Chapter 12, Problem 8WIO

Summary Introduction

(a)

To determine:

The frequency of the gray allele $\text{f}$.

Introduction:

Evolution occurs due to change in allele frequencies. The frequency of an allele is calculated as: $\frac{\text{the number of copies of an allele}}{\text{total}\text{number}\text{of}\text{alleles}\text{in}\text{the}\text{population}}$

Explanation of Solution

It has been mentioned that there are $100$ Fraggles in the population out of which $84$ have green fur and $16$ have gray fur. A dominant allele $\text{F}$ confers green fur, and a recessive allele $\text{f}$ confers gray fur. The allelic frequency of the gray allele $\text{f}$ can be calculated as:

$\text{Allele}\text{frequencies:p+q}=\text{1}$,

where, $\text{p}=\text{frequency}\text{of}\text{dominant}\text{allele}$, $\text{q}=\text{frequency}\text{of}\text{recessive}\text{allele}$

$\text{Genotype}{\text{frequency}=\text{p}}^{\text{2}}{\text{+2pq+q}}^{\text{2}}=\text{1}$,

where, ${\text{p}}^{\text{2}}=\text{frequency}\text{of}\text{dominant}\text{genotype}\text{(FF)}$

$\text{2pq}=\text{frequency}\text{of}\text{heterozygous}\text{genotype}\text{(Ff)}$

${\text{q}}^{\text{2}}=\text{frequency}\text{of}\text{recessive}\text{genotype}\text{(ff)}$

$\begin{array}{l}\text{Genotype}\text{frequency=}\frac{\text{the number of individuals}\text{with}\text{the}\text{dominant/recessive}\text{genotype}}{\text{total}\text{size}\text{of}\text{the}\text{population}}\\ \text{Green}\text{fur}=\frac{84}{100}=0.84\\ \text{Gray}\text{fur}=\frac{16}{100}=0.16\\ \text{As}\text{Genotype}{\text{frequency}}^{=}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}\text{+2pq}=\text{0}\text{.84,}{\text{q}}^{\text{2}}=\text{0}\text{.16}\\ \text{therefore,}\text{q}=\text{0}\text{.4}\\ \text{Grey}\text{allele}\text{(f)=}0\text{.4}\end{array}$

Thus, frequency of the gray allele $\text{f}$ is $\text{0}\text{.4}$.

Summary Introduction

(b)

To determine:

The frequency of the green allele $\text{F}$.

Introduction:

According to Hardy-Weinberg equilibrium, the allele frequencies and genotype frequencies do not change from one generation to the next.

Explanation of Solution

$\text{Genotype}\text{frequency=}\frac{\text{the number of individuals}\text{with}\text{the}\text{dominant/recessive}\text{genotype}}{\text{total}\text{size}\text{of}\text{the}\text{population}}$

$\begin{array}{l}\text{Fraggles}\text{with}\text{Green}\text{fur}=\frac{84}{100}=0.84\\ \text{Fraggles}\text{with}\text{Gray}\text{fur}=\frac{16}{100}=0.16\\ \text{As}\text{Genotype}{\text{frequency}}^{=}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}=\text{frequency}\text{of}\text{dominant}\text{genotype}\text{(FF)}\\ \text{2pq}=\text{frequency}\text{of}\text{heterozygous}\text{genotype}\text{(Ff)}\\ {\text{q}}^{\text{2}}=\text{frequency}\text{of}\text{recessive}\text{genotype}\text{(ff)}\\ {\text{p}}^{\text{2}}\text{+2pq}=\text{0}\text{.84,}{\text{q}}^{\text{2}}=\text{0}\text{.16}\\ \text{therefore,}\text{q}=\text{0}\text{.4}\\ \text{Grey}\text{allele}\text{(f)=}0\text{.4}\\ \text{Allele}\text{frequencies:p+q}=\text{1}\\ \text{p}=\text{frequency}\text{of}\text{dominant}\text{allele}\\ \text{q}=\text{frequency}\text{of}\text{recessive}\text{allele}\\ \text{p+q}=\text{1}\\ \text{p}=\text{1-0}\text{.4}=\text{0}\text{.6}\\ \text{Thus,}\text{frequency}\text{of}\text{the}\text{green}\text{allele}\text{(F)=}0\text{.6}\end{array}$

Summary Introduction

(c)

To determine:

The number of Fraggles that are heterozygotes $\left(\text{Ff}\right)$.

Introduction:

Individuals that contain different alleles of a gene are called heterozygous. Individuals that carry two identical alleles of a gene are called homozygous. In case of homozygous individual, the identical alleles may be either dominant or recessive.

Explanation of Solution

There are $100$ Fraggles in a population out of which $84$ have green fur and $16$ have gray fur.

$\begin{array}{l}\text{Fraggles}\text{with}\text{Green}\text{fur}=\frac{84}{100}=0.84\\ \text{Fraggles}\text{with}\text{Gray}\text{fur}=\frac{16}{100}=0.16\\ \end{array}$

$\begin{array}{l}\text{Genotype}{\text{frequency}}^{=}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}=\text{frequency}\text{of}\text{dominant}\text{genotype}\text{(FF)}\\ \text{2pq}=\text{frequency}\text{of}\text{heterozygous}\text{genotype}\text{(Ff)}\\ {\text{q}}^{\text{2}}=\text{frequency}\text{of}\text{recessive}\text{genotype}\text{(ff)}\\ {\text{p}}^{\text{2}}\text{+2pq}=\text{0}\text{.84,}{\text{q}}^{\text{2}}=\text{0}\text{.16}\end{array}$

$\begin{array}{l}\text{therefore,}\text{q}=\text{0}\text{.4}\\ \text{Grey}\text{allele}\text{(f)=}0\text{.4}\end{array}$

$\begin{array}{l}\text{Allele}\text{frequencies:p+q}=\text{1}\\ \text{p}=\text{frequency}\text{of}\text{dominant}\text{allele}\\ \text{q}=\text{frequency}\text{of}\text{recessive}\text{allele}\\ \text{p+q}=\text{1}\\ \text{p}=\text{1-0}\text{.4}=\text{0}\text{.6}\\ \text{Thus,}\text{frequency}\text{of}\text{the}\text{green}\text{allele}\text{(F)=}0\text{.6}\\ \text{As,}{\text{p}}^{\text{2}}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}=0.36\\ {\text{q}}^{\text{2}}=0.16\\ \text{2pq}=2(0.6)(0.4)=0.48\\ \text{Thus,}\text{the}\text{number}\text{of}\text{heterozygotes}\text{(Ff)}=0.48\\ \end{array}$

Summary Introduction

(d)

To determine:

The number of Fraggles that are homozygous recessive ($\text{ff}$).

Introduction:

The study of population genetics is based on the relationship between allele frequencies and genotype frequencies. Some or all the assumptions of Hardy-Weinberg equilibrium are violated in a real population.

Explanation of Solution

$\begin{array}{l}\text{Genotype}{\text{frequency}}^{=}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}=\text{frequency}\text{of}\text{dominant}\text{genotype}\text{(FF)}\\ \text{2pq}=\text{frequency}\text{of}\text{heterozygous}\text{genotype}\text{(Ff)}\\ {\text{q}}^{\text{2}}=\text{frequency}\text{of}\text{recessive}\text{genotype}\text{(ff)}\end{array}$

$\begin{array}{l}\text{As}\text{it}\text{was}\text{calculated}\text{that}\text{p=}0\text{.6, q=}0\text{.4,}\text{2pq=}0.48\\ {\text{p}}^{\text{2}}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{q}}^{\text{2}}=0.16\\ \text{Thus,}\text{the}\text{number}\text{of}\text{homozygous recessive}\text{fraggles}\text{(ff)=}0\text{.16}\end{array}$

Summary Introduction

(e)

To determine:

The number of Fraggles that are homozygous dominant ($\text{FF}$).

Introduction:

Hardy-Weinberg equilibrium allows inferring characteristics of a population based on limited information. The genotype frequencies in a population can be estimated using allele frequencies.

Explanation of Solution

$\begin{array}{l}\text{Genotype}{\text{frequency}}^{=}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}=\text{frequency}\text{of}\text{dominant}\text{genotype}\text{(FF)}\\ \text{2pq}=\text{frequency}\text{of}\text{heterozygous}\text{genotype}\text{(Ff)}\\ {\text{q}}^{\text{2}}=\text{frequency}\text{of}\text{recessive}\text{genotype}\text{(ff)}\end{array}$

$\begin{array}{l}\text{As}\text{it}\text{was}\text{calculated}\text{that}\text{p=}0\text{.6, q=}0\text{.4,}\text{2pq=}0.48\\ {\text{p}}^{\text{2}}{\text{+2pq+q}}^{\text{2}}=\text{1}\\ {\text{p}}^{\text{2}}=0.36\\ \text{Thus,}\text{the}\text{number}\text{of}\text{homozygous dominant}\text{fraggles}\text{(FF)=}0\text{.36}\end{array}$