A study a population with genotypes BB, Bb, and bb that are in the following frequencies

Question

A study a population with genotypes BB, Bb, and bb that are in the following frequencies: f(BB) = 0.35, f(Bb) = 0.3, and f(bb) = 0.35. Is this population at Hardy-Weinberg equilibrium (HWE)? Justify. Choose best answer.

(A) Yes, the population is at HWE. Genotype frequencies add up to one.

(B) No, the correct genotype frequencies at HWE would be: f(BB) = 0.25, f(Bb) = 0.5, and f(bb) = 0.25.

(C) No, both homozygote genotypes in this population are too infrequent while the heterozygotes are too frequent, relative to predictions of HWE.

(D) No, both homozygote genotypes in this population are too frequent while the heterozygotes are too infrequent, relative to predictions of HWE.

(E) (B) and (D) are both true.

Answer 1

Hello, I'm happy to help.

The calculation to determine whether something is in Hardy-Weinberg Equilibrium is given by the equation:

p²+2pq+q²=1 and p+q=1

In this example, p²=0.35, 2pq=0.3, and q²=0.35

To find p and q, we must take the square root of p² and q^2. As such, p=0.59 and q=0.59.

While these add up to one in the first equation, the square root of 0.35 is 0.59. In the equation p+q=1, therefore, 0.59+0.59 does not add up to one. As such, we know that the sample is not in Hardy-Weinberg Equilibrium so we know B is true.

To determine whether statement D is true, we have to calculate what the estimated heterozygous frequency should be. To find this, we use the equation 2pq. The estimated frequency should then be 2(o.59)²=0.692. This is well under the value given in the question stem. Therefore, the answer is E.