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The phase problem in protein crystallography. The phase problem in protein crystallography. Bragg diffraction of X-rays (photon energy about 8 keV, 1.54 Å). Structure factors and electron density are a Fourier pair.

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The phase problem

in protein crystallography


The phase problem

in protein crystallography


Bragg diffraction of X-rays

(photon energy about 8 keV, 1.54 Å)



The problem is that the raw data are the squares of the modulus of the Fourier transform.

That´s the famous phase problem.



Molecular replacement the phases:

Mol A: GPGVLIRKPYGARGTWSGGVNDDFFH...

Mol B: GPGIGIRRPWGARGSRSGAINDDFGH...

?

Mol A

Mol B


If we have phases from a similar model... the phases:

Amplitudes: Manx

Phases: Manx

Amplitudes: Cat

Phases: Cat

Phases: Manx

Amplitudes: Cat

...we can combine them with the experimental amplitudes to get a better model.

we can use


Patterson maps can be used to find the phases:

.... the proper orientation (rotation)

.... the proper position (translation)

for the search model.

The density map

The Patterson map


The Patterson map is the Fourier transform of the intensities.

It can be calculated without the phases.


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


Flow chart of a typical molecular replacement procedure (AMORE)

rotfun (clmn)

sortfun

hklin (*.mtz)

hklpck0 (*0.hkl)

clmn0 (*0.clmn)

}

rotfun (cross)

rotfun (generate)

rotfun (clmn)

tabfun

xyzin1 (*1.pdb)

table1 (*1.tab)

hklpck1 (*1.hkl)

clmn1 (*1.clmn)

fitfun (rigid)

pdbset

trafun (CB)

rotfun (cross)

SOLUTF

SOLUTTF

solution.pdb

SOLUTRC


Poor phases yield self-fulfilling prophesies (AMORE)

Amplitudes: Karlé

Phases: Karlé

Amplitudes: Hauptmann

Phases: Hauptmann

Amplitudes: Hauptmann

Phases: Karlé

If Karlé phases Hauptmann, Hauptmann is Karléd!




Can we do holography with crystals? (AMORE)

In principle yes, but the limited coherence length requires a local reference scatterer.


For a particular h,k,l (AMORE)

FH2

FP

FPH1

FH1

FPH1

we can collect all knowledge about amplitudes and phases in a diagram

(the so-called Harker diagram)



Absorption is accompanied by dispersion. not strictly isomorphous.

This Kramers-Kronig equation is very general:

Its (almost) only assumption is the existance of a universal maximum speed (c) for signal propagation.


Which elements are useful for MAD data collection? not strictly isomorphous.

25 keV

0.5 Å

LIII

64-

7 keV

1.8 Å

K

26-46


The MAD periodic table not strictly isomorphous.

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

ActinidesTh Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr


All phasing can be done on one crystal. not strictly isomorphous.

F1,2

a

b

F-1,-2

F1,2 : scattering from b is phase  behind

F-1,-2 : scattering from b is phase  ahead

In the presence of absorption, Bijvoet pairs are nonequal.


assuming not strictly isomorphous.

with absorption:


Direct methods not strictly isomorphous.

?

Atomic resolution data

the best approach for small molecules


If atoms can be treated as point-scatterers, then not strictly isomorphous.

if all involved structure factors are strong


100 atoms in the unit cell not strictly isomorphous.

a small protein

The method is blunt for lower resolution or too many atoms.


Three-beam phasing not strictly isomorphous.

?

very low mosaicity data

an exciting, but not yet practical way


An example from our work not strictly isomorphous.

(solved by a combination of MAD and MR)

Metal ions


Can we tell from the fluorescence scans? not strictly isomorphous.

Compton

Zn

Cu

Fe

Ni

Co

Normally yes, but not in this case!


Can we tell from the anomalous signal? not strictly isomorphous.

order in the periodic table: Fe, Co, Ni, Cu, Zn


Here´s the maps! not strictly isomorphous.

2fo-fc map, 1.05 Å

anomalous map, 1.05 Å

anomalous map, 1.54 Å

Quantitatively:

f“ (1.05 Å) = 1.85  0.05 f“ (1.54 Å) = 2.4  0.2


Thanks to my group, particularly S. Odintsov and I. Saba not strictly isomorphous.ła

Thanks to Gleb Bourenkov, MPI Hamburg c/o DESY


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