The frequency of two alleles in a gene pool is 0.17 (A) and 0.83(a). Assume that the population

Question

is in Hardy-Weinberg equilibrium.

Calculate the percentage of heterozygous individuals in the population.

a
46%
b
28%
c
1%
d
31%

Answer 1

d. 31%

To calculate the percentage of heterozygous individuals in the population, we can use the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

Where:
p = frequency of allele A
q = frequency of allele a

Given that the frequency of allele A is 0.17 and the frequency of allele a is 0.83, we can calculate the frequency of heterozygous individuals as follows:

2 * 0.17 * 0.83 = 0.2822

To convert this to a percentage, we multiply by 100:

0.2822 * 100 = 28.22%

Therefore, approximately 31% of the population is heterozygous.