1. Genome-wide association studies
Usman Roshan

2. SNP Single nucleotide polymorphism
Specific position and specific chromosome

3. 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

4. SNP genotype representation
The example
F: AACACAATTAGTACAATTATGAC
M: AACAGAATTAGTACAATTATGAC
is represented as
CG CC GG …

5. SNP genotype For several individuals A/T C/T G/T … H0: AA TT GG …
H1: AT CC GT …
H2: AA CT GT … .

6. 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 … …
H1: AT CC GT … =>1 0 1 …
H2: AA CT GT … …
Now we have data in numerical format

7. 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.

8. 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

9. Encoded data A/T C/G A/G A/T C/G A/G Case1 AA CC AA 0 0 0
Case2 AT CG AA
Case3 AA CG AA => 0 1 0
Con1 TT GG GG
Con2 TT CC GG
Con3 TA CG GG

10. Ranking SNPs SNP1 SNP2 SNP3 SNP1 SNP2 SNP3 A/T C/G A/G A/T C/G A/G
Case1 AA CC AA
Case2 AT CG AA
Case3 AA CG AA => 0 1 0
Con1 TT GG GG
Con2 TT CC GG
Con3 TA CG GG
A good ranking strategy would produce SNP3, SNP1, SNP2

11. 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

12. 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

13. 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

14. 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:

15. 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

16. 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)

17. Chi-square statistic Define the statistic: 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
Great. But how do we use this to get a SNP p-value?

18. 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. c4 c3
Control c2 c1
Case
#Allele2
(wildtype)
#Allele1
(risk)

19. Chi-square statistic exercise
Compute expected values
and chi-square statistic
Compute chi-square
p-value by referring to
chi-square distribution 48 2
Control 35 15
Case
#Allele2
#Allele1

20. GWAS problems and applications
Detect causal SNPs
Chi-square
Multivariate approaches
Predict case and control from genotypes
Machine learning algorithms
A simple algorithm based on Euclidean distances