Final answer:
Using the principles of Hardy-Weinberg equilibrium, the calculation shows that the expected number of individuals with genotype aa in a population of 600 people where the frequency of allele A (p) is 20% would be 384. This result does not match any of the provided options.
Step-by-step explanation:
The subject in question is related to Hardy-Weinberg equilibrium, which is a principle of population genetics that asserts that the genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors. The genotype frequency of aa individuals in a population where p = 20% (which refers to the frequency of the A allele) can be determined using the Hardy-Weinberg equation: p² + 2pq + q² = 1. In this equation, q represents the frequency of the a allele, and q² represents the frequency of the homozygous recessive genotype aa.
Given that p + q = 1, if p = 0.2, then q = 1 - 0.2 = 0.8. The expected frequency of the genotype aa (q²) is 0.8². Multiplying the expected frequency of the aa genotype by the total population, we obtain (0.8²)(600) = (0.64)(600) = 384. Therefore, the expected number of individuals with the genotype aa is 384, which is not one of the options provided, suggesting either a mistake in the question options or a need to re-evaluate the initial given values and calculations.