Past exam of the biological physics course of the University of Oxford / 2015 / 5 / c / Solution

Interpret the irreversible limit as a large favorable drop for each forward step, so backward transitions during the observation can be neglected. Let the distinct forward rates be $k_1,k_2>0$. The first waiting time has probability density $k_1{e}^{-k_1{t}_1}$; conditional on its completion, the second has independent probability density $k_2{e}^{-k_2(t-t_1)}$. Integrating over the first completion time gives the convolutionHenceThe expression is symmetric under interchanging the rates, positive for either ordering, and integrates to one. Reverse rate constants do not enter this approximation; merely making all four constants different would not justify eliminating them without the downhill condition.