Assuming I have two FRET situations: (A) One donor 3 nm from acceptor; (B) Two donors both 3 nm from the acceptor (imagining acceptor in the center and two donors at two wings). How is the intensity difference of the acceptor while we excite the donor at the two situations?? Is it going to change at bulk from single-molecule level?
Very interesting, you should try both out to see!
For （B） two different kind of donors or two same kind of donors? Could be linker/space dependent also? May be!
Hi Yang,
Let's answer the question step by step:
1. You first excite the donor. In the system in B, you excite only donor 1 or donor 2. This is true only if you use a pulsed excitation system. In a cw excitation system, at some point, both donors will be excited.
2. If only one donor is excited at any given time, and it has a probability of x for FRET at 3 nm donor-acceptor distance, then the acceptor fluorescence intensity should be dependent on the donor as if there was only one donor. 3. If, however, there can be two excited donors at a given time, there will be occurrences in which FRET will occur from one donor, to the acceptor. If this is the case, then the acceptor cannot accept the energy transfer from the second donor, until it returns to ground-state.
4. Therefore, for the acceptor fluorescence intensity to increase due to cw excitation of 2 donors, the acceptor has to have a short fluorescence lifetime relative to a longer donor fluorescence lifetime. The short acceptor fluorescence lifetime promises that it will return to its ground-state fast enough so that there will be a high enough probability for it to be re-excited through FRET by the second donor.
To summarize, acceptor fluorescence intensity in a 2-donor-1-acceptor system should be higher than in a 1-donor-1-acceptor system only if:
1. cw excitation is used for donor excitation
2. donor excitation intensity is high enough
3. acceptor fluorescence lifetime is much shorter than the donor fluorescence lifetime
I have only one exclusion to the statements above - do not forget that a fraction of the signals you record in the detection channel reserved for the acceptor fluorescence is also influence by leakage of donor fluorescence (the long, redder, fluorescence spectral tail of the donor). Therefore, you will always have a slightly higher acceptor fluorescence intensity in the 2-donor-1-acceptor system, since instead of leakage of fluorescence of 1 donor into the acceptor channel, now you have leakage from 2.
I hope that helps,
Eitan
Thank you so much Dr. Lerner! Eitan Lerner It is very helpful. A follow-up question is that if we have a 1-donor-2-acceptor system then will the donor only be able to couple with one acceptor until the donor returns to ground state?
Almost. Just a few comments about the exact phrase.
In a 1-donor-2-acceptor system it is not that "the donor only be able to couple with one acceptor until the donor returns to ground state". The donor will transfer its energy only to one of the two acceptors, but almost immediately after transfer occurs, the donor will return to ground-state (within femtoseconds, right after internal conversion of the acceptor wins over donor-acceptor energy resonance). Therefore, the choice of wording ("until the donor returns to ground state") is not accurate. Now assume probability for FRET is equal from donor to acceptor 1 and from donor to acceptor 2 and that all three dyes rotate rapidly so that both donor-acceptor pairs have equally probable orientational freedoms - still the choice of acceptor for the energy transfer is random - so to say that "donor able to couple with one acceptor" is not exact. Eventually it will transfer the energy to one of the two acceptors, but the choice will stay random.
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