Chapter 25.4, Problem 37AQP

a.

Summary Introduction

To determine:

The frequency of “sickle cell alleles” to be found in next generation.

Explanation of Solution

The allele frequency for sickle cell hemoglobin for current generation $f(s)$ is $q$.

$q=0.028$

The allele frequency for normal cell hemoglobin for current generation $f(s)$ is $p$. The sum of $p$ and $q$ is $1$. Therefore, the value of $p$ is as follows:

$\begin{array}{c}p+q=1\\ p+0.028=1\\ p=1-0.028\\ =0.972\end{array}$

The genotypic frequency for present generation is as follows:

$\begin{array}{c}f(SS)=p^2\\ ={(}^{0.972}\\ =0.945\end{array}$

$\begin{array}{c}f(Ss)=2pq\\ =2(0.028)(0.972)\\ =0.054\end{array}$

$\begin{array}{c}f(ss)=q^2\\ ={(}^{0.028}\\ =0.001\end{array}$

The mean fitness is calculated as follows:

$\overline{w}=p^2{W}_{SS}+2pq{W}_{Ss}+q^2{W}_{ss}$

$\begin{array}{c}\overline{w}={(}^{0.972}(0.83)+2(0.972)(0.028)+{(}^{0.028}(0)\\ =(0.945)(0.83)+0.054+(0.0008)(0)\\ =0.784+0.054+0\\ =0.838\end{array}$

The genotypic frequency for next generation will be:

$\begin{array}{c}f(SS)=\frac{p^2{W}_{SS}}{\overline{w}}\\ =\frac{{(0.972)}^2(0.83)}{0.838}\\ =\frac{(0.945)(0.83)}{0.838}\\ =\frac{0.784}{0.838}\\ =0.936\end{array}$

$\begin{array}{c}f(Ss)=\frac{2pq{W}_{Ss}}{\overline{w}}\\ =\frac{2(0.972)(0.028)(1)}{0.838}\\ =\frac{(0.054)(1)}{0.838}\\ =\frac{0.054}{0.838}\\ =0.064\end{array}$

$\begin{array}{c}f(ss)=\frac{q^2{W}_{ss}}{\overline{w}}\\ =\frac{{(0.028)}^2(0)}{0.838}\\ =\frac{(0.0008)(0)}{0.838}\\ =\frac{0}{0.838}\\ =0\end{array}$

The frequency of new allele is:

$\begin{array}{c}q\text{'}=\frac{0.064}{2}\\ =0.032\end{array}$

Therefore, the frequency of allele for sickle cell is $\boxed{0.032}$.

b.

Summary Introduction

To determine:

The frequency of “sickle cell allele” at equilibrium.

Explanation of Solution

The allele frequency for sickle cell at equilibrium is $\widehat{q}=f(s)$.

$\begin{array}{c}f(s)=\frac{S_{SS}}{S_{SS}+S_{Ss}}\\ =\frac{0.17}{0.17+1.0}\\ =\frac{0.17}{1.17}\\ =0.145\end{array}$

The allele frequency for sickle cell at equilibrium is $\boxed{0.145}$.

Conclusion

Therefore, the frequency of allele for sickle cell is $0.032$ and the allele frequency for sickle cell hemoglobin at hemoglobin is $0.145$.