Chapter 9.3, Problem 36AQP

a.

Summary Introduction

To determine:

The order of the three genes on the phage chromosome.

Introduction:

The E.coli is infected by the two strains of phage $\lambda$, one with host genotype h st c and other strain with wild-type alleles of genotype h⁺st⁺ct^+. The cross between these two strains is used to determine the order of genes.

Explanation of Solution

Pictorial representation:

Figure 1(a)

Figure 1(b)

Figure 1(c)

In figure no 1(a), the F₁ generation obtained is used to find the order of genes. The possibilities that the genes could be placed in three orders are.

Gene c in the middle and it may be h c st

Gene h in the middle and it may be c hst

Gene st in the middle and it may be h st c

The chromosome which produces same set of alleles which the change in the middle locus is h+ st+ c+ and h stc . In other two chromosomes, two genes change their place. So, here the order of genes is shown in fig no,. 1(c)

b.

Summary Introduction

To determine:

The map distance between the genes.

Introduction:

The distance between the genes is mapped or measured by crossing over and determing the recombination frequency. The recombination may be singlecross over, double cross over, recombinant or non-recombinant.

Explanation of Solution

Pictorial representation:

Fig 2 (a) Crossing over between the genes h and st is as follows:

Fig: 2 (b) Crossing over between st and c genes is as follows:

As shown in fig no. 2 (a) and (b), the crossing is between h andst and st and c genes.

The crossing over between genes h and st and between st and c genes results into the formation of various recombination like 4 single crossover, 2 non-recombinants and 2 double cross over.

The table changes according to the order of genes as shown below:

Phage progeny genotype	Number of plagues	Recombination
h+st+c+	321	Non- recombinant
h st c	338	Non- recombinant
h+/stc	26	Single cross over
h /st+c+	30	Single cross over
h+st+/c	106	Single cross over
h st/ c+	110	Single cross over
h+/st/ c+	5	Double cross over
h /st+/c	6	Double cross over
Total	942

The formula used for recombinant frequency is:

$\text{RF=}\frac{\text{recombinant}\text{plagues}}{\text{total}\text{plagues}}\text{×100}$

The number of recombinant progeny that shows the crossing over in h gene and st gene is as follows:

= h⁺/st c , h /st⁺c⁺ , h⁺/st/ c⁺, h /st⁺/c.

= 26 + 30 + 5+ 6

= 67

$\text{RF=}\frac{\text{recombinant}\text{plagues}}{\text{total}\text{plagues}}\text{×100\%}$

$\begin{array}{c}\text{RF=}\frac{67}{942}\text{×100\%}\\ \text{=7}\text{.1\%}\end{array}$

Similarly, the recombinant frequency between st and c gene can be found by crossing over between st and c gene.

Total number of recombinant plagues is:

h⁺st⁺/c , h st/ c⁺, h⁺/st/ c⁺, h /st⁺/c

= 106 + 110 + 5 + 6

= 227

Now, the recombinant frequency between st and c gene is obtained as:

$\text{RF=}\frac{\text{recombinant}\text{plagues}}{\text{total}\text{plagues}}\text{×100\%}$

$\begin{array}{c}\text{RF=}\frac{227}{942}\text{×100\%}\\ \text{=24}\text{.1\%}\end{array}$

As the recombinant frequency is equal to the distance between the genes in map units, so the distance between the h andst gene is 7.1 map units and the distance between st andc gene is 24.1 map units.

c.

Summary Introduction

To determine:

The coefficient of coincidence and the interference.

Introduction:

The coefficient of coincidence is defined as the ratio of observed double crossover to the number of expected double cross over. Interference is defined as the extent up to which one crossover interferes with additional crossovers.

Explanation of Solution

coefficient of coincidence is determined as:

$\text{cofficient}\text{of}\text{coincidence=}\frac{\text{number}\text{of}\text{observed}\text{double}\text{crossover}}{\text{number}\text{of}\text{expected}\text{double}\text{crossovers}}$

$\text{Probability}\text{of}\text{recombination}=\frac{\text{recombinant}\text{plagues}}{\text{total}\text{plagues}}$

$\begin{array}{c}\text{=}\frac{67}{942}\\ =0.071\end{array}$

Similarly, the probability ofrecombination between st and c gene is obtained as:

$\text{Probability}\text{of}\text{recombination}=\frac{\text{recombinant}\text{plagues}}{\text{total}\text{plagues}}$

$\begin{array}{c}\text{RF=}\frac{227}{942}\\ \text{=0}\text{.241}\end{array}$

The total number of double crossovers is as follows:

= number of plagues of h⁺/st/ c⁺ and h /st⁺/c

= 5+6

=11

Number of expected double crossover can be obtained by

$\begin{array}{l}=\left(\text{probability}\text{of}\text{recombination}\text{between}h\text{and}st\right)\\ \times \left(\text{probability}\text{of}\text{recombination}\text{between}st\text{and}c\right)\times \left(\text{total}\text{number}\text{of}\text{plagues}\text{of}\lambda \text{phage}\right)\\ \end{array}$

$\begin{array}{l}=0.071\times 0.241\times 942\\ =16.11\end{array}$

Coefficient of coincidence $\begin{array}{l}=\frac{11}{16.11}\\ =0.68\end{array}$

Interference is determined as :

$\begin{array}{l}\text{Interference}=\text{1-coefficient}\text{of}\text{coincidence}\\ \text{=1}-\text{0}\text{.68}\\ =\text{0}\text{.32}\end{array}$

Conclusion

The recombination frequency obtained is used to determine the coefficient of coincidence and interference.