PS10-Sol - Genetics 320, Problem Set 12 Due Wednesay 8...

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Unformatted text preview: Genetics 320, Problem Set 12 Due Wednesay 8 December 2004 at 11 am 1 (1 point) : Consider a population of size N = 10 5 and a mutation rate of u = 10- 6 , (a) Assuming the Infinite alleles model, what is the probability that a random individual is a heterozygote? Here 4 Nu = 4 10 5 10- 6 = 0 . 4 , hence H = 4 Nu 1 + 4 Nu = . 4 1 . 4 = 0 . 286 (b) Recall that the number of mutations follows a Poisson distribution (See Problem 7 of PS 1). Assuming that the common ancestor to two random sequences occurs 2 N generations ago, what is the probability that two randomly-chosen sequences differ by one mutation? By two mutations? By five mutations? If the time back to the common ancestor is 2 N generations, then the expected number of new mutations is 2 (2 N ) u = 0 . 4 , and the probability of k mutations follows a Poisson distribution with parameter 0.4. Hence, Prob (1 mutation ) = (0 . 4) 1 e- . 4 / 1! = 0 . 268 . Prob (2 mutations ) = (0 . 4) 2 e- . 4 / 2! = 0 . 054 . Prob (5 mutations ) = (0 . 4) 5 e- . 4 / 5! = 0 . 000057 (c) What is the probability for this population size that two randomly-chosen alleles have there most recent common ancestor in the last 10,000 generations? 25,000 generations?...
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This homework help was uploaded on 04/03/2008 for the course BIO 320 taught by Professor Walsh,weinert during the Fall '07 term at University of Arizona- Tucson.

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PS10-Sol - Genetics 320, Problem Set 12 Due Wednesay 8...

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