Introduction to Linkage Disequilibrium

1 Introduction to Linkage DisequilibriumSorin IstrailSeptember 10, 2014 Sorin IstrailLinkage DisequilibriumLinkage DisequilibriumSuppose we have two genes on a single chromosomegene A and gene Bsuch that each gene has only two allelesAalleles:A1andA2 Balleles:B1andB2 Possible allele combinations:A1B1,A1B2,A2B1andA2B2 Sorin IstrailLinkage DisequilibriumLinkage DisequilibriumSuppose we have two genes on a single chromosomegene A and gene Bsuch that each gene has only two allelesAalleles:A1andA2 Balleles:B1andB2 Possible allele combinations:A1B1,A1B2,A2B1andA2B2 Sorin IstrailLinkage DisequilibriumLinkage Disequilibriump1= probability of seeing alleleA1p2= probability of seeing alleleA2p1+p2= 1By Hardy-Weinberg principle,probability of genotypeA1A1isp21probability of genotypeA1A2is 2p1p2probability of genotypeA2A2isp22HW equilibrium is about asinglelocus (with two alleles).

2 How dowe generalize to two loci??Sorin IstrailLinkage DisequilibriumLinkage Disequilibriump1= probability of seeing alleleA1p2= probability of seeing alleleA2p1+p2= 1By Hardy-Weinberg principle,probability of genotypeA1A1isp21probability of genotypeA1A2is 2p1p2probability of genotypeA2A2isp22HW equilibrium is about asinglelocus (with two alleles). How dowe generalize to two loci??Sorin IstrailLinkage DisequilibriumLinkage Disequilibriump1= probability of seeing alleleA1p2= probability of seeing alleleA2p1+p2= 1By Hardy-Weinberg principle,probability of genotypeA1A1isp21probability of genotypeA1A2is 2p1p2probability of genotypeA2A2isp22HW equilibrium is about asinglelocus (with two alleles).

3 How dowe generalize to two loci??Sorin IstrailLinkage DisequilibriumLinkage DisequilibriumLinkage Equilibrium: two sites/genes each with two allelesLinkage Equilibrium: Random AssociationLinkage Disequilibrium : correlation between two locip11= probability of seeing theA1B1haplotypep12= probability of seeing theA1B2haplotypep21= probability of seeing theA2B1haplotypep22= probability of seeing theA2B2haplotypeThe sites are in LinkageEquilibriumifp11=p1q1,p12=p1q2, IstrailLinkage DisequilibriumLinkage DisequilibriumLinkage Equilibrium.

4 Two sites/genes each with two allelesLinkage Equilibrium: Random AssociationLinkage Disequilibrium : correlation between two locip11= probability of seeing theA1B1haplotypep12= probability of seeing theA1B2haplotypep21= probability of seeing theA2B1haplotypep22= probability of seeing theA2B2haplotypeThe sites are in LinkageEquilibriumifp11=p1q1,p12=p1q2, IstrailLinkage DisequilibriumLinkage DisequilibriumLinkage Equilibrium: two sites/genes each with two allelesLinkage Equilibrium: Random AssociationLinkage Disequilibrium : correlation between two locip11= probability of seeing theA1B1haplotypep12= probability of seeing theA1B2haplotypep21= probability of seeing theA2B1haplotypep22= probability of seeing theA2B2haplotypeThe sites are in LinkageEquilibriumifp11=p1q1,p12=p1q2, IstrailLinkage DisequilibriumLinkage DisequilibriumLinkage Disequilibrium is adeviationfrom this equilibrium.

5 D=p11 p1q1 Note thatp11=p1q1+D,p12=p1q2 D,p21=p2q1 D,p22=p2q2+ IstrailLinkage DisequilibriumLinkage DisequilibriumLemmaD=p11p22 :p11p22= (p1q1+D)(p2q2+D)=p1q1p2q2+p1q1D+p2q2D+D2 p12p21= (p1q2 D)(p2q1 D)=p1q1p2q2 p2q1D p1q2D+D2 And by subtracting these, we obtainp11p22 p12p21=D(p1q1+p2q1+p2q2+p1q2)=D (1) =DSorin IstrailLinkage DisequilibriumLinkage DisequilibriumLemmaD=p11p22 :p11p22= (p1q1+D)(p2q2+D)=p1q1p2q2+p1q1D+p2q2D+D2 p12p21= (p1q2 D)(p2q1 D)=p1q1p2q2 p2q1D p1q2D+D2 And by subtracting these, we obtainp11p22 p12p21=D(p1q1+p2q1+p2q2+p1q2)=D (1) =DSorin IstrailLinkage DisequilibriumLinkage DisequilibriumLemmaD=p11p22.

6 P11p22= (p1q1+D)(p2q2+D)=p1q1p2q2+p1q1D+p2q2D+D2 p12p21= (p1q2 D)(p2q1 D)=p1q1p2q2 p2q1D p1q2D+D2 And by subtracting these, we obtainp11p22 p12p21=D(p1q1+p2q1+p2q2+p1q2)=D (1) =DSorin IstrailLinkage DisequilibriumLinkage DisequilibriumWhat is the range of D??LetDmin= max{ p1q1, p2q2}Dmax= min{p1q2,p2q1}Now define:D ={DDmax,D>0 DDmin,D< +D, andp1q1+D 0 (sincep11is aprobability), this impliesD p1q1(and similarlyD p2q2)Sorin IstrailLinkage DisequilibriumLinkage DisequilibriumWhat is the range of D??LetDmin= max{ p1q1, p2q2}Dmax= min{p1q2,p2q1}Now define:D ={DDmax,D>0 DDmin,D< +D, andp1q1+D 0 (sincep11is aprobability), this impliesD p1q1(and similarlyD p2q2)Sorin IstrailLinkage DisequilibriumLinkage DisequilibriumWhat is the range of D?}}

7 ?LetDmin= max{ p1q1, p2q2}Dmax= min{p1q2,p2q1}Now define:D ={DDmax,D>0 DDmin,D< +D, andp1q1+D 0 (sincep11is aprobability), this impliesD p1q1(and similarlyD p2q2)Sorin IstrailLinkage DisequilibriumLinkage DisequilibriumFor both to be satisfied, it must be thatD max{ p1q1, p2q2}Similarly,D min{p1q2,p2q1}For the two loci that we are considering, each lociindividuallyis inHardy-Weinberg equilibrium, buttogethera Disequilibrium IstrailLinkage DisequilibriumLinkage DisequilibriumFor both to be satisfied, it must be thatD max{ p1q1, p2q2}Similarly,D min{p1q2,p2q1}For the two loci that we are considering, each lociindividuallyis inHardy-Weinberg equilibrium.}

8 Buttogethera Disequilibrium IstrailLinkage DisequilibriumLinkage DisequilibriumFor both to be satisfied, it must be thatD max{ p1q1, p2q2}Similarly,D min{p1q2,p2q1}For the two loci that we are considering, each lociindividuallyis inHardy-Weinberg equilibrium, buttogethera Disequilibrium IstrailLinkage DisequilibriumLinkage DisequilibriumExample:Consider two snps in the coding region of glycoproteinA and glycoprotein B that change the amino acid sequence. Bothof the proteins are on chromosome 4 and are found on the outsideof red blood AminoAcidsFor Protein A:ASerinellGLeucineFor Protein B:TMethianinellCThreonineSorin IstrailLinkage DisequilibriumLinkage DisequilibriumWe have 1000 British people in the study (which means that thereare 2000 chromosomes).

9 The genotypes for each gene are asfollows:Protein A: 298AA Protein B:99TT489AG418TC213GG483 CCThe loci are individually in HW , we can estimate the allele frequencies:A:pA=2 298 + 4892000=.5425G:qa=489 + 2 2132000=.4575 Similarly, you can find that T:pB= .3080 and C:qb= . IstrailLinkage DisequilibriumLinkage DisequilibriumWe have 1000 British people in the study (which means that thereare 2000 chromosomes). The genotypes for each gene are asfollows:Protein A: 298AA Protein B:99TT489AG418TC213GG483 CCThe loci are individually in HW , we can estimate the allele frequencies:A:pA=2 298 + 4892000=.

10 5425G:qa=489 + 2 2132000=.4575 Similarly, you can find that T:pB= .3080 and C:qb= . IstrailLinkage DisequilibriumLinkage DisequilibriumIf the haplotypes are in Linkage Equilibrium, then the probability ofeach haplotype will bepApB,pAqb,qapB, OBSERVEDEXPECTEDAT474(.5425)(.3080)(2000 ) = (.5425)(.6920)(2000) = (.4575)(.3080)(2000) = (.4575)(.6920)(2000) = 2= (Oi Ei)2 Eiwhere the degrees of freedom =#categories-1-#other dependencies, in this case is 4-1-2 = 1. 2= with 1 , which yields a P-value of .0001, so wecan safely REJECT Linkage Equilibrium between the two IstrailLinkage DisequilibriumLinkage DisequilibriumWhat about the Linkage Disequilibrium ?