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# In a Hardy-Weinberg population with two alleles ( $A$ and $a$ ) that are in equilibrium the frequency of the allele $A$ is $0.2$, What is the percentage of the population that is heterozygous for this allele?a. $0.32$b. $4$c. $16$d. $32$

Last updated date: 16th Sep 2024
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Hint: Hardy – Weinberg Principle mathematically explains the occurrence and consistency of gene frequency for a particular gene. The principle expresses that the allelic frequency endures constant through generations and the gene pool remains constant. This phenomenon is known as genetic equilibrium. Also, all the allelic frequencies summarize to $1$.

As stated in the Hardy-Weinberg equation, the sum of or summarize allele frequencies for all the alleles at the locus should be $1$ , so $p + q = 1$ . Also, the Hardy-Weinberg equation is stated as: ${p^2} + 2pq + {q^2} = 1$ ;
Where $p$ is the frequency of the " $A$ " allele and $q$ is the frequency of the " $a$ " allele in the population.
In this equation, ${p^2}$ shows the frequency of the homozygous genotype $AA$ , ${q^2}$ shows the frequency of the homozygous genotype $aa$ , and $2pq$ shows the frequency of the heterozygous genotype $Aa$.
Here, $q = 0.2$ .Hence, $p = 1 - q = 1 - 0.2 = 0.8$
So now, population of heterozygous individual will be $2pq$ as raised that is $2 \times 0.8 \times 0.2 = 0.32$
It implies, there is $32\%$ of the heterozygous population.

Hence, the correct answer is option (D).