Download Logistic Regression using R - Wellcome Trust Centre for

Association Analysis,
Logistic Regression,
R and S-PLUS
Richard Mott
http://bioinformatics.well.ox.ac.uk/lectures/
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
What is R ?
•
•
•
•
•
•
Statistical analysis package
Free
Similar to commercial package S-PLUS
Runs on Unix, Windows, Mac
www.r-project.org
Many packages for statistical genetics,
microarray analysis available in R
• Easily Programmable
Modelling in R
• Data for individual labelled i=1…n:
– Response yi
– 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$m31)
>
The Logistic Model
• Prob(Yi=0) = exp(hi)/(1+exp(hi))
 hi = Sj xij bj - Linear Predictor
• xij – Design Matrix (genotypes etc)
• bj – Model Parameters (to be estimated)
• Model is investigated by
– estimating the bj’s by maximum likelihood
– testing if the estimates are different from 0
The Logistic Function
Prob(Yi=0) = exp(hi)/(1+exp(hi))
Prob(Y=0)
h
Types of genetic effect at a single
locus
AA
AG
GG
Recessive
0
0
1
Dominant
1
1
0
Additive
0
1
2
Genotype
0
a
b
Additive Genotype Model
• Code genotypes as
– AA
– AG
– GG
x=0,
x=1,
x=2
• Linear Predictor
 h = b0 + xb1
•
•
•
•
P(Y=0|x) = exp(b0 + xb1)/(1+exp(b0 + xb1))
PAA = P(Y=0|x=0) = exp(b0)/(1+exp(b0))
PAG = P(Y=0|x=1) = exp(b0 + b1)/(1+exp(b0 + b1))
PGG = P(Y=0|x=2) = exp(b0 + 2b1)/(1+exp(b0 + 2b1))
Additive Model: b0 = -2 b1 = 2
PAA = 0.12 PAG = 0.50 PGG = 0.88
Prob(Y=0)
h
Additive Model: b0 = 0 b1 = 2
PAA = 0.50 PAG = 0.88 PGG = 0.98
Prob(Y=0)
h
Recessive Model
• Code genotypes as
– AA
– AG
– GG
x=0,
x=0,
x=1
• Linear Predictor
 h = b0 + xb1
• P(Y=0|x) = exp(b0 + xb1)/(1+exp(b0 + xb1))
• PAA = PAG = P(Y=0|x=0) = exp(b0)/(1+exp(b0))
• PGG = P(Y=0|x=1) = exp(b0 + b1)/(1+exp(b0 + b1))
Recessive Model: b0 = 0 b1 = 2
PAA = PAG = 0.50 PGG = 0.88
Prob(Y=0)
h
Genotype Model
• Each genotype has an independent probability
• Code genotypes as (for example)
– AA
– AG
– GG
x=0, y=0
x=1, y=0
x=0, y=1
• Linear Predictor
 h = b0 + xb1+yb2 two parameters
•
•
•
•
P(Y=0|x) = exp(b0 + xb1+yb2)/(1+exp(b0 + xb1+yb2))
PAA = P(Y=0|x=0,y=0) = exp(b0)/(1+exp(b0))
PAG = P(Y=0|x=1,y=0) = exp(b0 + b1)/(1+exp(b0 + b1))
PGG = P(Y=0|x=0,y=1) = exp(b0 + b2)/(1+exp(b0 + b2))
Genotype Model: b0 = 0 b1 = 2 b2 = -1
PAA = 0.5 PAG = 0.88 PGG = 0.27
Prob(Y=0)
h
Models in R
response y
genotype g
AA
AG
GG
model
DF
Recessive
0
0
1
y ~ dominant(g)
1
Dominant
0
1
1
y ~ recessive(g)
1
Additive
0
1
2
y ~ additive(g)
1
Genotype
0
a
b
y ~ genotype(g)
2
Data Transformation
• g <- t$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'))
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')
> 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$m26)
Df Deviance Resid. Df Resid. Dev
1017
343.71
2
108.68
1015
235.03
2
1.14
1013
233.89
3
6.03
1010
227.86
> pchisq(6.03,2,lower.tail=F) calculate p-value
[1] 0.04904584
Adding the effects of Sex and other
Covariates
• Read in sex and other covariate data, eg.
age from a file into variables, say a$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)
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!
Haplotype Association
• 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