Chapter 25, Problem 1NST

Summary Introduction

To determine:

The frequency of the AA, Aa, and aa genotypes in the given population.

Introduction:

In 1908, G. H. Hardy and Wilhelm Weinberg evaluated the basic principle of population genetics. At equilibrium, the genetic variation of a population can be calculated by using a formula known as the Hardy-Weinberg equation. For a population to remain at equilibrium five conditions are required which includes no natural selection, no immigration, random mating, the population should be large and no change in allelic frequency.

Explanation of Solution

The population in the given problem is not to be in Hardy Weinberg equilibrium. The frequency of AA, Aa, and aa genotypes can be calculated by the Hardy-Weinberg equation. It is mathematically expressed as $p^2+2pq+q^2=1$.

Where “p” is the frequency of a dominant allele (AA), “q” is the frequency of a recessive allele  (aa) and “2pq” is the frequency of heterozygotes (Aa).

For finding the genotype frequency the frequency of p and the frequency of q need to be calculated. The frequency of p (A) is as follows:

$p=0.2+0.3=0.5$

The frequency of p is 0.5. This is applied in the equation $p+q=1$ to find the frequency of q. The frequency of q (a) is as follows:

$q=1-0.5=0.5$

The frequency of the genotype AA= $p^2$

                                                       $\begin{array}{l}={(0.5)}^2\\ =0.25\end{array}$

The frequency of the genotype Aa =$2pq$

                                                       $\begin{array}{l}=2(0.5)(0.5)\\ =0.50\end{array}$

The frequency of the genotype aa=$q^2$

                                                     $\begin{array}{l}={(0.5)}^2\\ =0.25\end{array}$

The initial population given in the problem is not in equilibrium. However, mating under the Hardy–Weinberg conditions the first generation of the population is in equilibrium and it will be conditioned to the next generation until it fails to mate under the Hardy–Weinberg conditions

Conclusion

Thus the frequency of the genotype AA, Aa and aa is 0.25. 0.50 and 0.25 respectively.