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I'm trying to calculate how many photons will a GFP emit per second. The calculation seem easy just multiply the number of absorbed photons by the quantum yield. But how can I obtain the number of absorbed photons?

I know this would be directly related to the absorbance of the protein and the light intensity.

Is there an equation that relates both parameters to obtain the number of photons absorbed?

GFP absorbance at 488 nm:

$A_{488}=5.6\times10^4M^{-1}cm^{-1}$.

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  • $\begingroup$ This paper is behind a paywall, but the supplementary material is available for free. In the last paragraph of page S-3 they seem to be calculating what you're after, referring to three different papers as their sources. Goes beyond my (bio-)physics knowledge to comment on the accuracy though. $\endgroup$
    – gaspanic
    Apr 15, 2022 at 21:25

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Since the quantum yield isn't the main focus here, I'll just take the Wikipedia figure of 0.79 photons emitted / photons absorbed

The rest can be worked out from first principles. The absorbance represents the number of log10 units by which light is reduced, but its units are designed to confuse: 5.6E4 (L/mol) / cm = 5.6E7 cm^3/mol / cm = 5.6E7 cm^2/mol GFP.* This represents the area you can paint with 1 mol of GFP to reduce light passing through by 90%. But let's substitute the value of a mole, to get 9.3E-17 cm^2 = 0.93 A^2 per GFP. That chromophore casts a nice little shadow.

But our units aren't giving us photons - why not? Well, the absorbance and fluorescence is proportional to the illumination. So we need some notion of the amount of light, preferably something like irradiance (W/m^2). The first figure I saw on a search was about 300 kW/m^2; you'll need to decide on your own. So take 9.3E-21 m^2 * 300 kW/m^2 = 2.8E-15 W absorbed per GFP assuming this 90% absorption level.**

Now we still have to figure out the joules per photon. The frequency of this light is (3.0E8 m/s)/488 nm = 6.1E14 / s = 610 THz (hmm, maybe a little more cyan than green). We take that times Planck's constant = 6.63E-34 Js and get 4E-19 J/photon. Divide 2.8E-15 (J/s)/GFP by that and we have 7000 photons per second passing through GFP, of which we're supposing 90% are absorbed. And then I have to also multiply by 0.79 quantum efficiency. Taking those together it is approximately 5000 photons

I encourage you to rework this since there's some chance of foul-ups here, and remember to get your own numbers for the microscope irradiance.

*Remember absorbance is -log10(transmittance), so an absorption of 1 means 90% of the light is blocked. So the 56,000 there indicates that 1 mole of GFP can cut the light by 90% ... 56,000 times over again. Or we can divvy it up into 56,000 different square centimeters and reduce light in all of them by 90%.

** 90% of this will be absorbed at a painted surface, such as a slide. In a cuvette the energy absorbed would be much less because most of the GFP is in shadow, while spreading the GFP out further would increase the photons absorbed a bit because the rear part of the 90% is still in shadow. I think 90% absorption isn't tremendously unreasonable to use since people will try to arrange for as bright a signal as they reasonably can, which may be localized to small features on a slide; but for example absorbing 68% of photons (reducing transmittance by the square root of 10 with an absorbance of 0.5) would cover twice the area, collecting 36% more photons total. The limit at 0% would be ln(10) = 2.3 times higher than the figure at 90% absorption, but you can't see GFP if it's spread out to 0% absorption.

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  • $\begingroup$ Hi @Mike_Serfas I think you nailed it. Just as a favor to people that will read this in the future could you please explain in more detail the way you get the per molecule absorption (specially how you go from absorbance to area). Also could you change the format so the equations are displayed in a diferent row? $\endgroup$
    – BPinto
    Apr 18, 2022 at 17:23
  • $\begingroup$ @BPinto - thanks for making me look at this again. Spreading out the GFP more varies the amount of light absorbed by more than I was thinking, though I should have known, because at 90% much of it is in shadow. I'm not sure it's wrong to assume 90%, but there's no two sigfigs to be had here. As for the equations ... well, I don't enjoy trying to code those formats when the text-based version seems so practical. $\endgroup$ Apr 18, 2022 at 21:03

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