Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S12 / 3

Cells store usable free energy in chemical disequilibria: reduced carbon compounds, redox carriers such as NADH, ATP relative to ADP and inorganic phosphate, ion concentration gradients and membrane voltage. These reservoirs are maintained away from equilibrium; the useful quantity is the chemical or electrochemical potential difference, not simply a bond's isolated energy. Some ATP-consuming reactions can be driven because a coupled overall reaction decreases Gibbs free energy, even when the uncoupled synthesis would not.

Oxidation of nutrients transfers reducing equivalents to electron carriers and then to an electron-transport chain. Coupled proton pumping converts part of the redox free-energy drop into a protonmotive force. ATP synthase can convert that force into ATP chemical potential, and ATP hydrolysis drives biosynthesis, transport and mechanical motion. Substrate-level phosphorylation is another ATP source; photosynthetic cells can initially supply the disequilibrium with absorbed light. These are flows with unavoidable dissipation and heat release, not independent energy-creation processes. A reservoir is depleted unless metabolism replenishes it.

The cylindrical part has length 2 micrometres and radius 0.5 micrometres. Including both hemispherical ends gives area $A=2\pi rL+4\pi r^2=3\pi \times 10^{-12}$ m${}^2$. The membrane capacitance is $C={ϵ}_0{ϵ}_{r}A/d\simeq 4.17\times 10^{-14}$ F. Define the signed inward electrochemical drop per unit positive charge asThere is a factor-of-two ambiguity in the wording of the charge data. If the stated excess means the inner capacitor charge $Q=-40000e$, then $\Delta \psi =Q/C=-0.154$ V and $\Delta p\simeq -0.183$ V at 298 K. If “inside minus outside” is taken literally as $Q_{\mathrm{in}}-Q_{\mathrm{out}}=-40000e$ for opposite capacitor charges, the inner plate has $-20000e$, giving $\Delta \psi =-0.0768$ V and $\Delta p\simeq -0.106$ V. The latter is the literal charge-difference interpretation; the former is the commonly intended excess-charge convention. In either case inward proton motion is favorable, and the positive available driving magnitude is $-\Delta p$.

An electron-transport chain couples favorable redox transfers to pumping protons outward, producing the voltage and concentration terms of the protonmotive force. The membrane's low proton permeability prevents immediate equilibration. The membrane-embedded $F_o$ part of ATP synthase provides a controlled return pathway: protonation and deprotonation of rotor sites couple proton flow to rotation of the c-ring and central shaft. The soluble $F_1$ catalytic head has sites that change affinity and conformation as this shaft turns. ADP and phosphate bind, ATP forms, and the conformational cycle permits ATP release; the mechanical coupling is essential because ATP synthesis is thermodynamically uphill under cellular concentrations.

If $n_H$ protons traverse the machine per ATP, reversibly available work requires $n_{H}e(-\Delta p)\ge \Delta G_{\mathrm{ATP}}$, with extra dissipation and transport costs in a real cell. Ring stoichiometry makes $n_H$ organism dependent, rather than a universal integer inferred from the membrane voltage alone. ATP synthase is reversible: sufficient ATP chemical potential can drive reverse rotation and proton pumping. Uncoupling agents dissipate the gradient without corresponding ATP production, illustrating that respiration alone does not guarantee phosphorylation.^[ref]