Chapter 3.3, Problem 34AQP

a.

Summary Introduction

To determine:

The proportion of the genotype $\text{Aa Bb Cc Dd Ee}$ among the progeny of the cross $\text{AaBbCcddEe}\times \text{AabbCcDdEe}$.

Introduction:

The cross between $\text{AaBbCcddEe}$ and $\text{AabbCcDdEe}$ will give rise to many combinations and the probability of specific genotype in progenies can be determined. Some simpler crosses can also help in determining the proportion of genotype in progeny from many combinations obtained from the cross. The cross occurs only between the genes found on the same locus of each parent’s chromosome.

Explanation of Solution

Tabular representation: Table.1: The proportion of possible genotypes formed in offsprings obtained by the cross $\text{AaBbCcddEe}\times \text{AabbCcDdEe}$.

	Genetic cross	Probability of possible genotype of offspring
Cross between genes present on locus 1 (A and a):	$\text{Aa}\times \text{Aa}$	$\frac{\text{1}}{\text{4}}\text{AA,}\frac{\text{1}}{\text{2}}\text{Aa,}\frac{\text{1}}{\text{4}}\text{aa}$
Cross between genes present on locus 2 (B and b):	$\text{Bb}\times \text{bb}$	$\frac{\text{1}}{\text{2}}\text{Bb}$, $\frac{\text{1}}{\text{2}}\text{bb}$
Cross between genes present on locus 3 (C and c):	$\text{Cc}\times \text{Cc}$	$\frac{\text{1}}{\text{4}}\text{CC,}\frac{\text{1}}{\text{2}}\text{Cc,}\frac{\text{1}}{\text{4}}\text{cc}$
Cross between genes present on locus 4 (D and d):	$\text{dd}\times \text{Dd}$	$\frac{\text{1}}{\text{2}}\text{Dd,}\frac{\text{1}}{2}\text{dd}$
Cross between genes present on locus 5 (E and e):	$\text{Ee}\times \text{Ee}$	$\frac{\text{1}}{\text{4}}\text{EE,}\frac{\text{1}}{\text{2}}\text{Ee,}\frac{\text{1}}{\text{4}}\text{ee}$

Table.1: The proportion of possible genotypes formed in offsprings obtained by the cross $\text{AaBbCcddEe}\times \text{AabbCcDdEe}$.

The probability of obtaining $\text{Aa Bb Cc Dd Ee}$ genotype can be calculated by multiplication rule in which the probability of each allele is multiplied. Probability of $\text{Aa}$ is $\frac{\text{1}}{\text{2}}$, probability of $\text{Bb}$ is $\frac{\text{1}}{\text{2}}$ , probability of $\text{Cc}$ is $\frac{\text{1}}{\text{2}}$ , probability of $\text{Dd}$ is $\frac{\text{1}}{\text{2}}$ and probability of $\text{Ee}$ is $\frac{\text{1}}{\text{2}}$. Therefore, the probability of obtaining $\text{Aa Bb Cc Dd Ee}$ genotype can be calculated as:

$\begin{array}{c}\text{Probability of Aa Bb Cc Dd Ee}=\frac{\text{1}}{\text{2}}\text{Aa}\times \frac{\text{1}}{\text{2}}\text{Bb}\times \frac{\text{1}}{\text{2}}\text{Cc}\times \frac{\text{1}}{\text{2}}\text{Dd}\times \frac{\text{1}}{\text{2}}\text{Ee}\\ =\frac{1}{32}\end{array}$

Thus, the probability of $\text{Aa Bb Cc Dd Ee}$ genotype is $\frac{1}{32}$.

b.

Summary Introduction

To determine:

The proportion of the genotype $\text{Aa bb Cc dd ee}$ among the progeny of the cross $\text{AaBbCcddEe}\times \text{AabbCcDdEe}$.

Introduction:

The cross between $\text{AaBbCcddEe}$ and $\text{AabbCcDdEe}$ will give rise to many combinations and the probability of specific genotype in progenies can be determined. Some simpler crosses can also help in determining the proportion of genotype in progeny from many combinations obtained from the cross. The cross occurs only between the genes found on the same locus of each parent’s chromosome.

Explanation of Solution

The probability of obtaining $\text{Aa bb Cc dd ee}$ genotype can be calculated by multiplication rule in which the probability of each allele is multiplied. Probability of $\text{Aa}$ is $\frac{\text{1}}{\text{2}}$, probability of bb is $\frac{\text{1}}{\text{2}}$ , probability of $\text{Cc}$ is $\frac{\text{1}}{\text{2}}$ , probability of dd is $\frac{\text{1}}{\text{2}}$ and probability of ee is $\frac{\text{1}}{4}$. Therefore, the probability of obtaining $\text{Aa bb Cc dd ee}$ genotype can be calculated as:

$\begin{array}{c}\text{Probability of Aa bb Cc dd ee}=\frac{\text{1}}{\text{2}}\text{Aa}\times \frac{\text{1}}{\text{2}}\text{bb}\times \frac{\text{1}}{\text{2}}\text{Cc}\times \frac{\text{1}}{\text{2}}\text{dd}\times \frac{\text{1}}{4}\text{ee}\\ =\frac{1}{64}\end{array}$

Thus, the probability of $\text{Aa bb Cc dd ee}$ genotype is $\frac{1}{64}$.

c.

Summary Introduction

To determine:

The proportion of the genotype $\text{aa bb cc dd ee}$ among the progeny of the cross $\text{AaBbCcddEe}\times \text{AabbCcDdEe}$.

Introduction:

The cross between $\text{AaBbCcddEe}$ and $\text{AabbCcDdEe}$ will give rise to many combinations and the probability of specific genotype in progenies can be determined. Some simpler crosses can also help in determining the proportion of genotype in progeny from many combinations obtained from the cross. The cross occurs only between the genes found on the same locus of each parent’s chromosome.

Explanation of Solution

The probability of obtaining $\text{aa bb cc dd ee}$ genotype can be calculated by multiplication rule in which the probability of each allele is multiplied. Probability of $\text{aa}$ is $\frac{\text{1}}{4}$, probability of bb is $\frac{\text{1}}{\text{2}}$ , probability of $\text{cc}$ is $\frac{\text{1}}{4}$ , probability of dd is $\frac{\text{1}}{\text{2}}$ and probability of ee is $\frac{\text{1}}{4}$. Therefore, the probability of obtaining $\text{aa bb cc dd ee}$ genotype can be calculated as:

$\begin{array}{c}\text{Probability of aa bb cc dd ee}=\frac{\text{1}}{4}\text{aa}\times \frac{\text{1}}{\text{2}}\text{bb}\times \frac{\text{1}}{4}\text{cc}\times \frac{\text{1}}{\text{2}}\text{dd}\times \frac{\text{1}}{4}\text{ee}\\ =\frac{1}{256}\end{array}$

Thus, the probability of $\text{aa bb cc dd ee}$ genotype is $\frac{1}{256}$.

d.

Summary Introduction

To determine:

The proportion of the genotype $\text{AA BB CC DD EE}$ among the progeny of the cross $\text{AaBbCcddEe}\times \text{AabbCcDdEe}$.

Explanation of Solution

In the given cross the genotype of parent 1 is $\text{AaBbCcddEe}$ and the genotype of parent 2 is $\text{AabbCcDdEe}$. Parent 1 will not contribute allele D to the off spring as it is homozygous for allele d. It can pass only allele d to the offspring.

Similarly parent 2 will not contribute to allele B as it is homozygous for allele b. It can pass only allele b to the offspring.