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

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.