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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.

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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