Question

Please Show Work!!

Calculate the number of individuals with a dominant phenotype (assuming dominance of allele A over allele a) considering the following: the population is in Hardy-Weinberg equilibrium, it has 100 individuals, and the frequency of the homozygote recessive genotype is 0.2. (round to the nearest integer).

Answer 1

According to hardy Weinberg equilibrium equation we know that p^2 + 2pq + q^2 = 1

Where q^2 is the frequency of homozygote recessive genotype.

So q^2 = 0.2. q = √0.2 = 0.45

We also know that p+q = 1. p = 1-0.45 = 0.55

Frequency of homozygous dominant individual is p^2 = (0.55)^2 = 0.3

Frequency of heterozygous dominant individual = 2pq = 2×0.55×0.45 = 0.5

So the total frequency of dominant individual is 0.5 + 0.3 = 0.8 or 80%

Therefore number of individual with dominant phenotype is 0.8 × 100 = 80 individuals