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| subject: | Re: Fixation rates for mu |
"phillip smith" wrote
>
> 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
>
A random neutral mutation has a 1/2N chance of being fixed. The chance that
all ten will be lost (asuming each starts with one mutation) is therefore
(1 - 1/2N) to the power ten.
It takes about 2N generations for an allele destined to be fixed to be
fixed. However don't quote me on the maths. What you are asking is the mean
time taken, presuming all mutations will not be lost, for no wild-type
organsims to be left. I don't know how you would work this out. A
mathematician would know.
If you want an estimate, rather than a mathematically correct formula,
remembering that the assumptions of perfectly random mating, constnat
population size, and so on won't be met in real populations, then it should
be easy enough to write a computer simulation.
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