A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA genotype? 0.2 0.4 0.64 0.04 0.8 Solution C). 0.64 Hardy and Weinberg mathematically proved that in a population, all dominant and recessive alleles comprise all alleles for that gene. This was mathematically represented as p+ q = 1.0 Where, p = frequency of dominant alleles q = frequency of recessive alleles. Given that, the frequency of a = 0.2 As the population is in H-W equilibrium, the frequency of A = 1-0.2 = 0.8 The frequency of AA genotype = 0.8*0.8 = 0.64.

1. A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg
equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA
genotype?
0.2
0.4
0.64
0.04
0.8
Solution
C). 0.64
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive
alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Given that, the frequency of a = 0.2
As the population is in H-W equilibrium, the frequency of A = 1-0.2 = 0.8
The frequency of AA genotype = 0.8*0.8 = 0.64