Final answer:
In the Hardy-Weinberg equation, the right answer is 'c. Only two alleles exist for that gene.' With p as the frequency of allele A at 0.4, the frequency of the AA genotype (p²) is 0.4², which is 0.16 or 16%.
Step-by-step explanation:
In the Hardy-Weinberg equation, p + q = 1, where p and q represent the frequency of the two alleles in a population of diploid organisms. The correct statement in this context is that c. Only two alleles exist for that gene.
To answer the Hardy-Weinberg question provided, if p = 0.4, then the frequency of the AA genotype, which is denoted as p², can be calculated.
Since p stands for the frequency of the dominant allele A, the calculation for the frequency of the AA genotype would be p² = 0.4² = 0.16, meaning that the frequency of the AA genotype is 0.16 or 16%.