
At a particular locus, the frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?
A. 0.36
B. 0.16
C. 0.24
D. 0.48
Answer
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Hint: According to the Hardy Weinberg principle, the allele frequency or genotypic frequency in a population will remain constant or in equilibrium from one generation to another, given that there is no evolutionary influence.
Complete answer: The Hardy Weinberg principle or equilibrium is a mathematical equation used in population genetics. It is used to identify the frequency of a given allele in a population given that there are no external evolutionary influences such as genetic drift, inbreeding, mutation, meiotic drive, etc. The equation is given as ${p^2} + 2pq + {q^2} = 1$, where $p$ stands for the dominant homozygous species AA and $q$ stands for the recessive homozygous species aa and $2pq$ stands for the heterozygous species “Aa”. Given there are no evolutionary influences on the population the collective frequency is always in equilibrium. Let us consider the given example, where the allele frequency for the dominant allele “A” is 0.6 and the allele frequency for the recessive allele “a” is 0.4, which means that $p = 0.6$ and $q = 0.4$. The question asks the frequency of the heterozygous species in the given mating population which is $2pq$ . Thus, substituting the values of $p = 0.6$ and $q = 0.4$ in $2pq$, we get;
$2pq = 2 \times (0.6) \times (0.4) = 0.48$.
Therefore, the correct answer is option D.
Note: This equation was proposed by G.H. Hardy and Willem Weinberg to disprove the idea that only dominant alleles have an increasing frequency in any given population. The Hardy Weinberg principle fails when certain assumptions are not considered, such as organisms are diploid, only sexual reproduction occurs, generations are nonoverlapping, mating is random, population size is infinitely large, allele frequencies are equal in the sexes, there is no migration, gene flow, admixture, mutation or selection.
Complete answer: The Hardy Weinberg principle or equilibrium is a mathematical equation used in population genetics. It is used to identify the frequency of a given allele in a population given that there are no external evolutionary influences such as genetic drift, inbreeding, mutation, meiotic drive, etc. The equation is given as ${p^2} + 2pq + {q^2} = 1$, where $p$ stands for the dominant homozygous species AA and $q$ stands for the recessive homozygous species aa and $2pq$ stands for the heterozygous species “Aa”. Given there are no evolutionary influences on the population the collective frequency is always in equilibrium. Let us consider the given example, where the allele frequency for the dominant allele “A” is 0.6 and the allele frequency for the recessive allele “a” is 0.4, which means that $p = 0.6$ and $q = 0.4$. The question asks the frequency of the heterozygous species in the given mating population which is $2pq$ . Thus, substituting the values of $p = 0.6$ and $q = 0.4$ in $2pq$, we get;
$2pq = 2 \times (0.6) \times (0.4) = 0.48$.
Therefore, the correct answer is option D.
Note: This equation was proposed by G.H. Hardy and Willem Weinberg to disprove the idea that only dominant alleles have an increasing frequency in any given population. The Hardy Weinberg principle fails when certain assumptions are not considered, such as organisms are diploid, only sexual reproduction occurs, generations are nonoverlapping, mating is random, population size is infinitely large, allele frequencies are equal in the sexes, there is no migration, gene flow, admixture, mutation or selection.
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