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.
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.
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.
Linkage Analysis A a Father’s chromosomes are D d aa Aa Aa Aa aa Aa aa Aa aa
Linkage Analysis A a Father’s chromosomes are d D aa Aa Aa Aa aa 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 handoutCalculating Gametes under Linkage on the handouts section of the course web page.)
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
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:
Probability Genotype Gametes d A .5(1 - ) A a d D A .5 D d .5 a .5(1 - ) D a
Linkage Analysis a a Mother’s chromosomes are d d Mother can only give gamete ad with probability = 1.0
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
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 - )
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.
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:
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.
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
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.
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.