Linkage and association Sarah Medland

Genotypic similarity between relatives IBS Alleles shared Identical By State “look the same”, may have the same DNA sequence but they are not necessarily derived from a known common ancestor - focus for association IBD Alleles shared Identical By Descent are a copy of the same ancestor allele - focus for linkage M1M1 Q1Q1 M2M2 Q2Q2 M3M3 Q3Q3 M3M3 Q4Q4 M1M1 Q1Q1 M3M3 Q3Q3 M1M1 Q1Q1 M3M3 Q4Q4 M1M1 Q1Q1 M2M2 Q2Q2 M3M3 Q3Q3 M3M3 Q4Q4 IBS IBD 2 1

In linkage analysis we will be estimating an additional variance component Q  For each locus under analysis the coefficient of sharing for this parameter will vary for each pair of siblings The coefficient will be the probability that the pair of siblings have both inherited the same alleles from a common ancestor

QACE P Twin1 ECAQ P Twin2 MZ=1.0 DZ=0.5 MZ & DZ = qace ecaq

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=0 p.ibd1=.6 p.ibd2=.2 Use all the information

Alternative approach Model each of the possible outcomes  IBD0 IBD1 IBD2 Weight each of the models by the probability that it is the correct model The pairwise likelihood is equal to the sum of likelihood for each model multiplied by the probability it is the correct model The combined likelihood is equal to the sum of all the pairwise likelihoods

DZ pairs * pIBD2 + * pIBD1 + * pIBD0

How to do this in mx? Script link_mix.mx G2 DZ TWINS Data NInput=124 NModel=3 Missing = Rectangular File=example3.dat Labels …. Select pheno1 pheno2 z0_20 z1_20 z2_20age1 sex1 age2 sex2; Definition_variables z0_20 z1_20 z2_20age1 sex1 age2 sex2; Tells Mx we will be using 3 different means and variance models Tells 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 case

How 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 0.5 Specify B z0_20 z1_20 z2_20! put ibd probabilities in B This script runs an AE model D is the QTL VC path coefficent We are placing the prob. Of being IBD 0 1 & 2 in the B matrix

How to do this in mx? Begin Algebra; T = X*X'+Z*Z'+D*D' ; ! total variance U = ; ! IBD 0 cov (=non-qtl cov) K = U + ; ! 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 matrix Pre-computing the total variance and the covariance for the diff. IBD groups Stacking the pre- computed covariance matrices

How 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 matrix

Summary Weighted likelihood approach more powerful than pi-hat Quickly becomes unfeasible  3 models for sibship size 2  27 models for sibship size 4  Q: How many models for sibship size 3? For larger sib-ships/arbitrary pedigrees pi-hat approach is method of choice

Association Introduction

Association Simplest design possible Correlate phenotype with genotype Candidate genes for specific diseases common practice in medicine/genetics Genome-wide association with millions available SNPs, can search whole genome exhaustively

Allelic Association chromosomeSNPstrait variant Genetic variation yields phenotypic variation More copies of ‘B’ alleleMore copies of ‘b’ allele

2a bbBBBb d midpoint GenotypeGenetic Value BB Bb bb a d -a Va (QTL) = 2pqa 2 (no dominance) Biometrical model

10 Basic premise of assoc. for qualitative trait Chose a phenotype & a candidate gene(s) Collect 2 groups - cases and controls  Unrelated individuals  Matched for relevant covariates Genotype your individuals for your gene(s) Count the % of cases & controls with each genotype Run a chi-square test

Y i =  +  X i + e i where Y i = trait value for individual i X i =1 if allele individual i has allele ‘A’ 0 otherwise i.e., test of mean differences between ‘A’ and ‘not-A’ individuals 10 The equivalent for a quantitative trait - run a regression Play with Association.xls

Practical – Find a gene for sensation seeking: Two populations (A & B) of 100 individuals in which sensation seeking was measured In population A, gene X (alleles 1 & 2) does not influence sensation seeking In population B, gene X (alleles 1 & 2) does not influence sensation seeking Mean sensation seeking score of population A is 90 Mean sensation seeking score of population B is 110 Frequencies of allele 1 & 2 in population A are.1 &.9 Frequencies of allele 1 & 2 in population B are.5 &.5

Sensation seeking score is the same across genotypes, within each population. Population B scores higher than population A Differences in genotypic frequencies Genotypic freq.

Suppose we are unaware of these two populations and have measured 200 individuals and typed gene X The mean sensation seeking score of this mixed population is 100 What are our observed genotypic frequencies and means? Calculating genotypic frequencies in the mixed population Genotype 11: 1 individual from population A, 25 individuals from population B on a total of 200 individuals: (1+25)/200=.13 Genotype 12: (18+50)/200=.34 Genotype 22: (81+25)/200=.53 Calculating genotypic means in the mixed population Genotype 11: 1 individual from population A with a mean of 90, 25 individuals from population B with a mean of 110 = ((1*90) + (25*110))/26 =109.2 Genotype 12: ((18*90) + (50*110))/68 = Genotype 22: ((81*90) + (25*110))/106 = 94.7

Now, allele 1 is associated with higher sensation seeking scores, while in both populations A and B, the gene was not associated with sensation seeking scores… FALSE ASSOCIATION Genotypic freq. Gene X is the gene for sensation seeking!

What if there is true association? allele 1 frequency 0.1 allele 2 frequency 0.9 allele 1 = -2, allele 2 = +2 Pop mean = 90 allele 1 frequency 0.5 allele 2 frequency 0.5. allele 1 = -2 allele 2 = +2 Pop mean = Genotypic freq.

Calculate: Genotypic means in mixed population Genotypic frequencies in mixed population Is there an association between the gene and sensation seeking score? If yes which allele is the increaser allele?

False positives and false negatives Posthuma et al., Behav Genet, 2004

How to avoid spurious association? True association is detected in people coming from the same genetic stratum Can check that individuals come from the same population using a large set of highly polymorphic genes – genomic control Can use family members as controls – family based association

Fulker (1999) between/within association model Fulker et al, AJHG, 1999

b ij as Family Control b ij is the expected genotype for each individual  Ancestors  Siblings w ij is the deviation of each individual from this expectation Informative individuals  To be “informative” an individual’s genotype should differ from expected  Have heterozygous ancestor in pedigree β b ≠ β w is a test for population stratification β w > 0 is a test for association free from stratification

BTW – this is on top of a linkage model

So… 4 tests for the price of one  Test for pop-stratification a w =a b  Robust test for association a w =0  Test so see if linked loci is the functional variant QTL≠ 0 in the presence of a w it is in LD with the variant but is not the casual variant  Test for dominance effects if dominance is also modeled

Combined Linkage & association Implemented in QTDT (Abecasis et al., 2000) and Mx (Posthuma et al., 2004) Association and Linkage modeled simultaneously: Association is modeled in the means Linkage is modeled in the (co)variances QTDT: simple, quick, straightforward, but not so flexible in terms of models Mx: less simple, but highly flexible

Implementation in Mx link_assoc.mx #define n 3! number of alleles is 3, coded 1, 2, 3 … G1: calculation group between and within effects Data Calc Begin matrices; A Full 1 n free! additive allelic effects within C Full 1 n free! additive allelic effects between D Sdiag n n free ! dominance deviations within F Sdiag n n free ! dominance deviations between I Unit 1 n ! one's End matrices; Specify A Specify C Specify D Specify F The locus has 3 alleles These 1*3 vector matrices contain the b/n & w/n additive effects of each of the 3 alleles These 3*3 off-diagonal matrices contain the b/n & w/n dominance effects of each of the 3 alleles

Sticking it together… Begin algebra; K = + ;! Within effects, additive L = D + D' ;! Within effects, dominance W = K+L ;! Within effects - additive and dominance in one matrix M = + ;! Between effects, additive N = F + F' ;! Between effects, dominance B = M+N ;! Between effects - additive and dominance in one matrix End algebra ;

W = K+L ; K+L = 2*a1 a1+a2+ d12 a1+a3+ d13 a1+a2+ d12 2*a2 a2+a3+ d23 a1+a3+ d13 a2+a3+ d23 2*a3 This is the 1/2 genotype mean composed of the simple additive effects of allele 1 + allele 2 and any deviation from these simple additive effects (dominance effects)

W = K+L ; K+L (parameter numbers) = Between effects stick together in the same way

K = + ;! Within effects, additive L = D + D' ;! Within effects, dominance W = K+L ;! Within effects total K = + = a11 [1 1 1] + [a1 a2 = a31 a1 a1 a1 a1 a2 a3 a1a1 a1a2 a1a3 a2 a2 a2 + a1 a2 a3 = a2a1 a2a2 a2a3 a3 a3 a3 a1 a2 a3 a3a1 a3a2 a3a3 I = [ 1 1 1], A = [a1 a2 a3] D = d d31 d32 0 W = K+L = a1a1 a1a2 a1a3 0 d21 d31 a1a1 a1a2d21 a1a3d31 a2a1 a2a2 a2a3 + d21 0 d32 = a2a1d21 a2a2 a2a3d32 a3a1 a3a2 a3a3 d31 d32 0 a3a1d31 a3a2d32 a3a3 L = D + D' = d21 d31 0 d21 d31 d d32 = d21 0 d32 d31 d d31 d32 0 M = + ;! Between effects, additive N = F + F' ;! Between effects, dominance B = M+N ;! Between effects - total

We have a sibpair with genotypes 1,1 and 1,2. μ1 = Grand mean + pair mean + ½ pair difference μ2 = Grand mean + pair mean - ½ pair difference To calculate the between-pairs effect, or the mean genotypic effect of this pair, we need matrix B: ((c1c1) + (c1c2f21)) / 2 To calculate the within-pair effect we need matrix W and the between pairs effect: For sib1: (a1a1) + ((c1c1) + (c1c2f21)) / 2 For sib2: (a1a2d21) - ((c1c1) + (c1c2f21)) / 2 W = a1a1 a1a2d21 a1a3d31 a2a1d21 a2a2 a2a3d32 a3a1d31 a3a2d32 a3a3 B = c1c1 c1c2f21 c1c3f31 c2c1f21 c2c2 c2c3f32 c3c1f31 c3c2f32 c3c3

G3: datagroup: sibship size two, DZ … Definition_variables tw1a1 tw1a2 tw2a1 tw2a2 … Begin Matrices; … G Full 1 nvar = G2! grand mean B Computed n n = B1 ! spurious and genuine genotypic effects (between) W Computed n n = W1! genuine genotypic effects (within) K Full 1 4 Fix! Will contain first and second allele of twin1 L Full 1 4 Fix ! Will contain first and second allele of twin2 S Full 1 1 Fix! Will contain 2 (for two individuals per family) … End Matrices; Matrix K Matrix L Specify K tw1a1 tw1a2 tw1a1 tw1a2 Specify L tw2a1 tw2a2 tw2a1 tw2a2 Alleles 1 and 2 for twins 1 and 2 respectively We are going to put the allele numbers here These matrices must be initialized – given default values

Sticking it together using part For sibpairs 1,1 and 1,2 To calculate the between-pairs effect, or the mean genotypic effect of this pair, we need matrix B: ((c1c1) + (c1c2f21)) / 2 We can use the part function to draw out a specified element of a matrix B = c1c1 c1c2f21 c1c3f31 c2c1f21 c2c2 c2c3f32 c3c1f31 c3c2f32 c3c3

Sticking it together using part We can use the part function to draw out a specified element of a matrix So to draw out c3c1f31 we could say: \part(B, K) where K is a matrix containing B = c1c1 c1c2f21 c1c3f31 c2c1f21 c2c2 c2c3f32 c3c1f31 c3c2f32 c3c3

Sibpair with genotypes: 1,1 and 1,2 Specify K tw1a1 tw1a2 tw1a1 tw1a2 = Specify L tw2a1 tw2a2 tw2a1 tw2a2 = V = (\part(B,K) + \part(B,L) ) %S ; (c1c1 + c1c2f21)/2 C = (\part(W,K) + \part(W,L) ) %S ;(a1a1 + a1a2d21)/2 Means G + F*R '+ V + (\part(W,K)-C) | G + I*R' + V +(\part(W,L)-C); = G + F*R’ + (c1c1 + c1c1f21)/2 + (a1a1 - (a1a1 + a1a2d21)/2) | G + I*R' + (c1c1 + c1c1f21)/2 + (a2a1 - (a1a1 + a1a2d21)/2) W = a1a1 a1a2d21 a1a3d31 a2a1d21 a2a2 a2a3d32 a3a1d31 a3a2d32 a3a3 B = c1c1 c1c2f21 c1c3f31 c2c1f21 c2c2 c2c3f32 c3c1f31 c3c2f32 c3c3

Constrain sum additive allelic within effects = 0 Constraint ni=1 Begin Matrices; A full 1 n = A1 O zero 1 1 End Matrices; Begin algebra; B = \sum(A) ; End Algebra; Constraint O = B ; end Constrain sum additive allelic between effects = 0 Constraint ni=1 Begin Matrices; C full 1 n = C1! O zero 1 1 End Matrices; Begin algebra; B = \sum(C) ; End Algebra; Constraint O = B ; end

!1.test for linkage in presence of full association Drop D end !2.Test for population stratification: !between effects = within effects. Specify 1 A Specify 1 C Specify 1 D Specify 1 F end !3.Test for presence of dominance end !4.Test for presence of full association end !5.Test for linkage in absence of association Free D end

ModelTest-2lldfVs modelChi^2Df-diffP-value Linkage in presence of association 2B=W 3Dominance 4Full association 5Linkage in absence of association