Chapter 7.3, Problem 29AQP

a.

Summary Introduction

To determine:

(a). The order of these genes on the chromosomes.

Explanation of Solution

Cross between $\frac{\text{P}\text{Sh-1}{\text{Hb}}^{\text{2}}}{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}\text{×}\frac{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}$

The  following table represents the resulted progeny of the cross:

Genotype	Number
$\frac{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}$	274
$\frac{\text{P}\text{Sh-1}{\text{Hb}}^{\text{1}}}{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}$	320
$\frac{\text{P}\text{Sh-1}{\text{Hb}}^{\text{2}}}{\text{P}\text{sh-1}{\text{Hb}}^{\text{2}}}$	527
$\frac{\text{P}\text{Sh-1}{\text{Hb}}^1}{\text{P}\text{h-1}{\text{Hb}}^{\text{2}}}$	45
$\frac{\text{P}\text{Sh-1}\text{H}{\text{b}}^{\text{2}}}{\text{p}\text{sh-1}{\text{Hb}}^{\text{2}}}$	53.1
$\frac{\text{p}\text{Sh-1}{\text{Hb}}^{\text{2}}}{\text{p}\text{sh-1}{\text{Hb}}^{\text{2}}}$	6
$\frac{\text{P}\text{Sh-1}{\text{Hb}}^{\text{1}}}{\text{p}\text{sh-1}{\text{Hb}}^{\text{2}}}$	5
$\frac{\text{P}\text{Sh-1}{\text{Hb}}^2}{\text{p}\text{sh-1}{\text{Hb}}^{\text{2}}}$	0
$\frac{\text{p}\text{Sh-1}{\text{Hb}}^{\text{1}}}{\text{p}\text{sh-1}{\text{Hb}}^{\text{2}}}$	1
Total	708

From the result it is clear that all the progeny have $\text{p}\text{sh-1}{\text{Hb}}^2$ in the lower chromosome from the parents with homozygous recessive genotype, whereas the upper chromosome has genes from the heterozygous parents. Non recombinants are obtained from single crossover and recombinants are obtained from the double crossover.

$\text{P}\text{sh-1}{\text{Hb}}^{\text{1}}$, and $\text{p}\text{sh-1}{\text{Hb}}^2$ are the genotype in progeny that are present on upper chromosome with ONE crossover. Mostly they are non-recombinant.

$\text{P}\text{sh-1}{\text{Hb}}^{\text{1}}$, and $\text{p}\text{sh-1}{\text{Hb}}^2$ are the genotype in progeny that are present on upper chromosome. Mostly they are recombinant.

On comparing the genotype of progeny non-recombinants and recombinants vary at $\text{sh-1}$ position. Thus, $\text{sh-1}$ must be present in the middle. Thus, the order of gene is: $\text{p}\text{sh-1}\text{Hb}$

b.

Summary Introduction

To determine:

The map distance between the genes.

Explanation of Solution

Distance between the genes can be calculated as

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)

Distance between the genes $\text{p}\text{and}\text{sh-1}$

The $\text{single}\text{crossover}\text{between p}\text{ and}\text{sh-1}$

$\begin{array}{l}\text{P}{\text{Sh-1Hb}}^{\text{1}}=\text{57}\\ \text{p}{\text{Sh-1Hb}}^{\text{2}}=\text{45}\\ \end{array}$

$\begin{array}{l}\text{The}\text{double}\text{crossover}\text{between p}\text{ and}\text{sh-1}\\ \text{P}{\text{Sh-1Hb}}^1=\text{0}\\ \text{p}{\text{Sh-1Hb}}^{\text{2}}=\text{1}\end{array}$

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

$\begin{array}{c}\text{Recombination frequency}\left(\text{R}\text{.F}\text{.}\right)\text{=}\frac{\text{Number of recombinant progeny}}{\text{Number}\text{of}\text{total progeny}}\\ =\frac{\left(\text{57+45+1+0}\right)}{\text{708}}\\ =\frac{\text{103}}{\text{708}}\\ =\text{0}\text{.145}\end{array}$

The percentage of recombinant frequency is calculated as,

$\begin{array}{c}\text{Recombinant frequency }\left(\text{R}\text{.F}\text{.\%}\right)=\text{0}\text{.145×100}\\ =\text{14}\text{.5\%}\end{array}$

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

Therefore, the number of map units $\begin{array}{c}\text{1}\text{\%}\text{R}\text{.F}\text{.= }\left(\text{1×14}\text{.5}\right)\text{m}\text{.u}\text{.}\\ \text{=14}\text{.5}\text{m}\text{.u}\text{.}\end{array}$

Hence, the distance between $\text{p}\text{and}\text{sh-1}$=14.5 map units.

Distance between the genes$\text{sh-1}\text{and }\text{Hb}$

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 $\text{single}\text{crossover}\text{between p}\text{ and}\text{sh-1}$

$\begin{array}{c}\text{P}{\text{Sh-1Hb}}^2=6\\ \text{p}{\text{Sh-1Hb}}^1=\text{5}\end{array}$

The $\text{double}\text{crossover}\text{between p}\text{ and}\text{sh-1}$

$\begin{array}{c}\text{P}{\text{Sh-1Hb}}^2=\text{0}\\ \text{p}{\text{Sh-1Hb}}^1=\text{1}\end{array}$

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

$\begin{array}{c}\text{Recombinant frequency}\left(\text{R}\text{.F}\text{.}\right)=\frac{\left(\text{6+5+1+0}\right)}{\text{708}}\\ =\frac{\text{103}}{\text{708}}\\ =0.\text{017}\end{array}$

The percentage of recombinant frequency is calculated as,

$\begin{array}{c}\text{Recombinant frequency }\left(\text{R}\text{.F}\text{.\%}\right)=\text{0}\text{.017×100}\\ =\text{1}\text{.7\%}\end{array}$

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

Hence, the distance between $\text{sh-1}\text{and }\text{Hb}$=$\text{1}\text{.7}\text{m}\text{.u}\text{.}$

c.

Summary Introduction

To determine:

The coefficient of coincidence and inference among the genes

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 $\text{p}\text{and}\text{sh-1}$=0.145.

Recombination frequency of crossover between $\text{sh-1}\text{and}\text{Hb}$= 0.017

Sum of total progeny =708

Formula for expected double cross over is

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

$\begin{array}{c}\text{Expected}\text{numberof}\text{double}\text{crossover}=\text{0}\text{.145×0}\text{.017×708}\\ \text{=1}\text{.7}\end{array}$

Formula for coefficient of coincidence is:

$\begin{array}{c}\text{Coffeicient}\text{of}\text{coincidence}=\frac{\text{Observed}\text{number}\text{of}\text{double}\text{crossover}}{\text{Expected}\text{number}\text{of}\text{double}\text{crossoover}}\\ =\frac{\text{1+0}}{\text{1}\text{.7}}\\ =\frac{\text{1}}{\text{1}\text{.7}}\\ =\text{0}\text{.59}\end{array}$

Interference is calculated as:

$\begin{array}{c}\text{Interfrence}=\text{1}-\text{Cofficient of}\text{coincidence}\\ =\text{1}-\text{0}\text{.59}\\ =\text{0}\text{.41}\end{array}$

The value of Coefficient of coincidence = 0.59 and interference =0.41

Conclusion

The order of gene is: $\text{p}\text{sh-1}\text{Hb}$ and coefficient of coincidence = 0.59 and interference =0.41.