Question

A gene locus has two alleles A and a. The frequency of the dominant allele A is 0.4. What is 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)

The frequency of dominant allele is given as 0.4
So p = 0.4
p + q = 1
So, 0.4 + q = 1
q = 1 - 0.4
q = 0.6

the frequency of homozygous dominant, heterozygous, and homozygous recessive individuals in the population is calculated just by applying the $p^2+2\mathrm{pq}+q^2=1$ equation where $p^2$ is the frequency of homozygous dominant, 2pq is the frequency of heterozygous individuals, and $q^2$ is the frequency of homozygous recessive individuals.

So homozygous dominant = $p^2$
$p^2={(0.4)}^2$
$p^2=0.16$


Heterozygous = 2pq
2pq = $2\times 0.4\times 0.6$
2pq = 0.48


Homozygous recessive = $q^2$
$q^2={(0.6)}^2$
$q^2=0.36$

So the frequencies are 0.16 (AA); 0.48 (Aa); 0.36 (aa)