What is the expected percentage of heterozygous individuals under Hardy-Weinberg equilibrium when the frequency of allele A is 0.10 and allele a is 0.38?

To determine the expected percentage of heterozygous individuals under Hardy-Weinberg equilibrium, we utilize the genotype frequency calculations derived from allele frequencies.

In this scenario, the frequencies of the alleles A and a are given, where the frequency of allele A (p) is 0.10 and the frequency of allele a (q) can be calculated as 1 - p - (the frequency of any additional alleles). However, in this example, it appears we only have two alleles, so we will assume q is calculated as 1 - 0.10 = 0.90.

The expected frequency of heterozygous individuals (genotype Aa) in a population at Hardy-Weinberg equilibrium can be calculated using the formula 2pq, where p is the frequency of the dominant allele A and q is the frequency of the recessive allele a.

Substituting into the equation gives:

• p = frequency of A = 0.10
• q = frequency of a = 0.90

Calculating 2pq: 2pq = 2 * (0.10) * (0.90) = 2 * 0.09 = 0.18 or 18