Genome-wide association
studies
Usman Roshan
SNP
• Single nucleotide polymorphism
• Specific position and specific chromosome
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: AACACAATTAGTACAATTATGAC
M: AACAGAATTAGTACAATTATGAC
SNP genotype representation
The example
F:
M:
AACACAATTAGTACAATTATGAC
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 becomes
A/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
Case 1
Case 2
Case 3
Control 1
Control 2
Control 3
A/T
AA
AT
AA
TT
TT
TA
C/G
CC
CG
CG
GG
CC
CG
A/G …
AA
AA
AA
GG
GG
GG
Encoded data
Case1
Case2
Case3
Con1
Con2
Con3
A/T
AA
AT
AA
TT
TT
TA
C/G
CC
CG
CG
GG
CC
CG
A/G
AA
AA
AA =>
GG
GG
GG
A/T
0
1
0
2
2
1
C/G
0
1
1
2
0
1
A/G
0
0
0
2
2
2
Ranking SNPs
Case1
Case2
Case3
Con1
Con2
Con3
SNP1
A/T
AA
AT
AA
TT
TT
TA
SNP2
C/G
CC
CG
CG
GG
CC
CG
SNP3
A/G
AA
AA
AA
=>
GG
GG
GG
SNP1
A/T
0
1
0
2
2
1
SNP2
C/G
0
1
1
2
0
1
A good ranking strategy would produce SNP3, SNP1, SNP2
SNP3
A/G
0
0
0
2
2
2
Chi-square test
• Gold standard is the univariate nonparametric 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
#Allele1
#Allele2
Case
10
90
Control
2
98
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
#Allele1
(risk)
#Allele2
(wildtype)
Case
10
90
Control
2
98
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?
n k
nk
 p (1 p)
k 
• Binomial applet:
http://www.stat.tamu.edu/~west/applets/binomialdemo.html

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.
#Allele1
(risk)
#Allele2
(wildtype)
Case
c1 (X1)
c2 (X2)
Control
c3 (X3)
c4 (X4)
Chi-square statistic
n
2
(c

e
)
2
i
i

=

Define the statistic:
ei
i=1
where
ci = observed frequency for ith outcome
ei = expected
frequency for ith outcome
n = total outcomes
The probability distribution of this statistic is given by the
chi-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?
Null hypothesis for case
control contingency table
•
We have two random variables:
–
–
•
•
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
–
#Allele1
(risk)
#Allele2
(wildtype)
Case
c1
c2
Control
c3
c4
D: disease status
G: allele type.
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 pvalue).
SNPs with very small probabilities deviate
significantly from the independence
assumption and therefore considered
important.
Chi-square statistic exercise
• Compute expected values
and chi-square statistic
• Compute chi-square
p-value by referring to
chi-square distribution
#Allele1
#Allele2
Case
15
35
Control
2
48
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:
log(
•
Pr(G | D  case )
)  wT G  w0
Pr(G | D  control )
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 T G w 0 )
How do we get the odds ratio
from logistic regression? (I)
• Using Bayes rule we have
log(
Pr(G | D  case)
)  wT G  w0
Pr(G | D  control)
By exponentiating both sides we get
T
Pr(G | D  case)
 ew G  w0
Pr(G | D  control)
And by taking the ratio with G=1 and G=0 we get
Pr(G  1 | D  case)
wT 1 w0
Pr(G  1 | D  control) e
 wT 0  w  e w
0
Pr(G  0 | D  case)
e
Pr(G  0 | D  control)
How do we get the odds ratio
from logistic regression? (II)
Continued from previous slide: by rearranging the terms in the numerator
and denominator we get
Pr(G  1 | D  case)
Pr(G  1 | D  case)
Pr(G  1 | D  control)
Pr(G  0 | D  case)

Pr(G  0 | D  case)
Pr(G  1 | D  control)
Pr(G  0 | D  control) Pr(G  0 | D  control)
By symmetry of odds ratio this is
Pr(G  1 | D  case)
Pr(D  case | G  1)
Pr(G  0 | D  case)
Pr(D  control | G  1)

 OR (odds ratio)
Pr(G  1 | D  control)
Pr(D  case | G  0)
Pr(G  0 | D  control) Pr(D  control | G  0)
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.
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