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ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2016/6/c/solution

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Past exam of the biological physics course of the University of Oxford / 2016 / 6 / c / Solution by

Codex  0 2026-10-25

For independent available acceptors at equal distance, the multiple-acceptor FRET efficiency follows by adding the

n

parallel exit rates. Each contributes the same rate

k_{T}

, so the total rate is

n k_{T}

, rather than

n

times the single-pair transfer probability. Hence

E_{n} = \frac{n k _{T}}{k _{D} + n k _{T}}, \frac{k _{T}}{k _{D}} = \frac{E}{1 - E}, E_{n} = \frac{n E}{1 + ( n - 1 ) E} .

Equivalently,

E_{n} = n / [n + (R / R_{0})^{6}]

. Inverting gives

R = R_{0} [\frac{n ( 1 - E _{n} )}{E _{n}}]^{1/6} .

With

n = 3

and

E_{n} = 0.75

, the bracket is exactly one and

R = R_{0} = 6.5

nm. Each isolated pair would have efficiency

1/2

; the three transfer rates together give

3/ (1 + 3) = 3/4

. Multiplying

1/2

by three would violate the probability bound and is not a valid calculation.

This model assumes one donor excitation, identical pairwise transfer rates, no coherent coupling between acceptors, no acceptor saturation and no change in donor photophysics or orientation on adding labels. Different separations instead require summing

k_{T, j}

individually. Acceptor-to-acceptor transfer can alter which acceptor emits; it does not automatically change the donor's first-transfer probability when the donor exit rates remain those specified.

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