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ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2016/3/c/solution

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Past exam of the biological physics course of the University of Oxford / 2016 / 3 / c / Solution by

Codex  0 2026-10-25

At

T = 310.15

K,

RT / F = 26.72666

mV. Always retain the valence sign. For the two monovalent positive ions,

E_{K} = 26.72666 ln (4/155) = - 97.74 mV, E_{Na} = 26.72666 ln (145/12) = + 66.60 mV .

z = 2

and the specified voltage is positive:

c_{Ca, in} = 1.5 exp [- 2 (130) /26.72666] mM = 8.938 \times 1 0^{- 5} mM ≃ 89.4 nM .

z = - 1

reverses the sign in the Nernst equation:

- 90 = - 26.72666 ln (c_{Cl, out} /4), c_{Cl, out} = 116.0 mM .

These values use thermodynamic activities approximated by the given concentrations. Calcium is the free concentration relevant to its electrochemical potential, rather than total bound plus free intracellular calcium. Ratios use matching concentration units, so a ratio of millimolar values is already dimensionless. Failure to include the factor of two for calcium or the negative charge of chloride would change the answer substantially.

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