Transcript No Slide Title

Recombination
(Crossing Over)
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.
A
AA
a
D
d
E
e
E
e
c
d
D
e
E
a
d
C
A
D
C
c
c
b
b
C
B
B
c
b
C
C
B
E
b
a
D
B
A
d
a
e
A
The probability of recombination
depends upon DISTANCE
A
a
B
b
C
c
D
E
d
e
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.
F
f
G
g
Linkage Analysis
Key Idea:
Cosegregation of a trait (disorder) with marker
gene(s) within families.
Common sense explanation:
1) 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.
Linkage Analysis
A
D
Father’s chromosomes are
aa
Aa
Aa
aa
Aa
a
d
Aa
aa
Aa
aa
Linkage Analysis
A
d
Father’s chromosomes are
aa
Aa
Aa
aa
Aa
a
D
Aa
aa
Aa
aa
Linkage Analysis
A
a
D
d
Father’s chromosomes are
aa
Aa
Aa
aa
Aa
Aa
aa
Aa
aa
Linkage Analysis
(To calculate gametes under linkage, see the handout
Calculating Gametes under Linkage on the handouts
section of the course web page.)
Linkage Analysis
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
A
d
a
D
a
d
a
d
Linkage Analysis
Father’s chromosomes are
Father’s Gametes:
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
A
d
a
D
Genotype
Gametes
A
d
a
D
.5(1 - )
A
A
d
Probability
.5
D
d
a
.5
.5(1 - )
a
D
Linkage Analysis
Mother’s chromosomes are
Mother can only give gamete ad with
probability = 1.0
a
d
a
d
Linkage Analysis
Father’s Gametes:
Ad
M
o
t
h
e
r
s
G
a
m ad
e 1.0
t
e
aD
.5(1 - )
.5(1 - )
Ad
ad
aD
ad
.5(1 - )
normal
.5(1 - )
disorder
AD
ad
.5
.5
AD
ad
ad
ad
.5
disorder
.5
normal
Linkage Analysis
Offspring Phenotypes
and Probabilities:
Marker
Phenotype
Disorder
Phenotype
Probability
Aa
normal
.5(1 - )
Aa
affected
.5
aa
normal
.5
aa
affected
.5(1 - )
Linkage Analysis
Summary of empirical results of
linkage analysis
•Very successful for single gene disorders.
•Successful for Mendelian forms for DCGs.
•Not very successful for risk genes for DCGs
and behavioral phenotypes.
Haplotypes
Haplotype = Series of alleles along a very
short section of the same chromosome.
Examples:
A
b c
D
a
b C
d
= the AbcD Haplotype
= the abCd Haplotype
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.
Haplotypes
Haplotypes expected by chance
B .4
A
A .7
.28
a
a .3
B
B
.12
b .6
A
b
.42
a
b
.18
Haplotypes
Test for Linkage Disequilibrium = Compare
observed frequencies with those expected by
chance.
Expected
Observed
Haplotype
by Chance
Frequency
A B
.28
.37
A
b
.42
.33
a
B
.12
.03
a
b
.18
.27
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.
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.