Re: [ccp4bb] Dose in diffraction patterns?

[James Holton]

One more correction:

For example, if you see an average pixel value of 20 photons on a 
Pilatus 6M, then that is P=120e6 photons.  If that was a t=0.1 s 
exposure from a sample 100 microns thick, then the beamline flux was 
about 1e12 photons/s.  Note that this is the flux after any attenuation, 
not before.



On 5/7/2020 9:06 AM, James Holton wrote:
Ah!  I did that last formula wrong.  Never do algebra in your head 
without checking. It should be:


The equation then becomes:

f = P/t/L/1.2e-5

Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux 
and t=exposure (as above).


For example, if you see an average pixel value of 20 photons on a 
Pilatus 6M, then that is P=12e6 photons.  If that was a t=0.1 s 
exposure from a sample 100 microns thick, then the beamline flux was 
about 1e12 photons/s.  Note that this is the flux after any 
attenuation, not before.


Oh, and if you want a reference for that 2000 ph/um^2 = 1 Gy rule, it 
is here:

https://doi.org/10.1107/S0909049509004361

And, of course, if you are lucky enough to have accurate flux, size 
and shape information for the beam and sample, plus chemical 
composition the most accurate dose you'll get from raddose-3D: 
https://www.raddo.se/


-James Holton
MAD Scientist

On 5/6/2020 4:04 PM, James Holton wrote:

In general?  No.

I believe a few places put "flux" into the header, but as Andreas 
just mentioned that is only one of the bits of information you need 
to calculate dose.  If all you want is a rough estimate, then the 
numbers you need are:

f = flux (photons/s)
t = exposure time (s)
w = wavelength (A)
a = beam area (um^2)

The dose (D) to a sample of protein/water/plastic under a given beam 
will be roughly:

D = f*t*w^2/a/2000  (in Gy)

For example: a crystal under a square 100 um x 100 um beam at 1 A 
wavelength with flux 1e12 ph/s will get 1 MGy dose in 20 s.


The 2000/w^2 is a fudge factor that fits the true curve of metal-free 
protein crystals to within 15% for the wavelength range 0.5 < w < 3.  
The error induced by not knowing if the beam was round or square and 
just multiplying together the width and height to get the area (a), 
is 21%.  The error from not realizing you had 100 mM uranium in your 
sample is about a factor of two at 1 A.  Smaller concentrations and 
lighter atoms have less impact on accuracy.


If you don't know the flux, or beam size, you can try looking them up 
at http://biosync.sbkb.org/ . I scraped these for my little dose 
calculator here:

bl831.als.lbl.gov/xtallife.html
Some of the biosync numbers are more accurate than others, however, 
depending on how often beamline scientists remember to update the 
site, and how well they know themselves.  And attenuation is not 
always written into the header either.


In a pinch, you can estimate the flux by the total number of photons 
on the image (P).  This is assuming that you know the sample 
thickness (L) in microns.  You must also assume that the total 
scattering cross section of the atoms in the sample is close to that 
of oxygen (0.2 cm^2/g), that the sample density is 1.2 g/cm^3 and 
that about 50% of the scattered photons reach the detector.  None of 
these are terrible assumptions. The equation then becomes:


f = P/t/L*1.2e-5

Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux 
and t=exposure (as above).


Getting P from a pixel array is easy: you just add up all the pixel 
values.  From a CCD you want to be careful to subtract the baseline 
value from each pixel first (40 on ADSC, 10 on Mar/Rayonix), and then 
divide by the "gain", which near 1 A is ~1 on Mar/Rayonix, 0.6 for 
ADSC Q315 (swbin) and 1.8 for Q315r (hwbin).  A few considerations, 
yes, but it can be a good sanity check.


-James Holton
MAD Scientist


On 5/5/2020 11:48 AM, Murpholino Peligro wrote:
Do diffraction patterns publicly accessible contain information 
about the x-ray absorbed dose?




Thanks



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Re: [ccp4bb] Dose in diffraction patterns?

[James Holton]

Ah!  I did that last formula wrong.  Never do algebra in your head 
without checking. It should be:


The equation then becomes:

f = P/t/L/1.2e-5

Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux 
and t=exposure (as above).


For example, if you see an average pixel value of 20 photons on a 
Pilatus 6M, then that is P=12e6 photons.  If that was a t=0.1 s exposure 
from a sample 100 microns thick, then the beamline flux was about 1e12 
photons/s.  Note that this is the flux after any attenuation, not before.


Oh, and if you want a reference for that 2000 ph/um^2 = 1 Gy rule, it is 
here:

https://doi.org/10.1107/S0909049509004361

And, of course, if you are lucky enough to have accurate flux, size and 
shape information for the beam and sample, plus chemical composition the 
most accurate dose you'll get from raddose-3D: https://www.raddo.se/


-James Holton
MAD Scientist

On 5/6/2020 4:04 PM, James Holton wrote:

In general?  No.

I believe a few places put "flux" into the header, but as Andreas just 
mentioned that is only one of the bits of information you need to 
calculate dose.  If all you want is a rough estimate, then the numbers 
you need are:

f = flux (photons/s)
t = exposure time (s)
w = wavelength (A)
a = beam area (um^2)

The dose (D) to a sample of protein/water/plastic under a given beam 
will be roughly:

D = f*t*w^2/a/2000  (in Gy)

For example: a crystal under a square 100 um x 100 um beam at 1 A 
wavelength with flux 1e12 ph/s will get 1 MGy dose in 20 s.


The 2000/w^2 is a fudge factor that fits the true curve of metal-free 
protein crystals to within 15% for the wavelength range 0.5 < w < 3.  
The error induced by not knowing if the beam was round or square and 
just multiplying together the width and height to get the area (a), is 
21%.  The error from not realizing you had 100 mM uranium in your 
sample is about a factor of two at 1 A.  Smaller concentrations and 
lighter atoms have less impact on accuracy.


If you don't know the flux, or beam size, you can try looking them up 
at http://biosync.sbkb.org/ . I scraped these for my little dose 
calculator here:

bl831.als.lbl.gov/xtallife.html
Some of the biosync numbers are more accurate than others, however, 
depending on how often beamline scientists remember to update the 
site, and how well they know themselves.  And attenuation is not 
always written into the header either.


In a pinch, you can estimate the flux by the total number of photons 
on the image (P).  This is assuming that you know the sample thickness 
(L) in microns.  You must also assume that the total scattering cross 
section of the atoms in the sample is close to that of oxygen (0.2 
cm^2/g), that the sample density is 1.2 g/cm^3 and that about 50% of 
the scattered photons reach the detector.  None of these are terrible 
assumptions. The equation then becomes:


f = P/t/L*1.2e-5

Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux 
and t=exposure (as above).


Getting P from a pixel array is easy: you just add up all the pixel 
values.  From a CCD you want to be careful to subtract the baseline 
value from each pixel first (40 on ADSC, 10 on Mar/Rayonix), and then 
divide by the "gain", which near 1 A is ~1 on Mar/Rayonix, 0.6 for 
ADSC Q315 (swbin) and 1.8 for Q315r (hwbin).  A few considerations, 
yes, but it can be a good sanity check.


-James Holton
MAD Scientist


On 5/5/2020 11:48 AM, Murpholino Peligro wrote:
Do diffraction patterns publicly accessible contain information about 
the x-ray absorbed dose?




Thanks



To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/webadmin?SUBED1=CCP4BB&A=1








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Re: [ccp4bb] Dose in diffraction patterns?

[James Holton]

In general?  No.

I believe a few places put "flux" into the header, but as Andreas just 
mentioned that is only one of the bits of information you need to 
calculate dose.  If all you want is a rough estimate, then the numbers 
you need are:

f = flux (photons/s)
t = exposure time (s)
w = wavelength (A)
a = beam area (um^2)

The dose (D) to a sample of protein/water/plastic under a given beam 
will be roughly:

D = f*t*w^2/a/2000  (in Gy)

For example: a crystal under a square 100 um x 100 um beam at 1 A 
wavelength with flux 1e12 ph/s will get 1 MGy dose in 20 s.


The 2000/w^2 is a fudge factor that fits the true curve of metal-free 
protein crystals to within 15% for the wavelength range 0.5 < w < 3.  
The error induced by not knowing if the beam was round or square and 
just multiplying together the width and height to get the area (a), is 
21%.  The error from not realizing you had 100 mM uranium in your sample 
is about a factor of two at 1 A. Smaller concentrations and lighter 
atoms have less impact on accuracy.


If you don't know the flux, or beam size, you can try looking them up at 
http://biosync.sbkb.org/ . I scraped these for my little dose calculator 
here:

bl831.als.lbl.gov/xtallife.html
Some of the biosync numbers are more accurate than others, however, 
depending on how often beamline scientists remember to update the site, 
and how well they know themselves.  And attenuation is not always 
written into the header either.


In a pinch, you can estimate the flux by the total number of photons on 
the image (P).  This is assuming that you know the sample thickness (L) 
in microns.  You must also assume that the total scattering cross 
section of the atoms in the sample is close to that of oxygen (0.2 
cm^2/g), that the sample density is 1.2 g/cm^3 and that about 50% of the 
scattered photons reach the detector.  None of these are terrible 
assumptions. The equation then becomes:


f = P/t/L*1.2e-5

Where 1.2e-5 = 0.2 cm^2/g * 1.2 g/cm^3 * 1e-4 cm/micron * 50%, f=flux 
and t=exposure (as above).


Getting P from a pixel array is easy: you just add up all the pixel 
values.  From a CCD you want to be careful to subtract the baseline 
value from each pixel first (40 on ADSC, 10 on Mar/Rayonix), and then 
divide by the "gain", which near 1 A is ~1 on Mar/Rayonix, 0.6 for ADSC 
Q315 (swbin) and 1.8 for Q315r (hwbin).  A few considerations, yes, but 
it can be a good sanity check.


-James Holton
MAD Scientist


On 5/5/2020 11:48 AM, Murpholino Peligro wrote:
Do diffraction patterns publicly accessible contain information about 
the x-ray absorbed dose?




Thanks



To unsubscribe from the CCP4BB list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/webadmin?SUBED1=CCP4BB&A=1






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Re: [ccp4bb] Dose in diffraction patterns?

[Andreas Förster]

Dear Murpholino,

all diffraction patterns contain information on absorbed dose in the form
of radiation damage.

You might be more interested in a quantitative description of dose.  For
this, you need information on the size and shape of your crystal; the
sequence of your protein + cofactors, ligands, metals and crystallization
buffer; the size, shape and flux of your beam; and the position of the
rotation axis relative to the beam (does the crystal rotate or wobble in
the beam?).  In a perfect world, all this information would be part of the
metadata that's saved with your diffraction data.  Currently, none of this
is.  (Correct me if I'm wrong.  I'd like to know.)  With it, you can
estimate dose with the current or future versions of Raddose 3D.

All best.


Andreas



On Tue, May 5, 2020 at 8:49 PM Murpholino Peligro 
wrote:

> Do diffraction patterns publicly accessible contain information about the
> x-ray absorbed dose?
>
>
>
> Thanks
>
> --
>
> To unsubscribe from the CCP4BB list, click the following link:
> https://www.jiscmail.ac.uk/cgi-bin/webadmin?SUBED1=CCP4BB&A=1
>


-- 

Andreas Förster, Ph.D.
Application Scientist Crystallography, Area Sales Manager Asia & Pacific
Phone: +41 56 500 21 00 | Direct: +41 56 500 21 76 | Email:
andreas.foers...@dectris.com
DECTRIS Ltd. | Taefernweg 1 | 5405 Baden-Daettwil | Switzerland |
www.dectris.com







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[ccp4bb] Dose in diffraction patterns?

[Murpholino Peligro]

Do diffraction patterns publicly accessible contain information about the
x-ray absorbed dose?



Thanks



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