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I measure the intensity and polarization of light emitted from a blood sample that contains excited fluorescein (experiment known as fluorescence polarization assay). The excitation is done, as usual, with vertically polarized light.

In case the sample has certain macromolecules of interest (an antibody, in this case), the fluorescein links to the much-larger macromolecules, yielding a higher polarization factor, giving a positive- or "sick"- result. Alternatively, if there is no antibody, the excited fluorescein randomly rotates and then emits light, and the polarization factor is smaller giving a negative-or "healthy"- result.

The question:

I find that the light intensity (in both vertical and perpendicular polarizations) decreases for the positive case. Why is that so? Could it be that a non-radiative process like FRET acts stronger for fluorescein linked to macromolecules than for free fluorescein?

EDIT 1: I have used two different intensity detection systems, PMTs and SiPMs, and got similar behaviours.

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    $\begingroup$ There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission. When bound to the protein fluorphores are in a different environment and this can affect their fluorescence yield, ANS (anilino naphthalene sulphonate) is a good example. The same is probably happening in your case, you know the motion is restricted as the polarisation is greater so there has to be some interaction. $\endgroup$
    – porphyrin
    May 19, 2021 at 19:36
  • $\begingroup$ Think that many good fluorescent molecules cease to be so when in solid state or strongly interacting with a solvent. Non radiative channel are opened. To me, what you've described and taking aside fine details, should be a quite general behavior. In fact looking at polarisation and not only intensity should be just for better sensitivity. $\endgroup$
    – Alchimista
    May 20, 2021 at 13:26
  • $\begingroup$ @Alchimista Thank you. The fluorescein-solvent interaction for sure gets stronger with the sample positivity - the more antibodies, the more linking. I am not sure what are these "non-radiative channels"? $\endgroup$
    – nicolas005
    May 20, 2021 at 18:06
  • $\begingroup$ @nicolas005 this is far more general than in your biological oriented experiments. An excited species can collide and be de-excited. Now think of something adsorbed or solvated or anyway interacting. Decay can happen through vibrational levels instead that radiatively. At the end the energy goes off as heat. $\endgroup$
    – Alchimista
    May 21, 2021 at 13:50
  • $\begingroup$ Specifically to the point about FRET see below. I upvote that. $\endgroup$
    – Alchimista
    May 21, 2021 at 13:52

1 Answer 1

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There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission. When bound to the protein fluorphores are in a different environment (e.g. pH change, hydrophobic pocket, tight binding, polar/non-polar nearby residues etc.) and this can affect their fluorescence yield and lifetime up or down, ANS (anilino naphthalene sulphonate) is a good example. The same type of thing is probably happening in your case, you know the motion is restricted as the polarisation is greater so there has to be some interaction. Fluorescein emits at a relatively long wavelength so aromatic protein residues cannot act via FRET as they absorb in the UV. You would need a chromophore (such as heam/chlorin/porphyrin) that absorbs in the green to red region to be a FRET acceptor from the fluorescein and so reduce its intensity.

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  • $\begingroup$ Good to know that FRET should not be the cause of the intensity decay. Thank you for the answer. $\endgroup$
    – nicolas005
    May 20, 2021 at 18:07

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