136 4.4  Fluorescence Correlation Spectroscopy

line with a positive gradient at low time interval values, but at higher time interval

values, the gradient would decrease until reaching a plateau at the very highest time

intervals. What do you think could explain this?

d

With increasing incubation of the cells in the toxin, it was found that the frequency of

detection of the tracks in the droplet increased, that the initial gradient of the average

mean squared displacement versus time interval plot was lower, and the plateau was

reached at a lower level of mean squared displacement. Explain these observations.

Answers

a

The localization precision in 1D for a single photon is given simply by the optical

resolution limit, which is the PSF width, assuming that the characteristic wave­

length can be best approximated by the peak emission wavelength here, using

Equation 4.3:

Δx =​ 0.61 × 595 × 10^–​9/​1.4 =​ 2.59 × 10^–​7 or 259 nm

However, with Gaussian fitting there is a better estimate for the center of the

dye fluorescence intensity profile, such that the localization precision improves

with the number N of fluorescence photons sampled (Equation 4.4). If the non-​

photon associated localization error (i.e., dark noise and localization uncer­

tainty error associated with finite pixel size on the camera) is negligible, then

the 1D localization precision σx from Gaussian fitting scales as ~s/​√N where s

approximates to the single photon localization precision, or ~Δx. To determine N,

we need to estimate the number of fluorescence photons captured in total from

a single PAmCherry molecule and detected by the camera. This total efficiency

ε of photon detection equals 0.50 multiplied by 0.88 multiplied by the geomet­

rical photon capture efficiency (the solid angle Ω subtended at the objective lens

for detecting photons divided by all solids angle of 4π steradians). Ω is given by

Equation 3.21 Ω=​2π(1 –​ cosθ), where NA=​nsinθ, assuming the refractive index

n =​ 1.515. Thus:

θ =​ sin^–​1(1.4/​1.515) =​ 67.5˚

ε =​ (2π × (1 –​ cos(67.5))/​4π) × 0.30 × 0.80 =​ 0.07 or 7%

The total number of photons therefore detected by the camera from one

PAmCherry molecule prior to photobleaching is 10⁵ × 0.07 =​ 7000. However,

these photons are detected over a total of 50 ms on average, with a sampling

time per image frame of 5 ms, so the number of photons detected by the camera

per image frame is N =​ 7000 × (5/​50) =​ 700. Thus:

σx =​ 259/​√(700) =​ 9.8 nm

But by combining the total sigma x and y values and by symmetry the 2D local­

ization is:

σ2D =​ √ (σx

2+​ σy

2) =​ σx.√2 =​ 13.8 nm

b

The increase in the number of detected tracks is expected since the intensity

of activation increases the activation probability. The observation of two-​fold

increases in brightness of the tracks may indicate that two photoactive PAmCherry

molecules are colocalized to within the localization precision, which might mean

that the protein in not a single monomer but is a dimer. This will at most increase

the number of detected photons by a factor of two, thus will improve (i.e., reduce)