Final answer:
In a population where there are 58 AA individuals and 120 aa individuals, the frequency of Aa individuals is 0.41 after calculating 1 minus the combined frequencies of AA and aa from the total population. The answer is option B.
Step-by-step explanation:
The student is asking about the frequency of individuals with an Aa genotype in a population using Hardy-Weinberg principles. In this population, 58 individuals are homozygous for the red color allele (AA), and 120 are homozygous for the blue color allele (aa). Because the sum of genotype frequencies in a population should be 1 (p² + 2pq + q² = 1), and given that the number of homozygous dominant (AA) plus homozygous recessive (aa) individuals is 178 (58+120), we can infer that remaining individuals are heterozygous (Aa).
To calculate the frequency of the heterozygous genotype (Aa), we subtract the combined frequencies of the homozygous genotypes from 1 (total population). Hence, frequency of Aa = 1 - ((58/300) + (120/300)). The frequency of Aa would therefore be 1 - (0.1933 + 0.4) = 1 - 0.5933 = 0.4067, which when rounded to two decimal places is 0.41, corresponding to option B.