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
1%
b
28%
c
31%
d
46%
b) 28%
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 (0.17)
q = frequency of allele a (0.83)
2pq = frequency of heterozygous individuals
Substitute the values in the equation:
(0.17)^2 + 2(0.17)(0.83) + (0.83)^2 = 1
0.0289 + 0.2834 + 0.6889 = 1
0.0289 + 0.2834 + 0.6889 = 1
Therefore, the frequency of heterozygous individuals (2pq) in the population is 0.2834, which is equivalent to 28.34%. So the correct answer is 28%.