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FLIM-FRET

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FLIM-FRET

Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

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

Tel: +32 (0)9 264.80.78

Fax: +32 (0)9 264.81.89

E-mail: [hidden email]

 
David Stanek-3 David Stanek-3
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Re: FLIM-FRET

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
Tel.: +420-296443118
Fax: +420 224 310 955


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

Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

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

Tel: +32 (0)9 264.80.78

Fax: +32 (0)9 264.81.89

E-mail: [hidden email]

 

Mario-2 Mario-2
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Re: FLIM-FRET

Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal Re: FLIM-FRET
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
Tel.: +420-296443118
Fax: +420 224 310 955
email: [hidden email]
web: www.img.cas.cz


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

Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
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
Tel: +32 (0)9 264.80.78
Fax: +32 (0)9 264.81.89
E-mail: [hidden email]


--
________________________________________________________________________________
Mario M. Moronne, Ph.D.

ph  (510) 528-8076
cell (510) 367-8497
Kevin Braeckmans Kevin Braeckmans
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Re: FLIM-FRET

Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal Re: FLIM-FRET

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 .

 

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

Tel.: +420-296443118

Fax: +420 224 310 955

email: [hidden email]

web: www.img.cas.cz

 

 

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



Search the CONFOCAL archive at http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

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

Tel: +32 (0)9 264.80.78

Fax: +32 (0)9 264.81.89

E-mail: [hidden email]

 

 

--

________________________________________________________________________________

Mario M. Moronne, Ph.D.

 

ph  (510) 528-8076

cell (510) 367-8497

 
Aryeh Weiss Aryeh Weiss
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Re: FLIM-FRET

Search the CONFOCAL archive at
http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

Kevin Braeckmans wrote:

> Search the CONFOCAL archive at
> http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
>
> 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.
>
>  

Think of it this way -- that subpopulation that is not FRETting does not consist
of a fixed population of molecules. Rather -- it consists of those molecules
that managed to emit their photon before FRET could "steal" the energy from
them. So what you see are those potential donors that emit quickly -- the others
have lost their energy to the acceptors and will never be seen in the donor channel.

--aryeh
--
Aryeh Weiss
School of Engineering
Bar Ilan University
Ramat Gan 52900 Israel

Ph:  972-3-5317638
FAX: 972-3-7384050
Kevin Braeckmans Kevin Braeckmans
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Re: FLIM-FRET

Search the CONFOCAL archive at
http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

Dear Aryeh,

That makes sense, thank you.

That also provides me with an answer to my second original question: with
increasing FRET efficiency, the precision of the FLIM measurements will
decrease since less and less photons coming from the FRETting donor
molecules will be available (for a fixed observation time). It also follows
that FLIM cannot measure (the theoretical case of) 100% FRET efficiency.

Thanks to all who have responded. As always, the combined knowledge of this
list proves to be an invaluable resource.

Kind regards,

Kevin


Kevin Braeckmans, Ph.D.
Lab. General Biochemistry and Physical Pharmacy
Ghent University
Harelbekestraat 72
9000 Ghent
Belgium
Tel: +32 (0)9 264.80.78
Fax: +32 (0)9 264.81.89

> -----Oorspronkelijk bericht-----
> Van: Confocal Microscopy List [mailto:[hidden email]]
> Namens Aryeh Weiss
> Verzonden: donderdag 29 november 2007 16:34
> Aan: [hidden email]
> Onderwerp: Re: FLIM-FRET
>
> Search the CONFOCAL archive at
> http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
>
> Kevin Braeckmans wrote:
> > Search the CONFOCAL archive at
> > http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
> >
> > 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.
> >
> >
>
> Think of it this way -- that subpopulation that is not FRETting does
> not consist
> of a fixed population of molecules. Rather -- it consists of those
> molecules
> that managed to emit their photon before FRET could "steal" the energy
> from
> them. So what you see are those potential donors that emit quickly --
> the others
> have lost their energy to the acceptors and will never be seen in the
> donor channel.
>
> --aryeh
> --
> Aryeh Weiss
> School of Engineering
> Bar Ilan University
> Ramat Gan 52900 Israel
>
> Ph:  972-3-5317638
> FAX: 972-3-7384050
Aryeh Weiss Aryeh Weiss
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Re: FLIM-FRET

Search the CONFOCAL archive at
http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal

Kevin Braeckmans wrote:

> Search the CONFOCAL archive at
> http://listserv.acsu.buffalo.edu/cgi-bin/wa?S1=confocal
>
> Dear Aryeh,
>
> That makes sense, thank you.
>
> That also provides me with an answer to my second original question: with
> increasing FRET efficiency, the precision of the FLIM measurements will
> decrease since less and less photons coming from the FRETting donor
> molecules will be available (for a fixed observation time). It also follows
> that FLIM cannot measure (the theoretical case of) 100% FRET efficiency.
>

In the event of 100% FRET efficiency (I should only be so lucky...) you might
detect it due to a change in acceptor lifetime. The acceptor emission will be
non-exponential, but on average, FRET excited acceptors should have longer
emission times than directly excited acceptors.

As for acceptor photobleaching, 100% efficiency would be great, because the
donor will be zero prebleach, and will go up to whatever postbleach. The limit
will be how precisely you can measure the zero.

--aryeh
--
Aryeh Weiss
School of Engineering
Bar Ilan University
Ramat Gan 52900 Israel

Ph:  972-3-5317638
FAX: 972-3-7384050
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