Question

The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium and
(a) Calculate the percentage of heterozygous individuals in the population.
(b) Calculate the percentage of homozygous recessives in the population.

• A. 60% and 40%
• B. 66% and 31%
• C. 40% and 60%
• D. 31% and 66%

Answer 1

Answer: (D)

According to the Hardy-Weinberg Equilibrium equation, the percentage of heterozygous individuals in the population is represented by the 2pq. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%.
The percentage of homozygous recessives individuals in the population is represented by the q2, which equals to 0.81 × 0.81 = 0.66 or 66%.