pqs Question 3 18 marks8 minutes] This question is about calculating the change

Question

pqs Question 3 18 marks8 minutes] This question is about calculating the change in allele frequency for the special case of h a) Simplify the equation given for Asp (the change in allele frequency p due to selection) for the special case of h-%. Make sure you show the steps in the simplification. (Do not proceed to the approximation step.) 4 marks 1 mark] Use the approximate simplified formula to find the allele frequency in the next generation when the b) Explain how w can be approximated to 1 when q is very small. c) initial value of p is 0.1 and s- 0.1 [3 marks]

Explanation / Answer

a). p1 = p2 + pq(1-hs) / w

w = p2 + 2pq(1-hs) + q2 (1 - s)

p = p2 + pq (1 - hs)/w - p

p is the expression we want to find.

To simplify this equation, we will multiply p by w/w so we can have a common denominator then expand w and start cancelling terms.

p = p2 + pq(1-hs) - pw/w

p = p2 + pq(1-hs) - p [ p2 + 2pq (1- hs) + q2 (1 - s) ] / w

p = p2 +pq - pqhs - p [ p2 + 2pq - 2 pqhs + q2 - q2s ] /w

Notice that inside the [ ] there is p2 + 2pq + q2 which we know equals 1.

p = p2 + pq - pqhs - p [1 - 2pqhs - q2s ] / w

p = p [ (p + q - qhs ) - [ 1- 2pqhs - q2s ] ]

Now remove the brackets, changing signs as needed. Also note that p + q = 1. So..

p = p [ 1- qhs - 1+ 2pqhs + q2s ] / w

Here the +1 and -1 cancel. Also there is a +2pqhs and a - qhs we get

p = p [ ( 2pq - q) hs + q2s ] / w

Move the q and s out of the brackets to get

p = pqs[ ( 2p - 1) h + q ] / w

( 2p - 1), is equal to p+p-1 so

p =pqs[ ph + q (1- h)] / w

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