Question

The following numbers of genotypes were observed in a population: 264 AA, 452 Aa and 105 aa. Determine if this population is in Hardy-Weinberg equilibrium

Answer 1

Genotype AA = 264

Genotype Aa = 452

Genotype aa = 105

Total individuals (N) = 264+452+105 = 821

Allele frequency of A = [2(AA) + (Aa)]/2N = [2x264 + 452]/2x821 = 0.596 = 0.6

Allele frequency of a = [2(aa) + Aa]/2xN = [2x105 + 452]/2x821 = 0.403 =0.4

Say frequency of A = p and frequency of a = q

If the allels are in Hardy Weinberg equilibrium, then p² + 2pq + q² = 1

Putting the values we get, p² + 2pq + q² = 0.6² + 2x0.6x0.4 + 0.4² = 0.36 + 0.48 + 0.16 = 1.00

Hence the population is in Hardy Weinberg equilibrium.