In a population that is in Hardy-Weinberg equilibrium, there are two possible alleles for a certain gene, A and a. If the frequency of allele A is 0.4, what fraction of the population is heterozygous?

Question

(1 point)

0.40

0.60

0.16

0.48

Answer 1

To solve this question, we need to use the Hardy-Weinberg equation, which is:

p^2 + 2pq + q^2 = 1

Where p is the frequency of allele A, q is the frequency of allele a, and p^2, 2pq, and q^2 represent the frequencies of the three possible genotypes (AA, Aa, and aa) respectively.

Given that the frequency of allele A is 0.4, we know that p = 0.4.

To find the frequency of heterozygotes (Aa), we need to calculate 2pq.

p = 0.4, q = 1 - p = 1 - 0.4 = 0.6

2pq = 2(0.4)(0.6) = 0.48

Therefore, the fraction of the population that is heterozygous is 0.48.

Answer: 0.48