Linkage Disequilibrium Granovsky Ilana and Berliner Yaniv Computational Genetics

What is Linkage Disequilibrium? When the occurrence of pairs of specific alleles at different loci on the same haplotype is not independent, the deviation form independence is termed linkage disequilibrium In general, linkage disequilibrium is usually seen as an association between one specific allele at one locus and another specific allele at a second locus

LinkageDisequilibrium Coefficient Definitions Linkage Disequilibrium Coefficient Definitions Marker 2 Marker1 Allele1 (probability = p2) Allele2 (probability = 1-p2) Allele1 (probability = p1) X1 p1*p2+D11 X2 p1*(1-p2)-D11 Allele2 (probability = 1-p1) X3 (1-p1)*p2-D11 X4 (1-p1)*(1-p2)+D11 Xi -number of observations in cell i (X1+X2+X3+X4)=n D11 -coefficient of gametic linkage disequilibrium between allele 1 at locus 1 and allele 1 at locus 2 D11=E[X1X4-X2X3|n=1]

Population-based sampling and the EH program We wish to test the absence of disequilibrium between allele A at locus 1 and allele B at locus 2 (D AB =0) The sample of individuals we have consist of genotyping data with no possibility to fully distinguish all of the haplotypes in each individual

Table of all possible two-locus genotypes Locus2 AAAaaa BBk1k2k3 Bbk4k5k6 bbk7k8k9 In cell 5 there can be either of two phases, AB/ab or Ab/aB

Analysis of likelihood We maximize the log likelihood of the data observed: For cell 1: p 1 =[P(A B)] For cell 4: p 4 =2P(A B)P(A b) For cell 5: p 5 =P(A B/a b)+P(A b/a B) = =2P(A B)P(a b)+2P(A b)P(a B) 2 2

Table of probabilities in each cell Locus 1 Locus 2 AAAaaa BBp(A B) 2p(A B)p(a B) P(a B) Bb 2p(A B)p(A b)2P(A B)P(a b)+ +2P(A b)P(a B) 2p(a B)p(a b) bbP(A b) 2p(A b)p(a b) P(a b)

Analysis of likelihood We maximize the likelihood above over the possible haplotype frequencies (p(A), p(B) and D AB. This likelihood is then compared with the maximum likelihood when D AB is set equal to 0 (absence of linkage disequilibrium)

Example Locus 1 Locus 2 AAAaaa BB K1=10K2 = 10K3=3 Bb K4=15K5=50K6=13 bb K7=5K8=13K9=10 Aa B4529 b3846 Aa B b *When censoring k5 all the haplotypes can be uniquely determined

Example cont. P(A) = = P(B) = = D AB = p(A B) –p(A)p(B) = 0.28 – 0.525*0.468 = * Biased example due to the elimination of the 50 observations in k 5.

EH program input file format EH = estimated haplotype. –Input file EH.dat Line 1: Number of alleles at each of the two loci Line 2: k1 k4 k7 Line 3: k2 k5 k8 Line 4: k3 k6 k9

EH program output file Output – Estimates of Gene Frequencies (including k 5) Allele Locus # of typed Individuals: 129

EH program output file Allele at locus 1 Allele at locus 2 Haplotype frequency Independent w/association

Chi square test dfLn(L)Chi-square H0: No association H1: Allelic association allowed The difference between the 2 chi-square is 8.89 The P-value associated with chi-square (with 1 df) is It is clear the k5 contributes siginificant information

Haplotype frequencies Without k5With k5 HaplotypeIndepe ndent associateIndepe ndent associate A B A b a B a B p(A) p(B) D ab Summary

Multiallelic genotype information in EH program Locus 2 Locus 11/11/22/21/32/33/3 1/1a1b1c1d1e1f1 1/2a2b2c2d2e2f2 2/2a3b3c3d3e3f3 1/3a4b4c4d4e4f4 2/3a5b5c5d5e5f5 3/3a6b6c6d6e6f6 Line 1: Number of alleles at each locus Subsequent lines :

Multilocus genotype data Locus 3 Locus 1Locus 21/11/22/2 1/1 a1b1c1 1/2a2b2c2 2/2a3b3c3 1/21/1a4b4c4 1/2a5b5c5 2/2a6b6c6 2/21/1a7b7c7 1/2a8b8c8 2/2a9b9c9

Ex. 23 Full data Solution file:Solution file: Censored data solution file.Censored data solution file Censored data 1/1 haplotype data Locus 2 Locus 1 1/11/21/31/42/22/32/43/33/44/4 1/ / / / / /

Haplotypes from censored genotype data Allele at locus 2 Allele at locus Allele at locus 2 Allele at locus

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