Question

the hardy - weinberg equation
$(p + q)^2 = p^2 + 2pq+q^2 = 1$
$p=$ frequency of
$p^2=$ equilibrium frequency of
$q=$ frequency of
$q^2=$ equilibrium frequency of
$2pq=$ equilibrium frequency of

Explanation:

The Hardy - Weinberg equation \((p + q)^2=p^{2}+2pq + q^{2}=1\) is used in population genetics. In this equation:

• \(p\) represents the frequency of the dominant allele in a population.
• \(q\) represents the frequency of the recessive allele in a population.
• \(p^{2}\) is the equilibrium frequency of the homozygous dominant genotype.
• \(2pq\) is the equilibrium frequency of the heterozygous genotype.
• \(q^{2}\) is the equilibrium frequency of the homozygous recessive genotype.

Answer:

\(p=\) frequency of dominant allele in a population; \(p^{2}=\) equilibrium frequency of homozygous dominant genotype; \(q=\) frequency of recessive allele in a population; \(q^{2}=\) equilibrium frequency of homozygous recessive genotype; \(2pq=\) equilibrium frequency of heterozygous genotype.