1. 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 B C D E a b c d e C A B D E a b c d e A B C d e a b c D E A B C d e a b c D E

2. The probability of recombination
depends upon DISTANCE A B C E F G D a b c e f g d
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.

3. 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.

4. Linkage Analysis
Father’s chromosomes are A D a d Aa aa

5. Linkage Analysis
Father’s chromosomes are A d a D Aa aa

6. Linkage Analysis Father’s chromosomes are Aa aa Aa aa Aa Aa aa Aa aa AD a d
Father’s chromosomes are Aa aa Aa aa Aa Aa aa Aa aa

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

8. 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 d a D
A father’s who is a d
and a mother who is

9. Linkage Analysis Father’s chromosomes are Father’s Gametes:d a D
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

10. A d a D
.5(1 - )
.5
Genotype
Gametes
Probability

11. Linkage Analysis Mother’s chromosomes ared
Mother can only give gamete ad with
probability = 1.0

12. Linkage Analysis Father’s Gametes: Ad aD AD ad Ad aD AD ad ad ad ad ad
.5(1 - )
.5(1 - ) M o t h e r s
.5
.5 G a m e t Ad aD AD ad ad ad ad ad ad
1.0
.5(1 - )
.5(1 - )
.5
.5
normal
disorder
disorder
normal

13. Linkage Analysis Offspring Phenotypes and Probabilities: Aa Aa aa aa
Marker
Phenotype
Disorder
Phenotype
Probability Aa
normal
.5(1 - ) Aa
affected
.5 aa
normal
.5 aa
affected
.5(1 - )

14. 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.

15. Haplotypes
Haplotype = Series of alleles along a very short section of the same chromosome.
Examples: A b c D
= the AbcD Haplotype a b C d
= the abCd Haplotype

16. 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.

17. A .7 a .3 B .4 b .6 Haplotypes Haplotypes expected by chance A B a b
.28
.42
.12
.18

18. Haplotypes
Test for Linkage Disequilibrium = Compare observed frequencies with those expected by chance. A B a b
.28
.42
.12
.18
Expected by Chance
Haplotype
Observed Frequency
.37
.33
.03
.27

19. 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.

20. DNA region of interest:
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.