How would you test for deviation from Hardy-Weinberg equilibrium using observed and expected genotype counts?

• A. A chi-square test comparing observed counts with expected counts p^2 for AA; 2pq for Aa; q^2 for aa to compute chi-square and p-value
• B. A regression of genotype frequencies on generation
• C. A t-test comparing observed counts to expected counts
• D. A Fisher exact test on allele counts

Answer: (A)

Testing deviation from Hardy-Weinberg equilibrium relies on comparing what you actually observe for genotypes with what you’d expect if the population were in equilibrium given the allele frequencies.

First, estimate the allele frequencies from the data: p is the frequency of one allele, q = 1 − p for the other. Then compute the expected number of individuals with each genotype in your sample of size n: AA would be n p^2, Aa would be n 2 p q, and aa would be n q^2.

Next, use a chi-square goodness-of-fit test by summing (observed − expected)² divided by the expected for the three genotypes. Compare that statistic to a chi-square distribution with 1 degree of freedom (since there is one independent allele frequency parameter to estimate for a two-allele locus). A small p-value suggests the observed counts deviate from Hardy-Weinberg expectations beyond what random sampling would produce.

This is the standard approach because it directly tests whether genotype counts fit the expected proportions under HWE. Other options aren’t appropriate here: regression across generations isn’t testing a single-population equilibrium, a t-test isn’t suitable for count data, and Fisher’s exact test on allele counts doesn’t assess the fit of genotype frequencies to p^2 : 2pq : q^2 proportions.