Principles of Fluorescence Spectroscopy

904 ANSWERS TO PROBLEMS
where f is the fraction not complexed. Hence
f τ0 Q(FAD)
τ Q(FMN)
(8.60)
83% of the FAD exists as a nonfluorescent complex.
A8.8. Using the data provided one can calculate the following
quantities needed for the Stern-Volmer plots:
[I – ], M F 0 /F ∆ F F 0 /∆F [I – ] –1 ,M –1
0.0 1.000 0 – –
0.01 1.080 0.074 13.51 100
0.03 1.208 0.172 5.814 33.3
0.05 1.304 0.233 4.292 20.0
0.10 1.466 0.318 3.145 10.0
0.20 1.637 0.389 2.571 5.0
0.40 1.776 0.437 2.288 2.5
The downward curvature of the Stern-Volmer plot
indicates an inaccessible fraction (Figure 8.80). From
the intercept on the modified Stern-Volmer plot one
finds f a = 0.5. Hence one tryptophan residue per subunit
is accessible to iodide quenching. The slope on
the modified Stern-Volmer plot is equal to (f a K) –1 .
Thus K = 17.4 M –1 . By assumption, the quenching
constant of the inaccessible fraction is zero using
these results one can predict the quenching plots for
each tryptophan residue.
[I – ], M [I – ] –1 ,M -1 (F 0 /F) b (F 0 /F) a + (F 0 /F) b (F 0 /∆F) a ++
0.0 0 1.0 1.0 – –
0.01 100 " 1.174 " 6.747
0.03 33.3 " 1.522 " 2.916
0.05 20.0 " 1.870 " 2.149
0.10 10.0 " 2.740 " 1.575
0.20 5.0 " 4.480 " 1.287
0.40 2.5 " 7.96 " 1.144
+ Calculated from F0 /F = 1 + 17.4 [I – ].
++Calculated from F 0 /∆F = 1/K[Q] + 1.
(4.6)(0.09)
(2.4)(1.0)
0.17
For the accessible fraction the Stern-Volmer plot is
linear and the apparent value of f a = 1 (Figure 8.81).
Hence if the quenching data were obtained using 300nm
excitation, where only the accessible residue was
excited, all the fluorescence would appear to be accessible.
Since the inaccessible fraction is not quenched,
F 0 /F = 1 for this fraction. One cannot construct a modified
Stern-Volmer plot since ∆F = 0 for this fraction.
The bimolecular quenching constant can be calculated
using K = 17.4 M –1 and τ = 5 ns, yielding a bimolecular
quenching constant k q = 0.35 x 10 10 M –1 s –1 .
A8.9. Quenching of Endo III by poly(dAdT) displays saturation
near 20 µM, which indicates specific binding of
Figure 8.80. Predicted Stern-Volmer plots for the accessible and inaccessible
tryptophan residues.
poly(dAdT) to Endo III. Assume the quenching is
dynamic. Then K D is near 10 5 M –1 , resulting in an
apparent value of k q = 2 x 10 13 M –1 s –1 . This is much
larger than the diffusion controlled limit, so there must
be some specific binding.
About 50% of the fluorescence is quenched. In Section
8.9.1 we saw that both residues were equally fluorescent.
Hence the titration data (Figure 8.75) suggests
that one residue, probably 132, is completely
quenched when poly(dAdT) binds to Endo III.
A8.10. The structure of wild-type tet repressor is shown in
Figure 8.39. The W75F mutant contains a phenylalanine
in place of Trp 75, and thus only one Trp at position
43. This tryptophan is immediately adjacent to
bound DNA, which quenches the Trp 43 emission.
The extent of quenching is over 50% because there is
only one type of tryptophan. The wild-type protein
would be expected to show less quenching because
Trp 75 will probably not be quenched by DNA.
A8.11. The relative intensities can be calculated from the
Stern-Volmer equation:
CHAPTER 9
F 0
F τ 0
τ 1 k qτ 0
(8.61)
For DBO with a lifetime of 120 ns, and k q = 9 x 10 6
s –1 , the relative intensity is F/F 0 = 0.48. The DBO is
about 50% quenched. For τ 0 = 2 ns, F/F 0 = 0.98 and
the quenching would probably not be detectable. For
τ 0 = 2 ms, F/F 0 = 5.6 x 10 –5 and the fluorophore would
be completely quenched.
A9.1. In order for PET to occur ∆G < 0. We can use the
Rehm-Weller equation to estimate the oxidation