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**Association Analysis, Logistic Regression, R and S-PLUS**

Richard Mott

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**Logistic Regression in Statistical Genetics**

Applicable to Association Studies Data: Binary outcomes (eg disease status) Dependent on genotypes [+ sex, environment] Aim is to identify which factors influence the outcome Rigorous tests of statistical significance Flexible modelling language Generalisation of Chi-Squared Test

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**What is R ? Statistical analysis package Free**

Similar to commercial package S-PLUS Runs on Unix, Windows, Mac Many packages for statistical genetics, microarray analysis available in R Easily Programmable

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**Modelling in R Data for individual labelled i=1…n: Response yi**

Genotypes gij for markers j=1..m

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**Coding Unphased Genotypes**

Several possibilities: AA, AG, GG original genotypes 12, 21, 22 1, 2, 3 0, 1, 2 # of G alleles Missing Data NA default in R

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**Using R Load genetic logistic regression tools**

> source(‘logistic.R’) Read data table from file > t <- read.table(‘geno.dat’, header=TRUE) Column names names(t) t$y response (0,1) t$m1, t$m2, …. Genotypes for each marker

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**Contigency Tables in R ftable(t$y,t$m31) prints the contingency table**

>

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**Chi-Squared Test in R > chisq.test(t$y,t$m31)**

Pearson's Chi-squared test data: t$y and t$m31 X-squared = , df = 2, p-value = Warning message: Chi-squared approximation may be incorrect in: chisq.test(t$y, t$m31) >

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**The Logistic Model Prob(Yi=0) = exp(hi)/(1+exp(hi))**

hi = Sj xij bj - Linear Predictor xij – Design Matrix (genotypes etc) bj – Model Parameters (to be estimated) Model is investigated by estimating the bj’s by maximum likelihood testing if the estimates are different from 0

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**The Logistic Function Prob(Yi=0) = exp(hi)/(1+exp(hi))**

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**Types of genetic effect at a single locus**

AA AG GG Recessive 1 Dominant Additive 2 Genotype a b

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**Additive Genotype Model**

Code genotypes as AA x=0, AG x=1, GG x=2 Linear Predictor h = b0 + xb1 P(Y=0|x) = exp(b0 + xb1)/(1+exp(b0 + xb1)) PAA = P(Y=0|x=0) = exp(b0)/(1+exp(b0)) PAG = P(Y=0|x=1) = exp(b0 + b1)/(1+exp(b0 + b1)) PGG = P(Y=0|x=2) = exp(b0 + 2b1)/(1+exp(b0 + 2b1))

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**Additive Model: b0 = -2 b1 = 2 PAA = 0.12 PAG = 0.50 PGG = 0.88**

Prob(Y=0) h

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**Additive Model: b0 = 0 b1 = 2 PAA = 0.50 PAG = 0.88 PGG = 0.98**

Prob(Y=0) h

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**Recessive Model Code genotypes as Linear Predictor**

AA x=0, AG x=0, GG x=1 Linear Predictor h = b0 + xb1 P(Y=0|x) = exp(b0 + xb1)/(1+exp(b0 + xb1)) PAA = PAG = P(Y=0|x=0) = exp(b0)/(1+exp(b0)) PGG = P(Y=0|x=1) = exp(b0 + b1)/(1+exp(b0 + b1))

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**Recessive Model: b0 = 0 b1 = 2 PAA = PAG = 0.50 PGG = 0.88**

Prob(Y=0) h

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**Genotype Model Each genotype has an independent probability**

Code genotypes as (for example) AA x=0, y=0 AG x=1, y=0 GG x=0, y=1 Linear Predictor h = b0 + xb1+yb2 two parameters P(Y=0|x) = exp(b0 + xb1+yb2)/(1+exp(b0 + xb1+yb2)) PAA = P(Y=0|x=0,y=0) = exp(b0)/(1+exp(b0)) PAG = P(Y=0|x=1,y=0) = exp(b0 + b1)/(1+exp(b0 + b1)) PGG = P(Y=0|x=0,y=1) = exp(b0 + b2)/(1+exp(b0 + b2))

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**Genotype Model: b0 = 0 b1 = 2 b2 = -1 PAA = 0.5 PAG = 0.88 PGG = 0.27**

Prob(Y=0) h

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**Models in R response y genotype g**

AA AG GG model DF Recessive 1 y ~ dominant(g) Dominant y ~ recessive(g) Additive 2 y ~ additive(g) Genotype a b y ~ genotype(g)

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**Data Transformation g <- t$m1**

use these functions to treat a genotype vector in a certain way: a <- additive(g) r <- recessive(g) d <- dominant(g) g <- genotype(g)

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**Fitting the Model Equivalent models: genotype = dominant + recessive**

afit <- glm( t$y ~ additive(g),family=‘binomial’) rfit <- glm( t$y ~ recessive(g),family=‘binomial’) dfit <- glm( t$y ~ dominant(g),family=‘binomial’) gfit <- glm( t$y ~ genotype(g),family=‘binomial’) Equivalent models: genotype = dominant + recessive genotype = additive + recessive genotype = additive + dominant genotype ~ standard chi-squared test of genotype association

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Parameter Estimates > summary(glm( t$y ~ genotype(t$m31), family='binomial')) Coefficients: Estimate Std. Error z value Pr(>|z|) b0 (Intercept) <2e-16 *** b1 genotype(t$m31) b2 genotype(t$m31) --- Signif. codes: 0 `***' `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 >

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**Analysis of Deviance Chi-Squared Test**

> anova(glm( t$y ~ genotype(t$m31), family='binomial')) Analysis of Deviance Table Model: binomial, link: logit Response: t$y Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev NULL genotype(t$m31)

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Model Comparison Compare general model with additive, dominant or recessive models: > afit <- glm(t$y ~ additive(t$m20)) > gfit <- glm(t$y ~ genotype(t$m20)) > anova(afit,gfit) Analysis of Deviance Table Model 1: t$y ~ additive(t$m20) Model 2: t$y ~ genotype(t$m20) Resid. Df Resid. Dev Df Deviance >

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**Scanning all Markers > logscan(t,model=‘additive’)**

Deviance DF Pval LogPval m e e m e e m e e m e e m e e m e e m e e m e e m e e m e e m e e …

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Multilocus Models Can test the effects of fitting two or more markers simultaneously Several multilocus models are possible Interaction Model assumes that each combination of genotypes has a different effect eg t$y ~ t$m10 * t$m15

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Multi-Locus Models > f <- glm( t$y ~ genotype(t$m13) * genotype(t$m26) , family='binomial') > anova(f) Analysis of Deviance Table Model: binomial, link: logit Response: t$y Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev NULL genotype(t$m13) genotype(t$m26) genotype(t$m13):genotype(t$m26) > pchisq(6.03,2,lower.tail=F) calculate p-value [1]

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**Adding the effects of Sex and other Covariates**

Read in sex and other covariate data, eg. age from a file into variables, say a$sex, a$age Fit models of the form fit1 <- glm(t$y ~ additive(t$m10) + a$sex + a$age, family=‘binomial’) fit2 <- glm(t$y ~ a$sex + a$age, family=‘binomial’)

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**Adding the effects of Sex and other Covariates**

Compare models using anova – test if the effect of the marker m10 is significant after taking into account sex and age anova(fit1,fit2)

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Multiple Testing Take care interpreting significance levels when performing multiple tests Linkage disequilibrium can reduce the effective number of independent tests Permutation is a safe procedure to determine significance Repeat j=1..N times: Permute disease status y between individuals Fit all markers Record maximum deviance maxdev[j] over all markers Permutation p-value for a marker is the proportion of times the permuted maximum deviance across all markers exceeds the observed deviance for the marker logscan(t,permute=1000) slow!

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**Haplotype Association**

Different from multiple genotype models Phase taken into account Haplotype association can be modelled in a similar logistic framework Treat haplotypes as extended alleles Fit additive, recessive, dominant & genotype models as before Eg haplotypes are h = AAGCAT, ATGCTT, etc y ~ additive(h) y ~ dominant(h) etc

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