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

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

Step-by-Step Explanation:
We can solve this problem using the Hardy-Weinberg principle, which describes the genetic equilibrium in a population. According to this principle, the allele and genotype frequencies in a population remain constant from generation to generation in the absence of other evolutionary influences.
Let the frequency of the dominant allele (A) be represented by p, and the frequency of the recessive allele (a) be represented by q.

According to the Hardy-Weinberg equation for allele frequencies:
$$p+q=1$$
Given in the question, the frequency of the dominant allele A is:
$$p=0.4$$

To find the frequency of the recessive allele a (represented by q), we rearrange the equation:
$$q=1-p$$
Substituting the value of p:
$$q=1-0.4=0.6$$

The Hardy-Weinberg equation for genotype frequencies in a population is:
$$p^2+2pq+q^2=1$$
Where:
- p² is the frequency of homozygous dominant individuals (AA)
- 2pq is the frequency of heterozygous individuals (Aa)
- q² is the frequency of homozygous recessive individuals (aa)

Now, we calculate the individual genotype frequencies:
1. Frequency of homozygous dominant individuals (AA):
$$p^2={\left(0.4\right)}^2=0.16$$
2. Frequency of heterozygous individuals (Aa):
$$2pq=2\times 0.4\times 0.6=0.48$$
3. Frequency of homozygous recessive individuals (aa):
$$q^2={\left(0.6\right)}^2=0.36$$

Thus, the frequencies are 0.16 for AA, 0.48 for Aa, and 0.36 for aa.