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E. coli strain B is doubly infected with two rII mutants of phage T4. 0.2 ml of a 108 dilution of the progeny is plated on E. coli B and 0.1 ml of a 104 dilution of the progeny is plated on E. coli K. 16 plaques appeared on strain B, 400 on strain K. Calculate the recombination frequency between these two mutations.

Relax

Respuesta :

To Find :

The recombination frequency between these two mutations.

Solution :

Formula of calculating recombination frequency is :

[tex]R.F = \text{Number of mutations}\times \dfrac{\text{number of k plaques} \times D.F }{\text{Volume used}}\times \dfrac{Volume'}{\text{Number of B plaques} \times D.F' }}[/tex]

Putting all given values in above equation, we get :

[tex]R.F = \dfrac{2\times 400\times 0.2 \times 10^4}{16\times 0.1\times 10^8}\\\\R.F = 0.01[/tex]

Therefore, the recombination frequency between these two mutations is 0.01 or 1 %.

Cricetus

The recombination frequency between these two mutations will be "0.01".

Given:

  • No. of mutation = 2
  • Volume, [tex]V = 0.1[/tex]

                     [tex]V'= 0.2[/tex]

  • k plaques = 400

The formula:

→ [tex]R.F = No. \ of \ mutations\times \frac{No. \ of \ k \ plaques\times D.F }{Volume}\times \frac{Volume'}{No. \ of \ B \ plaques\times D.F'}[/tex]

By putting the values in the above formula, we get

→         [tex]= \frac{2\times 400\times 0.2\times 10^4}{16\times 0.1\times 10^8}[/tex]

→         [tex]= \frac{160\times 10^4}{1.6\times 10^8}[/tex]

→         [tex]= 0.01[/tex]

Thus the solution above is the right approach.

Learn more about frequency here:

https://brainly.com/question/22387406

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