Genetics: A Conceptual Approach
Genetics: A Conceptual Approach
6th Edition
ISBN: 9781319127121
Author: Pierce
Publisher: MAC HIGHER
Question
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Chapter 25.2, Problem 23AQP

a.

Summary Introduction

To determine:

The calculation of genotypic and allelic frequency for the population.

a.

Expert Solution
Check Mark

Explanation of Solution

Table 1 Represents the frequency for M locus encodes for enzyme malate dehydrogenase are given:

GenotypeNumber of individuals in population
M1M120
M1M245
M2M242
M1M34
M2M38
M3M36
TOTAL125

Frequencies of different genotypes are as FOLOW:

Genotypic frequency=Number of individual with genotypeTotal number of individualf(M1M1)=20125= 0.16f(M1M2)=45125= 0.36f(M2M2)=42125= 0.34f(M2M3)=8125= 0.064f(M3M3)=6125= 0.48

The genotypic frequency of :

f(M1M1)= 0.16, f(M1M2)=0.36, f(M2M2)= 0.34, f(M2M3)= 0.064, and f(M3M3)= 0.48

In Hardy-Weinberg equilibrium the numbers of expected individuals are:

f(M1)= F(M1M2)+(M1M2)2+(M1M3)2=(0.16)+0.362+0.322= 0.16+0.18+0.016p = 0.356

f(M2)= F(M2M2)+(M1M2)2+(M2M3)2=(0.34)+0.362+0.642= 0.34+0.18+0.032q = 0.552

f(M3)= F(M3M3)+(M1M3)2+(M2M3)2=(0.48)+0.322+0.642= 0.048+0.16+0.032r = 0.096

The frequencies of alleles are M1, M2, andM3 0.356, 0.552, and 0.096.

b.

Summary Introduction

To determine:

The expected number of genotype if the population is in Hardy Weinberg equilibrium.

b.

Expert Solution
Check Mark

Explanation of Solution

The expected frequency for the genotypes in the population in Hardy Weinberg equilibrium:

f(M1M1)p2=(0.356)2= 0.127Expected number of Leopards with genotype (M1M1)= 0.127×125= 16f(M1M2)2pq= 2×(0.356)(0.552)= 0.393Expected number of Leopards with genotype (M1M2)= 0.393×125= 49

f(M2M2)q2=(0.552)2= 0.305Expected number of Leopards with genotype (M2M2)= 0.305×125= 38

f(M1M3)2pr= 2(0.352)(0.096)= 0.68Expected number of Leopards with genotype (M1M3)= 0.168×125= 8.5

f(M2M3)2qr= 2(0.552)(0.096)= 0.106Expected number of Leopards with genotype(M2M3)= 0.106×125= 13

f(M3M3)r2=(0.096)2= 0.009Expected number of Leopards with genotype (M3M3)= 0.009×125= 1

Table 1: Represent the chi-square test:

GenotypeObserved number (O)Expected number (E)(O-E)(O-E)2(O-E)2E
M1M120164161
M1M24549-4160.33
M2M2423841660.42
M1M348-41162
M2M3813-5251.9
M3M36152525

Calculation for chi-square:

Chi-square=(O-E)2E=30.65Degree of freedom= Numberof(genotypes-alleles) =6-3=3

From the above table of chi-square it is clear that value of p is less than 0.05. Thus, the population is in hardy –Weinberg equilibrium but locus is rejected.

Conclusion

Population of bear was found to be at Hardy-Weinberg equilibrium.

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