Chapter 9.3, Problem 29AQP

Summary Introduction

To determine:

The order of the genes and the distance between genotypes on the basis of two point crosses.

Introduction:

A series of mutations affects the biosynthetic pathway leading to tryptophan in the bacterium Bacillus subtilis. The seven strains are studied along with their mutation.

Explanation of Solution

The two points cross between the genotypes is as follow:

Recipient	Donor	Cross	Genotypes	The percent of recombination.
T3	168PT	$\text{T3×168PT}$	T-I-	12.7
T3	TII	$\text{T3×TII}$	T-I-	11.8
T3	T8	$\text{T3×T8}$	T-A-	43.5
T3	H25	$\text{T3×H25}$	T-H-	28.6
168	H25	$168\times \text{H25}$	I-H-	44.9
TII	H25	$\text{TII}\times \text{H25}$	I-H-	41.4
TI	H25	$\text{TI}\times \text{H25}$	I-H-	31.3
T8	H25	$\text{T8}\times \text{H25}$	A-H-	67.4
H25	T3	$\text{H25}\times \text{T3}$	H-T-	19.0
H25	T11	$\text{H25}\times \text{TII}$	H-I-	26.3
H25	T1	$\text{H25}\times \text{TI}$	H-I-	13.4
H25	T8	$\text{H25}\times \text{T8}$	A-H-	45.0

The average of observed recombination is found by:

$\text{Average=}\frac{\text{Sum}\text{of}\text{observations}}{\text{Number}\text{of}\text{observations}}$

The sums of observations need to be taken as

(a)

$\begin{array}{c}{T}^{-}{I}^{-}=12.7\left(T3\times 168PT\right)+11.8\left(T3\times T11\right)\\ =24.5\%\\ \text{Average}=\frac{24.5}{2}\\ =12.25\%\end{array}$

(b)

$\begin{array}{c}{T}^{-}{A}^{-}=43.5\%\left(T3\times T8\right)\\ =43.5\%\\ \text{Average}=\frac{43.5}{2}\\ =21.75\%\end{array}$

(c)

$\begin{array}{c}{A}^{-}{H}^{-}=45.0\left(\text{H25}\times \text{T8}\right)+67.8\left(T8\times H25\right)=112.4\%\\ \text{Average}=\frac{112.4}{2}\\ =56.2\%\end{array}$

(d)

$\begin{array}{c}{T}^{-}{H}^{-}=19.0\left(\text{H25×T3}\right)+28.6\left(\text{T3×H25}\right)\\ =\text{47}\text{.6\%}\\ \text{Average}=\frac{47.6}{2}\\ =23.8\%\end{array}$

(e)

$\begin{array}{c}{I}^{-}{H}^{-}=\left(\begin{array}{c}44.9\left(168\times \text{H25}\right)\text{+41}\text{.4}\left(\text{TII}\times \text{H25}\right)\text{+}\\ \text{31}\text{.3}\left(\text{TI}\times \text{H25}\right)\text{+26}\text{.3}\left(\text{H25}\times \text{TII}\right)\text{+13}\text{.4}\left(\text{H25}\times \text{T1}\right)\end{array}\right)\\ \text{=157}\text{.3\%}\\ \text{Average}=\frac{157.5}{5}\\ =31.5\%\end{array}$

The distance between the genes is as follows:

Gene pair	Distance
A and H	56.2 m.u.
T and A	43.5 m.u
I and H	31.5 m.u.
T and H	23.8 m.u.
T and I	12.3 m.u.

Summary Introduction

To draw:

A map of the genes on the chromosome.

Introduction:

A series of transformation is carried out on strains having the different mutations. The percentage of recombinants is determined by the crosses between the recipient and the donor.

Explanation of Solution

The genetic map drawn based on the genetic distance is as follow:

The distance between A and I is determined as:

Distance between A and I = (Distance between A and T)-(Distance between I and T)

Distance between A and I = (43.5-12.3) m.u

Distance between A and I = 31.2 m.u

Conclusion

The distance between the genes is determined by the crosses in between the genotypes.