Question

The frequency of allele a is 0.45 for a population in Hardy-Weinberg equilibrium. What are the expected frequencies of genotypes AA, Aa, and aa?

Short answer

The frequency of AA is 0.3025, Aa is 0.495, and aa is 0.2025.

Step-by-step solution

Hardy-Weinberg equilibrium

The equation which is used to calculate the allele frequencies of the non-evolving population is called the Hardy-Weinberg equation. The propositions of the model are as follows:

• When the population is large, the transfer of genes or gene flow does not take place.
• The mutation that takes place in a population is negligible in a large population.
• Random mating occurs between individuals in a large population.
• The force of natural selection does not act on a large population.

The equation which is used to find out the allele frequency is\(p + q = 1\) and \({p^2} + 2pq + {q^2}\). The p and q represent the frequency of ‘A’ and ‘a’ allele, respectively.

Frequency of allele ‘A’

The frequency of allele ‘a’ is 0.45.

\(\begin{aligned}{l}p + q &= 1\\p + 0.45 &= 1\\p &= 1 - 0.45\\p &= 0.55\end{aligned}\)

Hence, the frequency of allele ‘A’ is 0.55.

Frequency of genotype ‘AA’

According to Hardy-Weinberg, the law \({p^2}\) is equal to AA.

The frequency of allele ‘A’ is 0.55

Therefore, the frequency of genotype ‘AA’ is \({p^2} = {\left( {0.55} \right)^2} = 0.3025\)

Frequency of genotype ‘Aa’

The frequency of allele ‘A’ is 0.55, and ‘a’ is 0.45.

According to the Hardy- Weinberg equation, 2pq is equal to Aa.

Therefore, the frequency of genotype ‘Aa’ is \(2pq = 2\left( {0.55 \times 0.45} \right) = 0.495\)

Frequency of genotype ‘aa’

The frequency of allele ‘a’ is 0.45.

According to Hardy-Weinberg, the equation \({q^2}\) is equal to aa

Therefore, the frequency of genotype ‘aa’ is \({q^2} = {\left( {0.45} \right)^2} = 0.2025\)