Question

In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygous genotype of a certain trait is 0.16. What is the percentage of individuals homozygous for the dominant allele?

• A. 48%
• B. 16%
• C. 25%
• D. 36%

Answer 1

The correct option is D 36%

Hardy Weinberg law states that the genotype and allele frequencies will remain constant from generation to generation if the following conditions are met:

The population should be large.
The population should mate randomly without selecting certain genotypes.
No evolutionary influence should occur such as natural selection, mutation, immigration (Individuals entering the population) or emigration (Individuals leaving the population).

Consider a gene with two alleles, A and a.
Let us assume that the dominant allele for the gene is ‘A’ and the recessive allele is ‘a’
The genotype of the homozygous dominant individuals would be AA, heterozygous dominant individuals would be Aa and recessive individuals would be aa.

According to the Hardy-Weinberg equation,
p + q = 1 (Eq. 1)
Where, p = frequency of A allele
And q = frequency of a allele

Also, $p^2+q^2+2pq=1$ (Eq. 2)
p² = frequency of individuals with AA genotype in a population
2pq = frequency of individuals with Aa genotype in a population
q² = frequency of individuals with aa genotype in a population

Given in the question:
Frequency of individuals carrying recessive genotype (aa) in the population = q² = 0.16

To be found out:
Percentage of individuals carrying homozygous dominant genotype (AA) in the population = p²=?

We know that the frequency of the recessive homozygous genotype is q² and it is given in the question as 0.16.
∴ $q=\sqrt{0.16}$
q = 0.4

we also know that p + q = 1
Thus, p = 1 – q
Therefore, p = 1 – 0.4 = 0.6
Therefore, the frequency of homozygous dominant individuals represented by p²= (0.6) = 0.36 or 36%
And
The frequency of heterozygous individuals, 2pq= 2 (0.6) (0.4) = 0.48 or 48 %