The absolute fitness (W) of the genotypes A 1 A 1 , A 1 A 2 , and A 2 A 2 .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.

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Answer 1

Question

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

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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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Answer 2

Question

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

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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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Answer 3

Question

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

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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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Answer 4

Question

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

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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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Answer 5

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

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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.

Answer 6

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

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Answer 7

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.

Answer 8

Expert Solution

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:

$Absolute fitness=average fecundity×the probability of survival to adulthood.$

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

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

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

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

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

Answer 13

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.

Answer 14

Expert Solution

Explanation of Solution

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

The formula to calculate relative fitness is:

$Relative fitness=absolute fitness×mean absolute fitness in the population.$

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

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

Answer 15

Expert Solution

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

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

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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.

Answer 19

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.

Answer 20

Expert Solution

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:

$The frequency of allele A1 is calculates as=q=1-0.5=0.5.$

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.

Answer 21

Expert Solution

Answer 22

Expert Solution

Answer 23

Expert Solution

Answer 24

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

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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.

Answer 25

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.

Answer 26

Expert Solution

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.

$p=1(0.143+1)=0.87$

$Frequency of allele A1=1−p=1−0.87=0.13$

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.

Answer 27

Expert Solution

Answer 28

Expert Solution

Answer 29

Expert Solution

Answer 30

Ch. 5 - Prob. 1PDT Ch. 5 - Prob. 2PDT Ch. 5 - Prob. 3PDT Ch. 5 - Prob. 4PDT Ch. 5 - Prob. 5PDT Ch. 5 - Prob. 6PDT Ch. 5 - Prob. 7PDT

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Explanation of Solution

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

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

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

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Chapter 5 Solutions

Concept explainers

Videos

Explanation of Solution

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

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

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

Want to see more full solutions like this?

Chapter 5 Solutions

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.

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.

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.