Question

Using the Hardy-Weinberg equation and assuming that the frequency of the A1 allele is 0.8, which of the following is not true?

A) The frequency of the A2 allele is 0.2.
B) The frequency of A1A1 is 0.16.
C) The frequency of P(A1A2) is 0.32.
D) The frequency of A2A2 is 0.04.
E) Both B and D.

Answer 1

The correct answer is E) Both B and D are not true. The frequency of A1A1 should be 0.64, not 0.16, and the frequency of A2A2 is 0.04.

Step-by-step explanation:

The Hardy-Weinberg equation is used to describe the equilibrium population and is expressed as p^2 + 2pq + q^2 = 1. Given that the frequency of the A1 allele is 0.8, we can calculate the frequencies of the other alleles and genotypes. The correct answer is option E) Both B and D are not true.

Option B states that the frequency of A1A1 is 0.16, which is incorrect. To calculate the frequency of A1A1, we need to square the frequency of the A1 allele. Therefore, p^2 = (0.8)^2 = 0.64. So, the frequency of A1A1 should be 0.64, not 0.16. This means that option B is not true.

Option D states that the frequency of A2A2 is 0.04, which is also incorrect. To calculate the frequency of A2A2, we need to square the frequency of the A2 allele. Since the frequency of A1 is 0.8, the frequency of A2 would be 0.2. Therefore, q^2 = (0.2)^2 = 0.04. So, the frequency of A2A2 is indeed 0.04. This means that option D is true.

Thus, the correct answer is E) Both B and D are not true. The frequency of A1A1 should be 0.64, not 0.16, and the frequency of A2A2 is 0.04, which is true.