Hey I was hoping for help, specifically on c, but help on all would be appreciat
ID: 535081 • Letter: H
Question
Hey I was hoping for help, specifically on c, but help on all would be appreciated. I dont see any blurs on my side? but I expanded the immage if that helps?
In a fluorescence lifetime experiment a very short pulse of light is used to excite a sample. Then the intensity of the fluorescence is recorded as a function of time. Suppose we study a dansyl fluorophore. The intensity of the fluorescent light versus time is given below. (a) What is the observed fluorescence lifetime? (b) The fluorescence quantum yield under the same conditions was determined to be 0.7. What is the intrinsic rate constant for radiative fluorescence decay? (c) If has been determined separately that when the dansyl chromophore is 20 A away from the 11-cis-retinal chromophore in rhodopsin, the efficiency of fluorescence energy transfer is 50%. In a second experiment rhodopsin is covalently labeled with a dansyl chromophore and the observed fluorescence lifetime for the dansyl-rhodopsin complex is 6 ns. What is the distance between the dancyl label and the retinal chromophore in rhodopsin?Explanation / Answer
Decay of excited state by fluorescence is a first order process.
[F] = [F]0 e –kt
Take [F] 0 = 1.1 x 104
[F] = 4.9 x 103
4.9 x 103 = (1.1 x 104)e-5k (here unit of t is in ns: I have taken values for 5 ns)
ln (4.9 x 103) = ln (1.1 x 104) -5k
8.496 = 9.305 -5k
k = 0.809/5 = 0.1618 ns-1
Lifetime = 1/k = 6.18 ns
b) we know
quantum yield = kr/(kr + knr)
the observed lifetime = 6.18 ns = 1/(kr + knr) =
0.7 = 6.18 kr
kr = 07/6.18 = 0.113 ns-1.
c.
kFRET = (1/lifetime of the donor) (R0/R)6
Efficiency of the FRET =1-( Lifetime with acceptor/lifetime without acceptor = 6/6.18)= 029
In terms of k
Efficiency = kFRET/ (kFRET + (1/lifetime without acceptor))
Efficiency = kFRET/ (kFRET + 0.1618)
0.029 ((kFRET + 0.1618))= kFRET
0.971 kFRET = 0.1618
kFRET = 0.166 ns-1
Give that R0 is 2 nm (the distance at which 50% efficiency for FRET)
kFRET = (1/lifetime of the donor) (R0/R)6
0.166 x 109 = 1/(6.18 x 109 ) (2 x 10-9/R) 6
1.02 x 1018 = 64 x 10-54 (1/R) 6
R6 = 62.74 x 10-72
6 log R = -72 + 1.79
6 log R = -70.21
log R = -11.7
R = 1. 99 x 10 -12 m