review problems 1905 william bateson and r.c. punnett red petals, round pollen (rr,ss) purple...

Review Problems

1905 William Bateson and R.C. Punnett

Red petals, round pollen(rr,ss)

Purple petals, long pollen(RR,SS)

X

F1 Purple petals, long pollen (Rr,Ss)

Question

• If two genes are tightly linked, such that no crossing over occurs between them:

a. All progeny will be parentals.

b. All progeny will be nonparentals.

c. All progeny will be recombinants.

d. Progeny will be 50% parental, 50% nonparental.

e. Progeny will be 25% nonrecombinant, 75% recombinant.

a. All progeny will be parentals.

Xyellow (Gg), round (Ww) yellow (Gg), round (Ww)

+oo

GWGGWW

GgWw

GgWwGgWw

GgWw

GgWw

GgWwGGWW

GGWW

ggww

ggwwggww

ggwwGgWw

GGWW GgWw

Generation

F2

GW GW gw gw

gw

gw

GW

F1 selfed (Rr,Ss) X (Rr,Ss)

215Purple, long

24red, round

71red, long

71Purple, round

Expected F2

If they assort independently (they are not linked)

9 3 3 1

Xyellow (Gg), round (Ww) yellow (Gg), round (Ww)

GW Gw gW gw+o

o

GW

Gw

gW

gw

GGWW

GgWw

GgWwGgWW

GgWw

GgWW

GgWwGGWw

GGWw

ggWw

ggWwggWW

ggwwGgww

GGww Ggww

Generation

F2

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:

a. Are homologsb. Are genetically linked and would cosegregate during

meiosisc. Are separated by no more than 1 cMd. Are located on the same chromosomee. Are located on separate chromosomes

d. Are located on the same chromosome

284Purple, long

55red, round

21red, long

21Purple, round

Results

F1 selfed (Rr,Ss) X (Rr,Ss)

215Purple, long

24red, round

71red, long

71Purple, round

Expected F2

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

1905 William Bateson and R.C. Punnett

Red petals, round pollen(rr,ss)

Purple petals, long pollen(RR,SS)

X

F1 Purple petals, long pollen (Rr,Ss)

The observation

284Purple, long

55red, round

21red, long

21Purple, round

Results

F1 selfed (Rr,Ss) X (Rr,Ss)

216Purple, long

24red, round

72red, long

72Purple, round

Expected F2

Symbols and terminology

Cis AB/ab or ++/abTrans Ab/aB or +b/a+

AB alleles on the same homolog, no punctuationA/a alleles on different homologs, slashA/a; B/b genes known to be on different

chromosomes, semicolonA/a . B/b genes of unknown linkage, use a period

Thomas Hunt Morgan & Drosophilia

Red eyes, normal(pr+/pr+ . vg+/vg+)

Purple eyes, vestigal(pr/pr . vg/vg)

X

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)

Purple eyes, vestigal(pr/pr . vg/vg)

X

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+)

305/2839 = 10.7 percentpr+

vg

vg+

pr

cis or trans ?

Red eyes, vestigal(pr+/pr+ . vg/vg)

Purple eyes, normal(pr/pr . vg+/vg+)

X

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+)

Initial cross

304/2335 = 12.9 percentpr+

vg+

vg

pr

Morgan proposes Linkage and Crossing Over

Fig. 4-3

Crossing-over of the chromosomes.

A chiasma is formed.

Genetic recombination.

Occurs at Prophase I (tetrad stage)

Microscopic evidence for chromosome breakage and gene recombination

Harriet Creighton and Barbara McClintock, 1931

Wx

wx c

C Wx

wx

c

C

Fig. 4-8

For linked genes, recombinant frequencies are less than 50% 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.

Map distances are additive.

Fig. 4-9

QuestionYou 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:

a. DNA sequencing of the region in questionb. Recombination frequenciesc. Measuring the distance in a scanning EM

micrographd. 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?

a. Z-X-Yb. Y-X-Zc. X-Y-Zd. a and b

a. Z-X-Yb. Y-X-Z

Summary

• Gene linkage

• Crossing over

• Recombinant mapping

Morgan proposes Linkage and Crossing Over

Fig. 4-3

Fig. 4-8

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

Map distances are additive.

Fig. 4-9

Review Problems

A B

a b

1. A plant of genotype

is test crossed.

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 B14% for B and C24% for A and C 4% for B and D10% for C and D 14% for A and D

Provide a linear map for the chromosome.

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 B14% for B and C24% for A and C 4% for B and D10% 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 MarkersChapter 4, continued.

What is a molecular marker

SNP = single nucleotide polymorphisms

• Silent SNPs• SNP that cause phenotype• SNP in polygenes• SNP in intergenic regions• RFLPs (restriction fragment length polymorphisms)

AAGGCTCATTTCCGAGTA

AAGACTCATTTCTGAGTA

RFLPs

SNPs that introduce a restriction enzyme site.

GGATTCCCTAAG

GAATTCCTTAAG

EcoR1 site

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 nucleotidesACACACACACACAC

Minisatellites

Fig. 4-18

Microsatellites

Amplified by polymerase chain reaction.

CACACACACAGTGTGTGTGT

primer 1

primer 2St.

CACACACACACACAGTGTGTGTGTGTGT

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

142 A.B parental133 a.b parental113 A.b recombinant112 a.B recombinant

Total 500

Observed Expected

Using recombinant maps with physical maps

Summary

• Mapping using molecular markers– SNPs, RFLP mapping, haplotypes– SSLP

• Minisatellites

• Microsatellites