Genetics: From Genes to Genomes
Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 21, Problem 10P

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

Chapter 21, Problem 10P, Two hypothetical lizard populations found on opposite sides of a mountain in the Arizonan desert

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?
Expert Solution
Check Mark
Summary Introduction

a.

To determine:

The allele frequency of AF 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 AFAF = 38

Genotype frequency of AFAS = 44

Genotype frequency of ASAS = 18

Total genotype frequencies=(Genotype frequency of AFAF+Genotype frequency of AFAS+Genotype frequency of ASAS)=38+44+18=100

Each genotype is composed of two alleles.

Total number of alleles=Total genotype frequencies×2=100×2=200

The formula to be used is as follows:

Allele frequency of genotype=No. of population with a particular genotypeTotal number of alleles

Allele Frequency of AFin population I=(AFAF×2)+12(AFAS×2)Total number of alleles=(38×2)+12(44×2)200=76+12(88)200

Allele frequency of AFin population I=76+44200=120200=0.6

For population II:

Genotype frequency of AFAF = 0

Genotype frequency of AFAS = 80

Genotype frequency of ASAS = 20

Total genotype frequencies=(Genotype frequency of AFAF+Genotype frequency of AFAS+Genotype frequency of ASAS)=0+80+20=100

Each genotype is composed of two alleles.

Total number of alleles=Total genotype frequencies×2=100×2=200

Allele Frequency of AFin population II=(AFAF×2)+12(AFAS×2)Total number of alleles=(0×2)+12(80×2)200=0+12(160)200

Allele frequency of AFin population II=0+80200=80200=0.4

Thus, the allele frequency of AF in population I is 0.6 and allele frequency of AF in population II is 0.4.

Expert Solution
Check Mark
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 AF

q is the allele frequency of AS

For population I:

The allele frequency of AF (p) = 0.6

Allele Frequency of ASin population I=(ASAS×2)+12(AFAS×2)Total number of alleles=(18×2)+12(44×2)200=36+12(88)200

Allele frequency of ASin population I=36+44200=80200=0.4

The allele frequency of AS (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:

p+q=10.6+0.4=1L.H.S=R.H.S

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

For population II:

The allele frequency of AF (p) = 0.4

Allele Frequency of ASin population II=(ASAS×2)+12(AFAS×2)Total number of alleles=(20×2)+12(80×2)200=40+12(160)200

Allele frequency of ASin population II=40+80200=120200=0.6

The allele frequency of AS (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:

p+q=10.4+0.6=1L.H.S=R.H.S

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

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

Expert Solution
Check Mark
Summary Introduction

c.

To determine:

The frequency of genotypes (AF AF, AF AS, and AS AS ) 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.

Total number of alleles in a single new population=Total genotype frequencies in single population×2=200×2=400

The formula to be used is as follows:

Allele frequency of genotype=No. of population with a particular genotypeTotal number of alleles in single new population

Allele Frequency of AFin a single new population =(AFAF×2)+12(AFAS×2)Total number of alleles=(38×2)+12(124×2)400=76+12(248)400

Allele frequency of AFin  a single new population=76+124400=200400=0.5

The allele frequency of AF is represented as “p”.

Allele Frequency of ASin a single new population =(ASAS×2)+12(AFAS×2)Total number of alleles in single new population=(38×2)+12(124×2)400=76+12(248)400

Allele frequency of ASin  a single new population=76+124400=200400=0.5

The allele frequency of AS is represented as “q”.

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

AFAF(p)2=(0.5)2=0.5×0.5=0.25

AFAs(2pq)=2×0.5×0.5=0.5

ASAS(q)2=(0.5)2=0.5×0.5=0.25

Thus, the genotype frequency of AF AF in the next generation is 0.25, AF AS is 0.5, and AS AS is also 0.5.

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Genetics: From Genes to Genomes

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