Chapter 7.2, Problem 24AQP

a.

Summary Introduction

To determine:

The phenotype of progeny and their expected proportion when a female of each genotype is test crossed with the male $\left(\frac{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}{\text{e}\text{ro}}\right)\left(\frac{{\text{f}}^{\text{+}}}{\text{f}}\right)$

Introduction:

Test cross is the genetic cross between an individual that has dominant phenotype in specific trait to homozygous recessive for these specific traits. Test cross is used to identify the genotype of individual carrying dominant phenotype, and also used to check the linkage between the genes. In drosophila melanogaster, ebony eyes (e), and rough eyes (ro) are encoded by autosomal recessive gene that is present on chromosome they are separated by 20 map units. The gene that code for fork bristle (f) is X-linked recessive and assort independently of (e), and (ro).

Explanation of Solution

Pictorial representation: The cross between $\left(\frac{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}{\text{e}\text{ro}}\right)\left(\frac{{\text{f}}^{\text{+}}}{\text{f}}\right)$ is represented as below:

Fig.1: The cross between $\left(\frac{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}{\text{e}\text{ro}}\right)\left(\frac{{\text{f}}^{\text{+}}}{\text{f}}\right)$.

Two autosomal genes in drosophila melanogaster, for body type and eyes are located 20 map unit apart from each other on chromosome 3.

$\begin{array}{c}\text{1m}\text{.u}\text{.=1\%}\text{recombinant}\text{frequency (R}\text{.F}\text{.)}\\ \text{Thus,}\text{20}\text{m}\text{.u}\text{.=20}\text{\% R}\text{.F}\text{.}\end{array}$

Therefore, total percentage of recombinant would be 20 and non recombinant would be 80.

In females eight gametes are possible which will have four parental and four recombinants. Total percentage of recombinants is 20% so for each recombinant 5%. Similarly total percentage of non recombinant is 80% so each non recombinant will have 20%.

Given information is that, when drosophila is heterozygous for autosomal genes that are present in cis configuration with X-linked gene for heterozygous bristle. This female is crossed with homozygous recessive male for two autosomal genes and hemizygous for the X-linked gene.

The female are able to produce eight types of different gametes and these gametes are:

$\begin{array}{l}\underset{\_}{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}\underset{\_}{{\text{f}}^{\text{+}}}\\ \underset{\_}{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}\text{f}\\ \underset{\_}{\text{e}\text{ro}}\underset{\_}{{\text{f}}^{\text{+}}}\\ \underset{\_}{\text{e}\text{ro}}\underset{\_}{\text{f}}\end{array}]\begin{array}{c}\\ \text{Non recombinant}\\ \end{array}$

$\begin{array}{l}\underset{\_}{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}\underset{\_}{{\text{f}}^{\text{+}}}\\ \underset{\_}{\text{e}{\text{ro}}^{\text{+}}}{\text{f}}^{+}\\ \underset{\_}{{\text{e}}^{+}\text{ro}}\underset{\_}{\text{f}}\\ \underset{\_}{\text{e}{\text{ro}}^{+}}\underset{\_}{\text{f}}\end{array}]\begin{array}{c}\\ \text{Recombinant}\\ \end{array}$

The result of the cross is:

The following table represents the expected proportion from the cross:

Genotype	Body color	Eyes	Bristles	Proportion
${\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}{\text{f}}^{\text{+}}$	Normal	Normal	Normal	20%
${\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}\text{f}$	Normal	Normal	Forked	20%
$\text{e}\text{ro}{\text{f}}^{\text{+}}$	Ebony	Rough	Normal	20%
$\text{e}\text{ro}\text{f}$	Ebony	Rough	Forked	20%
${\text{e}}^{\text{+}}\text{ro }{\text{f}}^{\text{+}}$	Normal	Rough	Ebony	5%
$\text{e }{\text{ro}}^{\text{+}}{\text{f}}^{\text{+}}$	Ebony	Normal	Normal	5%
${\text{e}}^{\text{+}}\text{ro }\text{f}$	Normal	Rough	Forked	5%
$\text{e }{\text{ro}}^{\text{+}}\text{f}$	Ebony	Normal	Forked	5%

Table 1: The expected proportion from the cross.

b.

Summary Introduction

To determine:

The phenotype of progeny and their expected proportion when a female of each genotype is test crossed with the male $\left(\frac{{\text{e}}^{\text{+}}\text{ro}}{\text{e}{\text{ro}}^{+}}\right)\left(\frac{{\text{f}}^{\text{+}}}{\text{f}}\right)$.

Introduction:

Test cross is the genetic cross between an individual that has dominant phenotype in specific trait to homozygous recessive for these specific traits. Test cross is used to identify the genotype of individual carrying dominant phenotype, and also used to check the linkage between the genes. In drosophila melanogaster, ebony eyes (e), and rough eyes (ro) are encoded by autosomal recessive gene that is present on chromosome they are separated by 20 map units. The gene that code for fork bristle (f) is X-linked recessive and assort independently of (e), and (ro).

Explanation of Solution

Pictorial representation: The cross between $\left(\frac{{\text{e}}^{\text{+}}\text{ro}}{\text{e}{\text{ro}}^{+}}\right)\left(\frac{{\text{f}}^{\text{+}}}{\text{f}}\right)$ is represented as below:

Fig. 2: The cross between $\left(\frac{{\text{e}}^{\text{+}}\text{ro}}{\text{e}{\text{ro}}^{+}}\right)\left(\frac{{\text{f}}^{\text{+}}}{\text{f}}\right)$

Given information is that, when drosophila is heterozygous for autosomal genes that are present in Tran’s configuration with X-linked gene for heterozygous bristle. This female is crossed with homozygous recessive male for two autosomal genes and hemizygous for the X-linked gene.

The female are able to produce eight types of different gametes and these gametes are:

$\begin{array}{l}\underset{\_}{{\text{e}}^{\text{+}}\text{ro}}\underset{\_}{{\text{f}}^{\text{+}}}\\ \underset{\_}{\text{e}{\text{ro}}^{\text{+}}}\underset{\_}{{\text{f}}^{+}}\\ \underset{\_}{{\text{e}}^{+}\text{ro}}\underset{\_}{{\text{f}}^{\text{+}}}\\ \underset{\_}{\text{e}{\text{ro}}^{+}}\underset{\_}{\text{f}}\end{array}]\begin{array}{c}\\ \text{Non recombinant}\\ \end{array}$

$\begin{array}{l}\underset{\_}{{\text{e}}^{\text{+}}{\text{ro}}^{\text{+}}}\underset{\_}{{\text{f}}^{\text{+}}}\\ \underset{\_}{{\text{e}}^{+}{\text{ro}}^{\text{+}}}\underset{\_}{\text{f}}\\ \underset{\_}{\text{e}\text{ro}}\underset{\_}{{\text{f}}^{+}}\\ \underset{\_}{\text{e}\text{ro}}\underset{\_}{\text{f}}\end{array}]\begin{array}{c}\\ \text{Recombinant}\\ \end{array}$

The result of the cross is:

The following table represents the expected proportion from the cross:

Genotype	Body color	Eyes	Bristles	Proportion
${\text{e}}^{\text{+}}\text{ro}{\text{f}}^{\text{+}}$	Normal	Rough	Normal	20%
$\text{e }{\text{ro}}^{\text{+}}{\text{f}}^{+}$	Ebony	Normal	Normal	20%
${\text{e}}^{+}\text{ro}\text{f}$	Normal	Rough	Forked	20%
$\text{e}{\text{ro}}^{+}\text{f}$	Ebony	Normal	Forked	20%
${\text{e}}^{\text{+}}{\text{ro}}^{+}{\text{f}}^{\text{+}}$	Normal	Normal	Normal	5%
${\text{e}}^{+}{\text{ro}}^{\text{+}}\text{f}$	Normal	Normal	Forked	5%
$\text{e }\text{ro }{\text{f}}^{+}$	Ebony	Rough	Normal	5%
$\text{e }\text{ro }\text{f}$	Ebony	Rough	Forked	5%

Table 2: The expected proportion from the cross.