Re: Fixation rates for mutations by genetic drift

From: phillip smith (deletethis-phills_at_ihug.co.nz)
Date: 09/23/04 

Date: Thu, 23 Sep 2004 17:50:36 +0000 (UTC)

in article cis9gi$29fk$1@darwin.ediacara.org, Anon. at
bob.ohara@SOD.OFF.Spammers.helsinki.fi wrote on 23/9/04 4:33 AM:

> 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
Thanks Bob
I'm a bit of a mathematical cripple but seem to be attracted to it like a
moth to a candle. I think this sounds right and seems to fit the
observations

cheers

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