 Solution

ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2019/4/b/solution

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_{b} B = k_{f} - (k_{f} + k_{b}) B .

With

B (0) = 0

, the solution is

B (t) = \frac{k _{f}}{k _{f} + k _{b}} [1 - e^{- (k_{f} + k_{b}) t}] .

It rises monotonically from zero, initially with slope

k_{f}

, towards

B_{eq} = k_{f} / (k_{f} + k_{b})

(k_{f} + k_{b})^{- 1}

. The first panel of the original figure below shows this shape for illustrative equilibrium probability

0.8

A_{eq} = k_{b} / (k_{f} + k_{b})

, and their ratio is

k_{f} / k_{b}

. Applying detailed balance from part (a) gives

\frac{B _{eq}}{A _{eq}} = e^{- (G_{B} - G_{A}) / (k_{B} T)},

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.

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