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 Incoming links: Two-emitter photobleaching completion time

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S12 / 2 / 3 / Solution 2026-10-25

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Under the simplest interpretation, each spot contains two equally bright, independently bleaching green fluorescent protein emitters. Losing one halves the fluorescence and losing the other removes it. A stable protein dimer carrying one active label per subunit is consistent with this result. The conclusion is two active emitters in the observed spot; it does not by itself rule out unresolved coincident proteins, incomplete fluorescent maturation or blinking. Permanent membrane binding makes loss by dissociation less plausible, but lateral motion out of the observation region would still need experimental control.^[ref]

Let their independent photobleaching times be

T_{A}, T_{B} \sim Exp (k_{b})

. Then

t_{1} = min (T_{A}, T_{B})

has rate

2 k_{b}

, while the memoryless remaining lifetime

t_{2} - t_{1}

has rate

k_{b}

and mean

1/ k_{b}

, matching the stated single-emitter calibration. The two-emitter photobleaching completion time is the maximum, not one single exponential and not a sum of two equal-rate exponential variables. Its cumulative distribution function is

P (t_{2} \leq t) = P (T_{A} \leq t, T_{B} \leq t) = (1 - e^{- k_{b} t})^{2} (t \geq 0) .

(1)

Differentiating gives its probability density function:

f_{t_{2}} (t) = 2 k_{b} e^{- k_{b} t} (1 - e^{- k_{b} t}) = 2 k_{b} (e^{- k_{b} t} - e^{- 2 k_{b} t}), t \geq 0,

(2)

with zero density for

t < 0

. It integrates to one, and its mean is

1/ (2 k_{b}) + 1/ k_{b} = 3/ (2 k_{b})

. This uses two active fluorophores at the start of observation, constant illumination and independent irreversible bleaching with the calibrated rate.

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