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A population is in Hardy- weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele $A$ is $0.7$, the genotype frequency of $Aa$ is
$0.21$
$0.42$
$0.36$
$0.7$
Solution
For a gene with two alleles, $A$ (dominant) and $a$ (recessive), if the frequency of $A$ is $p$ and the frequency of $a$ is $q$, then the frequencies of the three possible genotypes ( $AA,\,Aa,$ and $aa$ ) can be expressed by the Hardy-Weinberg equation :
$p^2 + 2pq + q^2 = 1$
where, $p^2 =$ frequency of $AA$ (homozygous dominant) individuals, $2pq =$ frequency of $Aa$ (heterozygous) individuals and $q^2 =$ frequency of $aa$ (homozygous recessive) individuals. The equation can be used to calculate allele frequencies if the numbers of homozygous recessive individuals in the population is known.
Here, $p = 0.7$ and $q = 0.3$ (given)
$\therefore $ $2pq$ (frequency of heterozygote)
$= 2 \times 0.7 \times 0.3 = 0.42$
