Chapter 19, Problem 8QP

How Can We Measure Allele Frequencies in Populations?

In a population where the females have the allelic frequencies A = 0.35 and a = 0.65 and the frequencies for males are A = 0.1 and a = 0.9, how many generations will it take to reach Hardy–Weinberg equilibrium for both the allelic and the genotypic frequencies? Assume random mating and show the allelic and genotypic frequencies for each generation.

Summary Introduction

To determine: The number of generations taken to reach Hardy-Weinberg equilibrium law in the given case.

Introduction: The Hardy-Weinberg law states that in the absence of other evolutionary influences, the frequency of allele and genotypes or genetic variation within a population will remain constant across generations.

Explanation of Solution

In order to find the number of generations required to attain equilibrium, the allelic and genotypic frequencies need to be calculated. The number of generations that posses the same frequency of allele and genotypes are said be in equilibrium with one another.

The allele frequencies on a given population are as follows:

Females, f(A)= 0.35 and f(a)= 0.65

Males, f(A)= 0.1 and f(a)= 0.9

The probabilities of mating of an egg with the sperm are:

A egg with A sperm= (0.35)(0.1)= 0.035

A egg with a sperm= (0.35)(0.9)= 0.315

a egg with A sperm= (0.65)(0.1)= 0.065

a egg with a sperm= (0.65)(0.9)= 0.585

So, in the second generation, the expected genotypes are 0.035, 0.315, 0.065, and 0.065

The allele frequency can be calculated as:

The frequency of a dominant allele (A),

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.035)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.380}\right)\\ \text{=}\text{(0}\text{.035)+}\text{(0}\text{.190)}\\ \text{=}\text{0}\text{.225}\end{array}$

The frequency of a dominant allele (A) is 0.225.

The frequency of a recessive allele (a)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.585)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.380}\right)\\ \text{=}\text{(0}\text{.585)+}\text{(0}\text{.190)}\\ \text{=}\text{0}\text{.775}\end{array}$

The frequency of a recessive allele (a) is 0.775.

In the third generation, the expected frequencies for a recessive allele (a) is 0.775 and dominant allele (A) is 0.225.

The probabilities of mating of an egg with the sperm are:

A egg with A sperm= (0.225)(0.225)= 0.05

A egg with a sperm=  (0.225)(0.775)= 0.174

a egg with A sperm= (0.775)(0.225)= 0.174

a egg with a sperm= (0.775)(0.775)= 0.6

So, in the third generation, the expected genotypes are 0.05, 0.35, and 0.6.

The allele frequency can be calculated as:

The frequency of dominant allele (A)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.05)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.35}\right)\\ \text{=}\text{(0}\text{.05)+}\text{(0}\text{.175)}\\ \text{=}\text{0}\text{.225}\end{array}$

The frequency of a dominant allele (A) is 0.225.

The frequency of recessive allele (a)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.6)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.35}\right)\\ \text{=}\text{(0}\text{.6)+}\text{(0}\text{.175)}\\ \text{=}\text{0}\text{.775}\end{array}$

The frequency of a recessive allele (a) is 0.775.

Hence, two generations are required for the populations to reach Hardy-Weinberg law.

Summary Introduction

To determine: The allelic and genotypic frequencies for each generation.

Introduction: The allele frequency refers to the number of times an allele occurs at a specific locus within a population. It can be represented as a percentage or in fraction. Genotype frequency refers to the number of times a particular genotype occurs within a given population.

Explanation of Solution

The genotypic and allelic frequencies are calculated as follows:

In the second generation,

The allele frequencies for a given population are:

Females, f(A)= 0.35 and f(a)= 0.65

Males, f(A)= 0.1 and f(a)= 0.9

The probabilities of mating of an egg with the sperm are:

A egg meeting with A sperm= (0.35)(0.1)= 0.035

A egg meeting with a sperm=  (0.35)(0.9)= 0.315

a egg meeting with A sperm= (0.65)(0.1)= 0.065

a egg meeting with a sperm= (0.65)(0.9)= 0.585

So, in the second generation, the expected genotypes are 0.035, 0.315, 0.065 and 0.065

The frequency of allele can be calculated as:

The frequency of dominant allele (A)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.035)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.380}\right)\\ \text{=}\text{(0}\text{.035)+}\text{(0}\text{.190)}\\ \text{=}\text{0}\text{.225}\end{array}$

The frequency of a dominant allele (A) is 0.225.

The frequency of recessive allele (a)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.585)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.380}\right)\\ \text{=}\text{(0}\text{.585)+}\text{(0}\text{.190)}\\ \text{=}\text{0}\text{.775}\end{array}$

The frequency of a recessive allele (a) is 0.775.

In the third generation, the expected frequencies for recessive allele (a) are 0.775 and dominant allele (A) is 0.225.

The probabilities of mating of an egg with the sperm are:

A egg with A sperm= (0.225)(0.225)= 0.05

A egg with a sperm=  (0.225)(0.775)= 0.174

a egg with A sperm= (0.775)(0.225)= 0.174

a egg with a sperm= (0.775)(0.775)= 0.6

So, in the third generation, the expected genotypes are 0.05, 0.35, and 0.6.

The allele frequency can be calculated as:

The frequency of dominant allele (A)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.05)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.35}\right)\\ \text{=}\text{(0}\text{.05)+}\text{(0}\text{.175)}\\ \text{=}\text{0}\text{.225}\end{array}$

The frequency of a dominant allele (A) is 0.225.

The frequency of recessive allele (a)

$\begin{array}{c}\text{(frequency}\text{of}\text{Homozygous}\text{A)}\text{+}\frac{\text{1}}{\text{2}}\text{frequency}\text{of}\text{heterozygous}\text{alleles}\\ \text{=}\text{(0}\text{.6)}\text{+}\left(\frac{\text{1}}{\text{2}}\text{0}\text{.35}\right)\\ \text{=}\text{(0}\text{.6)+}\text{(0}\text{.175)}\\ \text{=}\text{0}\text{.775}\end{array}$

The frequency of a recessive allele (a) is 0.775.