Chapter 26, Problem 11EQ

In a donor population, the allele frequencies for the common $\left(H{b}^A\right)$ and sickle cell $\left(H{b}^S\right)$ alleles are 0.9 and 0.1, respectively. A group of 550 individuals from this population migrates to another population containing 10,000 individuals; in the recipient population, the allele frequencies are  $H{b}^A=0.99\text{ and} H{b}^S=0.01$.

A. Calculate the allele frequencies in the conglomerate population.

B. Assuming the donor and recipient populations are each in Hardy-Weinberg equilibrium, calculate the genotype frequencies in the conglomerate population prior to any mating between the donor and recipient populations.

C. What will be the genotype frequencies of the conglomerate population in the next generation, assuming it achieves Hardy-Weinberg equilibrium in one generation?

Summary Introduction

To review:

The following mention frequencies of a population of 10000 migrants with allele frequencies Hb^A is 0.99 and Hb^S is 0.01.

A. Allele frequency in the conglomerate population.

B. Genotype frequency in conglomerate population prior to mating between the recipient and donor populations.

C. The genotype frequency in conglomerate population, when the population achieves Hardy Weinberg equilibrium in a generation.

Introduction:

The variant form of a gene is known as an allele. The allele frequency is a ratio of thenumber of copies in a population of a specific allele, to the total number of alleles in a population. This is expressed in terms of fraction or percentage. The natural populations can get the allele frequency changed by gene flow, mutation, or natural selection. The genetic variation in a population can be calculated by the Hardy-Weinberg equation, which was discovered by Wilhelm Weinberg and G. H. Hardy

Explanation of Solution

The donor population contains allele frequency for the common (Hb^A) and sickle cell (Hb^S) as 0.9 and 0.1 respectively. Almost 550 individuals migrated to another population, which is 10,000 individuals in number. The recipient population, the allele’s frequency is Hb^A is 0.99 and Hb^S is 0.01.

A. The allele frequency in the conglomerate population can be calculated by the following method:

$\Delta p_C=m(p_D–p_R)$

Where $\Delta p_C$ is the total change in the population of the individuals, $p_D$ is the allele frequency of the donor population for the sickle cell, $p_R$ is the allele frequency of the recipient population for the sickle cell.

On substituting the values:

$\begin{array}{l}\Delta p_C=\left(\frac{550}{10,550}\right)(0.1–0.01)\\ \Delta p_C=0.0047\end{array}$

It is known that,

$\begin{array}{l}{p}_C=p_R+\Delta p_C\\ =0.01+\text{ 0}\text{.0047}\\ =0.0147\end{array}$

Hence, allele frequency in the conglomerate population is 0.0147.

B. Genotype frequency in conglomerate population, prior to mating between recipient and donor populations are shown below:

For, 550 individuals who are migrating, the allele frequency is calculated separately as follows:

a.

$\begin{array}{l}H{b}^{A}H{b}^A={(}^{0.9}\\ =0.81\end{array}$

$\begin{array}{c}\text{The expected value is}=\text{(0}\text{.81)(550)}\\ =\text{445}\text{.5}\end{array}$

b.

$\begin{array}{l}H{b}^{A}H{b}^S=2(0.9)(0.1)\\ =0.18\end{array}$

$\begin{array}{c}\text{The expected value is}=(0.18)(550)\\ =99\end{array}$

c.

$\begin{array}{l}H{b}^{S}H{b}^S={(}^{0.1}\\ =0.01\end{array}$

$\begin{array}{c}\text{The expected value is}=(0.01)(550)\\ =5.5\end{array}$

Calculations for, individuals of the original recipient population,

$\begin{array}{c}H{b}^{A}H{b}^A={(}^p\\ ={(}^{0.99}\\ =0.98\end{array}$

Hence, 9,801 individuals are expected to have dominant genotype.

Calculations for, individuals having heterozygote genotype,

$\begin{array}{c}H{b}^{A}H{b}^S=2(p)(q)\\ =2(0.99)(0.01)\\ =0.0198\end{array}$

So, 198 individuals are expected to have heterozygote genotype.

Calculations for, individuals having recessive genotype,

$\begin{array}{c}H{b}^{S}H{b}^S={(}^q\\ ={(}^{0.01}\\ =0.0001\end{array}$

Only 1 individual is expected to have recessive genotype.

The overall population can be calculated by the following method,

$\begin{array}{c}\text{For homozygote dominant}=\frac{\text{(445}\text{.5+9801)}}{\text{10,550}}\\ =\text{0}\text{.971}\text{of}{\text{Hb}}^{\text{A}}{\text{Hb}}^{\text{A}}\text{homozygotes}\end{array}$

$\begin{array}{c}\text{For heterozygote}=\frac{(99+198)}{10,550}\\ =0.028H{b}^{A}H{b}^S\text{ heterozygotes}\end{array}$,

$\begin{array}{c}\text{For homozygote recessive}=\frac{(5.5+1)}{10,550}\\ =0.00062H{b}^{s}H{b}^S\text{homozygotes}\end{array}$

C. According to the Hardy Weinberg principle, the expected genotypic frequency in the conglomerate population is calculated as:

Allele frequency of genotype Hb^A is given as 0.0147 and of genotype Hb^S is 0.985,

$\begin{array}{c}H{b}^{A}H{b}^A={(}^{0.985}\\ =0.97\\ H{b}^{A}H{b}^S=2(0.985)(0.0147)\\ =0.029\end{array}$

$\begin{array}{c}H{b}^{S}H{b}^S={(}^{0.0147}\\ =0.0002\end{array}$

Hence, allele frequency for the dominant is 0.97, for the heterozygote is 0.029 and for the recessive allele is 0.0002.

Conclusion

Therefore, it can be concluded that:

A. the allele frequency of the conglomerate population is $p_C=0.0147$.

B. Genotype frequency of the migrating and the recipient population are 0.971 for homozygotes, 0.028 for heterozygotes and 0.00062 for recessive homozygote.

C. If the population follows the Hardy Weinberg principle, then the allele frequencies for Hb^AHb^A, Hb^AHb^S, and Hb^SHb^S are 0.97, 0.029 and 0.0002 respectively.