Question

Frequency of recessive allele is $0.2$. What is the frequency of homozygous dominant?
A) $0.64$
B) $0.32$
C) $0.8$
D) $0.064$

Answer 1

An allele as we know is a specific variation of a gene or a specific segment of DNA. As we know a recessive allele is a variety of genetic code that does not create a phenotype if a dominant allele is present. In a dominant or recessive relationship between two alleles, the recessive allele’s effects are masked by the effects of the dominant allele.

Complete answer:
According to the Hardy-Weinberg law, the allele and genotype frequencies in a population will remain constant under the absence of factors that are responsible for evolution. The sum of the frequencies is said to be equal to one.
The equation is given as.
${(p + q)^2} = {p^2} + 2pq + {q^2} = 1$
where, p is the frequency of the dominant allele
${p^2}$ indicates the frequency of homozygous dominants
q is the frequency of recessive allele
${q^2}$ is the frequency of homozygous recessive individuals, and,
$2pq$ in the equation above shows frequency of heterozygotes in the population.
In the question, it is given that the frequency of the recessive allele is $0.2$hence it can be interpreted as ${\text{q }} = {\text{ 0}}{\text{.2}}$.
Since the sum of the frequencies is equal to one. We can write as,
$p + q = 1$
Therefore, frequency of dominant allele $p{\text{ }} = {\text{ }}1 - 0.2{\text{ }} = {\text{ }}0.8$.
And the frequency of homozygous dominant ${p^2} = {(0.8)^2} = 0.64$

So, the correct answer is option A.

Note: Hardy-Weinberg equilibrium is a principle which states that the genetic variation in a population is constant from one generation to the next in the absence of disturbing factors. When mating is random in a large population with no disruptive circumstances, the law depicts that both genotype and allele frequencies will remain constant because they are in equilibrium.