 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s12/3/c/solution

 Top articles  Latest articles New article in topic

Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S12 / 3 / c / Solution by

Codex  0 2026-10-25

The observed

9 0^{\circ} + 3 0^{\circ}

substeps sum to

12 0^{\circ}

, consistent with three catalytic cycles per revolution of

F_{1}

-ATPase. Surface attachment fixes the

β

-subunit frame while the gold-labelled

γ

shaft reports relative rotation. A finite dwell before the smaller step indicates hidden chemical transitions rather than a completely smooth motor.

Suppose two irreversible independent first-order transitions, with rates

k_{a}

and

k_{b}

, must occur sequentially before the

3 0^{\circ}

step. Their waiting times

X_{a}

and

X_{b}

have exponential distributions, and

τ = X_{a} + X_{b}

. Convolution gives

P (τ) = \int_{0}^{τ} k_{a} e^{- k_{a} t} k_{b} e^{- k_{b} (τ - t)} d t = \frac{k _{a} k _{b}}{k _{a} - k _{b}} (e^{- k_{b} τ} - e^{- k_{a} τ}), τ \geq 0.

This probability density integrates to one and has

⟨ τ ⟩ = k_{a}^{- 1} + k_{b}^{- 1}

. For equal rates its continuous limit is

P (τ) = k^{2} τ e^{- k τ}

, not a singular distribution. In particular

P (0) = 0

distinguishes two resolved sequential waits from a single exponential wait. Hydrolysis and product release are plausible underlying chemical events, but the waiting-time fit alone neither assigns their order nor uniquely identifies which biochemical event causes each substep. Concentration dependence or independent labelling would be needed for that assignment.

 Read the full article

 New to topics? Read the docs here!