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Past exam of the biological physics course of the University of Oxford

2019

Write $B(t)$ for the <probability> of state B, so $A(t)=1-B(t)$. Gain minus loss gives
$$
\dot B=k_f(1-B)-k_bB=k_f-(k_f+k_b)B.
$$
With $B(0)=0$, the solution is
$$
\boxed{B(t)=\frac{k_f}{k_f+k_b}\left[1-e^{-(k_f+k_b)t}\right].}
$$
It rises monotonically from zero, initially with slope $k_f$, towards $B_{\rm eq}=k_f/(k_f+k_b)$ 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 $A_{\rm eq}=k_b/(k_f+k_b)$, and their ratio is $k_f/k_b$. Applying <detailed balance> from part (a) gives
$$
\frac{B_{\rm eq}}{A_{\rm eq}}=e^{-(G_B-G_A)/(k_BT)},
$$
which 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.

\Image[/past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2019-motor-and-polymer.png]
{title=Two-state equilibration, a tilted periodic motor, sequential-step times and freely jointed chain elasticity}
{description=Original model plots, not reconstructions of the source graphs. The motor potential uses sinusoidal coefficient $A=7.9\,\mathrm{pN\,nm}$ and forward tilt $F=5\,\mathrm{pN}$; other panels use dimensionless axes.}

Solution

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

Past exam of the biological physics course of the University of Oxford / 2019 / 5 / c / Solution

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

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