs have been identified as relatively common in the human population, and 300,000 to 600,000 of these are typically used in the search for SNPs that may be associated with complex diseases such as diabetes or hypertension (see Genetic Linkage and Chapter Chromosome Mapping and The genetic basis of complex traits). Identification of the specific nucleotide present in each of a million SNPs is possible through the use of DNA microarrays containing millions of tiny dots on a glass slide the size of a postage stamp. Each tiny particle contains a unique sequence of DNA oligonucleotides, present in millions of copies, which are microchemically synthesized in the fabrication of the microarray. Each oligonucleotide sequence is designed to hybridize specifically to small fragments of genomic DNA containing one or another pair of nucleotides present in a SNP. The microarrays also include several controls for each hybridization. Controls consist of oligonucleotides with intentional mismatches to avoid being misled by certain nucleotide sequences that are particularly "sticky" and hybridize very easily with genomic fragments and other sequences that form poorly or non-hybridizing structures. These microarrays are sometimes called SNPs

The chips make it possible to determine an individual's SNP genotype with nearly 100% accuracy. The principles behind oligonucleotide hybridization are shown in FIGURE 2.27. In the figure, the length of each oligonucleotide is seven nucleotides; in practice, this is very short, as typical SNP chips consist of oligonucleotides at least 25 nucleotides in length. Part A shows the two types of DNA duplexes that can form an SNP. In this example, some chromosomes carry a DNA molecule

with a T-A base pair in the position shown in red, whereas the DNA molecule in other chromosomes has a C-G base pair in the corresponding position. Small fragments of genomic DNA are labeled with a fluorescent tag and then the single strands are hybridized with an SNP chip containing the complementary oligonucleotides and numerous controls. The duplex containing T-A hybridizes only to the two oligonucleotides on the left and the one containing C-G hybridizes only to the two oligonucleotides on the right.

FIGURE 2.27(A) Oligonucleotides attached to a glass slide on an SNP chip can be used to identify duplex DNA molecules that contain alternative base pairs for an SNP—in this example, a T-A base pair versus a T-A base pair. C-G bases. (B) The SNP genotype of an individual can be determined by hybridization, since DNA samples from genotypes that are homozygous TA/TA, homozygous CG/CG or heterozygous TA/CG give different fluorescent patterns.

After hybridization, the SNP chip is examined using fluorescence microscopy to detect the spots that fluoresce due to the tag in the genomic DNA. Possible patterns are shown in Part B. Genomic DNA from an individual whose chromosomes contain two copies of the TA form of the duplex (TA/TA homozygote) causes the leftmost colon to fluoresce, but the rightmost colon remain unmarked. Likewise genomic DNA

a homozygous CG/CG individual will fluoresce the rightmost colon, but not the leftmost colon. Finally, genomic DNA from a heterozygous TA/CG individual will cause all four dots to fluoresce because the TA duplex marks the leftmost two dots and the CG duplex marks the rightmost two dots. The use of SNP chips or other available technologies for high-throughput genotyping of millions of SNPs in thousands of individuals allows the identification of genetic risk factors for disease. A typical study compares the genotypes of patients with certain diseases with the genotypes of healthy people who are matched with patients for factors such as gender, age, and ethnic group. Comparison of SNP genotypes between these groups often reveals which SNPs in the genome mark the location of genetic risk factors, as discussed in more detail in Section 2.7.

Tandem Repeat Polymorphisms An important type of DNA polymorphism results from differences in the number of copies of a DNA sequence that can be repeated many times in a row at a given location on a chromosome. Any given chromosome can have any number of copies of the tandem repeat, typically ranging from tens to a few hundred. FIGURE 2.28 shows DNA molecules that differ in the number of tandem repeats. In this case, the number of repetitions varies from 1 to 10.

FIGURE 2.28 A genetic polymorphism in which the alleles in a population differ in the number of copies of a DNA sequence (usually 2 to 60 bp) that is repeated together along the chromosome. This example shows alleles where the repeat count ranges from 1 to 10. Amplification using primers flanking the repeat produces a unique fragment length for each allele.

LATEST STATUS: High-throughput SNP genotyping

Naturally occurring enzymes are often used for critical steps in molecular genetic analysis. Use of restriction enzymes and

DNA polymerase for DNA marker analysis and PCR are two examples, and the enormous biodiversity (especially microbes) on Earth continues to be a source of new tools. A good example is SNP detection. The basic SNP chip approach, as shown in Figure 2.27, depends on having enough genomic DNA available to be analyzed; also requires precise hybridization of genomic DNA fragments to oligonucleotides

the chips Both limitations have been addressed by the latest techniques. To increase the amount of genomic DNA, the PCR-free whole genome amplification technique was developed. PCR itself has two problems. First, it requires thermal cycling for the denaturation, renaturation, and elongation steps. Second, Taq polymerase (the one most commonly used in PCR) is error-prone: an incorrect base is inserted every 800 bases. To get around these problems, scientists have developed a technique for amplifying entire genomes called multiple displacement amplification. As shown in Figure A, random sequences of six bases (hexamers) are used as primers. DNA polymerase is derived from bacteriophage φ29, an enzyme that replicates DNA with high fidelity. Furthermore, DNA polymerase will shift this strand when it encounters the 5' end of a strand that is paired with its template; the resulting single strand can then hybridize to another of the hexamer primer sequences present in the reaction, thus serving as a template for further replication. In this way, all genomic DNA can be amplified from a single cell, and the process occurs without temperature cycling (denaturation and renaturation are not necessary).

FIGURE A

After this phase, the amplified DNA is cut into small fragments, denatured and hybridized to oligonucleotides (FIGURE B). Multiple copies of different oligonucleotides were attached to beads with free 3' ends and these beads were embedded in a solid matrix. However, the SNP to be tested is not part of the sequence contained in the oligonucleotide; instead, the SNP is the next base in the sequence after the 3' end of the oligonucleotide. DNA polymerase is added along with modified nucleoside triphosphates that can be detected by fluorescence under conditions where a base is added to the oligonucleotide probe. The identity of this base depends on the SNP allele in the hybridized genomic DNA and can be determined by fluorescence scanning on the chip. In the case shown in Figure B, the individual tested is homozygous for the A allele of SNP 1, heterozygous for the A and G alleles of SNP 2, and homozygous for the G allele of SNP 3.

FIGURA B

These cutting-edge methods make SNP genotyping fast and efficient, even with a very small amount of starting material (such as the saliva samples used by personal genomics company 23andMe), and can generate enormous amounts of data. In fact, up to 48 samples of 300,000 or more SNPs can be genotyped in just three days. While these methods require some sophisticated instrumentation (many of which can be performed robotically), they remain essentially based on the biological processes of base pairing and DNA replication.

When one of the DNA molecules is cleaved by a restriction endonuclease that cleaves at sites flanking the tandem repeat, the size of the resulting restriction fragment is determined by the number of repeats it contains. A duplex DNA molecule containing the repeats can also be amplified by the polymerase chain reaction using primers that flank the tandem repeat. Whether obtained by endonuclease digestion or PCR, the resulting DNA fragments increase in size according to the number of repeats they contain. Therefore, as shown in the figure on the right

2.28, two DNA molecules that differ in the number of copies of the tandem repeat can be distinguished, since each produces a fragment of DNA of different size that can be separated by amplification of that fragment. The acronym SSR stands for Simple Sequence Repeat (these sequences are also known as microsatellite loci). An example of an SSR is the repeated sequence 5'-…ACACACAC…-3'. SSRs are important in genetic analysis for two reasons: ■ SSRs are abundant in the genomes of most eukaryotes. ■ SSRs are often highly polymorphic.

To illustrate their frequency in the human genome, the most common SSRs are tabulated in TABLE 2.3. There are many more dinucleotide repeats than trinucleotide repeats, but there is wide variation in frequency within each class. The most common dinucleotide repeat is 5'-…ACACACAC…-3', present in over 80,000 locations throughout the human genome. On average, the human genome has one SSR for every 2 kb of human DNA, or about 1.5 million SSRs in total.

TABLE 2.3 Some single repeats in the SSR repeat unit of the human genome

Number of non-human genome SSRs

5'-AC-3'

80.330

5'-AT-3'

56.260

5'-AG-3'

23.780

5'-GC-3'

290

5'-AAT-3'

11.890

5'-AAC-3'

7.540

5'-AGG-3'

4.350

5'-AAG-3'

4.060

5'-ATG-3'

2.030

5'-CGG-3'

1.740

5'-ACC-3'

1.160

5'-AGC-3'

870

5'-ACT-3'

580

Data from the International Human Genome Sequencing Consortium, Nature. 2001; 409:860-921.

Not only do SSRs tend to be polymorphic, but most polymorphisms tend to have a large number of alleles. Each "allele" corresponds to a DNA molecule with a different number of copies of the repeat unit, as shown in Figure 2.28. In other words, a tandem repeat number polymorphism usually has multiple alleles in the population. Even with multiple alleles, however, none special

The chromosome can only carry one of the alleles defined by the number of tandem repeats, and any single genotype can carry at most two different alleles. However, a large number of alleles implies an even greater number of genotypes. For example, even with only 10 alleles in a population, there could be 10 different homozygous genotypes and 45 different heterozygous genotypes.

More generally, given n alleles, there are a total of n(n + 1)/2 possible genotypes, of which n are homozygous and n(n − 1)/2 are heterozygous. Thus, if an SSR has a relatively large number of alleles in a population, but no allele is exceptionally common, then each of the many genotypes in the population will have a relatively low frequency. If genotypes at 6-8 highly polymorphic loci are considered simultaneously, any possible multilocus genotype is extremely rare. The very low frequency of a multiloci genotype gives tandem repeat polymorphisms their usefulness in DNA typing (sometimes called DNA fingerprinting) for individual identification and for assessing the degree of genetic relatedness between individuals. For example, forensic DNA profiling determines genotypes for 13 SSR loci distributed throughout the human genome; The probability that two individuals (other than unrelated twins) will have identical genotypes at all 13 loci is less than one in a billion. The use of DNA typing in criminal investigations is illustrated in Problem 2.24 and Challenge Problem 2 at the end of this chapter and discussed further in the Genes in Populations chapter. Technological advances have done much to make DNA typing easier. Up to this point, we have only considered electrophoretic analyzes performed manually. Indeed, SSR genotyping is routinely performed using automated systems in which multiple loci can be genotyped in a single reaction. In this method (FIGURE 2.29), PCR is performed using multiple primers and the primers simultaneously

are labeled with chemical markers that can be detected by fluorescence. The resulting products are then separated electrophoretically in a gel contained in a capillary tube. As molecules exit the tube, they are captured by a photodetector and their size recorded. Figure 2.29B shows the results of this analysis. In this case, 24 SSR loci were genotyped from an initial sample of 1.2 picograms of human genomic DNA - comparable to the amount of DNA that can be obtained from a single swab.

FIGURE 2.29 Genotyping of SSR using automated multiplex PCR. (A) Schematic diagram showing four loci that are amplified, each with primers containing different fluorescent markers (blue, green, purple, and red) and producing fragments of different sizes. (B) An example of SSR genotyping results. In this case, 24 different sets of SSR locus primers were used in combination with four different fluorescent markers. Shaded boxes indicate loci names. If an individual is homozygous (in this case D2S1338 and D12S391) a single peak is observed. If an individual is heterozygous, two peaks will be seen. (B) Courtesy of Promega Corporation. retrieved from

http://www.promega.com/~/media/images/resources/figures/1090010999/10911ma_800px.jpg?h=650&la=en&w=800.

Copy Number Variation In addition to the subtle copy number variation represented by genomic features such as short sequence tandem repeats (SSRs), a substantial part of the human genome can be duplicated or deleted into much larger but still submicroscopic pieces.

ranging from 1 kb to 1 Mb (Mb means mega base pairs or 1 million base pairs). This type of variation is known as Item Number Variation (CNV). The extra or missing copies of the genome in CNVs can be detected by hybridization with oligonucleotides on DNA microarrays. Since each spot on the microarray consists of millions of identical copies of a given oligonucleotide sequence, the number of these sequences that undergo hybridization depends on how many copies of the complementary sequence are present in the genomic DNA. a typical

The region exists in two copies (one inherited from the mother and one from the father). If an individual has an extra copy of the region, the hybridization ratio and therefore the fluorescence intensity is 3:2; in comparison, if an individual is missing one copy, the hybridization ratio and therefore the fluorescence intensity is 1:2. These differences can be easily detected with DNA microarrays. Furthermore, because CNVs are relatively large, a microarray typically contains many different oligonucleotides complementary to sequences at intervals throughout the CNV; therefore, the CNV leads to an increase or decrease in the signal intensity of all oligonucleotides contained in the CNV. Current SNP chips also contain approximately 1 million oligonucleotide probes designed to detect known CNVs. CNVs are larger than 1 kb by definition, but many are much larger. In a study of about 300 people with proven descent from Africa, Europe or Asia, about 1500 CNVs were discovered by microarray hybridization. These averaged 200-300 kb in length. Overall, the CNVs spanned 300 to 450 million base pairs, or 10 to 15% of the nucleotides in the entire genome. Many of the CNVs were located in regions close to known mutated genes associated with inherited diseases. CNVs in the hemoglobin alpha and beta genes are known to be associated with malaria resistance, and CNVs in the HIV-1 receptor gene CCL3

are associated with AIDS resistance. CNVs have been reported

Risk factors for complex diseases such as Alzheimer's, autism and schizophrenia.

2.7 Applications of DNA Markers Why are geneticists interested in DNA markers and DNA polymorphisms? Your interest can be justified by several reasons. In this section, we look at the most commonly cited reasons.

Genetic Markers, Genetic Mapping and “Disease Genes”

Perhaps the main objective in the study of DNA polymorphisms in human genetics is to identify the chromosomal location of genes that have mutated alleles associated with hereditary diseases. Associated with disorders caused by the interaction of various genetic and environmental factors, such as B. heart disease, cancer, diabetes, depression, etc., it is important to think of a deleterious allele as a risk factor for the disease that increases the likelihood of occurrence of the disease rather than as the sole pathogen. This point needs to be emphasized, especially since genetic risk factors are often referred to as disease genes. An important "disease gene" for breast cancer in women, for example, is the BRCA1 gene. For women who carry a mutated BRCA1 allele, the lifetime risk of breast cancer is about 60 percent. By comparison, the lifetime risk of breast cancer in women who are not carriers is about 12%. Therefore, without the genetic risk factor, many women develop breast cancer. In fact, BRCA1 mutations are found in only 16% of affected women with a family history of breast cancer. The importance of a genetic risk factor can be expressed quantitatively as a relative risk, which corresponds to the ratio of affected to unaffected among carriers of the risk factor divided by the ratio of affected to unaffected.

unaffected people among those who do not use it. For example, in the case of BRCA1, the relative risk is (0.60/0.40 = 1.5) divided by (0.12/0.88 = 0.136), which equals 11. DNA in disease location and identification

Genes result from genetic linkage, or the tendency for genes that are close enough together on a chromosome to be inherited together. To see how this works, look at Figure 2.1. In this hypothetical example, the DNA marker is not actually in the PAH gene, but a marker allele is associated with the disease allele as a result of a genetic linkage. Genetic linkage is discussed in detail in the chapter Genetic Linkage and Chromosomal Mapping, but the main concepts are summarized in FIGURE 2.30, which shows the location of many DNA polymorphisms along a chromosome that also carries a genetic risk factor called D ( for the disease gene). 🇧🇷 🇧🇷 Each DNA polymorphism serves as a genetic marker for its own position on the chromosome. The importance of genetic linkage is that DNA marker alleles close enough to the disease gene tend to be inherited along with the disease gene in pedigrees - and the closer the markers, the stronger this association. Therefore, the initial approach to identifying a disease gene is to find DNA markers that are genetically linked to the disease gene to pinpoint its chromosomal location, a process known as genetic mapping. Once the chromosomal location is known, other methods can be used to locate the disease gene itself and study its functions.

FIGURE 2.30 Concepts of genetic location of genetic risk

disease factors. Polymorphic DNA markers (indicated by the vertical lines) next to a genetic risk factor (D) on the chromosome tend to be inherited along with the disease itself. The genomic location of the risk factor is determined by examining the known genomic locations of associated DNA polymorphisms.

If genetic linkage seems like a detour to identifying disease genes, consider the alternative. The human genome contains approximately 30,000 genes. If there was no genetic link, we would have to examine 30,000 DNA polymorphisms, one in each gene, to identify a disease gene. However, the human genome has only 23 pairs of chromosomes, and because of genetic linkage and the power of genetic mapping, only a few hundred DNA polymorphisms are actually needed to identify the chromosome and approximate location of a genetic risk factor.

Other Uses for DNA Markers DNA polymorphisms are widely used in all aspects of modern genetics as they provide a large number of readily available genetic markers for genetic mapping and other purposes. Some other uses of DNA polymorphisms are discussed in this section.

Individual identification. We have already mentioned that DNA polymorphisms are used as a means of DNA typing (DNA fingerprinting) to identify different individuals in a population. DNA typing in other organisms is used to determine individual animals in endangered species and to identify the degree of genetic relatedness between individual organisms living in packs or herds. For example, DNA typing in wild horses has shown that the wild stallion responsible for a mare's harem sires less than a third of the foals.

epidemiology and science of food safety. DNA typing has important applications in tracking the spread of viral and bacterial epidemics, as well as identifying the source of contamination in contaminated food.

human population history. DNA polymorphisms provide important information that allows anthropologists to reconstruct the evolutionary origin, global expansion, and diversification of human populations. We will cover this topic in detail in the chapter Molecular and Human Evolutionary Genetics.

Improvement of domesticated plants and animals. Plant and animal breeders have turned to DNA polymorphisms as genetic markers in pedigree studies to identify genes associated with favorable traits through gene mapping. These genes can then be incorporated into currently used plant varieties and animal breeds.

history of domestication. Plant and animal breeders also study genetic polymorphisms to identify the wild ancestors of domesticated crops and animals and to infer the artificial selection practices that led to genetic changes in these species during domestication.

DNA polymorphisms as ecological indicators. DNA polymorphisms are evaluated as biological indicators of genetic diversity in key indicator species found in biological communities exposed to chemical, biological or physical stress. They are also used to monitor the genetic diversity of endangered and captive-bred species.

Evolutionary Genetics. DNA polymorphisms are studied to describe the patterns in which different types of genetic variation occur throughout the genome, to deduce the evolutionary mechanisms by which genetic variation is maintained, and to shed light on the processes by which genetic polymorphisms occur within cells. species and are converted into genetic genes. differences between species.

population studies. Population geneticists use DNA polymorphisms to assess the extent of genetic variation in different populations of organisms that differ in genetic organization (prokaryotes, eukaryotes, organelles), population size, brood structure, or life history characteristics. They use genetic polymorphisms within subpopulations of a species as indicators of population histories, migration patterns, and so on.

Evolutionary relationships between species. Differences in DNA sequences between species serve as the basis of molecular phylogenetics, in which sequences are analyzed to determine the ancestral history (phylogeny) of species and to trace the origin of morphological, behavioral, and other adaptations that arose in the course of history. evolution.

SUMMARY ■ The analysis of RFLPs was an important tool for the initial mapping of the human genome. ■ Single nucleotide polymorphisms average to one base in 1000, providing a wealth of potential genetic markers. ■ DNA microarray technology is used to genotype individuals simultaneously at hundreds of thousands of SNP loci. ■ Single sequence repeats are also ubiquitous in the human genome and are often used in forensic DNA analysis.

■ The use of DNA markers for genetic analysis has been employed in areas ranging from epidemiology to evolutionary genetics.

CHAPTER SUMMARY ■ A strand of DNA is a polymer of deoxyribonucleotides A, T, G, and C joined in the 3'-to-5' direction by phosphodiester bonds. ■ The two strands of DNA in a duplex are held together by hydrogen bonds between the AT and G–C base pairs and by stacking the paired bases. ■ Each type of restriction endonuclease enzyme cleaves double-stranded DNA at a specific base sequence, usually four or six nucleotides in length. ■ DNA fragments produced by a restriction enzyme can be electrophoresed, isolated, sequenced, and otherwise manipulated. ■ Separate strands of DNA or RNA that are complementary in nucleotide sequence can spontaneously join together (hybridize) to form duplexes. ■ DNA replication occurs only by extending the growing strand in the 5'-3' direction by adding consecutive nucleotides to the 3' end.

■ The polymerase chain reaction uses short oligonucleotide primers in consecutive rounds of DNA replication to selectively amplify a specific region of a DNA duplex. ■ Genetic markers in DNA provide a large number of easily accessible locations in the genome that can be used to identify the chromosomal locations of disease genes, DNA typing in individual identification, genetic improvement of crops and domesticated animals, and many other applications .

REVIEW THE BASICS ■ Define and give an example of each of the following genetic terms: locus, allele, genotype, heterozygote, homozygote, phenotype. ■ What is a DNA marker? Explain how harmless DNA markers can be used to help identify disease genes through gene mapping. ■ What are the four bases commonly found in DNA nucleotides? Which form base pairs? ■ What chemical groups are found at the 3' and 5' ends of a single polynucleotide strand? ■ What does it mean to say that a single strand of DNA has a polarity? What does it mean to say that the DNA strands in a duplex molecule are antiparallel? ■ What are restriction enzymes and why are they important for studying specific DNA fragments? What does it mean to say that most restriction sites are palindromes? ■ Describe how a Southern blot is performed. What is it used for? What is the role of the probe?

■ How does the polymerase chain reaction work? What is it used for? What information about the target sequence needs to be known in advance? What is the role of oligonucleotide primers? ■ What is a DNA microarray? How is one used for SNP genotyping? What are the possible sources of error in the process? ■ What is a DNA marker? Explain how harmless DNA markers can be used to help identify disease genes through gene mapping.

TROUBLESHOOTING GUIDE ISSUE 1 A geneticist plans to use polymerase chain reaction (PCR) to amplify a portion of the DNA sequence shown below using oligonucleotide primers that hybridize to the regions highlighted in red. (These are for illustration only and are too short for practice.) Provide the sequence of primers to be used, including polarity, and deduce the sequence of the amplified DNA fragment.

ANSWER Primers must be able to anneal at the chosen primer sites and must be oriented with their 3' ends facing each other. Therefore, the "forward" primer (which extends from left to right) must have the sequence 5'-ACGAT-3', and the "reverse" primer (which extends from right to right). -left direction) must have the sequence 3'-ACTAG-5'. Since the convention for writing nucleic acid sequences is to place the 5' end to the left, the reverse primer is 5'-GATCA-3'. PROBLEM 2 The genetic material of bacteriophage M13 is a single-stranded DNA molecule of 6,407 nucleotides, whose complete sequence can be found in publicly available databases, such as GenBank. After M13 DNA is introduced into a bacterial cell, a complementary strand of DNA is synthesized by bacterial enzymes, resulting in a double-stranded replicative form of the phage genome. How would you know if both strands of the replicative form are being transcribed inside an infected cell? ANSWER To determine whether both strands of DNA are templates for RNA synthesis, one can isolate the replicative form of phage DNA, separate the two strands, and test each strand

separately for the ability to hybridize to mRNA found in M13-infected cells. Alternatively (and less conclusively), one could examine the M13 genome sequence in GenBank and search both DNA strands for open reading frames that might encode proteins. Coincidentally, only one strand of M13 contains such open reading frames, and therefore it is likely that only one strand of DNA is transcribed. PROBLEM 3 A 6200 base pair plasmid contains at least three complete copies of a viral gene arranged in tandem. Cleavage of the plasmid with the restriction enzyme HindIII leads to fragments of 900, 1300 and 4000 bp. The tandem copies of the gene are contained in only one of the HindIII fragments. The gene encodes a protein of 405 amino acids. Which HindIII fragment likely contains the copies of the gene?

ANSWER To encode a protein with 405 amino acids, at least 1215 bp are needed. So three tandem copies of the gene would require 3645 bp and four copies 4860 bp. The smallest HindIII fragment is not long enough to contain even one copy of the gene. The 1300 bp fragment can contain one copy, but not two or more. Only the largest fragment could contain three copies of the gene, and it could not contain more than three copies. PROBLEM 4 Factor V is a protein involved in the blood clotting cascade. Men of European descent who are heterozygous for a specific rare allele of the F5 gene that encodes it have a 52% chance of developing venous thromboembolism, a blood clot that forms in a leg vein. In the general population, 12% of all men of the same type develop this problem. What risk factor is associated with this specific genotype?

ANSWER If 52 percent of men with the risk-associated genotype develop the problem, then 100 percent - 52 percent or 48 percent do not. Therefore, we can calculate the proportion of males with this genotype developing the disease as 52/48, or 1.08. Next, we need to be able to characterize the risk of venous thromboembolism independent of genotype. We make the same calculation for the general population, where 12% of men develop the disease and 88% do not. The corresponding ratio is therefore 12/88 or 0.136. Finally, we can determine the increased risk associated with the F5 genotype as

Thus, the risk of developing venous thromboembolism for a European male heterozygous for the risk-associated F5 allele is 7.9 times greater than for the European male population as a whole.

ANALYSIS AND APPLICATIONS

2.1 What chemical groups are found at the 3' and 5' ends of a single polynucleotide molecule? 2.2 Which carbon atom contains the phosphate group and which contains the 3'-hydroxyl group in the deoxyribonucleotide shown below?

2.3 Many restriction enzymes produce restriction fragments with “sticky ends”. What does that mean?

2.4 Which of the following sequences are palindromes and which are not? explain your answer (a) 5'-CCGG-3' (b) 5'-TTTT-3' (c) 5'-GCTAGC-3' (d) 5'-CCGCTC-3' (e) 5'-AAGGTT -3'

2.5 The list below gives half of each set of palindromic restriction sites. Substitute Ns to complete the sequence of each restriction site. (a) 5'-AGNN-3' (b) 5'-ATGNNN-3' (c) 5'-ATTNNN-3' (d) 5'-NNNAGC-3'

2.6 In addition to the nucleotide sequence, what is different about the ends of the restriction fragments produced by the following restriction enzymes? (The down arrow represents the cleavage site on each strand.) (a) ScaI (5'-AGT↓ACT-3') (b) NheI (5'-G↓CTAGC-3') (c) CfoI ( 5' -GCG↓C-3')

2.7 A solution contains 4 kb, 8 kb, 10 kb, and 13 kb double-stranded DNA fragments, which are separated on an electrophoresis gel. On the attached gel chart, match the fragment sizes to the correct bands.

2.8 The linear DNA fragment shown here has cleavage sites for AatII (A) and XhoI (X). On the attached diagram of an electrophoresis gel, indicate the positions where the bands would be found after digestion with: (a) AatII alone. (b) XhoI alone. (c) AatII and XhoI together.

The dashed lines on the right indicate the positions to which the 1-12 kb bands would migrate.

2.9 The circular DNA molecule shown here has cleavage sites for AatII and XhoI. in follow-up

Diagram of an electrophoresis gel, indicating the positions where bands would be found after digestion with: (a) AatII alone. (b) XhoI alone. (c) AatII and XhoI together.

The dashed lines indicate the positions to which the 1-12 kb bands would migrate.

2.10 Consider the attached diagram of a region of duplex DNA in which the Bs represent bases in Watson-Crick pairs. Specify as precisely as possible the identity of: (a) B5, assuming B1 = A. (b) B6, assuming B2 = C.

2.11 In the DNA duplex precursor nucleotides described in Exercise 2.10, what base was each of the 1–4 phosphate groups associated with before being incorporated into the polynucleotide strand?

2.12 What is the probability that, in a random sequence consisting of equal parts of all four nucleotides, a certain short sequence of nucleotides corresponds to a restriction site for: (a) a restriction enzyme with a cleavage site of 4 bases? (b) A restriction enzyme with a 6 base cleavage site? (c) A restriction enzyme with an 8 base cleavage site?

2.13 In a random sequence composed of equal portions of all four nucleotides, what is the average distance between restriction sites for: (a) a restriction enzyme with a 4-base cleavage site?

(b) A restriction enzyme with a 6 base cleavage site? (c) A restriction enzyme with an 8 base cleavage site?

2.14 If Escherichia coli DNA were essentially a 4.6 x 106 bp random sequence with equal proportions of all four nucleotides (this is an oversimplification), approximately how many restriction fragments would you expect from digestion with: (a) a "4-cutter" restriction enzyme? (b) A “6-cut” restriction enzyme? (c) An 8-Cutter Restriction Enzyme?

2.15 A circular DNA molecule is cleaved with AfeI, NheI, or both restriction enzymes together. The attached diagram shows the resulting electrophoresis gel with the indicated band sizes. Draw a circular DNA diagram showing the relative positions of the AfeI and NheI sites.

2.16 In the diagrams of DNA fragments presented here, the marks indicate the positions of the restriction sites for a given restriction enzyme. A mixture of the two types of molecules is digested and analyzed in a Southern blot using probe A or probe B, which hybridize to the fragments at the positions indicated by the squares. in closed gel

Diagram indicating the bands that would result from each of these probes. (The scale on the right shows the expected positions of fragments from 1 to 12 kb.)

2.17 In the accompanying diagram, the marks indicate the positions of restriction sites on two alternative DNA fragments that may be present at one location on a human chromosome. RFLP analysis is performed using a DNA probe that hybridizes to the fragments in the position shown by the rectangle. How many genotypes are possible for this RFLP? (Use the symbol A1 to refer to the allele that gives the upper DNA fragment and A2 to refer to the allele that gives the lower DNA fragment.) On the attached gel diagram, indicate the above genotypes and phenotype (band position ) at or positions) expected from each genotype. (The scale on the right shows the expected positions of fragments from 1 to 12 kb.)

2.18 If the pentamers were long enough to serve as PCR-specific oligonucleotide primers (in practice, they are very short), which DNA fragment would be labeled with the "forward" primer 5'-AATGC-3' and the "reverse" primer "3'-GCATG-5' acting on the double-stranded DNA molecule shown here?

2.19 Would the 3'-AATGC-5' and 5'-GCATG3' primer pairs amplify the same fragment described in Exercise 2.18? explain your answer 2.20 Suppose that a 3 kb fragment of human DNA is amplified by PCR. The total genome size is 3 × 106 kb, which is 3 × 109 base pairs. (a) What proportion of the total DNA is the target sequence before amplification? (b) What fraction does it represent after 10 cycles of PCR? (c) After 20 PCR cycles? (d) After 30 PCR cycles?

2.21 The two allelic DNA fragments shown below are amplified by PCR and the products digested with PstI and Hind3. (a) What will be the total sizes of the two PCR products?

(b) Draw the gel pattern you would expect from double-digested DNA.

2.22 Graphically represented below are six allelic DNA fragments showing a variety of SNP polymorphisms. Three of these fragments also carry an allele of a gene associated with an inherited disease (indicated by D). As a clinical geneticist, you want to use these SNP markers to determine whether a person is at increased risk of developing the disease. Which of these SNPs would help you in this determination? What not? Justify your answers.

2.23 Psoriasis is a skin disease that has been shown to have a significant genetic component, although, like breast cancer, multiple genes contribute to its development. An allele of one of these genes, known as PSORS1, was located on chromosome 6, as was an SNP marker (rs10484554, a C/T polymorphism); the T allele of this SNP is associated with the PSORS1 risk allele. In a population study, 22.4% of heterozygous individuals (CT genotype) developed psoriasis, whereas in a control population of individuals with CC genotype, the incidence of psoriasis was 11.4%. What is the relative risk associated with this genotype? 2.24 A cigarette butt found at the scene of a burglary has enough epithelial cells adhered to the paper for DNA to be extracted and analyzed by DNA typing. Here are the results of typing three probes (Locus 1-Locus 3) of evidence (X) and cells of seven suspects (A-G). Which of the suspects can be ruled out? Which ones can't be ruled out? Can you identify the culprit? Explain your reasoning.

2.25 A woman is not sure which of the two men is the father of her child. DNA typing is performed on the blood of the child (C), the mother (M), and the two males (A and B), using probes for a highly polymorphic DNA marker on two different chromosomes ("locus 1" and "locus 1 "2" are used). The result is shown in the attached diagram. Does one or both of the tested loci exclude one of the males as a possible father? Explain your reasoning.

2.26 Diesterase from snake venom cleaves the chemical bonds shown in red in the accompanying diagram,

What remains are mononucleotides phosphorylated at the 3' position. If phosphate numbers 2 and 4 are radioactive, which mononucleotides will be radioactive after cleavage with snake venom diesterase?

2.27 A DNA sample collected at the crime scene is believed to have come from the perpetrator. It is genotyped for four SSR loci. Four suspects are also being genotyped. All five genotypes are shown below as fluorometric traces, with colors corresponding to different loci. (a) Who can be excluded as a suspect and why? (b) Which person could be the perpetrator? Do you think the evidence is conclusive? Why or why not?

CHALLENGING PROBLEMS CHALLENGING PROBLEMS 1 The Drosophila melanogaster genome is 180 × 106 bp and a 1.8 kb fragment must be amplified by PCR. How many PCR cycles are required for the amplified target sequence to represent at least 99% of the total DNA?

CHALLENGE PROBLEM 2 The body of a young homicide victim is in an advanced state of decomposition and cannot be identified. Police suspect the victim is one of five people reported missing by her parents. DNA typing is performed on tissues from the victim (X) and the five pairs of parents (AE), using probes for a highly polymorphic DNA marker on two different chromosomes (“locus 1” and “locus 2”). The result is shown in the attached diagram. How do you interpret the fact that the genome

Does each individual's DNA produce two bands? Can you identify the victim's parents? Explain your reasoning.

CHALLENGE PROBLEM 3 The snake venom diesterase enzyme described in Problem 2.26 was originally used in a procedure called nearest neighbor analysis. In this method, a strand of DNA is synthesized in the presence of all four nucleoside triphosphates, one of which carries a radioactive phosphate in the (innermost) position. Then the DNA is fully digested with snake venom diesterase and the resulting mononucleotides are separated and analyzed for radioactivity. Examine the diagram in Exercise 2.26 and explain:

(a) How this procedure reveals the "nearest neighbors" of the radioactive nucleotide. (b) Whether the “nearest neighbor” is on the 5' or 3' side of the labeled nucleotide.

FOR FURTHER READING Botstein, D., White, R.L., Skolnick, M., & Davis, R.W. (1980). Construction of a human genetic linkage map using restriction fragment length polymorphisms. American Journal of Human Genetics, 32(3), 314-31. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/6247908 A classic paper describing how RFLP mapping can be used in human genetic analysis.

Kelly, T.J. & Smith, H.O. (1970). A restriction enzyme from Haemophilus influenzae: II. Base sequence of the recognition site. Journal of Molecular Biology, 51(2), 393-409. http://doi.org/10.1016/00222836(70)90150-6 One of the first reports of a restriction enzyme recognition sequence. Rower, L. (2013). DNA fingerprinting in forensics: past, present, future. Investigative Genetics, 4(1), 22. http://doiorg/10.1186/2041-2223-4-22

A current overview of the status of forensic DNA testing. Southern, E.M. (1975). Detection of specific sequences between DNA fragments separated by gel electrophoresis. Journal of Molecular Biology, 98(3), 503-517. http://doi.org/10.1016/S00222836(75)80083-0 The original Southern blotting procedure. Steemers FJ, Chang W, Lee G, Barker DL, Shen R, & Gunderson KL. (2006). Whole-genome genotyping using the single-base extension assay. Nature Methods, 3(1), 31-33. http://doi.org/10.1038/nmeth842

A description of the basis for genome-wide SNP genotyping.

UNIT 2 Transmission Genetics Chapter 3 Mendelian Genetics: The Principles of Segregation and Assortment Chapter 4 The Chromosomal Basis of Inheritance Chapter 5 Genetic Linkage and Chromosome Mapping

Chapter 6 Human Karyotypes and Chromosomal Behavior Chapter 7 The Genetic Basis of Complex Traits

Chapter 8 Genetics of bacteria and their viruses

CHAPTER 3

Mendelian Genetics: The Principles of Segregation and Assortment

CHAPTER OVERVIEW 3.1 Morphological and Molecular Phenotypes 3.2 Single Gene Assortment 3.3 Segregation of Two or More Genes 3.4 Human Lineage Analysis 3.5 Incomplete Dominance and Epistasis 3.6 Probability in Genetic Analysis 3.7 Conditional Probability and Lineages ROOTS OF DISCOVERY: What Gregor Mendel thought you had figured it out? Gregor Mendel (1866) Experiments with hybrid plants ROOTS OF DISCOVERY: This country is your country The Huntington's Disease Research Collaborative Group (1993)

A new gene containing an expanding and unstable trinucleotide repeat in HD chromosomes

LEARNING OBJECTIVES AND SCIENTIFIC SKILLS When you learn the fundamentals of transmission genetics, including segregation and independent classification, and are able to combine these with elementary principles of probability, you will gain important scientific skills that will enable you to solve problems such as the following: ■ For a gene, given the genotypes or phenotypes of the parents, apply the principle of segregation to predict the possible types of offspring and their expected proportions. ■ For two or more genes, given the genotypes or phenotypes of the parents, use the principle of independent classification to predict the possible types of offspring and their expected proportions.

■ For all specified deviations from Mendelian assumptions about dominance or epistasis, predict how the expected proportions of offspring types will change in a given mating. ■ Identify inheritance patterns of Mendelian dominant or simple recessive traits in human pedigrees. ■ Infer the genotypes of individuals in family trees based on their own phenotypes and those of their close relatives. In this chapter, we will look at how genes are passed from parent to offspring and how this determines the distribution of genes.

Genotypes and phenotypes among related individuals. The study

the inheritance of traits is the genetics of heredity. This topic is also called Mendelian genetics because the underlying principles were first derived from experiments with peas (Pisum sativum) conducted in 1856-1863 by Gregor Mendel, a monk at the monastery of St. Thomas in Brno (Brno) in the Czech Republic. Mendel reported his experiments to a local natural history society, published the results and their interpretation in their scientific journal in 1866, and began

Correspondence with one of the leading botanists of the time. His experiments were meticulous and exceptionally well documented, and his work contains the first clear exposition of the statistical rules governing the intergenerational transmission of genes. Despite this, Mendel's essay was largely ignored for 34 years, until its importance was finally appreciated.

3.1 Morphological and Molecular Phenotypes Until the advent of molecular genetics, geneticists were primarily concerned with morphological traits, in which differences between organisms can be expressed in terms of characteristics such as color, shape, or size. Mendel examined seven contrasting morphological characters in the seed

shape, seed color, flower color, pod shape and so on (FIGURE 3.1). Perhaps the best-known example of a contrasting Mendelian trait is round wrinkled seeds. When pea seeds dry out, they lose water and shrink. Round seeds are round because they shrink evenly; wrinkled seeds are wrinkled because they shrink irregularly. The shriveled phenotype is due to the lack of a branched chain form of starch known as amylopectin, which is not synthesized in shriveled seeds due to a defect in starch branching enzyme I (SBEI).

FIGURE 3.1 The seven different traits of peas studied by Mendel. The phenotype shown on the far right is the dominant trait that appears in the hybrid produced by the cross.

The non-mutant or wild-type allele of the gene for SBEI is designated W and the mutant allele is designated w. Seeds that are heterozygous Ww have only half the SBEI enzyme as wild-type homozygous WW seeds, but up to half the normal amount

The enzyme produces enough amylopectin for heterozygous Ww seeds to shrink evenly and remain phenotypically round. Therefore, in terms of seed shape, the WW and Ww genotypes have the same (round) phenotype. The W allele is called the dominant allele and w the recessive allele. The molecular basis of the wrinkled mutation is that the SBEI gene has been disrupted by inserting a DNA sequence called a transposable element into the gene. Such DNA sequences

are able to move from one place to another within a chromosome or between chromosomes (transposition). The molecular mechanism of transposition is discussed in the chapter on mutation, repair, and recombination, but for our present purposes it is only necessary to know that transposable elements are present in most genomes, particularly large genomes of eukaryotes, and that many are spontaneous. the insertion of transposable elements into a gene. FIGURE 3.2 provides a diagram of the DNA structure of the wild-type W and mutant w alleles, showing the insertion of DNA that interrupts the w allele. Highlighted are the polymerase chain reaction (PCR) primer sites present on both alleles flanking the insertion site. The graph in part C shows the pattern of bands that would be expected if one were to amplify genomic DNA and separate the resulting products by electrophoresis. The product of the W allele would be smaller than that of the w allele due to the DNA inserted into the w allele, so it would migrate faster than the corresponding fragment of the w allele and would move closer to the bottom of the lie gels. Homozygous WW genomic DNA would produce a single, faster migrating band; that of ww homozygotes a single slower migrating band; and that of heterozygous Ww-2 bands with the same electrophoretic mobilities observed in homozygous genotypes. At the

In Figure 3.2C, the band in the homozygous genotypes is shown to be slightly thicker than in the heterozygous genotype because the only band in each homozygous genotype comes from the two copies of the homozygous allele. Thus, this band contains more DNA than the corresponding DNA in the

heterozygous genotype in which only one copy of each allele is present.

FIGURE 3.2 (A) W (round) is an allele of a gene that specifies the amino acid sequence of starch branching enzyme I (SBEI). (B) w (wrinkled) is an allele that encodes an inactive form of the enzyme, as its DNA sequence is disrupted by the insertion of a transposable element. (C) At the morphological phenotype level, W dominates over w: the WW and Ww genotypes have rounded seeds, while the ww genotype has wrinkled seeds. The molecular difference between the alleles can be detected as a restriction fragment length polymorphism (RFLP) using the enzyme EcoRI and a probe that hybridizes to the site shown. At the molecular level, alleles are codominant: DNA from each genotype gives rise to a different molecular phenotype—a single band of different size for WW and ww homozygotes, and both bands for Ww heterozygotes.

In this small garden property next to the monastery of St. Thomas, Gregor Mendel grew over 33,500 pea plants over an eight-year period, including over 6,400 plants in just one year. He received some help from two monk friends who helped him with this.

Try it.

As shown in Figure 3.2C, PCR analysis clearly distinguishes between the WW, Ww, and ww genotypes because the heterozygous Ww genotype exhibits both of the bands seen in the homozygous genotypes. This situation is described by saying that the W and w alleles are codominant in terms of molecular phenotype. However, as indicated by the seed shapes in Figure 3.2C, W dominates w in terms of morphological phenotype. In the following discussion, we use PCR analysis of the W and w alleles to highlight the importance of molecular phenotypes in modern genetics and to demonstrate experimentally the principles of genetic transmission.

SUMMARY ■ Mendel studied seven morphological characters in peas. ■ Dominant alleles are those that result in identical phenotypes in homozygotes and heterozygotes. ■ For every trait Mendel studied, there were two alleles—one dominant and one recessive.

■ Using molecular markers can allow scientists to identify all genotypes - homozygous or heterozygous. Such markers are therefore codominant.

3.2 Single gene splitting Mendel chose peas for his experiments for two main reasons. First, he had access to cultivars that differed in contrasting characteristics, such as round versus wrinkled seeds and yellow versus green seeds. Second, their previous studies showed that peas normally reproduce by self-pollination, which uses pollen produced in a flower to fertilize eggs in the same flower. Production of hybrids by

Outcrossing required painstaking surgery on immature flowers to expose receptive female structures, excise and discard undeveloped pollen-producing male structures, and expose female structures to mature pollen from another plant. After cross-fertilization, each dissected flower was placed in a fine mesh bag to prevent stray pollen from accessing the internal female structures. The relatively small amount of space required to grow each plant and the relatively large number of progeny that could be obtained gave Mendel the ability, as he says in his article, to "determine the number of different forms in which the hybrid progeny occurs". . to "determine their numerical relations". For example, plants with round seeds have the WW genotype and those with wrinkled seeds have the ww genotype. Homozygous genotypes are indicated experimentally by observing that the hereditary traits in each variety are purebred, meaning that plants only produce offspring like themselves if they can self-pollinate normally. A hybrid results from the crossing of plants that differ in one or more characteristics. If the parents differ in one, two or three characteristics, the hybrid is a monohybrid, dihybrid or trihybrid. Keep an eye

Parents and their hybrid offspring, we say that the parents form the P1 generation and their hybrid offspring form the F1 generation. As peas are sexual organisms, any cross can be accomplished in two ways, depending on which phenotype is present in the female parent and which is present in the male parent. For example, in round versus wrinkled, the female parent can be round and the male wrinkled, or vice versa. These are called reciprocal crosses. Mendel was the first to demonstrate the following important principle: the outcome of a genetic cross does not depend on which trait is present in the male and which is present in the female; reciprocal crosses give the same result.

This principle is illustrated for round and wrinkled seeds in FIGURE 3.3. The gel symbols show the PCR products that would produce genomic DNA from each seed type in these crosses. The genotypes of the crosses and their offspring are as follows:

FIGURE 3.3 Morphological and molecular phenotypes showing equivalence of reciprocal crosses. In this example, the hybrid seeds are round and give a two-band PCR pattern, regardless of cross direction.

In both reciprocal crosses, the offspring have the morphological phenotype of round seeds, but as shown by the PCR patterns on the right, all offspring genotypes are actually Ww heterozygotes and therefore different from either parent. The genetic equivalence of reciprocal crosses illustrated in Figure 3.3 is a principle quite general in its applicability, but there is one important exception having to do with sex chromosomes, discussed in the chapter entitled The Chromosomal Basis of Sex Heritage.

In the following section, we examine some of Mendel's original experiments in the context of molecular analysis to relate morphological phenotypes and their relationships to expected molecular phenotypes.

Phenotypic relationships in the F2 generation Although the offspring of the crosses in Figure 3.3 show the dominant round seed phenotype, PCR analysis shows that they are indeed heterozygous. The w allele is occluded with respect to the morphological phenotype because w is recessive to w. Despite this, the shriveled phenotype reappears in the next generation when the hybrid offspring self-pollinate. For example, if the round F1 seeds from cross A in Figure 3.3 were grown into sexually mature plants and subjected to self-pollination, some of the resulting seeds would be round and some wrinkled. The progeny seeds produced by selfing of the F1 generation constitute the F2 generation. When Mendel did this experiment in the F2 generation, he was

observed the results shown in the following graph:

Note that the ratio of 5474:1850 is approximately 3:1.

A 3:1 ratio of dominant to recessive forms in F2 offspring is characteristic of simple Mendelian inheritance. Mendel's data demonstrating this point are shown in TABLE 3.1. Notice that the first two traits in the table (round seeds versus wrinkled seeds and yellow versus green seeds) have far more observations than any of the others. The reason for this is that these characteristics can be classified directly in seeds, while the others can only be classified in mature plants, and Mendel was able to analyze many more seeds than mature plants. Key observations from the data in Table 3.1 are as follows: ■ F1 hybrids express only the dominant trait (because F1 offspring are heterozygous - eg, Ww). ■ In the F2 generation, plants are present with either the dominant or recessive trait (meaning that some F2 offspring are homozygous - eg ww). ■ The F2 generation has about three times more

many plants with a dominant phenotype than plants with a recessive phenotype.

TABLE 3.1 Results of Mendel's monohybrid experiments Parental traits

F1-Hub

Number of children F2

F2 ratio

Round × wrinkled (seeds)

(Video) Inside Genetics: Analysis of Genes and Genomes, Ninth Edition

Time

5474 round, 1850 wrinkled

2,96 : 1

Yellow × Green (seeds)

Gelb

6022 yellow, 2001 green

3.01: 1

Purple × white (flowers)

takes away

705 purple, 224 white

3,15 : 1

Inflated × constricted (sleeves)

Inflated

882 inflated, 299 contracted

2,95 : 1

Green × Yellow (green pods)

Verde

428 green, 152 yellow

2,82 : 1

Axial × terminal (flower position)

Axial

651 axial, 207 clamp

3,14 : 1

Long × short (trunk)

Lang

787 long, 277 short

2,84 : 1

The 3:1 ratio observed in the F2 generation is the key to understanding the mechanism of genetic transmission. In the next section, we use a PCR analysis of the W and w alleles to explain why this ratio is established.

The Principle of Segregation The 3:1 ratio can be explained using FIGURE 3.4. This is the core of Mendelian genetics. You must master it and be able to derive the types of offspring resulting from crosses. The diagram illustrates these key features of single-gene inheritance: 1. Genes come in pairs, which means that a cell or individual has two copies (alleles) of each gene. 2. For each pair of genes, the alleles can be identical (homozygous WW or homozygous ww) or different (heterozygous ww).

3. Each reproductive cell (gamete) produced by an individual contains only one allele of each gene (ie, W or w). 4. In the formation of gametes, each given gamete has an equal probability of containing one of the alleles (thus, in a heterozygous Ww genotype, half of the gametes contain W and the other half

w included). 5. The joining of male and female reproductive cells is a random process that pairs alleles together.

FIGURE 3.4 A schematic explanation of the 3:1 ratio of dominant:recessive morphological phenotypes observed in the F2 generation of a monohybrid cross. (A) Two purebred parent lines, one round and one wrinkled, are crossed to give F1 offspring with round seeds. (B) These offspring are selfed to give the F2 generation. Due to dominance, a 3:1 phenotype ratio is observed in the F2 generation. Note that the ratio of WW:Ww:ww genotypes in the F2 generation is 1:2:1, as seen in the restriction fragment phenotypes.

The essential feature of transmission genetics is the separation (technically called segregation) in unmodified form of the two alleles in an individual during the formation of its reproductive cells. Segregation corresponds to points 3 and 4 in the list above. The principle of separation is sometimes referred to as Mendel's first law. Principle of Segregation: In the formation of gametes, the paired hereditary determinants separate (segregate) in such a way that each gamete has an equal probability of containing one of the two members of the pair. Another key feature of transmission genetics is that hereditary determinants are present in pairs in both parent and offspring organisms, but as single copies in reproductive cells. This feature corresponds to points 1 and 5 in the list above.

Figure 3.4 illustrates the biological mechanism underlying the important Mendelian ratios in the F2 generation of 3:1 for phenotypes and 1:2:1 for genotypes. To understand these relationships, first consider the parental generation in which the original cross is WW × ww (Figure 3.4A). The sex of the parents is not given, as the cross gives identical results. (However, there is a convention in genetics that, unless otherwise specified, crosses are given with the female parent listed first.) In the original cross, the WW parent produces only W-containing gametes, while the ww parent produces only W-containing gametes. gametes containing w produce gametes. Segregation continues to occur in both homozygous and heterozygous genotypes, even though all gametes carry the same allelic type (W from the WW homozygote and w from the ww homozygote). When W-bearing and W-carrying gametes come together in fertilization, the genotype is a hybrid.

Ww heterozygote shown by the bands on the gel icon next to the F1 progeny. In terms of seed shape, Ww hybrid seeds are round because W is dominant over w. Now consider Figure 3.4B. If heterozygous F1 progeny form gametes, segregation implies that half of the gametes will contain the W allele and the other half will contain the W allele. These gametes come together randomly when an F1 individual is selfed or when two F1 individuals are crossed. the result of

Random fertilization can be derived from the type of cross multiplication square shown in the figure below, where female gametes and their frequencies are arranged at the top and male gametes and their frequencies are arranged on the left. This computing device, widely used in genetics, is called a Punnett square, after its inventor, Reginald C. Punnett (1875–1967). The Punnett square in Figure 3.4B shows that random combinations of F1 gametes result in an F2 generation with the genotypic composition 1/4 ww, 1/2 ww, and 1/4 ww. This can be confirmed by the PCR banding patterns in the gel symbols due to the codominance of W and w with the molecular phenotype. But since W is dominant over w in terms of morphological phenotype, the WW and Ww genotypes have round seeds and the ww genotypes have wrinkled seeds, giving the phenotypic ratio of round:wrinkled seeds of 3:1. Therefore, it is a combination of segregation, random association of gametes, and dominance that results in a 3:1 ratio.

ROOTS OF DISCOVERY What did Gregor Mendel think he discovered? Gregor Mendel (1866) Monastery of St. Thomas, Brno [then Brno], Czech Republic

Experiments on Hybrid Plants Mendel's article is remarkable and well worth reading. Although credited with discovering segregation, he never uses the term. His description of segregation is in italics in the first passage. (All text not in italics is taken from the original.) In his description of the process, Mendel carefully guides us through the separation of A and a in gametes and their accidental reunion in fertilization. One error in the description is Mendel's occasional confusion between genotype and phenotype, illustrated by his spelling of A instead of AA and a instead of aa in the presentation near the end of the passage. Most early geneticists did not consistently distinguish between genotype and phenotype until 1909, when the terms themselves were coined.

"Whether the plan according to which the individual experiments were set up and carried out was appropriate to the task at hand

”

be decided by a favorable decision.

The experiments described here were triggered by the artificial pollination of ornamental plants to obtain new color variants. The astonishing regularity with which the same hybrid forms appeared whenever fertilization took place between the same species suggested further experiments aimed at tracing this development of the hybrids in their offspring. This article discusses attempting such a detailed experiment…. Whether the plan according to which the individual experiments were set up and carried out corresponded to the task set should be judged benevolently.

[The results of the experiment are described in detail here.] The experiment also justifies the assumption that pea hybrids form germ and pollen cells that correspond in their composition to all constant forms, which result from the combination of characters united by fertilization , in equal numbers. The difference in form between the offspring of hybrids and the circumstances in which they are observed find an adequate explanation in the principle [of segregation] just derived. The simplest case is the series for a pair of distinct features. will be shown

that this series is described by the expression: A + 2Aa + a, where A and a denote the forms with constant distinct characters and Aa the hybrid form for both. The series contains four people on three different terms. In its production, pollen and germ cells of forms A and a participate in fertilization in equal proportions on average; therefore, each form manifests itself twice as four individuals are produced. Are involved in fertilization:

Which of the two types of pollen connects with each individual germ cell is purely coincidental. According to the laws of probability, however, it will always happen, on average, in many cases, that each form of A and a pollen unites with equal frequency with each form of A and a germ cell; therefore, on fertilization, one of the two A pollen cells encounters an A germ cell, the other encounters an a germ cell, and likewise one a pollen cell becomes associated with an A germ cell and the other a.

The result of fertilization can be visualized by writing the labels of the associated germ and pollen cells on the form

of fractions, pollen cells above the line, germ cells below. In the discussed case, we obtain

In the first and fourth terms, germ cells and pollen cells are the same;

therefore, the products of your association must be constants, that is, A and a. In the second and third, however, there is again a union of the two different parental traits; therefore, the forms resulting from such pollination are absolutely identical to the hybrid from which they are derived. Thus, repeated hybridization takes place. The surprising phenomenon that hybrids can produce offspring similar to themselves beyond the two types of parents is explained with this: Aa and aA both give the same association, Aa since, as already mentioned, it makes no difference the fertilization sequence, which of the two characteristics belong to the pollen and which to the germ cell. And so,

This represents the average self-fertilization course of hybrids when two different traits are associated with them. With individual flowers and individual plants, however, the rate at which line members are formed can vary considerably. the ratio of the variability of their descendants. Source: G. Mendel, Negotiations of the Natural Research Society at Brno 4 (1866): 3–47.

The proportion of F2 genotypes is as important as the proportion of F2 phenotypes. The Punnett square in Figure 3.4B also shows that the ratio of WW:Ww:ww genotypes is 1:2:1, which can be directly confirmed by PCR analysis.

Evidence of segregation Behind the round seeds in Figure 3.4B lies a genotype ratio of 1 WW: 2 WW. In other words, among F2 seeds that are round (or more generally among organisms that exhibit the dominant morphological phenotype), 1/3 are homozygous (WW in this example) and 2/3 are heterozygous (WW in this example). This conclusion is evident from the PCR patterns in Figure 3.4B, but not apparent from the morphological phenotypes. If you didn't already know about genetics, this would be a very bold hypothesis, as it implies that two organisms with the same morphological phenotype (round seeds in this case) can still differ in molecular phenotype and genotype.

But this is exactly what Mendel suggested. But how could this hypothesis be tested experimentally? Mendel realized that he could be tested by selfing the F2 plants. In selfing, plants from homozygous WW genotypes must be pure for round seeds, while those from heterozygous Ww genotypes must produce round and wrinkled seeds in a 3:1 ratio. In contrast, plants grown from wrinkled seeds must be wrinkled because these plants are homozygous ww. Mendel's results are summarized in FIGURE 3.5. As predicted by the genetic hypothesis, plants bred from F2 curly seeds were true curly seeds, producing only curly seeds in the F3 generation. But some of the plants grew

of round seeds showed signs of segregation. Among 565 plants grown from round F2 seeds, 372 plants produced round and wrinkled seeds in a ratio very close to 3:1, while the remaining 193 plants produced only round seeds in the F3 generation. The 193:372 ratio corresponds to 1:1.93, very close to the 1:2 ratio of the WW:WW genotypes predicted by the genetic hypothesis.

FIGURE 3.5 Summary of F2 phenotypes and offspring produced by selfing.

An important feature of the homozygous round and homozygous wrinkled seeds produced in the F2 and F3 generations is that the phenotypes are exactly the same as those observed in the original parents of the P1 generation. This makes sense in terms of DNA, as the DNA of each allele remains unchanged unless a new mutation occurs. Mendel described this result in a letter, saying that in the offspring of crosses, "the two parental characters appear separate and unaltered, and there is nothing to suggest that one of them has inherited or inherited anything from the other." that hereditary determinants for traits were passed down parental lines as

two different elements that maintain their purity in hybrids. In other words, hereditary determinants do not "blend" or "contaminate" each other. In modern terminology, this means that genes are passed unchanged from generation to generation, with rare but important exceptions. In fact, this concept, sometimes called the particulate gene, is considered by many to be Mendel's most important contribution.

The test cross and the back cross

Another simple way to test the genetic hypothesis in Figure 3.4 is with a test cross, a cross between an organism that is heterozygous for one or more genes (for example, Ww) and an organism that is homozygous for the recessive allele is (for example, example, ww ). The result of this test cross is shown in FIGURE 3.6. Since the heterozygous parent is expected to produce W and w gametes in equal numbers, while the homozygous recessive parent produces only w gametes, the expected offspring are 1/2 with ww genotype and 1/2 with ww genotype. The former have the dominant phenotype because W is dominant over w and the latter have the recessive phenotype. A testcross is often extremely useful in genetic analysis: In a testcross, the phenotypes of the offspring show the relative frequencies of the different gametes produced by the heterozygous parent, since the recessive parent contributes only recessive alleles. Furthermore, in contrast to the F2 progeny of a dihybrid cross, knowledge of the phenotypes allows drawing an exact conclusion about the genotype.

FIGURE 3.6 In a testcross of a heterozygous ww parent to a homozygous recessive ww parent, the offspring are ww and ww in a 1:1 ratio. A testcross shows the result of segregation.

Mendel conducted a series of test crosses with various traits. The results are shown in TABLE 3.2. In all cases, the ratio of phenotypes among the test cross offspring is very close to the 1:1 ratio expected from allele segregation in the heterozygous parent.

TABLE 3.2 Testcross results according to Mendel's testcross (F1 heterozygote × homozygote

descendants of

recessive)

cross test

Runde × Faltige Together

193 round, 192

1.01:

Baltic

196 yellow, 189 green

1.04:

Yellow × green seeds

Relationship

1 purple × white flowers

85 purple, 81 white

1,05 : 1

Long × short stems

87 long, 79 short

1.10: 1

Another valuable type of cross is backcross, in which hybrid organisms are crossed with one of the parental genotypes. Backcrosses are commonly used by geneticists and plant and animal breeders. Note that the testcrosses in Table 3.2 are also backcrosses, as in each case it is the heterozygous F1 parent.

originates from a cross between homozygous dominant and homozygous recessive.

SUMMARY ■ Mendel's experiments began by crossing true lines (P1 generation) that differed for a particular phenotype to create hybrids (F1). The hybrids were then crossed and the resulting progeny (F2) analyzed. ■ The 3:1 ratio observed in the F2 generation led Mendel to infer the principle of segregation. ■ Mendel confirmed the principle of segregation by showing that in F2 progeny, 2/3 of the plants with the dominant phenotype are homozygous and the remaining 1/3 are heterozygous.

■ The frequency of gametes produced by a heterozygote can be determined directly with a testcross.

3.3 Splitting Two or More Genes The results of many genetic crosses depend on splitting the alleles of two or more genes. Genes can be on different chromosomes or on the same chromosome. Although in this section we consider the case of genes that are on two different chromosomes, the same principles apply to genes that are on the same chromosome but are so far apart that

separate independently. The case of linkage of genes on the same chromosome is examined in the chapter Genetic Linkage and Chromosomal Mapping. To illustrate the principles, consider again a cross between homozygous genotypes, but in this case homozygous for the alleles of two genes. A concrete example is a pure pea variety with wrinkled and green seeds (genotype ww gg) compared to a variety with round and yellow seeds (genotype WW GG). As suggested by the use of uppercase and lowercase letters for alleles, dominant alleles are W and G, recessive alleles are w and g. The mutated gene responsible for Mendel's green seeds has been identified. It is an inborn error of metabolism that prevents the metabolic pathway from breaking down the green pigment chlorophyll. Homozygous mutant seeds cannot break down their chlorophyll and therefore remain green, while wild-type seeds break down their chlorophyll and turn yellow, like the leaves of certain trees in autumn. The mutated gene is officially called staygreen. A cross of WW GG plants with ww gg plants gives F1 seeds with the genotype Ww Gg which are phenotypically round and yellow due to dominance ratios. When F1 seeds are grown into mature, self-pollinated plants, the F2 progeny show the result of simultaneous segregation of the W, w, and o allele pair.

Pair of G alleles, g. When Mendel performed this cross, he got the following number of F2 seeds:

With this data, the first thing to look at is the expected 3:1 ratio for each feature taken individually. The round:wrinkled ratio (clustered over yellow and green) is:

The ratio of yellow to green (via round and wrinkled) is:

These two proportions agree satisfactorily at 3:1. (Testing of fitness to a predicted ratio is described in the chapter entitled The Chromosomal Basis of Inheritance.) Furthermore, in the F2 progeny of the dihybrid cross, the separate 3:1 ratios for the two traits were randomly combined. In random combinations, as shown in FIGURE 3.7, of the 3/4 round progeny, 3/4 are yellow and 1/4 are green; Likewise, of the 1/4 of the offspring that are wrinkled, 3/4 are yellow and 1/4 are green. Therefore, the general ratios of round yellow to round green to crinkled yellow to crinkled green would be expected to be as follows:

The observed ratio of 315:108:101:32 corresponds to 9.84:3.38:3.16:1, which is a good fit for the ratio of 9:3:3:1 derived from the Punnett square in Figure 3.7.

Figure 3.7 The 3:1 round:wrinkled ratio when arbitrarily combined with the 3:1 yellow:green ratio results in the 9:3:3:1 ratio found in the F2 progeny of the dihybrid cross.

The Principle of Independent Classification The independent sorting of the W, w, and G, g allele pairs is shown in FIGURE 3.8. What independence means is that if a gamete contains W, then it has an equal probability of containing either G or g; and if a gamete contains w, it has an equal probability of containing either G or g. The implication is that the four gametes are formed with equal frequency:

FIGURE 3.8 Independent assortment of the W,w and G,g allele pairs means that between each of the W and w gamete classes the ratio of G:g is 1:1. Likewise, between each of the G and g gamete classes, the W:W ratio is 1:1.

The result of independent sorting when the four types of gametes randomly combine to form the next generation zygote is shown in FIGURE 3.9. Note that the expected ratio of phenotypes among F2 offspring is:

FIGURE 3.9 The independent variety is the biological basis for the 9:3:3:1 ratio of F2 phenotypes resulting from a dihybrid cross. (A) A cross of round yellow and wrinkled green purebred parent lines gives round and yellow F1 progeny. (B) The results of selfing of the F1 progeny are presented as a Punnett square.

However, as the Punnett square also shows, the proportion of genotypes in the F2 generation is more complex. The reason for this relationship is shown in FIGURE 3.10. Among the seeds that have the WW genotype, the proportion of

FIGURE 3.10 Genotypes and phenotypes of F2 progeny from a dihybrid cross for seed shape and color.

For seeds with the Ww genotype, the ratio is:

(which is a 1:2:1 ratio multiplied by 2 because there are twice as many Ww genotypes as WW or ww). For seeds with the ww genotype, the proportion of

The combination of these results in an overall genotypic ratio of

Seed phenotypes are shown below the genotypes and the combined phenotypic ratio is:

The principle of independent sorting of two pairs of alleles on different chromosomes (or sufficiently far apart on the same chromosome) came to be known as the principle of independent sorting. It is sometimes referred to as Mendel's second law. Principle of Independent Selection: The separation of members of any pair of alleles is independent of the separation of other pairs in the formation of reproductive cells.

The testcross with independently sorted genes is therefore in F2 progeny when one of each segregating allele is present.

dominantly, nine different genotypes appear as only four distinguishable phenotypes. This makes the results of such crosses difficult to interpret. Alternatively, test crossing can be used:

The result is shown in FIGURE 3.11. Since plants with doubly heterozygous genotypes produce four types of gametes - W G, W g, w G and w g - with equal frequency, while plants with ww gg genotypes produce only w g gametes, the possible offspring genotypes are Ww Gg, Ww gg , ww Gg and ww gg, and these are expected to have the same frequency. Due to the dominance relationships – W over w and G over g – the offspring phenotypes are expected

round yellow, round green, rough yellow and rough green in the ratio 1:1:1:1.

FIGURE 3.11 Genotypes and phenotypes resulting from a testcross of the double heterozygote Ww Gg.

As in a single-gene testcross, in a two-gene testcross, the ratio of offspring phenotypes is a direct demonstration of the proportion of gametes produced by the doubly heterozygous parent, meaning that there is a 1:1 correspondence between the genotype and the phenotype. In his current cross, Mendel has produced 55 round yellow, 51 round green, 49 wrinkled yellow and 53 wrinkled green offspring, which is in line with the predicted ratio of 1:1:1:1.

Three or more genes A Punnett square of the type in Figure 3.9 is a good way to show the logic behind the genotype and phenotype frequencies for two genes submitted to independent classification. As a method to

However, solving problems is not very efficient. When working on a problem, especially when time is limited (for example, during an exam), drawing and filling in the entire square takes a lot of time and offers a lot of opportunity to make mistakes. Alternatively, we can use the forked line approach. To illustrate this method, consider an example of a trihybrid cross with pairs of alleles (W, w), (G, g) and (P, p), where P is a dominant allele for purple flowers and p is a recessive mutation Pro

White flowers. A Punnett square is a 64-cell cube that is difficult to draw and tedious to complete. However, if the three genes sort independently, rather than trying to analyze the variety of all three loci simultaneously, we can look at how the genes sort individually and then combine those results. Furthermore, we will only see what we can see - the phenotypes. This approach is illustrated in FIGURE 3.12, called the bifurcated diagram. The underlying principle is simple: if we were to look at the three genes individually, we would expect to see a 3:1 ratio of dominant to recessive phenotype. So in the diagram, we first look at the phenotypes associated with the W locus, where we would expect 3/4 of the offspring to be round and 1/4 to be wrinkled. Then, after separating the progeny by first phenotype, we would ask what the expected ratio of yellow to green seeds should be for plants with round versus wrinkled seeds. For each and assuming an independent assortment, this ratio is 3/4 : 1/4. Finally, we separate each of the four resulting phenotypic classes in the expected ratio - 3/4:1/4 - for purple versus white flowers, resulting in eight phenotypic classes.

FIGURE 3.12 A bifurcated line plot showing how the phenotypic ratios in a trihybrid cross can be predicted without resorting to the use of a Punnett square.

So what are the expected proportions of these eight classes? They can be obtained by multiplying the fractions along each of the particular paths leading to one of the classes. For example, the proportion of offspring expected to be round, yellow, and purple is as follows:

What we do is ask, "What proportion of the offspring do we expect to be round (3/4), wrinkled (3/4), and yellow (3/4)?" We can then do the same for the remaining seven phenotypes make categories. The results of this analysis are shown in TABLE 3.3. It is worth taking the time to study this table carefully. For phenotypes resulting from dominant genotypes, the genotype is given as the dominant allele followed by a hyphen (eg, W-). This indicates that we don't know what the other allele of the genotype is - it could be the dominant allele or the recessive allele.

TABLE 3.3 Independent assortment in a tri-hybrid cross phenotype

genotype

Expected

observed

Expected

27/64

269

270

Fraction Round, yellow, purple

W— G— P—

round, yellow, white

W—G—S

3/64

round, green, purple

W— gg P—

9/64

round, green, white

W-gg pp

3/64

crumpled, yellow,

ww G— P—

9/64

ww G-S

3/64

ww gg P—

3/64

ww gg p

1/64

Purple Wrinkled, yellow, white Wrinkled, green, purple Wrinkled, green, white Expected phenotypes in a trihybrid cross. A dash in a genotype symbol indicates that the dominant or recessive allele is present; B. refers to W

common to the WW and Ww genotypes. (Expected numbers are 640 instead of 639 due to a rounding error.)

Mendel actually did this three gene crossover. The results for the phenotypes shown in Table 3.3 agree well with the ratio of 27:9:9:9:3:3:3:1 expected from an independent range. The eight types of offspring represent 33 = 27 different genotypes. Mendel performed a testcross on each offspring with the dominant phenotype for one or more traits to determine whether its genotype was homozygous or heterozygous for each of the genes. this effort

alone required 632 test crosses. No wonder he complained that "of all the experiments [this one] required the most time and effort". We emphasize the efficiency and usefulness of a more conceptual approach to genetic computations than the Punnett square. We will return to this topic when we consider how probability theory can be applied more formally to genetic reasoning. Before doing that, though, let's address the issue of genetic analysis of human heredity.

SUMMARY ■ In dihybrid crosses with genes that affect two different traits, each gene follows the principle of segregation. ■ The 9:3:3:1 phenotype ratio observed in the F2 generation can be explained by the principle of independent classification. ■ Testcrosses produce a 1:1 genotype-phenotype match in the offspring.

■ The principle of independent sorting can be applied to crosses with three or more genes, most efficiently using the forked line approach.

3.4 Human pedigree analysis Large deviations from expected genetic relationships are often found in individual human families and in large domesticated animals due to the relatively small number of offspring. However, the effects of segregation are evident when examining phenotypes across multiple generations of related individuals. A diagram of a pedigree showing the phenotype of each individual in a group of relatives is a pedigree. In this section, we introduce the basics of family tree analysis.

dominant and recessive traits

Inheritance FIGURE 3.13 defines the standard symbols used to represent human family trees. Women are represented by circles and men by squares. (A diamond is used when sex is unknown – as in miscarriage.) Individuals with the phenotype of interest are indicated by colored or shaded icons. For recessive alleles, heterozygous carriers are sometimes represented with half-filled symbols. A mating between a female and a male is indicated by connecting their symbols with a horizontal line that connects vertically to a second horizontal line below the one that connects their progeny symbols. Descendants within a sibling, called siblings or siblings, are presented from left to right in birth order.

FIGURE 3.13 Traditional symbols used to represent human family trees.

A pedigree for the Huntington's disease trait caused by a dominant mutation is shown in FIGURE 3.14. The numbers in the pedigree are for easy reference to specific individuals. Successive generations are indicated by Roman numerals. Within a generation, all individuals are numbered consecutively from left to right. The pedigree begins with female I-1 and male I2. The man has Huntington's disease, a progressive form of nerve degeneration that usually begins in middle age. This leads to severe physical and mental impairments and eventually death. The dominant allele, HD, that causes Huntington's disease is rare. All affected individuals in the pedigree have the heterozygous HD hd genotype, while unaffected individuals have the homozygous normal hd hd genotype. In many families, the disease has complete penetrance. The penetrance of a genetic disorder is the proportion of individuals with the risk genotype who have the trait; total penetrance means that the trait is expressed in 100% of individuals with that genotype. The genealogy shows the

the inheritance features due to a rare dominant allele with complete penetrance are as follows: 1. Females and males are equally likely to be affected. 2. The affected offspring have an affected parent (except for rare new mutations), and the affected parent is equally likely to be the mother or the father. 3. On average half of those in siblings with affected parents are affected.

FIGURE 3.14 Lineage of a human family showing inheritance of the dominant gene for Huntington's disease. Women and men are represented by circles and squares, respectively. Red symbols denote people affected by the disease.

A pedigree for a trait due to a homozygous recessive allele is shown in FIGURE 3.15. The characteristic is albinism, the absence of pigment in the skin, hair and iris of the eyes. This pedigree illustrates inheritance features due to a rare recessive allele with complete penetrance: 1. Females and males are equally likely to be affected. 2. Affected individuals, if they breed, usually have unaffected offspring.

3. Most affected people have unaffected parents. 4. Parents of affected persons are usually relatives. 5. Among the siblings of those affected, the proportion of those affected is around 25 percent.

FIGURE 3.15 Family tree of albinism. In recessive inheritance, affected individuals (solid symbols) usually have unaffected parents. The double horizontal line indicates a mating between relatives - in this case, first cousins.

In the rare recessive mode of inheritance, the partners of homozygous affected individuals are usually homozygous for the normal allele, so that all offspring are heterozygous and unaffected. Heterozygous carriers of the mutant allele are significantly more common than homozygotes because an individual is more likely to inherit only one copy of a rare mutant allele than two copies. Most homozygous recessive genotypes therefore result from matings between carriers (heterozygotes × heterozygotes), in which each offspring has a 1/4 chance of being affected. Another important feature of rare recessive inheritance is that the parents of affected individuals are often relatives (consanguineous). In a pedigree, a cross between relatives is indicated by a double line connecting the partners, as in the first

Prime pairing in Figure 3.15. Mating between relatives is important for recessive alleles to become homozygous; That is, if a recessive allele is rare, it is more likely to become homozygous through inheritance from a common ancestor than from completely unrelated parents. The reason for this pattern is that the carrier of a rare allele can have many offspring who are also carriers. If two of these carriers mate with each other (for example, in a first cousin mating), the hidden recessive allele can become homozygous with a probability of 1/4. Mating between relatives constitutes inbreeding, and the consequences of inbreeding are discussed further in the Genes in Populations chapter. As an affected individual reports that both parents are heterozygous carriers, the expected proportion of affected individuals among siblings is approximately 25%, but the exact number depends on the details of how affected individuals are identified and included in the analysis.

Most Human Genetic Variations Are Not "Bad" Before the advent of molecular methods, there were many practical obstacles to the study of human genetics. With the exception of traits such as ABO and other blood types, few traits were known to show simple Mendelian inheritance. Most of these were linked to genetic diseases and presented specific challenges: ■ Most of the genes that cause simple Mendelian genetic diseases are rare, seen only in a small number of families. ■ Many genes for simple Mendelian diseases are recessive, ie they are not detected in heterozygous genotypes. ■ The number of descendants per human family is relatively small, so segregation into individual siblings usually cannot be determined.

■ The human geneticist cannot perform testcrosses, backcrosses, or other experimental matings. This text contains numerous examples of how molecular genetics has revolutionized the study of human genetics. For example, in the chapter on DNA structure and genetic variation, we discussed the prevalence of single nucleotide polymorphisms (SNPs) and copy number variations (CNV) in the human genome. Only a very small number of these common polymorphisms are associated with disease, and even they serve as genetic risk factors that interact with other risk factors, including other genes and the environment, to determine whether a person will actually develop the disease. Although most SNPs and CNVs do not have adverse effects, both types of variation show simple Mendelian inheritance, which is why human geneticists use them for family and population studies.

However, some seemingly innocuous, simply Mendelian features were discovered in the pre-molecular era. One of the best known was linked to the ability to taste a chemical known as phenylthiocarbamide (PTC), which has the molecular structure shown here:

PTC is a synthetic chemical developed by an industrial chemist in the early 1930's. The flavor polymorphism was discovered one day while carelessly releasing a cloud of powdered PTC into the air. The PTC powder didn't bother the chemist at all, but his lab partner complained loudly about the bitter taste it left in his mouth. Out of curiosity, the chemist began testing the PTC-tasting ability of family and friends and recruited a geneticist to get started.

Study the situation. Family studies soon showed that the ability to taste PTC is an inherited trait as a simple Mendelian dominant. In the European population, about 70 percent of people are tasters and 30 percent are non-tasters, but these proportions vary widely between ethnic groups. For people of African or Asian descent, the tasting frequency is around 90%, while for Australian Aborigines it is only 50%. However, the ability to taste PTC varies. the most sensitive

Tasters can taste concentrations as low as 0.001 millimoles (mM), while the most insensitive non-tasters cannot detect concentrations as high as 10 mM. An arbitrary cut-off is used to classify subjects as "tasters" or "non-tasters", typically at a concentration of 0.2 mM PTC. The greatest variation in taste ability between tasters and non-tasters is due to taster major polymorphism, but some differences are due to other genes, gender, and likely environmental factors. The result of the other variables is that about 5 percent of heterozygous tasters are classified as "non-tasters" and at least 5 percent of homozygous non-tasters are classified as "tasters". It is now known that the molecular basis of the taste polymorphism lies in a taste receptor protein known as hTAS2R38. There are several alleles of the gene, but the most common forms of the protein differ by three amino acids at scattered positions along the protein. The allelic forms are known as PAV and AVI because the three main amino acids in the PAV protein are proline, alanine, and valine, while these three positions in the AVI protein are occupied by alanine, valine, and isoleucine. The PAV form is the one that gives the possibility of tasting PTC. If you think about it, a polymorphism in PTC tasting doesn't make sense. PTC is a completely artificial chemical found in the

Laboratory, so why should there be polymorphism in the ability to taste it? One clue comes from the observation that the chemical structure of PTC resembles a large and heterogeneous class of molecules called glucosinolates. These compounds are synthesized by some plants, including some human food plants, and their synthesis probably evolved as a chemical defense against insect herbivores. Among the plants that produce glucosinolates is one pointed out by former President George H.W. Bush, who removed broccoli from the White House menu in 1989, proclaiming, "I don't like broccoli. And I didn't like it as a kid and my mom made me eat it. And I'm the president of the United States and I'm not going to eat broccoli from new!" In gleeful protest, broccoli growers across the United States sent Bush tons of the product.Seventeen years later, new studies have shown that individuals carrying the PAV form of the hTAS2R38 taste receptor actually find that broccoli tastes significantly more bitter than homozygous individuals. for the allele encoding the AVI form Tasters also report a greater perception of bitterness from kale, turnips, swedes, and horseradish.

Molecular Markers in Human Pedigrees Because DNA manipulation techniques allow direct access to DNA, modern genetic studies of human pedigrees are performed primarily using genetic markers present in the DNA itself, rather than the phenotypes produced by mutated genes. Different types of DNA polymorphisms have been discussed in the chapter DNA Structure and Genetic Variation, along with the methods used to detect and study them. An example of DNA polymorphism spanning three generations in a human family tree is shown in FIGURE 3.16. The type of polymorphism is a single sequence repeat polymorphism (SSRP) in which each allele differs in size according to the number of copies of a short DNA it contains

Sequence repeated in tandem. Size differences are detected electrophoretically by PCR after amplification of the region. SSRPs usually have many codominant alleles, and most individuals are heterozygous for two different alleles. In the example in Figure 3.16, all parents are heterozygous, as are all children.

FIGURE 3.16 Human lineage showing segregation of SSRP alleles. Six alleles (A1-A6) are present in the pedigree, but each individual can have only one allele (if homozygous) or two alleles (if heterozygous).

Six alleles are shown in Figure 3.16, labeled A1 through A6. On the gel, band numbers correspond to allele indices. Generation II mating occurs between two heterozygous genotypes: A4A6 × A1A3. Due to the separation in each parent, four genotypes are possible in the offspring (A4A1, A4A3, A6A1 and A6A3); These would conventionally be written minor index first as A1A4, A3A4, A1A6, and A3A6. In the case of accidental insemination, the genotypes of the offspring are equally likely, as can be verified using a Punnett square for mating. Figure 3.16 illustrates some of the main advantages of multiple codominant alleles for human genealogy analysis:

■ Heterozygous genotypes can be distinguished from homozygous genotypes. ■ Many individuals in the population are heterozygous and many matings are instructive in terms of segregation. ■ Each segregating genetic marker produces up to four distinct offspring genotypes.

SUMMARY ■ Human heredity can be analyzed through pedigree analysis. ■ If traits are dominant, affected individuals usually have affected parents. ■ In recessive traits, affected individuals usually have unaffected parents. ■ Heterozygotes for recessive traits are called carriers and are more common in populations than homozygotes for the trait.

■ Simple Mendelian inherited disorders are rare.

3.5 Incomplete Dominance and Epistasis Dominance and codominance are not the only possibilities for allele pairs. There are situations of incomplete dominance where the phenotype of the heterozygous genotype is between the phenotypes of the homozygous genotypes. A classic example of incomplete dominance concerns flower color in snapdragon Antirrhinum majus (FIGURE 3.17). In wildflowers, a red type of anthocyanin pigment is formed through a series of enzymatic reactions. A wild-type enzyme encoded by the I allele is the rate-limiting factor for the overall reaction, so the amount of red pigment is determined by how much enzyme the I allele produces. The alternative allele i encodes an inactive enzyme and flowers ii are ivory. As the amount of the critical enzyme is reduced in Ii heterozygotes, the amount of red pigment in the flowers is also reduced and dilution causes the flowers to turn pink. A cross between plants with different flower colors therefore provides a direct phenotypic indication of segregation (Figure 3.21). Crossing II (red) × ii (ivory) gives F1 plants with genotype II and pink flowers. In F2 progeny obtained by selfing of F1 hybrids, one experiment resulted in 22

Plants with red flowers, 52 with pink flowers and 23 with ivory flowers, which corresponds to the expected ratio of 1:2:1.

FIGURE 3.17 Red versus white flower color in snapdragons indicates incomplete dominance.

ROOTS OF DISCOVERY This land is your land The Huntington's Disease Collaborative Research Group (1993) Comprises 58 authors from 9 institutions

A new gene containing a trinucleotide repeat

Scattered and Unstable in Huntington's Disease Chromosomes Modern genetics research is sometimes carried out by large collaborative groups at various research institutes spread across many countries. This approach is illustrated by the search for the gene responsible for Huntington's disease. The search was widely publicized due to the severity of the disease, the late age of onset, and the fact that it was well known.

dominant heritage. Famed folk singer Woody Guthrie, who wrote "This Land Is Your Land" and other well-known songs, died of the disease in 1967. When the gene was identified, it was discovered that it encoded a protein (now called huntingtin) with a function unknown that is expressed in many cell types throughout the body and not exclusively in nervous tissue, as expected. Within the coding sequence of this gene is a trinucleotide repeat (5'-CAG-3'), which is repeated several times in a row according to the general formula (5'-CAG-3')n. Among normal alleles, the number n of repeats ranges from 11 to 34, with a mean of 18; for mutant alleles, the number of repeats ranges from 40 to 86. This tandem repeat is genetically unstable, as it can increase (“expand”) the number of copies by an unknown mechanism. In two cases where a new mutant allele was analyzed, the number of repeats increased from 36 to 44 in one and from 33 to 49 in the other. This mutational mechanism is quite common in some human genetic diseases. The excerpt gives several other examples. The authors also emphasize that their discovery raises important ethical issues, including those related to genetic testing, confidentiality and informed consent.

“We think it's extremely important that the current Hartl, Daniel L. and Bruce Cochrane. Genetics: Analysis of Genes and Genomes: Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created by utah on 2021-08-09 19:18:21.

“We believe it is extremely important that current internationally recognized guidelines and advisory protocols for testing individuals at risk continue to be adhered to and that samples from unaffected loved ones are not inadvertently tested.

”

without full consent.

The genetic defect that causes Huntington's disease has been mapped to chromosome 4

in one of the first successful linkage analyzes using DNA markers in humans. Since then, we have pursued an approach to isolate and characterize the HD gene based on progressive refinement of its location…. [We found that a 500 kb segment is the most likely site of the genetic defect. [The abbreviation “kb” stands for kilobase pairs; 1 kb corresponds to 1000 base pairs.] In this region, we identified a large gene, spanning approximately 210 kb, encoding a previously undescribed protein. The reading frame contains a polymorphic trinucleotide repeat (CAG)n with at least 17 alleles in the normal population, ranging from 11 to 34 CAG copies. In HD chromosomes, the length of the trinucleotide repeat is substantially increased…. The ability to directly monitor trinucleotide repeat size in individuals “at risk” for HD is expected to revolutionize testing for the disease…. The authors then point out the importance of adhering to strict ethical standards in testing programs, particularly with regard to testing unaffected relatives of affected individuals without their fully informed consent. This is particularly important in Huntington's disease, as the alleles associated with the disease are dominant and only manifest phenotypically later in life. For example, an apparently healthy young adult may learn, as a result of testing, that they will develop this debilitating, debilitating condition.

eventual fatal illness later in life - information some would prefer not to know. Source: MacDonald ME, Ambrose CM, Duyao MP. A new gene containing an expanding and unstable trinucleotide repeat in HD chromosomes. The Huntington's Disease Collaborative Research Group, Cell 72 (1993): 971-983.

Multiple Alleles The occurrence of multiple alleles is illustrated by the A1-A6 alleles of the SSR marker in the human pedigree in Figure 3.16.

Multiple alleles are relatively common in natural populations and, as in this example, can be detected more easily by molecular methods. In the DNA of a gene, each nucleotide can be A, T, G, or C, so a gene with n nucleotides could theoretically mutate at any one of the positions to any of the other three nucleotides. Therefore, the number of possible single nucleotide differences in a gene of length n is 3 × n. For example, if n = 5000, there are potentially 15,000 alleles (not counting possibilities with more than one nucleotide substitution). Most potential alleles never actually existed. Some are absent because they did not occur, others occurred but were eliminated accidentally or because they were harmful, and others are still present but so low in frequency that they go unnoticed. However, at the DNA sequence level, most genes in most natural populations have multiple alleles, all of which can be considered "wild type". Multiple wild-type alleles are useful in applications such as DNA typing, since two unrelated individuals are unlikely to have the same genotype, especially when there are many different loci, each with multiple

alleles are examined. Many deleterious mutations also exist in multiple forms. Recall from the Genes, Genomes, and Genetic Analysis chapter that more than 400 mutant forms of the phenylalanine hydroxylase gene have been identified in patients with phenylketonuria. The A1–A6 alleles in Figure 3.16 also illustrate this.

that although a population of organisms can contain any number of alleles, any organism or cell can carry no more than two, and any gamete can carry no more than one. In some cases, the multiple alleles of a gene exist by sheer coincidence, reflecting the history of mutations that have occurred in the population and the spread of these mutations among population subgroups through migration and interbreeding. In other cases, biological mechanisms favor the maintenance of a large number of alleles. For example, the genes that control self-sterility in certain flowering plants can have a large number of allelic types. This type of self-sterility is found in species of red clover, which grow wild in many grasslands. Self-sterility genes prevent self-fertilization because a pollen grain can undergo pollen tube growth and fertilization only if it contains a self-sterility allele that differs from any of the alleles present on the flower on which it lands. In other words, a pollen grain that contains an allele already present in a flower will not work on that flower. Since all pollen grains produced by a plant must contain one of the self-sterility alleles present in the plant, pollen cannot act on the same plant that produced it, and self-fertilization cannot occur. Under these conditions, any plant with a new allele can fertilize all but the flowers on the same plant. Over the course of evolution, red clover populations have accumulated hundreds of alleles of the self-sterility gene, many of which have been isolated and their DNA sequence determined. Many of the alleles differ at several nucleotide sites, implying that the alleles are very old in the population.

Human ABO Blood Groups In a multiple allelic series, there can be different dominance relationships between different pairs of alleles. An example is found in the human ABO blood groups, which are determined by three alleles designated IA, IB, and IO. (There are actually two slightly different variants of the IA allele.) Each person's blood type can be A, B, AB, or O, depending on the type of polysaccharide (polymer of sugars) present on the surface of red blood cells. One of two different polysaccharides, A or B, can be formed from a parent molecule modified by the enzymatic product of the IA or IB allele. The gene product is a glycosyltransferase enzyme that attaches a sugar moiety to the precursor (FIGURE 3.18). The IA and IB alleles encode different forms of the enzyme with changes at four amino acid sites; these alter the substrate specificity so that each enzyme binds to a different sugar. People with the IAIA genotype produce red blood cells containing only polysaccharide A and are said to have type A blood. People with the IBIB genotype have red blood cells with only polysaccharide B and have blood type B. People who are heterozygous IAIB

have erythrocytes with A and B polysaccharides and have type AB blood. The third allele, IO, encodes an enzymatically inactive protein that leaves the parent unchanged; neither type A nor type B polysaccharide is produced. Thus, IOIO homozygous individuals lack the A and B polysaccharides and are said to have type O blood.

FIGURE 3.18 The ABO antigens on the surface of human red blood cells are carbohydrates.

In this multiple allelic series, the IA and IB alleles are codominant: the heterozygous genotype has the characteristics of both homozygous genotypes - the presence of both A and B carbohydrates in red blood cells. In contrast, the IO allele is recessive to both IA and IB. Therefore, heterozygous IAIO genotypes produce polysaccharide A and have blood type A, and heterozygous IBIO genotypes produce polysaccharide B and have blood type B. The genotypes and phenotypes of the ABO blood group system are summarized in the first three columns of the TABLE 3.4.

TABLE 3.4 Genetic control of the human ABO blood group genotype

Antigen

antibody

blood

blood groups

Gift

blood

present in

faces that

that can

in red

group

blood fluid

maybe

to accept

blood

phenotype

tolerated in

blood to

Transfusion

Cell HIM

ONE

Type A

Anti-B

A & AB

IAIO

ONE

Type A

Anti-B

A & AB

IBIB

Type B

Anti-A

B&O

B&AB

IBIO

Type B

Anti-A

B&O

B&AB

IAIB

A & B

Type AB

A, B, AB e

AB only

Anti A yet

Anti-BJA

Weder A

still b

Type the

Anti-A e Anti-B

oh just

A, B, AB and O

ABO blood types are important in medicine because of their need for blood transfusions. A key feature of the ABO system is that most human blood contains antibodies to polysaccharide A or B. An antibody is a protein produced by the immune system in response to a stimulating molecule called antigen and capable of binding to antigen. An antibody is usually specific because it recognizes only one antigen. Some antibodies combine with the antigen and form large molecular aggregates that can precipitate. Antibodies protect the body against invading viruses and bacteria. Although antibodies normally do not form without prior antigen stimulation, humans are capable of producing anti-A and

They produce anti-B. The production of these antibodies can be stimulated by antigens similar to the A and B polysaccharides, which are present on the surface of many common bacteria. However, a mechanism called tolerance prevents an organism from producing antibodies against its own antigens. This mechanism ensures that the A antigen or B antigen only elicits antibody production in people whose own red blood cells do not contain A or B, respectively. In summary: people with type O blood produce anti-A and anti-B antibodies, people with type A blood produce anti-B antibodies, people with type B blood produce anti-A antibodies, and people with type AB blood do not produce any of either type. of antibody.

FIGURE 3.19 is an example of how antibodies can be used to determine blood type.

FIGURE 3.19 The reaction between the indicated blood groups and anti-A (left column) and anti-B (right column). © Chatsikan Tawanthaisong/Shutterstock.

The antibodies found in the blood fluid of humans with each of the ABO blood groups are listed in the fourth column of Table 3.4. The clinical importance of the ABO blood groups lies in the transfusion of

Blood that contains A or B red blood cell antigens in people who develop antibodies against them causes an agglutination reaction, which causes the donor's red blood cells to clump together. In this reaction, the anti-A antibody agglutinates type A or type AB red blood cells, as both carry the A antigen. Likewise, the anti-B antibody agglutinates type B or type AB red blood cells. When blood cells clump together, many blood vessels become blocked and the transfusion recipient goes into shock and may die. Conversely, incompatibility, when the donor's blood contains antibodies against the recipient's red blood cells, is generally acceptable because the donor's antibodies are diluted quickly enough to avoid clumping. Compatible blood transfusion types are listed in the last two columns of Table 3.4. Note that a person of the AB blood group can receive blood from a person of any other ABO group; Type AB is called a universal receiver. On the other hand, a person of blood group O can donate blood to a person of any ABO group; Type O is referred to as a universal donor.

Epistasis In the Genes, Genomes, and Genetic Analysis chapter, we saw how the products of multiple genes may be required to carry out all the steps in a biochemical pathway. In genetic crosses, where two mutations that affect different steps in a single pathway segregate, the typical F2 ratio of 9:3:3:1 is not observed. A gene interaction that disrupts normal Mendelian relationships is called epistasis. One type of epistasis is illustrated by the interaction of the C, c and P, p allele pairs that affect flower color in peas. These genes encode enzymes in the biochemical pathway for the synthesis of

Anthocyanin pigment, and anthocyanin production requires the presence of at least one dominant wild-type allele of each gene. We can denote this genotype as

Suppose we make a cross between CC pp and cc PP as in FIGURE 3.20A. The phenotype of flowers of the F1 generation is wild-type purple. Why? Because the C allele is dominant over c and the P allele is dominant over p, the F1 plant is a double heterozygote of the Cc Pp genotype and therefore has purple flowers. The result is curious, however, because the original cross involved two homozygous recessive mutants, each with white flowers. Since we are told that the mutant c and mutant p alleles are on different genes, finding wild-type flowers in F1

The generation is logical.

FIGURE 3.20 Epistasis in determining flower color in peas. Formation of the purple pigment requires the dominant allele of the C and P genes. (A) Parental and F1 cross. (B) Bifurcated line plot showing that in this type of epistasis the F2 dihybrid

The ratio is changed to 9 purple: 7 white.

But what if we didn't know if the mutated alleles were on different genes? What if both mutants are found in a mutant screen or discovered in different labs and we don't know whether the c allele and the p allele are alleles from the same gene or alleles from different genes? The answer is the phenotype of the F1 offspring in a cross like the one in Figure 3.20. On the one hand, if the phenotype of the F1 offspring is wild-type, as in Figure 3.20, this observation indicates that c and p are alleles of different genes. On the other hand, if the phenotype of the F1 progeny is mutated (in this case, a plant with white flowers),

This result informs that c and p are alleles of the same gene. In the latter case, it would be better to change their names to denote their allelism, for example B. c = c1 and p = c2 because then the genotype of the F1 plant would be written as c1/c2 and it would be obvious from the genotype symbol that the phenotype is mutated, since all alleles of the gene are mutated. This discussion might remind you of the Beadle-Tatum experiments—specifically the use of the complementation test to determine whether or not two nutritional mutants in Neurospora were mutants of the same gene. The principle in Figure 3.20 is exactly the same. The difference is that Beadle and Tatum used Neurospora heterocaryae in which the mutant alleles are brought together by the formation of hybrid strands with two types of mutant haploid nuclei, whereas in diploid organisms, such as the peas in Figure 3.20, the hybrid nuclei are brought together by the cross produced by two are homozygous recessive mutants. The key point is that complementation tests are also used in sexually diploid organisms to identify which recessive mutations are alleles and which are not, and the phenotype of the F1 generation in Figure 3.20 illustrates the underlying principle. Whereas the F1 generation is shown in Figure 3.20A Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

Complementarily, the F2 generation presents a type of epistasis. Assume that the plants of the F1 generation are self-pollinated (indicated by the circled cross sign) and assume that the allele pairs (C,c) and (P,p) sort independently. The split line graph in FIGURE 3.20B represents the expected phenotypes of the F2 generation. Since only the C—P— progeny have violet flowers, the ratio of violet flowers to white flowers in the F2 generation is 9 to 7. Epistasis does not change the result of independent assortment; just obscures the fact

that the underlying ratio of the C—P— : C—pp : cc P— : cc pp genotypes is 9 : 3 : 3 : 1. For a trait determined by the interaction of two genes, each with a dominant allele, there is only one Limited number of ways to modify the 9:3:3:1 dihybrid ratio. The possibilities are shown in FIGURE 3.21. In the absence of epistasis, the F2 ratio of phenotypes is 9:3:3:1. In each row, different colors indicate different phenotypes. For example, in the proportion modified in the

Below, the phenotypes of the "3:3:1" classes are indistinguishable, resulting in a ratio of 9:7. This ratio is seen in the separation of the C,c and P,p alleles in Figure 3.20 with the 9:7 ratio of purple to white flowers. Taking all possible modified ratios in Figure 3.21 together, there are nine possible dihybrid ratios when both genes show complete dominance. Examples of each of the modified ratios are known. The types of epistasis that lead to these modified ratios are illustrated in the following examples, drawn from a variety of organisms. See the tasks at the end of the chapter for more examples.

FIGURE 3.21 Modified ratios of F2 dihybrids. In each line, different colors indicate different phenotypes.

9:7 is observed when a homozygous recessive mutation in one or both of the different genes results in the same mutated phenotype, as in Figure 3.20. 12:3:1 results when the presence of a dominant allele at one locus masks the genotype at another locus, e.g. B. the A genotype, making the B and bb genotypes indistinguishable. For example, in a genetic study of the hull color of oat seeds, a white-hulled variety was crossed with a black-hulled variety. The F1 hybrid seeds had black hulls. Among 560 offspring of the F2 generation, 418 black, 106 gray and 36 white trunk phenotypes were observed. The phenotype ratio is 11.6:3.9:1, or nearly 12:3:1. A genetic hypothesis to explain these results states that the black envelope phenotype indicates the presence of a dominant A allele and the gray envelope phenotype is due to another dominant allele.

B, whose effect is only evident in the aa genotypes. Based on this hypothesis, the original cultivars had genotypes aa bb (white) and AA BB (black). The F1 generation has the Aa Bb (black) genotype. If the allele pair A, a and the allele pair B, b are subjected to independent sorting, the F2 generation is expected to have the genotypic and phenotypic composition 9/16 A – B – (black envelope), 3/16 A - bb (black body), 3/16 aa B— (grey body), 1/16 aa bb (white body) or 12:3:1 white Wyandotte chickens (cc ii genotype). Both breeds have white feathers because the C allele is required for feather coloring, but the I allele in White Leghorns is a dominant inhibitor of feather coloration. The F1 generation of a dihybrid cross between these breeds has the Cc Ii genotype, which is expressed as white feathers due to the inhibitory effect of the I allele.

Generation only the C-ii genotype has colored feathers, so there is a 13:3 white: motley ratio. 9:4:3 is observed when homozygosity for a recessive allele for one gene masks expression of the genotype of another gene. For example, if genotype aa has the same phenotype whether it is genotype B- or bb, then the ratio is 9:4:3. For example, in mice, the grayish "agouti" coat color results from a horizontal band of yellow pigment just below the tip of each hair. The agouti pattern is due to a dominant A allele and in aa animals the coat color is black. A second dominant allele, C, is required for the formation of any type of hair pigment, and cc animals are albino (white). In an AA CC (Aguti) × aa cc (albino) cross, the F1 progeny are Aa Cc and phenotypically agouti. Crosses between F1 males and females produce F2 offspring in the proportions 9/16 A—C— (aguti), 3/16 A—cc (albino), 3/16

aa C— (black), 1/16 aa cc (albino) or 9 agouti : 4 albino : 3 black. 9:6:1 implies that homozygosity for either of the two recessive alleles gives the same phenotype, but the phenotype of the double homozygotes is different. In Duroc-Jersey pigs, red coat color requires the presence of two dominant alleles, R and S. Pigs of the R genotype - ss and rr S - have a sandy coat and rr ss pigs are white. The F2 ratio is therefore 9/16 R— S— (red), 3/16 R— ss (sandy), 3/16 rr S— (sandy), 1/16 rr ss (white), or 9 red: 6 sandy: 1 white.

SUMMARY ■ Incomplete dominance occurs when heterozygotes for one gene can be distinguished phenotypically from both homozygotes. ■ More than two alleles for a population can exist in a population, and their paired dominance relationships can differ. ■ Epistatic interactions between genes disrupt normal Mendelian relationships.

■ Epistatic genes usually encode different enzymes involved in a specific biochemical pathway.

3.6 Probability in genetic analysis Mendel's laws of genetic transmission are essentially laws of chance (probability). Mendel surpassed all his contemporaries in understanding that his principles of heredity were responsible for the different types of offspring he observed and the conditions in which they were found. No discoveries in genetics since Mendel's time have challenged the fundamental role of chance.

Heredity, which he was the first to recognize. To fully understand Mendelian genetics, we must understand the elementary principles of probability. Every probability problem starts with an experiment, which can be real or imaginary. In genetics, the experiment is typically a cross. Associated with the experiment is a set of possible outcomes of the experiment. In genetics, possible outcomes are typically genotypes or phenotypes. The possible outcomes are called elementary outcomes. They are elementary results in the sense that none of them can be reduced to a combination of the others. In the case of genetic crosses, for example, the elementary results are the different possible phenotypes in the offspring. In our probability applications, the number of elementary outcomes is usually relatively small, or in any case can be enumerated. Probability principles can also deal with conceptual experiments where there are an infinite number of elementary outcomes, but this requires some technical details that are beyond the scope of this text. Each elementary outcome is assigned a probability that is proportional to its probability of occurrence. In principle, probabilities can be assigned arbitrarily. There are only two rules. First, the probability of each elementary outcome must be a non-negative number between 0 and 1 and can be either 0 or 1.

An elementary outcome with probability 0 cannot occur, and one with probability 1 must occur. Second, the sum of the probabilities of all elementary outcomes must equal 1. This rule guarantees that at least one of each elementary outcome must occur. These two rules also address an annoying question that is often asked about tossing a coin: what happens if it lands on the edge? The answer is that this elementary result is given a probability of 0, so we don't need to worry about that. Here it is useful to consider a concrete example. Consider the conceptual experiment of selfing the F1 progeny of a

Cross between pea plants homozygous for the W round and G yellow alleles and plants homozygous for w and g. Recall that the allele pairs W, w, and G, g are sorted independently, so there are 16 elementary results. Genotypes are shown in the Punnett square in Figure 3.9, but shown differently in Figure 3.22A. Each of the elementary outcomes is equally likely, so the probability of each elementary outcome is assigned a value of 1/16. Note that the Ww Gg offspring genotype is listed four times, reflecting the fact that there are four possible combinations of parental gametes that can produce the Ww Gg offspring genotype.

FIGURE 3.22 (A) Sample space for possible offspring in a cross for pairs of alleles W, w for round versus wrinkled seeds and G, g for yellow versus green seeds. (B) Event A includes all genotypes whose phenotype is wrinkled. (C) Event B includes all genotypes whose phenotype is yellow. (D) The union and intersection (yellow) of A and B.

The enumeration of elementary outcomes and their probabilities form the so-called sample space of the probability problem. For the self-pollinated progeny of Ww Gg, the sample space is shown in Figure 3.22A. This is the default space in which all probability considerations related to this conceptual experiment occur.

Elementary results and events

Each combination of elementary outcomes forms an event in the sample space. In FIGURE 3.22B, the circle labeled A contains four elementary outcomes that define event A as "the offspring genotype is ww." Event A could equivalently be defined as "the phenotype of the offspring is wrinkled". Alternative ways of defining A in words show how subsets of elementary results can relate genotypes and phenotypes. Each event corresponds to a probability of that event, symbolized in this case by Pr{A}. A basic principle of probability is that the probability of an event is equal to the sum of the probabilities of all elementary outcomes involved in the event.

In the example in Figure 3.22B, Pr{A} = 4/16 because event A consists of four elementary outcomes (possible offspring genotypes), each with a probability of 1/16. An event cannot contain elementary outcomes, in which case its probability is 0; or it may contain all elementary outcomes, in which case its probability is 1. The elementary outcomes outside the circle defining event A in Figure 3.22B define another event, which we label A'. This event consists of all elementary outcomes not present in A; in other words, it consists of all offspring whose genotype is not ww, or equivalently, it consists of offspring whose phenotype is round. The event A' is called the complement of A and can also be denoted as AC, A or not-A. The probability of A′ is again equal to the sum of the probabilities of the elementary outcomes that make up A′, so in this case Pr{A′} = 12/16.

Events can also be composed of other events. To see how this is done, consider event B in FIGURE 3.22C. Event B is defined as all offspring of the cross whose genotype is gg, which can also be defined as all offspring whose seeds are green. Since B contains four elementary outcomes, its probability is Pr{B} = 4/16. Again there is a complementary event B', defined as all elementary results not contained in B; to define B' differently, all progeny whose genotype is GG or Gg. There are 12 elementary outcomes in B', so Pr{B'} = 12/16.

Look again at Figures 3.22B and C and consider the event consisting of elementary outcomes contained in A, B, or both. This event is called the union of A and B. We will refer to the union of A and B as A + B, but in many probability books you can find it symbolized as A B, where the symbol is pronounced "cup". The event A + B consists of seven elementary outcomes where the genotype of the offspring is ww, gg or both, ie Pr{A + B} = 7/16. Another important way in which two events A and B can combine is called the intersection of A and B. It consists of all the elementary results contained in A and B. An example is shown in Figure 3.22D, where the intersection of A and B is shaded in yellow. We'll refer to the intersection of A and B as AB, but in probability books the intersection is often represented as A∩B, with the ∩ symbol pronounced "cap".

Probability of the Union of Events As with all events, the probability of the union of events A + B is equal to the sum of the probabilities of the elementary outcomes contained in A or B or both. Likewise, the probability of the intersection of the AB events is equal to the sum of the probabilities of the events

elementary results contained in A and B. In general, an equation for the probability of union of A and B is:

The reason for subtracting Pr{AB} becomes clear when we look at Figure 3.22D. Since the genotype of the ww gg progeny is contained in both A and B, when the probabilities of the elementary outcomes in A are added to those in B, the ww gg genotype will contain twice. To correct for overcounting, the probability of that outcome must be subtracted from the total. As the double-counted elementary results are exactly those present in A and B, they form the intersection of A and B. Therefore, Pr{AB} is the quantity being subtracted in Equation 1.

An important special case of Equation 1 occurs when A and B do not intersect - that is, when they have no elementary results in common. In this case, A and B are said to be mutually exclusive (or disjoint) and Pr{AB} = 0. Therefore, if A and B are mutually exclusive, Equation 1 becomes:

This equation is sometimes called the addition rule for mutually exclusive events. Events are mutually exclusive when the occurrence of one event precludes the occurrence of the other. In other words, mutually exclusive events are incompatible in the sense that no elementary outcome can exist in both. An example, again based on the Ww Gg × Ww Gg crossing experiment, is shown in FIGURE 3.23. Here we define the events A* and B* as modified versions of A and B that do not overlap. In other words, the A* event can be defined as "the offspring phenotype is wrinkled but not green" and the B* event

can be defined as "the phenotype of the offspring is green but not wrinkled". Defined this way, A* and B* are mutually exclusive, and therefore Equation 2 implies that Pr{A* + B*} = Pr{A*} + Pr{B*} = 6⁄16.

FIGURE 3.23 Events A* and B* are defined so that they each exclude genotypes whose phenotype is wrinkled and green. A* and B* do not overlap and are therefore mutually exclusive.

Probability of overlapping events

Going back to events A and B in Figure 3.22D, consider the intersection probability AB. This event consists of all elementary outcomes contained in A and B, which in this example consist of just one genotype - i.e. ww gg. The probability of AB is therefore Pr{AB} = 1/16. This is a special case where the events are independent of each other, which means that knowing whether event A occurs says nothing about the occurrence of event B. The probability of the joint occurrence of independent events has the special property of being proportional to the product of the probabilities of the individual events. In the example of Figure 3.22D, we saw that Pr{A} = 4/16 and that Pr{B} = 4/16; in this case Pr{AB} = 4/16 × 4/16 = 16/256 = 1/16.

Events A and B in this example illustrate an important general principle: if events A and B are independent, their probability is given by

This equation is sometimes called the independent event multiplication rule. The choice of term for “independent events”

and "independent classification" is not accidental, as independent classification means that knowing the genotype of an offspring for the allele pair W, w tells nothing about the genotype for the allele pair G, g. This principle is illustrated in FIGURE 3.24A.

Figure 3.24 In genetics, two important types of independence are (A) independent assortment of alleles, which show independent arrangement, and (B) independent fertilizations, which result in contiguous offspring.

Another situation in genetics where independence is the rule is illustrated in Figure 3.24B, which deals with consecutive offspring of a cross. Consecutive offspring are independent because the genotypes of the early offspring have no effect on the probabilities of later offspring. The independence of consecutive descendants contradicts the widely held belief that in any human family the ratio of girls to boys should be "balanced" at about 1:1. Thus, a family with four girls would be more likely to have a boy next time. In reality, this belief is not supported by theory or actual data. Data shows that human families are equally likely to have a girl or a boy

birth, regardless of gender distribution in previous births. Although statistics guarantee that the sex ratio will balance on average in a large number of cases, this does not mean that it will balance in all individual cases. To make this principle more concrete, consider that in families with five children, those with five boys are equal to those with five girls, giving an overall sex ratio of 1:1. However, these two sex ratios are unusual.

SUMMARY ■ An experiment has elementary outcomes, each with a certain probability of occurring. ■ An event is a combination of elementary outcomes and its probability is the sum of the probabilities of these events occurring. ■ The probability of occurrence of any two mutually exclusive events is equal to the sum of the probabilities of those two events.

■ The probability of two independent events occurring simultaneously is the product of the probabilities of those two events.

3.7 Conditional Probability and Pedigrees Let us now return to pedigree analysis and consider how probabilistic reasoning can be applied to it. The analysis of human family trees creates many situations in which information about an individual is known and can affect assessments of the likelihood that the individual has a specific genotype or phenotype. Think of Huntington's disease, caused by an autosomal dominant allele. The incidence of this disease is approximately 2.7 cases per 100,000 people worldwide, which means that the probability of a given person contracting the disease is 2.7/100,000, or 2.7 × 10−5 if none other information is available. But what if we have some background information about the family tree? Consider the following possibilities:

1. Neither of the person's parents will develop Huntington's disease. In this case, ignoring the possibility of a new mutation occurring for the time being, the individual's chance of developing the disease is zero. 2. One of the person's parents actually develops Huntington's disease. If this is the case, the probability that the individual offspring received the disease allele from the parent and that he or she develops the disease is now 1/2. These are examples of conditional probabilities. A conditional probability can be thought of as the probability that a given event A will occur if event B occurs. For example, we could rephrase Case 2 as: "Assuming one of a person's parents develops HD (event A), what is the probability that he or she will also develop the disease (event B)?" In formal probability notation

This example is relatively simple, as Huntington's disease alleles are dominant and most often fully pervasive. But what if those conditions don't apply to a gene or phenotype of interest? The following two problems provide hypothetical examples.

Problem 1. In the pedigree of FIGURE 3.25, female I-1 is heterozygous for a mutant M gene that increases the risk of a specific disease from 5% in mm genotypes to 50% in Mm genotypes. The M allele is associated with the largest DNA band on the gel (the band closest to the top). The I-2 male is homozygous for the non-mutated form of the gene, and the m allele is associated with the smallest DNA band (the one closest to the bottom). Her daughter II-1 has the disease. What is the probability that II-1 has the Mm genotype?

FIGURE 3.25 The rare M allele in this pedigree, associated with the DNA closest to the top of the gel, increases the risk of disease from 5 percent in homozygous mm genotypes to 50 percent in heterozygous mm genotypes. A typical problem with conditional probability is to calculate the probability that the II-1 daughter has the Mm genotype given that she is affected.

To solve this problem, suppose there are 100 family trees as in Figure 3.25 instead of just one. A simple Mendelian inference tells us that we would expect 50% of their offspring to be Mm and 50mm. If we then consider what we know about the expected frequency of the trait in each of the two genotypes, we would have

Expect 5 of the mm people and 50 of the mm people to be affected (55 people in total). Therefore, the probability that an affected person is Mm is simply the fraction of all affected people who are MM - that is, 50/55 or 0.91. This probability is much greater than the 1/2 = 0.50 expected from Mendelian segregation, and the discrepancy shows that using conditional probability based on prior knowledge that the daughter is affected has a large impact on assessing how likely it is that she has a genotype Mm. The following problem also uses conditional probability, but in a slightly different way.

Exercise 2. Look again at the M mutation in Figure 3.25. Consider a human population where the proportion of people with the Mm genotype is 1% and where homozygous MM individuals are so rare that they can be ignored. Given that one person in this population is affected by the disease, what is the probability that the affected person has the Mm genotype? The individual in question is already known to have the disease state, and given that condition, we need to infer the probability that the genotype of the affected individual is Mm. We will take the same approach, but instead of having prior knowledge based on a pedigree, we will only know the frequency of the allele in the population. To do this, imagine a sample of 2,000 people from this population. Of these, 20 (ie 1 percent) have the Mm genotype, of which 10 (ie 50 percent) will be affected. The remaining 1980 individuals have the mm genotype, of which 99 (ie 5 percent) will be affected. In those affected, the probability of the Mm genotype is 10/(10 + 99) = 10/109 = 0.092. Note that although the Mm strong genotype

increases the risk of disease, but the vast majority of affected people (99 out of 109) still have the mm genotype. For problem 1, the probability that an affected person has the Mm genotype is 0.91, while for problem 2 it is 0.092. These probabilities differ by a factor of about ten. Where does the difference come from? It comes from the very different background information we apply to the problem. In problem 1, we know that the mother has the Mm genotype, so the probability that a child will have the Mm genotype is known in advance to be 0.5 (this is called the prior probability). In Task 2, we considered the population as a whole, and in this case, the prior probability of Mm is given as 0.01 (1 percent). However, despite the very different previous odds, we were able to navigate the problem and get the right answers. The next section discusses conditional probability reasoning in more formal terms.

Bayes' theorem In the two previous examples we used the so-called method of natural frequencies. That is, we generate hypothetical populations in which we calculate posterior probabilities - that is, the chances that affected individuals have the Mm genotype based on their expected frequency in these populations. In fact, it is possible to go directly from prior information, such as we are given in these problems, to posterior probabilities using conditional probabilities. This insight is attributed to Thomas Bayes (1702–1761), a Presbyterian minister in Tunbridge Wells, England. His ideas were summarized in his essay Towards Solving a Problem in the

Doctrine of Chances, published in 1763, two years after his death. To understand his idea, let's return to the concept of conditional probability, which can be formally defined as the probability that A and B occur together (symbolized Pr{AB}) divided by the probability of event B (symbolized Pr{B} ), provided that Pr{B} is not equal to 0. Written as an equation

where Pr{B|A} is expressed in words like "the probability of B given A".

For example, suppose in problem 1 A is the event that individual II-1 in Figure 3.25 is affected and B is the event that individual II-1 has the Mm genotype. We received prior information that the probability of II-1 being affected is 0.50 given its Mm genotype; therefore Pr{A|B} = 0.50. However, what the problem asks us to calculate is the probability that II-1 has the Mm genotype, given that it is affected; this probability is symbolized as Pr{B|A} and is not the same as Pr{A|B}. How can we apply our knowledge to calculate Pr{B|A} using Equation 4? We start with the denominator. It is the probability that A occurs with B or that A occurs with B'; this sum must equal the probability that A occurs, because if A occurs, it must occur in combination with B or with B'. Consequently,

Now we turn to the numerator in Equation 4. It is the probability that A will occur, given that B has occurred (Pr{A|B}, multiplied by the probability that B will occur (Pr{B}). Therefore, we can rewrite Equation 4 this way:

Based on our previous information about the trait and our knowledge of the pedigree, we can solve the problem:

In fact, Equation 5 is Bayes' theorem. To further illustrate its usefulness, consider Problem 2. In this case, we use the previous information that P{an individual is affected | genotype Mm} = 0.50 and Pr{an individual is affected | genotype mm} = 0.05, which tells us to choose A = "an individual is affected" and B = "the individual has genotype Mm". In this problem, the parental genotypes are not informed, but the population frequencies of Mm as Pr{B} = 0.01 and mm as Pr{B′} = 0.99. These considerations provide the necessary factors for the right-hand side of Bayes' theorem (Equation 5):

This result agrees with the previously calculated response using natural frequencies.

Problem 3. As another example of using conditional probability in genetic calculations, consider a problem that uses only the denominator in Equation 5. Consider the human population from Problem 2, which consists of 1% individuals of genotype Mm and 99% of mm . If a person is randomly selected from this population, what is the probability that he or she will be affected? We have already calculated this in a sample of 2000 people in this area

population, 10 with the Mm genotype will be affected and 99 with the mm genotype will be affected. Therefore, the overall probability of an affected person is 109/2000 = 0.0545. Alternatively, we can use the denominator of Equation 5, with A = "an individual is affected" and B = "the individual has genotype Mm", so Pr{A|B} = 0.50, Pr{A|B′} = 0.05 , Pr{B} = 0.01 and Pr{B'} = 0.99. Then,

Note that since the genotype is mm, 0.0545 is not much greater than 0.05, which is the probability that an individual will be affected. The reason the overall risk of being affected is so close to 0.05 is because the Mm genotypes represent such a small proportion of the total population.

For many students, the main problem with Bayes' theorem (Equation 5) is that it looks so intimidating. A frequently asked question is, "How do I know what to choose as event A and what to choose as event B?" The choice of A and B is determined by choosing them so that Pr{A|B} is prior information and Pr{B|A} is the probability you want to know. Which method should you use when solving conditional probability problems — natural frequencies or Bayes' theorem? You must use the method that suits you best. Some students prefer the informal natural frequencies style, while others prefer the more formal approach to Bayes' theorem. Whichever method you choose, you should try to understand both. This is an extremely important concept in relation to genetic risk assessment, especially when the penetrance of a trait is less than 100% or the trait is not a simple Mendelian trait.

SUMMARY ■ Determining conditional probabilities is a critical part of pedigree analysis. ■ Conditional probabilities can be calculated based on the expected frequencies of different outcomes in large hypothetical populations. ■ Bayes' theorem provides a formal means by which conditional probabilities can be calculated based on prior information.

CHAPTER SUMMARY • Hereditary traits are determined by the genes present in reproductive cells mating at fertilization. ■ Genes are usually inherited in pairs: one from the mother and one from the father. ■ Genes in a pair can differ in DNA sequence and in their effect on expression of a specific inherited trait.

■ The genes inherited from the mother and father are not altered because they are together in the same organism. ■ When germ cells are formed, the paired genes separate again into different cells. ■ Random combinations of reproductive cells containing different genes result in proportions of Mendelian traits appearing among the offspring. ■ The actual observed proportions for a trait are determined by the type of dominance and gene interaction. ■ The laws of probability form the basis for predictions based on genetic assumptions.

CHECK OUT THE BASICS ■ What is the principle of segregation and how does this principle show up in the results of a single gene cross (monohybrid)? ■ What is the principle of independent sorting and how is this principle reflected in the results of a cross between two genes (dihybrids)? ■ Explain why the random mating of male and female gametes is necessary for Mendelian separation and independent sorting to be observed in the offspring of a cross. ■ If two pairs of alleles show independent variety, under what conditions is a 9:3:3:1 ratio of phenotypes not observed in the F2 generation? ■ Explain the following statement: “Among the F2 offspring of a dihybrid cross, the genotype ratio is 1:2:1, but among the offspring expressing the dominant phenotype, the genotype ratio is 1:2.” ■ What are the main features of human family trees in which a rare dominant allele segregates? What does a rare recessive allele segregate into?

■ What is a mutant screen and how is it used in genetic analysis? ■ Explain the following statement: “In genetics, a gene is identified experimentally by a set of mutated alleles that lack complementation”. What is complementation? How does a complementation test allow a geneticist to determine whether two different mutations are mutations in the same gene or not? ■ What is meant by the statement that epistasis leads to a “modified dihybrid F2 ratio”? Give two examples of modified dihybrid F2 ratio and explain the gene interactions that result in the modified ratio.

■ What is the difference between mutually exclusive events and independent events? How are the probabilities of these two types of events combined? Give two examples of mutually exclusive genetic events and two examples of independent genetic events. ■ What is a conditional probability? How does Bayes' theorem allow us to calculate one?

TROUBLESHOOTING GUIDE PROBLEM 1 Complete the table by substituting 0, 1/4, 1/2, or 1 for the probability of each genotype of the offspring of each mating type. In what mating are the parents identical in genotype but the offspring as variable in genotype as can be for a single locus? In what mating are the parents as different as can be for a single locus, but the offspring are identical and distinct from either parent?

mating of offspring genotypes

AA × AA AA × Aa AA × aa Aa × Aa Aa × aa aa × aa

ANSWER This type of table is essential to be able to solve almost any type of quantitative problem in transmission genetics.

mating of offspring genotypes

AA × AA

1/2

AA × AA

E × E

1/4

1/2

1/4

Aa × aa

1/2

aa × aa

The mating in which the parents are identical in genotype but the offspring are as variable in genotype as possible for a single locus is Aa × Aa. A mating where the parents are as different as possible for a single locus, but the offspring are identical and distinct from both parents is AA × aa. PROBLEM 2 In domesticated chickens, a dominant C allele is required for feather color, but a dominant I allele of an unlinked gene is a color inhibitor, negating the effect of C. White Leghorns have the CC II genotype, while White Wyandottes have the CC II genotype. cc genotype ii. Both races are white, but for different reasons. In the F2 generation of a cross between White Leghorns and White Wyandottes: (a) What is the white:color phenotypic ratio? (b) What is the proportion of Cc Ii in white F2 chicks?

ANSWER (a) The initial cross of CC II x cc ii gives F1 chickens with genotype Cc II. Because the genes are not linked, the offspring of the F2 generation have a genotype ratio of 9 C— I— : 3 C— ii : 3 cc I— : 1 cc ii. Only chickens with the C-ii genotype will be colored because C is required for colored feathers, but I inhibits the action of C. This type of gene interaction or epistasis results in a phenotypic ratio of 13 white:3 colored. (b) In any dihybrid cross with unlinked genes, the probability that an F2 offspring is heterozygous for both genes is (1/2) × (1/2) = 1/4. Since only 13/16 of the chicks in this example are white, the ratio of Cc Ii among the white chicks is (1/4)/(13/16) = 4/13. You can also get this result by counting squares on the Punnett square for a dihybrid cross: A total of 13 genotypes produce white chicks, and of these four are heterozygous for both genes. PROBLEM 3 In Duroc-Jersey pigs, animals with genotype R—S—are red, those with genotype R—ss or rr S—sand, and those with genotype rr ss are white. Genes show independent variety. In the F2 generation, a cross

between the RR ss and rr SS genotypes: (a) What is the phenotypic ratio of red:non-red? (b) What is the proportion of Rr Ss in F2 red piglets? ANSWER (a) The initial cross RR ss × rr SS gives F1 pigs with genotype Rr Ss. The offspring of the F2 generation will have them

genotypic ratio 9 R— S— : 3 R— ss : 3 rr S— : 1 rr ss. Only pigs with the R—S— genotype will be red, while all others will not be red (some sandy and some white, in a 6:1 ratio). This type of two-gene interaction (epistasis) leads to a phenotypic ratio of 9 red: 7 non-red. (b) 4/9. The proportion of pigs with the Rr Ss genotype is the frequency of those that are heterozygous for both genes, i.e. 1/2 × 1/2 = 1/4. Since only 9/16 of the pigs are red, the ratio of Rr Ss among them is (1/2 × 1/2)/(9/16) = 4/9. You can also get this answer by counting squares on the Punnett square for a dihybrid cross: A total of 9 genotypes produce red pigs, 4 of which are heterozygous for both genes.

PROBLEM 4 The following pedigree shows the inheritance of a rare single dominant Mendelian mutation with a penetrance of 1/3. (A penetrance of 1/3 means that 1/3 of individuals with the mutant genotype actually express the mutant phenotype.) Female I-1 necessarily has the Mm genotype, where M is the mutant allele. Since the mutant allele is rare, you can assume that the I-2 male has the mm genotype. The II-1 person is unaffected. What is the probability that II-1 has the Mm genotype?

ANSWER This is a typical problem using Bayes' theorem. Let A be the event that individual II-1 has the Mm genotype and let B be the event that individual II-1 is unaffected. Then event A' is the event that individual II-1 has the mm genotype. Since I-2 has the Mm genotype, Pr{A} = 1/2 and also Pr{A′} = 1/2. Now we can apply Bayes' theorem:

where Pr{B|A} is the probability that an individual of genotype Mm is not affected and Pr{B|A′} is the probability that an individual of genotype mm is not affected. As the penetrance is 1/3, Pr{B|A} = 2/3. Since the mm genotype is never affected, P{B|A ′} = 1. Putting this all together, we have the answer we are looking for:

An alternative approach avoids the machinery of Bayes' theorem. Since individual I-1 has the genotype Mm, the possible offspring of I-1 are of three types: (1) Mm and affected, with a probability of 1/2 x 1/3 = 1/6; (2) Mm and unaffected, with probability 1/2 × 2/3 = 1/3; and (3) mm and unaffected, with probability 1/2. Since we know that II-1 is not affected, possibility (1) can be ruled out. Thus, the probability that II-1 has the Mm genotype, assuming that she is unaffected, is given by (1/3)/[(1/3 + 1/2)] = 2/5, which agrees with the obtained Answer agrees with Bayes' theorem.

ANALYSIS AND APPLICATIONS 3.1 In an Aa Bb Cc Dd × Aa Bb Cc Dd cross, in which all genes are independently sorted, what percentage of the offspring would be expected to be heterozygous for all four genes? 3.2 Consider a gene with four alleles, A1, A2, A3, and A4. In a cross between A1A2 and A3A4 what is

Probability that a given offspring will inherit A1 or A3 or both? 3.3 Assuming equal numbers of boys and girls, if a mating has already produced a girl, what is the probability that the next child will be a boy? If a mating has already produced two girls, what is the probability that the next child will be a boy? What kind of probability argument are you basing your answers on? 3.4 A cross is performed between the genotypes Aa BB Cc dd Ee and Aa Bb cc DD Ee. How many genotypes of offspring are possible?

3.5 An individual of the AA Bb Cc DD Ee genotype is tested in a cross. Assuming that the loci undergo independent sorting, what proportion of the offspring should have the Aa Bb Cc Dd Ee genotype? 3.6 Homozygous peas with round seeds are crossed with homozygous peas with wrinkled seeds. The F1 offspring undergo self-pollination. A single round seed is randomly selected from the F2 progeny and its DNA analyzed by electrophoresis as described in the text. What is the probability that the observed gel pattern is B?

3.7 Suppose that the trihybrid cross MM SS tt × mm ss TT is produced in a plant species where M and S are dominant, but there is no dominance between T and t.

Consider the F2 progeny of this cross and assume independent reach. (a) How many phenotypic classes are expected? (b) What is the probability of the MM SS tt genotype? (c) What proportion would you expect to be homozygous for all the genes?

3.8 Shown here is a pedigree and a gel graph indicating the clinical phenotypes related to phenylketonuria and the molecular phenotypes related to an RFLP superimposed on the PAH gene for phenylalanine hydroxylase. The II-1 individual is affected. (a) Indicate the expected molecular phenotype of II-1.

(b) Indicate the possible molecular phenotypes of II-2.

3.9 What mode of inheritance does the following family tree suggest? Based on this hypothesis, and assuming that the trait is rare and has complete penetrance, what are the possible genotypes of all individuals in this pedigree?

3.10 Huntington's disease is a rare human neurodegenerative disease characterized by a dominant allele, HD. The disorder usually manifests after age 45. A young man discovered that his father had the disease. (a) What is the probability that the youth will later develop the disorder? (b) What is the probability that one of the young man's children will be a carrier of the DH allele?

3.11 The following gel graph shows the banding patterns observed in the offspring of a cross between the double heterozygous genotypes shown on the left (P1 and P2). Banding patterns observed in the progeny are shown on the right. The bands are numbered from 1 to 4.

(a) Identify which pairs of numbered bands correspond to the two segregating pairs of alleles. (b) Assuming that the two pairs of alleles undergo independent sorting, what is the probability that one

Descending from the cross shows the pattern of stripes in the D band?

3.12 Suppose that the trait in the attached pedigree is due to simple Mendelian inheritance. (a) Is it likely to be due to a dominant allele or a recessive allele? Explain. (b) What does the double horizontal line connecting III-1 and III-2 mean? (c) What is the biological relationship between III-1 and III-2?

(d) If the allele responsible for the disease is rare, what are the most likely genotypes of all individuals in the family tree in generations I, II, and III? (Use A and a for dominant and recessive alleles, respectively.)

3.13 In Drosophila, the mutant allele bwdts, which causes brown eyes (normal eyes are red), is sensitive to temperature. In flies reared at 29°C, the mutant allele is dominant, but in flies reared at 22°C, the mutant allele is recessive. In a cross of bwdts/+ × bwdts/+, where the + sign denotes the bwdts wild-type allele, what is the expected ratio of brown-eyed flies to red-eyed flies when the pups are reared at 29°C? In 22 degrees?

3.14 Pedigree analysis tells you that a given parent can have either the AA BB or AA Bb genotype, each with equal probability. Assuming an independent distribution, what is the probability that this parent will produce an Ab gamete? What is the probability that the parent will produce an AB gamete? 3.15 The pedigree shown here refers to a rare autosomal recessive trait with complete penetrance. You can assume that no one in the pedigree has the recessive allele unless that person inherits it from I-1 or II-4 or both.

(a) Leaving aside siblings in Generation IV, what is the probability that both parents in the first cousin mating are heterozygous? (b) Considering siblings in Generation IV, what is the probability that both parents in the first cousin mating are heterozygous?

3.16 A clinical test for a given disease is 100% accurate in affected individuals, but will also give a false positive result in 0.2% of healthy individuals. If the proportion of people affected in a population is 0.002, what is the overall probability

that a randomly selected person gives a positive test result? 3.17 The accompanying pedigree chart and gel chart show the parental phenotypes for an SSRP with multiple alleles. What are the possible offspring phenotypes and in what proportions are they expected?

3.18 Round pea seed pollen obtained from the F2 generation of a cross between a purebred round seed strain and a purebred shriveled seed line was collected and used en masse to fertilize plants of the shriveled purebred line. What proportion of the offspring should have shriveled seeds? (Assume equal fertility across all genotypes.) 3.19 Cp-cp heterozygous chicks exhibit a condition called creepers, in which the leg and wing bones are shorter than normal (cp-cp). The dominant Cp allele is lethal when homozygous. Two alleles of an independently secreted gene determine white (W-) versus yellow (ww) skin color. From crosses between chickens heterozygous for these two genes, the phenotypic classes are represented

among viable offspring, and what are their expected relative abundances? 3.20 The F2 progeny of a given cross have a modified dihybrid ratio of 9:7 (instead of 9:3:3:1). What phenotypic ratio would you expect from a test cross of F1 offspring? 3.21 In the Aa × Aa mating, what is the smallest number of offspring, n, for which the probability of at least one aa offspring exceeds 95%?

3.22 A female is affected by a trait due to a dominant mutant allele that has 50% penetrance. If she has a child, what is the likelihood that he will be affected? 3.23 The coat color pattern in dogs is determined by the alleles of a single gene, with S (solid) dominating over s (spotted). Black coat color is determined by the dominant A allele of a second gene, and homozygous recessive aa animals are light brown. A female with a solid brown coat mates with a male with a solid black coat and produces a litter of six pups. The puppies' phenotypes are 2 solid brown, 2 solid black, 1 spotted brown, and 1 spotted black. What are the genotypes of the parents? 3.24 Consider a phenotype in which the N allele is dominant over the n allele. An Nn x Nn mating is performed and an individual with the dominant phenotype is randomly selected. This individual is tested and the mating produces four offspring, each with the dominant phenotype. What is the probability that

Does the parent with the dominant phenotype have the Nn genotype?

3.25 Cystic fibrosis is a genetic disease resulting from homozygosity for one of a large number of mutated alleles. See accompanying family tree. Person 2-2 wants to know if they are a carrier (ie, heterozygous) for a cystic fibrosis allele, so they undergo a genetic test that detects 90% of these alleles in a population. This test is negative. What is the probability that she is a carrier despite these results?

3.26 The following gel graph shows the phenotype of two parents (X and Y), each homozygous for two RFLPs that undergo independent sorting. Parent X has the A1A1 B1B1 genotype, with the A1 allele giving a 4 kb band and the B1 allele giving a 6 kb band. Parent Y has the A2A2 B2B2 genotype, with the A2 allele giving an 8 kb band and the B2 allele giving a 2 kb band. Show the expected phenotype of the F1 offspring and all possible phenotypes of the F2 offspring along with their expected proportions.

3.27 Complementation tests of recessive mutant genes a through f produced the data in the attached matrix. Circles represent missing data. Assuming that all missing mutant combinations would produce data consistent with the known inputs, complete the table by filling each circle with + or − as appropriate.

3.28 In plants, certain mutated genes are known to affect the ability of gametes to participate in fertilization. Assume that an A allele is one such mutation, and that pollen cells that carry the A allele are only half as likely to survive and participate in fertilization as pollen cells that carry the a allele. Complete the Punnett square for the F2 generation in a monohybrid cross. What is the expected ratio of AA:Aa:aa plants in the F2 generation?

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 Meiotic drive is an unusual phenomenon in which two alleles do not show Mendelian separation from the heterozygous genotype. Examples are known from mammals, insects, fungi and other organisms. The usual mechanism is that both types of gametes are formed, but one of them fails to function normally. The excess of one driver allele over the other can vary from a small amount to almost 100%. Assume that D is an allele that exhibits a meiotic drive against its alternative allele d, and assume that heterozygous Dd produce functional D-carrying and d-carrying gametes in a 3/4:1/4 ratio. When crossing Dd × Dd: (a) What are the expected proportions of the DD, Dd, and dd genotypes? (b) If D is dominant, what are the expected proportions of the D− and dd phenotypes? (c) What is the DD : Dd ratio for the D− phenotypes?

(d) Answer parts (a) through (c) assuming that the meiotic drive occurs in only one sex. CHALLENGE PROBLEM 2 The attached table summarizes the effect of inherited tissue antigens on the acceptance or rejection of transplanted tissues, such as skin grafts, in mammals. Tissue antigens are codominantly determined, so tissue taken from an Aa genotype donor carries both the A antigen and the a antigen. In the table, the + sign means that a donor tissue graft was accepted by the recipient and the − sign means that a donor tissue graft was rejected by the recipient. The rule is:

Any transplant will be rejected if the donor's tissue contains an antigen that is not present in the recipient. In other words, any transplant will be accepted if and only if the donor's tissue does not contain antigen different from those already present in the recipient.

The diagram shown here shows all possible skin grafts between inbred (homozygous) strains of mice (P1 and P2) and their F1 and F2 offspring. Assume that the inbred lineages P1 and P2 differ only in this

a tissue compatibility gene. For each of the arrows, what is the probability of accepting a transplant where the donor is a randomly selected animal from the population indicated at the bottom of the arrow and the recipient is a randomly selected animal from the population indicated by the arrow?

CHALLENGE QUESTION 3 A woman carrying one copy of the BRCA1 mutant allele has a 60% chance of developing breast cancer during her lifetime, while a woman who is homozygous for the normal allele has a 12% risk (mutant homozygotes are not viable). In the accompanying pedigree, individual II-3 has developed breast cancer and carries a mutant allele

of BRCA1. Her brother (II-2) is deceased but married and had a daughter (III-1) with an unrelated female who carries two normal alleles based on genotype. Based on the pedigree and risk factors associated with different genotypes, what is the probability that the III-1 person will develop breast cancer?

FOR FURTHER READING

Mendel, Gregor Johann. (1866). Experiments on hybrid plants. In Proceedings of the Natural History Society of Brno, IV(1865), 3–47. Reprinted in Fundamenta Genetica, editor Jaroslav Krˇízˇenecky', 15–56. Prague, Czech Republic: Czech Academy of Sciences, 1966. http://www.mendelweb.org/Mendel.html Original article by Mendel. Reid, J.B., & Ross, J.J. (2011). Mendel's genes: Towards a complete molecular characterization. Genetics, 189(1), 3-10. http://doi.org/10.1534/genetics.111.132118

What have we learned about the molecular basis of the seven genes characterized by Mendel. Sandell, M.A. & Breslin, P.A.S. (2006). Variability in a taste receptor gene determines whether we taste toxins in food. Current Biology, 16(18), R792-R792. http://www.cell.com/currentbiology/pdf/S0960-9822(06)02061-6.pdf A brief note describing the relationship between the PTC taste allele of hTASR38 and the perception of bitterness in foods. Shoemaker, J.S., Painter, I.S., & Weir, B.S. (1999). Bayesian statistics in genetics: a guide for the uninitiated. Trends in Genetics, 15(9), 354-358. http://doi.org/10.1016/S0168-9525(99)01751-5

A brief introduction to the application of Bayesian probability to genetic problems.

CHAPTER 4 The Chromosomal Basis of Inheritance CHAPTER OVERVIEW 4.1 The Stability of Chromosomal Complements 4.2 Mitosis 4.3 Meiosis 4.4 Sex Chromosomal Inheritance 4.5 Probability in Predicting the Distribution of Descendants

4.6 Testing fit to a genetic hypothesis ROOTS OF DISCOVERY: grasshopper, grasshopper

E. Eleanor Carothers (1913) The Mendelian Ratio to Certain Orthopteran Chromosomes ROOTS OF THE DISCOVERY: The White-Eyed Man Thomas Hunt Morgan (1910) Sex-Limited Inheritance in Drosophila

LEARNING OBJECTIVES AND SCIENTIFIC SKILLS Understanding how chromosomes and sex chromosomes are inherited and the basic principles of probability and statistical hypothesis testing discussed in this chapter will enable you to accomplish the following science skills: ■ Predict which products of mitosis or meiosis would result from chromosomes normal behavior or chromosomal misconduct. ■ Recognize the distinct pattern of X-linked inheritance and be able to trace X chromosomes as they are passed down from generation to generation, sex to sex.

■ In a genetic cross, use the binomial distribution to calculate the probability of a given combination of genotypes or phenotypes among the offspring. ■ Be able to formulate a genetic hypothesis to predict the expected results of a cross and compare the expected results with the observed results using a chi-square test of goodness of fit.

It wasn't a big reveal that genes are located on chromosomes. The parallel between their properties made this clear:

1. Genes come in pairs; Chromosomes come in pairs. 2. Segregate alleles of a gene; separate homologous chromosomes.

3. Unconnected genes undergo independent screening; non-homologous chromosomes undergo independent sorting. These parallels were recognized by Walter Sutton in 1902 on the basis of his observations of spermatogenesis in grasshoppers (Brachystola magna); in his words: “The pairwise union of paternal and maternal chromosomes and their subsequent separation during reductive division, as indicated above, may represent this

the physical basis of Mendel's law of inheritance. There was little doubt thereafter that chromosomes are the cellular carriers of genes—but the parallels do not constitute scientific evidence, nor is there widespread consensus among scientists. In this chapter, we examine some of the experimental evidence that was—and still is—considered sufficient to support the chromosomal theory of inheritance.

Nucleus INTERPHASE (left) of a cell in the bell (right) (genus Hyacinthoides). Source: (Left) © Pr. G. Giménez-Martín/Science Source, (Right) ©

Select Photo/Dreamstime.com.

4.1 The stability of chromosome complements The nucleus of the cell was first discovered in 1831, but it was not until the late 1860s that it was understood that nuclear division almost always coincided with cell division. The importance of the cell nucleus in heredity was reinforced by the almost simultaneous discovery

that the nuclei of two gametes fuse in fertilization. The next big breakthrough came in the 1880s with the discovery of chromosomes, which could be easily visualized under a light microscope using certain dyes. A few years later, it was discovered that chromosomes divide by an orderly process into daughter cells formed by cell division and gametes formed by the division of reproductive cells. Three important regularities have been observed in the chromosome set (the complete set of chromosomes) of plants and animals: 1. The nucleus of each somatic cell (a body cell, as opposed to a germ cell or germ cell) contains a fixed number of specific chromosomes for the species in question. This number varies enormously between species, and the number of chromosomes bears little relation to the complexity of the organism (TABLE 4.1). 2. The chromosomes in the nuclei of somatic cells are usually found in pairs. For example, the 46 human chromosomes are made up of 23 pairs (FIGURE 4.1). Cells with nuclei containing two similar sets of chromosomes are called diploid. A diploid individual carries two allelic copies of each gene present on each pair of chromosomes. Chromosomes come in pairs, as one chromosome of each pair comes from the organism's maternal father and the other from the organism's paternal father.

3. Gametes, which unite at fertilization to produce diploid somatic cells, have nuclei that contain only one set of chromosomes, consisting of one member of each pair. Gametic nuclei are haploid.

TABLE 4.1 Somatic (diploid) chromosome numbers of some plant and animal species Organism

chromosome

Body

horsetail numerical field

216

Cromossomenzahl

Hefe (Saccharomyces

cerevisiae) Adlerfarn

116

A fruit fly (Drosophila

melanogaster) giant sequoia

nematode

11♂, 12♀

(Cenorhabdite

elegans) wheat pasta

housefly

wheat bread

Scorpion

fava

Geometrid-Motte

224

peas

common toad

mustard cress

Chicken

More (Zea Mays)

mausoleum

Lily

doublet

Löwenmaul

Human

(Arabidopsis thaliana)

FIGURE 4.1 Male chromosome complement. There are 46 chromosomes that exist in 23 pairs. At the stage of the division cycle at which these chromosomes were observed, each chromosome consists of two identical halves that are juxtaposed longitudinally. With the exception of the members of a chromosome pair (the pair that determines sex), the members of all chromosome pairs are the same color because they contain DNA molecules labeled with the same mixture of fluorescent dyes. Colors vary from pair to pair as the dye blends are different colors. In some cases, the long and short arms were marked with different colors. Courtesy of Michael R. Speicher, Institute of Human Genetics, Medical University of Graz.

In multicellular organisms that develop from single cells, the presence of the diploid number of chromosomes in somatic cells and the haploid number of chromosomes in germ cells indicates that there are two processes of nuclear division that differ in their outcome. One of them (mitosis) maintains the number of chromosomes; the other (meiosis) cuts the number in half. These two processes are examined in the following sections.

SUMMARY ■ The behavior of chromosomes in the formation of gametes is analogous to that of alleles in Mendelian inheritance.

■ Somatic cells are typically diploid, while germ cells are haploid.

4.2 Mitosis Mitosis is a process of nuclear division that ensures that each of the two daughter cells receives a diploid complement of chromosomes identical to the diploid complement of the parent cell. Mitosis is usually accompanied by cytokinesis, the process by which the cell divides to give rise to two daughter cells. The basic process of mitosis is remarkably uniform across all organisms:

1. Each chromosome already exists as a duplicate structure at the start of nuclear division. (The duplication of each chromosome coincides with the replication of the DNA molecule it contains.) 2. Each chromosome splits lengthwise into identical halves that pull away from each other. 3. The separated chromosome halves move in opposite directions and each is enclosed in one of the two daughter nuclei formed. In a cell that does not undergo mitosis, the chromosomes are invisible under a light microscope. Each consists of an elongated thread that is too fine to see. This phase of the cell cycle is called interphase. In preparation for mitosis, the DNA in chromosomes is replicated during a phase of interphase called S (FIGURE 4.2), which stands for DNA synthesis. DNA replication is accompanied by the duplication of chromosomes. Before and after S are periods, called G1 and G2, respectively, in which no DNA replication occurs. The cell cycle (the life cycle of a cell) is usually described in terms of these three periods of interphase, followed by mitosis, M. The order of events is therefore G1 → S → G2 → M, as shown in Figure 4.2 . In this representation, the period M also includes the division of the cytoplasm (cytokinesis) into two approximately equal parts, each containing a daughter nucleus.

FIGURE 4.2 The cell cycle of a typical mammalian cell growing in tissue culture with a generation time of 24 hours.

The time required for a complete life cycle varies by cell type; for most cells in higher eukaryotes it is 18 to 24 hours. The relative lengths of different cycle periods also vary considerably with cell type. Mitosis is usually the shortest time, lasting 0.5-2 hours. The cell cycle is an active, regulated process governed by mechanisms that are essentially identical in all eukaryotes. These mechanisms are discussed in detail in the chapter Molecular Genetics of the Cell Cycle and Cancer. The transitions from G1 to S and from G2 to M are called checkpoints because they are delayed if key processes do not complete (Figure 4.2). For example, at the G1/S checkpoint, either (in some cell types) enough time must have elapsed since the previous mitosis, or (in other cell types) the cell must have reached sufficient size to begin DNA replication. Likewise, the G2/M checkpoint requires

DNA replication and repair of DNA damage are complete by the onset of M phase. FIGURE 4.3 shows the main features of chromosomal behavior in mitosis. Mitosis is conventionally divided

into four stages: prophase, metaphase, anaphase, and telophase. It is important to emphasize that DNA replication (S phase) precedes meiosis, so mitosis requires that these replicated chromosomes be correctly sorted into two daughter nuclei.

FIGURE 4.3 Chromosomal behavior during mitosis in an organism with two pairs of chromosomes (red/pink versus green/blue). At each stage, the smaller inner diagram represents the entire cell and the larger diagram is an exploded view showing the chromosomes at that stage.

Prophase At interphase, the chromosomes are in the form of elongated filaments and cannot be seen as discrete bodies under a light microscope. In addition to one or more conspicuous dark bodies (nucleolus), the nucleus has a diffuse granular appearance.

The beginning of prophase is marked by the condensation of chromosomes to form visibly distinct strands within the nucleus. Chromatin condensation is caused by large protein complexes known as condensins. At this stage, each chromosome is already duplicated lengthwise and consists of two closely linked subunits called chromatids. The longitudinally bipartite nature of each chromosome is easily recognized later in prophase. Each pair of chromatids is the product of the duplication of a chromosome in the S period of interphase. The chromatids of a pair are held together in a specific region of the chromosome called the centromere. As prophase progresses, chromosomes become shorter and thicker due to complicated coiling (see chapter Molecular Organization of Chromosomes and Genomes). At the end of prophase, the nucleoli disappear and the nuclear envelope, a membrane surrounding the nucleus, abruptly dissolves.

PROPHASE von Hyacinthoides © Pr. G. Giménez-Martín/Science Source

Metaphase At the beginning of metaphase, the mitotic spindle forms. This spindle is an elongated, football-shaped bundle of spindle fibers composed primarily of microtubules formed by polymerization of the protein tubulin. Many other proteins and at least one RNA-protein complex regulate tubulin polymerization and microtubule organization. The ends or poles of the spindle, where microtubules converge, mark the sites of centrosomes, which are the microtubule organizing centers where tubulin polymerization is initiated. Each pair of centrosomes results from the duplication of a single centrosome, which occurs at interphase, followed by the migration of daughter centrosomes to opposite sides of the nuclear envelope.

Metaphase of Hyacinthoides

The spindle has three types of microtubules: (1) those that anchor the centrosome to the cell membrane, (2) those that protrude between centrosomes, and (3) those that attach to chromosomes. The way they are established illustrates an important organizing principle in biology known as exploration and stabilization. As microtubules form, new tubulin subunits are added to the growing end of the polymer, but the growing end can also become unstable and initiate a depolymerization process that shrinks the microtubules. multiple proteins

help regulate the balance between polymerization and depolymerization. In spindle formation, microtubules grow from the poles of the spindle in essentially random directions (the exploratory part of the process), but unless something is done to stabilize the growing end, each polymer will eventually depolymerize and disappear. For microtubules that attach to chromosomes, the event that stabilizes the growing end is contact with a structure known as the kinetochore, which coincides with the position of the centromere. The process of random exploration and stabilization leads to the fact that only the chromosomal microtubules that are in contact with a kinetochore are stabilized, while the others depolymerize. Analogous types of stabilization are also likely responsible for microtubules that attach to the cell membrane and those that protrude between centrosomes.

After the spindle fibers are attached to the chromosomes, each chromosome moves to a position near the center of the cell, where its kinetochore lies in an imaginary plane approximately equidistant from the spindle poles. This imaginary plane is called the metaphase plate. When aligned on the metaphase plate, the chromosomes reach their maximum compaction and are easier to count and examine for differences in shape and appearance. Proper chromosome alignment is an important cell cycle checkpoint at metaphase, both in mitosis and meiosis. In a cell where a chromosome is attached to only one pole of the spindle, completion of metaphase is delayed. The signal for proper chromosomal alignment comes from the kinetochore, and the chemical nature of this signal appears to be the dephosphorylation of certain kinetochore-associated proteins. Through the signaling mechanism when all kinetochores are energized and

aligned at the metaphase plate, the metaphase checkpoint is passed and the cell continues the division process.

Anaphase In anaphase, the proteins that hold the chromatids together are broken down. The centromeres separate and the two sister chromatids of each chromosome move to opposite poles of the spindle. Once the centromeres separate, each becomes a sister chromatid

is considered a separate chromosome. Chromosomal movement results in part from a progressive shortening of the spindle fibers attached to the centromeres, pulling the chromosomes in opposite directions toward the poles, and often also from a transient elongation of the dividing cell in a direction parallel to the spindle. At the end of anaphase, the chromosomes are in two groups near opposite poles of the spindle. Each cluster contains the same number of chromosomes as were present in the original interphase nucleus.

ANAPHASE of Hyacinthoides. ©Pr. G. Gimenez-Martin/Science Source

Telophase In telophase, a nuclear envelope forms around each compact group of chromosomes, nucleoli form, and the spindle disappears. Chromosomes undergo a condensation reversal until they are no longer visible as separate entities. The two daughter nuclei slowly assume a typical interphase appearance as the cell's cytoplasm divides in two by a furrow that gradually deepens around the periphery. (In plants, a new cell wall is made between the daughter cells, separating them.)

TELOPHASE de Hyacinthoides © Pr. G. Gimenez-Martin/Science Source

SUMMARY ■ Mitosis, the process of somatic cell division, results in the production of two identical daughter nuclei. ■ In the cell cycle, DNA replication precedes mitosis. ■ Mitosis has four stages – prophase, metaphase, anaphase and telophase.

4.3 Meiosis Meiosis is the mode of cell division that results in haploid daughter cells containing only one member of each chromosome pair. Unlike mitosis, where daughter cells are genetically identical, meiosis produces genetic diversity because each daughter cell contains a different set of alleles. Meiosis consists of two consecutive nuclear divisions. Overview of the Chromosome

The behavior is sketched in FIGURE 4.4.

FIGURE 4.4 Overview of the behavior of a single pair of homologous chromosomes in meiosis. The main differences for

Mitosis is the pairing of homologous chromosomes (A) and the two consecutive nuclear divisions (B and D) that reduce the number of chromosomes by half. These two stages are commonly referred to as meiosis I and meiosis II. For clarity, this diagram does not include crossing over, an exchange of chromosome segments that occurs at the stage shown in part A.

1. Before the first nuclear division, the members of each pair of homologous chromosomes become tightly joined along their length (Figure 4.4). Each member of the pair is already replicated and consists of two sister chromatids joined at the centromere. The pairing of homologous chromosomes therefore creates a four-strand structure. 2. In the first nuclear division, homologous chromosomes are separated from each other, with members of each pair going to opposite poles of the spindle (Figure 4.4B). Each daughter chromosome consists of two chromatids attached to a common centromere (Figure 4.4C), so that both resulting nuclei contain a haploid set of chromosomes. (Chromosomes are counted by counting the number of centromeres, not the number of chromatids.) 3. The second nuclear division closely resembles a mitotic division, but there is no DNA replication. At metaphase, the chromosomes line up at the metaphase plate; In anaphase, the chromatids of each chromosome are separated into opposite daughter nuclei (Figure 4.4D). The net effect of the two divisions in meiosis is the formation of four haploid daughter nuclei, each containing the equivalent of a single sister chromatid from each pair of homologous chromosomes (Figure 4.4E).

Figure 4.4 does not show that paired homologous chromosomes can exchange genes. The exchange results in the formation of chromosomes, which are composed of segments of one homologous chromosome mixed with segments of the other. In Figure 4.4, the swapped chromosomes would be represented as segments of alternating colors. The exchange process, which is one of the critical features of meiosis, is discussed in the next section. In animals, meiosis occurs in specific cells called meiocytes.

a general term for the primary oocytes and spermatocytes in gamete-forming tissues (FIGURE 4.5). Oocytes form eggs and spermatocytes form sperm. Although the process of meiosis is similar in all sexually reproducing organisms, in female animals and plants only one of the four products develops into a functioning cell, while the other three decay. In animals, the products of meiosis form sperm or eggs.

FIGURE 4.5 The life cycle of a typical animal. The number n is the number of chromosomes in the haploid chromosome set. In males, the four products of meiosis develop into functional sperm; in females, only one of the four products develops into an egg.

In plants, the situation is a little more complicated: 1. The products of meiosis normally form spores that undergo one or more mitotic divisions to produce a haploid gametophyte organism. The gametophyte produces gametes by mitotic division of a haploid nucleus (FIGURE 4.6). 2. The fusion of haploid gametes creates a diploid zygote which develops into the sporophyte plant which undergoes meiosis to produce spores, thus restarting the cycle.

FIGURE 4.6 The life cycle of maize, Zea mays. As is typical of higher plants, the diploid spore-producing generation (sporophyte) is conspicuous, while the gamete-producing generation (gametophyte) is microscopic. The egg-producing spore is the megaspore and the sperm-producing spore is the microspore. Nuclei participating in meiosis and fertilization are shown in yellow and green, respectively.

In higher plants, such as corn and Mendel's peas, it is the sporophyte generation that makes up most of the plant material; The gametophyte is microscopic. In other plants, such as ferns, mosses, and liverworts, the situation is reversed, so that most of the plant consists of haploid gametophyte tissue. Meiosis is a more complex and significantly longer process than mitosis, typically taking days or even weeks to complete. FIGURE 4.7 gives an overview of the process. The essence of the process is that meiosis consists of two departments

the nucleus of the cell, but only a replication of the chromosomes. The nuclear divisions – referred to as the first meiotic division and the second meiotic division – can be divided into a sequence of stages similar to those used to describe mitosis. The characteristic events of this important process, which occurred during the first division of the nucleus, are described below.

FIGURE 4.7 Chromosomal behavior during meiosis in an organism with two homologous pairs of chromosomes (red/pink and green/blue). At each stage, the small diagram represents the entire cell and the larger diagram is an expanded view of the chromosomes at that stage.

The first meiotic division: reduction The first meiotic division (meiosis I) is sometimes called a reductional division because it halves the number of chromosomes. By analogy with mitosis, the first meiotic division can be divided into four stages called prophase I, metaphase I, anaphase I, and telophase I.

more complex than their mitotic counterparts. The steps and sub-steps can be visualized through Figure 4.7 and Figure 4.8.

FIGURE 4.8 Substages of prophase I in lily (Lilium longiflorum) microsporocytes. (A) Leptotes: Chromosomal condensation is initiated and pearly chromomeres are visible along the length of the chromosomes. (B) Zygotene: pairing (synapsis) of homologous chromosomes occurs (paired and unpaired regions are particularly evident in the lower left corner of this photograph). (C) Precocious pachytene: synapse is complete and crossing over between homologous chromosomes occurs. (D) Late pachytene: Continuous shortening and thickening of chromosomes. (E) Diplotene: Mutual rejection of paired homologous chromosomes that remain joined at one or more crossing points (chiasmata) along their length. Diakinesis (not shown) ensues: the chromosomes reach their maximum contraction. (F) Zygotene (at higher magnification in another cell) showing paired homologues and matching chromomeres across the synapse. Parts A, B, C, E, and F courtesy of Marta Walters and Santa Barbara Botanic Garden, Santa Barbara, California. Part D courtesy of Herbert Stern. Used with permission from Ruth Stern.

Prophase I.

This long phase lasts several days in most higher organisms. It is commonly divided into five substages: leptotene, zygotene, pachytene, diplotene, and diakinesis. These terms describe what the chromosomes look like at each substage. In the leptotene period, which literally means "thin thread," chromosomes first appear as long thread-like structures. Sister chromatid pairs can be distinguished by electron microscopy. At this early stage of the chromosome

Condensation, numerous dense grains appear at irregular intervals along its length. These localized contractions, called chromomeres, have a characteristic number, size, and position on a given chromosome (Figure 4.8A).

The period of zygotene is characterized by lateral pairing or synapsis of homologous chromosomes, starting at the ends of the chromosomes. (The term zygotene means “paired strands.”) As pairing progresses in a zipper-like fashion along the length of the chromosome, it results in precise chromomer-by-chromomere association (Figure 4.8B and F). The synapse is facilitated by the synaptonemal complex, a protein structure that helps keep homologous chromosomes aligned together. Any pair of synapsed homologous chromosomes is called a bivalent.

During the pachytene period (Figure 4.8C and D), whose name literally means "thick thread", the chromosomes continue to shorten and thicken (Figure 4.7). In late pachytenes, it can sometimes be seen that each bivalent (i.e., each set of paired chromosomes) actually consists of a tetrad of four chromatids, but the two sister chromatids of each chromosome are usually very close together. Genetic exchange by mating occurs during pachytene. In Figure 4.7, the exchange sites are indicated by the spots where the chromatids are of different types.

the colors intersect.

At the beginning of the diplotene period, the synaptonemal complex breaks down and the synapsed chromosomes begin to separate. Diplotene means "double stranded" and diplotene chromosomes are now clearly duplicated (Figure 4.8E). Pairs of homologous chromosomes are held together at intervals by cross-connections that result from crossover. Each cross-connection is called a chiasm (plural, chiasms) and is formed by the breakup and reunion of non-sister chromatids. As shown by the chromosome and diagram in FIGURE 4.9, a chiasm results from physical exchanges between chromatids of homologous chromosomes. In normal meiosis, each bivalent usually has at least one chiasm, and long-chromosome bivalents usually have three or more chiasms.

FIGURE 4.9 Photomicrograph (A) and interpretation drawing (B) of a bivalent composed of a pair of homologous chromosomes. This bivalent was photographed in a spermatocyte of a salamander, Oedipina poelzi, at the end of Diplotene. It shows two chiasmata in which the chromatids of homologous chromosomes appear to exchange pairing partners.

Courtesy of Dr. James Kezer. With the permission of Dr. Stanley K. Used sessions.

The final phase of prophase I is diakinesis, in which homologous chromosomes appear to repel each other and segments not connected by chiasmata move apart. (Diakinesis means “divergent.”) In this substage of prophase I, the chromosomes reach their maximum compactness. The homologous chromosomes in each bivalent remain connected by at least one chiasm, which persists until the first meiotic anaphase. At the end of diakinesis, a spindle begins to form and the nuclear envelope breaks down.

Metaphase I. Each bivalent is maneuvered into a position spanning the metaphase plate with the centromeres of homologous chromosomes aligned at opposite poles of the spindle (FIGURE 4.10A). Centromere orientation determines which member of each bivalent subsequently moves to each pole and whether the maternal or paternal centromere is aligned.

a specific pole is pure coincidence.

FIGURE 4.10 Posterior meiotic stages in microsporocytes of the lily Lilium longiflorum: (A) Metaphase I; (B) anaphase I; (C) metaphase II; (D) anaphase II; (E) Telophase II In telophase, cell walls are formed that will lead to the formation of four pollen grains. Courtesy of Herbert Stern. Used with permission from Ruth Stern.

As shown in Fig. 4.11, bivalents formed from non-homologous pairs of chromosomes can be oriented in two ways at the metaphase plate. If each of the nonhomologous chromosomes is heterozygous for a pair of alleles,

then one type of alignment gives rise to A B and a b gametes, while the other type gives rise to A b and a B gametes. Because metaphase alignment is random, the two types of alignment—and therefore the four types of gametes—are equally common. The ratio of the four types of gametes is 1:1:1:1, which means that the allele pairs A, a and B, b are subject to independent classification. That is, genes on different chromosomes are subject to

independent assortment, since nonhomologous chromosomes are randomly arranged on the metaphase plate in meiosis I.

FIGURE 4.11 The independent arrangement of genes (A or B) on nonhomologous chromosomes results from random alignment of nonhomologous chromosomes at metaphase I.

Roots of the Grasshopper Discovery, Grasshopper E. Eleanor Carothers (1913) University of Kansas, Lawrence, Kansas

The Mendelian relationship with regard to specific Orthoptera chromosomes As a graduate student, Carothers showed that non-homologous chromosomes undergo independent sorting in meiosis. To do this, she studied a species of grasshopper in which a pair of homologous chromosomes had links of unequal length. In the first anaphase of meiosis in males, she could tell by looking at whether the longer or shorter chromosome was going in the same direction as the X chromosome. As described in this article, she found 154 of the former and 146 of the latter - a result that agrees very well with the 1:1 ratio expected from an independent assortment. The Y chromosome is not mentioned because in the grasshopper she studied, the females have the XX sex chromosome constitution, while the males have the X sex chromosome constitution. In the males she studied, the X chromosome had no mating partner.

“Another link is added to the already long chain of evidence from Hartl, Daniel L. and Bruce Cochrane. Genetics: Analysis of Genes and Genomes: Analysis of Genes and Genomes, Jones & Bartlett Learning, LLC, 2017. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/utah/detail.action?docID=5208967. Created by utah on 2021-08-09 19:18:30.

"Another link is added to the already long chain of evidence that chromosomes are morphologically distinct individuals that are continuous from generation to generation and, as such, the

”

Carrier of hereditary characteristics.

The instrument, known as a camera lucida, was widely used at the time to examine chromosomes and other microscopic objects. It is an optical instrument containing a prism, or array of mirrors, which when mounted on a microscope reflects an image of the microscopic object onto a piece of paper where it can be traced.

The objective of this work is to describe the behavior of an unequal bivalent in the primary spermatocytes of certain grasshoppers. The distribution of the chromosomes of this bivalent in relation to the X chromosome follows the laws of chance; and therefore provides direct cytological support for Mendel's laws. This distribution is easy to follow due to a very clear difference in the size of homologous chromosomes. Thus, one more link is added to the already long chain of evidence that chromosomes are morphologically distinct individuals, continuous from generation to generation and, as such, are carriers of hereditary characteristics... This work is based mainly on Brachystola magna [a shorthorn grasshopper]…. The entire chromosome complex can be divided into two groups, one containing six small chromosomes and the other seventeen larger ones. [One of the largest is the X chromosome.] Examination shows that this group of six small chromosomes consists of five chromosomes of approximately the same size and one much larger. [One of the small ones is the counterpart of the much larger one, making this pair of chromosomes unequal in size.] … In early metaphases, the

The chromosomes appear as twelve separate individuals [the bivalents]. Side views show the X chromosome in its characteristic position near a pole…. Three hundred cells were drawn under the camera lucida to determine the distribution of chromosomes in asymmetric binary with respect to the X chromosome…. In the 300 cultured cells, the minor chromosome went to the same nucleus as the X chromosome 146 times, or 48.7% of the time; and the largest 154 times, or 51.3 percent of the time…. Considering the limited number of chromosomes and the large number of traits in any animal or plant, it is clear that each chromosome must control many different traits....

Since the rediscovery of Mendelian laws, increasing knowledge has reconciled facts that at first seemed totally incompatible with them. There is no cytological explanation for any other form of inheritance…. It seems to me likely that all inheritance is actually Mendelian. Note that Carothers made no connection between the X chromosome and sex - that understanding would come later. Important was his work's contribution to the growing body of evidence on the chromosomal basis of heredity. Source: E.E. Carothers, J. Morphol. 24 (1913): 487-511.

Anaphase I. At this stage, the homologous chromosomes, each consisting of two chromatids held together by an undivided centromere, separate and move to opposite poles of the spindle (Figure 4.10B). Chromosomal segregation at anaphase is the cellular basis of allele segregation:

The physical separation of homologous chromosomes at anaphase is the physical basis of Mendel's principle of separation.

Note, however, that the centromeres of the sister chromatids are closely linked and behave as one. A specific protein acts as the glue that holds the sister centromeres together. This protein appears in centromeres and adjacent chromosome arms during S phase and persists until meiosis I. It disappears only in anaphase II, when sister centromere cohesion is lost and the sister centromeres separate.

Telophase I. At the end of anaphase I, near each pole of the spindle, there is a haploid set of chromosomes consisting of one homolog of each bivalent (Figure 4.6). At telophase, the spindle collapses and, depending on the species, a nuclear envelope briefly forms around each group of chromosomes, or the chromosomes enter the second meiotic division after only limited unfolding.

The second meiotic division: equation The second meiotic division (meiosis II) is sometimes called division in equation because the number of chromosomes in each cell remains the same before and after the second division. In some species, chromosomes pass directly from telophase I to prophase II without loss of condensation; in others, there is a brief pause between the two meiotic divisions and the chromosomes may "decondense" (uncoil) a bit. Chromosomal replication never occurs between the two divisions; The chromosomes present at the beginning of the second division are identical to those at the end of the first. After a brief prophase (prophase II) and the formation of two-division spindles, the centromeres of the chromosomes in each nucleus line up in the midplane of the spindle at metaphase II (Figure 4.10C). In anaphase II, protein retention

collapse of the sister centromeres. As a result, the sister centromeres appear to split longitudinally and the chromatids of each chromosome move to opposite poles of the spindle (Figure 4.10D). Since the centromeres divide in anaphase II, each chromatid is considered a separate chromosome.

Telophase II (Figure 4.10E) is characterized by a transition to the interphase state of the chromosomes in the four haploid nuclei, accompanied by a division of the cytoplasm. Thus, the second meiotic division superficially resembles a mitotic division. However, there is an important difference: the chromatids of a chromosome are usually not genetically identical along their entire length, as the crossing occurs during the prophase of the first division, which is accompanied by the formation of chiasmata.

SUMMARY ■ Meiosis, the process of gamete formation, results in the production of four haploid cell nuclei. ■ In meiosis, there are two cycles of nuclear division: reduction (meiosis I) and equation (meiosis II). ■ Meiosis I leads to the production of two haploid nuclei and the independent arrangement of genes on different chromosomes. ■ Meiosis II parallels mitosis in the division of centromeres, resulting in the production of haploid daughter cells with identical genetic content.

■ During prophase of meiosis I, homologous chromosomes form chiasmata, resulting in the exchange of genetic material between them.

4.4 Inheritance of sex chromosomes The first rigorous experimental evidence that genes are parts of chromosomes was obtained in experiments that looked at the transmission pattern of sex chromosomes, the chromosomes responsible for determining the different sexes in some plants and in almost all plants. animals. We will examine these results in this section.

(Video) DNA, Chromosomes, Genes, and Traits: An Intro to Heredity

Chromosomal sex determination

Sex chromosomes are an exception to the rule that all chromosomes of diploid organisms exist in pairs of morphologically similar homologues. As early as 1891, microscopic analyzes showed that one of the chromosomes in males of some insect species had no homologue. This unpaired chromosome was called the X chromosome and was present in all male somatic cells but in only half of the sperm cells. The biological significance of these observations became clear when females of the same species were shown to have two X chromosomes. In other species in which females have two X chromosomes, the male has one X chromosome along with a morphologically incompatible chromosome. The mismatched chromosome is called the Y chromosome and pairs with the X chromosome during meiosis in males, usually only along part of its length due to a limited region of homology. The difference in chromosomal constitution between males and females is a chromosomal mechanism for determining sex at the time of fertilization. While each egg contains an X chromosome, half of the sperm contain an X chromosome and the rest contain a Y chromosome. Fertilization of an X-bearing egg by an X-

carrying sperm results in an XX zygote, which normally develops into a female; fertilization by a Y-bearing sperm results in an XY zygote, which normally develops into a male (FIGURE 4.12). The result is a cross-X chromosome inheritance pattern, in which a male gets his X chromosome from his mother and

it only goes to the daughters.

Figure 4.12 The chromosomal basis of sex determination in mammals, many insects, and other animals.

The XX-XY type of chromosomal sex determination is found in mammals, including humans, in many insects and other animals, and in some flowering plants. The female is called the homogametic sex because only one type of gamete (X-lager) is produced. The male is called the heterogametic sex

because two different types of gametes (X-carrier and Y-carrier) are produced. If the union of gametes at fertilization is random, a 1:1 sex ratio at fertilization is expected, since males produce equal numbers of X- and Y-bearing sperm. The X and Y chromosomes together form the sex chromosomes; this term distinguishes them from other pairs of chromosomes called autosomes. Although the sex chromosomes control the developmental switch that determines the first stages of female or male development, the developmental process itself requires many genes scattered throughout the genome, including genes on autosomes. The X chromosome also contains many genes with functions unrelated to sexual differentiation, as we will see in the next section. In most organisms, including humans, the Y chromosome carries few genes other than those related to male determination.

X-Linked Inheritance Convincing evidence that genes are located on chromosomes came from the study of a Drosophila white eye gene, which was found to be present on the X chromosome. Remember that in Mendelian crosses it did not matter which trait was present. in the male parent and which was present in the female parent: reciprocal crosses gave the same result. One of the first exceptions to this rule was found by Thomas Hunt Morgan in an early study of a white-eyed mutant fly in 1910 (see Roots of Discovery: The White-Eyed Male). Wild-type eye color is a brick-red combination of red and brown pigments (FIGURE 4.13). While white eyes can result from certain combinations of autosomal genes that knock out individual pigments, the white eye mutation that Morgan studied results in a metabolic blockade that turns off both pigments simultaneously. (The gene encodes a transmembrane protein that

it is necessary to transport pigment precursors from the eye to the pigment cells.)

FIGURE 4.13 Drawings (A) of a male and female fruit fly, Drosophila melanogaster. Photographs (B) show the eyes of a wild red-eyed male and a white-eyed mutant male. (A) Illustrations © Carolina Biological Supply Company. Used with permission. Photos courtesy of E.R. Lozovsky.

Thomas Hunt Morgan (1910) Columbia University New York, New York

Sex-Restricted Inheritance in Drosophila Morgan's genetic analysis of the white-eye mutation marks the beginning of Drosophila genetics. It is the nature of science that as knowledge increases, so do the terms used to describe things. This article provides an example of this evolution, as the term sex-restricted inheritance is now used with a completely different meaning than Morgan's. what

Morgan was now referring to X-linked or sex-linked inheritance. To avoid confusion, we've taken the liberty of substituting the modern equivalent whenever appropriate. Morgan also did not know that Drosophila males had a Y chromosome. He thought that females were XX and males X, as in grasshoppers (see Carother's article). We also provide the missing Y chromosome. Morgan's gene symbols have been kept as in the original. It uses R for the wild-type red eye allele and W for the recessive white eye allele. This is a strange departure from the convention established by Mendel that dominant and recessive alleles must be represented by the same symbol. Today we use w for the recessive allele and w1 for the dominant allele.

“No white-eyed females showed up.” The experiments Morgan performed were based on the spontaneous appearance of a single white-eyed male in a purebred wild-type (red-eyed) strain. When this male mated with his wild-type sisters, all the offspring had red eyes. When these F1 offspring were crossed:

[D] F1 hybrids, inbred, produced

No white-eyed female appeared. The new character turned out to be gender-linked in the sense that he was only passed on to the grandchildren. But was the white-eyed phenotype sex-restricted in the modern sense, restricted in expression, and found only in males?

This hypothesis was ruled out by Morgan's later discovery: the white-eyed male (mutant) was later crossed and produced with some of his daughters (F1).

The results show that the new trait, white eyes, can be transmitted to females by proper mating and is therefore not limited to one sex in this sense. Note that the four classes of individuals occur in approximately equal numbers (25 percent)....

Based on these results, Morgan hypothesized the existence of "factors" that determine gender (X and what we now know as Y) and eye color (R for red and W for white). He also found that: [D] the white-eyed man's sperm carries the white-eyed "W factor"; ... half of the sperm carry an "X" sex factor, [and] the other half do not, ie the male is heterozygous for the sex. [The male is actually XY.] Thus, the symbol for the male is "WXY" and for his two types of sperm WX—Y. Assume that all red-eyed female eggs carry the red-eye "factor" R; and that all the eggs (after meiosis) each carry an X, the symbol for the red-eyed female will therefore be RRXX and the symbol for her eggs will be RX…. Morgan then described four additional crosses (mostly backcrosses) where the results obtained agreed with those predicted by the hypothesis. These findings led him to the critical conclusion:

To obtain these results, it is necessary to assume that when the two classes of spermatozoa are formed in the RXY male, R and X go together.... The fact is that these R and X are combined and never existed separately. Interestingly, Morgan never uses the word "chromosome" in the original article. Genetic elements are referred to as "factors", which is reflected in the nomenclature he uses. In modern terminology, the X chromosomes that carry the alternative alleles would be referred to as Xw+ (red) and Xw (white). Combined with Carothers' cytological observations, Morgan's hypothesis suggests that his X "factor" is indeed the X chromosome as we know it today. Source: TH Morgan, Science 32 (1910): 120-122.

Morgan's study began with a single white-eyed male emerging from a wild-type laboratory population maintained for many generations. When this male was crossed with wild-type females, all F1 offspring of both sexes had red eyes, showing that the white eye allele is recessive. on F2

Offspring of mating F1 males with females, Morgan observed 2,459 red-eyed females, 1,011 red-eyed males, and 782 white-eyed males. It became clear that the white-eyed phenotype was somehow sex-linked because all white-eyed flies were male. However, white eyes were not limited to men. For example, when red-eyed F1 females from the wild-type ♀×white♂ cross to their white-eyed parents were backcrossed, the progeny consisted of red-eyed white-eyed females and red-eyed white-eyed males in approximately equals.

One important observation came from the mating of white-eyed females with wild males. All female offspring had wild-type eyes, but all male offspring had white eyes. This is the reciprocal of the original wild-type ♀ × white ♂ cross, which only produced wild-type females and wild-type males, so the reciprocal crosses gave different results. Morgan realized that if the white eye allele were present on the X chromosome, reciprocal crosses would produce different results. A gene on the X chromosome is called X-linked. The normal chromosomal complement of Drosophila melanogaster is shown in FIGURE 4.14. Females have an XX chromosome complement, while males have an XY chromosome. Morgan's hypothesis was that an X chromosome contains either a wild-type w+ allele or a mutant w allele, while the Y chromosome contains no white gene counterpart. If we use the white allele present on an X chromosome to represent the entire X chromosome, we can write the genotype of a white-eyed male as wY and that of a wild male as w+Y. Because the w allele is recessive, white-eyed females of the ww genotype and wild-type females are w+w heterozygotes or w+w+ homozygotes. the effects of this

Models for reciprocal crossings are shown in FIGURE 4.15. The wild-type ♀ × white ♂ cross is the A cross, and the white ♀ × wild-type ♂ cross is the B cross.

FIGURE 4.14 The diploid chromosome complements of a male and female Drosophila melanogaster. The centromere of the X chromosome is nearly terminal, but that of the Y chromosome divides the chromosome into two unequal arms. The large autosomes (chromosomes 2 and 3, shown in blue and green) are not easily distinguished in these cell types. The tiny autosome (chromosome 4, shown in yellow) appears as a dot.

FIGURE 4.15 Chromosomal interpretation of results obtained in F1 and F2 progeny in Drosophila crosses. Cross A is a cross between a wild (red-eyed) female and a white-eyed male. The B-cross is the reciprocal mating of a white-eyed female with a red-eyed male. On the X chromosome, the wild-type w+ allele is shown in red,

the mutant W allele blank. The Y chromosome does not carry any of the w gene alleles.

The X-linked mode of inheritance explains the different phenotypic ratios observed in the F1 and F2 offspring of the crosses. The characteristics of X-linked inheritance can be summarized as follows: 1. Reciprocal matings resulting in different phenotypic relationships between the sexes usually indicate X-linked inheritance.

white eyes in Drosophila, crossing a red-eyed female with a white-eyed male gives all red-eyed offspring (Figure 4.15, Cross A), while crossing a white-eyed female with a red-eyed male gives red-eyed female offspring. red and white-eyed male offspring (Figure 4.15, Cross B). 2. Heterozygous females transmit each X-linked allele to about half of their daughters and half of their sons; this is shown in the F2 generation of cross B in Figure 4.15. 3. Males who inherit an X-linked recessive allele exhibit the recessive trait because the Y chromosome does not contain a wild-type counterpart of the gene. Affected males transmit the recessive allele to all of their daughters but none of their sons; this principle is illustrated in Figure 4.16 in the F1 generation of the A cross. Any unaffected male should carry the wild-type allele on his X chromosome.

The essence of X-linked inheritance is captured in the Punnett square in FIGURE 4.16: A man passes his X chromosome only to his daughters, while a woman passes an X chromosome to offspring of both sexes.

Figure 4.16 The Punnett square shows that, in chromosomal sex determination, each child receives its X chromosome from its mother and its Y chromosome from its father.

X-Linked Human Inheritance Traits An example of a human trait with an X-linked inheritance pattern in humans is hemophilia A, a severe blood clotting disorder caused by a recessive allele. Those affected lack a blood clotting protein called Factor VIII, which is necessary for normal blood clotting, and subsequently experience excessive, often life-threatening, bleeding.

Wounds. A famous hemophilia family tree begins with Queen Victoria of England (FIGURE 4.17). One of her sons, Leopold, was a hemophiliac and two of her daughters were heterozygous carriers of the gene. Two of Victoria's granddaughters were also carriers and introduced the gene into the royal families of Russia and Spain through marriage. The heir to the Russian throne of the Romanovs, Tsarevich Alexis, was stricken with the disease. He inherited the gene from his mother, Tsarina Alexandra, one of Victoria's granddaughters. The Tsar, Tsarina, Alexis and her four sisters

they were all executed by the Russian Bolsheviks in the 1918 revolution. England's current royal family is descended from a normal son of Victoria and is free of the disease.

FIGURE 4.17 Genetic transmission of hemophilia A among the descendants of Queen Victoria of England, including her granddaughter, Tsarina Alexandra of Russia, and Alexandra's five children. The photo is of Tsar Nicholas II, Tsarina Alexandra and her children. Tsarevich Alexis suffered from hemophilia. (Insert) Photo courtesy of Boston Public Library Trustees, Department of Printing.

X-linked inheritance in human pedigrees has several characteristics that distinguish it from other forms of genetic transmission: ■ For every rare trait due to an X-linked recessive allele, affected individuals are exclusively or almost exclusively males. There is an excess of males because females carrying the rare X-linked recessive virus are almost exclusively heterozygous and therefore do not express the mutated phenotype. ■ Affected males who reproduce have normal children. This stems from the fact that a man only passes his X chromosome to his daughters. ■ A female whose father is affected has a 1:1 ratio of normal sons to affected sons, because any daughter of an affected male must be heterozygous for the recessive allele.

Heterogametic Females In some organisms, the homogametic and heterogametic sexes are reversed; that is, males are XX and females are XY. This type of sex determination is found in birds, some reptiles and fish, moths and butterflies. The reversal of XX and XY in the sexes results in an opposite pattern of non-reciprocal inheritance of X-linked genes. In order to differentiate sex determination in these organisms from the usual XX-XY mechanism, the constitution of the sex chromosome is designated ZZ for the homogametic sex and WZ for heterogametic sex. Therefore, in organisms with heterogametic females, the chromosome constitution of females is denoted by WZ and that of males by ZZ. A specific example of Z-linked inheritance in chickens is shown in FIGURE 4.18. Some breeds of chickens have feathers with alternating light and dark colored cross bands, resulting in

Phenotype known as blocked. In other breeds, the feathers are uniformly colored and not striped. Reciprocal crosses between barred purebreds and non-barred purebreds produce the results shown in Figure 4.18, indicating that the gene that determines barring must be dominant and located in the Z

Chromosome.

Fig. 4.18 Mottled feathers in chickens - a classic example showing that chromosomal sex determination in birds is inverse to that in mammals. In birds, females are the heterogametic sex. The Z chromosome carrying the dominant Barred mutation is highlighted in red.

Nondisjunction as Evidence for the Chromosome Theory of Inheritance The parallel between the inheritance of the Drosophila white mutation and the genetic transmission of the X chromosome supported the chromosomal theory of inheritance that genes are parts of chromosomes. Other Drosophila experiments provided definitive proof.

One of Morgan's students, Calvin Bridges, found rare exceptions to the expected pattern of inheritance in crosses involving multiple X-linked genes. For example, when white-eyed Drosophila females were crossed with red-eyed males, most of the offspring it consisted of the expected red-eyed females and white-eyed males. However, around 1 in 2,000 F1 flies was an exception - i.e. a white-eyed female or a red-eyed male. Bridges showed that these rare and extraordinary offspring resulted from the occasional failure of the mother's two X chromosomes to separate during meiosis - a phenomenon called nondisjunction. The consequence of non-disjunction of the X chromosome is the formation of some eggs with two X chromosomes and others without. Four classes of zygotes are expected from fertilization of these abnormal eggs (FIGURE 4.19). Animals without an X chromosome are not recognized because embryos without an X chromosome are not viable; Likewise, most offspring with three X chromosomes die early in development. Microscopic examination of the chromosomes of the exceptional offspring from the White♀×wildtype♂ cross showed that the exceptionally white-eyed females had two X chromosomes plus one Y chromosome, and the exceptionally red-eyed males had a single X but lacked a Y. the latter with a sex chromosome constitution designated as XO were infertile men.

FIGURE 4.19 Results of X chromosome nondisjunction in the first meiotic division in a female Drosophila.

These and related experiments conclusively demonstrated the validity of the chromosomal theory of inheritance. Chromosome Theory of Inheritance: Genes are physically located within chromosomes. Bridges' proof of chromosome theory was that the extraordinary behavior of chromosomes is precise.

parallel, an exceptional heritage of its genes. This proof of chromosome theory is one of the most important and elegant experiments in genetics.

Sex determination in Drosophila In the XX-XY mechanism of sex determination, the Y chromosome is associated with the male. In some organisms, including humans, this association occurs because the presence of Y

The chromosome triggers events in embryonic development that lead to male sexual characteristics (see chapter Human Karyotypes and Chromosomal Behavior). Drosophila is unusual among organisms with XX-XY sex determination because the Y chromosome, although associated with maleness, is not male-determined. We have already seen (Figure 4.19) that individuals with the XXY genotype are female; In fact, they are fully viable and fertile. TABLE 4.2 shows the relationship between sex chromosome genotype, number of autosome sets, and phenotypic sex. XXY embryos in Drosophila develop into morphologically normal fertile females, whereas XO embryos develop into morphologically normal but sterile males. (The O is written in the XO formula to emphasize that it lacks a sex chromosome.) The sterility of XO males shows that the Y chromosome, while not necessary for male development, is essential for male fertility; In fact, the Drosophila Y chromosome contains six genes necessary for the formation of normal sperm.

TABLE 4.2 Relationship between sex chromosome genotype, sets of autosomes and phenotype

Sex on Drosophila sex chromosomes

sets of autosomes

X:A ratio

phenotypic sex

Feminine

0,5

Macho

Feminine

0,5

male (sterile)

0,67

intersexual

Genetic sex determination in Drosophila relies on an X-linked gene known as lethal sex (Sxl) because some mutant alleles are lethal when present in males. As shown in FIGURE 4.20, the Sxl gene is active in normal females and inactive in normal males. The product of Sxl is the sex-killing protein SXL - an RNA-binding protein that binds to the RNA transcripts of various genes and causes female-specific RNA processing. In the absence of SXL, these transcripts undergo male-specific processing. Therefore, SXL targets genes whose protein products are normally produced only in females. One of the targets of SXL is the Sxl gene itself, so a small burst of SXL activity early in female development results in self-sustaining production of SXL as there is feedback in processing more Sxl transcripts.

FIGURE 4.20 Early steps in the genetic control of sex determination in Drosophila through activity of the lethal Sxl gene.

Contributing to the initial burst of SXL is the amount of product produced by four X-linked genes known as X-linked signal elements (XSE) (Figure 4.20). If the amount of XSE is high, early burst occurs and the organism becomes female. In contrast, if the amount of XSE is low, the early burst does not occur and the organism becomes male (Figure 4.20). For nearly 100 years, geneticists believed that sex was determined by the ratio of the number of X chromosomes (X) to the number of sets of autosomes (A). As we see in Table 4.2, the correlation is perfect. In normal women (XXAA), the ratio is equal to 1; in normal men (XAA), it is equal to 1/2; and in rare individuals of XXAAA composition, the ratio is 2/3 and the organism develops as an intersex. Furthermore, rare XA haploids initially develop as females, though they eventually die, which again supports the idea that an embryo with an X:A ratio of 1/2 develops as a male, an embryo with an X:A ratio: A 1 develops as a female and an embryo with an X : A ratio between 1/2 and 1 develops as an intersex.

Although the correlation with X : A is perfect, recent data shows that this assumption is actually wrong. Sex in Drosophila is based on the Sxl response to the amount of XSE gene products, which in normal flies is determined by the number of X chromosomes. Autosomes play a role, but not in sex determination. Its role focuses on determining when, early in development, commitment to sexual differentiation occurs. In normal XXAA or XAA embryos with two sets of autosomes, binding occurs after 15 division cycles. In haploid XA embryos with a set of autosomes, binding occurs one cell cycle later, when the amount of XSE is sufficient to induce Sxl. In XXAAA embryos, some cells commit at cycle 13 and have insufficient XSE to induce Sxe, while other cells commit at cycle 14 and have sufficient XSE to induce Sxl. The result is an embryo that is a mosaic of male and female cells, and the phenotypic result is an intersex.

Drosophila sexing is an excellent example of why correlation does not necessarily imply causation. It also illustrates the principle that ideas in science - even those long held and commonly accepted - can sometimes be proven wrong.

SUMMARY ■ Sex chromosomes are dimorphic. ■ In species with chromosomal sexing, one sex is homogametic (XX females in humans) and the other heterogametic (XY males in humans). The opposite is true in some other organisms (including birds): males are homogametic and females are heterogametic. ■ Genes on the X chromosome show a 'crossover' pattern of inheritance. ■ In pedigrees, X-linked recessive traits occur mainly in male offspring of unaffected parents. ■ The occurrence of nondisjunction of X chromosomes in Drosophila provided evidence that genes reside on chromosomes.

■ In Drosophila, sex is determined by a cascade of events triggered by the number of X chromosomes in the developing embryo.

4.5 Probability in predicting distributions of offspring Genetic transmission involves a large random component. Depending on chance, a given gamete from an Aa organism may or may not contain the A allele. A given gamete from an Aa-Bb organism may or may not contain both A and B alleles, depending on the random orientation of the chromosomes on the metaphase I plate. Genetic relationships result not only from the random arrangement of genes in gametes, but also from the random combination of gametes in zygotes. Although exact predictions of a specific event are not possible, it is possible to determine the probability of occurrence of a specific event, as we saw in the chapter Mendelian Genetics: The Principles of Segregation and Assortment. In this section, we consider some additional probability methods used in interpreting genetic data.

Using the Binomial Distribution in Genetics The probability addition rule deals with mutually exclusive outcomes of a genetic cross. Results are mutually exclusive when they are incompatible in the sense that they cannot occur simultaneously. For example, the possible sex distributions among three children consist of four mutually exclusive outcomes: zero, one, two, or three girls. These outcomes have probabilities 1/8, 3/8, 3/8, and 1/8, respectively. The addition rule states that the total probability of any combination of mutually exclusive events is equal to the sum of the probabilities of the events taken separately. For example, the probability that a sibling community of size three contains at least one girl contains the outcomes one, two, and three girls, so the overall probability of at least one girl is 3/8 + 3/8 + 1/8 = 7/8.

The probability multiplication rule deals with results of a genetic cross that are independent. Two possible outcomes are independent when knowing that one outcome will actually be realized does not tell you whether the other will also be realized. For example, in a birth sequence, the sex of a given child is not affected by the sex of a sibling born earlier; moreover, it has no influence on the gender distribution of siblings that may be born later. Each subsequent birth is independent of all others. When possible outcomes are independent, multiplication

The rule states that the probability of any combination of outcomes being realized is equal to the product of the probabilities of all individual outcomes, taken separately. For example, the probability that a brother with three sons will consist of three girls is 1/2 × 1/2 × 1/2 because the probability that each birth will produce a girl is 1/2 and consecutive births are independent of each other. Probability calculations in genetics often use addition and multiplication rules together. For example, the probability that all three children in a family are of the same sex uses addition and multiplication rules. The probability that all three are girls is (1/2)(1/2)(1/2) = 1/8, and the probability that all three are boys is also 1/8. Since these outcomes are mutually exclusive (a sibling community of size three cannot have three boys and three girls), the probability of three girls or three boys is the sum of the two probabilities, or 1/8 + 1/8 = 1/ 4 . The other possible outcomes for siblings of size three are that two of the children are girls and one is a boy, and two are boys and one is a girl. For each of these outcomes, three different birth orders are possible – for example, GGB, GBG, and BGG – each with a probability of 1/2 × 1/2 × 1/2 = 1/8. The probability of two girls and one boy, regardless of birth order, is the sum of the probabilities of the three possible orders, or 3/8; also the probability of two

Boys and a girl are also 3/8. Therefore, the sex ratio probability distribution in families with three children is:

The sex ratio information in this display can be obtained more directly by expanding the binomial expression (p + q)n where p is the probability of having a girl (1/2) and q is the probability of having a boy (1/2 ), and n is the number of children. In this example

where the red digits represent the possible number of birth orders for each sex distribution. Likewise, the binomial distribution of sex ratio probabilities in families with five children is:

Each term tells us the probability of a particular combination. For example, the third term is the probability of three girls (p3) and two

Boys (q2) in a family with five children:

There are n + 1 terms in a binomial expansion to the power of n. The exponents of p decrease by one from n in the first term to 0 in the last term, and the exponents of q increase by one from 0 in the first term to n in the last term The coefficients produced by successive values of n can be arranged in a triangle regular known as Pascal's triangle, shown in FIGURE 4.21 for n = 0 to 10. The horizontal lines of the triangle are symmetrical; each line begins

and ends with 1, and all other entries can be taken as the sum of the two numbers on either side in the line above.

Figure 4.21 Pascal's Triangle. The numbers in the nth row are the coefficients of the expansion terms of the polynomial (p + q)n.

To generalize a bit, if the probability of event A is p and that of event B is q, and if the two events are independent and mutually exclusive, then the probability that A will happen four times and B twice (in a given order) it conforms to the p4q2 multiplication rule. Usually we are interested in a combination of events, regardless of their order, e.g. For example, "four A's and two B's". In this case, we multiply the probability that the 4A : 2B combination is performed in any order by the number of possible orders. The number of different combinations of six events, four of type A and two of type B, is given by the coefficient of p4q2 in the expansion of (pA + qB)6. This coefficient can be found on the line for n = 6 in Pascal's triangle as the fifth entry from the left. (It is fifth because the consecutive entries are the coefficients of p0q6, p1q5, p2q4, p3q3, p4q2, p5q1, and p6q0.) So the total probability is four

The realizations of A and two realizations of B in six trials are given by 15p4q2. The rule of thumb for repeated trials of events with constant probabilities is as follows: if the probability of event A is p and the probability of alternative event B is q, then the probability that event A will occur in n trials is s times and event B will be performed t times

where s + t = n and p + q = 1.

The character n! is read as "n factorial" and represents the product of all positive integers from 1 to n (that is, n! = 1 × 2 × 3 × ... × n). Factorial n values from n = 0 to n = 15 are given in TABLE 4.3. The value 0! = 1 is arbitrarily defined to generalize its use in mathematical formulas. The size of n! increases very quickly; fifteen! is more than 1 trillion.

TABLE 4.3 Colleges n

40.320

362.880

3.628.800

39.916.800

479.001.600

120

6.227.020.800

720

87.178.291.200

5040

fifteen

1.307.674.368.000

Equation 1 also holds when s or t is equal to 0, because 0! = 1. (Also remember that every number to the power of zero is 1.) Each individual term in the binomial expansion (p + q)n is given by Equation 1) for the appropriate values of s and t. In Pascal's triangle, consecutive entries in the nth row are the values of n!/(s!t!) for s = 0, 1, 2, …, n. Consider a specific application of Equation 1, where we use Calculate the probability that the four kittens shown in the photograph are the result of mating between two heterozygous parents that produced exactly the expected 3:1 ratio of dominant (gray) and recessive traits ( white) in siblings of a certain size. The probability p that a kitten will have the dominant trait is 3/4 and the probability q that a kitten will have the recessive trait is 1/4. This is the "expected" Mendelian proportion. In this case, n = 4, s = 3, t = 1, and the probability of this combination of events is:

That is, in only 42% of feline families with four kittens would the offspring exhibit the expected phenotypic ratio of 3:1; the other siblings would go one way or the other due to random variations. The importance of this example lies in its demonstration that, although a ratio of 3:1 is the "expected" result (and the most likely result), the majority of families (58%) actually have a distribution of descendants other than 3 :1:1 .

Importance of the Binomial Coefficient The factorial part of the binomial expansion in Equation 1 that equals n!/(s! t!) is called the binomial coefficient. As we have already noted, this ratio lists all the ways in which the s elements of one type and the t elements of another type can be arranged in an order, provided that the s elements and the t elements are not distinguished from each other. 🇧🇷 A specific example might include s yellow peas and t green peas. Although yellow peas and green peas can be distinguished from each other by being different colors, yellow peas are indistinguishable from each other (because they are all yellow); The same caveat applies to peas (because they are all green).

The reasoning behind the factorial formula starts with the observation that the total number of elements is s + t = n. If there are n different elements, they can be arranged in different ways:

Why? Since the first element can be chosen in n ways, and

Once chosen, the next one can be chosen in n − 1 ways (because there are only n − 1 to choose from). Once the first two are chosen, the third can be chosen n − 2 ways, and so on. Finally, when n − 1 elements have been selected, there is only one way to select the last element. The elements s + t can be arranged in n! Wise, as long as the elements are all distinct from each other. However, if we reapply the argument we just used, each of the n! special agreements must contain s! different arrangements of elements s and t! different arrangements of t or s elements! × t! a total of. Division of n! of s! × t! therefore, it gives the binomial coefficient for the number of ways in which the s elements and the t elements can be arranged if the elements of each type are not distinguished from each other.

SUMMARY ■ The binomial distribution forms the basis for predictions based on genetic assumptions. ■ The probability of occurrence of independent events is the product of the probability of each event occurring independently. ■ The probability of simultaneous occurrence of mutually exclusive events is the sum of the events occurring separately.

■ The binomial equation can be used to predict the exact probability of occurrence of a given outcome of a genetic cross.

4.6 Testing the Goodness of Fit for a Genetic Hypothesis Geneticists often need to decide whether an observed relationship agrees well with a theoretical prediction. Just reviewing the data is unsatisfactory, as different investigators may have different opinions. What we need is to be able to establish a threshold - that is, an objective standard for determining whether the observed results agree with our hypothesis. To understand how this happens, we need to look a little deeper into the implications of probability theory.

Random Variables and Distributions Before we get into testing genetic hypotheses, let's consider the simplest possible example of a binomial variable — the coin toss. Suppose we flip a coin 20 times and it comes up heads 18 times. Of course, assuming it's fair coin, our expected number of employees is 10; The question is whether our observed result deviates so much from this expectation that we would reject the hypothesis that it is a fair coin.

One approach to this question would be to apply Equation 1, as we did for the four kittens. For this example we know that

Therefore, the probability of observing this result is

Simply put, the probability of getting exactly this result is quite small. But let's ask the question a little differently. Suppose we wanted to know how the results would be distributed if we repeated the experiment many times (say 1000). Of course, no one wants to waste time playing 20 coins that often, so as an alternative, we ask a computer to simulate this scenario for us. The result is presented in FIGURE 4.22 as a histogram of the results of the 1000 simulated experiments. Note the bottom and top parts of the distribution in red, where the

The number of heads observed is 6 or less, or 14 or more. In fact, these results together make up the 5% of the distribution that deviates the most from the expected result (in this case, 10 heads).

FIGURE 4.22 Histogram showing the results of 1000 simulations of tossing a coin 20 times. The results shown in red represent the 5% of the total that deviate the most from the expected number of 10.

Why is it important? Let's go back to our original experiment and formulate it as an evaluation of two mutually exclusive hypotheses: H0: The coin is fair (Pr{Heads} = 0.5}. H1: The coin is not fair (Pr{Heads ≠ 0, 5} ). H0 is called the null hypothesis and H1 the alternative hypothesis. Statisticians would then ask whether the deviation of observed results from expected results under the null hypothesis is too large to be explained by chance. What is "too large"?

In this case (and other common cases in genetics), the cutoff is defined as the deviation from the null hypothesis that is expected to occur by chance less than 5% of the time. We call this deviation the critical value. In the case of the experiment shown in Figure 4.22, it would be all results that were within the values shown in red (less than 7 or more than 13 heads). Since our original result (18 heads) falls into this category, we would reject the null hypothesis. Such a test is called a goodness-of-fit test, where the word "fit" means how much the observed results "fit" or agree with the expected results based on the null hypothesis. For example, in the case of genetics, we would ask how well the results of a cross match the expectations of a Mendelian hypothesis.

The chi-square method Suppose we cross a purple flowering plant with a white flowering plant and observe 14 purple flowering plants and 6 white flowering plants among the offspring. Is this result close enough to be accepted as a 1:1 ratio? What if we looked at 15 purple flowering plants and 5 white flowering plants? Do any of these results correspond to a 1:1 ratio? There were inevitably random variations in the observed results from one experiment to another. Who says which results are consistent with a particular genetic hypothesis? A conventional measure of goodness of fit is a value called the chi-square (its symbol is χ2), which is calculated from the number of offspring observed in each of the different classes compared to the number expected in each of the classes based on under some genetic hypotheses. For example, we might be interested in a cross between plants with purple flowers and those with white flowers.

when testing the hypothesis that the purple-flowered parent is heterozygous for a pair of alleles that determine flower color and that the white-flowered parent is homozygous recessive. Suppose further that we examine 20 plants descended from the mating and find that 14 are purple and 6 are white. The procedure for testing this genetic hypothesis (or any other genetic hypothesis) using the chi-square method is as follows: 1. Provide the details and specification of the genetic hypothesis

Genotypes and phenotypes of parents and possible offspring. In the flower color example, the genetic hypothesis implies that the genotypes in the purple × white cross can be symbolized as Pp × pp. The possible genotypes of the offspring are Pp and pp. 2. Use probability rules to make explicit predictions about the species and proportions of offspring that must be observed if the genetic hypothesis is true. Convert the proportions to the number of descendants (percentages are not allowed in a χ2 test). If the flower color crossing hypothesis is true, then we should expect the progeny genotypes Pp and pp to appear in a 1:1 ratio. Since the hypothesis is that pp flowers are purple and pp flowers are white, we expect the offspring phenotypes to be either purple or white in a 1:1 ratio. Out of 20 offspring, 10 are expected to be purple and 10 to be white. 3. Subtract the expected number from the observed number for each descendant class in succession. Square this difference and divide the result by the expected number. In our example, the calculation for the purple offspring is (14 − 10)2/10 = 1.6 and for the white offspring (6 − 10)2/10 = 1.6. 4. Sum the result of numbers calculated in step 3 for all descendant classes. The sum is the value of χ2 for

those dates. The sum of the purple and white descent classes is 1.6 + 1.6 = 3.2, which is the χ2 value for the experiment calculated assuming our genetic hypothesis is correct. In symbols, the calculation of χ2 can be represented by the expression

where ∑ means the sum of all descendant classes. Note that χ2 is calculated using the observed and expected numbers - not the parts, proportions or percentages. Using anything other than the real numbers is the most common beginner's mistake when using the χ2 method. The value of χ2 is useful as a measure of goodness of fit because the closer the observed numbers are to the expected numbers, the smaller the value of χ2 will be. A value of χ2 = 0 means that the observed numbers perfectly match the expected numbers. Also note that χ2 cannot be negative.

But what constitutes a value of χ2 large enough to reject our null hypothesis? This is where the logic of distribution comes in. Similar to the case of flipping a coin, we can ask what results we would expect given the expectation of the null hypothesis less than 5% of the time, measuring the difference by the value of χ2 calculated from the data. However, before we can do that, we need to consider one more factor – the number of degrees of freedom of the specific χ2 test, which is related to the number of classes in the data. The more data classes there are, the more values there are that can deviate from the expected value, and therefore the greater the value of χ2 that we can simply observe by chance.

The number of degrees of freedom is usually the number of classes in the data minus . The reason for deducting 1 is that when we calculate the expected offspring count, we ensure that the total number of offspring matches the number actually observed. Because of this, none of the data classes are really "free" to contain any number we might specify; Since the expected number in a class needs to be adjusted for the sum to come out correctly, a "degree of freedom" is lost. Analogous χ2 tests with three data classes have 2 degrees of freedom and those with four data classes have 3 degrees of freedom. In the case described earlier, the number of degrees of freedom is 2 − 1 or 1. With this information and our calculated value of χ2, we can do a simulation similar to the one shown in Figure 4.22, except that we calculate χ2 for 1000 "experiments" with 1 degree of freedom. The results shown in FIGURE 4.23 show that when χ2

is greater than the critical value of 3.84, then we would expect to deviate as large or greater from the expected result less than 5% of the time.

FIGURE 4.23 Histogram showing the value of χ2 obtained at 1000

simulated experiments with 1 degree of freedom. Results shown in red represent the 5 percent that most deviate from null hypothesis expectations (χ2 > 3.84; see Table 4.5).

As another example of how χ2 is calculated, let's assume that the offspring of an F1 × F1 cross contain two contrasting phenotypes seen at numbers 99 and 45. In this case, the genetic hypothesis could be that the trait is determined by a pair of alleles from a single gene, in which case the expected ratio of dominant:recessive phenotypes in F2 offspring is 3:1 Ask whether the observed ratio of 99 :45 agrees satisfactorily with the expected 3 to 1. The calculation of the χ2 value is shown in TABLE 4.4. The total number of descendants

is 99 + 45 = 144. The expected numbers in the two classes are calculated as (3/4) × 144 = 108 and (1/4) × 144 based on the genetic assumption that the true ratio is 3:1 = 36. Since there are two classes of data, there are two terms in the calculation of χ2:

TABLE 4.4 Calculation of χ2 for a monohybrid ratio phenotype

observed

Expected

deviation from

(Class)

number

expected

Wildtyp

108

−9

0,75

mutant

2.25

No total

144

x2 = 3,00

After calculating the χ2 value, the next step is to interpret whether this value represents a good or bad fit to the expected numbers. This assessment is made using the values in TABLE 4.5. In the table, the rows are degrees of freedom and the columns are the probabilities P that by chance a worse (or equally bad) fit would be obtained, assuming the genetic hypothesis is true. If the genetic hypothesis is true, the observed numbers should be quite close to the expected numbers. If the observed χ2 is so large that the probability of fitting bad or worse is very small, the observed results will not match theoretical expectations. In this case, the genetic hypothesis for calculating the expected number of offspring must be rejected because the observed number of offspring deviates too much from the expected number.

TABLE 4.5 Critical values of χ2

probability d.f.

0,5

0,1

0,05

0,025

0,01

0,45

2.71

3,84

5.02

6.63

1.39

4.61

5,99

7.38

9.21

2.37

6.25

7.81

9:35

11.34

3.36

7.78

9.49

11.14

13.28

4.35

9.24

11.07

12.83

15.09

5.35

10.64

12.59

14h45

16.81

6.35

12.02

14.07

16.01

6:48 p.m

7.34

13.36

15.51

17.53

20.09

8.34

14.68

16.92

19.02

21.67

9.34

15,99

18.31

20.48

23.21

In practice, the critical P values are conventionally chosen to be 0.05 (the 5 percent level) and 0.01 (the 1 percent level). For P-values in the range of 0.01 to 0.05, the probability that chance alone will result in a fit as bad or worse is between 1 in 20 experiments and between 1 in 100, respectively. If the P-value is within this range, the validity of the genetic hypothesis is considered very doubtful and the result is considered statistically significant at the 5% level. For P values less than 0.01, the probability that chance alone will result in a fit as bad or worse is less than 1 in 100 experiments. In this case, the result is said to be highly significant at the 1% level, and the genetic hypothesis is

categorically rejected. If the terminology of statistical significance seems backwards, it is because the term significant refers to the size of the difference between observed and expected numbers; A statistically significant result is a large ("significant") difference between what is observed and what is expected. After determining the appropriate number of degrees of freedom, we can interpret the value of χ2 in Table 4.5. In the case of our F2 descendants, because there are two classes in the data, there are 2 − 1 = 1 degrees of freedom, so we focus on the first row of the table. If we remember that our calculated χ2 value was 3.00, we can compare it with the values in the table. We see that it is less than the value associated with p=0.05, which means that chance alone would result in a fit as bad or worse more than 5% of the time. Using the patterns described earlier, we could conclude that the observed results are consistent with our null hypothesis of a 3:1 ratio.

As a second illustration of the χ2 test, we will determine the goodness of fit of the Mendel data back versus fold for the expected 3:1 ratio. Of the 7,324 seeds that Mendel observed, 5,474 were round and 1,850 were wrinkled. The expected numbers are (3/4) × 7324 = 5493 round and (1/4) × 7324 = 1831 wrinkled. The value of χ2 is calculated as follows:

The fact that χ2 is less than 1 already implies that the fit is very good. To find out how good, notice that since there are two classes of data (round and wrinkled), the number of degrees of freedom equals 2 − 1 = 1. From Table 4.5, the P-value for χ2 = 0.26

with 1 degree of freedom it is much greater than 0.05; in fact, it's about 0.65. In about 65% of all such experiments, an equally bad or worse fit would be expected purely by chance. Only about 35% of all experiments would yield a better fit.

Are Mendel's dates too good to be true? Many of Mendel's experimental results are very close to expected values. For the proportions listed in the chapter titled Mendelian Genetics: The Principle of Segregation and Assortment, the χ2 values are 0.26 (round versus wrinkled seeds), 0.01 (yellow versus green seeds), 0.39 (purple versus white flowers ), 0.06 (swollen versus constricted pods), 0.45 (green versus yellow pods), 0.35 (axial versus terminal flowers) and 0.61 (long versus short peduncles). (As an exercise in dealing with χ2, you should check these calculations yourself.) All tests of χ2

have P-values of 0.45 or greater, indicating that the reported results are in excellent agreement with theoretical expectations. In 1936, statistician Ronald Fisher pointed out that Mendel's results were suspiciously close to theoretical expectations. With a large number of experiments, it can be expected that some of them will produce fits that look dodgy simply because of random variations from one experiment to another. Doubtful expected values seem to be missing from Mendel's data. Figure 4.24 shows the observed deviations in Mendel's experiments compared to the randomly expected deviations. (The deviation measure is the square root of the χ2 value, which is assigned a plus or minus sign depending on whether the dominant or recessive phenotype class exceeds the expected number.) For each deviation measure, the height of the right bar gives the number of experiments observed by Mendel on such a scale

Detour; the height of the left bar indicates the number of experiments likely to deviate from this amount purely by chance. There are clearly few experiments with deviations less than -1 or greater than +1. This kind of discrepancy could be explained if Mendel discarded or repeated a

few experiments with large discrepancies that led him to suspect that the results were unreliable.

FIGURE 4.24 Distribution of observed deviations in Mendel's 69 experiments (yellow bars) compared to expected values (orange bars). There is no evidence that the data were adjusted to improve the fit, but there are fewer experiments with large deviations than expected. Several experiments with large discrepancies may have been discarded or repeated.

Did Mendel cheat? Did he deliberately falsify his data to make it look better? Mendel's paper reports wildly different proportions of individual plants as well as repeated experiments

The first results were doubtful. These are not the things a dishonest person would admit. Only a small bias is needed to explain the excessive goodness of fit in Figure 4.24. When counting individual seeds or plants, only about 2 phenotypes per 1000 would need to be miscategorized to explain the bias in the 91% of data generated by the monohybrid proportions test. Overfitting can also be explained if three or four entire experiments are discarded or repeated because discrepant results were attributed to pollen contamination or some other accident. After carefully reviewing Mendel's 1966 data, evolutionary geneticist Sewall Wright concluded:

Mendel was the first to count segregants. It is too much to hope that he is aware of the precautions now known to be necessary to obtain fully objective data…. Checking scores you dislike but not others can lead to a systematic tendency to agree. I doubt that there are many geneticists today whose data, if voluminous, would appear entirely satisfactorily under the text χ2... In short, I am confident that no deliberate falsification has taken place. Mendel's data are among the first and most complete "raw data" ever published in genetics (in recent years, publishing raw data has become common practice). Additional investigations of the data will certainly be undertaken as new statistical approaches are developed. The bottom line, however, is that, to date, no reputable statistician has claimed this.

that Mendel consciously and intentionally corrected his data in favor of theoretical expectation.

SUMMARY ■ A test of quality determines whether differences between a set of observations and the expected results of a null hypothesis can be explained by chance. ■ The χ2 test provides a means of testing the goodness of fit to a genetic hypothesis. ■ The probability of observing a given value of χ2 depends on the number of degrees of freedom in the data. ■ Deviations from a null hypothesis are statistically significant when the probability of observing the χ2 value calculated by chance is less than 5 percent and highly significant when it is less than 1 percent.

■ Mendel's data show exceptional agreement with the random assortment hypothesis, but this result is unlikely to result from intentional misrepresentation of his results.

CHAPTER SUMMARY • Chromosomes in eukaryotic cells usually come in pairs. ■ The chromosomes of each pair separate in meiosis, with one chromosome going to each gamete. ■ During meiosis, chromosomes from different pairs undergo independent sorting because nonhomologous chromosomes move independently. ■ In many animals, sex is determined by a special pair of X and Y chromosomes. ■ The 'cross' inheritance pattern of X-linked genes is determined by the fact that a male receives his X chromosome only from his mother and the transmits only to his daughters. ■ Irregularities in the inheritance of an X-linked gene in Drosophila provided experimental evidence for the theory of chromosomal inheritance. ■ The offspring of genetic crosses follow the binomial probability formula.

■ The chi-square statistical test is used to determine how closely the observed genetic data agree with expectations derived from a hypothesis.

BASICS REVIEW ■ What is the genetic significance of the fact that gametes contain half the chromosome set of somatic cells? ■ The term mitosis derives from the Greek mitos, which means thread. The term meiosis derives from the Greek meioun, which means "to make smaller". What feature or features of this type of nuclear division could have led to the choice of these terms?

■ Explain the meaning of the terms reduction division and equation division. What is reduced or kept the same? Which core departments do the terms refer to? ■ Explain the following statement: “The independent orientation of nonhomologous chromosomes at metaphase I of meiosis is the physical basis for the independent arrangement of genes on different chromosomes.” ■ What are some of the important differences between the first meiotic division and the second meiotic division? ■ Draw a diagram of a bivalent and identify the following parts: centromere, sister chromatids, non-sister chromatids, homologous chromosomes, chiasm. ■ What are the main features of human pedigrees in which a rare X-linked recessive allele segregates? ■ How is the inheritance of a Y-linked gene different from that of an X-linked gene? ■ How does nondisjunction "prove" the chromosome theory of inheritance? ■ Why is a statistical test necessary to determine whether a set of observed data provides an acceptable fit for the expected result of a specific genetic hypothesis?

■ What statistical test is commonly used for this purpose? ■ What are the traditional P values for significant and highly significant and what do these numbers mean?

TROUBLESHOOTING GUIDE PROBLEM 1 A recessive mutation in an X-linked gene leads to hemophilia, characterized by prolonged prolonged blood clotting time. Assume two phenotypically normal parents

produce three healthy daughters and a son suffering from hemophilia. (a) What is the probability that all daughters are heterozygous carriers? (b) If one of the daughters marries a normal man and produces a son, what is the probability that the son will be affected? ANSWER (a) Because phenotypically normal parents have an affected child who has inherited the mother's only X chromosome, the mother must be a carrier of the mutation. Therefore, the probability that a given daughter is a carrier is ½. The probability that all three daughters are carriers is (½)3, as their births are independent events.

(b) If the daughter is not a carrier, the probability of an affected son is 0; if the daughter is a carrier, the chance of an affected son is ½. Since the probability that the daughter is a carrier is ½ [from part (a)], the overall probability of an affected son is (½) × 0 + (½) × (½) = ¼. PROBLEM 2 What is the probability that a brother with seven children will have four boys and three girls or three boys and four girls? Assume that every child has an equal chance of being a boy or a girl. ANSWER The probability that the sibling consists of four boys and three girls in any birth order is [7!/(4!3!)](½)4(½)3, where the factor in square brackets is the number of births possible orders of four boys and three girls. This probability is 35/128.

Likewise, the probability that the brother has three boys and four girls is [7!/(3!4!)](½)3(½)4, which is also 35⁄128. The question asks the probability of four boys and three girls or three boys and four girls, and these events are mutually exclusive; therefore, the overall probability is 35/128 + 35/128 = 35/64, or about 55 percent. PROBLEM 3 A geneticist performs a cross between two strains of fruit flies that are heterozygous for a recessive allele of each of two genes, st (scarlet) and bw (brown), which affects eye color. Flies homozygous for st have bright red (scarlet) eyes, while those with the st+/st+ and st+/st genotypes have wild-type brick-red eyes. Bw homozygous flies have brown eyes, while flies with bw+/bw+ and bw+/b genotypes have wild-type brick-red eyes. The double homozygous st/st, bw/bw genotype leads to white eyes. In a test to independently assemble these two genes, a geneticist crosses st+/st, bw+/bw females with st+/st, bw+/bw males. Among the 240 young, there are 150 flies with wild eyes, 36 with scarlet eyes, 46 with brown eyes and 8 with white eyes.

(a) What are the expected numbers in each phenotypic class under the null hypothesis that these genes are independently sorted? (b) What is the chi-square value on a test of goodness of fit between observed values and expected values based on the null hypothesis of independent ordering? (c) How many degrees of freedom does this chi-square value have? (d) What is the p-value for the chi-square value in this quality test? Does this P-value support the null hypothesis of independent assortment, or should the hypothesis be rejected?

(e) Would a larger chi-square value increase or decrease the value of P? ANSWER (a) Since this is a st+/st, bw+/bw × st+/st, bw+/bw dihybrid cross, a 9:3:3:1 ratio of phenotypes in the offspring should be expected if the genes are independent of each other others. We can calculate the expected number of flies in each class of offspring: phenotype

offspring

expected number

Wildtyp

st+/−, bw+/−

(9/16) × 240 = 135

scarlet

st/st, sw+/−

(3/16) × 240 = 45

braun

st+/−, sw/sw

(3/16) × 240 = 45

Branco

st/st, sw/sw

(1/16) × 240 = 15

(b) The chi-square value is calculated as follows:

where the sum covers all descendant classes. In that case

(c) The number of degrees of freedom equals the number of data classes minus 1. In this case, there are four data classes; so it has 3 degrees of freedom. (d) The p-value for a chi-square value of 6.76 with 3 degrees of freedom equals 0.08. This P-value is greater than 0.05.

Therefore, we should not reject the null hypothesis of independent assortment. (e) A larger chi-square value would decrease the P value. PROBLEM 4 The bar-eyed mutation in Drosophila melanogaster has the following genetic transmission characteristics: (a) Mating of bar-eyed males with wild-type females produces offspring of wild-type and wild-type daughters. (b) The bar-eyed females from the mating in (a) give a 1:1 ratio of Bar:wild-type sons and a 1:1 ratio of Bar:wild-type daughters when mated with wild-type males.

What mode of inheritance do these traits suggest? ANSWER Since the sexes are unequally affected in the offspring of the mating in (a), some association with sex chromosomes is suggested. The offspring of the mating (a) provides the important clue. Since a male receives his X chromosome from his mother, the observation that all males are wild-type suggests that the gene for streaky eyes is on the X chromosome. Furthermore, the fact that all daughters are affected is consistent with the X-linkage when the bar mutation is dominant. Mating (b) supports the hypothesis of an X-linked dominant gene, since females from mating (a) have the bar/+ genotype and therefore would produce the observed offspring. Thus, the data are consistent with the slash mutation being an X-linked dominant.

ANALYSIS AND APPLICATIONS 4.1 The accompanying diagrams illustrate a pair of homologous chromosomes in prophase I of meiosis. Which diagram corresponds to which stage:

Leptotene, Zygotene, Pachytene, Diplotene and Diacinese?

4.2 The accompanying diagrams illustrate anaphase in an organism that has two pairs of homologous chromosomes. Identify the stages as mitosis, meiosis I, or meiosis II.

4.3 What is the probability that four offspring of the Aa × Aa mating consist of exactly 3 A— and 1 aa? 4.4 A female who is heterozygous for a PKU mutation and an X-linked hemophilia mutation has a child with a phenotypically normal male who is also heterozygous for a PKU mutation. What is the probability that the child has both diseases? (Assume the couple is equally likely to have a boy or a girl.) 4.5 In the attached pedigree chart, the purple symbols represent individuals affected by X-linked

Duchenne muscular dystrophy. What is the probability that the III-3 female is a heterozygous carrier?

4.6 A chromosomally normal female and a chromosomally normal male have a child whose sex chromosome constitution is XYY. In which parent and in which meiotic division did nondisjunction occur? 4.7 Drosophila virilis is a diploid organism with 6 pairs of chromosomes (12 chromosomes in total). How many chromatids and chromosomes are present in the following stages of cell division: (a) metaphase of mitosis? (b) Metaphase I of meiosis? Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

4.8 In a trihybrid cross with genes subjected to independent selection: (a) What is the expected fraction of triply heterozygous offspring in the F2 generation? (b) What is the expected proportion of triple homozygous offspring in the F2 generation?

4.9 The Venezuelan diploid caconid lizard Gonatodes taniae has a somatic chromosome number of 16. If the centromeres of the 8 homologous pairs are denoted as Aa, Bb, Cc, Dd, Ee, Ff, Gg, and Hh: (a) How many different combinations of centromeres can they be produced during meiosis? (b) What is the probability that a germ cell contains only the capitalized centromeres?

4.10 For brothers with six children and an assumed sex ratio of 1:1: (a) What is the percentage without girls? (b) What is the relationship with exactly one girl? (c) What is the ratio if there are exactly two girls? (d) What is the proportion if there are exactly three girls?

(e) What is the proportion with three or more boys?

4.11 A litter of cats contains eight kittens. What is the probability that it contains males and females in an even number? Assume that male and female kittens are equally likely, and for the purposes of this problem, consider 0 to be an even number. Does the answer surprise you? Why? 4.12 A gardener has crossed a true onion plant with red onion and a true onion plant with white onion. All F1 plants had white bulbs. When the seeds resulting from the self-fertilization of the F1 plants were cultivated, the onion bulbs were obtained in the proportion of 12 white onions : 3 red onions : 1 yellow onion. Propose a hypothesis to explain these observations.

4.13 Among 160 offspring in the F2 generation of a dihybrid cross, a geneticist observes four different phenotypes in the ratio 91:21:37:11. She believes this result may be consistent with a 9:3:3:1 ratio. To test this hypothesis, she calculates the chi-square value. Does the test support her hypothesis or should she reject it? 4.14 What is the chi-square value that tests the fit between observed numbers 40:60 compared to expected numbers 50:50? 4.15 What is the probability that a sibling with seven children will have at least one boy and at least one girl? Assume the gender ratio is 1:1.

4.16 How many genotypes are possible for an autosomal gene with six alleles? How many genotypes are possible for an X-linked gene with six alleles? 4.17 The growth habit of peanut Virgínia Arachis hypogaea can be either "corridor" (scattering) or "cluster" (compact). Two purebred peanut strains with the habit of growing in clusters are crossed. F1 plants have a stolid growth habit. If the plants are able to self-pollinate, the F2 progeny ratio is 9 stolons: 7 bunches. What genetic hypothesis might explain these observations? 4.18 In some human pedigrees, the blue/brown eye color variation separates as a single genetic difference, with the brown allele being dominant over the blue allele. Two brown-eyed individuals, each with a blue-eyed parent, mate. Provided that

the trait differs as a single gene difference in this family tree: (a) What is the probability that the first child will have brown eyes? (b) If the result is a brown-eyed child, what is the probability that the child is heterozygous? (c) What is the probability that this couple will have three blue-eyed children? That none of the three children will have blue eyes?

4.19 In the attached pedigree, the I-2 male is affected by red-green blindness due to an X-linked mutation. What is the probability that the IV-1 male is colorblind? (Assume that the only possible source of the color blindness mutation in the pedigree is from the I-2 male.)

4.20 The II-1 male in the accompanying pedigree is affected by a trait due to a rare X-linked recessive allele. Use the principles of conditional probability to calculate the probability that the III-1 female is a heterozygous carrier.

4.21 The attached family tree is the same as in Exercise 4.20, but with additional information based on molecular analysis. The II-1 male, on the other hand, is affected by a trait due to a rare X-linked recessive allele. The bands on the gel are PCR products of an SSRP that identifies the wild-type allele (the shorter S allele corresponds to the shorter band at the bottom of the gel) or the mutant allele (the longer L allele corresponds to the band at the top of the gel). Using the information from the pedigree and the gels, calculate the probability that the III-1 female is a heterozygous carrier.

4.22 In the attached pedigree, males I-2 and III-2 are affected by a trait based on a common X-linked recessive allele. What is the probability that the individual is a III-1 heterozygote?

4.23 Yellow body color in Drosophila is determined by the recessive y allele of an X-linked gene, and wild-type gray color is determined by the y+ allele. What ratios of genotypes and phenotypes would you expect from the following crosses? (a) Yellow male × wild female (b) Yellow female × wild male (c) Daughter of mating in part (a) × wild male

(d) Daughter of mating in part (a) × yellow male

4.24 The accompanying pedigree and gel diagram show the molecular phenotypes for a double allelic SSRP that yields PCR products with the indicated base pair sizes. What mode of inheritance does the family tree suggest? Based on this hypothesis and using A1 to represent the SSRP allele associated with the 3 kb band and A2 to represent the allele associated with the 8 kb band, the genotype of each individual in the pedigree is deduced.

4.25 Large crocuses with yellow flowers combine with short crocuses with white flowers. Both varieties are purebred. All F1 plants are backcrossed with the fast flowering variety. This backcross produces 800 offspring in the following proportions: 234 tall yellow plants, 203 tall white plants, 175 short yellow plants, and 188 short white plants. Does the observed result fit the genetic hypothesis of 1:1:1:1 segregation evaluated by a χ2 test? 4.26 Linked X chromosomes in Drosophila are formed by two X chromosomes linked at a common centromere. Females of the C(1)RM/Y genotype, in which C(1)RM denotes the attached X chromosomes, produce equal proportions of C(1)RM-bearing and Y-carrying gametes. between a male carrying the X-linked allele for corpus luteum and an X-linked female with a wild-type corpus? How does this result differ from the typical pattern of X-linked inheritance? (Note: Drosophila zygotes containing three X chromosomes or no X chromosomes do not survive.)

4.27 A rare autosomal dominant W mutation results in curly and woolly hairs in some European pedigrees. A woman with woolly hair type O marries a man with straight hair (ww) blood type AB. The genes are on different chromosomes. (a) What are the chances that the mating will produce a group B woolly child? (b) What are the chances that the mating will produce a straight-haired child in group B? (c) If three straight-haired children with type A blood are born to these parents, what are the chances that the next child born will be hairy and with type B blood?

4.28 Ducklings of the domesticated wild duck Anas platyrhynchos may have a dark brown plumage known as Mallard, an almost black dorsal plumage known as Dusky, or a pattern known as Restricted, in which the black is limited to spots on the head and tail. These phenotypes result from the action of three alleles of a single autosomal gene. Three types of crosses were made with the following results: 1. Constrained × Mallard: All F1s are constrained; F1×F1 crosses result in an F2 generation restricted to the 3:1 Mallard ratio. 2. Mallard × dark: All F1s are wild ducks; F1 × F1 crosses result in an F2 generation with a ratio of 3 Mallard: 1 Dark. 3. Restricted × Dark: All F1 are restricted; F1 × F1 crosses produce an F2 generation restricted to the 3:1 ratio

sinister. (a) Suppose an F1 male from cross 1 is crossed with an F1 female from cross 2. List the phenotypes and their expected frequencies among the offspring of this cross. (b) Suppose an F1 male from cross 3 is crossed with an F1 female from cross 2. List the phenotypes and their expected frequencies among the offspring of this cross.

4.29 In cattle, an allele for no horns shows complete dominance: HH and Hh are hornless or hornless, and hh is horned. In contrast, the effect of the allele that produces red fur (R) shows incomplete dominance with the allele that produces white fur (r). The heterozygous Rr genotype is roan - an intermediate color in which white hairs are mixed with red hairs. H and R undergo independent classification. (a) What would be the phenotype of an F1 offspring from the RR HH × rr hh mating?

(b) What would be the expected phenotypes and their ratios in the F2 progeny of the F1 × F1 cross in part (a)? (c) What would be the phenotypes and their expected proportions among the offspring resulting from crossing the F1 individuals from part (a) with the original whitehorn stock?

4.30 Familial Mediterranean Fever (FMF) is a hereditary inflammatory disease that affects populations

from across the Mediterranean, particularly populations of Armenian and non-Ashkenazi Jews, Anatolian Turks, and Levantine Arabs. FMF is inherited in an autosomal recessive manner. It is not known whether the gene that causes the disease is the same in these different populations. To investigate this possibility, we examined the allele's binding to a variety of SSRPs. An SSRP on chromosome 16 was particularly informative. For this SSRP, two Armenian extended families from the same geographic region provided the results shown in the attached family trees and gels. Green symbols represent data subjects. The numbers between the gels correspond to the SSRP alleles. (a) Is there evidence linking the FMF allele to any SSRP allele?

(b) Does any of the data conflict with your statement? If yes, how do you calculate?

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 The attached pedigree and gel graph refer to a morphological phenotype (green symbols in the pedigree) and a molecular phenotype (a two-allele SSRP; sizes

of PCR products in base pairs are shown). What mode of inheritance do these data suggest for each trait?

CHALLENGE PROBLEM 2 In mice, the dominant T allele results in a short tail, but the homozygous T/T genotype is lethal. A cross between two short-tailed mice will produce a litter of five pups. What is the expected distribution from tailless mice to tailed mice? Why are the two most likely outcomes equally likely?

CHALLENGE QUESTION 3 X-linked hemophilia is present in the companion's pedigree, as indicated by the red icons. Given that she had two healthy children, what is the probability that the woman carries III-5? What is the likelihood that their next child will be affected?

FOR FURTHER READING Fisher, R.A. (1936). Has Mendel's work been rediscovered? Annals of Science, 1(2), 115-137. http://doi.org/10.1080/00033793600200111 The eminent geneticist and statistician R. A. Fisher addresses the question of whether Mendel's circumstances were "too good." Salz, H. (2016). Sex determination in Drosophila: the top view. Fly, 4(1), 60–70. http://doi.org/10.4161/fly.4.1.11277 A somewhat technical but very complete overview of the sex determination pathway in Drosophila. Sutton, W.S. (1902). On the morphology of the chromosome group in Brachystola magna. Biological Bulletin, 4, 24-39.

Sutton, W.S. (1903). Chromosomes in heredity. Biological Bulletin, 4, 231-251. Two papers by Walter Sutton provide some of the first experimental evidence of a link between chromosomes and heredity.

Martin Shields/Alamy Stockfotos.

CHAPTER 5 Genetic linkage and chromosome mapping CHAPTER 5.1 OVERVIEW Linkage and recombination of genes on a chromosome

5.2 Genetic mapping 5.3 Genetic mapping in a three-point test cross 5.4 Mapping by tetrad analysis 5.5 Genetic mapping in humans 5.6 Peculiarities of recombination ROOTS OF DISCOVERY Genes all in one line

Alfred H. Sturtevant (1913) The linear array of six sex-linked factors in Drosophila as demonstrated by their mode of association ROOTS OF DISCOVERY Mapping Markers in the Human Genome David Botsein, Raymond White, Mark Skolnick, and Ronald Davis (1980) Construction of a genetic linkage map in humans using restriction fragment length polymorphisms

LEARNING OUTCOMES AND SCIENTIFIC SKILLS The principles of genetic linkage and chromosome mapping are essential for the discovery of disease genes and genetic risk factors for disease. By understanding these principles, you will be able to apply the following scientific skills: ■ Use a genetic map to predict the types and relative abundances of gametes that would be produced by an individual of a given genotype. ■ Analyze the results of a genetic cross with three linked genes to deduce the genotypes of the parents, the order of the genes along the chromosome, the map distances between the genes, and the degree of interference between the crosses.

■ Identify the connection between a gene and its centromere in an organism with ordered tetrads and estimate the map distance between the gene and its centromere. ■ Calculate and interpret Lod values based on simple family trees. ■ Interpret the results of a genome-wide association scan. Recall in the chapter Medelian Genetics: The Principles of Segregation and Assortment that Mendel examined seven traits in pea plants and found that they all ranked independently. However, as we saw in The Chromosomal Basis of Inheritance, chromosomes are the opposite.

to individual genes, they are the basic units of segregation and classification. Mendel didn't know this, of course, but later researchers were able to expand the genetic analysis of peas. We now know they contain seven pairs of autosomes; moreover, the genes characterized by Mendel were located on these chromosomes (TABLE 5.1).

TABLE 5.1 Trait of Mendel's genes

Keimblatt

Dominant

recessive

genetic function

chromosome

phenotype

Gelb

Verde

Stay-Green-Gen

takes away

Branco

bHLH transcript

Farbe Samenschale/Blume

Factor

colored rod length

hoch

Dwarf

GA-3-Oxidase1

III

Pod-Form

Inflated

Restricted

Sklerenchym

III

Formation in position of pods of

Axial

terminal

meristem function

Time

Baltic

starch branching

Flor

same way

Enzyme 1 capsule color

Verde

Gelb

chloroplast

Structure on the wall of the cocoon

At first glance, these results seem somewhat confusing. Since homologous chromosomes behave as units during meiosis, genes can be expected to be located within them.

The chromosome would not undergo the same independent sorting pattern seen in non-homologous chromosomes. So how is it that genes on the same chromosome (such as those that determine seed shape and pod color, both on the V chromosome) were subjected to independent classification in Mendel's experiments? To understand this result, we now turn to the topic of linkage and its application to genetic mapping. Genes that are always transmitted together must appear

full link. Pioneering studies on this topic were conducted by Thomas Hunt Morgan, who studied mutations in the Drosophila X chromosome. He looked at linkage, but what he saw was an incomplete form of linkage, evidenced by offspring that exhibited a combination of traits not found in the parents. The linkage between genes is usually incomplete because homologous chromosomes can experience a segment exchange when paired. Thus, while the alleles present on each chromosome tend to stick together when inherited, some chromosomes that have new combinations of alleles can be transferred. An exchange event (crossover) between homologous chromosomes results in the recombination of genes, resulting in daughter chromosomes that carry combinations of alleles not present on the parental chromosomes. What we will see in this chapter is that the probability of switching between any two genes serves as a measure of the genetic distance between the genes. This makes it possible to create a genetic map - a diagram of a chromosome that shows the relative position of genes. Genetic mapping of linked genes is an important research tool in genetics because it allows a new gene to be assigned to a chromosome and even to a precise position in relation to other genes on the same chromosome. Genetic mapping is often the first step

identification and isolation of a new gene. It is essential in

Human genetics to identify genes associated with hereditary diseases, such as B. genes, whose presence predisposes carrier women to the development of breast cancer.

5.1 Linkage and recombination of genes on a chromosome As we saw in the chapter entitled The chromosomal bases of inheritance, a direct test of independent selection consists of making a testcross between a double F1 heterozygote (Aa Bb) and the double recessive homozygotes (aa bb ). Shown in FIGURE 5.1 are

the expected gametes from the Aa Bb parent if the genes are on different chromosomes. Because the possible metaphase I orientations of the homologous chromosomes are equally likely, the double heterozygote produces all four possible types of gametes in equal proportions. The alleles present in a gamete are typically written on a line with a space between each gene; in this case, the four equally likely types of gametes are A B, A b, a B, and a b.

Figure 5.1 The physical basis for the independent arrangement of genes on different chromosomes is that nonhomologous chromosomes orient independently at metaphase I of meiosis. In this example, the left orientation leads to gametes A B and a b, while the right orientation leads to gametes A b and a B.

white eyes versus normal red eyes and the m versus m+ alleles, which determine miniature versus normal wing size. When genes are linked, it is important to keep track of which alleles are present together on the same chromosome. This is usually done using a slash (also called a comma) to separate alleles present on homologous chromosomes. Morgan's first cross was between females with white eyes and normal wings and males with red eyes and miniature wings. Using the slash, we would write this cross like this:

As expected, the resulting F1 progeny consisted of wild-type females (w m+/w+m genotype) and white-eyed unreduced males (w m+/Y genotype). Then the F1 offspring were crossed:

In this cross, the X chromosomes that carry the wm+ allelic combination or the w+m allelic combination are referred to as parental chromosome types; this terminology reflects the fact that the parents involved in the cross carry the alleles in these configurations. On the one hand, if the w and m genes were fully linked, all offspring would inherit one or another of the parental chromosome types. On the other hand, if recombination occurred between the w and m genes, some offspring would inherit an X chromosome with the allelic combination w+m+ or the allelic combination wm. These are known as recombinant chromosome types because they contain combinations of alleles that differ from those present on the parental chromosomes. When the cross shown above was performed, the female offspring were all non-miniaturized with a 1:1 ratio of red:white eyes.

(if you don't understand why, write the Punnett square), and the male offspring were as follows:

Since every male receives his only X chromosome from his mother, a male's phenotype reveals the genotype of the X chromosome he inherited. The results of the experiment show a significant deviation from the 1:1:1:1 ratio of the four expected male phenotypes when classified independently, but the w:w+ and m:m+ ratios are both approximately 1:1. This type of independent assortment bias indicates that parental combinations of alleles for the w and m genes tend to stick together in inheritance but are not fully linked. In this experiment, allele combinations on parental chromosomes were present in 428/644 = 66.5 percent of F2 males, and recombinant (non-parental) chromosomes were present in 216/644 = 33.5 percent of F2 males. The value of 33.5% for recombinant chromosomes is called the recombination frequency; can also be written as 0.335. This recombination frequency should be contrasted with the 50% recombination expected with an independent sort and the 0% recombination expected with a complete linkage. This results in the recombinant X chromosomes w+m+ and wm

Crossing-over at meiosis in F1 females. In this example, the frequency of recombination between the linked w and m genes is 33.5 percent, but for other linked gene pairs it ranges from nearly 0 to 50 percent. Even genes on the same chromosome, if far enough apart, can undergo independent sorting, meaning their recombination frequency is 50%. This implies the following principle:

Genes with less than 50% recombination frequency are present (turned on) on the same chromosome. Two genes that undergo independent sorting, as indicated by a recombination frequency of 50%, are on non-homologous or distant chromosomes on a single chromosome. The possibility that two genes on the same chromosome could be unlinked creates a terminology problem. We cannot say that all genes on the same chromosome are "linked", because linkage means a recombination frequency of less than 50%, and genes that are far enough apart are not actually linked. Instead, the term synteny is used: two genes are considered synthesizers when they are on the same chromosome, regardless of whether they have an independent arrangement or linkage. If we look again at Table 5.1, this may explain why all of Mendel's traits were ranked independently. Those pairs of genes (those that affect seed shape and pod color, for example) that classify independently are actually syntenic, but the frequency of recombination between them is 50%.

Coupling versus rejection of synthetic alleles Geneticists use a slash notation for linked genes, which has the general form w+m/w m+. This notation is a simplification of the following more significant but complicated form:

In this notation, the horizontal bar replaces the slash when separating the two homologous chromosomes. For consistency, linked genes are always written in the same linear order. This convention allows the wild-type allele of a gene to be indicated with a plus sign at the appropriate position. For example, the genotype w m+/w+m could also be clearly written as w +/+ m. This genetic notation system is used for Drosophila and many other organisms. A genotype that is heterozygous for each of the two linked genes can have alleles in one of two possible configurations, as shown in FIGURE 5.2 for the w and m genes. In a so-called trans or rejection configuration, the mutant alleles are on opposite chromosomes and the genotype is written w+/+m (Figure 5.2A). In the alternative configuration, called the cis or linkage configuration, the mutant alleles are present on the same chromosome and the genotype is written as w m/++.

FIGURE 5.2 Possible configurations of mutant alleles in a genotype heterozygous for two mutations. (A) The trans or rejection configuration has the mutated alleles on opposite chromosomes. (B) The cis or linkage configuration has the mutant alleles on the same chromosome.

Morgan's investigation of the connection between the white and miniature alleles began with the trans configuration. He also studied the descent

the cis configuration of the w and m alleles resulting from crossing miniature white females with non-miniature red males. This cross is represented in the "+" notation as follows:

In this case, the F1 females were phenotypically wild-type double heterozygotes and the males had white eyes and miniature wings. A cross of the F1 progeny, again using the "+" notation, is written as follows:

This cross produced the following offspring:

This result shows that the frequency of recombination with w and m in the cis configuration (coupling) is the same observed with w and m in the trans configuration (repulsion) - 37.7% versus 33.5%. The difference is within the expected range of sample variation between experiments. In this case, however, the chromosomes that make up the parental and recombinant classes of the offspring are reversed. They are reversed because the original parents of the F1 female were different: rejection (w +/+ m) in one case and mating (w m/+ +) in

the other. Repeatedly finding equal recombination frequencies in such experiments leads to the following conclusion: recombination between linked genes occurs with the same frequency whether the gene alleles are in the trans configuration (repulsion) or in the cis configuration (coupling); is the same no matter how the alleles are arranged.

The chi-square test for binding

Each estimated value of recombination frequency (r) is subject to sample variation from one experiment to the next. How to test the statistical significance of an observed value of r < 1/2 against the alternative hypothesis that r = 0.5? To give a concrete example, how can we be sure that the r = 0.377 value observed when testing wm in the coupled configuration is not a statistical coincidence? Perhaps the true value is r = 0.5 and the genes, although synthenic, are still disconnected. Statistical significance for linkage can be tested on the observed numbers using the chi-square test discussed in the chapter The Chromosomal Basis of Inheritance. As an example, we use the wm experiment, which gives the estimate r = 0.377. The total chi-square value in the cross-test data is given in the top row of TABLE 5.2. With four descendant classes, there are 4 − 1 = 3 degrees of freedom. Only one of these classes is relevant to the linkage hypothesis; the others refer to the separation of w versus w+ and m versus m+. This point is indicated by the chi-square values in the lower rows of the table. Testing the 1:1 split of w versus w+ involves combining the m- and m+ data to obtain the observed ratio:

TABLE 5.2 Chi-square linkage test Chi-square test

Calculation

graduation

von

Wert

total freedom chi-square

(395 − 311,75)2/311,75 + (382 −

76.775

311,75)2/311,75 + (223 −

1,5 × 10−16

311.75)2/311.75 + (247 − 311.75)2/311.75 = separation of w

(605 − 623,5)2/623,5 + (642 −

contra w+

623,5)2/623,5 =

separation of m

(629 − 623,5)2/623,5 + (618 −

against m+

623,5)2/623,5 =

shortcut

(470 − 623,5)2/623,5 + (777 −

(recombinant

623,5)2/623,5 =

1.098

0,295

0,097

0,755

75.580

3,5 × 10−18

versus non-recombinant) Sum of 1-d.f. Thu

76.775

quadratic values

To test the 1:1 m- versus m+ split, we pool the w- and w+ data:

The non-linkage hypothesis (r = 1/2) predicts a 1:1 ratio of recombinant:non-recombinant progeny, so in this case we pooled the data as follows:

The three individual chi-square values, each with 1 degree of freedom (d.f.), are shown in Table 5.2. Every chi-square with 3 d.f. has a P-value that can be calculated using numerical methods such as 1.5 × 10−16; this leaves no doubt that the underlying ratio of the four progeny classes deviates from 1:1:1:1. Neither w versus w+ segregation nor m versus m+ segregation indicate a 1:1 difference, with P values of 0.295 and 0.755, respectively. However, the chi-square value for recombinants compared to non-recombinants is highly significant, with a P-value that can be calculated as 3.5 × 10−18. From this test, we can safely conclude that the cause of the

Discrepancy is linkage, and we have already estimated the recombination frequency as r = 470/1247 = 0.377. Also note in Table 5.2 that the 1-d.f. The chi-square values add up to the total chi-square value of 76.775. Additivity means that the three tested hypotheses are independent in the sense that the truth or falsity of one of them does not affect the validity of the others. In practice, we are often not interested in the detailed segregation analysis described in Table 5.2. Usually, we just want to test for binding, and this test is based on the chi-square value for recombinant versus non-recombinant: The chi-square test for binding is a test for an equal number (1:1 ratio) of recombinant versus non-recombinant progeny , which has 1 degree of freedom. Testing of individual 1:1 splits is generally not required.

Each pair of linked genes has a characteristic frequency of recombination. The recessive y allele of another X-linked gene in Drosophila results in a yellow body color instead of the usual gray color determined by the y+ allele. In one experiment, white-eyed females were mated with yellow-bodied males, and then wild-type F1 females were cross-tested with white-eyed, yellow-bodied males:

The offspring was

In a second experiment, yellow-bodied, white-eyed females were crossed with wild-type males, and then F1 wild-type females and yellow-bodied, white-eyed F1 males were crossed:

In this case, 98.6 percent of the F2 offspring had parental chromosomes and 1.3 percent had recombinant chromosomes. Parental and recombinant types were reversed in the

reciprocal crosses, but the recombination frequency was practically the same. Females with the trans y+/+w genotype produced about 1.4 percent of the recombinant offspring carrying the recombinant yw or ++ chromosome; likewise, females with the cis y w/++ genotype produced about 1.3 percent of recombinant offspring carrying the recombinant y + or + w chromosomes. However, the frequency of recombination between the yellow body and white eyes genes was much lower than that between the white eyes and miniature wings genes (1.4% versus about 35%). These and other experiments support the following conclusions: ■ Recombination frequency is a characteristic of a given gene pair. ■ Recombination frequencies are the same in cis (mating) and trans (rejection) heterozygotes.

Recombination in Females Versus Males Drosophila is unusual in that no recombination occurs in males. Although it is not known how (or why) mating is prevented in males, the lack of recombination in Drosophila males means that all synthetic genes (those located on the same chromosome) show complete linkage in the male. For example, the cn (red) and bw (brown eyes) genes are both on chromosome 2, but so far apart that 50% of recombination occurs in females. then the cross

produces offspring of genotype + +/cn bw and cn bw/cn bw (the non-recombinant types), as well as cn +/cn bw and + bw/cn bw (the

recombinant types) in the ratio of 1:1:1:1. However, since there is no crossing over in males, reciprocal crossing occurs.

produces only offspring of the non-recombinant genotypes ++/cn bw and cn bw/cn bw in equal proportions. The lack of recombination in Drosophila males is a convenience that experimental designs take advantage of. As shown in the case of cn and bw, all alleles present on any chromosome in a male must be transmitted as a group without being recombined with alleles present on the homologous chromosome.

The complete lack of mating in Drosophila males is atypical, but it is not uncommon for the frequency of recombination to differ between the sexes. The human genome, for example, also shows more recombination in females than in males, but the difference is not as extreme as it is in Drosophila. Across the human genome, the frequency of recombination between genetic markers in males averages about 60% of the value observed between the same genetic markers in females.

SUMMARY ■ Gametes produced as a result of meiosis can be parental or recombinant. ■ If recombination between two genes occurs less than 50% of the time, the genes are linked. ■ Recombinant gene-linked gametes arise as a result of mating in meiosis. ■ Each pair of loci has a characteristic recombination frequency, regardless of whether the alleles at these loci are cis or trans genotypes in the parental generations. ■ When two genes are syntenic but undergo recombination more than 50% of the time, they arrange themselves independently, as if they were on different chromosomes.

■ Recombination rates can differ between males and females

5.2 Genetic Mapping The linkage of genes in a chromosome can be represented in the form of a genetic map, showing the linear arrangement of genes along the chromosome, with the distances between adjacent genes proportional to the frequency of recombination between them. A genetic map is also known as a linkage map or chromosomal map. The concept of genetic mapping was the first

Developed in 1913 by Morgan's student Alfred H. Sturtevant (see Roots of Discovery: Genes All in a Row). Geneticists first understood that recombination between genes occurs through an exchange of segments between homologous chromosomes in the process now called crossing over. Each crossover manifests itself physically as a chiasm, or cruciform configuration, between homologous chromosomes; Chiasmata are seen in prophase I of meiosis. Each chiasm results from the breakage and reunion of non-sister chromatids, resulting in the exchange of corresponding segments between them. The crossover theory holds that each chiasm results in a new association of genetic markers. This process is shown in FIGURE 5.3. If there is no crossing over (Figure 5.3A), the alleles present on each homologous chromosome remain in the same combination. When crossing over occurs (Figure 5.3B), the outermost alleles on two of the chromatids are exchanged (recombined).

FIGURE 5.3 (A) Alleles on the same chromosome stick together if there is no crossover between them. (B) In crossing-over, the result is two recombinant and two non-recombinant products, because the exchange occurs between only two of the four chromatids.

Map Distance and Recombination Frequency The unit of distance in a genetic map is called the map unit. At short intervals, 1 map unit equals 1 percent recombination. For example, two genes that recombine with a frequency of 3.5% are said to be 3.5 map units apart. A card unit is

Also called Centimorgan (abbreviated cM) in honor of T. H. Morgan. Therefore, a distance of 3.5 map units corresponds to 3.5 centimorgans and indicates 3.5% recombination between the genes. For simplicity, we list the four fully equivalent ways in which a genetic distance between two genes can be represented: ■ as a frequency of recombination (0.035 in the example above) ■ as a percentage of recombination (3.5%) ■ as a distance of map in map units (3.5 map units) ■ As map distance in centimorgans (3.5 cM)

ROOTS OF THE DISCOVERY Genes in sequence Alfred H. Sturtevant (1913) Columbia University, New York, New York

The linear array of six sex-linked factors in Drosophila as shown by their mode of association Genetic mapping remains the cornerstone of genetic analysis. It is the most important technique used in modern human genetics to identify the chromosomal location of mutated genes associated with hereditary diseases. The genetic markers used in human genetics are usually single nucleotide polymorphisms (SNPs) that differ from person to person, but the basic principles of genetic mapping are the same as originally formulated by Sturtevant. In this snippet, we replace the symbols

currently in use for the y (yellow body), w (white eyes), v (red eyes), m (miniature wings), and r (rudimentary wings) genes. (The sixth gene mentioned is another mutant white allele, now called eosin white.) In this article, Sturtevant uses the term "crossing over" instead of "recombination" and "crossovers" instead of "recombinant chromosomes". We have kept their original terms, but in some cases we have bracketed the modern equivalent.

“These results provide a new argument in favor of the chromosomal view of inheritance, as they strongly suggest it.

”

that the examined factors are arranged in a linear series.

After summarizing Morgan's observations on w and m, Sturtevant noted: "It has become apparent that some of the sex-linked factors are associated, d in F1 flies are much more common in

the F2 then are new combinations of the same characters. From a chromosomal perspective, this means that chromosomes, or at least certain segments of them, are more likely to remain intact during meiosis than to exchange materials…. It seems, if this hypothesis is correct, that the ratio of "crossovers" [recombinant chromosomes] could be used as an index of the distance between any two factors. So, by determining the distances (in the above sense) between A and B and between B and C, one should be able to predict AC…. Does this prediction come true? In the data in the table below, the first column is the recombination frequency, which was predicted based on the sum of recombination frequencies between intermediate markers (i.e., the sum of observed frequencies

Frequencies between markers A and B and between markers B and C). The second column is the observed frequency of recombination between A and C, ignoring the B factors

calculated distance

Observed percentage of crosses

s-v

30.7

32.2

j-m

33,7

35,5

j–r

57,6

37.6

w-m

32,7

33,7

w-r

56,6

45.2

Reprinted from Sturtevant, Z.H. (1913). The linear arrangement of six sex-linked factors in Drosophila as shown by their mode of association. Journal of Experimental Zoology, 14(1), 43–59. Copyright (c) 2005 by John Wiley and Sons, Inc. Reprinted with permission from John Wiley & Sons, Inc.

Sturtevant's notes:

It is immediately noted that the longer distances y-r and w-r result in a smaller percentage of crossings than the calculation requires. This is a point that was expected and probably due to the occurrence of two breaks in the same chromosome, or "double crossing over". The final conclusion of the work is the most significant, as it establishes the conceptual basis for mapping genes based on recombination frequencies, which is the paradigm so far: it was discovered that it is possible to map six factors linked to sex in Drosophila in a linear fashion Arrange in series, using the number of crossovers per 100 cases [the frequency of recombination] as an index of the distance between any two factors. A source of error in predicting the strength of association between untested factors is found in

Double cross. The occurrence of this phenomenon is demonstrated…. These results provide a new argument in favor of the chromosomal view of inheritance, as they strongly suggest that the studied factors are arranged in a linear series. Source: A.H. Sturtevant, J. Exp. Zool. 14 (2005): 43-59.

Physically, 1 map unit can be defined as the chromosome length at which, on average, 1 crossover is formed every 50 meiotic cells. This principle is illustrated in FIGURE 5.4. If 1 meiotic cell in 50 has a crossover, the crossover frequency is 1/50 or 2 percent. However, the frequency of recombination between genes is 1 percent. The correspondence of 1% recombination with 2% crossover is a bit confusing until you realize that each crossover results in 2 recombinant chromatids and 2 non-recombinant chromatids (Figure 5.4). A crossover frequency of 2% means that among the 200 chromosomes that result from meiosis in 50 cells, exactly 2 chromosomes (those involved in the exchange) are recombinant for genetic markers flanking the crossover. In other words, 2% crossover equals 1% recombination because only half of the chromatids in each cell are involved.

an exchange are actually recombinant.

FIGURE 5.4 Chromosome configurations in 50 meiotic cells in which 1 cell shows crossover between two genes. (A) The 49 cells without mating produce 98 AB and 98 AB chromosomes; these are all non-recombinant. (B) The cell with a crossover gives rise to chromosomes that are A B , A b , a B , and a b , the middle two types of which are recombinant chromosomes. (C) The recombination frequency is 2/200 or 1 percent (also referred to as 1 card unit or 1 cM). Therefore, 1% recombination means that 1 in 50 meiotic cells will crossover in the region between the genes.

In situations where genetic markers are present along the chromosome, such as the allele pairs A,a and B,b in Figure 5.4, recombination between marker genes occurs only when the genes cross. FIGURE 5.5 illustrates a case where the crossover occurs between the A gene and the centromere rather than between the A and B genes. The crossover results in a physical exchange of segments between the innermost chromatids. However, because the exchange is outside the region between A and B, all resulting gametes must carry either the A B or a b allele.

Combination. These are non-recombinant chromosomes. The presence of the crossover goes unnoticed because it does not occur in the region between the genetic markers.

FIGURE 5.5 Out-of-region crossover between two genes is not detectable by recombination. Although a chromosomal segment is swapped, the genetic markers remain in non-recombinant configurations - in this case, A B and a b.

In some cases, the region between the genetic markers is large enough that two (or even more) junctions can form in a single meiotic cell. A possible configuration for two crossovers is shown in FIGURE 5.6. In this example, both crossovers are between the same pair of chromatids. This results in a physical exchange of a chromosomal segment between the marker genes, but the double crossover is not detected because the markers themselves do not recombine. The lack of recombination arises because the second cross, in terms of recombination between A and B, reverses the effect of the first. The resulting chromosomes are either A B or a b, neither of which is recombinant.

FIGURE 5.6 If two crossovers occur between marker genes and both involve the same pair of chromatids, no crossover will be detected because all of the resulting chromosomes are not recombinant A B or a b.

Since a double crossover in a region between two genes may go unnoticed because it does not result in recombinant chromosomes, there is an important difference between the distance between two genes, measured by recombination frequency, and the distance measured in map units. The map units measure how much crossing over occurs between genes. For any two genes, the map distance between them is defined as follows:

In contrast, recombination frequency reflects how much recombination is actually observed in a given experiment. When double crosses do not result in recombinant gametes, as in Figure 5.6, they contribute to map distance but not to recombination frequency. This distinction is only important if the region in question is large enough for double crossing to occur. When the region between genes is so short that no more than one crossover can be made in the

region in any meiosis, then the map units and recombination frequencies are the same (because there are no multiple crossovers that can reverse). This is the basis for saying that, for short distances, 1 map unit equals 1 percent recombination. Over an interval so short that it gives 1 percent observed recombination, the frequency of multiple crossovers is usually negligible, so in this case the map distance is equal to the recombination frequency.

For a set of linked genes, if the chromosomal region between each adjacent pair is short enough that multiple crossovers are not formed in the region, then recombination frequencies (and therefore map distances) between genes are additive. This important feature of recombination, which is also the rationale used in genetic mapping, is shown in FIGURE 5.7. The genes are all located on the Drosophila X chromosome: y (yellow body), rb (ruby red eye color), and cv (shortened cross-wing vein). The recombination frequency between y and rb genes is 7.5% and between rb and cv is 6.2%. The genetic map can be one of three options depending on whether y, cv or rb is in the middle. Map A, which has y in the middle, can be excluded because it implies that the recombination frequency between rb and cv must be greater than that between rb and y, which contradicts the observed data.

FIGURE 5.7 In Drosophila, the y (yellow body) and rb (ruby red eyes) genes have a recombination frequency of 7.5 percent, and rb and cv (shortened winged cross vein) have a recombination frequency of 6.2 Percent. There are three possible genetic maps depending on whether y is in the middle (A), cv is in the middle (B) or rb is in the middle (C). Map A can be excluded because it implies that rb and y are closer than rb and cv, while the observed recombination frequency between rb and y is greater than that between rb and cv. Maps B and C are compatible with the given data.

Maps B and C correspond to recombination frequencies. They differ in their predictions of recombination frequency between y and cv. On Map B, the predicted distance is 1.3 map units; in comparison, on Map C the predicted distance is 13.3 map units. In fact, the observed recombination frequency between y and cv is 13.3 percent. Map C is correct. There are actually two genetic maps that correspond to the C map. They differ only in whether y is placed on the left or on the right. A card is:

The other card is:

These two ways of presenting the genetic map are completely equivalent. By this kind of reasoning, all known genes on a chromosome can be assigned a position on a genetic map of the chromosome by considering consecutive subsets of three markers. Each set of synthetic genes forms a linkage group. The number of linkage groups equals the number of haploid chromosomes in the species. For example, cultivated maize (Zea mays) has 10 pairs of chromosomes and 10 linkage groups. A partial genetic map of chromosome 10 is shown in FIGURE 5.8 along with the phenotypes exhibited by some of the mutants. The peaks in Figures 5.8C and 5.8F demonstrate the result of Mendelian segregation. Figure 5.8C shows a 3:1 segregation of yellow:orange nuclei produced by the recessive allele pericarp orange 2 (orp2) in a cross between two heterozygous genotypes, and Figure 5.8F shows a 1:1 segregation of marbled:white nuclei produced by the dominant allele R1-mb in a cross between a heterozygous genotype and a normal homozygous genotype. As an exercise, try counting the cores in Figures 5.8C and 5.8F to see if these relationships hold.

FIGURE 5.8 Genetic map of chromosome 10 in maize, Zea mays. The map distance for each gene is given in standard map units (centimorgans) relative to a position 0 for the telomere of the short arm (lower left corner). (A) Mutations in the oil Yellow-1 gene (oy1) result in a yellow-green plant. The plant in the foreground is heterozygous for the dominant Oy1 allele; Behind is a normal plant. (B) Mutations in the lesion-16 gene (les16) result in many

discolored patches irregularly distributed on the leaf blade and leaf sheath. The photo shows the phenotype of a heterozygote for Les16, a dominant allele. (C) The orp2 allele is a recessive allele expressed as the orange pericarp, a maternal tissue that surrounds the nuclei. The ear shows orp2 segregation in a cross between two heterozygous genotypes, resulting in a 3:1 ratio of yellow:orange seeds. (D) The zn1 gene is Zebra Necrotic-1, in which dying tissue appears in bands of longitudinal leaflets. The left leaf is homozygous zn1; the one on the right is the wild type. (E) Mutations in the Teopod-2 (tp2) gene result in many small, partially clipped ears and a single tassel. An ear from a plant heterozygous for the dominant Tp2 allele is shown. (F) The R1-mb mutation is an allele of the r1 gene, resulting in a red or purple color in the aleurone layer of the seed. Note the marbled color in the cores of a peak secretion for R1-mb. Data from an illustration by E. H. Coe. Photos courtesy of M.G. Neuffer, College of

Agriculture, Food and Natural Resources, University of Missouri.

Crossing Over The orderly arrangement of genes represented by a genetic map is consistent with the conclusion that each gene occupies a well-defined location (locus) on the chromosome. In a heterozygote, the alleles occupy corresponding positions on homologous chromosomes. The passage is provoked by a physical

Exchange of segments resulting in a new association of genes on the same chromosome. This process has the following characteristics: 1. The exchange of segments between the parental chromatids occurs in the first meiotic prophase after chromosome duplication. At this stage, the four chromatids (strands) of a pair of homologous chromosomes are closely linked. Crossing over is a physical exchange between chromatids in a pair of homologous chromosomes. 2. The exchange process consists of the breakage and union of the two chromatids, resulting in the mutual exchange of equal and corresponding segments between them (Figure 5.3). 3. Crossover points are more or less random along the length of a chromosome pair. Therefore, the probability of switching between two genes increases as the physical distance between the genes along the chromosome increases. This principle is the basis of genetic mapping.

Recombination between genes results from a physical exchange between chromosomes Classic evidence that recombination involves a physical exchange of segments between homologues

The chromosomes used two structurally altered Drosophila X chromosomes that made it possible to detect parental and recombinant chromosomes under a microscope. This experiment, performed by Curt Stern in 1936, is sketched in FIGURE 5.9. One of the abnormal X chromosomes was missing a segment that was attached to chromosome 4; this X chromosome can be identified by its missing terminal segment. The second aberrant X chromosome could be identified because it had a small second arm made from a piece of the Y

Chromosome. The mutant alleles auto (a recessive allele resulting in clove color instead of wild-type red) and B (a dominant allele resulting in striped eyes instead of round eyes) were present on the first abnormal X chromosome and wild-type alleles of these genes were located in the second abnormal X.

FIGURE 5.9 (A) Diagram of a cross where the two X chromosomes in a female Drosophila are morphologically distinguishable from each other and from a normal X chromosome. One X chromosome has an end segment missing and the other has a second arm made from a fragment of the Y chromosome. (B) Result of cross. Offspring that are a slash but not a slash (single asterisks) contain a structurally normal X chromosome, and offspring that are a slash but not a slash (double asterisks) contain an X chromosome with both morphological markers. The result shows that genetic recombination between marker genes is associated with a physical exchange between homologous chromosomes. Segregation of the missing terminal segment of the X chromosome attached to chromosome 4 is not shown.

Females with both structurally and genetically marked X chromosomes were crossed with males who had a normal X carrying the recessive alleles of the genes (Figure 5.9). Among the offspring of this cross, flies with parental or recombinant chromosomes can be identified by eye color and shape. In the genetically recombinant progeny of the cross, the X chromosome had the appearance that would be expected if the recombination of the genes were accompanied by a switch that recombined the chromosomal markers. In particular, wild-type rodent offspring had an X chromosome with a terminal segment missing and the Y arm attached; in contrast, the round-eyed, carnation-colored offspring had a structurally normal X chromosome, with no missing terminal and no Y arm. As expected, the non-recombinant offspring were found to have an X chromosome structurally identical to one of the X chromosomes of the mom.

The crossing takes place in four

Strand Stage of Meiosis So far, we have asserted, without citing experimental evidence, that crossing over in meiosis occurs after the chromosomes have duplicated, at the stage when each strand of divalent chromatid has four strands of chromatids. Experimental evidence that mating occurs after chromosome duplication comes from a study of laboratory stocks of D. melanogaster, in which the two X chromosomes in a female are joined at a common centromere to form a variant chromosome shape, which is referred to as appended -X. or X-compound, chromosome. The normal X chromosome in Drosophila has a centromere near the end of the chromosome. Attaching two of these chromosomes to a single centromere results in a chromosome with two equal arms, each consisting of a nearly complete X. women with a

The compound X chromosome usually also contains a Y chromosome, and they produce two classes of viable offspring: females, who have the maternal compound X chromosome along with a paternal Y chromosome, and males, who have the maternal Y chromosome along with an X chromosome. paternal. have a chromosome (FIGURE 5.10).

FIGURE 5.10 Attached X chromosomes (Compound X) in Drosophila. (A) A structurally normal X chromosome in a female. (B) A linked X chromosome, in which the long arms of two normal X chromosomes are attached at a common centromere. (C) Typical females with an attached X also contain a Y chromosome. (D) Result of a cross between a female with an attached X and a normal male. Genotypes with three X chromosomes or no X chromosomes are lethal. In this case, a male fly gets its X chromosome from its father and its Y chromosome from its mother—the opposite of the usual situation in Drosophila.

Attached X chromosomes are commonly used to study X-linked genes in Drosophila. If an X-attached female is mated with a male carrying an X-linked mutation, the sons will get the mutated X chromosome and the daughters will get the attached X chromosome. In matings with attached X females, then the

Inheritance of an X-linked gene in males is from father to son to grandson, etc., which is the opposite of normal X-linked inheritance. On an attached X chromosome, where one X carries a recessive allele and the other carries the non-mutant wild-type allele, crossing between the arms of the X chromosome can result in attached X products in which the recessive allele is present in both. it is the attached X chromosome (FIGURE 5.11). Therefore, heterozygous females with the X attached can produce some female offspring that are homozygous for the recessive allele. The frequency with which homozygosity is observed increases as the distance from the gene map to the centromere increases. From the diagrams in Figure 5.11, it is clear that homozygosity can only arise if the crossover between the gene and the centromere occurs after chromosome duplication. Finding homozygous female offspring with an appended X, therefore, implies that crossing over occurs in the four-chain stage of meiosis. If this were not the case and crossing over occurred before chromosome duplication (at the second-strand stage), this would result in an exchange of alleles between the arms of the chromosome and would never produce the homozygous products actually observed.

Figure 5.11 Crossover must occur at the four-strand stage of meiosis to produce a homozygous X chromosome linked from a heterozygous chromosome for an allele. To achieve homozygosity, exchange between the centromere and the gene must occur.

Multiple crossovers When two genes are far apart along a chromosome, more than one crossover can occur between them in a single meiosis, and the probability of multiple crossovers increases with the distance between the genes. Multiple crossing complicates the interpretation of recombination data. The problem is that some of the multiple switches between genes do not lead to recombination and therefore go unnoticed. As we saw in Figure 5.6, the effect of one crossover can be canceled out by another.

Crossover further down the chromosome. If there are two exchanges between the A and B genes between the same two chromatids, the net effect is that all chromosomes are non-recombinant, whether A B or a b. Two of the products of this meiosis have an exchange of their intermediate segments, but the chromosomes are not recombinant for the genetic markers, so they are genetically indistinguishable from uncrossed chromosomes. The possibility of such cancellation means that the true map distance defined in Equation 1 is underestimated by setting it equal to the observed recombination frequency. In higher organisms, double crossover on sufficiently short chromosome sections is effectively ruled out. Therefore, by using recombination data for closely linked markers to build genetic linkage maps, we can avoid multiple crosses that cancel each other out. For genes far apart along a chromosome, the true map distance is estimated by summing the recombination frequencies in shorter subintervals within which multiple crossovers do not occur.

The minimum recombination frequency between two genes is 0. The recombination frequency also has a maximum: no matter how far apart two genes may be from each other, the maximum recombination frequency between any two genes is 50 percent. This maximum recombination frequency presumably applies to the synthetic genes analyzed by Mendel (Table 5.1); 50% recombination is the same value that would be observed if the genes were on non-homologous chromosomes and sorted independently.

The maximum frequency of recombination is observed when genes are so far apart on the chromosome that at least one crossover almost always forms between them. Figure 5.3B shows that a single change in each meiosis would result in half the products with parental combinations and the other half with recombinant combinations of genes. Two switches between two genes have the same effect as shown in FIGURE 5.12. Figure 5.12A shows a double-stranded double junction in which the same chromatids participate in both exchanges; without recombination

of marker genes can be detected. When the two alterations share a chromatid (three-strand double junction, Figures 5.12B and C), the result is indistinguishable from a single alteration: two products are produced with parental blends and two with recombinant blends. Note that there are two types of three-strand doublets, depending on which three chromatids are involved. The final possibility is that the second switch affects chromatids that did not participate in the first switch (double four-strand crossover, Figure 5.12D), in which case all four products are recombinant.

Figure 5.12 The result of two crosses in the gap between two genes is indistinguishable from an independent ordering of the genes, since the chromatids randomly participate in the exchange. (A) A double splice of two strands. (B,C) The two types of three-strand double junctions. (D) A four-strand double junction.

In most organisms, when double crossovers are formed, the chromatids participating in the two exchange events are randomly selected. In this case, the expected proportions of the three types of double exchanges are 1/4 double two-strand, 1/2 double three-strand, and 1/4 double four-strand. This means on average

Recombinant chromatids are found between the four chromatids

produced from meiosis with two exchanges between a pair of genes. This is the same ratio obtained in a single exchange between genes. Additionally, a maximum of 50 percent recombination is achieved for any number of exchanges. In the discussion of Figure 5.12, we emphasized that, in most organisms, the chromatids that participate in double replacement events are chosen at random. In this case, the maximum recombination frequency is 50%. When there is a non-random selection of chromatids in consecutive transitions, the phenomenon is called chromatid interference. Note in Figure 5.12 that, compared to a random selection of chromatids, excess four-strand double crossing (positive chromatid interference) results in a maximum recombination frequency of more than 50%; likewise, excess double-stranded crossover (negative chromatid interference) results in a maximum recombination frequency of less than 50 percent. Therefore, the finding that the maximum frequency of recombination between two genes on the same chromosome is not 50% can be taken as evidence of chromatid interference. So far, positive chromatid interference has not been observed in any organism; Negative chromatid interference has been reported in some fungi. Double crossing is detectable in recombination experiments using three-point crosses involving three pairs of alleles. If there is a third pair of alleles, c+ and c, between the two we were looking at (the outermost genetic markers), double changes can be detected in the region if the crosses flank the c gene (FIGURE 5.13). The two crossings - which in this example take place between the same +

Pair of chromatids – would result in a reciprocal exchange of the c+ and c alleles between the chromatids. A three-point cross is an efficient way to obtain recombination data; it's also a simple way to determine the order of the three genes, as we'll see in the next section.

Figure 5.13 A double cross of two strands spanning the intermediate pair of alleles in a triplet heterozygote results in a reciprocal exchange of the intermediate pair of alleles between the two chromatids involved.

SUMMARY ■ The crossover frequency between specific pairs of genes is proportional to the physical distance between them on the chromosome. ■ A map unit, or centimorgan, is the distance over which crossing produces 1% recombinant gametes. ■ A single crossover occurring within one meiosis produces two non-recombinant and two recombinant gametes. ■ Two or more crossovers can occur between pairs of loci.

■ Crossover occurs after DNA replication during prophase I of meiosis.

5.3 Genetic Mapping in a Three-Point Test Cross The data in TABLE 5.3 results from a test cross in maize with three linked genes. We will use this data to illustrate the analysis of a three-point crossover. The recessive alleles are lz (for slow or prostrate habit), gl (for glossy leaf), and su (for sugary endosperm), and the multiple heterozygous parent in the cross had the following genotype:

TABLE 5.3 Progeny of a three-point testcross in maize testcross phenotype

Hybrid gamete genotype

number

offspring

pai

Normal (wild type)

Lz Gl Su

286

Lazy

lz Gl Su

Brillant

Lz gl Su

sugary

Lz Gl see below

lazy, bright

lz gl Su

Lazy, sweet as candy

lz Gl see below

Bright, sweet like sugar

Lz gl su

Lazy, bright, sugar candy

lz gl su

272

Thus, the two classes of offspring that inherit uncrossed gametes (parent type) are normal plants and plants with the lazy-shiny-sugar phenotype. These classes are much larger than any cross-classes. If the combination of dominant and

recessive alleles on the chromosomes of the heterozygous parent were unknown, so from their relative frequencies among the offspring we could deduce that the uncrossed gametes were Lz Gl Su and lz gl Su. This point is important enough to state as a general principle: in any genetic cross with linked genes, however complex they may be, the two most common gamete types for each gene pair are

non-recombinant; these provide the linkage phase (cis versus trans) of the gene alleles in the multiple heterozygous parent. In mapping experiments, the order of genes along the chromosome is usually not known in advance. In Table 5.3, the order in which the three genes are presented is completely arbitrary. However, there is an easy way to determine the correct order from the three-point data: just identify the genotypes of the double-crossed gametes produced by the heterozygous parent and compare them to the non-recombinant gametes. Since the probability of two simultaneous exchanges is significantly less than the probability of a single exchange, double mated gametes will be the least common types. Table 5.3 shows that the classes consisting of four plants with the sugary phenotype and two plants with the lazy glossy phenotype (products of gametes Lz Gl su and lz gl Su, respectively) are the least abundant and therefore the double offspring - crossover. As shown in Figure 5.13, the double crossover causes members of the intermediate pair of alleles to be switched between chromosomes. Therefore, if the parental chromosomes are:

and the double crossed chromosomes are:

so Su and su are interchanged by the double cross and must be the middle pair of alleles. Therefore, the genotype of the heterozygous parent in the cross should be written as follows:

which is now correctly represented both in terms of gene order and the configuration of alleles on homologous chromosomes. A two-strand double cross between chromatids of these parental types is plotted below, and the products can be seen to correspond to the two gamete types identified as double crosses in the data.

From this diagram it can also be seen that the reciprocal products of a single exchange between lz and su would be Lz su gl and lz su eq and that the products of a single exchange between su and gl would be su gl and lz su eq. We can now summarize the data more informatively by noting the genes in the correct order and identifying the numbers of the different types of chromosomes produced by the heterozygous parent and present in the offspring.

Note that each class of individual recombinants contains two reciprocal products and that these are found at approximately equal frequencies (40 versus 33 and 59 versus 44). This observation illustrates an important principle: the two reciprocal products resulting from a cross or a combination of crosses are expected to occur with approximately equal frequency among offspring.

When calculating the recombination frequency from the data, remember that double recombination chromosomes result from two changes - one in each of the chromosomal regions defined by the three genes. Therefore, the recombinant chromosomes between lz and su are represented by the following chromosome types:

Since 79/740 or 10.7 percent of the chromosomes recovered in the offspring between the lz and su genes are recombinants, the map distance between these genes is estimated to be 10.7 map units or 10.7 centimorgans.

Likewise, chromosomes that are recombinant between su and gl are represented by the following chromosome types:

The recombination frequency between the su and gl genes is 109/740, or 14.7 percent, so the map distance between them is estimated to be 14.7 map units, or 14.7 centimorgans. The genetic map of the section of the chromosome where the three genes are located is therefore:

One mistake many students make when analyzing three-point crosses is forgetting to include duplicate recombinants when calculating the recombination frequency between adjacent genes. You can avoid falling into this trap by remembering that double recombinant chromosomes show single recombination in both regions.

Chromosomal interference at double junctions The ability to detect double junctions allows you to determine whether the switch occurs independently in two different regions. Using information from the corn example, we know from the recombination frequencies that the probability of recombination between lz and su is 0.107 and between su and gl is 0.147. If the crossover in the two regions is independent (the

means that forming a swap does not change the probability of the second swap), so the probability of a swap in both regions is the product of these separate probabilities, or 0.107 × 0.147 = 0.0157. This implies that in a sample of 740 gametes, the expected number of double-crosses would be 740 × 0.0157 = 11.6, while the actual observed number is only 6. Such deficiencies in the observed number of double-crosses are common and identify a phenomenon called chromosomal interference, in which mating in one region of one chromosome reduces the likelihood of a second mating in a nearby region. Because chromosomal interference is nearly universal, while chromatid interference is virtually unknown, the term interference, when used without qualification, almost always refers to chromosomal interference.

(Video) Heredity: Crash Course Biology #9

The coincidence coefficient is the observed number of duplicated recombinant chromosomes divided by the expected number. Its value provides a quantitative measure of the degree of interference, defined as:

For our example data, the coefficient of coincidence is 6/11.6 = 0.51, which means that the observed number of duplicate crosses was only 51% of the number we would expect if the crosses in the two regions were independent. The degree of interference depends on the distance between the genetic markers and the species. In some species, interference increases as the distance between the two outer markers decreases, until reaching a point where double crossings are eliminated - that is, no double crossings are found and the coincidence coefficient is 0 (or for say the same, the interference is equal to 1). In Drosophila this distance is approx.

10 units of card. For most organisms, interference essentially disappears when the total distance between genetic markers is greater than about 30 map units and the coincidence coefficient approaches 1.

Genetic Mapping Functions The effect of interference on the relationship between genetic map distance and recombination frequency is shown in FIG.

FIGURE 5.14. Each curve is an example of a mapping function that represents the mathematical relationship between the genetic distance over an interval in mapping units (centimorgans) and the observed frequency of recombination over the interval. In other words, a mapping function converts a map distance between genetic markers into a recombination frequency between the markers. As we have seen, when the map spacing between markers is small, the recombination frequency is equal to the map spacing. This principle is reflected in the curves of Figure 5.14 in the region where the map spacing is less than about 15 cM. Unless this distance, all curves are almost straight, which means that the image distance and the recombination frequency are the same: 1 image unit corresponds to 1 percent recombination and 10 image units correspond to 10 percent of recombination.

Figure 5.14 A mapping function is the relationship between the genetic map distance over an interval and the observed frequency of recombination over the interval. Map distance is defined as half of the average number of intersections converted to a percentage. The three mapping functions correspond to different assumptions about disturbances, i. In the first curve, i = 1 (complete interference); in this case, the curve ends abruptly, as there can be at most one crossover per chromosome. On the third curve, i = 0 (no disturbance). The mapping function in the middle is based on the assumption that i decreases as a linear function of distance.

For distances greater than 15 map units, the curves differ. The upper curve is based on the assumption of complete interference i, so i = 1. With this imaging function, the linear relationship applies up to an imaging distance of 50 cM. The turn then ends abruptly at a map spacing of 50 map units. It ends because complete interference means that a chromosome cannot have more than one crossover along its length; Since a crossover has 50 map units, this is the maximum chromosome length.

The other two curves in Figure 5.14 represent different interference patterns along the chromosome. The bottom curve is commonly called the Haldane mapping function in honor of its inventor. No interference is assumed (i = 0) and the mathematical form of the function is r = (1/2)(1 − e−d/50), where d is the map distance in centimorgans. Any mapping function for which i is between 0 and 1 must be in the range between the upper and lower curves. The specific example shown in Figure 5.14 is the Kosambi mapping function, which assumes that interference decreases according to i = 1 − 2r as a linear function of distance. Although the underlying assumptions are simple, the formula for the Kosambi function is not simple.

Most mapping functions are nearly linear near the origin, like those in Figure 5.14. This quasi-linearity implies that, for map distances less than about 15 cM, regardless of the chromosomal interference pattern, there are so few duplicate recombinants that the percent recombination frequency is essentially equal to the map distance. Therefore, the map distance between two widely separated genetic markers can be estimated with some certainty by summing the map distances in smaller segments between the markers, provided each of the smaller segments is less than 15 map units in length.

Genetic Map Distance and Physical Distance Generally speaking, the greater the physical separation between genes along a chromosome, the greater the map distance between them. On the one hand, physical distance and genetic map distance are usually correlated, as a greater distance between genetic markers provides a greater chance of mating to occur.

On the other hand, the general correlation between physical distance and genetic map distance is not absolute. We have already established that the frequency of recombination between genes can differ in males and females. Unequal recombination frequencies mean that the sexes can have different spacing in their genetic maps, even though the physical chromosomes of the two sexes are the same and the genes must be in the same linear order. An extreme example is found in males of Drosophila, where crossing over does not occur. In Drosophila

males, the map distance between any pair of genes located on the same chromosome is 0; however, genes on different chromosomes undergo independent sorting. The general correlation between physical distance and genetic map distance can be broken even on a single chromosome. For example, crossing over is much less common in some regions of the chromosome than in other regions. The term heterochromatin refers to specific regions of the chromosome that exhibit a dense and compact structure at interphase; these regions accommodate many of the standard dyes used to visualize chromosomes. The rest of the chromatin, which only becomes visible after the chromosomes condense in mitosis or meiosis, is called euchromatin. In most organisms, the main heterochromatic regions are adjacent to the centromere; smaller blocks are located at the ends of the arms of chromosomes (the telomeres) and are interspersed with euchromatin. In general, crossing over is much less common in regions with heterochromatin than in regions with euchromatin. Because there is less crossover with heterochromatin, a given length of heterochromatin appears much shorter on the genetic map than an equal length of euchromatin. Therefore, in heterochromatic regions, the genetic map gives a distorted image of the physical map. An example of such a distortion appears in FIGURE

5.15 Comparing the Physical Map and the Genetic Map of Chromosome 2 in Drosophila. The physical map is shown as the chromosome appears in metaphase of mitosis. Two genes near the tips and two genes near the euchromatin-heterochromatin junction are indicated on the genetic map. The card distances in the euchromatic arms are 54.5 and 49.5 card units, respectively, for a total euchromatic card distance of 104.0 card units. However, heterochromatin, which makes up about 25% of the entire chromosome, is only 3.0 map units in genetic length. That one

The distorted length of heterochromatin in the genetic map results from the reduced frequency of crossover in heterochromatin. Despite the distortion of the genetic map through heterochromatin, in regions of euchromatin there is a good correlation between the physical distance between genes and their distance in map units in the genetic map.

FIGURE 5.15 Chromosome 2 in Drosophila as it appears in metaphase of mitosis (physical map, top) and genetic map (bottom). Heterochromatin and euchromatin have contrasting colors. The genes are net (liquid wing veins), pr (purple eye color), cn (cinnabar eye color), and sp (wing pigment spot). The total length of the map is 54.5 + 49.5 + 3.0 = 107.0 map units. Heterochromatin makes up 3.0/107.0 = 2.8 percent of the total map length, but accounts for about 25 percent of the physical length of the metaphase chromosome.

SUMMARY ■ A three-point testcross can be used to determine the order of three genes on a chromosome and the map distances that separate them. ■ In the progeny of this cross, the progeny of the two reciprocal classes with parental gametes represent the largest class, while the double recombinants represent the smallest class. ■ Comparing progeny with parental gametes with progeny with double recombinants allows us to determine the order of genes on chromosomes. ■ Chromosomal interference occurs when the observed number of duplicated matings differs from the expected number, since mating events occur independently. It is quantified by the coincidence coefficient. ■ The genetic map functions estimate the true genetic map distance between genes based on the observed frequency of recombination.

■ Genetic maps are generally correlated with physical chromosome maps, although there are significant differences in the relative distances between loci.

5.4 Mapping by tetrad analysis Up to this point, all of our genetic analyzes have inferred descent from crosses, which obviously resulted from the occurrence of many meiosis (two for each individual). We were able to draw some conclusions about meiotic events based on the process, as occurs in the case of X chromosome abnormalities (Figures 5.9 and 5.11), but these remain limited and indirect.

Some fungal species allow us to directly examine the results of individual meiosis. They are haploid, so recessive mutations are expressed and produce a large number of offspring. Most importantly, each meiotic tetrad is contained in a sac-like structure called an ascus (plural: asci) and can be recovered as an intact group. Each product of meiosis is contained in a separate reproductive cell called an ascospore, and all ascospores formed from a meiotic cell remain together in the ascus (FIGURE 5.16). The advantage of using these organisms to study recombination is the potential to analyze all the products of each meiotic division. In this section, we show how analysis of meiotic products in Asci (tetrad analysis) can be used to study genetic linkage.

Fig. 5.16 Formation of an ascus containing all four products of a single meiosis. Each product of meiosis forms a reproductive cell called an ascospore; these cells are held together in the ascus. The segregation of a pair of chromosomes is shown.

The life cycle of these organisms is relatively short. In budding yeast, diploid is induced to meiosis by growth on a medium that is nutritionally deficient. The resulting haploid meiotic products that form ascospores germinate to give the vegetative stage of the organism (FIGURE 5.17). In some species, each of the four products of meiosis appears in sequence

undergoes mitotic division, with the result that each member of the tetrad produces a pair of genetically identical ascospores. In most organisms, meiotic products or their derivatives are arranged in no particular order in the ascus. However, bread molds of the genus Neurospora and related species have the useful property that meiotic products are arranged in a specific order that is directly related to the levels of meiotic divisions. We'll briefly discuss unordered tetrads, and then move on to the ordered ones, which are particularly valuable when using

Genetic inference to understand meiosis.

FIGURE 5.17 Life cycle of the yeast Saccharomyces cerevisiae. Mating type is determined by the a and α alleles. Both haploid and diploid cells normally reproduce by mitosis (budding). Depletion of nutrients in the growth medium induces meiosis and sporulation of cells in the diploid state. Diploid nuclei are orange; haploid nuclei are yellow.

Disordered Tetrads In tetrads, when two pairs of alleles separate, three patterns of separation are possible. For example on the cross

There are three types of tetrads:

Referred to as the parental ditype (PD) tetrad. Only two genotypes are

shown, and the alleles have the same combinations found in the parents.

ab ab ab

Called an NPD tetrad (non-parental ditype). Only two genotypes are

shown, but the alleles have no parental matches.

aB aB AB

Referred to as a tetratype (TT) tetrad. All four possible genotypes are

Gift.

The formation of tetrads, each containing all four products of a single meiosis, provides experimental evidence for two fundamental principles of transmission genetics. The first principle is that of Mendelian segregation. Note that each type of tetrad has a ratio of 2A:2a and 2B:2b. These ratios imply that Mendelian segregation is not just a statistical average of ratios over a large number of meiotic subdivisions. Instead: Mendelian allelic segregation occurs at each individual meiotic division and results in exactly two daughter cells carrying one allele and exactly two daughter cells carrying the alternative allele.

The second principle concerns crossing over. Note that tetratypic tetrads consist of AB, A b, a B, and a b allelic configurations. In other words, two of the products are parental type and two are recombinant type. The equality of parental and recombinant types implies the following: The crossover occurs at the four-strand stage of meiosis and results in two chromosomes that are recombinants and two chromosomes that are non-recombinant. Tetrads have played an important role in fungal genetics because the connection between genes can be easily identified by tetrad analysis. Linkage evidence is based on the relative number of parental ditype (PD) and non-parental ditype (NPD) tetrads:

If the genes are not linked, the parental ditype tetrads and the non-parental ditype tetrads are expected to have the same frequency (PD = NPD). Linkage is indicated when non-parental ditype tetrads occur at a much lower frequency than parental ditype tetrads (NPD < PD). The reason for the equality PD = NPD for unlinked genes is shown in FIGURE 5.18 (Panel A) for two pairs of alleles located on different chromosomes. In the absence of crossover between any gene and its centromere, the two chromosome configurations are equally likely at metaphase I, i.e. PD = NPD. When one of the genes crosses to its centromere (Figure 5.18B), a tetratypic tetrad results, but this does not change the fact that PD = NPD.

Figure 5.18 Types of disordered asci produced with two genes on different chromosomes. (A) Without crossover, the random arrangement of chromosome pairs at metaphase I results in two different combinations of chromatids—one producing PD tetrads and the other NPD tetrads. (B) When crossing-over occurs between a gene and its centromere, the two chromosome sets result in TT tetrads. When both genes are tightly linked to their centromere (so that mating between each gene and its centromere is rare), few TT tetrads are produced.

In contrast, parental ditypes are much more common than non-parental ditypes when the genes are linked. To see why, assume that the genes are linked and consider the events required to produce the three types of tetrads. Figure 5.19 shows that when there is no crossover between the genes, a PD tetrad is formed.

educated. A simple crossover between the genes leads to a TT tetrad. The formation of a two-strand, three-strand, or four-strand double junction results in a PD, TT, or NPD tetrad, respectively. With linked genes, meiotic cells without junctions will always outperform those with double four-strand junctions. This leads to

the NPD inequality 1.0, which implies that amino acid changes occur faster than neutrality - a result that would be expected if amino acid changes were favorable. Examples with Ka/Ks > 1.0 are the sex-determining gene Sry, some homeodomain proteins, vertebrate growth hormone, some olfactory and gustatory receptors in certain Drosophila species, and histone proteins specialized in spermatogenesis. We will discuss other examples related to human evolution later in this chapter. The estimation and interpretation of the Ka/Ks ratio is not as

simple as it may seem at first glance. Some of the complications are as follows: ■ Most first codon positions in a coding sequence consist of a combination of nondegenerate and doubly degenerate sites, and most third codon positions are a combination of nondegenerate, doubly, and put quadruple. Correct enumeration of non-synonymous sites and synonymous sites therefore requires an averaging procedure. In practice, enumeration is usually performed by computer programs. ■ The mutation process can be biased. For example, most organisms show a mutational propensity for transitions versus transversions. Any distortion can change the neutral expectation value of Ka/Ks. ■ Over long enough evolutionary periods, mutations can occur at the same location, changing a nucleotide

switch to another and then another, or go back to the original. Correct estimates of Ka/Ks must be corrected for such "multiple hits". ■ A Ka/Ks value sufficiently greater than 1.0 to be statistically significant requires strong selection among many amino acids distributed throughout the molecule. Amino acid changes can therefore be selected for but do not produce Ka/Ks significantly greater than 1.0. ■ Ka/Ks only provides information on possible selection at the amino acid level; it is completely insensitive to changes in gene expression levels. Thus, a gene may be under strong positive selection for regulatory changes but not have an increased level of Ka/Ks.

Origins of new genes: orthologs and

Paralogs In the course of evolution, new genes often emerge from existing genes, and new gene functions develop from functions of earlier genes. The raw material for new genes comes from duplications of regions of the genome that may contain one or more genes (described in the chapter Human Karyotypes and Chromosomal Behavior). Duplications are relatively common. Analysis of genomic sequences from a variety of eukaryotes suggests that a eukaryotic genome of 30,000 genes should experience about 60 to 600 duplications per million years. From an evolutionary point of view, two types of duplication must be distinguished. The first is typified by beta-globin in the genetic tree of FIGURE 19.1. Each time a speciation event occurred, represented by a branch of the tree, the beta-globin gene

has been duplicated in the sense that each derived species has a copy of the beta globin gene that existed in the parent species. Genes that are duplicated concurrently with speciation and retain the same function are called orthologous genes. New gene functions can arise from duplications that occur in the genome of a single species. Duplications within a genome result in paralogous genes. An example of paralogous genes are the beta globin gene and the alpha globin gene. although the genes

now distinct, they are sufficiently similar in sequence to show conclusively that they descended from a single gene in a distant common ancestor. Therefore, the beta globin and alpha globin genes are paralogs. (To take this discussion one step further, the beta genes in any two mammalian species are orthologous, as are the alpha genes. However, the beta gene in one species and the alpha gene in another are paralogous.) If a duplication of gene occur. First, the paralogs are superfluous and you can go ahead anyway. Probably the most common event is that one of the paralogs undergoes a mutation that destroys its function or a deletion that eliminates it. Occasionally, however, mutations occur that cause divergence in the functions of the paralogs. For example, they may develop different pH optima so that one gene product works optimally in relatively basic compartments of the cell and the other in relatively acidic compartments. In another case, genetic rearrangements can merge two unrelated genes and give rise to a new activity. The Molecular Genetics of the Cell Cycle and Cancer chapter provides examples of how translocations that fuse a transcription factor to an oncogene can lead to acute leukemia. These special rearrangements are extremely deleterious, but the creation of chimeric genes provides an example of how new functions can be acquired.

Gene duplications also allow paralogous copies to develop more specialized functions. An example is shown in FIGURE 19.5. At the top is a gene in a multicellular eukaryote that has enhancers for expression in tissue types 1 (yellow) and 2 (green). The other panels show gene duplication followed by sequential mutations that delete either the type 2 enhancer elements (left) or the type 1 enhancer elements (right). The result is that the left paralog is expressed only in type 1 tissue and the right only in type 2 tissue.

Functional skills are referred to as subfunctionalization. This can often be advantageous, as each of the specialized genes is free to evolve toward optimal function in its own range of expression.

FIGURE 19.5 Specialization of paralogous genes through subfunctionalization. In this example, the paralogs are specialized for tissue-specific expression.

SUMMARY ■ DNA and protein sequences can be used to infer evolutionary history. ■ Genetic trees that trace the history of individual genes are not necessarily identical to the pedigrees of the taxa involved. ■ Various methods are available for deriving genetic trees; Bootstrapping provides a means of evaluating the reliability of genetic trees. ■ The concept of a “molecular clock” is useful for dating divergence times in phylogenies. ■ The rates of evolution of certain nucleotides are influenced by their functional role, mainly in the coding regions.

■ Gene duplication is an important source of new genes and functions.

19.2 Ancient DNA Up to this point, all of our evolutionary conclusions have been derived from the analysis of modern sequences. Once considered a novelty, DNA sequencing of ancient material (aDNA) is rapidly emerging as a powerful tool for evolutionary genetic analysis. The first such sequence to be reported was a 279 bp fragment of mitochondrial DNA from the skin of a 140-year-old museum specimen of a quagga, an extinct species of zebra. The first reported supposedly ancient human DNA sequence came from an Egyptian mummy, believed to be from around 500 BC. it was dated. It was a sequence of a 3.4 kb cloned fragment; In fact, it probably comes from a modern contamination. We then learned that sequencing DNA presents several challenges: ■ The DNA must be less than approximately 1 million years old and must have been preserved under conditions suitable for preservation.

■ Most aDNA is broken down into fragments of between 30 and 80 nucleotides. As such, it is difficult to work with, and assembling such fragments into longer contigs is a challenge. ■ The bases of aDNA are degraded and this degradation can affect sequencing accuracy. In particular, the deamination of cytosine to uracil (a topic explored in the Mutation, Repair, and Recombination chapter) means that, in the polymerase chain reaction (PCR) method used to amplify aDNA fragments, C-to-T and G- to- A transition will occur, introducing errors (beyond those that occur in any PCR reaction) in the determined sequences. ■ Much of the DNA in typical ancient human samples is of microbial origin. Typically, in an ancient human sample, less than 20% of the DNA present is endogenous human DNA; the rest is mostly microbial.

■ Sample contamination is a critical issue, especially when working with human DNA. Since most DNA methods involve PCR amplification as a first step, any modern contaminating DNA (perhaps from an examiner's skin cell) is an undamaged target for amplification. In fact, several methods have been developed to solve these problems, some of which really exploit the nature of the

The Problem Itself In this section, we describe a series of steps that an investigator can take to obtain high-quality sequence information from old samples. Depending on the nature of the material and the questions addressed, the goal can range from a set of single nucleotide polymorphism (SNP) markers sufficient for phylogenetic reconstruction to a draft genome sequence that can be used to identify multiple biologics to answer a question. Questions. As shown in FIGURE 19.6, ancient DNA normally has single-stranded ends, and it is at these ends that the uracil bases resulting from cytosine deamination are concentrated. The single-stranded ends can be made double-stranded with DNA polymerase, and the uracil bases can be removed (with DNA polymerase filling in the resulting gap) by treatment with uracil-N-glycosylase. The double-stranded oligonucleotide adapters are then ligated to the ends of the fragments (FIGURE 19.6), which can then be used as primer sequences for PCR amplification, resulting in an amplified library.

FIGURE 19.6 Amplification of ancient DNA. (A) DNA is normally found as double-stranded molecules with long single-stranded ends containing numerous deaminated cytokines. (B) These ends are treated with uracil-N-glycosylase and then filled in with DNA polymerase. (C) Double-stranded oligonucleotide linkers are added to the ends of molecules to serve as initiation sites for PCR. Reprinted from L. Orlando, TP Gilbert and E. Willersley, 2015, Nature Reviews:

Genetik, 16, S. 395–405.

At this point, with massively parallel sequencing, it is technically possible to sequence the entire amplified library and then use bioinformatics tools to identify endogenous sequences. More commonly (and particularly when the goal is not to assemble a complete genomic sequence), the library is enriched by a method such as that shown in FIGURE 19.7. Essentially, this method involves selecting an amplified aDNA library of those sequences that hybridize with RNA sequences generated by in vitro transcription of modern human DNA sequences. The hybridized RNA can then be digested with RNase, leaving an enriched pool of single-stranded DNA that can be used as a template for massively parallel sequencing.

FIGURE 19.7 Enrichment of endogenous sequences in aDNA. (A) Amplified aDNA containing both endogenous and exogenous sequences is denatured. (B) An amplified library obtained from modern DNA is prepared using phage T7 promoters as terminal adapters. This amplified DNA is used as a template for transcription of complementary RNA in the presence of biotinylated uracil. (C) In vitro transcribed RNA can hybridize to the denatured DNA library, and RNA:DNA hybrids are separated on the basis of binding of biotin to streptavidin. Data from L. Orlando, TP Gilbert and E. Willersley, 2015, Nature Reviews: Genetics, 16, pp. 395-405.

Even when the most careful laboratory techniques are used, some contamination of DNA with modern material is inevitable. A variety of methods exploiting genetic and evolutionary inferences can be used to determine the level of contamination. For example,

When analyzing Neanderthal samples, the level of contamination can be determined by the frequency of reads of sequences containing bases known to be of H. sapiens and Neanderthal origin, and these reads can be ruled out as contaminants. Furthermore, the presence of Y chromosome markers in DNA derived from a female and evidence of the presence of multiple X chromosome haplotypes in male remains are indicators of contamination. To some extent, once massively parallel sequencing is complete, standard bioinformatics pipelines can be used to assemble the short reads into contigs. In aDNA analysis, even with extensive enrichment of endogenous sequences, there are readings of sequences from other sources that need to be filtered. Furthermore, statistical approaches can correct incorrect embeddings resulting from the damaged points present in the original sample. Computer algorithms to deal with these problems are rapidly evolving.

Ancient DNA is now widely used in the study of human evolution, and we will mention some of its uses in this context later in this chapter. Molecular analysis of ancient material also offers other exciting possibilities: ■ RNA-seq (described in the chapter Manipulating Genes and Genomes) was used to identify mRNA transcripts present in 700-year-old corn kernels (unfortunately, RNA molecules not preserved in tissue animal). ■ DNA associated with nucleosomes is protected from post-mortem damage, so DNA sequences associated with nucleosomes are over-represented in DNA samples. Therefore, the depth of coverage of these sequences will be greater than those in non-nucleosomal chromatin.

■ If aDNA can be used to reconstruct entire genomes, could these genomes serve as the basis for 'wiping out' extinct species? The completion of a design sequence for the woolly mammoth has led to considerable speculation about this possibility. Obviously, the transition from a genome to an organism presents many technical and ethical challenges, and even if they can be overcome, the construction of the genome requires the availability of a reference sequence of a modern species (which prevents the return of dinosaurs, at least least for now). 🇧🇷

CONCLUSION ■ DNA analysis plays an increasingly important role in evolutionary molecular genetics. ■ One of the challenges associated with DNA sequencing is that it is often degraded and can be contaminated by DNA from both ancient microbes and modern sources.

■ aDNA sequencing is typically performed by constructing a library that can be PCR amplified and sequenced using high-throughput methods.

19.3 Where Humans Fit in the Tree of Life We now turn to applying the tools of molecular evolution to understand human evolution. Humans are primates, one of nearly 20 orders of placental mammals. As early as 1863, Thomas Henry Huxley recognized that humans belong in a natural grouping with chimpanzees and gorillas due to morphological similarities. Hominoids are the group of great apes with small

Bodies (gibbons, siamangs) and big bodies (humans, chimpanzees, bonobos, gorillas, orangutans). Since the 1960s, it has been recognized from immunological data that orangutans are genetically distinct from other large hominoids. However, the relationship between humans, chimpanzees and gorillas was not determined for another 30 years.

Evidence that humans are closest to chimpanzees Chimpanzees are the species closest to humans. There are two types of chimpanzee - the common chimpanzee (Pan troglodytes) and the bonobo or pygmy chimpanzee (Pan paniscus). They are more closely related than any other primate species. Together, the two types of Pan

form a clade (a group on an evolutionary tree that includes a common ancestor and all of its descendants), and both species are genetically close to humans. What surprised some naturalists was that chimpanzees are not the closest relatives of gorillas (Gorilla gorilla), despite similarities between chimpanzees and gorillas in some characteristic traits not found in humans, such as: affecting the head and skull) and post-human proportions. cranial bumps (affecting the rest of the body excluding the skull) and protruding body hair. Humans have so many unique traits (upright walking, large brain for body size, language, culture, cuisine, ability to run long distances, covert ovulation, slow development, longevity, relative hairlessness) that many scientists assume humans must be genetically very different from chimpanzees and gorillas. Two main types of genetic evidence link humans and chimpanzees evolutionarily. The first comes from DNA hybridization experiments and the second from DNA sequence studies of multiple independently evolving genetic elements.

Similarities in Genomic DNA As discussed in the chapter DNA Structure and Genetic Variation, increasing temperature can cause strands of DNA to form a double strand

The molecule breaks down (denatured). When the duplex consists of strands of different species, the temperature at which the strands separate depends on the extent of base pairing. Yarns with a higher proportion of mismatches denature at a lower temperature than those with a better match. The thermal stability of DNA duplexes formed by hybridizing the unique (non-repeating) DNA of species pairs, therefore, provides a measure of how similar their genomic DNA is in nucleotide sequence.

When such hybrid DNA molecules are formed between many pairs of different species of great apes and humans, the hybrid that requires the most heat to denature is between humans and chimpanzees (FIGURE 19.8A). Greater thermal stability indicates that humans and chimpanzees share the greatest degree of DNA sequence similarity in their genomes, which is most easily explained if they share a younger common ancestor than any other pair of species in the experiment. Differences in the thermal stability of hybrid DNA molecules can serve as a measure of distance between species, and from these data a species tree can be derived, as shown for primates in FIGURE 19.8B.

Figure 19.8 Evidence for humans and chimpanzees as closest evolutionary relatives from DNA hybridization studies. (A) Melting temperatures Tm are shown for different double-stranded DNA molecules. Each of these is then expressed as a change in melting temperature ΔTm relative to the baseline melting temperature for the human double-stranded DNA molecule. (B) ΔTm values are genetic distances used to infer the primate evolutionary tree. Data from A. Caccone and J.R. Powell. 1989, Evolution, 43, pp. 925-942.

A limitation of DNA hybridization is that it provides a general measure of sequence similarity without actually identifying sequence differences. What particular nucleotide substitutions that humans and chimpanzees share are absent in gorillas? This issue will be discussed below.

Multiple Genetic Element Analysis As an alternative to DNA hybridization, we can use gene trees to derive species trees, although we must keep in mind two complicating factors related to gene trees. The first is the problem of incomplete ancestral classification due to the presence of polymorphisms in related species that predate the split of the two species. Fortunately, incomplete ancestry classification only occurs in a minority

of cases. Second, as is the case with any character, a given string may be ancestral or derived, depending on whether or not it is identical to the corresponding string of an immediate ancestor. To classify the bases in this way, we need to have a reasonable idea of which sequences are likely to occur in the ancestor. This is not always possible, especially when comparing rapidly evolving genes. Despite these limitations, genetic tree analysis is popular with many

Genes in different regions of the genome that evolve independently often point to the true tree of the species. Independence between genes is important because analysis of closely linked genes can produce genetic trees different from the species tree if the linked genes share the same misleading pattern of incomplete classification. However, when a sufficient number of gene trees from independently evolving genetic regions support a single tree, cumulative support increases confidence that the dominant tree is the true tree of the species (the actual phylogeny of the species). Studies of this type have found overwhelming statistical support for a tree - the one with humans and chimpanzees as the closest relatives and gorillas as a more distant species (a sister group) to the human-chimpanzee clade. More conclusive evidence came from a different approach - an analysis of the distribution of highly repetitive Alu sequences. Recall that this family includes repetitive retroposon-derived sequences present in nearly 1 million copies in human (and other primate) genomes. Importantly, most Alu sequences are transpositionally inactive. Therefore, if we examine a specific location in two genomes and find an Alu sequence present in one but not the other, we can conclude that the absence of the sequence is the ancestral state and the presence of it is the derived state. Mark Batzer and his colleagues studied the distribution of 101 Alu sequences in 8

Primate species and economic analysis used to infer phylogeny. The results of their analysis are shown in FIGURE 19.9. Note the boot support levels: all but one of the nodes are 100% supported (and the exception node can be explained by an unusual recombination event). Most importantly, these results provide unequivocal evidence that humans and chimpanzees are indeed sister species.

FIGURE 19.9 Hominid phylogeny based on analysis of the absence or presence of Alu sequences in the genomes studied, where absence is the ancestral trait and presence is the derived trait. The numbers below the branches indicate the number of Alu insertions that have occurred on the branch; Numbers above branches indicate boot support. Reproduced from Salem et al., 2005, Proc. national Akad, Sci, 100, pp. 12787-12791.

Differences Between Human and Chimpanzee Genomes Chimpanzees differ from humans in many traits, including the occurrence of disease, so the idea of comparing genomes to uncover the genetic basis for such differences is appealing. To give an example, chimpanzees experimentally infected with the human immunodeficiency virus (HIV) do not fully develop.

inflated cases of acquired immunodeficiency syndrome (AIDS), as well as most people without antiretroviral therapy. Another example is that humans suffer from liver complications due to advanced stage hepatitis B or C, but chimpanzees do not. The question of what genomic differences had accumulated over the past 6 to 8 million years of evolution between two closely related hominoid species provided the scientific impetus for sequencing the chimpanzee genome. The nature and extent of the differences between

the human and chimpanzee genomes are summarized in FIGURE 19.10. About 35 million base pairs of DNA differ between humans and chimpanzees because of single nucleotide substitutions, accounting for 1.2% of the genome. Furthermore, species are characterized by a surprising number of insertion or deletion events - that is, stretches of DNA that are present in one species but not in the other. The amount of DNA involved in these losses or gains is significant: approximately 45 million base pairs of DNA present in humans are not found in chimpanzees, and another 45 million base pairs of DNA present in chimpanzees are not present in humans. Overall, when counting nucleotide substitutions and insertion or deletion events, humans and chimpanzees differ in about 4.2% of their genomes.

FIGURE 19.10 Genomic differences between humans and chimpanzees. The genome of the common chimpanzee Pan troglodytes is used for this comparison. Data from the Chimpanzee Sequencing and Analysis Consortium. 2005, Nature, 437, pp. 69-87.

Most protein-coding genes show only subtle differences between the two species. The average gene has only acquired two amino acid changes since species divergence. About 30% of the proteins are identical in humans and chimpanzees. When focusing on individual chromosomes, most autosomes show about the same amount of genetic differences between the two species. The sex chromosomes are different: the Y chromosome differs more between humans and chimpanzees than the autosomes, while the X chromosome differs less than the autosomes. Other genetic differences revealed by the chimpanzee genome are intriguing, such as the discovery that some alleles known to cause disease in humans are the wild-type allele in chimpanzees. These include alleles for common human diseases such as diabetes and coronary artery disease. For example, the PPARG

Gene is important for lipid metabolism and fat cell development; a PPARG allele with a proline at the 12th residue in the encoded amino acid is wild-type in chimpanzees but increases the risk of type 2 diabetes in humans. A possible explanation for this difference is that in the common ancestor of humans and

In chimpanzees, disease-causing alleles were originally benign and generalized, but later physiological changes in humans (possibly in response to changes in ecology, environment, or behavior) have modified the phenotypic effects of these alleles. Six major regions of the human genome are relatively invariant in all humans (exhibiting far fewer than the normal number of polymorphisms), but they differ between humans and chimpanzees as much as any other region. It is likely that these chromosomal regions selectively contain advantageous genetic variants that arose as new mutations during hominid evolution. These variants resulted in selective sweeps (see the Genes in Populations chapter), resulting in the low level of haplotype variation seen in modern human populations. The six almost immutable regions of the human genome are distributed on different chromosomes, and each is millions of base pairs (TABLE 19.1). While most regions contain multiple genes, one contains no known genes. While these intriguing chromosomal regions may contain some of the genes responsible for the unique phenotype of all living humans, much remains to be done to determine which genes or genetic elements are responsible. In one case, there's a clue: This region on chromosome 4 has been associated with obesity in two different studies, suggesting that humans have been selecting for traits related to body weight or weight since we diverged from our metabolism-related common chimpanzee ancestor.

TABLE 19.1 Six Regions of the Human Genome That Have Been Selectively Scanned in the Last 250,000 Years Chromosome

Size

Gene

(MB) 1

400

Fourteen known genes from ELAVL4 to GPX7

4.12

ARHGAP15 (parcial), GTDC1, ZFHX1B

3.20

No known gene or predicted gene based on experimental evidence

2.63

UNC5D is FKSG2

4.32

Ten known genes from PAMCI to ATP2B1

4.07

Fifty-seven known genes from CARD10 to PMM1

Daten des Chimpanzee Sequencing and Analysis Consortium. 2005, Nature, 437, S. 69–87.

RESUME

■ DNA-DNA hybridization between human and chimpanzee DNA indicates a close evolutionary relationship between the two taxa. ■ Phylogenetic analysis based on insertion sites of the Alu sequence provided definitive evidence that humans and chimpanzees are sister taxa. ■ Sequence differences between humans and chimpanzees are subtle and much remains to be learned about their functional significance.

19.4 What do the genetic differences between humans and chimpanzees mean? How do genotypic differences translate into phenotypic differences? Most traits are determined by multiple genetic and environmental factors that interact in complex ways, leaving the relationship between genotype and phenotype unclear for most traits. Any

Some of the observed genomic differences between humans and chimpanzees may contribute significantly to phenotypic differences, while others make little or no detectable difference. Humans differ from chimpanzees in many important ways (FIGURE 19.11), and examining genetic differences to find those responsible for phenotypic traits unique to humans is the subject of ongoing research.

Figure 19.11 Some of the major phenotypic differences between humans and chimpanzees (Pan troglodytes).

Human Molecular Adaptations As we saw earlier in this chapter, the Ka/Ks measure—the ratio of nucleotide differences that change amino acids in a protein to those that do not—can be used to assess molecular adaptations in genes encoding identified proteins. An abnormally high ratio indicates that natural selection due to the modified protein actively promoted a change in the amino acid sequence between two species.

confers some advantage in the reproduction or survival of the organisms that carry it. What molecular adaptations were discovered when comparing the human and chimpanzee genomes? The Chimpanzee Genome Project analyzed sets of genes with related functions. The set of genes with the highest Ka/Ks value, which evolved most rapidly between humans and chimpanzees, is involved in epidermal differentiation (maturation of skin cells). Keratin related genes

Hair proteins also evolved rapidly. Therefore, the genes that code for features of our body's outer covering have changed greatly over the last 6 to 8 million years. Other classes of genes that diverge rapidly include those associated with pregnancy, smell, immune function and the regulation of physiological processes. Another class of genes that show great variation between species are transcription factors. Human transcription factors underwent 47% more amino acid changes than chimpanzees. Many of these genes play critical roles in early development. Although the types of genes that underwent accelerated evolution in the human lineage have been identified, much more research is needed to understand how specific genes of each type altered function in humans compared to chimpanzees and how they are associated with related relevant biomedical traits. .

FOXP2: A Gene Linked to Language Humans and chimpanzees differ in the FOXP2 gene, which is linked to language ability. A mutated form of the gene has been found in several people in three generations of a family with language impairments and in a similarly affected but unrelated individual. The types of mutations found in FOXP2 are

Diagram in Fig. 19.12. Affected individuals had two distinct types of problems - difficulty articulating words with appropriate movements of the mouth and face, and difficulty with language competence, including some aspects of grammar, syntax, word comprehension and language processing. The speech of those affected is described as "quite incomprehensible to the naive listener" with regard to the quality of their voice. Despite this, these individuals can still speak, implying that FOXP2 is not "the" language gene, but just one of several genes related to the production of competent language.

Figure 19.12 Structure of part of the FOXP2 gene, showing known human mutations that affect language and human differences from chimpanzees and other mammals. SE data Fisher, et al. 2003, year. Rev. Neurosci., 26, pp. 57-80.

Cross-species comparison of DNA sequences of the FOXP2 gene found it to be in the top 5% of the most highly conserved proteins. The specific amino acid mutated in the language-impaired family does not vary in FOXP2 proteins in organisms spanning much of the tree of life, from yeast to humans. Natural selection allowed few amino acids

FOXP2 changes over millions of years, and the few changes that have occurred are concentrated along the line leading to humans. Over the past 6-8 million years, humans have acquired two amino acid changes in the FOXP2 protein (Figure 19.12). In contrast, only one amino acid change occurred in the approximately 130 million years between the divergence of the mouse and the last common ancestor of humans and chimpanzees (FIGURE 19.13). Since the initial discovery that FOXP2 was mutated in some people with speech difficulties, another FOXP2 mutation has been found

by screening other people with speech impairments. However, the two FOXP2 differences between humans and chimpanzees do not occur at amino acid positions known to affect language.

FIGURE 19.13 Molecular evolution of the FOXP2 gene. Above the branches are the numbers of deduced amino acid changes that have occurred. The four variable amino acid sites are shown on the right for each species.

Data from W. Enard, et al. 2002 , Nature , 418 , pp . 869-8

FOXP2 is a highly expressed transcription factor in the human fetal brain. Human FOXP2 mutations that cause language disorders result in reduced levels of the encoded protein or a localized amino acid change in a crucial structural part of the protein. Lower than normal FOXP2 protein levels or altered FOXP2 proteins can cause abnormal brain development in fetuses, resulting in speech and language problems. Mice genetically engineered to have two copies of the human FOXP2 gene show behavioral differences from wild-type mice. They show reduced exploratory behavior and differ in

Ultrasonic vocalizations that play a social role mainly between mother and child. Physiologically, mice produce less dopamine in the brain. Dopamine is a neurotransmitter that affects many aspects of behavior and cognition, including voluntary movement, mood, learning, attention, problem solving, and working memory. Anatomically, FOXP2 transgenic mice exhibit cellular changes in the basal ganglia that increase the efficiency of neural circuits between the basal ganglia and the cortex. These circuits are known to control a variety of human characteristics, including language, visual memory, word and sentence comprehension, mental arithmetic ability, and cognitive flexibility. One view is that changes in FOXP2 during hominid evolution not only facilitated the evolution of various aspects of language, but also laid the groundwork for human creativity.

Differences in gene expression between

Humans and Chimpanzees Just as biomedical researchers study differences in gene expression between two types of human tissues (for example, between cancerous and normal liver cells, or between fetal and adult brain cells), evolutionary biologists study differences in gene expression between tissues of closely related species. Microarray technology and RNA-seq (see the Manipulating Genes and Genomes chapter) make it possible to compare millions of gene transcripts from two different sources in a single experiment, revealing quantitative differences in the mRNA levels of many genes simultaneously. The data for many individual genes can be summarized as a distance matrix of gene expression differences, and the distance matrix data can then be used to build a tree. When mRNA sources are from different species, the evolutionary tree shows the general divergence of gene expression between species.

Microarray technology was used to compare gene expression in blood, liver and brain tissue in humans, chimpanzees and orangutans. The tree of gene expression differences in the brain has a relatively longer branch leading from its common ancestor to humans than to chimpanzees. In contrast, trees based on blood and liver gene expression show that both branches are approximately equal in length. This difference suggests that genes expressed in the human brain have changed in expression level much more than in the human brain over the last 6 to 8 million years of evolution.

chimpanzee brain. As indicated in FIGURE 19.14, genes expressed in the liver show the greatest changes in overall expression, whereas genes expressed in the brain experienced accelerated changes in expression in the human lineage. However, the rates of change in expression of liver-expressed genes were approximately the same in both strains.

FIGURE 19.14 Differences in gene expression between humans and chimpanzees. Data from W. Enard et al. 2002, Science, 296, pp. 340-343 and W.-P. Hsieh, et al. 2003,

Genetics, 165, S. 747-757.

Transcription factor genes not only undergo a greater number of amino acid changes in humans than in chimpanzees, but their gene expression levels also undergo a greater degree of alteration. Among this class of genes, most human alterations involve an increase in gene expression compared to chimpanzees. Since transcription factors help regulate other genes, this observation underscores the importance of changes in gene regulation in human evolution.

SUMMARY ■ Comparison of specific genes based on the Ka/Ks ratio suggests that the evolution of the outer envelope of the body was very rapid. ■ FOXP2, a transcription factor, evolves very slowly but has been considered a key factor in the evolution of language in humans. ■ Transcriptome analysis shows that gene levels are changing

Expression occurred much more rapidly in human brains than in chimpanzee brains.

19.5 A Summary of Human Evolution Our knowledge of human phenotypic evolution comes primarily from studying the physical remains of early hominids—their fossilized bones (the focus of paleoanthropology) and their tools and other cultural remains (the focus of archaeology). Reconstructing human evolution is difficult because the remains are ancient (some up to 7 million years old) and rare, and there are different types of evidence.

obtained differently depending on their physical properties. Fossil teeth are more likely to be salvaged than fossil wrist bones, and stone and bone tools outlast wicker baskets or the remains of ancient fires. As a result, we have incomplete and biased evidence to reconstruct the past. To compensate for these problems, recent technological developments allow us to study ancient remains in new ways. Aspects of tooth development can now be examined using an imaging method that does not destroy fossil material during analysis. Ancient DNA from bones can also be isolated and its DNA sequenced. This section provides a summary of human evolution and lays the groundwork for discussing molecular genetic evidence for various models of human evolution and distinctive human characteristics. To reconstruct the common ancestor of humans and chimpanzees, we need to be able to distinguish between ancestral and derived characters. The common ancestor of humans and chimpanzees was likely ape-like because the most likely phylogeny based on genetic evidence has gorillas as the closest relatives of the Homo/Pan clade and because gorillas resemble chimpanzees in many physical traits. Application of the principle that evolution is most likely to follow a path with the least number of changes (economics) suggests an ape-like ancestor, similar to a mixture of features of gorillas and common chimpanzees. This economic reconstruction of

The common ancestor of humans and chimpanzees seems reasonable, given that some of the unique characteristics that make humans have many intricate and interconnected functional parts, such as the ability to walk upright (requiring a remodeled pelvis, spine and lower limbs) or our great increase in brain (which supports language). , complex reasoning and culture). It is unlikely that these complex features evolved more than once or were lost in a lineage after it evolved.

The cast of characters in human evolution

Our summary aims to capture some of the key features of hominid evolution. It is simplified and does not include all species or species that are still in dispute. The following summary is based on paleontological and archaeological evidence; the genetic evidence will be discussed later. Broadly speaking, hominids fall into two groups (FIGURE 19.15). Early hominids had small bodies and brains, the size of ape brains. Unlike the great apes, they walked upright on two legs (bipedal locomotion), although they probably did not walk as efficiently as later hominids. These primitive forms were relatively primitive and more ape-like in the brain and face, but from the neck down they displayed new, relatively derived features and were less ape-like. These creatures lived exclusively in Africa in forest habitats. They ate fruit when it was seasonally available and turned to roots and tubers as a second option when fruit was not available.

Figure 19.15 A simplified family tree of hominid evolution. Although not shown here, after Neanderthals and modern humans split from their common ancestor Homo heidelbergensis in Africa, a third spread outside Africa must have occurred along the line leading to Neanderthals. Hypothetical positions of Homo floresensis and Homo denisova are shown as dashed lines. Data from the Earth System Context Committee for Hominid Evolution. 2010. Understanding the Impact of Climate on Human Evolution. National Academy of Sciences, Washington, DC.

Evolutionarily, these early hominids likely had relatively rapid rates of maturation – more like great apes and unlike modern humans. This conclusion assumes that tooth development is representative of general development. The first permanent molars of early hominids emerged at 3-4 years of age, as in chimpanzees, compared to around 6 years of age in modern humans. During the approximately 5 million years of this first phase of early human evolution, teeth grew (except the male canines, which became smaller) and tooth enamel thickened. Adult males were 1.5 to 2 times larger than

adult females, and this degree of difference between the sexes (sexual dimorphism) is greater than that found in living humans and tends to have a more ape-like pattern. The oldest known species (Sahelanthropus tchadensis) lived until 7 million years ago (Ma), while the most recent known species of this early group of hominids (Paranthropus boisei) lived until just over 1 Ma (Figure 19.15). The last organisms are called "megadons" because of their huge teeth. Temporarily sandwiched between these species were those of the genus Australopithecus, famous for leaving at least 3.6 Ma footprints in volcanic ash at Laetoli, Tanzania. Australopithecus spawned the lineage that became the first homo species in Africa sometime before 2.5 Ma to 3 Ma. The term australopithecin collectively refers to members of Australopithecus and Paranthropus and reflects their parentage. The second group of later hominids consists of species of the genus Homo (Fig. 19.15). Compared to earlier species, later hominids are characterized by larger bodies, larger brains and longer legs. They were able to walk on two legs in addition to walking. This change in overall body shape poses a puzzle: here we have a larger animal with an enlarged, metabolically expensive brain that requires more energy.

overall, however, able to maintain this phenotype with smaller teeth than its ancestors. This is a clear sign that early Homo were able to obtain food much more efficiently than their ancestors. The pattern of increased body and brain size associated with decreased tooth size observed in the hominid fossil record is real and requires explanation because it requires innovation in energy use. Later hominids are sometimes divided into three subgroups:

Transitional hominids, premodern homos and anatomically modern humans. Homo habilis was a transitional form (Fig. 19.15). His name - literally "handy man" - derives from the structure of his hand bones and his likely ability to manipulate objects more skillfully than his ancestors. The first stone tools are found at 2.6 Ma, but it is not clear which species made them. Among premodern Homo were Homo erectus and Homo neanderthalensis. H. erectus was a premodern species that first appeared around 1.9 Ma and lasted until about 200,000 years ago. It resembled modern H. sapiens in size and body proportions, but H. erectus had a brain only two-thirds the size. Its teeth and jaws were barely larger than those of a living human with large teeth. Adult females were not much smaller than adult males, in sharp contrast to the high degree of sexual dimorphism observed in australopithecines. Evolutionarily, H. erectus resembled apes and australopithecines in that the first permanent molars appeared at around 4 to 5 years of age, which is earlier than the 6 years required in modern humans, and likely had a diet similar to that of modern humans. Notably, H. erectus was the first hominid species to leave Africa, with this migration taking place approximately 1.8 million years ago. Its fossils have been found in Africa, Europe, Southeast Asia and South Asia. The relationship of H. erectus to H. sapiens and, in particular, the fate of H. erectus

Populations that left Africa and their descendants are the subject of debates to which molecular evidence can contribute. Descendants of H. erectus who are not members of our own species are often referred to as archaic humans. This is a very general term that encompasses several species within the genus Homo, including Homo heidelbergensis, which is generally considered the common ancestor of Neanderthals and modern humans (Figure 19.15). Fossil remains of H. heidelbergensis date from 600,000 to 400,000 years ago and indicate that its brain was larger than that of H. erectus.

Homo neanderthalensis first appeared in the fossil record about 200,000 to 400,000 years ago and went extinct about 28,000 years ago. In our summary, it is considered an archaic species separate from H. sapiens, although it has been classified as a subspecies of ours, named Homo sapiens neanderthalensis. (“Neanderthals” is the modern spelling of “Neanderthals.”) Neanderthals lived in Europe, the Middle East, and West Asia, but no specimens are known from Africa. They arose from the descendants of H. erectus - more precisely from those H. heidelbergensis that migrated from Africa. Compared to H. erectus, Neanderthals evolved more slowly and their first permanent molars took longer to emerge. Compared to anatomically modern humans, Neanderthals were bonier, stockier, had a wider pelvis, and larger joints (FIGURE 19.16). Neanderthals' chests were more barrel-shaped and their faces were larger, with powerful eyebrows, prominent front teeth, large noses, and receding chins. They had a similar brain size to anatomically modern humans who lived around the same time, but their bodies were a little heavier. Humans living today have smaller brains and bodies compared to Neanderthals and Neanderthals

ancestors of modern humans. Neanderthal limbs were relatively short, likely reflecting an adaptation to cold climates – a phenomenon seen in many other mammals. Neanderthal skeletons reflect greater musculature and strength compared to living humans, which likely correlates with energy load

Lifestyle. They hunted efficiently, made spears and intricate stone tools, and were the first hominids to bury their dead. The archaeological record indicates that Neanderthals lived only in small social groups.

FIGURE 19.16 Some of the major phenotypic differences between modern humans and Neanderthals. (Left) © Ralph Hutchings/Visuals Unlimited; (Right) © Martin Shields/Science Source.

Several types of evidence now support the hypothesis that two other archaic hominin species existed in Eurasia. On the Indonesian island of Flores, fossil remains of hominids identified as Homo floresiensis based on anatomical features have been found (FIGURE 19.17). Considerable uncertainty remains as to its origin and the timing of its extinction, but the

The most recent evidence suggests that the youngest fossils are about 60,000 years old, predating the arrival of H. sapiens in Southeast Asia. Individuals of H. floresiensis were of short stature (adults were approximately 1 meter tall) and had cranial capacities comparable to those of H. erectus.

Fig. 19.17 Artist's impression of what Homo floresiensis looks like compared to a modern human female. © Mauricio Anton/Science Source.

Even more remarkable was the discovery of the Denisova people. In 2004, a 30,000 to 50,000-year-old finger bone was discovered in Denisova Cave in southern Siberia. This very limited fossil evidence made anatomical analysis impossible, and as a result, its owner's status as a member of another species remains an open question. However, as we will see, it is the ancient DNA evidence

supports the designation of this species as a sister group to H. neanderthalis. Our own species, Homo sapiens, otherwise known as anatomically modern humans, first appeared in Africa just under 200,000 years ago. At that time, there were Neanderthals and Denisovans in Europe and H. erectus (and possibly H. floresiensis) in Southeast Asia. Thus, 200,000 years ago, no less than five species of hominids existed on Earth. Much later, about 60,000 years ago,

some populations of H. sapiens dispersed from Africa to inhabit other parts of the Old World in a second hominin diaspora from Africa; other populations of H. sapiens remained in Africa (Figure 19.15). As H. sapiens spread across Eurasia, they began to inhabit the same geographic regions as Neanderthals, Denisovans, and other archaic human populations. H. sapiens began to spread in Europe 45,000 years ago. At that time or earlier, they reached East Asia and Australia. In morphology, H. sapiens differs from other hominids in having a relatively flat face, including smaller frontal ridges and a powerful forehead. Considering tooth development as an indication of general development, the slower development of teeth in H. sapiens suggests that modern humans mature at a slightly slower rate than Neanderthals. The origin of H. sapiens may have coincided with the origin of symbolic language. By the time of the oldest known art (rock paintings, sculptures) 30,000 to 60,000 years ago, symbolic expression and language were undoubtedly part of the human phenotype. About 10,000 to 12,000 years ago, the domestication of plants and animals marked a major shift in the culture and biology of modern humans, allowing humans to live in even larger and more sedentary social groups than their predecessors. That one

The increase in group size eventually led to the development of complex civilizations and cities. Larger population sizes meant that infectious diseases could be maintained in populations; Thus, infectious diseases became an important selective force at a time when cultural and technological innovations were also unleashed.

Models of modern human origins An ongoing point of debate is the relationship of Neanderthals to

our own species: were Neanderthals the closest relatives of H. sapiens that went extinct without interbreeding with us, or was there gene flow due to hybridization between the two species? A similar question about hybridization concerns H. sapiens and H. erectus and their archaic descendant populations. Figure 19.18 describes three main models (hypotheses) that have been proposed to explain how H. sapiens arose and how it is related to earlier hominins. All models include the fact that H. erectus was the first hominid to leave Africa and inhabit Eurasia (the first "out of Africa" migration). Models differ in their interpretation of what happened to these geographically diverse populations of H. erectus. In the multiregional model (Figure 19.18A), H. sapiens evolved in several geographic areas at approximately the same time from several dispersed populations of H. erectus in Africa and Eurasia. This model assumes sufficient gene flow between populations of H. sapiens to maintain the biological integrity of the new species. He predicts that the last common ancestor of all living humans was probably relatively old, perhaps 1.8 million years old, coinciding with the time of the last common ancestor of H. erectus populations leaving Africa.

Fig. 19.18 Three classic models of modern human origin. The red arrows represent the origin of Homo sapiens, which occurs several times in different locations in the multiregional model (A), but occurs only once in Africa in the other two models. The dotted lines indicate gene flow. In the hybridization model (C), gene flow represents a mixture (hybridization) between archaic hominids and H. sapiens. The T bars in the African replacement model (B) represent the extinction of archaic hominids (descendants of H. erectus) and their replacement by unmixed H. sapiens.

In contrast, the African surrogate model (Figure 19.18B) proposes that H. sapiens evolved only once in Africa and quite recently (about 200,000 years ago), and some of them left Africa to colonize other parts of the world (the second Africa event). The surrogate model assumes that the two species did not interbreed when H. sapiens populations encountered archaic human populations descended from the first migration from Africa. Instead, modern humans replaced archaic humans, and the latter disappeared. In this model, the last common ancestor of living human populations would have existed no more than 200,000 years ago.

The third model, called the hybridization model (Figure 19.18C), assumes that H. sapiens evolved only once in Africa and that, after its spread from Africa, hybridization between modern humans and archaic humans occurred when their populations entered into contact with each other. he came. The hybridization model predicts that the last common ancestor of modern humans would have been ancient due to gene flow from archaic humans into our species.

SUMMARY ■ An evolutionary analysis based on the fossil record requires distinguishing between ancestral and derived characters. ■ Hominids - members of the human lineage - are usually divided into early and later hominids. ■ Homo erectus and its descendants are often referred to as archaic humans. ■ Homo neanderthalis existed in Asia and Europe between 200,000 and 400,000 years ago and became extinct about 28,000 years ago.

■ Fossil and DNA evidence suggest the existence of two other species of the genus Homo that coexisted with Homo sapiens: H. floresiensis and H. denisova. ■ Three models for the origin of Homo sapiens have been proposed.

19.6 Genetic Evidence for Modern Human Origins The sequence of nucleotides in DNA and amino acids in proteins can be effectively used to infer the evolutionary history of humans and their populations. Advances in sequencing technology have also made it possible to analyze ancient DNA samples - particularly those extracted from the remains of Neanderthals, Denisovans and prehistoric humans. The results support hybridization between the ancestors of some humans, Neanderthals and Denisovans, as sequences from the latter two can be found in modern humans.

Tracing human history through

Mitochondrial DNA Inferring ancestral history is easier for DNA molecules that do not undergo recombination, such as B. mitochondrial DNA (mtDNA). These organelle genomes are also useful for tracking ancestors, as they evolve at a much faster rate than nuclear genes in many species. In humans, differences in mitochondrial DNA sequences between lineages accumulate according to a molecular clock at an average rate of about one change per mitochondrial lineage every 3,800 years. For example, the people of Papua New Guinea have been relatively genetically isolated from the indigenous peoples of Australia since these areas were colonized approximately 40,000 and 30,000 years ago, respectively. The total evolutionary time between populations is therefore 70,000 years (40,000 years in Papua New Guinea and 30,000 years in Australia). If one nucleotide change accumulates on average every 3800 years, the expected number of differences in the mitochondrial DNA of present-day Papua New Guineans and Australian Aborigines is estimated to be 18.4

nucleotides (calculated as 70,000/3800). These calculations are very rough, as a molecular clock can be quite variable over a given time interval, and estimates of settlement times can change with new archaeological finds. (Since this mtDNA replacement rate was first estimated, new evidence has pushed Australia's colonial age to 50,000 years.) Still, this example shows how the rate of mtDNA evolution can be used to predict the number of differences among populations. In practice, the calculation is often done the other way around, using the number of observed differences between pairs of populations to estimate the number of years since the populations separated geographically.

Probable historical relationships between human populations can be reconstructed based on differences in mtDNA nucleotide sequences. FIGURE 19.19 shows the inferred mtDNA ancestral relationships based on the complete mtDNA sequences of 53 individuals representing human populations worldwide rooted in a chimpanzee mtDNA sequence. Because mtDNA does not recombine, the tree is the genetic history of just a single DNA molecule; Also, since mtDNA is inherited from the mother, it is the genetic history of women. However, the mtDNA tree exhibits several notable features: ■ African populations were the first to diverge from the root of the human tree. ■ Much of the mtDNA diversity in African populations is not found in non-Africans; On average, African mtDNA has about twice as much genetic variation as non-African mtDNA. ■ Three of the four major mtDNA lineages are unique to sub-Saharan Africans (green).

■ The age of the most recent common ancestor of all human mtDNA sequences (green circle) is approximately 170,000 ± 50,000 years. This common ancestor of all living human mitochondrial types has been named "Mitochondrial Eve" in recognition of the fact that mitochondrial DNA is inherited from the mother. Technically, the genetically common ancestor period is known as the coalescence period. ■ A restricted subset of mtDNA lineages is found among non-Africans (purple and red); These sequences share more recent common ancestry with six African mtDNAs (purple) than these African mtDNAs with other Africans.

■ The age of the most recent common ancestor of the mtDNA lineage connecting African and non-African populations (red dot) is approximately 50,000 ± 25,000 years. This is an upper estimate for the spread of modern humans from Africa.

FIGURE 19.19 Inferred evolutionary relationships between human mitochondrial DNA molecules based on analysis of the complete nucleotide sequence of mtDNA (16,000 base pairs) from 53 humans from geographically diverse populations. Data from M. Ingman et al., 2000, Nature, 408, pp. 708-713.

These mtDNA results support the African surrogate model as the data are consistent with an African origin for modern humans based on a tree topology and a common ancestor of modern humans

that was more recent than one that was 1 or 2 million years old. There were also earlier hominin migrations from Africa, but according to mtDNA evidence, the descendants of these earlier migrants appear to have been replaced by later humans. The age of the oldest Homo sapiens fossils is ± 195,000 years

5000 years of Ethiopia. This time agrees with 170,000 ± 50,000 years for the time of the most recent common ancestor of all the mtDNA lineages in FIGURE 19.19. Even if a molecule evolves according to a molecular clock, paleontological or archaeological evidence is needed to anchor events in real time. Molecular data can only provide relative branch lengths (for example, one branch is twice as long as another). Converting relative times in a molecular tree to absolute dates in years requires at least one well-dated fossil associated with a node in the tree to serve as a calibration point. In FIGURE 19.19, the calibration is based on a human-chimpanzee divergence time of 5 million years. Current evidence supports a value closer to 7 million years, so the dates in FIGURE 19.19 would be slightly more accurate when multiplied by 7/5. A corrected coalescence time for mitochondrial Eve would then be 238,000 ± 70,000 years. This slightly earlier date is still consistent with the fossil evidence and a more recent African origin, being much smaller than the 1.8 million-year-old estimate for the common ancestor of modern humans predicted by the multiregional model. Studies of mitochondrial DNA from Neanderthals also raise the question of the origin of modern humans. These studies revealed that Neanderthal mtDNA is very different from that of modern humans. The various mtDNA types that have been sequenced from Neanderthals form a group that represents a separate sister group to the group of modern human mtDNA types. Assuming a date for the human-chimpanzee divergence of 6 million years and using the molecular clock, a

It is estimated that the common genetic ancestor of modern humans and Neanderthals existed 466,000 years ago (in the range of 321,000 to 618,000 years with 95% confidence). This discovery of separate clades for Neanderthal mtDNAs and human mtDNAs suggests that neither Neanderthal mtDNA found its way into modern humans. The implication is that if hybridization occurred, it did not between Neanderthal females and modern human males. A coalescence period represents the time since the last genetically common ancestor between two or more groups. As the time of coalescence occurs before mutational divergence, it necessarily precedes the moment when the groups descended from this ancestor separated from each other. This discrepancy explains why coalescence times estimated from genetic data are often older than population split times based on archaeological or paleontological data. The time of population split between Neanderthals and modern humans was probably between 270,000 and 440,000 years.

The Neanderthal Genome Using ancient DNA sequencing techniques, a draft Neanderthal genome sequence was assembled using three fossils found in Vindija, Croatia. Two more genomes – one from the Caucasus and one from Siberia – were then characterized. The surprising result is that living humans from Eurasia (samples from France, China and Papua New Guinea) have about 2 to 3% of their genomes descended from Neanderthals, while the genomes of living humans from sub-Saharan Africa lack Neanderthal DNA. As indicated in Figure 19.20, the direction of gene flow was from Neanderthals to modern humans, not the other way around. pairings between

Species likely varied between Neanderthal males and non-African human females, as modern human mtDNAs show no trace of a Neanderthal contribution. Note that this result does not support any of the three simple models for modern human origins shown in Figure 19.18. The data come closer to supporting the African surrogate model, but with a low level of hybridization between Neanderthals and the common ancestral population of H. sapiens that left Africa. Since all non-Africans share the same genomic contributions as Neanderthals, no matter where they live,

The hybridization event must have occurred very early in the migration of H. sapiens from Africa, but before modern humans spread to different parts of the world, probably 50,000 to 80,000 years ago. For the vast majority of our nuclear genome (96-99 percent), we last had a common ancestor that recently lived in Africa.

Figure 19.20 Hybridization between archaic hominins and modern humans has occurred at least twice, involving different ancestors of modern human populations. These mixing events were discovered through analyzes of genome data from modern humans, Neanderthals and Denisovans. Melanesians are represented here by Papuans and Bougainville Islanders from New Guinea.

Dados de D. Reich, et al. 2010, Nature, 468, pp. 1053-1060.

The Neanderthal genome also gives us insight into our own genomes. Certain segments of the genomes of non-Africans can be identified as likely descendants of Neanderthals. These genomic segments are notable for having greater genetic diversity among non-Africans than among Africans. The genetic regions that were likely favored by selection along the lineage that led to modern humans are revealed by comparing the genomes of modern humans, Neanderthals, and chimpanzees. The most prominent example of a selected region contains the THADA gene associated with type 2

Diabetes, suggesting that modern human physiology has undergone selection for changes in energy metabolism. Several genes expressed in the skin have also undergone unique changes across the human lineage, highlighting the skin as an organ that has undergone adaptive change. Genes associated with cognitive development were also preferentially selected. One of these genes is in the critical region for Down syndrome, another has been linked to schizophrenia and two more to autism. Taken together, these results provide strong evidence for this

Modern human cognitive ability was a trait selected on a more recent evolutionary timescale. Neanderthal nuclear DNA studies also revealed the presence of two unique amino acid changes in the language-related FOXP2 gene that are fixed in humans but not found in chimpanzees or other species; However, the presence of these mutations does not necessarily mean that Neanderthals could speak. In terms of physical appearance, a characteristic and unique allelic variant of the melanocortin-1 receptor (MC1R) gene, known to regulate skin and hair color in many animals, was found in two Neanderthals and indicates the nature of the allelic variant that some Neanderthals may have red hair and pale skin. No modern humans share this particular allele, but other mutations in the human gene produce a similar phenotype. Some Neanderthals were probably also able to detect the bitter taste of PTC (phenylthiocarbamide), as some modern humans do due to variation in their TAS2R38 gene (see the chapter Mendelian Genetics: The Principles of Segregation and Selection).

The Denisova Genome The only surviving tooth found in Denisova Cave was enough to construct a draft genomic sequence. The surprising result is this

Although closer to Neanderthals than modern humans, Denisovans are genetically distinct from Neanderthals. As indicated in Figure 19.20, they are a sister group to Neanderthals. Denisovans may represent a new hominid species previously unknown to paleontologists, and they achieved this status by characterizing their ancient DNA. Thus, by the time Homo sapiens left Africa, three archaic groups were already living in Eurasia—Neanderthals, Homo erectus, and Denisovans.

Among modern humans, a small amount of Denisovan DNA has been found in East Asians, South Asians, and Native Americans. Surprisingly, this percentage is higher (4-6 percent) in populations in Oceania (Papua New Guinea and Melanesia; FIGURE 19.20). The origin of Denisovan DNA in non-Melanesians remains uncertain. Gene flow between Denisovans and Oceanian ancestors likely occurred after Neanderthal DNA hybridized with that of non-African human ancestors, but before Melanesian ancestors left mainland Asia (at most 50,000 years ago) to enter the colonized islands of the northeast from Australia and Papua New Guinea. At first glance, Denisovan mitochondrial DNA is intriguing because it is so different from all known Neanderthals and modern humans. On the mtDNA tree, Denisovans are an outgroup of humans and Neanderthals, as opposed to the tree based on their nuclear genome. Denisovans appear to be more archaic in their mtDNA than in their nuclear genome. The mtDNA coalescence time for Denisovans, Neanderthals and humans is approximately 1 million years, while the mitochondrial coalescence time for Neanderthals and modern humans is half that, 500,000 years. It is possible that Denisovans retained their archaic mtDNA type, while living Neanderthals and Melanesians lost all traces of it through incomplete ancestral classification. Another possibility is Denisovans.

they themselves hybridized with another, as yet unknown, archaic hominid. The discovery of new fossil hominin specimens (and their genomes) will help clarify these alternatives as our knowledge of hominid evolution, which we have long suspected to be incomplete, becomes more complex.

SUMMARY ■ Ancient DNA analyzes have found evidence of hybridization between Homo sapiens, Neanderthals, and Denisovans. ■ Mitochondrial DNA analysis supports the African surrogate human origin model. ■ Numerous cases of Neanderthal genes that may have undergone positive selection in humans have been reported. ■ Denisovan genome differs from that of Neanderthals.

■ Genes of Denisovan origin have been found in several human populations, mainly in Oceania.

19.7 Measuring Human Diversity We present the 1000 Genomes Project in the chapter Genes, Genomes and Genetic Analysis. This project sequenced 2,504 diploid genomes from 26 global populations, divided into 5 "superpopulations". Based on these sequences, over 80 million SNPs have been identified. Furthermore, the genotypes were "phased" using pedigree and statistical analysis, meaning that the SNPs alleles in them were assigned to the maternal or paternal chromosomes. Adjacent SNPs in a region of a chromosome tend to be inherited together as a block and are therefore in linkage disequilibrium. In particular, only about 500,000 SNPs need to be examined to detect disease associations, as neighboring SNPs tend to be inherited as "haplotype blocks". In this section, we'll look at how this type of genomic data can be used to learn more about the genetic history and relationships between different human populations.

Tracing human history with genetics

Marker SNP data can be used to understand genetic relatedness between human populations. A large study examined 650,000 SNPs in 938 unrelated individuals from 51 human populations worldwide. About 97 percent of the SNPs were autosomal, 2 percent linked to the X chromosome and 1 percent to the Y chromosome or mitochondrial DNA. Subjects lived in places where their grandparents and ancestors had lived, researchers purposefully excluded immigrants, and subjects were analyzed as individuals rather than initially considered members of specific groups. FIGURE 19.21 shows the tree that summarizes the genetic relatedness between these individuals, with several outstanding features:

■ The SNP-based tree shows higher resolution across human populations than the mtDNA tree in Figure 19.19. The higher resolution is not surprising considering that more than 40 times more data collected about each chromosome was used to build the SNP tree. ■ Individuals from the same continent or continental subregion often occur together in a clade or in closely related clades. Non-Africans, in particular, tend to genetically resemble their neighbors. However, some African groups are more closely related to non-African groups than to other Africans, as seen in the mtDNA tree. ■ Sub-Saharan African groups appear near the root of the tree, followed (in that order) by populations from the Middle East, Europe, Central and South Asia, Oceania, East Asia, and the Americas. Populations that are geographically more distant from Africa tend to be farther from the root of the tree, and the order of their appearance on the tree roughly corresponds to the pattern of human expansion from Africa reconstructed from paleontological and archaeological evidence.

■ Human populations have fewer heterozygous genes, indicating reduced diversity depending on distance from a reference point in Africa.

FIGURE 19.21 Evolutionary tree derived from modern human populations based on over 650,000 SNPs from 938 individuals from 51 populations. Data from J.Z.Li, et al. 2008, Science, 319, pp. 1100-1104.

The tree in Fig. 19.21, based primarily on nuclear DNA data from modern humans, supports an African origin for modern humans. The reduced heterozygosity pattern was interpreted as a model of serial colonization in which successive

Colonies are created from a subset of humans from the previous colony (FIGURE 19.22). In this model, the first population of modern humans to leave Africa was a small subsample of the original population with less genetic diversity. Over time, the small colony grew, and eventually another small subgroup of people broke away and colonized a second geographic region, even further away from Africa. This second founding group had even less genetic diversity than the first. This process continued, creating a series of populations that became increasingly distant from one another.

Africa and less heterozygous at each step of the chain. FIGURE 19.23 presents a scenario that describes how modern Homo sapiens colonized the world based on a combination of genetic, paleontological, and archaeological evidence.

Figure 19.22 The serial colonization model for the spread of modern humans from Africa is characterized by a series of founder events in which only a small subsample of each initial population gives rise to a new population (colony) in a geographically distinct location. Each founder event represents a genetic bottleneck, where the genetic diversity of the new colony is reduced compared to the diversity of the initial population.

FIGURE 19.23 A possible scenario for modern Homo sapiens migrations based on genetic, paleontological, linguistic, and archaeological data. The date of 200,000 years (200 ka) represents an estimate of the first appearance of Homo sapiens in South Africa. Humans began to disperse from Africa around 60 ka. Not shown are the two known hybridization events with Neanderthals and Denisovans. Hybridization with Neanderthals likely occurred immediately after modern humans left Africa, when they arrived in the Middle East, and before they spread to other parts of the world. How long it took humans to reach different continents and which routes they took is still a matter of debate, and that could change as new fossil evidence is discovered.

Recent evidence from ancient DNA suggests that although serial colonization events did occur, later interbreeding or interbreeding between individuals played a crucial role even in prehistoric times. For example, consider the relationship of Native Americans to East Asians. Under the serial colonization model, we would expect the gene pool of modern Native Americans (excluding the portion resulting from recent contact with European Americans and African Americans) to comprise a subset

that of modern Siberians, and the modern Siberian gene pool is similar to that of the ancestral population from which Native Americans emerged. When the genomes of two 24,000-year-old Siberian fossils were characterized, they turned out to be very different from modern Siberians and more closely related to the natives.

Americans (and surprisingly for modern people in West Asia). Modern Siberians appear to have arisen as a result of the migration from East Asia that occurred after North America was colonized. FIGURE 19.24 shows a commonly used technique for analyzing and visualizing genomic variation, in this case using the same populations as in Figure 19.21. In this computational analysis, a specified number of ancestral groups are specified (seven in this case) and then marker genotypes within individuals are assigned to each of the groups in order to minimize Hardy Weinberg deviations and linkage equilibria ( see Genes in Populations chapter). Thus, Figure 19.24 consists of a series of 938 vertical lines, each representing the multilocus genotype of an individual in the study. The colors represent the parts of the genome derived from each of the seven ancestral populations. As we can see, particularly in the Middle East, many modern genomes are highly mixed, suggesting a common ancestor with individuals from various geographic areas. Even in areas such as northern Europe, where there is little evidence of mixing of Project 1000 Genomes data, the picture becomes much more complicated as more individuals are genotyped and older DNA analyzes are included.

FIGURE 19.24 Analysis of geographic differentiation and admixture in human populations. Based on the assumption that there are seven ancestral populations (see Figure 19.23), a computational analysis was performed to assign the genotypes of 938 individuals to each of these populations. Each genotype is a vertical line; the colors of the lines indicate the origins of the corresponding fractions of these genomes. Reprinted from J.Z.Li, et al. 2008, Science, 319, pp. 1100-1104. reprinted with

Permission from the AAAS.

The distribution of variability within and between groups The human species is hierarchically structured. At the lowest level are individual populations, above that are population clusters within a given geographic region, and finally the hierarchy is crowned by the clustered population groups of major continental regions that together make up our species. When

human genetic variation based on more than 1 million autosomal SNPs in the 1000 Genomes Project data is divided into these levels, 87% of genetic variation is contained in populations, 1% in populations in superpopulations, and 12% in superpopulations. Thus, the overwhelming majority of genetic variation within our species is variation observed at the lowest hierarchical level - that is, within populations. X-chromosome variation (based on over 32,000 SNPs) paints a similar picture: 80% of the variation is found in the within-population component, 2% in populations within geographic regions, and 18% across geographic regions. These genome-wide studies, showing that most human genetic variation occurs within local populations, demonstrate why the population is by far the most important focus for studying human genetic diversity.

Tracing human history through the Y chromosome Because the Y chromosome does not undergo recombination for most of its length, the genetic markers on the Y chromosome are fully linked and remain together as the chromosome is passed down from generation to generation. The genetic relationship between the Y chromosomes can therefore be traced: chromosomes

that are closely related share more alleles along their length than more distantly related chromosomes and therefore have closely related haplotypes. Single sequence repeat polymorphisms (SSR) are suitable targets for many genealogical studies of the Y chromosome because of their relatively high mutation rate due to replication errors and the large number of alleles. The rationale is that Y chromosomes with haplotypes that share alleles on each of the 20-30 SSRs along the chromosome must essentially descend from the same ancestral Y chromosome.

recently. In haplotypes that differ at a single locus, the genetic relationship is less close; for those that differ at two loci, it is even less close; and so on. This simple logic is the basis for tracing population history through Y chromosome polymorphisms. Haplotypes that share many alleles have a younger common ancestral Y chromosome than haplotypes that share fewer alleles. Furthermore, as the SSR mutation rate can be estimated, the point in time when the ancestral chromosome existed can be deduced. This line of reasoning forms the basis for estimating that the most recent common ancestor of all extant human Y chromosomes existed 50,000 to 150,000 years ago. Such estimates are not very accurate and require a lot of guesswork. However, much can be learned about the history of human populations through studies of the Y chromosome, as the following example shows.

THE CUTTING EDGE: The Colonization of Western Europe

Look again at FIGURE 19.21 and focus on the "European" group. Remember that these relationships were based on

Comparisons of SNP genomes in modern Europeans seem to imply that these people arose as a result of migration from the Middle East. But what information do we have about the history of humans in Europe that might enrich our understanding of the history of species in the region? What if we could do population genomics on ancient populations in Europe and elsewhere?

Traditionally, evidence of prehistoric human populations has come from physical (fossil) remains and associated cultural artifacts. We know, for example, that the oldest European traces of modern Homo sapiens date to around 45,000 years ago, and that humans have existed in Europe continuously since then. We also know from cultural artifacts that the "Neolithic transition" - from a hunter-gatherer lifestyle to an agrarian lifestyle - took place in the Middle East approximately 10,000 to 12,000 years ago and spread to Europe over the next few thousand years. years, following two routes – one along the Mediterranean coast and the other up the Danube (FIGURE A). But what can genetics tell us?

FIGURE A Proposed pathways of Neolithic and post-Neolithic gene flow to Western Europe. Shown in red are those from the Near East around 7500 BC. BC, which led to a mix between migrant agricultural peoples and resident hunter-gatherers. Shown in blue is gene flow from the steppes of Asia around 4000 BC. BC, which is also supposed to be the origin of the Indo-European languages in Europe. Proposed routes of gene flow to Western Europe in the Neolithic and Post-Neolithic periods. Shown in red are those from the Near East around 7500 BC. BC, which led to a mix between migrant agricultural peoples and resident hunter-gatherers. Shown in blue is gene flow from the steppes of Asia around 4000 BC. BC, which is also supposed to be the origin of the Indo-European languages in Europe.

While technically challenging, sequencing ancient DNA provides informative results. However, as shown in FIGURE 19.21, much can be learned from robust SNP genotyping, in contrast to whole-genome sequencing. In 2015, a large group of researchers from multiple labs reported doing just that – genotyping 69 ancient specimens from across Europe and Central Asia, dating between 1000 and 5000 BC. (determined by radiocarbon dating). What made this large-scale analysis possible was the use of SNP-specific oligonucleotides known as hybridization "decoys" used to enrich aDNA libraries for regions containing SNPs. The approach was

success: In total, the 69 samples were successfully genotyped for an average of 212,000 SNPs. Key findings are summarized in FIGURE B and placed in historical and geographic context in Figure A. In Panel B, each bar represents samples from a population of individuals, showing their inferred ancestry. From these data, the authors were able to draw several important conclusions: ■ The earliest samples from Hungary in south-central Europe were almost entirely Neolithic in character, consistent with other data that point to a migration of agrarians from the Near East by around 7500 BC. ■ Over time (until about 4500 BC), individuals of Neolithic descent intermingled with the hunter-gatherers that occupied Europe in pre-Lithic years, increasing the proportion of individual genomes derived from them.

■ Around 4500 BC. There was a rapid influx of genes from the steppe region of present-day Russia around 1000 BC, leaving evidence of little or no ancestry in some populations, whether from early Neolithic migrants or older hunter-gatherers.

FIGURE B Estimate of miscegenation of ancient populations studied. Note the dramatic increase of Yamnaya (steppe) ancestors in the German population around 4,000 years ago.

Daten von Haak e outros 207–211.

So in this case, the serial colonization model serves as a starting point, but a fuller picture requires that we include the mixing of settlers with existing populations, as well as additional waves of expansion and mixing in later epochs. These dates, in turn, raise intriguing questions about the origins of Indo-European languages in Europe. Linguists have hypothesized that they originated in Anatolia, in present-day Turkey, and spread across Europe during the Neolithic period, or later with the advent of wheeled vehicles from the steppes north of the Black and Caspian Seas. The abrupt genetic transition that began around 4500 B.C. occurred supports the second hypothesis.

Literature Haak, W., Laziridis, I., Patterson, N., Rohland, N., Mallik, S., Llamas, B…. Reich, D. (2015). Massive migration from the steppes was a source of Indo-European languages in Europe. Nature, 522, 207-211.

In the 13th century, the Mongol Empire became the largest land empire known in history. At its greatest extent, it stretched from China, through Russia, to the Middle East and then to Eastern Europe. The founder was born around 1162 and given the name Temujin. As a young man, he organized a confederation of tribes who, around 1200, running on their small Mongol ponies, equipped with tall wooden saddles and stirrups and armed with bows and arrows, began to conquer their neighbors. Soon after, Temujin took the name Genghis Khan, meaning Universal Ruler. He used to be merciless, killing men and boys from rebellious cities and kidnapping women and girls. In response to a question about the source of happiness, he reportedly said: "The greatest happiness is to defeat your enemies, chase them before you, rob them of their riches, see your loved ones in tears, save your wives and embrace your daughters in your bosom.” With countless wives, concubines and countless unrecorded sexual conquests, Genghis Khan and his descendants were prolific, his eldest son Tushi had 40 recognized children and his grandson Kubilai Khan (under whom the Mongol Empire reached its fullest extent) had 22 recognized sons. Although the legacy of Genghis Khan is well documented in history, it was hardly expected to show up in Y-chromosome studies. large region.

Asia provided the remarkable result in FIGURE 19.25. Each circle represents a population sample, with its area proportional to the sample size. The red sectors denote the relative frequencies of each group of nearly identical Y chromosome haplotypes, while the white sectors represent the relative frequencies of other haplotypes that are genetically much more diverse. The most recent common ancestor of the closely related haplotypes is estimated to have existed 1000 ± 300 years ago. In addition, the geographic region where closely related haplotypes cluster

it is mostly included in the Mongol Empire (shading). The only exception is Population 10, made up of the Hazara ethnicity of Pakistan. This population provides a clue to the origin of the closely related Y chromosomes, as the Hazara consider themselves to be of Mongolian origin and many claim to be direct male descendants of Genghis Khan. Regardless of their origin, closely related Y chromosomes are found in about 8% of males in a large region of Asia (populations 1 to 16).

FIGURE 19.25 Distribution of Y-chromosome (red) haplotypes believed to be descended from Genghis Khan or his close male relatives among populations near and bordering the former Mongol Empire. The specific populations are (1) Mongolian, (2) Han [Gansu], (3) Kazakh Chinese, (4) Han [Xinjiang], (5) Xib, (6) Uighur, (7) Kyrgyz, (8) Kazakh , (9) Uzbek, (10) Hazara, (11) Hezhe, (12) Daur, (13) Evenki, (14) Han [Inner Mongol], (15) Inner Mongol, (16) Manchu, (17) Oroqen , (18) Han [Heilongjiang], (19) Korean Chinese, (20) Korean, (21) Japanese, (22) Shezu, (23) Han [Guangdong], (24) Yaozu [Liannan], (25) Lizu , (26) Buyi, (27) Yaozu [Bama], (28) Huizu, (29) Han [Sichuan], (30) Hani, (31) Qiangzu, (32) Chinese Uighur, (33) Tibetan, (34) ) Brush, (35) Balti, (36) Kalash, (37) Tajik, (38) Baloch, (39) Parsi, (40) Makrani Negroid, (41) Makrani Baloch, (42) Brahui, (43) Turkmen, (44) Kurds, (45) Azenians, (46) Armenians, (47) Lezgis, (48) Georgians, (49) Ossetians, and (50) Svans. Data from T. Zerjal, 2003, Am. J. Hmm. Genet., 72, pp. 717-721.

A direct proof of the connection with Genghis Khan could, in principle, be made by determining the Y chromosome haplotype in

Material recovered from his tomb. He died in 1227 as a result of a fall from a horse and is said to have been secretly buried in a hidden spot near his birthplace.

SUMMARY ■ Analysis of approximately 650,000 SNPs in different populations resulted in a deduced phylogeny with much higher resolution than that obtained from mitochondrial DNA. ■ This analysis supports a serial colonization model that starts in Africa and extends to other parts of the world. ■ Although genetic differentiation is observed in modern human populations, it affects only about 12% of the genome.

■ The Y chromosome can be used to trace paternal lineages, such as B. descending from Genghis Khan.

19.8 Unique Human Genetic Adaptations As human populations colonized different regions of the world (FIGURE 19.23), they faced new environments with different climates and ecosystems, and humans adapted in response to these physiological challenges. Exposure to local parasites and other environmental diseases exert selective forces on the human immune system that differ geographically. New environments also provided opportunities for the exploration of new local food resources, while some staples were lost. We have already seen how the domestication of dairy animals in Europe and Africa led to selection for the lactase persistence phenotype (section "Molecular Signatures of Selection" in the chapter "Genes in Populations"). Culture also played an important role - for example, in the development of new foraging tools and strategies. In the following sections, we look at some of the genetic adaptations discovered so far.

Amylase and Dietary Starch With the development of agriculture, human diets changed as people took advantage of local plants and turned them into cultivated food sources. Plant species such as wheat in the Middle East, rice in Asia and corn in the Americas were domesticated. Plant foods differ in the amount of nutritional starch. Domesticated plants generally contain more starch than their wild ancestors, and human groups vary in the degree to which they depend on starchy plants. A starchy diet is characteristic of almost all farmers and hunter-gatherers who live in arid conditions and rely on roots and tubers. Low-starch diets are common among pastoralists, among hunter-gatherers in humid conditions such as

than the rainforests and among Arctic inhabitants with a diet rich in meat or fish. Different diets have probably exerted different selection pressures on human physiology since the advent of agriculture.

Digestive enzymes are therefore prime candidates as potential targets for selection among human populations. Amylase is an enzyme present in saliva and the pancreas that hydrolyses starch. Salivary amylase performs the first step in starch digestion when food is chewed. Studies of copy number variation across the human genome (discussed in the chapter DNA Structure and Genetic Variation) have shown that people vary in the number of amylase genes present in their genomes. As indicated in FIGURE 19.26, examination of populations on low-starch and high-starch diets has shown that variation in copy number correlates with dietary starch (mean of 5.4 copies in low-starch populations vs. mean of 6.7 copies in high-starch populations). starch populations). Since populations with high- and low-starch diets can live together, geographic ancestry is not an alternative explanation for the pattern. Furthermore, chimpanzees have only one copy of amylase per haploid genome, and bonobos appear to have two non-functional copies; Therefore, the most likely ancestral condition for hominids is low gene copy number. This suggests that the number of copies of the amylase gene probably increased during human evolution. While the advent of agriculture may have encouraged an increase in copies of the amylase gene, it is also possible that selection began when hominids shifted their habitat from forests and forests to drier savannahs, relying on starchy roots and tubers.

Figure 19.26 The copy number of amylase varies with the amount of starch in the human diet. Data from G.H. Perry et al. 2007, Nature Genetics, 39, pp. 1256-1260.

George H. Perry1,2, Nathaniel J. Dominy3, Katrina G. Claw1, Arthur S. Lee2, Heike Fiegler4, Richard Redon4, John Werner and six other authors (2007) 1. Arizona State University, Tempe, Arizona. 2. Brigham and Women's Hospital, Boston, Massachusetts. 3. University of California, Santa Cruz, California. 4. The Wellcome Trust Sanger Institute, Hinxton, UK.

Nutrition and evolution of human amylase gene copy number variation

The development of increased enzyme activity can occur through regulatory mutations that increase transcription of a single gene or through increases in gene copy number. This study reports a strong correlation between the amount of starch in the diet of the human population and the number of copies of a gene that encodes salivary amylase, an enzyme that breaks down starch. Humans aren't the only primates to develop higher levels of amylase activity. A group of Old World monkeys called cercopithecines, which also includes macaques and mangays, produce even more salivary amylase than humans. Cercopithecins are unique among primates in storing starchy foods, such as the seeds of immature fruits, in a cheek pouch, and it is a plausible hypothesis that increased amylase facilitates starch digestion. It is not known whether the increased amylase production in cercopithecins is due to copy number variations or some other mechanism.

"We prefer a model in which the AMY1 copy number in at least some highly positive or biased selections

”

starch populations…

First, we can consider the variation in starch content in the diet of different human populations. A distinction can be made between "starchy" populations, for whom starchy food resources are a significant part of the diet, and "starchy" populations, with traditional diets that contain relatively few starchy foods [but] emphasize the resources protein (e.g. meat and blood) and simple saccharides (e.g. fruit, honey and milk)…. Then, the genomes of individuals from these populations can be analyzed to determine the number of copies of the AMY1 gene, which encodes salivary amylase.

We estimated AMY1 copy number in three high-starch and four low-starch population samples…. Remarkably, the proportion of high-starch individuals in the matched sample with at least six copies of AMY1 (70 percent) was nearly double that of the low-starch population (37 percent)…. But here, too, the scientific maxim that “correlation does not necessarily imply causation” applies. Are these differences really the result of evolutionary adaptation to dietary starches, or could something like genetic drift be responsible? The authors note that AMY1 copy number correlates more strongly with dietary starch than with the geographic distribution of the study population. They conclude that we prefer a model in which AMY1 copy number has undergone positive or directional selection in at least some high-resistance populations, but evolved neutrally (ie, by genetic drift) in low-resistance populations. Comparisons with other great apes indicate that AMY1 copy number was probably acquired in the human lineage... . The initial human-specific increase in AMY1 copy number may have coincided with a change in diet early in hominid evolutionary history.

Source: Green, R.E. etc. al. (2010). Science, 328(5979):710-722.

Adaptation to parasites and diseases Since the advent of agriculture about 10,000 years ago and the consequent increase in population density, infectious diseases have been a burden on the health of the human population. Natural selection, therefore, has favored alleles that protect against infectious disease, and no disease illustrates this point more clearly than malaria. Several mutations in different genes that confer partial protection against malaria have been identified and all have done so.

originated in the human population less than 10,000 years ago. The list of these changes includes the mutations in TABLE 19.2. ■ Mutations that alter the amino acid sequence of beta-hemoglobin. The sickle cell HbS allele provides the amino acid

The Glu6Val acid shift was discussed in the chapter "Genes in Populations". This allele is relatively common in much of central Africa, as well as in other geographic regions where malaria was or still is endemic. The diversity of HbS alleles suggests that this mutation occurred at least four times independently in Africa. Another allele, HbC, has a mutation in the same codon that produces Glu6Lys. The highest allele frequency of HbC is around 0.2 and is found in a limited region of west-central Africa. The HbC allele is less than 5,000 years old and both heterozygous and homozygous genotypes enjoy protective effects against malaria and are associated only with a low level of anemia. Yet another allele, HbE, has the amino acid substitution Glu26Lys. Heterozygous genotypes have high malaria resistance and no adverse health effects, while homozygous genotypes have lower protection and mild anemia. The HbE allele reaches high frequencies in eastern India, Southeast Asia and Myanmar, and is particularly common in northern Thailand and Cambodia, where 70% of people are heterozygous carriers. The HbE allele is estimated to have arisen only about 1000-4000 years ago. ■ Mutations that change the amount of beta-hemoglobin. The term "thalassemia" refers to a group of diseases in which globin chains are produced in reduced amounts. Beta thalassemia can result from mutations in the promoter or enhancer regions that decrease the level of transcription, but more commonly results from deletion of all or part of the beta-globin gene. Heterozygous genotypes have some protection against malaria with few adverse symptoms, while

homozygous genotypes have severe, life-threatening anemia that requires regular blood transfusions or bone marrow transplants. ■ Mutations that change the amount of alpha hemoglobin. Chromosome 16 normally carries two copies of the alpha globin gene. Deletion of one or both copies is associated with alpha thalassemia. In heterozygous genotypes, these deletions reduce the amount of hemoglobin and confer some protection against malaria; against it in

homozygous genotypes can cause severe disease. Individuals with three copies of the alpha globin gene are clinically normal, but individuals with two or fewer copies have anemia, the severity of which depends on the number of copies. Fetuses without all four copies die in utero. ■ Glucose-6-phosphate dehydrogenase (G6PD) deficiency. The stability of red blood cells requires the G6PD enzyme. When the activity of this enzyme is reduced, cellular stress can cause red blood cells to rupture. Infection of red blood cells by the malaria parasite is a form of cellular stress, and certain alleles of the X-linked gene for G6PD that reduce activity confer protection against serious infections. The most common mutant allele in Africa has two amino acid changes (Val68Met and Asn142Asp) that reduce G6PD activity by about 88%. The allele is approximately 4000–12,000 years old, and in heterozygous females and carrier males, the allele reduces the risk of serious infection by about 50%. Another more recent allele dates from about 2000 to 7000 years ago and has a Ser188Phe substitution that decreases enzyme activity by about 97%. Most individuals with these alleles do not have clinical symptoms; However, certain environmental factors can lead to the destruction of red blood cells and anemia. The most common environmental triggers are aspirin, nonsteroidal anti-inflammatory drugs

Inflammatory medications, viral or bacterial infection, ingestion of fava beans and exposure to the chemical in moth crystals. TABLE 19.2 Genetic adaptations that protect against malaria Gen

Main types of mutations

beta-globin

HbS (Glu6Val), HbC (Glu6Lys), HbE (Glu26Lys)

beta-globin

Partial or complete deletion (beta thalassemia)

Alpha-Depth

Deletion of one or both duplicates (alpha thalassemia)

G6PD

A– (Val68Met + Asn142Asp), Med (Ser188Phe)

Malaria is not unique when it comes to inducing genetic adaptations for disease resistance. Another example is found in two alleles for a blood apolipoprotein found only in regions of sub-Saharan Africa where sleeping sickness is endemic. This disease is caused by a single-celled parasite transmitted by the tsetse fly. While blood protein alleles appear to offer some protection against the parasite, they are also linked to kidney disease.

Skin Color Human skin color is a complex trait controlled by a combination of genes, environmental factors, and their interactions. The outermost layer of the skin contains melanocytes, specialized cells that contain subcellular organelles called melanosomes that produce and store melanin, the main pigment that absorbs light. All mammals produce two types of melanin - eumelanin, which is dark brown, and pheomelanin, which is red-yellow. The number of melanosomes, their size and shape, and the relative proportions of the two pigments, as well as other factors, including hormones,

Exposure to UV rays and aging produce the primary pigmentation phenotype of human skin. Human skin color is likely an adaptive response to ultraviolet (UV) radiation. The skin pigment absorbs ultraviolet light, which protects the organism from the mutagenic effects of ultraviolet radiation, described in the chapter Mutation, Repair and Recombination. UV exposure and human skin color correlate with latitude, and dark skin protects against skin cancer in high UV exposure environments.

However, skin cancer is not the only selective factor for skin color. Dark skin also protects against folic acid (folic acid), a water-soluble B vitamin that is essential for DNA synthesis, repair, and gene expression. Folic acid is particularly important in rapidly dividing cells, such as those involved in fetal development and spermatogenesis. Folate deficiency leads to defective neural tube development, spinal defects and early fetal loss. Folic acid is destroyed by UV radiation; however, skin pigmentation protects the molecule. Skin color also plays an important role in regulating the influence of UV radiation on vitamin D, which is synthesized in the skin through a sensitive reaction to UV rays. Vitamin D is essential for human health as it allows the absorption of calcium from food to help build strong bones. Vitamin D deficiency causes rickets, a disease in which the cartilaginous precursors of bone development fail to mineralize. Vitamin D is necessary for normal cell growth and the proper functioning of the immune system, as well as inhibiting the growth of cancer cells. Furthermore, it is vital for reproduction, as rickets in females can result in narrowing of the pelvic canal, creating an increased risk of infant and maternal mortality and morbidity. Female infertility and fetal loss are consequences of vitamin D deficiency in laboratory mice.

Although vitamin D is found in some foods, the diet is a minor source of vitamin D compared to the amount synthesized in the skin when exposed to UV radiation. of vitamin D synthesis. For a given exposure to UV rays over a given period of time, lighter skin allows for the synthesis of more vitamin D compared to darker skin. For this reason, lighter skin color is believed to be selectively beneficial at higher latitudes, where the population is less exposed.

sunlight and ultraviolet radiation. Human skin color therefore results from a balance of influences on human health, including protection against skin cancer, prevention of folic acid breakdown, and sufficient vitamin D production. evolution of hominin skin color. but principles of physiology and evolution can guide speculation about how this has changed over time. Chimpanzees have fair skin and are covered in body hair. Early hominids probably also had fair skin and body hair. At some point during human evolution, body hair was lost, perhaps to keep the body cool in response to higher levels of daily activity. Paleontologists believe this change occurred about 1.5 to 2.0 Ma ago, when hominid lower limbs became longer and their diurnal ranges increased. Exposed pale skin is a problem in areas of high UV radiation because of the increased risk of skin cancer. It is believed that darker skin was preferred in early tropical humans after loss of body hair. Later, when humans spread out of Africa, skin color selection changed again as humans continued to migrate from equatorial locations to regions with reduced exposure to UV rays. For some human skin color genes, evidence suggests that natural selection has altered their abundance in different populations. For example, the SLC24A5 gene shows a completely different allele

Frequencies in Europe versus the rest of the world. An allele with the amino acid substitution Ala111Thr occurs with a frequency of 98-100 percent in several European populations, while the ancestral allele with Ala111 occurs with a frequency of 93-100 percent in African, Asian, and Amerindian populations. The SLC24A5 polymorphism explains about 25% to 38% of the differences in skin color between Europeans and Africans; However, most human genetic polymorphisms do not show such extreme population differences.

The light skin coloration of Asians is at least partially influenced by genetic variants other than those responsible for the reduced skin pigmentation of Europeans. For example, an allele of the OCA2 gene that results in His615Arg is present with high frequency in East Asians but is absent in Europeans and West Africans. This allele leads to lower levels of melanin in East Asians. Individuals with zero, one, or two copies of the derived 615Arg allele have progressively lighter skin color, and there is evidence of selection for the allele in East Asian populations. Human skin color shows a convergent evolution. Even within a single human population, the convergence of genetic variants for skin color can be extensive. The MC1R gene, which controls what type of melanin is produced by melanocytes, has more than 30 variants in Europeans that affect its activity. Another MC1R variant, not found in living humans, was present in Neanderthals. These different mutations independently result in melanosomes that contain pheomelanin (red-yellow pigment) rather than eumelanin (brown-black pigment) and produce features such as red and blonde hair, freckles, sun sensitivity, and fair skin. This phenotype likely evolved independently many times over the course of human evolution. Therefore, the similarity of skin color does not necessarily indicate a genetic relationship between individuals.

SUMMARY ■ The number of copies of the human amylase gene is correlated with the amount of starch in the diet. ■ Mutations have been identified in several genes that confer resistance to malaria. ■ The skin color evolution brought several adjustments. ■ Differences in SLC24A5 genotypes explain at least 25% of skin pigmentation differences between Europeans and Africans.

CHAPTER SUMMARY ■ The ancestral history of protein and nucleic acid molecules can be deduced as sequences accumulate changes over the course of evolution.

■ Phylogenetic analysis of Alu sequence patterns shows that chimpanzees are the closest living species to humans. ■ Several species of early hominins evolved in Africa, and some later hominins of the genus Homo migrated from Africa to other parts of the world 1.8 million years ago. ■ Anatomically modern humans evolved in Africa about 200,000 years ago, and a subgroup migrated to the Middle East about 60,000 years ago. ■ Anatomically modern humans co-existed simultaneously with other later hominids. DNA sequencing shows that a small portion of non-African human genomes (2 to 3 percent) came from interbreeding with Neanderthals, who

it probably occurred in the Middle East shortly after humans left Africa. ■ Living Melanesians have an additional contribution to their genome (2–3 percent) from Denisovans, a second archaic hominid species. ■ Despite limited interbreeding with Neanderthals and Denisovans, modern humans likely displaced older hominid populations they encountered when colonizing the world. ■ Approximately 85 percent of the genetic diversity of human populations is present in any given local population. The genetic differences between the groups are relatively small. ■ The human genome shows many examples of evolutionary adaptations that affect traits such as increased resistance to infectious diseases, better digestion of starch in foods, and changes in skin color.

CHECK THE BASICS

■ What is a genetic tree? A kind of tree? Do you always have to agree? Explain why or why not. ■ Distinguish between orthologous and paralogous genes. What kind of duplication provides the raw material for the evolution of new gene functions? ■ What is the melting temperature in a DNA hybridization experiment, and what does the melting temperature tell us about the similarity between DNA strands? ■ Given a set of gene trees of independent genes in a species group, why would you expect most gene trees to have the same structure as the actual species tree? ■ What does it mean to say that FOXP2 is a language-related gene, but not “the language gene”?

■ What is the principle of evolutionary economics? How is it used to reconstruct common ancestor traits of humans and chimpanzees? Explain. ■ Would you consider Australopithecus an example of an early hominid, a transitional hominid, a premodern homo, or an anatomically modern human and why? ■ Was there ever a time when anatomically modern humans (Homo sapiens) coexisted with both Homo neanderthalensis and Homo erectus? ■ What is the multiregional model for the emergence of modern man? Do genomic sequences support this model? ■ Do you think the term “Mitochondrial Eve” is appropriate to describe the coalescence of human mitochondrial DNA sequences? Why or why not? ■ What is serial colonization and how does this process affect the geographic pattern of genetic diversity? ■ What is blending? What role has it played in recent human evolutionary history?

TROUBLESHOOTING GUIDE PROBLEM 1 The accompanying illustration is a highly simplified tree of human mitochondrial DNA. The dates of modern man's origin and recent expansion out of Africa are based on the calibrated mitochondrial molecular clock on the assumption that the time of the most recent common ancestor of humans and chimpanzees is 5 million years ago. What would happen to these estimates if, instead, the human-chimpanzee divergence time was assumed to be 6.5 million years? Would the estimate of the margin of error also change?

ANSWER The estimates themselves and the margins of error must be increased by a ratio of 6.5:5 or by a factor of 6.5/5 = 1.3. The 170,000 ± 50,000 year estimate therefore becomes 220,000 ± 65,000 years, and the 50,000 ± 25,000 year estimate becomes 65,000 ± 32,500 years. PROBLEM 2 The attached table shows the Ka and Ks values in the protein coding region of each of the seven AG genes. Ka is the number of non-synonymous substitutions (change of amino acids) per non-synonymous nucleotide site and Ks is the number of synonymous substitutions per synonymous nucleotide site. For which gene is the amino acid sequence most conserved? Does one of the genes show evidence of positive selection for amino acid changes? ANSWER With Ka and Ks data, the pattern of evolution

Change in protein coding regions is best assessed using the Ka/Ks ratio. Ka/Ks values < 1.0 represent a conserved amino acid sequence (and the smaller the ratio, the stronger the conservation), Ka/Ks = 1.0 is expected with selective neutrality, and Ka/Ks > 1.0 is expected with more positivity Selection of expected amino acid changes. For these genes the Ka/Ks values are 0.015 (Gen A), 0.018 (B), 0.094 (C), 0.025 (D), 1.16 (E), 0.062 (F) and 0.027 (G). The smallest value is 0.015; therefore, the A gene is the most conserved, although only slightly more so than the B gene. The only gene for which Ka/Ks > 1.0 is the E gene, for which the ratio is

1.16. This value would be consistent with positive selection affecting the E.

ANALYSIS AND APPLICATIONS 19.1 What is the UPGMA distance matrix for the tree shown here?

19.2 Which rate pair should be connected first in the distance matrix shown here, and what is the resulting UPGMA tree?

19.3 Shown here is the species tree of four species of Old World monkeys, identified as A–D. If the DNA

If hybridization were performed with non-repetitive DNA from these species, for which pair (or pairs) of species would the melting temperature be lower? For which pair (or pairs) of species would the melting temperature be higher?

19.4 In a DNA hybridization experiment, what relationship would you predict between the melting temperatures of non-repeating DNA from modern humans (H), Neanderthals (N), and the common chimpanzee Pan troglodytes (C)? Briefly justify your answer. In the following list, the symbol Tm(AB) represents the melting temperature between non-repeating genomic DNA from species A and species B. (a) Tm(HN) < Tm(HC) < Tm(NC) (b) Tm(HN) < Tm(NC) < Tm(HC)

19.5 In a DNA hybridization experiment, what relationship would you predict between the melting temperatures of non-repeating DNA from modern humans (H), the common chimpanzee Pan troglodytes (C), and the bonobo or pygmy chimpanzee Pan paniscus (B)? Briefly justify your answer. In the following list are the

The symbol Tm(AB) represents the melting temperature between non-repeating genomic DNA from species A and species B. The symbol "≈" means "is approximately the same". (a) Tm(HC) ≈ Tm(HB) < Tm(BC) (b) Tm(HC) ≈ Tm(HB) > Tm(BC) (c) Tm(BC) ≈ Tm(HB) < Tm(HC) ). ) (d) Tm(BC) ≈ Tm(HB) > Tm(HC)

19.6 The attached diagram shows the species tree of gorilla (Gorilla gorilla), chimpanzee (Pan troglodytes) and human (Homo sapiens). How do these types correspond to the A, B, and C designations?

19.7 The accompanying diagram is a species tree of three species, in which the numbered line segments correspond to different branches of the species tree. If a new mutation (giving a derived allele) occurs and is fixed within an interval corresponding to one of the lineage segments: (a) In which segment would the mutation give a derived allele common to all three species? (b) In which segment would the mutation result in a derived allele shared by species B and C?

19.8 If the evolutionary divide between humans and chimpanzees occurred 6.5 million years ago, what is the total evolutionary time that separates humans and chimpanzees?

19.9 A large population of primates that are randomly mated exhibit a copy number polymorphism consisting of a tandem duplication of a gene encoding a protein. In this population, 20% of the chromosomes carry one copy of the gene and 80% carry two copies of the gene. What are the expected proportions of people with 2, 3, and 4 copies of the gene? 19.10 A large population of primates that are randomly mated exhibit a copy number polymorphism consisting of tandem copies of a gene encoding a protein. In this population, 20% of the chromosomes carry one copy of the gene, 30% carry two copies of the gene, and 50% carry three copies of the gene. What is the expected proportion of people with five or more specimens? 19.11 In humans, chimpanzees, and gorillas, the gene encoding the protein opsin, a light-sensitive blue pigment, is autosomal, while red and green opsins are X-linked. The situation is more complicated for some New World monkeys, including marmosets, tamarins and spider monkeys. These species have only one

opsin gene on the X chromosome, but it is polymorphic. Some X chromosomes carry a green opsin, while others carry a red opsin. Sex determination in these monkeys is similar to sex determination in humans. Consider a species in which the frequency of the red opsin allele is 0.36 and the green opsin allele is 0.64. Assuming mating is random for this gene, what is the expected frequency of females able to perceive: (a) red but not green? (b) Green but not red? (c) Red and green?

19.12 Given the situation described in question 19.11, what is the expected frequency of men with the ability to perceive: (a) red, but not green? (b) Green but not red?

19.13 If the probability that a pair of nucleotides in human DNA differs from its counterpart in chimpanzee DNA is 1%, what is the probability that a 1 kb region has an identical nucleotide sequence in both species? (Note: Assume adjacent nucleotide sites are independent.) 19.14 Which of the following conditions makes it least likely that a gene tree will match the species tree? (a) A short period of time between speciation events

(b) a large ancestral population (c) a polymorphism in the common ancestor (d) all of these (e) none of these

19,15 derived SNPs (dissimilar to chimpanzee) have been identified in the Neanderthal genome. These inferred SNPs in Neanderthals are compared to SNPs found in three modern human populations shown in the following tree.

(a) If more SNPs in human populations 2 and 3 match Neanderthal SNPs than SNPs in human population 1, so that the degree of match is expressed as H2 = H3 > H1, what does this say about hybridization among Neanderthals what about modern people? (b) Alternatively, suppose the pattern of matching with Neanderthal-derived SNPs is H3 > H2 > H1. What does this mean for the mix?

19.16 Which model of modern human origin predicts the greatest similarity between currently existing human populations? Briefly justify your answer. (a) The multiregional model (b) The African surrogate model (c) The hybridization model

(d) The models are indistinguishable in this respect.

19.17 Rank the following species according to the probability that each contributed genes to the genome of modern humans (Homo sapiens) through hybridization. The options are "not possible", "very likely" and "possibly". (a) Australopithecus afarensis (b) Homo habilis (c) Homo neanderthalensis (d) Homo erectus

19.18 Interpret the following table of Ka and Ks values for a gene that is present in both humans and gorillas.

19.19 A 900 kb inversion on human chromosome 17, which is rare in Africans, has a frequency of about 0.20 in European populations. The inversion haplotype (that is, the particular combination of alleles present in the inversion) is identical from one inverted chromosome to another. Why is the inversion haplotype expected to be preserved? 19.20 Consider a population in which a gene is selectively scanned. Explain whether each of the

The following conditions would increase or decrease the size of the haplotype block associated with the preferred gene. (a) A higher selection coefficient (b) An increase in recombination frequency (c) An increase in mutation rate

19.21 A selective scan for a new preferred mutant allele results in a haplotype block (a region of connected alleles surrounding the mutant) that increases in frequency with the mutant. How would you expect the haplotype block size to depend on the recombination frequency in the region? How would you expect the size of the haplotype block to depend on the strength of selection?

19.22 Many of the genetic adaptations that protect against malaria are also associated with anemia due to defective or reduced red blood cell counts. Can you suggest an explanation? 19.23 An organism is polymorphic for two types of mutations that affect the reproductive period. Reproduction is allowed one year earlier, but reproduction ends at the normal time. The other allows reproduction for another year, but reproduction begins at normal time. Which of these mutants (if any) do you think would be favored by natural selection? 19.24 A species is subject to epidemics of two different infectious diseases. One kills mainly young people, the other mainly old people. For which infectious disease would you expect natural selection?

more favoring genetic adaptations for resistance, and why? 19.25 Which populations – ancestors or descendants – should have the greatest genetic diversity in the serial colonization model and why? 19.26 Which of the following functions do human skin pigments have? (a) To protect against the mutagenic effects of ultraviolet radiation (b) To promote vitamin D synthesis (c) To protect against the breakdown of folic acid, a B vitamin (d) All of the above

19.27 On the strictest definition of a species, a member of one species cannot produce viable, fertile offspring with a member of another species. Does this strict definition affect whether Neanderthals should be called a species Homo neanderthalensis or a subspecies Homo sapiens neanderthalensis? 19.28 The alleles of a gene are typed from three populations X, Y, and Z. Two types of alleles can be distinguished, A and A'. Each population is committed to one of the alleles. Which of the following findings agrees with the genetic tree shown here?

(a) X fixed for A, Y and Z fixed for A′ (b) X and Y fixed for A, Z fixed for A′ (c) X and Z fixed for A, Y fixed for A′ (d) X and Y and Z fixed to A'

19.29 Would you expect lactase persistence in Neanderthals? Extra copies of the salivary amylase gene? Why or why not? 19.30 Would you consider Hemoglobin S in Africa and Hemoglobin H in India to be an example of convergent evolution? Why or why not? 19.31 Why are regions of the genome with little or no recombination most useful for tracing the genetic ancestry of populations? 19.32 Has sexual dimorphism increased, decreased, or remained roughly the same throughout human evolution?

CHALLENGE PROBLEMS CHALLENGE PROBLEM 1 Species X, Y, and Z are related to the parent species W according to the species tree shown here. The population on the branch labeled 1 is polymorphic for the A and A' alleles. Alleles A and A' have equal frequencies and A' is the derived allele. Branch 2 is so short that fixing any one of the alleles on this branch has negligible probability. Suppose that one allele or another is independently fixed on each of branches 3, 4, and 5 in a process called ancestry sorting.

What is the probability that the alleles are ordered by lineage so that: (a) there is no gene tree (that is, the alleles at X, Y, and Z are identical)? (b) Does the genetic tree agree with the species tree? (c) Does the genetic tree not match the species tree? CHALLENGE PROBLEM 2 In Challenge Problem 1, assume that branch 2 is long enough that fixation of A or A' occurs on branch 2 with probability 2/3 and that fixation occurs independently on branch 3. With probability 1/3, fixation of A or A' occurs independently on branches 3, 4, and 5. What is the probability that the alleles undergo an ancestry sort such that: (a) there is no gene tree (that is, the alleles in X, Y, and Z are identical)? (b) Does the genetic tree agree with the species tree? Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

FOR FURTHER READING Li JZ, Absher DM, Tang H, Southwich AM, Casto A, Ramachandran S, … Myers RM (2008). Human relationships around the world derived from patterns of genome variation. Science,

319 (5866), 1100-1104. http://doi.org/10.1126/science.1153717 A seminal paper that used SNP variation to determine the distribution of genetic variation within and between human populations and relationships between populations. Nielsen, R., Akey, J.M., Jakobsson, M., Pritchard, J.K., Tishkoff, S., & Willerslev, E. (2017). Tracing the colonization of the world through genomics. Nature, 541 (7637), 302-310. http://doi.org/10.1038/nature21347 A current perspective on the recent evolutionary history of Homo sapiens.

Orlando L, Gilbert MTP and Willerslev E (2015). Reconstruction of ancient genomes and epigenomes. Nature Reviews Genetics, 16(7), 395-408. http://doi.org/10.1038/nrg3935 An excellent overview of the theory and practice of ancient DNA analysis. Salem AH, Ray DA, Xing J, Callinan PA, Myers JS, Hedges DJ, … Batzer MA (2003). Alu elements and hominin phylogenetics. Proceedings of the National Academy of Sciences of the United States of America, 100(22), 12787-12791. http://doi.org/10.1073/pnas.2133766100

Demonstrates how the use of Alu insertion points as phylogenetic traits unambiguously resolves phylogenetic relationships among hominids. Sarich, V.M. & Wilson, AC. (1967). Immunological time scale for hominid evolution. Science, 158 (3805), 1200-1203. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/4964406 An important application of molecular clock-based thinking to human evolution. In this case, immunological distance was used as a measure of protein divergence and led to the hypothesis that humans and chimpanzees diverged from each other around 5 million years ago. That number was controversial at the time, but later work has supported the proposed overall timeline.

Wall, JD, & Yoshihara Caldeira Brandt, D (2016). Archaic mixture in human history. Current Opinion in Genetics & Development, 41, 93–97. http://doi.org/10.1016/j.gde.2016.07.002 Recent review of genetic evidence for hybridization of Homo sapiens with Neanderthals and Denisovans.

Appendix A: Answers to Even Number Problems Chapter 1 1.2 With regard to replication, they concluded that each strand of the double helix was used as a template for the formation of a new daughter strand with a complementary base sequence. In terms of coding ability, they found that genetic information can be encoded by the sequence of bases along the DNA molecule, analogous to the letters of the alphabet printed on a strip of paper. Finally, with regard to mutations, they found that changes in genetic information can result from errors in replication, and the altered nucleotide sequence can be maintained. 1.4 A mixture of heat-killed S cells and live R cells could cause pneumonia in mice, but neither heat-killed S cells alone nor live R cells alone could do this.

1.6 Phenol would not destroy the conversion activity, strong alkali would. 1.8 Because the mature T2 phage contains only DNA and protein; the labeled RNA was left in the material released by the ruptured cells. 1.10 In this bacteriophage DNA, the amount of A does not equal T and the amount of G does not equal C. From this, you can deduce that the DNA molecule is single-stranded and not double-stranded. 1.12 5'-CTGAT-3'

1.14 The complementary strand has the sequence 5'TACTACTAC…-3'. 1.16 5'-AUCAG-3' 1.18 The complementary sequence is 5'-AUACGAUA-3'. 1.20 The codon for leucine must be 5'-UUA-3'. 1.22 Six possible reading frames would need to be examined. Any DNA strand can be transcribed, and any transcript can be translated into one of three reading frames. 1.24 The result means that an mRNA is translated into non-overlapping groups of three nucleotides: The genetic code is a triplet code. 1.26 The 5'-UGG-3' codon encodes Trp and in this random polymer the Trp codon is expected to have a frequency of 1/4 x 3/4 x 3/4 = 9/64. The Val amino acid can be specified by 5'-GUU-3' or 5'-GUG-3'; The former has an expected frequency of 3/4 × 1/4 × 1/4 = 3/64 and the latter 3/4 × 1/4 × 3/4 = 9/64, for a total of 12/64 or 3 /16. The amino acid Phe could only be specified by 5'-UUU-3' in this random polymer, and therefore Phe would have an expected abundance of (1/4)3 = 1/64.

1.28

1:30. The amino acid sequence is Met-Val-His-Leu-Thr-Pro-Glu-Glu-Lys-Ser. B. The resulting amino acid sequence would be Met-Val-His-Leu-Thr-Pro-Arg-Arg-Ser-Leu...; all codons after the deletion are shifted by -1 compared to the unmutated RNA due to the reading frame shift. ç. The resulting amino acid sequence is Met-Val-His-Leu-Thr-Pro. In this case, the translation for Pro ends

because the frameshift mutation creates the stop codon 5'-UGA-3'.

Chapter 2 2.2 (e) is the 5' carbon atom and contains the phosphate group; (c) is the 3' carbon atom and carries the 2-hydroxyl group.4 A DNA palindrome reads the same forward and backward, but along opposite strands. one. This is a palindrome because the partner strand is 3'-GGCC-5'; B. No, because the partner strand is 3'-AAAA-5'; ç. Yes, because the partner strand is 3'-CGATCG-5', i. No, because the partner strand is 3'-GGCGAG-5'; and. No, because the partner strand is 3'-TTCCAA-5'. 2.6 a. ScaI produces blunt ends; B. NheI creates a 5' overhang; ç. CfoI creates a 3' overhang.

2.8

2.10a. B5 = T; B. B6 = G; ç. B7 = the complementary pyrimidine; that is, B8 = T or A. 2.12 The probability of a restriction site is 1/4n, where n is the number of nucleotides in the site. one. 1/44 = 0.0039; B. 1/46 = 0.00024; ç. 1/48 = 0.000015. 2.14 a. 4.6×106/256 ≈ 1.8×104; B. 4.6×106/4096 ≈ 1.1×103; ç. 4.6 × 106/65536 ≈ 70.

2.16 Probe A provides 4 and 12 kb bands, Probe B provides 8 and 12 kb bands.

2.18 Sites complementary to the primers are in bold.

Therefore, the amplified fragment has the sequence

2,20 ein. 3 kb/3 × 106 kb = 0,0001 %; b. (3 × 210 kb)/(3 × 210 kb + 3 × 106 kb) = 0,1 %; c. (3 × 220 KB)/(3 × 220 KB + 3 × 106 KB)

= 51.2%; i.e. (3x230kb)/(3x230kb + 3x106kb) = 99.9%. 2.22 The third SNP (C/T) would help, as all fragments that carry the "C" SNP allele also carry the disease "D" allele. The other three SNPs would not because the SNP and disease alleles are not strictly associated. For example, an A fragment with the "A" allele of the first SNP (A/T) can carry either "D" (the top five fragments) or "d" (the bottom fragment). 2.24 As the test is based on all of an individual's genomic DNA, any bands found in the test must be consistent with the offender's bands. Anyone with one or more bands that do not match those on the evidence may be disassociated. To find the corresponding bands on suspects, draw straight horizontal lines through the X bands for each locus. For locus 1, suspects

C, D, E, F and G are excluded, but A and B are not excluded; for locus 2, suspects B, C, D, E, and G are excluded, but A and F are not excluded; and for locus 3, suspects C, D, E, and G are excluded, but A, B, and F are not excluded. By putting all three probes together, all suspects except suspect A can be ruled out. This does not necessarily mean that A is the culprit, just that A cannot be ruled out. However, a combination of three highly polymorphic DNA markers is highly suggestive. 2.26 After cleavage, phosphate 2 goes with B1 and phosphate 4 with B3, so mononucleotides B1 and B3 will be radioactive.

Chapter 3 3.2 3/4 3.4 3 × 2 × 2 × 1 × 3 = 36 3.6 2/3

3.8 Apparently, the 7 kb band is a marker for the recessive allele. one. The expected phenotype of II-1 is a single 7 kb band. B. As II-2 is unaffected, possible molecular phenotypes are a single 3 kb band or two 3 kb and 7 kb bands. 3.10a. As the trait is rare, it is likely that the affected parent is heterozygous HD/hd, with hd being the normal allele. Half of his gametes contain the DH allele, so there is a 1/2 chance that your child has acquired the allele and later develops the disease. B. We don't know if the young man is heterozygous HD/HD, but the probability is 1/2 that he is; if he is heterozygous, half of his gametes will carry the Hd allele. Therefore, the overall probability that your child has the DH allele is (1/2) × (1/2) = (1/4).

3.12a. The trait is likely due to a recessive allele, since the pedigree includes inbreeding (mating between relatives). B. The double line indicates a consanguineous mating. ç. III-1 and III-2 are first cousins. i.e. I-1, I-2, II-2 II-3, III-1 and III-2 are more likely Aa, II-1 and II-4 are more likely AA. 3.14 The probability that the father has the AA BB genotype is Pr{AA BB} = 1/2. The probability of producing an A b gamete is Pr{A b|AA BB} = 0, provided one parent has the AA BB genotype. The probability that the father has the AA Bb genotype is Pr{AA Bb} = 1/ 2. The probability of producing an A b gamete given that the father has the AA Bb genotype is Pr{ A b|AA Bb} = 1/ 2. Therefore, the overall probability of producing an A b gamete is Pr{A b}= (1/2)(0) + (1/2)(1/2) = 1/4. By the same reasoning, the probability of producing an A B gamete is Pr{A B} = (1/2)(1) + (1/2)(1/2) = 3/4. 3.16 (0.002)(1) + (1 - 0.002)(0.002) = 0.004 3.18 Parents have WW and ww genotypes. All F1 offspring from this cross must be Ww heterozygotes. The ratio of different phenotypes in the F2 generation is 3/4 round seeds

(WW × Ww) and 1/4 wrinkled (ww). Between the F2 round seeds are 2/3 WW and 1/3 WW. Therefore, among the progeny of the cross at masses of (2/3 ww × 1/3 ww) × ww, the expected ratio with wrinkled seeds will be (2/3) × (1/2) × 1 = 1/3 . 3.20 The 9:3:3:1 ratio is that of the AB-B-A-bb, aa B- and aa bb genotypes. The modified 9:7 implies that all of the last three genotypes have the same phenotype. The F1 testcross is between the Aa Bb and aa bb genotypes and offspring are expected in the proportions 1/4 Aa Bb, 1/4 Aa bb, 1/4 aa Bb and 1/4 a abb. The last three genotypes would be

again have the same phenotype, and therefore the ratio of phenotypes among testcross progeny is 1:3. 3.22 Since the female is affected by the trait, she has a 1/2 chance of passing the dominant allele to her child, and the child has a 1/2 chance of actually having the trait (because the trait shows 50% penetrance). Therefore, the probability that the child is affected is 1/2 × 1/2 = 1/4. 3.24 Use Bayes' theorem with event A as the event that the F1 individual with the dominant phenotype has the Nn genotype (probability 2/3). Event A' is that the individual is homozygous dominant NN (probability 1/3). Let B be the event that all four testcross offspring have the dominant phenotype. then

3.26 To avoid ambiguity, we list the bands in the order in which the alleles are written, not in order of size. The F1 progeny (A1A2 B1B2) each have four bands of 4, 8, 2 and 6 kb. The F2 progeny consists of 1/16 A1A1 B1B1 (4 and 6 kb bands), 2/16 A1A1 B1B2 (4, 6 and 2 kb bands), 1/16 A1A1 B2B2 (4 and 2 kb bands), 2 /16 A1A2 B1B1 (4 volumes, Copyright © 2017. Jones & Bartlett Learning, LLC. All rights reserved.

8 and 6 kb), 4/16 A1A2 B1B2 (4, 8, 6 and 2 kb tapes), 2/16 A1A2 B2B2 (4, 8 and 2 kb tapes), 1/16 A2A2 B1B1 (8 and 2 kb tapes) 6 kb), 2/16 A2A2 B1B2 (8, 6 and 2 kb tapes) and 1/16 A2A2 B2B2 (8 and 2 kb tapes). 3.28 Paternal/Maternal

1/2 A

1/2 a

1/3A

1/6 AA

2/3 a

2/6 Aa

2/6 AA

The expected ratio is 1/6 AA : 3/6 AA : 2/6 AA.

Chapter 4 4.2 Figure (A) is anaphase II of meiosis as the number of chromosomes has been reduced by half. Panel (B) is the anaphase of mitosis as the homologous chromosomes are not paired. Figure (C) is anaphase I of meiosis when homologous chromosomes are paired.

4.4 Since the two genes are on different chromosomes (the mutation for phenylcentenuria is autosomal, the one for hemophilia is X-linked), they will separate independently, so we can consider them separately. Since both parents are heterozygous for an autosomal recessive PKU mutation, there is a 1/4 chance that their child will be homozygous recessive. In the case of X-linked hemophilia, only a boy may be affected; a girl would have inherited a normal X chromosome from her fath

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