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

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.