Thursday, October 18, 2007

GFP bleaching rate as a function of laser power and pixel time

EGFP has a molar extinction coefficient (ε) of 55 000 L/(mole * cm) = 9.13 x 10-21 m2/molecule, which is also called its optical cross section (Am).

The percentage of molecules in the excited state at steady-state is ka / (ka * kf), where ka is the number of incoming photons per molecule and second from the laser, and kf is the rate of return to the ground state, which is 1 / τf where τf is the average seconds in the excited state.

For GFP, τf is 3,3 ns, making kf 3,03 * 108 molecules/s.

ka is the optical cross section times the number of photons per square meter and second. The number of photons per square meter and second is the the number of photons per second divided by the pixel area Ap. The number of photons per second is the laser power (p) in Watts (=J/s) divided by the photon energy in J. The photon energy is h c / λ, where h is Planck's constant, c is the speed of light, and λ is the light wavelength. So, ka = Am * p * λ / (Ap * h * c).

For a pixel spot with radius 0,25 µm, a laser wavelength of 488 nm, and a laser power of 0,3 mW, ka = 34 million photons / molecule and second, and x = 10%. At this rate there should not be much bleaching, especially considering that only about 22 photons will reach a molecule during a pixel time of 0,64 µs (the fastest on our confocal mic).

Same calculation for Venus (YFP variant): Extinction coefficient = 92 200 L / (mole cm) = 1,53 x 10-20 m2/molecule. λ= 515, τf = ?, ...

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Thursday, April 26, 2007

I'm analyzing images from confocal measurements of the a cell filled with fluorescent calcein, as it is exposed to osmotic change. I've made a 4D-recording: 3D stacks taken after each other.

I mark a cell, and for each stack calculate its volume (by counting the number of voxels whose intensity is greater than a threshold), and for each slice the intensity.

I correct for bleaching (about 0.35% / frame=5 seconds?) or Calcein degradation.

The following things are strange:
  • The average fluorescent intensity in a slice through the middle of the cell changes less than the cell volume (about half)
  • I've verified that the total amount of intensity in the cell is approximately constant for each stack.
  • So where does the intensity go when the cell shrinks?
Example: z=7, t=10:
  • Cell volume (normalized)=0.96
  • unbleached intensity=690 000
  • area=476
  • average intensity = 1449
t=29
  • Cell volume (normalized)=1.28 (+33%)
  • unbleached intensity in slice=568000 (-18%)
  • area=457 (-4%)
  • average intensity=1247 (-14%)
Found the problem! When I increased the intensity threshold for considering voxels to be in the cell, I got very good matches.

Tuesday, February 28, 2006

Apertures in cameras vs confocal microscopes

Question:

In a normal confocal microscope, a pinhole is used to get the image of only a thin slice through the specimen, eliminating light from all points which are not in focus.

In a normal camera, a small aperture is used to get a large depth of field, with a sharp imaging of objects both close and far away from the camera.

Why does a small hole give rise to these apparently opposite effects?


Answer (I think):

In a confocal microscope, the pinhole is in the focal plane, and the light reaching the sensor is actually "out of focus", whereas in a camera, the film (or digital sensor) is in the focal plane, while the aperture is in front of the focal plane.

Comments?

Tuesday, February 14, 2006

nm and THz: Photons, frequency and wavelength

Ok, so I think I figured out the photon wavelength / frequency thing. Pretty obvious in retrospect.

Photons have a constant frequency (f=E/h), but their wavelength varies with the speed of light (related to density?) in the matter through which the photons travel.

All this is pretty irrelevant to me, since in the end I'm interested in fluorescence, which will only depend on the photon's energy.

However, it's interesting to note that the practice of describing light i wavelength (nm) rather than frequency is not very good because the frequency is constant while the wavelength may vary for a given photon. So instead of talking about 500 nm, maybe we should say 600 THz (terahertz) light? This would also be more intuitive in that a higher frequency is equivalent to a higher energy (whereas a shorter wavelength is equivalent to a higher energy).

Monday, February 13, 2006

No Results

For the record:

I'll have to censor myself in this blog since scientific journals require that results in papers have not been published before.

I think I can manage this and still discuss details which will be interesting, but won't be significant parts of the results which I hope to publish some day.

Laser power, photons per second, speed of light in different mediums (media?)

One thing I'd like to have is a theoretical model of the confocal microscope, in order to understand noise sources and to be able to quantify the measurement results.

So, I started at the beginning, looking at the laser source.

Goal 1: to express the number of photons per second as a function of the laser power (in Watts) and the wavelength.

The energy per photon is E = h * f (where h is Planck's constant ≈ 6.6 * 10-34, f is the frequency in Hz, and the energy E is in Joules). Or more relevantly for us, using the wavelength rather than frequency (f = c / λ), E = h * c / λ (where c is the speed of light).

Question: What c should be used? The speed of light in vacuum? Or the speed of light in the gas in the laser? Or the speed of light in the air outside the laser?
Assuming that photons maintain their energy when passing through matter with different densities, does the wavelength of the light change to compensate? (That seems like the most likely event). But if so, is the wavelength of the laser specified for vacuum? Probably all this is academic because of the small change in speed of light in vacuum and air, but to be completely accurate I'd still like to know the answer.

Taking the power of the laser in plaser [Watts] = [Joules * second], multiplying it by one second, and dividing it by the photon energy should then give us the number of photons per second:

fphotons [Hz] = plaser [W] * c [m/s] / λ [m].

For a 1 W laser, this works out to be c / λ = f ! (In other words, about 600 * 1012 photons / second.

In the confocal microscope I think we're using voxel times of about 1 μs, so at 500 nm we should be getting 600*106 photons per voxel from the laser.