So this question is part math, part biology, but I'm asking about the biology part here (or maybe the chemistry part). Basically I'm asking about my attempt to calculate the possible brightness per unit mass of GFP, and I'm asking if it makes sense. Here's what I've done: (I'm ignoring the quantum yield which should be 0.8):

GFP molar mass=27,000 amus (27 kDa)

1 mole weighs 27 kilograms

therefore 1 gram=(6.02*10^23)/27000=2*10^19 molecules of GFP

Assumption that I'm asking about; I'm assuming in this calculation that GFP releases light from one excited electron (roughly 2.5 electron volts). How many electrons are excited in GFP fluorescence?

(2*10^19)/(6*10^18), (the number of electrons in a Coulomb/amp)=3.3 amps

Brightness=2.5*3.3 =8.25 joules per gram

From Researchgate: " The time constant of the fast, pH-dependent chemical process decreases with pH from 300 microseconds at pH 7 to 45 microseconds at pH 5..." Given that's the case, that's roughly 3000 emissions (or 3333) per second (at 7 Ph), so 1 gram of GFP would have a maximum brightness of 24,750 watts, if there was enough UV light, or 25 watts for a milligram.
That sounds plausible, although high. Does that make sense, or what part of this calculation was wrong?

  • $\begingroup$ According to jic.ac.uk/microscopy/more/T5_9.htm , only 80% of photons absorbed lead to photon emission. (It's "quantum efficiency".) In practice, numbers might be far lower for some other reasons. I recall that even one gram of eGFP was less bright than a 30 W fluorescent light bulb. I suspect that I could not excite all the GFP molecules as often as needed. They are quite sparse in solution, even at "high concentration". $\endgroup$
    – nvja
    Nov 21, 2016 at 5:44
  • $\begingroup$ But perhaps it could be brighter than that if the efficiency was maximized (i.e. the protein was spread out over a thin surface and completely concentrated, in theory, since it would have to be in some solution unless it was in powder form)? $\endgroup$
    – Tom
    Nov 21, 2016 at 5:49


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.