You wrote:
>We are not looking for substitutions that "pile up" we are looking for
>substitutions to various locations in a sequence as shown above.
>By the way, I don't see the population anywhere in your equeation. k is the
>kth mutation, p is probability and n is generation. Population is irrelevant
>even in your equation, isn't it?
You are right that I should use p = 10^-4. I realized that a few minutes
after I sent off the post. You are right that population is irrelevant in my
equation, because I am calculating relative frequencies.
Ok, let me summarize where I am to see if you agree.
I'm assuming a sequence that has a probability, p, of having one mutation
every time it is duplicated. Using Gish's numbers, we agree to use p = 10^-4.
The number of offspring that have k mutations, Ak, divided by the number of
offspring that have zero mutations, A0, after N generations is (and I'm
pretty sure of this)
Ak/A0 = (1/k!)(pN)^k
So, it takes about N = 1/p generations to have a large number of different
mutants running around. That's 10,000 generations of humans, or 200,000 years.
Your estimate of 1.2 million years is a little too high for the following
reason. You assume it takes 10,000 generations (1/p) to get the first
mutation in, and then another 10,000 generations to get the second one in,
etc... But, in fact you have waited until A1/A0 = 1, i.e. 1/2 of the
population has suffered a mutation until you allow a second mutation to start.
I have more to say, but I want to make sure we are in agreement on the math
before proceeding.
Jim
Jim Blake
Associate Professor
Department of Electrical Engineering
Texas A&M University
College Station, TX 77843