Principles of Fluorescence Spectroscopy

PRINCIPLES OF FLUORESCENCE SPECTROSCOPY 919
D. f 1 = 0.0004, f 2 = 0.9996. This result shows that
off-gating essentially eliminates the short-lived
component, decreasing its fractional contribution
from 0.296 to 0.0004.
A20.2. The oxygen bimolecular quenching constant can be
calculated using the decay times in the absence and in
the presence of 100% oxygen. The value of τ 0 /τ = 16.3
= 1 + k q τ 0 [O 2 ]. Using [O 2 ] = 0.001275 M and τ 0 = 3.7
µs one obtains k q = 3.24 x 10 9 M –1 s –1 . This value is
reasonably close to the diffusion-controlled limit and
indicates that the quenching by oxygen is highly efficient.
CHAPTER 21
A21.1. To answer this question we need to design quenching
or anisotropy measurements that could potentially be
used for sequencing. Consider sequencing with four
fluorescent ddNTPs. The Stern-Volmer quenching
constants could be different due to either different lifetimes
or different accessibilities to the collisional
quencher. Then the sequence could be determined by
the quenching constant for each fluorescent band on
the gel. Determination of the quenching constant
requires a minimum of two intensity measurements: in
the absence of quencher and in the presence of a
known concentration of quencher. Although such
measurements are possible, the use of two samples to
measure a single base is too complicated for largescale
sequencing of DNA.
Suppose that the sequencing reaction is performed
with a single fluorescent primer. Because anisotropy
measurements depend on molecular weight, in principle
each oligonucleotide will display a different
anisotropy. In practice the anisotropies for DNA
oligomers, differing by a single base pair, are likely to
be too similar in magnitude for useful distinction
between oligomers. If the adjacent base pair changes
the lifetime of the labeled oligomer, then the
anisotropy measurements may be able to identify the
base.
Consider the use of four fluorescent ddNTPs, each
with a different lifetime. In this case the anisotropy
would be different for each base pair, and the
anisotropy measurement could be used to identify the
base. This approach is more likely to succeed for
longer oligomers, where the anisotropy will become
mostly independent of molecular weight. For shorter
oligomers the anisotropy will depend on the fragment
length.
CHAPTER 22
A22.1. In Figure 22.15 it is clear that the F-actin is red and the
green color is where the mitochondria are expected to
be localized. In Figure 22.19 the colors are reversed:
F-actin is green and mitochondria are red. At first
glance it appears that the legend for one of the figures
is incorrect, or that the cells in Figure 22.15 were
labeled with fluorophores that stained actin red and
mitochondria green. The legends are correct. Both
images are created using pseudocolors. Figure 22.19
was created by an overlay of three intensity images.
The color of each image was assigned to be similar to
the emission maxima of the probes: DAPI is blue,
Bodipy-FL is green, and MitoTracker is red. These
assignments give the impression that Figure 22.19 is a
real color image. In Figure 22.15 the colors were
assigned according to lifetime. The lifetime in the
nucleus was assigned a blue color, which agrees with
Figure 22.19. However, in Figure 22.15 the lifetime of
F-actin was assigned to be red, and the lifetime of
MitoTracker was assigned to be green. The pseudocolor
assignments of the red- and green-emitting fluorophores
are opposite in Figures 22.15 and 22.19.
CHAPTER 23
A23.1. The intensity needed to excite the fluorophore can be
calculated using eq. 23.5. If there is no intersystem
crossing than S 1 = τσ I e S T. The intensity required is
given by S 1/S T = 0.5 τσ I e. The cross-section for
absorption can be calculated using eq 23.1, yielding σ
= 4 x 10 –16 = 4 Å 2 . In performing this calculation it is
important to use a conversion factor of 10 3 cm 2 /liter.
The number of photons per cm 2 per second can be
calculated from I e = 0.5 (τσ) –1 = 3.1 x 10 23 /cm 2 s. The
power can be calculated from the number of photons
per second per cm 2 and the energy per photon = hν/λ,
yielding 103 kW/cm 2, and an area of 1 µm 2 = 10 –8
cm 2, so the power needed to saturate the fluorophore
is 0.001 watts/cm 2.