Final answer:
In a Hardy-Weinberg equilibrium population with allele frequency of A being 0.7, the genotype frequency of Aa is calculated using the equation 2pq, which gives us 2 * 0.7 * 0.3 = 0.42. Therefore, the genotype frequency of Aa is 0.42.
Step-by-step explanation:
If a population is in Hardy–Weinberg equilibrium for a gene with only two alleles and the gene frequency of allele A is 0.7, then to find the genotype frequency of Aa, one needs to use the Hardy–Weinberg equation p² + 2pq + q² = 1. In this scenario, p is the frequency of the dominant allele A, and q is the frequency of the recessive allele a. Since we know that p + q = 1, we can infer that q = 0.3. The frequency of the heterozygote genotype Aa is thus represented by 2pq.
To calculate the frequency: 2pq = 2 * 0.7 * 0.3 = 0.42. So, the correct answer to the student’s question is C. 0.42.