Re: FLIM-FRET

Posted by Mario-2 on
URL: http://confocal-microscopy-list.275.s1.nabble.com/FLIM-FRET-tp591029p591031.html

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Hi All, Kevin, David,

Just a little comment:

By convention, for non-FRETing molecules the number of Ca molecules will be depleted (ignoring excitation, etc.) as in the absence of an acceptor, i.e.,  dCa/dt will be proportional to -kr Ca plus other terms. kr in this context is a rate constant which is related to the probability k. For practical purposes when there are any significant number of molecules, Ca, we can apply a Poisson statistical model and leave behind probabilities derived from  binomial combinatorial probability formalism. This is the case for virtually all fluorescence microscopy systems, which makes life simpler and we can apply rate constants rather than probabilities.

Anyway, ignoring all other quenching terms, kr = 1/tau, where tau is the decay time, or in half life terms,  t(1/2)= 0.693147...*tau->

Ca(t1/2) = Ca(t=0)/2 = Ca(t=0)*exp[-t1/2/tau], or  ->

Ca(t1/2) = Ca(t=0) * exp[-0.693147....*tau/tau] ->

Ca (tau(1/2)) = Ca(t=0) * exp[-0.693147...] ->

Ca (tau(1/2)) = Ca(t=0) * 1/2

Getting back to the more important FRET Lifetime issue, we can add an additional decay term such that

dCa/dt = -(kr Ca + kfret Ca)

dCa/Ca = -(kr + kfret) dt. Rearranging, integrating, rearranging, exponentiating and we get

Ca(t) = Ca(0) exp[-(kr + kfret) t ]

From the last expression it can be seen from this very very simplified analysis that the decay constants are additive such that the overall rate constant will in fact increase when there is both donor to ground state photon emission as well as energy transfer via FRET. Thus, the apparent lifetime of the donor population will decrease faster assuming at least some FRET activity. As far as getting 100% FRET, that is not going to happen. The above equation would seem to imply that it could be possible if kfret >> kr but this is a trivial model that doesn't include all the "stuff" even disregarding bleaching, which is in fact another way of doing the measurement. There is always the problem of getting the perfect match between donor and acceptor and because true FRET is very distance and orientation sensitive, vibrational motion of the donor-acceptor spacing and the respective dipole axes means there will always be some spread in the transfer probability. If the pair get too close then other processes come into play such as intramolecular energy transfer, which is not FRET at all.

For the FRET-FLIM experts out there I apologize for my oversimplifications but maybe some others might like having some simple rate equations to help examine the issues if not derive the answers. I do not consider myself an expert but I have had some practice over the years with physics and acceptor-donor chemistry.

Happy to continue the discussion online or offline. Happy Holidays Everyone,

Mario M

Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal I had the exact same problem and was asking people around. Finally one physicist explained that to me. Here is as I (as a biologist) got his explanation.

Yes, you are right that you are counting only photons from molecules that are not FRETing. BUT: Number of emitted photons in any given time depends on number of fluorochromes in an excited state:

n=k.Ca

where n - number of photons;

k - probability that they leave the excited state;

Ca - number of activated molecules .

In "NO FRET" situation excited molecules can relax only (forget about bleaching and other stuff) through emission of a photon. But when their are FRETing they have two ways how to relax - either emit a photon or give energy to the FRET partner. That results in faster depletion of molecules in excited state which we can measure as "faster" decay. As I understand it, you actually don't change k (which has halftime hidden somewhere inside) but Ca, which as a result looks like you have faster decay.

Hope it helps

David

David Stanek, PhD.

Department of RNA biology

Institute of Molecular Genetics AS CR

Videnska 1083

142 20 Prague 4

Czech Republic

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On Nov 26, 2007, at 10:24 AM, Kevin Braeckmans wrote:

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Dear fellow microscopists,

I would like to ask a question about FLIM-FRET. Having no practical experience with this technique, I was reading some reviews on this topic where it is explained that the FRET efficiency is deduced from a reduction in the lifetime of the donor fluorophore. I just want to make sure that I understand correctly what is exactly being measured here.

Let's say we have a donor-acceptor pair having 50% FRET efficiency. This means that on multiple excitation events, there will be FRET in 50% of the cases, and 50% relaxation of the donor through other relaxation mechanisms, one of which is fluorescence decay. However, when FRET occurs, there is no emission of a donor fluorescence photon. So, when it is said that it is the donor lifetime which is measured in FLIM-FRET, it must be the lifetime measured from the donor fluorescence photons emitted when actually no FRET is taking place. Is that correct?

It this is so, it also follows that FLIM-FRET is necessarily incapable of measuring 100% FRET efficiency (because there will be no donor fluorescence photons emitted anymore). Or more generally, the FLIM-FRET accuracy will decrease with increasing FRET efficiency since less photons will be available. Is that correct?

Any explanation/input is much appreciated.

Thanks and kind regards,

Kevin

Kevin Braeckmans, Ph.D.

Lab. General Biochemistry & Physical Pharmacy

Ghent University

Harelbekestraat 72

9000 Ghent

Belgium

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