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?
(1 point)
0.40
0.60
0.16
0.48
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