The frequency of two alleles in a gene pool is 0.17 (A) and 0.83(a). Assume that the population
is in Hardy-Weinberg equilibrium.
Calculate the percentage of heterozygous individuals in the population.
a
28%
b
1%
c
46%
d
31%
The percentage of heterozygous individuals in the population can be calculated using the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
where:
p = frequency of allele A = 0.17
q = frequency of allele a = 0.83
First, calculate the frequency of heterozygous individuals (2pq):
2 * 0.17 * 0.83 = 0.2822
Convert this to a percentage:
0.2822 * 100 = 28.22%
Therefore, the percentage of heterozygous individuals in the population is approximately 28%, which is closest to option a.