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The phase problem in protein crystallography. The phase problem in protein crystallography.

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Presentation on theme: "The phase problem in protein crystallography. The phase problem in protein crystallography."— Presentation transcript:

1 The phase problem in protein crystallography

2 The phase problem in protein crystallography

3 Bragg diffraction of X-rays (photon energy about 8 keV, 1.54 Å)

4 Structure factors and electron density are a Fourier pair

5 The problem is that the raw data are the squares of the modulus of the Fourier transform. That´s the famous phase problem.

6 In protein crystallography, there are several ways to get the phases: Molecular replacement Heavy atom methods Direct methods Non-standard methods

7 Mol A: GPGVLIRKPYGARGTWSGGVNDDFFH... Mol B: GPGIGIRRPWGARGSRSGAINDDFGH... Mol A Mol B ? Molecular replacement

8 If we have phases from a similar model... Amplitudes: Manx Phases: Manx Amplitudes: Cat Phases: Cat Amplitudes: Cat we can use Phases: Manx...we can combine them with the experimental amplitudes to get a better model.

9 Patterson maps can be used to find.... the proper orientation (rotation).... the proper position (translation) for the search model. The density mapThe Patterson map

10 The Patterson map is the Fourier transform of the intensities. It can be calculated without the phases.

11 The matching procedure requires a search in up to six dimensions Luckily, the problem can be factorized into first, a rotation search then, a translation search

12 Flow chart of a typical molecular replacement procedure (AMORE) xyzin1 (*1.pdb) table1 (*1.tab) hklpck1 (*1.hkl) clmn1 (*1.clmn) tabfun rotfun (generate) rotfun (clmn) hklin (*.mtz) hklpck0 (*0.hkl) clmn0 (*0.clmn) rotfun (clmn) sortfun rotfun (cross) } SOLUTRC trafun (CB) SOLUTTF fitfun (rigid) SOLUTF pdbset solution.pdb

13 Poor phases yield self-fulfilling prophesies Amplitudes: Karlé Phases: Karlé Amplitudes: Hauptmann Phases: Hauptmann Amplitudes: Hauptmann If Karlé phases Hauptmann, Hauptmann is Karléd! Phases: Karlé

14 Heavy atom methods ?

15 Can we do X-ray holography?

16 Can we do holography with crystals? In principle yes, but the limited coherence length requires a local reference scatterer.

17 For a particular h,k,l FPFP F H1 F H2 F PH1 we can collect all knowledge about amplitudes and phases in a diagram (the so-called Harker diagram)

18 Normally, there´s the problem that different crystals are not strictly isomorphous. Thus, the best is a reference scatterer that can be switched on and off.

19 Absorption is accompanied by dispersion. This Kramers-Kronig equation is very general: Its (almost) only assumption is the existance of a universal maximum speed (c) for signal propagation.

20 Which elements are useful for MAD data collection? 7 keV 25 keV 0.5 Å 1.8 Å K LIII 26-46 64-

21 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Ha Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr The MAD periodic table

22 All phasing can be done on one crystal. F 1,2 F -1,-2 a b F 1,2 : scattering from b is phase  behind F -1,-2 : scattering from b is phase  ahead In the presence of absorption, Bijvoet pairs are nonequal.

23 assuming with absorption:

24 Direct methods ? Atomic resolution data the best approach for small molecules

25 If atoms can be treated as point-scatterers, then if all involved structure factors are strong

26 100 atoms in the unit cell a small protein The method is blunt for lower resolution or too many atoms.

27 Three-beam phasing ? very low mosaicity data an exciting, but not yet practical way

28 An example from our work (solved by a combination of MAD and MR) Metal ions

29 Can we tell from the fluorescence scans? Normally yes, but not in this case! Co Zn Fe Ni Cu Compton

30 Can we tell from the anomalous signal? order in the periodic table: Fe, Co, Ni, Cu, Zn

31 2fo-fc map, 1.05 Å anomalous map, 1.05 Å anomalous map, 1.54 Å Here´s the maps! Quantitatively: f“ (1.05 Å) = 1.85  0.05 f“ (1.54 Å) = 2.4  0.2

32 Thanks to my group, particularly S. Odintsov and I. Sabała Thanks to Gleb Bourenkov, MPI Hamburg c/o DESY


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