# What are the requirements for FRET?

I have a question about fluorescence resonance energy transfer (FRET), and the requirements needed for this technique.

Q: Which of the following need to be met for FRET?

• A) Strong overlap between the donor emission spectrum and the acceptor absorption spectrum
• B) Molecules must be $<10\,\mathrm{nm}$ apart
• C) Molecules must be $>10\,\mathrm{nm}$ apart
• D) Low quantum yield of the donor
• E) Small extinction coefficient of the acceptor

I'm aware that A and B are true, and C is not a requirement. But what about a low quantum yield of the donor or a small extinction coefficient of the acceptor? Do the size of these numbers really matter?

One way to express the fluorescence quantum yield (QYf) is:

1) QYf = kf/sum(ki)

where kf is the kinetic constant of the fluorescence and sum(ki) is the sum of the kinetic constants for the overall deactivation processes, I.e. non-radiative processes and radiative. It is also true that:

2) 1/sum(ki) = tau

Where tau is the lifetime of the molecule. That means QYf is proportional to the lifetime (replace 2 in 1). High QYf high fluorescence lifetime. In the FRET scenario this means that a Donor with high QYf has more time to donate its energy to an Acceptor, while a Donor with low QYf would deactivate really quickly to the ground state because of faster non radiative processes, resulting in a poor FRET. Thus, a LOW QYf disfavors the FRET process.

The extinction coefficient is proportional to the absorption of a molecule. Thus, a high extinction coefficient for the acceptor will favor the FRET. Matematically, you can see that the extinction coefficient of the acceptor is proportional to J, i.e the integral overlap between donor and acceptor thus proportional to the FRET process.

Hope this helps

A low quantum yield would indicate that there are efficient intra-molecular non-radiative decay pathways, likely internal conversion or spin-orbit coupling. These decay pathways may be faster that FRET, rendering FRET essentially a non-competetive process.

A small extinction coefficient would indicate a small transition dipole moment. If the transition dipole moment is small it is difficult for the FRET partner to transfer energy since it is a weak transition resulting in a small coupling between the two chromophores.

The equation for the rate constant ($\pu{s^{-1}}$) for Forster (or Resonance of dipole-dipole ) energy transfer at separation R is $$k_R= \alpha\frac{\kappa^2\phi}{\tau R^6}\int_0^{\infty} \frac{F(\nu)\epsilon(\nu)}{\nu^4} d\nu$$ where the constant $\alpha =(9000\ln(10) )/(128\pi^5n^4N)$, n is the solution refractive index and N Avogadro's number. The quantum yield of the donor is $\phi$, and its excited state lifetime (in the absence of quencher) is $\tau$, the orientation term is $\kappa$ and R the separation of donor and acceptor.

In the (overlap) integral F is the fluorescence spectrum measure in frequency (not wavelength) and the area under $F(\nu)$ is normalised to unity. The molar extinction coefficient of the acceptor is $\epsilon(\nu)$ also measured on a frequency scale normally in units $\pu{dm^3mol^{-1}cm^{-1}}$.

The rate constant is more commonly written as $$k=\frac{1}{\tau} \left( \frac{R_0}{R} \right)^6$$

where $R_0$ is the critical distance at which energy transfer rate constant equals the fluorescence rate constant ($1/\tau$) and is also a measure of the overlap of fluorescence from the donor and absorption by the acceptor.

This the decay rate of a molecule that is fluorescing and undergoing energy transfer is at separation R equal to $k=k_f+k_{isc}+k{ic}+k_R$ where isc and ic are intersystem crossing and internal conversion respectively.

In you questions;

(A) the bigger this is the larger $k_R$ is.

(B) and (C) there is no real cut off, $R_0$ for chlorophyll is $\approx 8$ nm so transfer can occur beyond this distance it just falls off as $1/R^6$

(D) Clearly the smaller $\phi$ is the lower the rate of transfer and in direct proportion. If the molecule has a low fluorescence yield clearly the molecule cannot transfer energy as it is directed elsewhere.

(E) If $\epsilon$ is small then the overlap integral is small so the rate $k_R$ is small.