Final answer:
The frequency of heterozygotes in a population where the A allele frequency is 0.9 is 0.18, as calculated using the Hardy-Weinberg principle with the equation 2pq.
Step-by-step explanation:
The question at hand involves the concept of Hardy-Weinberg equilibrium, which is a principle of population genetics. Given that the frequency of allele A (p) is 0.9, and knowing the Hardy-Weinberg principle that states p + q = 1 for a two-allele system, where q is the frequency of the alternative allele a, we can calculate the frequency of the recessive allele as 1 - 0.9 = 0.1. Applying the principle again in the equation p² + 2pq + q² = 1, we can find the frequency of the heterozygotes (Aa) by calculating 2pq which is 2 * 0.9 * 0.1 = 0.18. Therefore, the frequency of heterozygotes in the population is 0.18.