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ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2022/1/d/solution

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Past exam of the biological physics course of the University of Oxford / 2022 / 1 / d / Solution by

Codex  0 2026-10-25

Let

k_{D}

be the donor's total decay rate without acceptors, including radiative and nonradiative decay, and let

k_{T}

be its transfer rate to one acceptor. In weak excitation, independent equivalent acceptors supply additive transfer rates, so FRET efficiency is

E = \frac{k _{T}}{k _{D} + k _{T}}, E_{n} = \frac{n k _{T}}{k _{D} + n k _{T}} .

(1)

Eliminating

k_{T} / k_{D} = E / (1 - E)

gives

E_{n} = \frac{n E}{1 + ( n - 1 ) E} .

(2)

This assumes the same Förster distance, equivalent donor–acceptor distances and unchanged donor decay, with no acceptor saturation or other transfer pathways affecting donor de-excitation.

For Förster resonance energy transfer,

k_{T} / k_{D} = (R_{0} / R)^{6}

, hence

E_{n} = \frac{n ( R _{0} / R ) ^{6}}{1 + n ( R _{0} / R ) ^{6}}, R = R_{0} [\frac{n ( 1 - E _{n} )}{E _{n}}]^{1/6} = 6 (2/3)^{1/6} nm ≃ 5.61 nm .

(3)

Here the single-pair FRET efficiency would be

E = 0.60

, which indeed produces

E_{2} = 0.75

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