Chapter 21, Problem 10P

Two hypothetical lizard populations found on opposite sides of a mountain in the Arizonan desert have two alleles (A^F, A^S ) of a single gene A with the following three genotype frequencies:

a.	What is the allele frequency of AF in the two populations?
b.	Do either of the two populations appear to be at Hardy-Weinberg equilibrium?
c.	A huge flood opened a canyon in the mountain range separating populations 1 and 2. They were then able to migrate such that the two populations, which were of equal size, mixed completely and mated at random. What are the frequencies of the three genotypes (AF AF, AF AS, and AS AS ) in the next generation of the single new population of lizards?

Summary Introduction

a.

To determine:

The allele frequency of A^F in the two populations.

Introduction:

The branch of genetics that studies the transmission of genetic material in a population is termed as population genetics. The proportion of gene copies that are of a common allele type in a population is termed as allele frequency. The allele frequency is important for understanding population genetics.

Explanation of Solution

The given information is as follows:

For population I:

Genotype frequency of A^FA^F = 38

Genotype frequency of A^FA^S = 44

Genotype frequency of A^SA^S = 18

$\begin{array}{rll}\text{Total genotype frequencies}& =& \left(\begin{array}{l}\text{Genotype frequency of }{A}^F{A}^F+\text{Genotype frequency of }{A}^F{A}^S+\\ \text{Genotype frequency of }{A}^S{A}^S\end{array}\right)\\ & =& \text{38}+\text{44}+\text{18}\\ & =& 100\end{array}$

Each genotype is composed of two alleles.

$\begin{array}{rll}\text{Total number of alleles}& =& \text{Total genotype frequencies}\times \text{2}\\ & =& \text{100}\times \text{2}\\ & =& \text{200}\end{array}$

The formula to be used is as follows:

$\text{Allele frequency of genotype}=\frac{\text{No}\text{. of population with a particular genotype}}{\text{Total number of alleles}}$

$\begin{array}{rll}\text{Allele Frequency of }{A}^F\text{in population I}& =& \frac{\left(A^F{A}^F\times 2\right)+\frac{1}{2}\left(A^F{A}^S\times 2\right)}{\text{Total number of alleles}}\\ & =& \frac{\left(38\times 2\right)+\frac{1}{2}\left(44\times 2\right)}{200}\\ & =& \frac{76+\frac{1}{2}\left(88\right)}{200}\end{array}$

$\begin{array}{rll}\text{Allele frequency of }{A}^F\text{in population I}& =& \frac{\text{76}+\text{44}}{\text{200}}\\ & =& \frac{\text{120}}{\text{200}}\\ & =& \text{0}\text{.6}\end{array}$

For population II:

Genotype frequency of A^FA^F = 0

Genotype frequency of A^FA^S = 80

Genotype frequency of A^SA^S = 20

$\begin{array}{rll}\text{Total genotype frequencies}& =& \left(\begin{array}{l}\text{Genotype frequency of }{A}^F{A}^F+\text{Genotype frequency of }{A}^F{A}^S+\\ \text{Genotype frequency of }{A}^S{A}^S\end{array}\right)\\ & =& 0+80+20\\ & =& 100\end{array}$

Each genotype is composed of two alleles.

$\begin{array}{rll}\text{Total number of alleles}& =& \text{Total genotype frequencies}\times \text{2}\\ & =& \text{100}\times \text{2}\\ & =& \text{200}\end{array}$

$\begin{array}{rll}\text{Allele Frequency of }{A}^F\text{in population II}& =& \frac{\left(A^F{A}^F\times 2\right)+\frac{1}{2}\left(A^F{A}^S\times 2\right)}{\text{Total number of alleles}}\\ & =& \frac{\left(0\times 2\right)+\frac{1}{2}\left(80\times 2\right)}{200}\\ & =& \frac{0+\frac{1}{2}\left(160\right)}{200}\end{array}$

$\begin{array}{rll}\text{Allele frequency of }{A}^F\text{in population II}& =& \frac{0+80}{\text{200}}\\ & =& \frac{80}{\text{200}}\\ & =& \text{0}\text{.4}\end{array}$

Thus, the allele frequency of A^F in population I is 0.6 and allele frequency of A^F in population II is 0.4.

Summary Introduction

b.

To determine:

Whether both the population appears to be at Hardy-Weinberg equilibrium.

Introduction:

Geoffrey H. Hardy was a scientist who proposed the concept of Hardy-Weinberg equilibrium. This concept is used to associate the allele frequency with the genotype frequency. The populations that have allele frequency and the genotypic frequency at equilibrium follow the concept of Hardy-Weinberg equilibrium.

Explanation of Solution

According to Hardy-Weinberg equilibrium:

$p+q=1$

Where:

p is the allele frequency of A^F

q is the allele frequency of A^S

For population I:

The allele frequency of A^F (p) = 0.6

$\begin{array}{rll}\text{Allele Frequency of }{A}^S\text{in population I}& =& \frac{\left(A^S{A}^S\times 2\right)+\frac{1}{2}\left(A^F{A}^S\times 2\right)}{\text{Total number of alleles}}\\ & =& \frac{\left(18\times 2\right)+\frac{1}{2}\left(44\times 2\right)}{200}\\ & =& \frac{36+\frac{1}{2}\left(88\right)}{200}\end{array}$

$\begin{array}{rll}\text{Allele frequency of }{A}^S\text{in population I}& =& \frac{36+44}{\text{200}}\\ & =& \frac{80}{\text{200}}\\ & =& \text{0}\text{.4}\end{array}$

The allele frequency of A^S (q) = 0.4

The formula to be used is as follows:

$p+q=1$

Substituting the value of p = 0.6 and q = 0.4 in the above formula gives the following result:

$\begin{array}{rll}p+q& =& 1\\ 0.6+0.4& =& 1\\ \text{L}\text{.H}\text{.S}& =& \text{R}\text{.H}\text{.S}\end{array}$

This indicates that the population I appear to be at Hardy-Weinberg equilibrium.

For population II:

The allele frequency of A^F (p) = 0.4

$\begin{array}{rll}\text{Allele Frequency of }{A}^S\text{in population II}& =& \frac{\left(A^S{A}^S\times 2\right)+\frac{1}{2}\left(A^F{A}^S\times 2\right)}{\text{Total number of alleles}}\\ & =& \frac{\left(20\times 2\right)+\frac{1}{2}\left(80\times 2\right)}{200}\\ & =& \frac{40+\frac{1}{2}\left(160\right)}{200}\end{array}$

$\begin{array}{rll}\text{Allele frequency of }{A}^S\text{in population II}& =& \frac{40+80}{\text{200}}\\ & =& \frac{120}{\text{200}}\\ & =& \text{0}\text{.6}\end{array}$

The allele frequency of A^S (q) = 0.6

The formula to be used is as follows:

$p+q=1$

Substituting the value of p = 0.4 and q = 0.6 in the above formula gives the following result:

$\begin{array}{c}p+q=1\\ 0.4+0.6=1\\ \text{L}\text{.H}\text{.S}=\text{R}\text{.H}\text{.S}\end{array}$

This reflects that population II appears to be at Hardy-Weinberg equilibrium.

Thus, both population I and population II are at Hardy-Weinberg equilibrium.

Summary Introduction

c.

To determine:

The frequency of genotypes (A^F A^F, A^F A^S, and A^S A^S ) in the next generation.

Introduction:

The set of the alleles in DNA that carries the information for the expression of a trait in an individual is known as its genotype. For example, genotype ‘TT’ expresses the tallness in plants. The genotypes are responsible for controlling the expression of traits.

Explanation of Solution

The following table represents the population number of a single population after a natural calamity:

Population	AF AF	AF AS	AS AS	Total
Population I	38	44	18	$38+44+18=100$
Population II	0	80	20	$0+80+20=100$
Single population	$38+0=38$	$44+80=124$	$18+20=38$	$38+124+38=200$

Each genotype is composed of two alleles.

$\begin{array}{rll}\text{Total number of alleles in a single new population}& =& \text{Total genotype frequencies in single population}\times \text{2}\\ & =& 2\text{00}\times \text{2}\\ & =& 4\text{00}\end{array}$

The formula to be used is as follows:

$\text{Allele frequency of genotype}=\frac{\text{No}\text{. of population with a particular genotype}}{\text{Total number of alleles in single new population}}$

$\begin{array}{rll}\text{Allele Frequency of }{A}^F\text{in a single new population }& =& \frac{\left(A^F{A}^F\times 2\right)+\frac{1}{2}\left(A^F{A}^S\times 2\right)}{\text{Total number of alleles}}\\ & =& \frac{\left(38\times 2\right)+\frac{1}{2}\left(124\times 2\right)}{400}\\ & =& \frac{76+\frac{1}{2}\left(248\right)}{400}\end{array}$

$\begin{array}{rll}\text{Allele frequency of }{A}^F\text{in a single new population}& =& \frac{\text{76}+12\text{4}}{\text{400}}\\ & =& \frac{200}{\text{400}}\\ & =& 0.5\end{array}$

The allele frequency of A^F is represented as “p”.

$\begin{array}{rll}\text{Allele Frequency of }{A}^S\text{in a single new population }& =& \frac{\left(A^S{A}^S\times 2\right)+\frac{1}{2}\left(A^F{A}^S\times 2\right)}{\text{Total number of alleles in single new population}}\\ & =& \frac{\left(38\times 2\right)+\frac{1}{2}\left(124\times 2\right)}{400}\\ & =& \frac{76+\frac{1}{2}\left(248\right)}{400}\end{array}$

$\begin{array}{c}\text{Allele frequency of }{A}^S\text{in a single new population}=\frac{\text{76}+12\text{4}}{\text{400}}\\ =\frac{200}{\text{400}}\\ =0.5\end{array}$

The allele frequency of A^S is represented as “q”.

The frequencies of three genotypes among zygotes due to random mating are as follows:

$\begin{array}{rll}{A}^F{A}^F{\left(p\right)}^2& =& {\left(0.5\right)}^2\\ & =& 0.5\times 0.5\\ & =& 0.25\end{array}$

$\begin{array}{rll}{A}^F{A}^s\left(2pq\right)& =& 2\times 0.5\times 0.5\\ & =& 0.5\end{array}$

$\begin{array}{rll}{A}^S{A}^S{\left(q\right)}^2& =& {\left(0.5\right)}^2\\ & =& 0.5\times 0.5\\ & =& 0.25\end{array}$

Thus, the genotype frequency of A^F A^F in the next generation is 0.25, A^F A^S is 0.5, and A^S A^S is also 0.5.