Question

A GWAS (Genome Wide Association Study) study is genotyping 1,000 individuals (300 controls and 700 cases) for 500,000 different SNPs in search for genes involved in the (non-fictional) disease lupus.

At position 45,897,592 of chromosome 2, the measurements are as follows for a C/T SNP:

-control individuals: 50 C : 550 T

-case individuals: 200 C : 1200 T

2.1) Is one allele more frequent among the cases compared to the controls? Which one?

2.2) To test if the difference in allele frequencies between case and controls is significant, you will now calculate the corresponding chi-square value. Make sure you give the details of your calculations.

2.3) Based on the chi-square value that obtained in the previous question, you conclude that there is a significant association between allelic state at this position of the genome and lupus. Using the table below, would you say that the odds of this association being spurious are:

Chi-square value	3.84	6.64	10.83
p-value	0.05	0.01	0.001

A) Greater than 5%

B) Between 5% and 1%

C) Between 1% and 0.1%

D) Less than 0.1%

That is all the data there is supposed to be, this is the complete question.

Thank you for your help!

Answer 1

2.1) Is one allele more frequent among the cases compared to the controls? Which one?

control individuals: 50 C : 550 T (total is 1000 individuals)

50 of 1000 = frequency = 5%

550 of 1000 = frequency = 55%

case individuals: 200 C : 1200 T

200 of 1000 = frequency - 20%

1200 of 1000 = frequency = 120%

According to the p-value cases are more frequent.

2.

Control	Observed	Expected	X = (O-E)^2/E		cases	Observed	Expected	X = (O-E)^2/E
	50	275	184.0909			200	700	357.1429
	500	275	184.0909			1200	700	357.1429
Total	550		368.18		Total	1400		714.29
Mean	275				Mean	700
Chi square X = 368.2					Chi square X = 714.2

Degree of freedom = n-1 = 2-1 = 1

Critical value is 6.635, p>0.01, rejected

3. A) Greater than 5%