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Recombination (Crossing Over). A. a. A. A. a. A. a. A. B. b. B. b. B. b. e. d. C. B. A. C. c. C. c. C. c. C. D. d. D. d. D. d. E. c. D. b. a. E. e. E. e. e. E. Homologous chromosomes (the one you received from

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No slide title 1277492

Recombination

(Crossing Over)

A

a

A

A

a

A

a

A

B

b

B

b

B

b

e

d

C

B

A

C

c

C

c

C

c

C

D

d

D

d

D

d

E

c

D

b

a

E

e

E

e

e

E

Homologous chromosomes (the one you received from

mother-- -- and the partner that you received from

father-- ) join.

They exchange genetic material.

The resulting chromosome passed to the offspring contains

some of the paternal and some of the maternal chromosome.


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The probability of recombination

depends upon DISTANCE

A

a

B

b

C

c

D

d

E

e

F

f

G

g

The further away two loci are, the more likely

the chromosomes will pair up somewhere

between the two loci and the less likely the

two will be transmitted together.

The closer together two loci are, the less likely

the chromosomes will pair up somewhere

between the two loci and the more likely the

two will be transmitted together.

The probability of a recombination is very low

for loci D and E. It is very high for loci A and G.


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Linkage Analysis

Key Idea:

Cosegregation of a trait (disorder) with marker gene(s) within families.

Common sense explanation:

  • Affected offspring within a family will tend to have the same genotype(s) on the marker.

  • 2) Unaffected offspring within a family will tend to have the same genotype(s) on the marker.

  • 3) Affected offspring will have different genotypes than unaffected offspring on the marker.


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Linkage Analysis

A

a

Father’s chromosomes are

D

d

aa

Aa

Aa

Aa

aa

Aa

aa

Aa

aa


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Linkage Analysis

A

a

Father’s chromosomes are

d

D

aa

Aa

Aa

Aa

aa

Aa

aa

Aa

aa


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Linkage Analysis

A

a

D

d

Father’s chromosomes are

aa

Aa

Aa

Aa

aa

Aa

aa

Aa

aa


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Linkage Analysis

(To calculate gametes under linkage, see the handoutCalculating Gametes under Linkage on the handouts section of the course web page.)


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Linkage Analysis

A

a

d

D

a

a

d

d

Problem: The gene for a dominant disorder

has two alleles, D which causes the disorder,

and d which is the normal allele. This gene is

located close to a marker gene with two

alleles, A and a. Give the probabilities for

the genotypes and phenotypes of the following

mating:

A father’s who is

and a mother who is


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Linkage Analysis

A

a

Father’s chromosomes are

d

D

1. No recombination: probability = (1 - )

1.a) Father gives Ad: prob = .5(1 - )

1.b) Father gives aD: prob = .5(1 - )

2. Recombination: probability = 

1.a) Father gives AD: prob = .5

1.b) Father gives ad: prob = .5

Father’s Gametes:


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Probability

Genotype

Gametes

d

A

.5(1 - )

A

a

d

D

A

.5

D

d

.5

a

.5(1 - )

D

a


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Linkage Analysis

a

a

Mother’s chromosomes are

d

d

Mother can only give gamete ad with

probability = 1.0


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Linkage Analysis

Father’s Gametes:

aD

Ad

AD

ad

.5(1 - )

.5(1 - )

.5

.5

M

o

t

h

e

r

s

G

a

m

e

t

e

ad

AD

aD

Ad

ad

ad

ad

ad

ad

1.0

.5(1 - )

.5

.5(1 - )

.5

normal

disorder

normal

disorder


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Linkage Analysis

Offspring Phenotypes

and Probabilities:

Marker

Phenotype

Disorder

Phenotype

Probability

Aa

.5(1 - )

normal

Aa

affected

.5

aa

normal

.5

aa

affected

.5(1 - )


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Linkage Analysis

Summary of empirical results oflinkage analysis

  • Very successful for single gene disorders.

  • Successful for Mendelian forms for DCGs.

  • Not very successful for risk genes for DCGs and behavioral phenotypes.


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A

b

c

D

a

b

C

d

= the AbcD Haplotype

= the abCd Haplotype

Haplotypes

Haplotype = Series of alleles along a very short section of the same chromosome.

Examples:


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Haplotypes

Linkage Disequilibrium: Some haplotypes occur more frequently than expected by chance.

Example:

Assume two linked loci, the first with alleles A and a and the second with alleles B and b. Assume that the frequency of allele A is .70 and the frequency of allele B is .40. If the two loci are in linkage equilibrium, then the frequency of the haplotypes can be predicted using simple probability theory.


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B.4

b.6

b

b

A

a

B

A

B

a

A.7

.42

.28

a.3

.12

.18

Haplotypes

Haplotypes expected by chance


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A

B

a

B

A

b

a

b

Expectedby Chance

ObservedFrequency

Haplotype

.28

.37

.42

.33

.12

.03

.18

.27

Haplotypes

Test for Linkage Disequilibrium = Compare observed frequencies with those expected by chance.


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Haplotypes

Haplotype Map Project (HapMap)= international collaborative effort to identify regions throughout the human genome in linkage disequilibrium.

  • Linkage disequilibrium is a rule rather than an exception.

  • Are recombination “hot spots” (thus, little linkage disequilibrium.

  • If there are, say, 5 genes in a haplotype group in strong linkage disequilibrium, then only test 1 gene.


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Haplotypes

DNA region of interest:

  • Instead of genotyping all 37 SNPs in the region, genotype one SNP from each of the 7 haplotype blocks.

  • If there is a “hit” for one block, then genotype the SNPs within the block to get closer to the disease gene.