Question

Let’s say at one SNP you observe the following genotype counts: AA=600; AT=100; TT=300.

a) Calculate the χ 2 -test statistic for Hardy-Weinberg proportions, the degrees of freedom, and its associated p-value.

(b) Can we reject the null hypothesis that the population is in HWE at this locus?

(c) Undaunted, you examine the data from the other SNPs in your study. You test for Hardy-Weinberg proportions and find that about 99,093 of your SNPs have P-values <0.1, 10,020 have P-values <0.01, and 1,050 have P-values <0.001. What do you make of these results? Do you believe your assay performed adequately? Do you get your royalties? Why or why not?

Answer 1

1) Here, alles p = 2*obs (AA) + obs (AT) / 2* (obs AA + obs TT+ obs AT) = 600*2+100/2000=0.65

q=1-0.65 = 0.35

Expected Frequencies

AA = p²*n =( 0.65) sq 1000 = 422.5

AT = 2pqn = 455

TT= q square n = 122.5

Chi square statistic = (o-e)²/e = 376.2

The degree of freedom of the Hardy Weinberg proportion is 1.

the p value is <0.00001

2) As the p value is very small the null hypothesis that the population is following the HWE is rejected

3)Only 0.9% of samples have a p value less than 0.001. Since our p value is even smaller it seems that sample is not a true representation of the population. The results are thus not reliable