Chapter 5, Problem 1PDT

Summary Introduction

To calculate: The absolute fitness (W) of the genotypes A₁A₁, A₁A₂, and A₂A₂ .if their fecundities values are given as 50, 55, and 70 eggs.

Introduction: Fitness refers to the number of offspring that are produced by an individual to the upcoming generations. There are two measures of biological fitness one is absolute fitness and second is relative fitness.

Explanation of Solution

The absolute fitness determines the change in abundance of any genotype.

If the egg-to-adult survival rates of genotypes, A₁A₁, A₁A₂, and A₂A₂ are 90 percent, 85 percent, and 75 percent respectively.

The fecundities are given as:

$\begin{array}{l}\begin{array}{l}{\text{A}}_{\text{1}}{\text{A}}_{\text{1}}=\text{50}\hfill \\ {\text{A}}_{\text{1}}{\text{A}}_{\text{2}}=\text{55}\hfill \end{array}\\ {\text{A}}_{\text{2}}{\text{A}}_{\text{2}}=\text{70}\end{array}$

The formula to calculate absolute fitness is:

$\text{Absolute}\text{}\text{}\text{}\text{}\text{}\text{fitness}=\text{average}\text{}\text{}\text{}\text{}\text{}\text{}\text{fecundity}\times \text{the}\text{}\text{}\text{}\text{}\text{}\text{probability}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{of}\text{}\text{survival}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{to}\text{adulthood}\text{.}$

$\begin{array}{c}{\text{So, the absolute fitness of A}}_{\text{1}}{\text{A}}_{\text{1}}\text{= 0}\text{.9}\times \text{50}=\text{45}\\ {\text{The absolute fitness of A}}_{\text{1}}{\text{A}}_{\text{2}}=\text{ 0}\text{.85}\times \text{55}=\text{46}\text{.7}\\ {\text{The absolute fitness of A}}_{\text{2}}{\text{A}}_{\text{2}}\text{= 0}\text{.75}\times \text{70}=\text{52}\text{.5}\end{array}$

Hence, absolute fitness of these genotypes is 45, 46.7, and 52.5 respectively.

Summary Introduction

To calculate: The relative fitness using A₁A₁ as the fitness reference.

Introduction: Fitness refers to the number of offspring that are produced by an individual to the upcoming generations. There are two measures of biological fitness one is absolute fitness and second is relative fitness.

Explanation of Solution

The relative fitness determines the change in frequency of any genotype.

The formula to calculate relative fitness is:

$\text{Relative}\text{}\text{}\text{}\text{}\text{}\text{}\text{fitness}=\text{absolute}\text{}\text{}\text{}\text{}\text{fitness}\times \text{mean}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{absolute}\text{}\text{fitness}\text{}\text{}\text{}\text{in}\text{}\text{the}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{population}\text{.}$

$\begin{array}{c}{\text{So, the realtive fitness of A}}_{\text{1}}{\text{A}}_{\text{1}}=\frac{\text{0}\text{.45}}{\text{0}\text{.525}}=\text{0}\text{.857}\\ {\text{The relative fitness of A}}_{\text{1}}{\text{A}}_{\text{2}}=\frac{\text{0}\text{.467}}{\text{0}\text{.525}}=\text{0}\text{.889}\\ {\text{The relative fitness of A}}_{\text{2}}{\text{A}}_{\text{2}}=\frac{\text{0}\text{.525}}{\text{0}\text{.525}}=\text{1}\end{array}$

So, the relative fitness of these genotypes is 0.857, 0.889, and 1.

Summary Introduction

To calculate: The frequency one generation later, if the frequency of the A₂ allele is 0.5 (p = 0.5).

Introduction: The allele that is located at a particular locus in a population, its relative frequency value is calculated as a fraction; this frequency is known as allele frequency.

Explanation of Solution

In a group of the population, there are different values of allele frequencies.

$\begin{array}{l}{\text{The frequency of the A}}_{\text{2}}\text{allele is given as}=\text{p}=\text{0}\text{.5}\text{. }\left(\text{Equation 1}\right)\hfill \\ \hfill \\ \text{According to the Hardy-Weinberg equation, p}+\text{q}=\text{1}\text{. }\left(\text{Equation 2}\right)\hfill \\ \hfill \end{array}$

Therefore, by putting the value of equation 1 into equation 2:

$\begin{array}{l}{\text{The frequency of allele A}}_{\text{1}}\text{is calculates as}=\text{q}=\text{1}-\text{0}\text{.5}=\text{0}\text{.5}\text{.}\hfill \\ \hfill \end{array}$

Allelic frequency of both the allele is the same, so the population has half of the gametes with allele A₁, and half of the gametes with A₂ allele.

Therefore,

$\begin{array}{l}{\text{The probability of A}}_{\text{1}}{\text{male gamete mating A}}_{\text{1}}\text{female gamete is 0}\text{.5}\times \text{0}\text{.5}=\text{0}\text{.25}\hfill \\ {\text{The probability of A}}_{\text{1}}{\text{male gamete mating A}}_{\text{2}}\text{female gamete is 0}\text{.5}\times \text{0}\text{.5}=\text{0}\text{.25}\hfill \end{array}$

$\begin{array}{l}{\text{The probability of A}}_{\text{2}}{\text{male gamete mating A}}_{\text{1}}\text{female gamete is 0}\text{.5×0}\text{.5}=\text{0}\text{.25}\hfill \\ {\text{The probability of A}}_{\text{2}}{\text{male gamete mating A}}_{\text{2}}\text{female gamete is 0}\text{.5×0}\text{.5}=\text{0}\text{.25}\hfill \end{array}$

From the above equations, allele A₂ is involving in three probabilities.

So, the frequency of A₂ allele, one generation later is:

${\text{A}}_{\text{2}}{\text{A}}_{\text{2}}+\frac{1}{2}{\text{A}}_{\text{1}}{\text{A}}_{\text{2}}=0.25+\frac{1}{2}0.50=0.25+0.25=0.5$

Hence, one generation later the frequency of A₂ allele is 0.5.

Summary Introduction

To calculate: The allele frequency when the population reaches equilibrium.

Introduction: The allele that is located at a particular locus in a population, its relative frequency value is calculated as a fraction, this frequency is known as allele frequency.

Explanation of Solution

To calculate the allelic frequency when the population reaches equilibrium, the selective coefficient is used.

Selection coefficient is represented by the alphabet‘s’ and it is the measure of differences in relative fitness acting against a genotype.

Selection coefficient (s) = 1-relative fitness

$\begin{array}{l}{\text{Selection coefficient of A}}_{\text{1}}{\text{A}}_{\text{1}}\text{=1-0}\text{.857=0}\text{.143}\hfill \\ {\text{Selection coefficient of A}}_{\text{1}}{\text{A}}_{\text{2}}\text{=1-0}\text{.889=0}\text{.111}\hfill \\ {\text{Selection coefficient of A}}_{\text{2}}{\text{A}}_{\text{2}}\text{=1-1=0}\hfill \end{array}$

If the population reaches equilibrium, then the frequency (p) of A₂ allele will be:

$p=\frac{t}{\left(s+t\right)}$

Where, t=generation.

$p=\frac{1}{\left(0.143+1\right)}=0.87$

${\text{Frequency of allele A}}_{\text{1}}\text{=1}-\text{p}=\text{1}-\text{0}\text{.87}=\text{0}\text{.13}$

So, when the population reaches equilibrium the frequency values will be:

The frequency of A₂ allele is 0.87.

The frequency of A₁ allele is o.13.