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scanner you can upload a photo of each page Question 1 [30 marks] Reversible mutation In a study on the bacterium Staphylococ

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Answer #1

a) There a formula for that.

A = u +

A = 1x10-4 / (1x10-4 + 5x10-5) = 0.66 = 2/3

The frequency of A under neutrality is 0.66

b) The formula now is:

p200 - = (po - - )(1 - H - v) A 200 tu

1.10 - 4 p200 - 5.10 - 5 + 1x10 – 4 1.r10 – 4) A 200 1.c 10 - 4 .r10 – 5 + 1x10 –4 10 )(1-5.610 - 5 - 5

p200 – 0.66 = (0 – 0.66) (0.99985) A200 = (-0.66) (0.97)

p200 = (-0.66) (0.99985) A200 = (-0.66) (0.97) = -0.64

p200 = (-0.66) 0.97) + 0.66 = 0.66 – 0.64 = 0.019 = 0.02

Frequency of A after 200 generations starting from 0 would be 0.02

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