Re: Fixation rates for mutations by genetic drift

From: phillip smith (deletethis-phills_at_ihug.co.nz)
Date: 09/22/04 

Date: Wed, 22 Sep 2004 16:33:21 +0000 (UTC)

in article cipq84$1gtb$1@darwin.ediacara.org, phillip smith at
deletethis-phills@ihug.co.nz wrote on 22/9/04 6:00 AM:

> 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
>
> I think I have made a mistake some where so any one who can put me right
> would be appreciated
>
>
> Cheers
This is wrong I think its simply t = un
We have evidence for this if you take the genome as 10^ 9 and mutation rate
as 10 exp -9 then then we would expect it to take one replication on average
for every cell to have at least 1 mutation.

So if we have 10 genes then the rate of fixation is 10^-8

But each gene has approx 1000 nucleotides so we are at 10^-5

Does this seem ok

Cheers

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