1Linkage and Association inMxSarah MedlandLinkage2T2AEA Ca ac eT11 111Ee1Cc11 or .51 In biometrical modeling A is correlated at 1for MZ twins and .5 for DZ twins .5 is the average genome-wide sharing of genesbetween full siblings (DZ twin relationship) In linkage analysis we will be estimating anadditional variance component Q For each locus under analysis the coefficient ofsharing for this parameter will vary for each pair ofsiblings The coefficient will be the probability that the pair ofsiblings have both inherited the same alleles from acommon ancestorˆ!3Genotypic similarity between relativesIBD Alleles shared Identical By Descent are a copy of the same ancestor allelePairs of siblings may share 0, 1 or 2 alleles IBDThe probability of a pair of relatives being IBD is known as pi-hatM1Q1M2Q2M3Q3M3Q4M1Q1M3Q3M1Q1M3Q4M1Q1M2Q2M3Q3M3Q4IBS IBD21ˆ( 2) .5* ( 1)p IBD p IBD!= +Q A C EPTwin1E C A QPTwin2MZ=1.0 DZ=0.5MZ & DZ = 1.01 1 1 11 1 1 1q a c ee c a qˆ!4Distribution of pi-hatPIHAT651.00.94.88.81.75.69.63.56.50.44.38.31.25.19.13.060.00STUDY: 2 Harold sample (middelb lft)403020100Std. Dev = .30 Mean = .45N = 117.00 Adult Dutch DZ pairs:distribution of pi-hatat 65 cM onchromosome 19 Model resemblance(e.g. correlations,covariances) betweensib pairs, or DZ twins,as a function of DNAmarker sharing at aparticular chromosomallocationˆ!DZ by IBD statusT2QEQ Fq qf eT11 111Ee1Ff1.511m3 m4T2QEQ Fq qf eT11 111Ee1Ff111m5 m6 Variance = Q + F + E Covariance = πQ + F + ET2QEQ Fq qf eT11 111Ee1Ff1111m1 m25Covariance StatementsG2: DZ IBD2 twins Matrix K 1 Covariance F+Q+E | F+K@Q _ F+K@Q | F+Q+E;G3: DZ IBD1 twins Matrix K .5 Covariance F+Q+E | F+K@Q _ F+K@Q | F+Q+E;G4: DZ IBD0 twins Covariance F+Q+E | F_ F | F+Q+E;DZ by IBD status + MZT2AEA Ca ac eT11 111Ee1Cc1.511m1 m2Q1Q1q q1T2AEA Ca ac eT11 111Ee1Cc1.511m5 m6Q1Q1q qT2AEA Ca ac eT11 111Ee1Cc1111m7 m8Q1Q1q q1T2AEA Ca ac eT11 111Ee1Cc1.511m3 m4Q1Q1q q.56Covariance Statements +MZG2: DZ IBD2 twins Matrix K 1 Covariance A+C+Q+E | H@A+C+K@Q _ H@A+C+K@Q | A+C+Q+E;G3: DZ IBD1 twins Matrix K .5 Covariance A+C+Q+E | H@A+C+K@Q _ H@A+C+K@Q | A+C+Q+E;G4: DZ IBD0 twins Covariance A+C+Q+E | H@A+C_ H@A+C | A+C+Q+E;G5: MZ twins Covariance A+C+Q+E | A+C+Q _ A+C+Q | A+C+Q+E;Using the full distributionPIHAT651.00.94.88.81.75.69.63.56.50.44.38.31.25.19.13.060.00STUDY: 2 Harold sample (middelb lft)403020100Std. Dev = .30 Mean = .45N = 117.00 More power if we useall the availableinformation So instead of dividingthe sample we will useas a continuouscoefficient that will varybetween sib-pair acrosslociˆ!7Q F EPTwin1E F QPTwin2DZ=11 1 11 1 1q f ee f qˆ!DZ twins using full distribution of pihatCovariance F+E+Q | F+P@Q_ F+P@Q | F+E+Q ;The example dataLipid data: apoB8Pihat.mx!script for univariate linkage - pihat approach!DZ/SIB#loop $i 1 4 1#define nvar 1#NGroups 1DZ / sib TWINS genotyped Data NInput=324 Missing =-1.0000 Rectangular File=lipidall.dat Labels sample fam ldl1 apob1 ldl2 apob2 … Select apob1 apob2 ibd0m$i ibd1m$i ibd2m$i; Definition_variables ibd0m$i ibd1m$i ibd2m$i ;This use of the loop command allowsyou to run the same script over andover moving along the chromosomeThe format of the command is:#loop variable start end incrementSo…#loop $i 1 4 1Starts at marker 1 goes to marker 4 andruns each locus in turnEach occurrence of $i within the scriptwill be replaced by the current numberie on the second run $i will become 2With the loop command the last endstatement becomes an exit statementand the script ends with #end loopPihat.mx!script for univariate linkage - pihat approach!DZ/SIB#loop $i 1 4 1#define nvar 1#NGroups 1DZ / sib TWINS genotyped Data NInput=324 Missing =-1.0000 Rectangular File=lipidall.dat Labels sample fam ldl1 apob1 ldl2 apob2 … Select apob1 apob2 ibd0m$i ibd1m$i ibd2m$i; Definition_variables ibd0m$i ibd1m$i ibd2m$i ;This use of the ‘definition variables’command allows you to specify whichof the selected variables will be used ascovariatesThe value of the covariate displayed inthe mxo will be the values for the lastcase read9Pihat.mx!script for univariate linkage - pihat approach!DZ/SIB#loop $i 1 2 1#define nvar 1#NGroups 1DZ / sib TWINS genotyped Data NInput=324 Missing =-1.0000 Rectangular File=lipidall.dat Labels sample fam ldl1 apob1 ldl2 apob2 … Select apob1 apob2 ibd0m$i ibd1m$i ibd2m$i; Definition_variables ibd0m$i ibd1m$i ibd2m$i ;Begin Matrices; X Lower nvar nvar free ! residual familial F Z Lower nvar nvar free ! unshared environment E L Full nvar 1 free ! qtl effect Q G Full 1 nvar free ! grand means H Full 1 1 ! scalar, .5 K Full 3 1 ! IBD probabilities (from Merlin) J Full 1 3 ! coefficients 0.5,1 for pihat End Matrices;Specify K ibd0m$i ibd1m$i ibd2m$i Matrix H .5 Matrix J 0 .5 1 Start .1 X 1 1 1 Start .1 L 1 1 1 Start .1 Z 1 1 1 Start .5 G 1 1 1Pihat.mx Begin Algebra; F= X*X'; ! residual familial variance E= Z*Z'; ! unique environmental variance Q= L*L';! variance due to QTL V= F+Q+E;! total variance T= F|Q|E;! parameters in one matrix S= F%V| Q%V| E%V;! standardized variance component estimates P= ???? ;! estimate of pihat End Algebra; Labels Row S standest Labels Col S f^2 q^2 e^2 Labels Row T unstandest Labels Col T f^2 q^2 e^2 Means G| G ; Covariance F+E+Q | F+P@Q_ F+P@Q | F+E+Q ; Option NDecimals=4 Option RSiduals Option Multiple Issat!End!test significance of QTL effect! Drop L 1 1 1Exit#end loopYou need to fix thisbefore you run thescript10Run the script across yourregion and add the -2LL to thexls fileConverting chi-squares to LOD scores For univariate linkage analysis(where you have 1 QTL estimate)Χ2/4.6 = LOD11Converting chi-squares to p values Complicated Distribution of genotypes and phenotypes Boundary problems For univariate linkage analysis(where you have 1 QTL estimate)p(linkage)=21/ 2!APOb00.20.40.60.811.21.41.61.80 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 104positionLod-scorelod_pihatFE12What does it mean?Alternative approach is a summary statistic Convenient Loss of information .5 can mean p.ibd0=0 p.ibd1=1 p.ibd2=0 or p.ibd0=.2 p.ibd1=.6 p.ibd2=.2 Use all the
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