Linkage and associationAim of QTL mapping…Location and IdentificationSlide 4LinkageSlide 6Slide 7Slide 8Slide 9Alternative approachSlide 11DZ pairsHow to do this in mx?Slide 14Slide 15Slide 16Your jobSummaryAssociationSlide 20DefinitionsAllelic AssociationBiometrical modelBasic premise of assoc. for qualitative traitThe equivalent for a quantitative trait - run a regressionWhere this goes wrong…Slide 27Obviously this is a spurious examplePractical – Find a gene for sensation seeking:Slide 30Slide 31Calculating genotypic frequencies in the mixed populationCalculating genotypic means in the mixed populationGene X is the gene for sensation seeking!What if there is true association?Calculate:False positives and false negativesHow to avoid spurious association?Slide 39Fulker (1999) between/within association modelSlide 41Slide 42bij as Family ControlBTW – this is on top of a linkage modelSo…Combined Linkage & association Implemented in QTDT (Abecasis et al., 2000) and Mx (Posthuma et al., 2004)Implementation in Mx link_assoc.mxSticking it together…K = (A'@I) + (A@I') ;Slide 50Slide 51L = D + D' ;W = K+L ;Slide 54Slide 55Slide 56Slide 57Sticking it together using partSlide 59Slide 60Slide 61Variance/covariance matrixSlide 63Slide 64Linkage and associationSarah MedlandAim of QTL mapping…LOCALIZE and then IDENTIFY a locus that regulates a trait (QTL)•Locus: Nucleotide or sequence of nucleotides with variation in the population, with different variants associated with different trait levels.Location and Identification•Linkage•localize region of the genome where a QTL that regulates the trait is likely to be harboured•Family-specific phenomenon: Affected individuals in a family share the same ancestral predisposing DNA segment at a given QTLLocation and Identification•Association•identify a QTL that regulates the trait•Population-specific phenomenon: Affected individuals in a population share the same ancestral predisposing DNA segment at a given QTLLinkageRevisitedGenotypic similarity between relativesIBS Alleles shared Identical By State “look the same”, may have thesame DNA sequence but they are not necessarily derived from a known common ancestor - focus for associationIBD Alleles shared Identical By Descent are a copy of the same ancestor allele- focus for linkageM1Q1M2Q2M3Q3M3Q4M1Q1M3Q3M1Q1M3Q4M1Q1M2Q2M3Q3M3Q4IBSIBD21T2AEA Ca ac eT11 111Ee1Cc11 or .51In biometrical modeling A is correlated at 1 for MZ twins and .5 for DZ twins.5 is the average genome-wide sharing of genes between full siblings (DZ twin relationship)In linkage analysis we will be estimating an additional variance component QFor each locus under analysis the coefficient of sharing for this parameter will vary for each pair of siblingsThe coefficient will be the probability that the pair of siblings have both inherited the same alleles from a common ancestor ˆpQ 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 qˆpAlternative approach is a summary statistic ConvenientLoss of information .5 can mean p.ibd0=0 p.ibd1=1 p.ibd2=0 or p.ibd0=0 p.ibd1=.6 p.ibd2=.2Use all the informationˆpˆpAlternative approachModel each of the possible outcomes IBD0 IBD1 IBD2Weight each of the models by the probability that it is the correct modelThe pairwise likelihood is equal to the sum of likelihood for each model multiplied by the probability it is the correct modelThe combined likelihood is equal to the sum of all the pairwise likelihoodsDZ pairsT2QEQ Fq qf eT11 111Ee1Ff1.511m3 m4T2QEQ Fq qf eT11 111Ee1Ff111m5 m6T2QEQ Fq qf eT11 111Ee1Ff1111m1 m2* pIBD2 +* pIBD1 +* pIBD0How to do this in mx?Script link_mix.mxG2 DZ TWINS Data NInput=124 NModel=3 Missing =-99.00 Rectangular File=example3.dat Labels ….Select pheno1 pheno2 z0_20 z1_20 z2_20 age1 sex1 age2 sex2;Definition_variables z0_20 z1_20 z2_20 age1 sex1 age2 sex2;Tells Mx we will be using 3 different means and variance modelsTells Mx that these variables will be used as covariates – the values for these variables will be updated for each case during the optimization – the mxo will show the values for the final caseHow to do this in mx?Begin Matrices; X Lower nvar nvar = X1 Z Lower nvar nvar = Z1 D Lower nvar nvar = D1 B full 3 1 ! will contain IBD probabilities (from Genehunter) def var… Matrix H 0.5Specify B z0_20 z1_20 z2_20 ! put ibd probabilities in BThis script runs an AE model D is the QTL VC path coefficentWe are placing the prob. Of being IBD 0 1 & 2 in the B matrixHow to do this in mx? Begin Algebra; T = X*X'+Z*Z'+D*D' ; ! total variance U = H@X*X' ; ! IBD 0 cov (=non-qtl cov) K = U + H@D*D' ; ! IBD 1 cov W = U + D*D' ; ! IBD 2 cov A = T|U_ U|T_ ! IBD 0 matrix T|K_ K|T_ ! IBD 1 matrix T|W_ W|T ; ! IBD 2 matrixPre-computing the total variance and the covariance for the diff. IBD groupsStacking the pre-computed covariance matricesHow to do this in mx? Means G+O*R'| G+S*R'_G+O*R'| G+S*R'_G+O*R'| G+S*R'; Covariance A ; Weights B ;The means matrix contains corrections for age and sex – it is repeated 3 times and vertically stacked Tells Mx to weight each of the means and var/cov matrices by the IBD prob. which we placed in the B matrixYour jobPick a number 0 - 35 (but not 20)Change the script to run linkage at the location you pickedTest for linkageQ: Do the MZ twins add anything in this analysis? – why – why notSummaryWeighted likelihood approach more powerful than pi-hatQuickly becomes unfeasible3 models for sibship size 227 models for sibship size 4Q: How many models for sibship size 3?For larger sib-ships/arbitrary pedigrees pi-hat approach is method of choiceAssociationIntroductionAssociationSimplest design possibleCorrelate phenotype with genotypeCandidate genes for specific diseasescommon practice in medicine/geneticsGenome-wide association with millions available SNPs, can search whole genome exhaustivelyDefinitionschromosomeSNPs trait variantPopulation DataAffection TraitA 10.3A 9.9U 15.8haplotypesgenotypesallelesAllelic AssociationchromosomeSNPs trait variantGenetic variation yields phenotypic variationMore copies of ‘B’ allele More copies of ‘b’ allele2abb BBBbdmidpointGenotype Genetic ValueBBBbbbad-aVa (QTL) = 2pqa2 (no dominance)Biometrical modelBasic premise of assoc. for qualitative
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