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Linear Modelling III
Richard Mott
Wellcome Trust Centre for Human
Genetics
Synopsis
• Logistic Regression
• Mixed Models
Logistic Regression and 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
Data
• Data for individual labelled i=1…n:
– Binary Response yi
– Unphased Genotypes gij for markers j=1..m
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
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$m2, …. Genotypes for each marker
Contigency Tables in R
• ftable(t$y,t$m31) prints the contingency table
> ftable(t$y,t$m31)
11 12 22
0
1
>
515 387
28 11
75
2
Chi-Squared Test in R
> chisq.test(t$y,t$m31)
Pearson's Chi-squared test
data: t$y and t$m31
X-squared = 3.8424, df = 2, p-value = 0.1464
Warning message:
Chi-squared approximation may be incorrect in:
chisq.test(t$y, t$m31)
>
The Logistic Model
• Effect of SNP k on phenotype
• Null model
The Logistic Model
• Prob(Yi=1) = exp(hi)/(1+exp(hi))
• hi = Sj xij bj - Linear Predictor
• xij – Design Matrix (genotypes etc)
• bj – Model Parameters (to be estimated)
• Model is investigated by
– estimating the bj s by maximum likelihood
– testing if the estimates are different from 0
The Logistic Function
Prob(Yi=1) = exp(hi)/(1+exp(hi))
Prob(Y=1)
h
Types of genetic effect at a single locus
AA
AG
GG
Recessive
0
0
1
Dominant
1
1
0
Additive
0
1
2
Genotype
0
a
b
Additive Genotype Model
• Code genotypes as
– AA
– AG
– GG
x=0,
x=1,
x=2
• Linear Predictor
– h = b0 + xb1
•
•
•
•
P(Y=1|x) = exp(b0 + xb1)/(1+exp(b0 + xb1))
PAA = P(Y=1|x=0) = exp(b0)/(1+exp(b0))
PAG = P(Y=1|x=1) = exp(b0 + b1)/(1+exp(b0 + b1))
PGG = P(Y=1|x=2) = exp(b0 + 2b1)/(1+exp(b0 + 2b1))
Additive Model: b0 = -2 b1 = 2
PAA = 0.12 PAG = 0.50 PGG = 0.88
Prob(Y=1)
h
Additive Model: b0 = 0 b1 = 2
PAA = 0.50 PAG = 0.88 PGG = 0.98
Prob(Y=1)
h
Recessive Model
• Code genotypes as
– AA
– AG
– GG
x=0,
x=0,
x=1
• Linear Predictor
– h = b0 + xb1
• P(Y=1|x) = exp(b0 + xb1)/(1+exp(b0 + xb1))
• PAA = PAG = P(Y=1|x=0) = exp(b0)/(1+exp(b0))
• PGG = P(Y=1|x=1) = exp(b0 + b1)/(1+exp(b0 + b1))
Recessive Model: b0 = 0 b1 = 2
PAA = PAG = 0.50 PGG = 0.88
Prob(Y=1)
h
Genotype Model
• Each genotype has an independent probability
• Code genotypes as (for example)
– AA
– AG
– GG
x=0, z=0
x=1, z=0
x=0, z=1
• Linear Predictor
– h = b0 + xb1+zb2 two parameters
•
•
•
•
P(Y=1|x) = exp(b0 + xb1+zb2)/(1+exp(b0 + xb1+zb2))
PAA = P(Y=1|x=0,z=0) = exp(b0)/(1+exp(b0))
PAG = P(Y=1|x=1,z=0) = exp(b0 + b1)/(1+exp(b0 + b1))
PGG = P(Y=1|x=0,z=1) = exp(b0 + b2)/(1+exp(b0 + b2))
Genotype Model: b0 = 0 b1 = 2 b2 = -1
PAA = 0.5 PAG = 0.88 PGG = 0.27
Prob(Y=1)
h
Models in R
response y
genotype g
AA
AG
GG
model
DF
Recessive
0
0
1
y ~ recessive(g)
1
Dominant
0
1
1
y ~ dominant(g)
1
Additive
0
1
2
y ~ additive(g)
1
Genotype
0
a
b
y ~ genotype(g)
2
Data Transformation
• g <- t$m1
• use these functions to treat a genotype vector
in a certain way:
–a
–r
–d
–g
<<<<-
additive(g)
recessive(g)
dominant(g)
genotype(g)
Fitting the Model
•
•
•
•
afit
rfit
dfit
gfit
<<<<-
glm(
glm(
glm(
glm(
t$y
t$y
t$y
t$y
~
~
~
~
additive(g),family=‘binomial’)
recessive(g),family=‘binomial’)
dominant(g),family=‘binomial’)
genotype(g),family=‘binomial’)
• Equivalent models:
– genotype = dominant + recessive
– genotype = additive + recessive
– genotype = additive + dominant
– genotype ~ standard chi-squared test of genotype
association
Parameter Estimates
> summary(glm( t$y ~ genotype(t$m31), family='binomial'))
Coefficients:
Estimate Std. Error z value Pr(>|z|)
b0 (Intercept)
-2.9120
0.1941 -15.006
<2e-16 ***
b1 genotype(t$m31)12 -0.6486
0.3621 -1.791
0.0733 .
b2 genotype(t$m31)22 -0.7124
0.7423 -0.960
0.3372
--Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
>
Analysis of Deviance
Chi-Squared Test
> anova(glm( t$y ~ genotype(t$m31), family='binomial'),
test="Chisq")
Analysis of Deviance Table
Model: binomial, link: logit
Response: t$y
Terms added sequentially (first to last)
NULL
genotype(t$m31)
Df Deviance Resid. Df Resid. Dev
1017
343.71
2
3.96
1015
339.76
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
1
1016
38.301
2
1015
38.124
1
0.177
>
Scanning all Markers
> logscan(t,model=‘additive’)
Deviance DF
Pval
LogPval
m1 8.604197e+00 1 3.353893e-03 2.474450800
m2 7.037336e+00 1 7.982767e-03 2.097846522
m3 6.603882e-01 1 4.164229e-01 0.380465360
m4 3.812860e+00 1 5.086054e-02 1.293619014
m5 7.194936e+00 1 7.310960e-03 2.136025588
m6 2.449127e+00 1 1.175903e-01 0.929628598
m7 2.185613e+00 1 1.393056e-01 0.856031566
m8 1.227191e+00 1 2.679539e-01 0.571939852
m9 2.532562e+01 1 4.842353e-07 6.314943565
m10 5.729634e+01 1 3.748518e-14 13.426140380
m11 3.107441e+01 1 2.483233e-08 7.604982503
…
…
…
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$m15
Multi-Locus Models
> f <- glm( t$y ~ genotype(t$m13) * genotype(t$m26) , family='binomial’,
test="Chisq")
> anova(f)
Analysis of Deviance Table
Model: binomial, link: logit
Response: t$y
Terms added sequentially (first to last)
NULL
genotype(t$m13)
genotype(t$m26)
genotype(t$m13):genotype(t$m26)
Df Deviance Resid. Df Resid. Dev
1017
343.71
2
108.68
1015
235.03
2
1.14
1013
233.89
3
6.03
1010
227.86
> pchisq(6.03,3,lower.tail=F) calculate p-value
[1] 0.1101597
Adding the effects of Sex and other
Covariates
• Read in sex and other covariate data, eg. age
from a file into variables, say a$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’)
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)
Random Effects
Mixed Models
• So far our models have had fixed effects
– each parameter can take any value independently of the
others
– each parameter estimate uses up a degree of freedom
– models with large numbers of parameters have problems
•
•
•
•
saturation - overfitting
poor predictive properties
numerically unstable and difficult to fit
in some cases we can treat parameters as being sampled from a
distribution – random effects
• only estimate the parameters needed to specify that distribution
• can result in a more robust model
Example of a Mixed Model
•
Testing genetic association across a large number of of families
–
–
–
–
yi = trait value of i’th individual
bi = genotype of individual at SNP of interest
fi = family identifier (factor)
ei = error, variance s2
– y=m+b+f+e
– If we treat family effects f as fixed then we must estimate a large number of parameters
– Better to think of these effects as having a distribution N(0,t2) for some variance t which must
be estimated
•
individuals from the same family have the same f and are correlated
–
•
cov = t2
individuals from different families are uncorrelated
– genotype parameters b still treated as fixed effects
•
mixed model
Mixed Models
• y=m+b+f+e
• Fixed effects model
– E(y) = m + b + f
– Var(y) = Is2
• I is identity matrix
• Mixed model
– E(y) = m + b
– Var(y) = Is2 + Ft2
• F is a matrix, Fij = 1 if i, j in same family
• Need to estimate both s2 and t2
Generalised Least Squares
• y = Xb + e
• Var(y) = V (symmetric covariance matrix)
• V = Is2 (uncorrelated errors)
• V = Is2 + Ft2 (single grouping random effect)
• V = Is2 + Gt2 (G = genotype identity by state)
• GLS solution, if V is known
T -1
-1
•
•
•
•
T -1
ˆ
b = (X V X) X V y
unbiased,
efficient (minimum variance)
consistent (tends to true value in large samples)
asymptotically normally distributed
Multivariate Normal Distribution
• Joint distribution of a vector of correlated observations
• Another way of thinking about the data
• Estimation of parameters in a mixed model is special case of likelihood
analysis of a MVN
• y ~ MVN(m, V)
– m is vector of expected values
– V is variance-covariance matrix
– If V = Is2 then elements of y are uncorrelated and equivalent to n
independent Normal variables
– probability density/Likelihood is
1
L(y | m,V) = exp(- (y - m)' V-1 (y - m))
2
2 n2
(2p V )
Henderson’s Mixed Model Equations
•
•
•
•
General linear mixed model.
b are fixed effect
u are random effects
X, Z design matrices
y = Xb + Zu + e
u » MVN(0,G)
(G = It 2 )
e » MVN(0,R)
(R = Is 2 )
y » MVN(Xb,V)
V = ZGZ T + R
æ XT R-1X
XT R-1Z öæbˆ ö æ XT R-1yö
ç ÷ = ç T -1 ÷
ç T -1
T -1
-1 ÷
è Z R X Z R Z + G øè uˆ ø è Z R y ø
Mixed Models
• Fixed effects models are special cases of mixed models
• Mixed models sometimes are more powerful (as fewer
parameters)
– But check that the assumption effects are sampled from a
Normal distribution is reasonable
– Differences between estimated parameters and statistical
significance in fixed vs mixed models
• Shrinkage: often random effect estimates from a mixed
model have smaller magnitudes than from a fixed effects
model.
Mixed Models in R
• Several R packages available, e.g.
– lme4 general purpose mixed models package
– emma
for genetic association in structured populations
• Formulas in lme4, eg test for association between genotype and trait
•
•
•
H0: E(y) = m
vs
H1: E(y) = m + b
•
Var(y) = Is2 + Ft2
–
–
–
h0 = lmer(y ~ 1 +(1|family), data=data)
h1 = lmer(y ~ genotype +(1|family), data=data)
anova(h0,h1)
Formulas in lmer()
• y ~ fixed.effects + (1|Group1) + (1|Group2)
etc
– random intercept models
• y ~ fixed.effects +(1|Group1) + (x|Group1)
– random slope model
Example: HDL Data
> library(lme4)
> b=read.delim("Biochemistry.txt”)
cc=complete.cases(b$Biochem.HDL)
> f0=lm(Biochem.HDL ~ Family , data=b,subset=cc)
> f1=lmer(Biochem.HDL ~ (1|Family) , data=b,subset=cc)
> anova(f0)
Analysis of Variance Table
Response: Biochem.HDL
Df Sum Sq Mean Sq F value
Pr(>F)
Family
175 147.11 0.84064 5.2098 < 2.2e-16 ***
Residuals 1681 271.24 0.16136
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> f1
Linear mixed model fit by REML
Formula: Biochem.HDL ~ (1 | Family)
Data: b
Subset: cc
AIC BIC logLik deviance REMLdev
2161 2178 -1078
2149
2155
Random effects:
Groups
Name
Variance Std.Dev.
Family
(Intercept) 0.066633 0.25813
Residual
0.161487 0.40185
Number of obs: 1857, groups: Family, 176
Fixed effects:
Estimate Std. Error t value
(Intercept) 1.60615
0.02263
70.97
> plot(resid(f0),resid(f1))
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