Question

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?

A. 0.40

B. 0.60

C. 0.16

D. 0.48

Answer 1

The fraction of the population that is heterozygous for alleles A and a in a Hardy-Weinberg equilibrium population, given that p (the frequency of A) is 0.4, is calculated as 2pq, which equals 2*(0.4)*(0.6) or 0.48. So the correcct option is D.

Step-by-step explanation:

In a population that is in Hardy-Weinberg equilibrium, the frequency of two alleles, A and a, are represented by p and q respectively, where p + q = 1. When given that the frequency of allele A (p) is 0.4, we can calculate the frequency of the allele a (q) as 1 - p which equals 0.6. To find the fraction of the population that is heterozygous (Aa), we use the 2pq part of the Hardy-Weinberg equation, which in this case is 2*(0.4)*(0.6), leading to an answer of 0.48.

Answer 2

I Think The answer would be c I hope This helps