Introduction to Genetic Analysis Griffiths • Wessler • Carroll • Doebley Introduction to Genetic Analysis ELEVENTH EDITION CHAPTER 4 Mapping Eukaryote Chromosomes by Recombination © 2015 W. H. Freeman and Company

CHAPTER OUTLINE 4.1 Diagnostics of linkage 4.2 Mapping by recombinant frequency 4.3 Mapping with molecular markers 4.4 Centromere mapping with linear tetrads 4.5 Using the chi-square test to infer linkage 4.6 Accounting for unseen multiple crossovers 4.7 Using recombination-based maps in conjunction with physical maps 4.8 The molecular mechanism of crossing over

A recombination-based map of one of the chromosomes of Drosophila

Why map the genome ? Gene position important to build complex genomes To determine the structure and function of a gene To determine the evolutionary relationships and potential mechanism.

Two types of maps ? Recombination-based maps* Physical maps

The observation 1905 William Bateson and R.C. Punnett Red petals, round pollen (rr,ss) X Purple petals, long pollen (RR,SS) F1 Purple petals, long pollen (Rr,Ss)

F1 selfed (Rr,Ss) X (Rr,Ss) Results 284 Purple, long 21 Purple, round 21 red, long 55 red, round 216 Purple, long 24 red, round 72 red, long Purple, round Expected F2

Symbols and terminology AB alleles on the same homolog, no punctuation A/a alleles on different homologs, slash A/a; B/b genes known to be on different chromosomes, semicolon A/a . B/b genes of unknown linkage, use a dot or period Cis AB/ab or ++/ab Trans Ab/aB or +b/a+

Thomas Hunt Morgan & Drosophilia Red eyes, normal (pr+/pr+ . vg+/vg+) X Purple eyes, vestigal (pr/pr . vg/vg) F1 Red eyes, normal wings (pr+/pr . vg+/vg) Instead of selfing the population, he did a test cross.

Test cross Red eyes, normal (pr+/pr . vg+/vg) X Purple eyes, vestigal 1339 Red eyes, normal wings (pr+ . vg+) 1195 Purple eyes, vestigal (pr . vg) 151 Red eyes, vestigal (pr+ . vg) 154 Purple eyes, normal wings (pr . vg+)

Test cross 1339 Red eyes, normal wings (pr+ . vg+) 1195 Purple eyes, vestigal (pr . vg) 151 Red eyes, vestigal (pr+ . vg) 154 Purple eyes, normal wings (pr . vg+) pr+ vg+ 305/2839 = 10.7 percent pr vg cis or trans ?

Initial cross Red eyes, vestigal (pr+/pr+ . vg/vg) X Purple eyes, normal (pr/pr . vg+/vg+) F1 Red eyes, normal wings (pr+/pr . vg+/vg) Test cross with pr/pr . vg/vg 157 Red eyes, normal wings (pr+ . vg+) 146 Purple eyes, vestigal (pr . vg) 965 Red eyes, vestigal (pr+ . vg) 1067 Purple eyes, normal wings (pr . vg+) 303/2335 = 13.0 percent pr+ vg+ vg pr

Linked alleles tend to be inherited together

Crossing over produces new allelic combinations

Chiasmata are the sites of crossing over Chiasmata are the sites of crossing over. Occurs in Prophase I of meiosis.

Occurs in Prophase I (tetrad stage) Crossing-over of the chromosomes. A chiasma is formed. Genetic recombination.

Microscopic evidence for chromosome breakage and gene recombination Harriet Creighton and Barbara McClintock, 1931 -studied two genes in corn (on chromosome 9) for seed color (C and c) and endosperm composition (Wx and wx) Wx C Wx c wx c wx C

Page 133 Note the knob on the C end and the longer piece of chromosome on the Wx end.

Recombinants are produced by crossovers

For linked genes, recombinant frequencies are less than 50 percent in a testcross

Mapping by Recombinant Frequency Morgan set his student Alfred Sturtevant to the project. “In the latter part of 1911, in conversation with Morgan, I suddenly realized that the variations in strength of linkage, already attributed by Morgan to differences in the spatial separation of genes, offered the possibility of determining sequence in the linear dimension of a chromosome. I went home and spent most of the night (to neglect of my undergraduate homework) in producing the first chromosome map.” Sturtevant

Frequency of crossing over indicates the distance between two genes on the chromosome.

A map of the 12 tomato chromosomes

Map distances are additive.

Question You construct a genetic linkage map by following allele combinations of three genes, X, Y, and Z. You determine that X and Y are 3 cM apart, and X and Z are 3 cM apart, and that Y and Z are 6 cM apart. These cM numbers are most likely based on: DNA sequencing of the region in question Recombination frequencies Measuring the distance in a scanning EM micrograph Independent assortment b. Recombination frequencies

Question Referring to the cM numbers in the last question, what is the relative gene order of these three genes? Z-X-Y Y-X-Z X-Y-Z a and b a. Z-X-Y b. Y-X-Z

Longer regions have more crossovers and thus higher recombinant frequencies

Double recombinants arising from two crossovers

Different gene orders give different double recombinants

Phenotypic ratios in progeny reveal the type of cross

Summary Gene linkage Crossing over Recombinant mapping Be sure to read p. 96-98 and p. 150-151 in order to do Chi-Square (c2) tests for some homework problems

Chi-Square (c2) Test We must apply the χ2 analysis to test a hypothesis of no linkage. If that hypothesis is rejected, we can infer linkage. (We cannot test a hypothesis of linkage directly because we have no way of predicting what recombinant frequency to test.) The calculation for testing lack of linkage is as follows: Observed (O) Expected (E) O–E (O – E)2 (O – E)2/E 60 50 10 100 2.00 37 –13 169 3.38 41 –9 81 1.62 62 12 144 2.88 χ2 = Σ (O – E)2/E for all classes = 9.88 Since there are four genotypic classes, we must use 4 − 1 = 3 degrees of freedom. Consulting the chi-square table in Chapter 3, we see our values of 9.88 and 3 df give a p value of ~0.025, or 2.5 percent. This is less than the standard cut-off value of 5 percent, so we can reject the hypothesis of no linkage.

Review Problems 1. A plant of genotype A B is test crossed. a b If the two loci are 14 m.u. apart, what proportion of progeny will be AB/ab ? 43% AB, 43% ab, 7% Ab, 7% aB

Review Problems 2. A plant of genotype A/a . B/b is test crossed. The progeny are 74 A/a . B/b 76 a/a . b/b 678 A/a . b/b 672 a/a . B/b Explain. A and B are linked in trans and are 10 m.u. apart.

Review Problems 3. You have analyzed the progeny of a test cross to a tetrahybrid. The results indicate that 10% of the progeny are recombinant for A and B 14% for B and C 24% for A and C 4% for B and D 10% for C and D 14% for A and D Provide a linear map for the chromosome.

Review Problems |----------|----|----------| A 10 B 4 D 10 C 3. You have analyzed the progeny of a test cross to a tetrahybrid. The results indicate that 10% of the progeny are recombinant for A and B 14% for B and C 24% for A and C 4% for B and D 10% for C and D 14% for A and D Provide a linear map for the chromosome. |----------|----|----------| A 10 B 4 D 10 C

Mapping with Molecular Markers Chapter 4, continued. SNPs-single nucleotide polymorphisms SSLPs-simple sequence length polymorphisms SSLPs also called VNTRs-variable number tandem repeats Molecular markers can be used instead of phenotype to map genes

What is a molecular marker SNP = single nucleotide polymorphisms AAGGCTCAT TTCCGAGTA AAGACTCAT TTCTGAGTA Silent SNPs SNP that cause phenotype SNP in polygenes SNP in intergenic regions RFLPs (restriction fragment length polymorphisms)

RFLPs SNPs that introduce a restriction enzyme site. EcoR1 site GGATTC CCTAAG GAATTC CTTAAG digest with EcoR1

Simple sequence length polymorphisms (SSLPs) Also known as VNTRs (variable number tandem repeats) Repeats of DNA sequence, with different numbers of repeats occurring in different individuals; used for DNA fingerprinting. Minisatellites (DNA fingerprints) – Repeating units of 15-100 nucleotides Microsatellites (DNA fingerprints) – Repeating units of 2-15 nucleotides [e.g., ACACACACACACAC]

A microsatellite locus can show linkage to a disease gene

Alignment of physical and recombination maps

Summary Mapping using molecular markers SNPs, RFLP mapping SSLPs/VNTRs Minisatellites Microsatellites