Suppose you observed 455 individual cockroaches in the infested apartment after 5 years of exposure to hydramethylnon + glucose bait. if the population had not evolved, how many individuals of each genotype would you expect? round your answers to the nearest whole number. (for getting started, see hint 1.)

This question is unfortunately incomplete. Let' say the allele for glucose- avoiding cockroaches is 'A' and let's say the allele for non glucose-avoiding cockroaches is 'B'. The previous part of this question determined that the frequency of 'A' in the population is 0.94 (a = 0.94) whereas the frequency of allele 'B' in the population is 0.06 (b = 0.06). Since the population has not evolved, you can assume the population is in Hardy-Weinberg equilibrium. Therefore, the frequencies of the genotypes AA, AB and BB can be calculated through: a^2 + 2ab + b^2 = 1 where a^2 is the frequency of AA2ab is the frequency of AB and b^2 is the frequency of BB.
Therefore, the frequency of AA = 0.8836, the frequency of AB = 0.1128 and the frequency of BB is 0.0036. To determine the number of individuals in the population for each genotype, multiply the frequencies by the total population.
Therefore, number of individuals of AA = 402, AB = 51, BB = 2 
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The answer is letter c