Question

In a population that is in Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype is 0.09. What is the frequency of individuals that are homozygous for the dominant allele?

a. 0.7
b. 0.21
c. 0.42
d. 0.49
e. .91

Answer 1

In a Hardy-Weinberg equilibrium, if the frequency of the homozygous recessive genotype is 0.09, the frequency of the homozygous dominant genotype is 0.49.

Step-by-step explanation:

If the frequency of the homozygous recessive genotype (aa) in a population that is in Hardy-Weinberg equilibrium is 0.09, we know this represents the q² term in the Hardy-Weinberg equation (p² + 2pq + q² = 1). To find the frequency of the homozygous dominant genotype (AA), we'll first need to calculate the frequency of the recessive allele 'a' (q), which is the square root of 0.09, thus q = 0.3. Now, we use p + q = 1 to find the frequency of the dominant allele 'A' (p), which is 1 - q, or 1 - 0.3 = 0.7. Finally, the frequency of the homozygous dominant genotype (AA) is p², which is 0.7² = 0.49. Therefore, the correct answer is d. 0.49.