Past exam of the biological physics course of the University of Oxford / 2022 / 4

Detailed balance means that at thermodynamic equilibrium each elementary transition and its reverse carry equal probability flux. For normalized state probabilities $A_{\mathrm{eq}}$ and $B_{\mathrm{eq}}$,With the printed convention $\Delta G_{AB}=G_A-G_B$, the sign is thereforeThe $G$ values are free energies per system, including internal degeneracies; for molar Gibbs free energies, replace $k_{B}T$ by $RT$. If B has lower free energy, its forward-to-reverse rate ratio exceeds one, as it should.

For constant rates, conservation gives $A=1-B$, so the master equation is $\dot{B}=k_1(1-B)-k_{2}B$. Its steady state is $B_0=k_1/(k_1+k_2)$, and separation of the decaying deviation yieldsfor the specified initial state A. The relaxation time constant is $(k_1+k_2{)}^{-1}$.

Take $v>0$ and the displayed potential energy to be the position-dependent mechanical free energy $G(x)$ of a bound motor. On the descending ramp $G^{\prime }=-G_0/x_1$, and on the ascending ramp $G^{\prime }=G_0/(P-x_1)$. Thus the force conditional on being bound isIt is periodic, piecewise constant and undefined at idealized sharp corners; the corners have zero measure in averages. An unbound motor has zero force.

A cohort transported through position $x$ obeys $d/dt=vd/dx$. Its bound probability therefore satisfiesDefine $B_0=k_1/(k_1+k_2)$, $\alpha =(k_1+k_2)/v$, $\beta =k_2/v$, $a=e^{-\alpha x_1}$ and $q=e^{-\beta (P-x_1)}$. The exact periodic steady state includes a small carry-over $b_{\mathrm{in}}=B(0)=B(P)$:These expressions follow by solving each first-order ordinary differential equation, requiring continuity at $x_1$, and imposing periodicity.

The stipulated $v\ll k_2(P-x_1)$ implies $q\ll 1$, so the incoming bound fraction is exponentially small. Dropping it gives the simpler intended formsThe occupancy rises towards $B_0$ on the binding interval, then decays over length $v/k_2$ outside it. The mean force per member of the full ensemble at a given position is $B(x)F_{\mathrm{bound}}(x)$, distinct from the force per bound motor. The right panels of c show both the occupancy and the conditional force.

For a uniform distribution of phases along one period, average the ensemble force $B{F}_{\mathrm{bound}}$:Using the negligible-carry-over form in b givesConsequentlywhere $\boxed{B_0=k_1/(k_1+k_2),\alpha =(k_1+k_2)/v,\beta =k_2/v}$. This is the printed expression to the accuracy justified by the velocity condition.

It is not an exact identity for the periodic problem at arbitrary $v$. Keeping $b_{\mathrm{in}}$ instead givesFor a concrete counterexample outside the small-velocity approximation, choose $k_1=k_2=P=G_0=v=1$ and $x_1=1/2$. Then the exact periodic value is about $+0.01976$, whereas the printed reset expression is about $-0.06478$. The discrepancy is due to nonzero incoming occupancy, not an integration sign change.

As $v\to 0^{+}$, both $\alpha$ and $\beta$ diverge, and both formulas giveThe bound population occupies the downhill binding region, while occupancy on the uphill region is confined to a vanishing boundary layer. A nonzero limiting stall force is compatible with zero mechanical power $v⟨F⟩$. The spatially restricted binding and uniform unbinding do not obey local detailed balance for this sawtooth free energy: a passive equilibrium model would require $k_1(x)/k_2(x)$ to reflect the local energy difference and not vanish on an interval of finite energy. This is a driven or prescribed-kinetics motor model; the equilibrium relation in a cannot be imposed on it without an energy supply and additional states.