### Questions

1. In Equations 15 to 18 of the lecture we wrote down chemical reactions for a model of protein production in which mRNA is produced from a source, protein is produced from mRNA and both protein and mRNA are degraded to a sink.

(a) Write down the chemical master equation for the probability *p*(*N*_{P},*N*_{M},*t*) of having *N*_{P} protein molecules and *N*_{M} mRNA molecules at time *t*, for this model, supposing that the rate of mRNA production (Equation 15) is *k*, the rate of mRNA degradation (Equation 16) is γ, the protein production per mRNA molecule (Equation 17) is *s*, and the rate of protein degradation (Equation 18) is μ.

(b) Explain why the steady-state solution of this master equation must obey the relation

Hint: you might want to refer to Equation 9 of the lecture.

2. Consider a Poisson process in which an mRNA molecule is produced

Let us suppose we are able to see the production of mRNA molecules, molecule by molecule, in time, as in Yu et al’s experiments (Slide 9 of the lecture). We start watching at time *t* = 0.

(a) What is the probability of an mRNA being produced in any very short time interval τ to τ+ dτ?

(b) What is the probability that no mRNA is produced in the interval τ to τ+ dτ?

(c) What is the probability that nothing happens for *N* time intervals, each of length dτ, then the first mRNA is produced?

(d) Explain based on the previous parts of the question why the distribution of waiting times between events for a Poisson process is

*p*(τ) = *ke*^{-kτ}

### Answers

1. (a)

(b)If we sum over all the possible values of *N*_{P} we obtain the probability of finding *N*_{M} mRNA molecules in the cell, regardless of the number of protein molecules. But we know the mRNA is produced and degraded in a simple Poisson birth-death process, and the steady-state solution of the chemical master equation for this is given by Equation 9 of the lecture notes, with k and γ as the birth and death rates.

2.

(a) *k* d*t*

(b) 1- *k* d*t*

(c) [1 - *k*dτ]^{N}*k*dτ

(d) Divide the waiting time *t* up into *N* intervals of length dτ = τ/*N*. Then we can use the result of part (c). Taking the limit of large *N* we arrive at the exponential wait time distribution

*p*(τ)dτ = *ke*^{-kτ}dτ