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

At thermal equilibrium, detailed balance equates the opposing transition fluxes: $k_f{p}_A^{\mathrm{eq}}=k_b{p}_B^{\mathrm{eq}}$. Define the state free energy difference explicitly by $\Delta G_{AB}=G_B-G_A$. The equilibrium Boltzmann distribution then givesAny state degeneracy is included in the state free energies. Detailed balance is stronger than simply requiring a time-independent total occupancy: it equates each pair of equilibrium transition fluxes.

Source illustration TODO: the source's reaction coordinate diagram is missing. The definition above fixes the sign independently; no barrier heights or ordering of drawn energy levels are inferred. Below, a large forward free-energy drop means $G_A-G_B\gg k_{B}T$. If instead $\Delta G_{AB}=G_B-G_A\gg k_{B}T$, forward transitions are unfavorable: for example $\Delta G_{AB}=10k_{B}T$ gives $k_f/k_b=e^{-10}$, so neglecting reverse transitions would be wrong.

Conservation of total probability gives $A(t)=1-B(t)$. The two-state master equation is thereforeSolving this linear ordinary differential equation givesThe middle panel in d sketches this solution for the explicitly illustrative ratio $k_f/(k_f+k_b)=0.6$. The curve starts at zero with slope $k_f$, is increasing and concave down, and approaches the horizontal limit $B_{\infty }=k_f/(k_f+k_b)$ on time constant $(k_f+k_b{)}^{-1}$. Its initial tangent meets the asymptote at that time constant. The complementary limit is $A_{\infty }=k_b/(k_f+k_b)$, so $B_{\infty }/A_{\infty }=k_f/k_b=e^{-\Delta G_{AB}/(k_{B}T)}$, the Boltzmann distribution ratio. These are occupancy probabilities; $B(t)$ is not the cumulative first-passage probability when reverse transitions remain possible.

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.

Taking $k_2\to k_1=k$ in the preceding probability density gives $p_2(t)=k^{2}t{e}^{-kt}$. Convolving once more with an independent exponential distribution givesMore generally, suppose $p_n(s)=k^n{s}^{n-1}{e}^{-ks}/(n-1)!$. ThenSince $p_1(t)=k{e}^{-kt}$, this proves the Erlang distribution formula by mathematical induction. Normalization follows either from the convolution or ${\int }_0^{\infty }{t}^{n-1}{e}^{-kt}dt=(n-1)!/k^n$.For $n>1$, differentiating $\log p_n$ gives $(n-1)/t-k=0$, and the derivative changes from positive to negative. For $n=1$ the maximum occurs at the boundary $t=0$. Source dimensional qualification: the printed expression without $dt$ is a probability density, with units of inverse time; it cannot itself be a dimensionless interval probability.

In the right panel of the original illustration in d, $p_2/k$ rises linearly near zero and peaks at $kt=1$ with height $1/e$. The curve $p_3/k$ rises quadratically and peaks later, at $kt=2$, with lower height $2e^{-2}$. Both have exponential tails, but the latter has the larger expected value $3/k$ and variance $3/k^2$, compared with $2/k$ and $2/k^2$. This distinguishes both the onset and the width of the requested sketches.