Question

A gene locus has two alleles A and a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

• A. 0.16 (AA); 0.36 (Aa); 0.48 (aa)
• B. 0.36 (AA); 0.48 (Aa); 0.16 (aa)
• C. 0.16 (AA); 0.24 (Aa); 0.36 (aa)
• D. 0.16 (AA); 0.48 (Aa); 0.36 (aa)

Answer 1

The correct option is D 0.16 (AA); 0.48 (Aa); 0.36 (aa)

According to the Hardy-Weinberg equation, if there are two alleles for a particular gene, A represents the dominant allele and a represents the recessive allele, then the frequency of A is p and a is q.

Combined frequencies of all alleles for a gene in a population is equal to 1.

Therefore, p+q=1.

The frequency of homozygous dominant individuals (AA) in the population would be equal to pxp or p2

The frequency of homozygous recessive individuals (aa) in the population would be equal to qxq or q2

Since there are two ways of forming the heterozygote Aa, (allele A from the male parent and a from the female parent and vice versa) the frequency of heterozygous dominant individuals (Aa) in the population would be 2pq.

The sum of all three genotypic frequencies is p2+2pq+ q2=1 which is a binomial expansion of (p+q)2

Therefore,
p = frequency of the dominant allele(A)
q = frequency of the recessive allele (a)
P2= frequency of the homozygous dominant individuals (AA)
q2= frequency of the homozygous recessive individuals (aa)
2pq = frequency of the heterozygous dominant individuals (Aa)

As per the data given in the question,

Frequency of dominant allele (p) = 0.4

Applying Hardy Weinberg equilibrium:

p + q = 1
q = 1 - 0.4 = 0.6

Frequency of homozygous dominant genotype, $p^{\mathrm{2\; (}}\mathrm{AA}$) = $0.4^2$= 0.16

Frequency of heterozygote, 2pq (Aa) = 2 × 0.4 × 0.6 = 0.48

Frequency of homozygous recessive genotype, $q^2$ (aa) = $0.6^2$ = 0.36