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At a particular locus, 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?
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$0.48$
In a stable population, for a gene with two alleles, ' $A$ ' (dominant) and 'a' (recessive), if the frequency of ' $A$ ' is $p$ and the frequency of ' $\mathrm{a}$ ' is $q$, then the frequencies of the three possible genotypes (AA, Aa and aa) can be expressed by the Hardy-Weinberg equation:
$p^2+2 p q+q^2=1$
where $p^2=$ Frequency of AA (homozygous dominant) individuals
$q^2=$ Frequency of aa (homozygous recessive) individuals
$2 p q=$ Frequency of Aa (heterozygous) individuals.
So, $p=0.6$ and $q=0.4$ (given)
$\begin{aligned} \therefore 2 p q(\text { frequency of heterozygote }) & =2 \times 0.6 \times 0.4 \\ & =0.48 .\end{aligned}$
$p^2+2 p q+q^2=1$
where $p^2=$ Frequency of AA (homozygous dominant) individuals
$q^2=$ Frequency of aa (homozygous recessive) individuals
$2 p q=$ Frequency of Aa (heterozygous) individuals.
So, $p=0.6$ and $q=0.4$ (given)
$\begin{aligned} \therefore 2 p q(\text { frequency of heterozygote }) & =2 \times 0.6 \times 0.4 \\ & =0.48 .\end{aligned}$
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