8. Calculate the distance from the donor (D) to the acceptor (A) in the protein represented. Assume that there is a single D-A distance, and that diffusion does not occur during the donor excited-state lifetime. Given: A.The F DA /F D ratio is 0.50 (that is, the ratio between the fluorescence intensity of the donor in the presence of the acceptor (F DA ) and the fluorescence of the respective control molecule which lacks the acceptor (F D )). B.The spectral overlap integral  ( J (l)) for this donor-acceptor pair is 4.5 x 10 13 M -1 cm -1 nm 4 . C.The refractive index of the solution is 1.4. D.The fluorescence quantum yield of the donor in the absence of acceptor is 0.3. Solution Fret E = 1- FDA/FD E = 1-0.50 = 0.50 Rn = R0 (n(1-E)/E)1/6 Rn is the distance between A and D E = 0.5 R0 = 4.5 X 1013 unit Hence Rn = 4.5 X 1013 X 1.4 (1-0.5/0.5)1/6 =4.5 X 1013 X (1.4)1/6 = 4.5 X 1013 X 8.4 = 38291.4 angstrom N = 1.4 .

1. 8. Calculate the distance from the donor (D) to the acceptor (A) in the protein represented.
Assume that there is a single D-A distance, and that diffusion does not occur during the donor
excited-state lifetime. Given:
A.The F DA /F D ratio is 0.50 (that is, the ratio between the fluorescence intensity of the donor in
the presence of the acceptor (F DA ) and the fluorescence of the respective control molecule
which lacks the acceptor (F D )).
B.The spectral overlap integral  ( J (l)) for this donor-acceptor pair is 4.5 x 10 13
M -1
cm -1
nm
4
.
C.The refractive index of the solution is 1.4.
D.The fluorescence quantum yield of the donor in the absence of acceptor is 0.3.
Solution
Fret E = 1- FDA/FD
E = 1-0.50
= 0.50
Rn = R0 (n(1-E)/E)1/6
Rn is the distance between A and D
E = 0.5
R0 = 4.5 X 1013 unit
Hence Rn = 4.5 X 1013 X 1.4 (1-0.5/0.5)1/6
=4.5 X 1013 X (1.4)1/6
= 4.5 X 1013 X 8.4

2. = 38291.4 angstrom
N = 1.4