SMLM1

Question 1

Why is it good/bad if the “off”-time
gets smaller?

Answer 1

A. Single molecule spots start to overlap. That’s bad.
C. Measurement time gets shorter. That’s good.

Question 2

PALM

Answer 2

Photo-Activated
Localization Microscopy
switching with
endogenously-expressed
fluorophores (no antibodies
needed)

These fluorophores need to be Photo-Activateable (duhh) e.g. mEOS. This means they are fluorescent only if activated by certain activation wavelength. Then they can also revert to dark state but be reactivated again.

Question 3

STORM

Answer 3

STochastical Optical Reconstruction Microscopy

Use of a standard fluorophores but in a dimer (e.g Cy5 with Cy3)

Then similiarly to FRET you depend on the resonant transfer of energy from Cy5 to Cy3. So then you again have activation of Cy3 but through a different mechanism. Then Cy3 is fluorescent until it reverts to dark state again. Then another activation from Cy5 can bring it back.

Question 4

dSTORM

Answer 4

Again STORM but instead of FRET-like mechanism relies on redox reaction. When reduced it becomes fluorescent, when oxidized: deactivated. The spontanoeus reaction with O2 allows for blinking then.

Question 5

PAINT

Answer 5

Point Accumulation for Imaging in Nanoscale Topography

Its just TIRF and fluorophores than can sponlatenously bind on the speciment you look on TIRF. The binding and unbinding creates blinking as the fluorophores away from the TIRF plane are not fluorescent.

Question 6

Why is it good/bad if the “on”-time
gets smaller?

Answer 6

B. You get less photons (Very bad)
C. You cycle faster through all molecules.
D. You have less probability for photobleaching/loss of
fluorescence.

Question 7

What is the second most important factor for resolution in light imaging after difraction?

Answer 7

Pixel size of the digital sensors being used.

Question 8

Types of photosensors

Answer 8

CCD (Charge Coupled Device) : after photons are converted to electrons the charge signal is amplified and ADC off the chip sensor
CMOS ((Complementary Metal-Oxide Semiconductor)
amplifiers and charge conversion to digital signal occur in each pixel seperately.

Question 9

How to choose sampling rate = pixel size?

Answer 9

Nyquist-sampling: pixel size (dx) should be below 1/2 the bandwidth of the signal. Only then perfect reconstruction of signal is possible.

dx </= 1/2B

whicih in microscopy terms

dx </= lambda/4NA as B = 2NA/lambda (remember OTF)

Question 10

Intensity quantization

Answer 10

Each (continuous) intensity level is replaced by an integer level
* Typical number of levels, (the bit of the image): 2, 64, 256, 1024, 2^b (b =1,6,8,10,…). *b is not the bit!

Question 11

Shot noise

Answer 11

• Light is composed of discrete quanta (photons)
• Number of detected photons
  p in a fixed time
  interval is a stochastic variable
• Photon count
  p follows Poisson distribution

Question 12

Readout noise

Answer 12

• Caused by reading the contents of the CCD wells and
  transferring the result off-chip.
• Solution: slow-scan rates (< 1 MHz) and good electronics.
• Root-Mean-Square error (RMS) ≈ 0 to 40 electrons / pixels

Question 13

Localization uncertainty: rule-of-thumb

Answer 13

∆x = Abbe/2 X 1/(sqrt(Nphotons)
=λ/(4NAsqrt(Nphotons)

Combines ray picture (NA of lens), wave picture (λ) and particle
picture (SNR of photon count) in one formula!

then for typical number (λ = 488 nm, NA = 1.25,
Nphotons = 400):
∆x = 5 nm

Question 14

uncertainty in momentum in light imaging

Answer 14

Fluorophore emits a photon that changes the momentum of the emitter in the focal plane +/-NAh/lambda. Thus

∆p =sqrt(Nphotons)NA*h/λ

Question 15

How do you estimate the centre of a PSF

Answer 15

Generally it would be an Airy distribution but its easyer to estimate Gaussian centre so ppl do that instead based on the noisy measurement provided.

Question 16

Find the maximum in the likelihood L({μk}|{nk}) as a function
of the unknown parameters (x0,y0,σ,N)

Answer 16

MLE is used to estimate the most likely mean value based on poisson distribution

Maximum of L({μk}|{nk}) -> find maximum of log-likelihood as it will be the same but easyer to estimate as it will be a sum not a product.
Then by taking derivative of this sum to be zero we can find for which mean value do we get maximum likelihood. From this we can also determine the uncertainty

Question 17

CRLB

Answer 17

The Cramér-Rao Lower Bound provides a theoretical lower limit on the variance of an unbiased estimator, based on the Fisher information. If an estimator achieves this bound, it is efficient, meaning it has the lowest possible variance among unbiased estimators for the parameter in question.