Genome-wide association studies

Usman Roshan

SNP

• Single nucleotide polymorphism• Specific position and specific chromosome

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

SNP genotype

Suppose this is the DNA on chromosome 1 starting from position 1.

There is a SNP C/G on position 5, C/T on position 14, and G/T on position 21. This person is heterozygous in the first SNP and homozygous in the other two.

F: AACACAATTAGTACAATTATGACM: AACAGAATTAGTACAATTATGAC

SNP genotype representation

The example

F: AACACAATTAGTACAATTATGAC

M: AACAGAATTAGTACAATTATGAC

is represented as

CG CC GG …

SNP genotype

• For several individuals

A/T C/T G/T …

H0: AA TT GG …

H1: AT CC GT …

H2: AA CT GT …

.

.

.

SNP genotype encoding

• If SNP is A/B (alphabetically ordered) then count number of times we see B.

• Previous example becomesA/T C/T G/T … A/T C/T G/T …

H0: AA TT GG … 0 2 0 …H1: AT CC GT … =>1 0 1 …H2: AA CT GT … 0 1 1 …

Now we have data in numerical format

Genome wide association studies (GWAS)

• Aim to identify which regions (or SNPs) in the genome are associated with disease or certain phenotype.

• Design:– Identify population structure– Select case subjects (those with disease)– Select control subjects (healthy)– Genotype a million SNPs for each subject– Determine which SNP is associated.

Example GWAS

A/T C/G A/G …

Case 1 AA CC AA

Case 2 AT CG AA

Case 3 AA CG AA

Control 1 TT GG GG

Control 2 TT CC GG

Control 3 TA CG GG

Encoded data

A/T C/G A/G A/T C/GA/G

Case1 AA CC AA 0 0 0

Case2 AT CG AA 1 1 0

Case3 AA CG AA => 0 1 0

Con1 TT GG GG 2 2 2

Con2 TT CC GG 2 0 2

Con3 TA CG GG 1 1 2

Ranking SNPs

SNP1 SNP2 SNP3 SNP1 SNP2 SNP3

A/T C/G A/G A/T C/G A/G

Case1 AA CC AA 0 0 0

Case2 AT CG AA 1 1 0

Case3 AA CG AA => 0 1 0

Con1 TT GG GG 2 2 2

Con2 TT CC GG 2 0 2

Con3 TA CG GG 1 1 2

A good ranking strategy would produce SNP3, SNP1, SNP2

Chi-square test

• Gold standard is the univariate non-parametric chi-square test with two degrees of freedom.

• Search for SNPs that deviate from the independence assumption.

• Rank SNPs by p-values

Case-control example

• Study of 100 people:– Case: 50 subjects with cancer– Control: 50 subjects without cancer

• Count number of alleles and form a 2x2 contingency table

• Relative risk:RR = Pr(disease | one copy of risk allele)/

Pr(disease | zero copies of risk allele)(Jewell ‘03)

• Due to sampling we cannot estimate the relative risk from a case-control study

• But we can estimate the odds-ratio

982Control

9010Case

#Allele2#Allele1

Odds ratio

• Odds of an event A is defined as Odds(A)= Pr(A)/Pr(~A)

• Odds ratio is the ratio of two odds. For example the ratio of odds of A and B is

OR = Odds(A)/Odds(B) = Pr(A)/Pr(~A) / Pr(B)/Pr(~B)

• Odds ratio of disease and exposed and unexposed groups would be

OR = Odds(D|G=1)/Odds(D|G=0) = = Pr(D|G=1)/Pr(~D|G=1) / Pr(D|G=0)/Pr(~D|G=0) = Pr(D|G=1)/Pr(D|G=0) x Pr(~D|G=0)/Pr(~D|G=1) = RR x Pr(~D|G=0)/Pr(~D|G=1)

Symmetry in odds ratio

• The odds ratio is symmetric in disease and genotype:

OR = Odds(D|G=1)/Odds(D|G=0) =

= Odds(G|D=1)/Odds(G|D=0)

• Great! Because we can estimate P(G|D) from a case control study. We can now use the OR as an estimate of one’s risk of disease.

Example

• Odds of risk allele in case = (10/100)/(90/100)=1/9

• Odds of risk allele in control = (2/100)/(98/100)=1/49

• Odds ratio of risk allele = 49/9

982Control

9010Case

#Allele2

(wildtype)

#Allele1

(risk)

What about significance?

• Okay, so the OR measures the risk. But is it significant? Perhaps it is due to chance.

• Let’s look at the chi-square test for measuring significance.

Statistical test of association (P-values)

• P-value = probability of the observed data (or worse) under the null hypothesis

• Example:– Suppose we are given a series of coin-tosses– We feel that a biased coin produced the tosses– We can ask the following question: what is the probability

that a fair coin produced the tosses?– If this probability is very small then we can say there is a

small chance that a fair coin produced the observed tosses.– In this example the null hypothesis is the fair coin and the

alternative hypothesis is the biased coin

Binomial distribution

• Bernoulli random variable: – Two outcomes: success of failure– Example: coin toss

• Binomial random variable:– Number of successes in a series of independent Bernoulli trials

• Example: – Probability of heads=0.5– Given four coin tosses what is the probability of three heads? – Possible outcomes: HHHT, HHTH HTHH, HHHT– Each outcome has probability = 0.5^4– Total probability = 4 * 0.5^4

Binomial distribution

• Bernoulli trial probability of success=p, probability of failure = 1-p

• Given n independent Bernoulli trials what is the probability of k successes?

• Binomial applet: http://www.stat.tamu.edu/~west/applets/binomialdemo.html

€

n

k

⎛

⎝ ⎜

⎞

⎠ ⎟pk (1− p)n−k

Hypothesis testing under Binomial hypothesis

• Null hypothesis: fair coin (probability of heads = probability of tails = 0.5)

• Data: HHHHTHTHHHHHHHTHTHTH• P-value under null hypothesis = probability

that #heads >= 15• This probability is 0.021• Since it is below 0.05 we can reject the null

hypothesis

Chi-square statistic

• Define four random variables Xi each of which is binomially distributed Xi ~ B(n, pi) where n=c1+c2+c3+c4 is the total number of subjects and pi is the probability of success of Xi.

• Each variable Xi represents the number of case and control subjects with number of risk and wildtype alleles.

• The expected value E(Xi) = npi since each Xi is binomial.

c4 (X4)c3 (X3)Control

c2 (X2)c1 (X1)Case

#Allele2 (wildtype)

#Allele1 (risk)

Chi-square statistic

Define the statistic:

where

ci = observed frequency for ith outcomeei = expected frequency for ith outcomen = total outcomes

The probability distribution of this statistic is given by thechi-square distribution with n-1 degrees of freedom.Proof can be found at http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-443Fall2003/4226DF27-A1D0-4BB8-939A-B2A4167B5480/0/lec23.pdf

Great. But how do we use this to get a SNP p-value?

€

χ 2 =(ci − ei )

2

eii=1

n

∑

Null hypothesis for case control contingency table

• We have two random variables:– D: disease status– G: allele type.

• Null hypothesis: the two variables are independent of each other (unrelated)

• Under independence – P(D,G)= P(D)P(G)– P(D=case) = (c1+c2)/n– P(G=risk) = (c1+c3)/n

• Expected values– E(X1) = P(D=case)P(G=risk)n

• We can calculate the chi-square statistic for a given SNP and the probability that it is independent of disease status (using the p-value).

• SNPs with very small probabilities deviate significantly from the independence assumption and therefore considered important.

c4c3Control

c2c1Case

#Allele2

(wildtype)

#Allele1

(risk)

Chi-square statistic exercise

482Control

3515Case

#Allele2#Allele1• Compute expected valuesand chi-square statistic• Compute chi-square p-value by referring tochi-square distribution

Logistic regression

• The odds ratio estimated from the contingency table directly has a skewed sampling distribution.

• A better (discriminative) approach is to model the log likelihood ratio log(Pr(G|D=case)/Pr(G|D=control)) as a linear function. In other words:

• Why:– Log likelihood ratio is a powerful statistic– Modeling as a linear function yields a simple algorithm to estimate

parameters

• G is number of copies of the risk allele

• With some manipulation this becomes

€

Pr(D = case |G) =1

1+ e−(w TG+w0 )

€

log(Pr(G |D = case)

Pr(G |D = control)) = wTG + w0

How do we get the odds ratio from logistic regression? (I)

• Using Bayes rule we have

log(Pr(G | D =case)

Pr(G |D =control)) =wTG +w0

Pr(G | D =case)Pr(G |D =control)

=ewT G+w0

Pr(G =1|D =case)Pr(G =1|D =control)Pr(G =0 |D =case)

Pr(G =0 |D =control)

=ewT 1+w0

ewT 0+w0=ew

And by taking the ratio with G=1 and G=0 we get

By exponentiating both sides we get

How do we get the odds ratio from logistic regression? (II)

Pr(G =1|D =case)Pr(G =1|D =control)Pr(G =0 |D =case)

Pr(G =0 |D =control)

=

Pr(G =1|D =case)Pr(G =0 |D =case)

Pr(G =1|D =control)Pr(G =0 |D =control)

Since the original ratio (see previous slide) is equal to ew and is equal to the odds ratio we conclude that the odds ratio is given by this value.

Continued from previous slide: by rearranging the terms in the numerator and denominator we get

Pr(G =1|D =case)Pr(G =0 |D =case)

Pr(G =1|D =control)Pr(G =0 |D =control)

=

Pr(D =case|G =1)Pr(D =control |G =1)Pr(D =case|G =0)

Pr(D =control |G =0)

=OR (odds ratio)

By symmetry of odds ratio this is

How to find w and w0?

• And so ew is our odds ratio. But how do we find w and w0?

– We assume that one’s disease status D given their genotype G is a Bernoulli random variable.

– Using this we form the sample likelihood

– Differentiate the likelihood by w and w0

– Use gradient descent