Principles of Fluorescence Spectroscopy

PRINCIPLES OF FLUORESCENCE SPECTROSCOPY 897
(3.8)
B. The fluorescence intensity (F 1 or F 2 ) observed
with each excitation wavelength (1 or 2)
depends on the intensity and the concentrations
(C F and C B ) of the free (S f1 or S f2 ), or bound (S b1
or S b2 ), forms at each excitation wavelength:
F 1 = S f1 C F + S b1 C B , (3.9)
F 2 = S f2 C F + S b2 C B . (3.10)
The term S i depends on the absorption coefficient and
relative quantum yield of Fura-2 at each wavelength.
Let R = F 1 /F 2 be the ratio of intensities. In the
absence and presence of saturating Ca 2+ ,
R min = S f1 /S f2 , (3.11)
R max = S b1 /S b2 . (3.12)
Using the definition of the dissociate constant,
one obtains
CHAPTER 4
A Ca2 F Fmax KD Fmax F
Ca2 CB KD CF Ca2 R Rmax KD Rmin R ( Sf2 fb2 (3.13)
(3.14)
Hence one can measure the [Ca 2+ ] from these ratios of
the emission intensities at two excitation wavelengths.
However, one needs control measurements, which are
the ratio of the intensities of the free and bound forms
measured at one excitation wavelength, as well as
measurements of R min and R max. 183
A4.1. Calculation of the lifetimes from intensity decay is
straightforward. The initial intensity decreases to 0.37
(=1/3) of the initial value at t = 5 ns. Hence, the lifetime
is 5 ns.
)
From Figure 4.2 the phase angle is seen to be
about 60 degrees. Using ω = 2π ⋅80 MHz and τ φ =
ω –1 [tan φ] one finds τ = 3.4 ns. The modulation of
the emission relative to the excitation is near 0.37.
Using eq. 4.6 one finds τ m = 5.0 ns. Since the phase
and modulation lifetimes are not equal, and since
τ m > τ φ , the intensity decay is heterogeneous. Of
course, it is difficult to read precise values from
Figure 4.2.
A4.2. The fractional intensity of the 0.62-ns component
can be calculated using eq. 4.28, and is found to be
0.042 or 4%.
A4.3. The short lifetime was assigned to the stacked conformation
of FAD. For the open form the lifetime of
the flavin is reduced from τ 0 = 4.89 ns to τ = 3.38
ns due to collisions with the adenine. The collision
frequency is given by k = τ –1 – τ 0 –1 = 9 x 10 7 /s.
A4.4. In the presence of quencher the intensity decay is
given by
I(t) = 0.5 exp(–t/0.5) + 0.5 exp(–t/5) (4.42)
The α 1 and α 2 values remain the same. The fact that
the first tryptophan is quenched tenfold is accounted
for by the α i τ i products, α 1 τ 1 = 0.25 and α 2 τ 2 = 2.5.
Using eq. 4.29 one can calculate τ = 4.59 ns and
= 2.75 ns. The average lifetime is close to the
unquenched value because the quenched residue (τ 1 =
0.5 ns) contributes only f 1 = 0.091 to the steady-state
or integrated intensity. If the sample contained two
tryptophan residues with equal steady-state intensities,
and lifetimes of 5.0 and 0.5 ns then τ = 0.5(τ 1 ) +
0.5(τ 2 ) = 2.75 ns. The fact that reflects the relative
quantum yield can be seen from noting that /τ 0 =
2.75/5.0 = 0.55, which is the quantum yield of the
quenched sample relative to the unquenched sample.
A4.5. The DAS can be calculated by multiplying the fractional
intensities (f i (λ)) by the steady-state intensity at
each wavelength (I(λ)). For the global analysis these
values (Figure 4.65) match the emission spectra of the
individual components. However, for the single-wavelength
data the DAS are poorly matched to the individual
spectra. This is because the α i (λ) values are not
well determined by the data at a single wavelength.
A4.6. The total number of counts in Figure 4.45 can be calculated
from the ατ product. The value of α is the
number of counts in the time zero channel or 10 4
counts. The total number of photons counted is thus 4