Thank you Mario and David for the clarifications. I am probably
overlooking something very trivial, but it is still not clear to me.
I totally understand that the decay of the total number of
excited molecules will be faster if there is an additional decaying pathway,
such as FRET. My question/problem, however, is how this is measured in a FLIM
experiment? As far as I understand, the photon signal that is measured is in
fact coming from that subpopulation of excited donor molecules which are not
FRETting. So in my simple understanding I would say that it is till k_f, i.e.
relaxation through the fluorescence decay pathway, which is selectively being measured
and not the decay rate of the entire population of excited molecules.
What am I missing here?
Thanks for your patience,
Kind regards,
Kevin
Van: Confocal
Microscopy List [mailto:[hidden email]] Namens Mario
Moronne
Verzonden: dinsdag 27 november 2007 0:51
Aan: [hidden email]
Onderwerp: Re: FLIM-FRET
Search the CONFOCAL archive at
http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
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 .