Re: Fixation rates for mutations by genetic drift

From: Anon. (bob.ohara_at_SOD.OFF.Spammers.helsinki.fi)
Date: 09/22/04 

Date: Wed, 22 Sep 2004 16:33:22 +0000 (UTC)

phillip smith wrote:
> I wonder if any one can help me? I need to check a calculation for a paper.
>
>>>From Kimura the time for a neutral mutation to be fixed is the mutation
> rate.
> If I have ten genes each of which can have a neutral mutation. What is the
> mean time taken for there to be no individuals which are free of mutation.
> That is the whole population carries at least one of the ten mutations
>
> I think its is t =(1-u)^n
>
> Where n is the number of genes that can have a mutation in this case 10 and
> u is the mutation rate t = replications
>
And all mutations rates are equal! :-)

> I think I have made a mistake some where so any one who can put me right
> would be appreciated
>
Yes you have! This would be the time taken if there was some rule that
mutation 1 had to go extinct first, and when it went extinct, only then
could mutation 2 could drift to extinction.

Did Kimura show what the distribution of times to extinction was? Under
some conditions, it'll probably an exponential distribution, in which
case the time for the extinction of all 10 mutants will follow a gamma
distribution (which is also a Chi-squared distribution). Check the
derivations of a gamma, and you should see where it's coming from.

So, the answer is that under the assumtion that the time to extinction
of one mutant is exponentially distributed with mean u, the mean time to
extinction of n muatants is un.

Bob 

-- 
Bob O'Hara
Dept. of Mathematics and Statistics
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