Practice question

Hardy-Weinberg principle predicts recessive homozygote frequency equals q squared for genotype aa. If observed frequency of aa equals 0.36, then q squared equals 0.36, so q equals square root of 0.36. Calculation gives q equals 0.6, with p equals one minus q equals 0.4 under two-allele model. This straightforward conversion from genotype count to allele frequency assumes random mating, large population, and absence of selection altering proportions. It underlies human genetics estimates where recessive phenotype incidence directly reveals allele frequency, subsequently allowing carrier prediction via two pq calculation.