Chapter 6, Problem 17RQ

Assume that you are sampling a trait in animal populations; the trait is controlled by a single allelic pair A and a, and you can distinguish all three phenotypes AA, Aa, and aa (intermediate inheritance). Your sample includes:

Calculate the distribution of phenotypes in each population as expected under Hardy-Weinberg equilibrium. Is population I in equilibrium? Is population II in equilibrium?

Summary Introduction

To determine: Whether Population I is in Hardy-Weinberg equilibrium.

Introduction: A population is in Hardy-Weinberg equilibrium where the allele and genotype frequencies remain constant over generations. Hardy-Weinberg equilibrium follows the given equation, p² + 2pq + q² = 1, where, p² represents the dominant homozygous genotype, q² represents the recessive homozygous genotype and 2pq represent the heterozygous genotype.

Explanation of Solution

The animal population has A is a dominant allele and a is a recessive allele.

Genotype	Number of individuals	Frequency
AA	300	300/1000=0.30
Aa	500	500/1000=0.50
aa	200	200/1000=0.20
Total	1000

Frequency of allele A

$\begin{array}{l}f\text{(A)}\text{= Fequency of genotype AA + }\frac{1}{2}\text{ frequency of genotype Aa}\hfill \\ \begin{array}{l}\text{= 0}\text{.30 + 0}\text{.25}\\ \text{= 0}\text{.55}\end{array}\hfill \end{array}$

$\begin{array}{l}f\text{(a)}\text{= Fequency of genotype aa + }\frac{1}{2}\text{ frequency of genotype Aa}\hfill \\ \begin{array}{l}\text{= 0}\text{.20 + 0}\text{.25}\\ \text{= 0}\text{.45}\end{array}\hfill \end{array}$

For a population in Hardy-Weinberg equilibrium $\left(p^2+2pq+q^2=1\right)$

In this case     $A^2+2Aa+a^2=1$

Putting the values in the above equation,

$\begin{array}{l}={\left(0.55\right)}^2+2\left(0.55\right)\left(0.45\right)+{\left(0.45\right)}^2\hfill \\ \begin{array}{l}=0.30+0.50+0.20\\ =1\end{array}\hfill \end{array}$

Distribution of population according to Hardy-Weinberg equilibrium is as follows:

Genotype	The phenotypic ratio at Hardy-Weinberg equilibrium	Number of individuals (ratio x total number of the individual)
AA	0.30	300
Aa	0.50	500
aa	0.20	200
Total	1	1000

Population I is at Hardy-Weinberg equilibrium as expected and the observed phenotype is the same.

Summary Introduction

To determine: Whether Population II is in Hardy-Weinberg equilibrium.

Explanation of Solution

The animal population has A is a dominant allele and a is a recessive allele.

Genotype	Number of individuals	Frequency
AA	400	400/1000=0.40
Aa	400	400/1000=0.40
aa	200	200/1000=0.20
Total	1000

Frequency of allele A

$\begin{array}{l}f\text{(A)}\text{= Fequency of genotype AA + }\frac{1}{2}\text{ frequency of genotype Aa}\hfill \\ \begin{array}{l}\text{= 0}\text{.40 + 0}\text{.20}\\ \text{= 0}\text{.60}\end{array}\hfill \end{array}$

Frequency of allele a

$\begin{array}{l}f\text{(A)}\text{= Fequency of genotype aa + }\frac{1}{2}\text{ frequency of genotype Aa}\hfill \\ \begin{array}{l}\text{= 0}\text{.20 + 0}\text{.20}\\ \text{= 0}\text{.40}\end{array}\hfill \end{array}$

For a population in Hardy-Weinberg equilibrium $\left(p^2+2pq+q^2=1\right)$

In this case $A^2+2Aa+a^2=1$

Putting the values in the above equation,

$\begin{array}{l}={\left(0.60\right)}^2+2\left(0.60\right)\left(0.40\right)+{\left(0.40\right)}^2\hfill \\ \begin{array}{l}=0.36+0.48+0.16\\ =1\end{array}\hfill \end{array}$

Distribution of population according to Hardy-Weinberg equilibrium is as follows:

Genotype	The phenotypic ratio at Hardy-Weinberg equilibrium	Number of individuals (ratio x total number of an individual)
AA	0.36	160
Aa	0.48	480
aa	0.16	160
Total	1	1000

Population II is not at Hardy-Weinberg equilibrium as expected and the observed phenotype is not the same.