Review Problems 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)

Question If two genes are tightly linked, such that no crossing over occurs between them: All progeny will be parentals. All progeny will be nonparentals. All progeny will be recombinants. Progeny will be 50% parental, 50% nonparental. Progeny will be 25% nonrecombinant, 75% recombinant. a. All progeny will be parentals.

X GW GW gw gw + F2 GW GGWW GGWW GgWw GgWw Generation GW GGWW GGWW GgWw yellow (Gg), round (Ww) yellow (Gg), round (Ww) GW GW gw gw o + o F2 GW GGWW GGWW GgWw GgWw Generation GW GGWW GGWW GgWw GgWw gw GgWw GgWw ggww ggww gw GgWw GgWw ggww ggww

If they assort independently (they are not linked) F1 selfed (Rr,Ss) X (Rr,Ss) Expected F2 215 Purple, long 24 red, round 71 red, long Purple, round 9 3 3 1

X GW Gw gW gw + F2 GGWW GGWw GgWW GW GgWw Generation Gw GGWw GGww GgWw yellow (Gg), round (Ww) yellow (Gg), round (Ww) GW Gw gW gw o + o F2 GGWW GGWw GgWW GW GgWw Generation Gw GGWw GGww GgWw Ggww GgWW GgWw gW ggWW ggWw GgWw gw Ggww ggWw ggww

Question If two nuclear genes in a diploid eukaryote are physically linked by DNA sequence data, but we have no additional data other than this, we can say with confidence that they: Are homologs Are genetically linked and would cosegregate during meiosis Are separated by no more than 1 cM Are located on the same chromosome Are located on separate chromosomes d. Are located on the same chromosome

F1 selfed (Rr,Ss) X (Rr,Ss) Expected F2 215 Purple, long 24 red, round 71 red, long Purple, round 284 Purple, long 55 red, round 21 red, long Purple, round Results

Gene linkage, Recombination and Mapping Chapter 4

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 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+) 304/2335 = 12.9 percent pr+ vg+ vg pr

Linkage and Crossing Over Morgan proposes Linkage and Crossing Over Fig. 4-3

Occurs at 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 Wx C Wx c wx c wx C

For linked genes, recombinant frequencies are less than 50% in a testcross. Fig. 4-8

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.

Map distances are additive. Fig. 4-9

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

Summary Gene linkage Crossing over Recombinant mapping

Linkage and Crossing Over Morgan proposes Linkage and Crossing Over Fig. 4-3

For linked genes, recombinant frequencies are less than 50% in a testcross. Fig. 4-8

Map distances are additive. Fig. 4-9

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.

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

RFLP analysis Fig 4-15a

RFLP analysis Fig 4-15b

RFLP analysis Fig 4-15c

Using combinations of SNPs A haplotype is a chromosomal segment defined by a specific array of SNP alleles.

Using haplotypes to deduce gene position Fig. 4-16

Simple sequence length polymorphisms (SSLPs) VNTRs (variable number tandem repeats) Repeats of DNA sequence, with different numbers of repeats occurring in different individuals. Minisatellites (DNA fingerprints) – Repeating units of 15-100 nucleotides Microsatellites – repeat of 2-3 nucleotides ACACACACACACAC

Minisatellites Fig. 4-18

Microsatellites Amplified by polymerase chain reaction. primer 1 CACACACACACACA GTGTGTGTGTGTGT CACACACACA GTGTGTGTGT primer 2 St. M M’

Fig. 4-19

Molecular markers can be used instead of phenotype to map genes. Chi-square A/A . B/B X a/a . b/b A/a . B/b Test cross to a/a . b/b Observed Expected A.B parental 133 a.b parental 113 A.b recombinant 112 a.B recombinant Total 500

Using recombinant maps with physical maps

Summary Mapping using molecular markers SNPs, RFLP mapping, haplotypes SSLP Minisatellites Microsatellites