Final answer:
The percentage of heterozygous individuals in a population in Hardy-Weinberg equilibrium with allele frequencies of 0.19 (A) and 0.81 (a) is calculated using the formula 2pq, resulting in about 31%.
Step-by-step explanation:
The question asks us to calculate the percentage of heterozygous individuals in a population that is in Hardy-Weinberg equilibrium, given the allele frequencies of A (p = 0.19) and a (q = 0.81). In a Hardy-Weinberg population, the sum of allele frequencies (p + q) equals 1, and genotype frequencies can be predicted using the equation p² + 2pq + q² = 1, where p² represents the frequency of homozygous dominant individuals (AA), 2pq represents the frequency of heterozygous individuals (Aa), and q² represents the frequency of homozygous recessive individuals (aa).
To find the frequency of the heterozygous genotype (Aa), we use the equation for 2pq, which is 2 * (0.19) * (0.81). This calculation gives us 0.3078 or 30.78%, meaning the percentage of heterozygous individuals in the population is approximately 31%.
Therefore, the correct answer to the student's question is c. 31%.