Past exam of the biological physics course of the University of Oxford / 2019 / 4

At equilibrium, detailed balance requires zero net probability current along each reversible link: $k_f{p}_A^{\mathrm{eq}}=k_b{p}_B^{\mathrm{eq}}$. The equilibrium probabilities satisfy $p_i^{\mathrm{eq}}\propto e^{-G_i/(k_{B}T)}$, where each state free energy includes its internal degeneracy. ThereforeIn the source sketch B is lower, and the labelled vertical drop is naturally $\Delta G_{AB}=G_A-G_B>0$. With that drop convention the ratio is $\boxed{k_f/k_b=e^{\Delta G_{AB}/(k_{B}T)}}$. If instead $\Delta G_{AB}$ means the conventional reaction change $G_B-G_A$, the exponent has the minus sign. For example a drop of $2k_{B}T$ gives a forward/backward ratio $e^2$, not $e^{-2}$.

Write $B(t)$ for the probability of state B, so $A(t)=1-B(t)$. Gain minus loss givesWith $B(0)=0$, the solution isIt rises monotonically from zero, initially with slope $k_f$, towards $B_{\mathrm{eq}}=k_f/(k_f+k_b)$ on time scale $(k_f+k_b{)}^{-1}$. The first panel of the original figure below shows this shape for illustrative equilibrium probability $0.8$.

The other equilibrium probability is $A_{\mathrm{eq}}=k_b/(k_f+k_b)$, and their ratio is $k_f/k_b$. Applying detailed balance from part (a) giveswhich is precisely the two-state Boltzmann distribution. It fixes the ratio but not the sum of the rates, so equilibrium measurements alone do not determine the relaxation time.

Take forward as increasing $x$, and specify the sinusoidal amplitude convention byThus the unforced barrier is $2A$, not $A$. Let $s=Fd/(2\pi A)=0.7051$ and $\theta =\arcsin s$. Stationary points exist because $F<2\pi A/d=7.09\mathrm{pN}$. A minimum is at $x_m=d\theta /(2\pi )$ and its forward saddle at $x_{+}=d(\pi -\theta )/(2\pi )$. The distances and activation barriers areUse an Arrhenius estimate with the supplied distance-squared waiting-time scaling, $\tau \propto {\ell }^2{e}^{\Delta U/(k_{B}T)}$. Calibrate at $F=0$ with ${\ell }_0=d/2$ and rate $k_0=1{\mathrm{s}}^{-1}$ in each direction. At the explicit room-temperature estimate $T=300\mathrm{K}$, $k_{B}T=4.142\mathrm{pN}\mathrm{nm}$, givingApplying the same distance rule separately backwards gives $k_{-}\simeq 0.00241{\mathrm{s}}^{-1}$ and hence the literal scaling-model estimateThermodynamic qualification: for isothermal diffusion in this translated tilted potential, adjacent wells differ by $Fd=35\mathrm{pN}\mathrm{nm}$, so local detailed balance instead requires $k_{+}/k_{-}=e^{Fd/(k_{B}T)}$. The two separate distance prefactors introduce an extra factor $({\ell }_{-}/{\ell }_{+}{)}^2\simeq 8.91$, inconsistent with that requirement. Keeping the forward estimate while enforcing the proper ratio givesBoth interpretations are shown because the supplied distance heuristic is not, by itself, a thermodynamically consistent prescription for both directional prefactors. Additional hidden chemical driving would require its own affinity, not an unexplained prefactor asymmetry. A common transition-state prefactor, as in a consistent Kramers model, preserves the Boltzmann rate ratio. Moreover the forward barrier is only about $0.58k_{B}T$, so rare-escape exponential dwell kinetics and the numerical rate should be regarded as rough estimates. If “amplitude” meant the entire unforced barrier height, the maximum restoring force would be only $\pi A/d=3.55\mathrm{pN}$ and no dwell wells would survive at $5\mathrm{pN}$; the amplitude convention is therefore essential.

Assume identical dwell states and memoryless forward/backward exits with constant rates $k_{+}$ and $k_{-}$; put $K=k_{+}+k_{-}$. The probability density for the first exit to be forward at $t_1$ is $k_{+}{e}^{-K{t}_1}$. For the very next exit also to be forward at time $t$, multiply by $k_{+}{e}^{-K(t-t_1)}$ and integrate over the unobserved first time:This density integrates to $(k_{+}/K{)}^2$, the probability that both directions are forward. Conditional on that event, the normalized total-time density is the Erlang distributionIts mean is $2/K$, so the most probable time is not the mean. Since both backward-rate conventions in part (c) have $k_{-}\ll k_{+}$, neglecting backwards steps gives the usual approximation $p(t)=k_{+}^{2}t{e}^{-k_{+}t}$ and mode $1/k_{+}$. If “two forward steps” instead permits intervening backwards events, thinning independent Poisson exits gives that latter rate-$k_{+}$ distribution directly; it describes a different conditioning. The low barrier noted in part (c) limits the physical accuracy of all these memoryless dwell assumptions.