Consider a gene in a population of fruit bats. There are two alleles for this gene, A and a, that occur in a population with the frequencies 0.8 and 0.2, respectively. If the population is in Hardy-Weinberg equilibrium, what genotype frequencies will be observed?Multiple choice question.0.64 AA, 0.32 Aa, and 0.04 aa0.80 AA, 0.16 Aa, and 0.04 aa0.80 AA and 0.20 aa0.25 AA, 0.50 Aa, and 0.25 aa
Question
Consider a gene in a population of fruit bats. There are two alleles for this gene, A and a, that occur in a population with the frequencies 0.8 and 0.2, respectively. If the population is in Hardy-Weinberg equilibrium, what genotype frequencies will be observed?Multiple choice question.0.64 AA, 0.32 Aa, and 0.04 aa0.80 AA, 0.16 Aa, and 0.04 aa0.80 AA and 0.20 aa0.25 AA, 0.50 Aa, and 0.25 aa
Solution
The Hardy-Weinberg equilibrium is a principle stating that the genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors. The formula to calculate the genotype frequencies is p^2 + 2pq + q^2 = 1, where p is the frequency of one allele (in this case A) and q is the frequency of the other allele (in this case a).
Given that the frequency of allele A (p) is 0.8 and the frequency of allele a (q) is 0.2, we can substitute these values into the equation:
The frequency of AA is p^2 = (0.8)^2 = 0.64. The frequency of Aa is 2pq = 2*(0.8)*(0.2) = 0.32. The frequency of aa is q^2 = (0.2)^2 = 0.04.
So, the genotype frequencies that will be observed are 0.64 for AA, 0.32 for Aa, and 0.04 for aa. Therefore, the correct answer is "0.64 AA, 0.32 Aa, and 0.04 aa".
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