Q1CC(2)
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?
Step-by-Step Solution
VerifiedThe frequency of AA is 0.3025, Aa is 0.495, and aa is 0.2025.
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.
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.
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 \({p^2} = {\left( {0.55} \right)^2} = 0.3025\)
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 \(2pq = 2\left( {0.55 \times 0.45} \right) = 0.495\)
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 \({q^2} = {\left( {0.45} \right)^2} = 0.2025\)