Evolution
Evolution
4th Edition
ISBN: 9781605356051
Author: Douglas Futuyma, Mark Kirkpatrick
Publisher: SINAUER
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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.

Expert Solution
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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:

A1A1=50A1A2=55A2A2=70

The formula to calculate absolute fitness is:

Absolutefitness=averagefecundity×theprobabilityofsurvivaltoadulthood.

So, the absolute fitness of A1A1= 0.9×50=45The absolute fitness of A1A2= 0.85×55=46.7The absolute fitness of A2A2= 0.75×70=52.5

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.

Expert Solution
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Explanation of Solution

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

The formula to calculate relative fitness is:

Relativefitness=absolutefitness×meanabsolutefitnessinthepopulation.

So, the realtive fitness of A1A1=0.450.525=0.857The relative fitness of A1A2=0.4670.525=0.889The relative fitness of A2A2=0.5250.525=1

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.

Expert Solution
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Explanation of Solution

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

The frequency of the A2allele is given as=p=0.5(Equation 1) According to the Hardy-Weinberg equation, p+q=1(Equation 2) 

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

The frequency of allele A1is calculates as=q=1-0.5=0.5. 

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,

The probability of A1male gamete mating A1female gamete is 0.5×0.5=0.25The probability of A1male gamete mating A2female gamete is 0.5×0.5=0.25

The probability of A2male gamete mating A1female gamete is 0.5×0.5=0.25The probability of A2male gamete mating A2female gamete is 0.5×0.5=0.25

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

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

A2A2+12A1A2=0.25+120.50=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.

Expert Solution
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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

Selection coefficient of A1A1=1-0.857=0.143Selection coefficient of A1A2=1-0.889=0.111Selection coefficient of A2A2=1-1=0

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

p=t(s+t)

Where, t=generation.

p=1(0.143+1)=0.87

Frequency of allele A1=1p=10.87=0.13

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.

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