Exact formula for total probability of double mutations

Here we calculate the probability of having a double mutant in a crypt, as a function of time. We focus on the progeny of one stem cell. The total number of cells in the crypt, N(n), is

Note that n < I represents the "development" stage of the crypt. After n = I, the size of the crypt remains constant. Each cell division may result in creating mutants. The probability of mutation is p\, so that with probability 2pi(l — pi) one mutant is created, and with probability pf two mutants are created. A mutated cell can acquire a second mutation with probability p?.

Let us trace mutations in one clone. The number of mutants at time n (with one mutation) is a random variable, whose values go from 0 to 2'"1. If a double mutant is created, we set this random variable to the value E (for "end"). This is an absorbing state because once a double mutant is creatjed, it is assumed to stay in the system. We have a time-inhomogeneous Markov process for the variable j, j 6 {0,1,..., 2i_1, E}. The probability distribution for the variable j at time n is a row-vector whose evolution is given by

The transition matrices, Ai1-"', can be easily written down. The first matrix corresponds to the asymmetric division of the SC, and is given by with all the rest of the elements being zero. The matrices corresponding to p(n) =p(n-l)Af(»)) „=1,2,..

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