Question

Imagine these genotype frequencies in a population: p2 = 0.49, 2pq = 0.42, q2 = 0.09. Now assume that there is nonrandom mating where individuals with one genotype will only mate with individuals that also have their genotype. Assume this pattern of mating goes on until the frequency of heterozygotes is effectively zero. In addition, there is also inbreeding depression such that individuals with the genotype represented by p2 die before they can reproduce. What will be the eventual frequency of allele q?

a. 0.3
b. 0.91
c. 0.49
d. 1.0
e. 0.7

Answer 1

d. 1.0

Step-by-step explanation:

According to Hardy-Weinberg equilibrium,

p² + 2pq + q² = 1

p + q = 1

where p = frequency of dominant allele

q = frequency of recessive allele

p² = frequency of homozygous dominant genotype

2pq = frequency of heterozygous genotype

q² = frequency of homozygous recessive genotype

Given that non random mating occurs in this population due to which the equilibrium will be disturbed. Homozygotes of one type will only mate with homozygotes of the same type due to which frequency of heterozygotes (2pq) will become 0. Inbreeding depression will also make homozygous dominant population perish making p² 0. So the equation will change to:

p² +2pq + q² = 1

0 + 0 + q² = 1

Hence, q² = 1 which means that the entire population will only have one type of homozygous members.