Purple petals, long pollen (Rr,Ss)
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Purple petals, long pollen (Rr,Ss). F 1. Review Problems. 1905 William Bateson and R.C. Punnett. X. Red petals, round pollen (rr,ss). Purple petals, long pollen (RR,SS). Question. If two genes are tightly linked, such that no crossing over occurs between them:

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Purple petals, long pollen (Rr,Ss)

F1

Review Problems

1905 William Bateson and R.C. Punnett

X

Red petals, round pollen

(rr,ss)

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.


o

o

+

X

yellow (Gg), round (Ww)

yellow (Gg), round (Ww)

GW

GW

gw

gw

F2

GW

GgWw

GGWW

GGWW

GgWw

Generation

GW

GGWW

GgWw

GGWW

GgWw

gw

GgWw

GgWw

ggww

ggww

gw

GgWw

ggww

GgWw

ggww


24

red, round

71

red, long

215

Purple, long

71

Purple, round

If they assort independently (they are not linked)

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

Expected F2

9 3 3 1


o

o

+

GgWW

GGWw

GGWW

GgWw

GGWw

GgWw

GGww

Ggww

GgWW

GgWw

ggWW

ggWw

GgWw

ggWw

Ggww

ggww

X

yellow (Gg), round (Ww)

yellow (Gg), round (Ww)

GW

gW

Gw

gw

F2

GW

Generation

Gw

gW

gw


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


Results

24

red, round

71

red, long

55

red, round

21

red, long

215

Purple, long

71

Purple, round

284

Purple, long

21

Purple, round

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

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


Purple petals, long pollen (Rr,Ss)

F1

The observation

1905 William Bateson and R.C. Punnett

X

Red petals, round pollen

(rr,ss)

Purple petals, long pollen

(RR,SS)


Expected F2

24

red, round

72

red, long

216

Purple, long

72

Purple, round

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

Results

55

red, round

21

red, long

284

Purple, long

21

Purple, round


Symbols and terminology

ABalleles on the same homolog, no punctuation

A/aalleles on different homologs, slash

A/a; B/b genes known to be on different chromosomes, semicolon

A/a . B/bgenes of unknown linkage, use a period

CisAB/ab or ++/ab

TransAb/aB or +b/a+


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

F1

Thomas Hunt Morgan & Drosophilia

X

Red eyes, normal

(pr+/pr+. vg+/vg+)

Purple eyes, vestigal

(pr/pr . vg/vg)

Instead of selfing the population, he did a test cross.


Test cross

X

Red eyes, normal

(pr+/pr . vg+/vg)

Purple eyes, vestigal

(pr/pr . vg/vg)

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

vg

pr

cis or trans ?


Test cross with pr/pr . vg/vg

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

F1

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+)

pr+

vg

304/2335 = 12.9 percent

vg+

pr

Initial cross

X

Red eyes, vestigal

(pr+/pr+. vg/vg)

Purple eyes, normal

(pr/pr . vg+/vg+)


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

c

wx

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


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

is test crossed.

A B

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

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

GAATTC

CTTAAG

GGATTC

CCTAAG

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


CACACACACACACA

GTGTGTGTGTGTGT

Microsatellites

Amplified by polymerase chain reaction.

primer 1

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


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