E

This formula can be simplified to give

where the upper limit of summation is Af(n) = n for n < I, and N{n) = I for n> I.

We would like to calculate the probability to have at least one double mutant in the system by time n. We consider a sequence of clones. There are n divisions of the SC. Let us denote the event of acquiring a mutation in the SC at time k, and no further mutations in the SC before time n, by Sk,n- We have

Given Sk,n, the probability to have no double mutant by time n is k n

Prob(E-n\Skrn) = U(l-P0{E-n-j + l)) (1 - P„(E; n - j + 1)).

Each clone is originated at time j, so its "age" is n — j + 1. Similarly, the probability to have no double mutant by time n, given that no SC mutations have occurred by time n, is n

and the probability to have no SC mutations by time n is (1 - p\)n. The total probability to have at least one double mutant by time n is given by n

This formula was compared with numerical simulations for the total probability of a double mutant, and with the approximate expression for xss + xsd + xdd derived in Appendix A. The agreement with the numeric is perfect. The agreement with the approximation is very good, if the following consideration is taken into account. Note that the expressions for xss, xsd and xdd (formulas (5.5), (5.8) and (5.9)) were derived assuming that the size of the crypt is always N. On the other hand, in the simulations and in formula (A.l) we have a "development" phase where the size of the crypt is growing. The growth is exponential and thus during the first I — 2 steps, the population size is significantly less than N, and thus mutation probabilities are negligible. This is why in order to get a perfect fit, the expressions for xss, xsd and xdd should be shifted by I — 2 steps.

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