choice of wavelength

From: Seungil Han
Date: 15 February 2012 22:23

All,
I am curious to hear what our CCP4 community thoughts are....
I have a marginally diffracting protein crystal (3-3.5 Angstrom resolution) and would like to squeeze in a few tenth of angstrom.
Given that I am working on crystal quality improvement, would different wavelengths make any difference in resolution, for example 0.9 vs. 1.0 Angstrom at synchrotron?
Thanks.
Seungil


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From: Jacob Keller


I would say the better practice would be to collect higher
multiplicity/completeness, which should have a great impact on maps.
Just watch out for radiation damage though. I think the wavelength
will have no impact whatsoever.

JPK

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From: Bosch, Juergen


No impact ? Longer wavelength more absorption more damage. But between the choices given no problem.
Spread of spots might be better with 1.0 versus 0.9 but that depends on your cell and also how big your detector is. Given your current resolution none of the mentioned issues are deal breakers.

Jürgen

......................
Jürgen Bosch

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From: Jacob Keller


Well, but there is more scattering with lower energy as well. The
salient parameter should probably be scattering per damage. I remember
reading some systematic studies a while back in which wavelength
choice ended up being insignificant, but perhaps there is more info
now, or perhaps I am remembering wrong?

Jacob

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From: Francis E Reyes


Acta Cryst. (1997). D53, 734-737    [ doi:10.1107/S0907444997007233 ]

The Ultimate Wavelength for Protein Crystallography?

I. Polikarpov, A. Teplyakov and G. Oliva

http://scripts.iucr.org/cgi-bin/paper?gr0657



may give some insights.


To the OP, have you solved the structure? In some cases, seeing the packing at low resolution can give you ideas on how to change the construct to obtain higher diffracting crystals.



F

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From: Bart Hazes


Diffracted intensity goes up by the  cube of the wavelength, but so does absorption and I don't know exactly about radiation damage. One interesting point is that on image plate and CCD detectors the signal is also proportional to photon energy, so doubling the wavelength gives 8 times diffraction intensity, but only 4 times the signal on integrating detectors (assuming the full photon energy is captured). So it would be interesting to see how the equation works out on the new counting detectors where the signal does not depend on photon energy. Another point to take into account is that beamlines can have different optimal wavelength ranges. Typically, your beamline guy/gal should be the one to ask. Maybe James Holton will chime in on this.

Bart

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From: John R Helliwell


Dear Colleagues,
I think the following paper will be of particular interest for some
aspects of this thread:-

J. Appl. Cryst. (1984). 17, 118-119    [ doi:10.1107/S0021889884011092 ]
Optimum X-ray wavelength for protein crystallography
U. W. Arndt
Abstract: If the diffraction pattern from crystalline proteins is
recorded with shorter wavelengths than is customary the radiation
damage may be reduced and absorption corrections become less
important.

Best wishes,
John
Professor John R Helliwell DSc

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From: A Leslie

On 15 Feb 2012, at 23:55, Bart Hazes wrote:

Diffracted intensity goes up by the  cube of the wavelength, but so does absorption and I don't know exactly about radiation damage. One interesting point is that on image plate and CCD detectors the signal is also proportional to photon energy, so doubling the wavelength gives 8 times diffraction intensity, but only 4 times the signal on integrating detectors (assuming the full photon energy is captured). So it would be interesting to see how the equation works out on the new counting detectors where the signal does not depend on photon energy.

You make a good point about the variation in efficiency of the detectors, but I don't think your comment about the "new counting detectors" (assuming this refers to hybrid pixel detectors) is correct. The efficiency of the Pilatus detector, for example, falls off significantly at higher energies simply because the photons are not absorbed by the silicon (320  microns thick). The DQE for the Pilatus is quoted as 80% at 12KeV but only 50% at 16KeV and I think this variation is entirely (or at least mainly) due to the efficiency of absorption by the silicon.

Andrew

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From: Bart Hazes


Hi Andrew,

I completely agree and it is what I meant by "(assuming the full photon energy is captured)". If the fraction of photons counted goes up at longer wavelengths than the relative benefit of using longer wavelength is even more pronounced on Pilatus. So for native data sets the wavelength sweet spot with a pilatus detector may be a bit longer then what used to be optimal for a given beamline on a previous generation detector.

Bart

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From: Colin Nave


Bart

>> Diffracted intensity goes up by the  cube of the wavelength, but so
>> does absorption and I don't know exactly about radiation damage.

I think this statement should be
"As an approximation, diffracted intensity (integrated) goes up by the square of the wavelength, but so
does absorbed energy."

See for example
Arndt, U. W. (1984) Optimum X-ray wavelength for protein crystallography. J. Appl. Cryst. 17, 118-119.
Fig. 1. Plot of Ie (not Ip). Note that this includes loss of signal due to absorption through the sample. Subsequent calculations have included Compton scattering and other factors.

I believe Zachariasen first specifically pointed out the wavelength dependence of the integrated intensity (one has to include a Lorentz factor). For the second factor, the absorption of a photon approximately follows the cube of the wavelength but the absorbed energy itself follows the square of the wavelength.

Regards
 Colin

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From: James Holton

The short answer is: no.  Wavelength does not matter.  Not for native data anyway.

I wrote a paper about this recently.  It is open access: http://dx.doi.org/10.1107/S0907444910007262

In particular, check out Figure 2.  The two solid lines are pretty darn flat, and that means the wavelength dependence of damage and scattering power almost exactly cancel.  More on the dotted lines in a bit...

It is easy to screw up the "scattering per damage" calculation as there are many mathematical pitfalls.  Perhaps the trickiest one is thinking that the longer detector distances that would be used at shorter wavelengths (keeping the resolution on the edge of the detector fixed) leads to a net reduction of background/spot.  However, if you carefully calculate the area occupied by a spot, you again find that the noise due to background balances out, and there is again no wavelength dependence.  Lots of people have made that mistake.  Including me!  But eventually I found the error.  Some assurance can be hand that the "no wavelength dependence" conclusion is correct because experimental studies (http://dx.doi.org/10.1107/S0907444993013101), also found no significant wavelength dependence to "signal/noise/dose", as expected.

This is not to say that moving the detector back at constant wavelength is not a good idea.  It is!  You will generally get a signal/noise increase proportional to the distance (for weak spots).  And yes, this is why we spend so much money on large-area detectors!

Of course, the wavelength dependence of detector sensitivity is a completely different story.  For most theoretical calculations you assume a perfect detector system where the only noise is photon counting (also called shot noise).  It is important to remember that no such detectors actually exist.  Even Pilatus has some calibration error, pile-up error, etc. as well as a finite "capture fraction".  In fact, pretty much any modern detector is designed to capture only 80-90% of the incident photons at most wavelengths.  I could go on and on, but since the OP was only asking about "1.0 A vs 0.9 A", the change in detector performance over such a narrow range will be negligible when compared to things like crystal-to-crystal variation.  Did you know that a 110 micron crystal is twice the volume of a 90 micron crystal?  And therefore can absorb twice as much energy before enduring the same dose?

The only other "wavelength dependence" that could be of practical importance is the escape of photoelectrons from the illuminated volume because these can carry away some energy that would otherwise cause damage.  This "build up region" effect has long been a trick of medical dosimetry using MeV-class photons (Johns & Cunningham, 1974).  It was only recently demonstrated experimentally on an MX beamline 10.1073/pnas.1017701108.  It may be possible to take advantage of this effect in a "real-world" data collection, but any real gain will require crystal volumes so small that you cannot get a complete dataset from just one.  That is, unless you are working with VERY small molecules, you will need to be in the "multi-crystal dataset regime" before you can take advantage of photoelectron escape.

So, for any "regular" native data collection, I'd say: no, wavelength doesn't matter.

-James Holton
MAD Scientist