Chapter 7.3, Problem 6TPSQ

a.

Summary Introduction

To determine:

The order of the genes on the chromosome.

Introduction:

Interference is the genetic term that is used in crossover. It is used to measure the degree by which one crossover is interfering with additional crossovers. The coefficient of coincidence is the ratio that is obtained by the division of number of observed double cross over to number of expected double crossover.

Three recessive mutations in Drosophila melanogaster are:

Roughed $\left(ru,smallrougheyes\right)$,

Javelin $\left(jv,cylindricalbristeles\right)$,

$\text{Sepia}\text{eyes}\left(\text{se,}\text{dark}\text{brown}\text{eyes}\right)$ are linked.

Explanation of Solution

The following table represents the resulted progeny of cross:

Genotype	Number	Recombinant or non recombinant
$j{v}^{+}r{u}^{+}s{e}^{+}$	37	Non-Recombinant
$j{v}^{+}r{u}^{+}se$	2	Recombinant with single crossover between $ruandse$
$j{v}^{+}ruse$	14	Recombinant with single crossover between $jvandru$
$j{v}^{+}rus{e}^{+}$	146	Recombinant with double crossover
$jvrus{e}^{+}$	2	Recombinant with single crossover between $ruandse$
$jvruse$	35	Non-Recombinant
$j{v}^{+}r{u}^{+}s{e}^{+}$	154	Recombinant with double cross over
$jvr{u}^{+}s{e}^{+}$	10	Recombinant with single crossover between $jv\text{and}ru$
Total	4000

Non recombinant and recombinant progeny obtained from single cross over and double crossover can be identified from the resulted progeny.

These are the $jv r{u}^{+},s{e}^{+}\text{and}j{v}^{+},ruse$ genotype in the progeny that are present on upper chromosome with one crossover mostly they are non-recombinant.

These are $j{v}^{+}rus{e}^{+}$ and $j{v}^{+}r{u}^{+}s{e}^{+}$ are the genotype in progeny that are present on upper chromosome mostly they are recombinant with double cross over.

On comparing the genotype of progeny non recombinants and recombinants vary at $\left(ru\right)$ position. Thus, $\left(tu\right)$ must be present in the middle. Thus the order of gene is $\left(jvruse\right)$.

b.

Summary Introduction

To determine:

The progeny that will contain single cross over and double crossover with the position.

Introduction:

Interference is the genetic term that is used in crossover. It is used to measure the degree by which one crossover is interfering with additional crossovers. The coefficient of coincidence is the ratio that is obtained by the division of number of observed double cross over to number of expected double crossover.

Three recessive mutations in Drosophila melanogaster are:

Roughed $\left(ru,smallrougheyes\right)$,

Javelin $\left(jv,cylindricalbristeles\right)$,

$\text{Sepia}\text{eyes}\left(\text{se,}\text{dark}\text{brown}\text{eyes}\right)$ are linked.

Explanation of Solution

Single crossover occur between the genes $jvr{u}^{+}s{e}^{+}$ at $jvandru$ position, and in $jvrus{e}^{+}$ at $ruandse$ position.

Double cross over will occur between genes $j{v}^{+}rus{e}^{+}$ at $jvandse$ position, and in $j{v}^{+}r{u}^{+}s{e}^{+}$ at $ruandse$ position.

c.

Summary Introduction

To determine:

The calculation map distances amongst the genes.

Introduction:

Interference is the genetic term that is used in crossover. It is used to measure the degree by which one crossover is interfering with additional crossovers. The coefficient of coincidence is the ratio that is obtained by the division of number of observed double cross over to number of expected double crossover.

Three recessive mutations in Drosophila melanogaster are:

Roughed $\left(ru,smallrougheyes\right)$,

Javelin $\left(jv,cylindricalbristeles\right)$,

$\text{Sepia}\text{eyes}\left(\text{se,}\text{dark}\text{brown}\text{eyes}\right)$ are linked.

Explanation of Solution

Distance between the genes can be calculated as:

Distance between the genes$juandru$

The recombination frequency is calculated by the formula,

$\text{Recombination frequency}\left(\text{R}\text{.F}\text{.}\right)=\frac{\text{Number of recombinant progeny}}{\text{Number}\text{of}\text{total progeny}}$ (1)

The single crossover between $juandru$:

$\begin{array}{l}jvr{u}^{+}s{e}^{+}=10\\ j{v}^{+}ruse=14\end{array}$

The double crossover between $juandru$:

$\begin{array}{l}j{v}^{+}r{u}^{+}s{e}^{+}=154\\ j{v}^{+}rus{e}^{+}=146\end{array}$

Substitute the value of the number of recombinant progeny and that of the number of total progeny in equation (1).

$\begin{array}{c}\text{Recombinant frequency}\left(\text{R}\text{.F}\text{.}\right)=\frac{\left(\text{10+14+154+146}\right)}{4000}\\ =\frac{324}{4000}\\ =\text{0}\text{.081}\end{array}$

The percentage of recombinant frequency is calculated as,

$\begin{array}{c}\text{Recombinant}\text{frequency}\text{percent}\left(\text{R}\text{.F}\text{.}\text{\%}\right)=0.081\times \text{100}\\ =8\text{\%}\end{array}$

The number of map units in $1\%\text{R}\text{.F}\text{.}$ $=1\text{m}\text{.u}\text{.}$

Therefore, the number of map units in $1\%\text{R}\text{.F}\text{.}$ $\begin{array}{l}=\left(1\times 8\right)\text{m}\text{.u}\text{.}\\ =8\text{m}\text{.u}\text{.}\end{array}$

Hence, the distance between $ju\text{and}ru$ is $=8\text{m}\text{.u}\text{.}$.

The distance between the genes $ru\text{and}se$ is calculated as,

The single crossover between $ru\text{and}se$ is

$\begin{array}{l}j{v}^{+}r{u}^{+}se=2\\ jvrus{e}^{+}=2\end{array}$

The double crossover between $ru\text{and}se$ is

$\begin{array}{l}j{v}^{+}r{u}^{+}s{e}^{+}=154\\ j{v}^{+}rus{e}^{+}=146\end{array}$

Substitute the value of the number of recombinant progeny and that of the number of total progeny in equation (1).

$\begin{array}{c}\text{Recombinant frequency}\left(\text{R}\text{.F}\text{.}\right)=\frac{\left(\text{2+2+154+146}\right)}{4000}\\ =\frac{304}{4000}\\ =\text{0}\text{.76}\end{array}$

The percentage of recombinant frequency is calculated as,

$\begin{array}{c}\text{Recombinant}\text{frequency}\text{percent}\left(\text{R}\text{.F}\text{.}\text{\%}\right)=0.076\times \text{100}\\ =7.6\text{\%}\end{array}$

The number of map units in $1\%\text{R}\text{.F}\text{.}$ $=1\text{m}\text{.u}\text{.}$

Therefore, the number of map units in $1\%\text{R}\text{.F}\text{.}$ $\begin{array}{l}=\left(1\times 7.6\right)\text{m}\text{.u}\text{.}\\ =7.\text{6}\text{m}\text{.u}\text{.}\end{array}$

Hence, the distance between $ru\text{and}se$ =.$\text{7}\text{.6}\text{m}\text{.u}\text{.}$

d.

Summary Introduction

To determine:

The calculation of coefficient of coincidence and interference.

Introduction:

Interference is the genetic term that is used in crossover. It is used to measure the degree by which one crossover is interfering with additional crossovers. The coefficient of coincidence is the ratio that is obtained by the division of number of observed double cross over to number of expected double crossover.

Three recessive mutations in Drosophila melanogaster are:

Roughed $\left(ru,smallrougheyes\right)$,

Javelin $\left(jv,cylindricalbristeles\right)$,

$\text{Sepia}\text{eyes}\left(\text{se,}\text{dark}\text{brown}\text{eyes}\right)$ are linked.

Explanation of Solution

Expected number of double cross over will be equal to the product of the recombination frequency of single crossover with the sum of total number of progeny.

Recombination frequency of crossover between $ju\text{and}ru$ =0.081

Recombination frequency of crossover between $ru\text{and}se$ = 0.076

Sum of total progeny =4000

Formula for expected double cross over is,

$\text{Expected}\text{numberof}\text{double}\text{crossover=Recombinant}\text{frequency×Sum}\text{of total}\text{progeny}$

$\begin{array}{c}\text{Expected number of}\text{double}\text{crossover}=\text{0}\text{.081×0}\text{.076×4000}\\ \text{=24}\text{.6}\end{array}$

Formula for coefficient of coincidence is:

$\begin{array}{c}\text{Coffeicient}\text{of}\text{coincidence}=\frac{\text{Observed}\text{number}\text{of}\text{double}\text{crossover}}{\text{Expected}\text{number}\text{of}\text{double}\text{crossoover}}\\ =\frac{\text{2+2}}{24.6}\\ =\text{0}\text{.16}\end{array}$

Interference is calculated as:

$\begin{array}{c}\text{Interfrence}=\text{1-}\text{Cofficient of}\text{coincidence}\\ =\text{1}-\text{0}\text{.16}\\ =\text{0}\text{.84}\end{array}$

The value coefficient of coincidence = 0.16, and interference =0.8