practice question #1\nin a population in hardy-weinberg equilibrium, if the genotype frequencies are 81% aa…

Explanation:

Step1: Calculate frequency of A allele

According to Hardy - Weinberg equilibrium, frequency of (A) allele (p) can be calculated as (p = \text{frequency of }AA+\frac{1}{2}\text{frequency of }Aa). Given (AA = 0.81) and (Aa=0.16), then (p = 0.81+\frac{1}{2}\times0.16) (p = 0.81 + 0.08=0.89)

Step2: Calculate frequency of a allele

Since (p + q=1) (where (q) is the frequency of (a) allele), then (q = 1 - p) Substitute (p = 0.89) into the formula, (q=1 - 0.89 = 0.11)

Answer:

The frequency of (A) allele is (0.89) and the frequency of (a) allele is (0.11)