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
The problem is that the raw data are
the squares of the modulus of the
Fourier transform.
That´s the famous phase problem.
In protein crystallography, there are several
ways to get the phases:
• Molecular replacement
• Heavy atom methods
• Direct methods
• Non-standard methods
Molecular replacement
Mol A: GPGVLIRKPYGARGTWSGGVNDDFFH...
Mol B: GPGIGIRRPWGARGSRSGAINDDFGH...
Mol A
?
Mol B
If we have phases from a similar model...
Amplitudes: Manx
Amplitudes: Cat
Phases:
Manx
Phases:
Phases:
Manx
Cat
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 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.
F
F
I
h , k ,l

i, j
h , k ,l

f
i
exp[ 2i (hxi  kyi  lz i )]

f
j
exp[ 2i (hx j  ky j  lz j )]
í
h , k ,l
j
f f
i
j
exp[ 2i(h( xi  x j )  k ( yi  y j )  l ( zi  z j ))]
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)
rotfun
(generate)
tabfun
xyzin1
(*1.pdb)
table1
(*1.tab)
rotfun
(cross)
trafun
(CB)
SOLUTRC
SOLUTTF
clmn0
(*0.clmn)
rotfun
(cross)
}
rotfun
(clmn)
hklpck1
(*1.hkl)
clmn1
(*1.clmn)
fitfun
(rigid)
pdbset
SOLUTF
solution.pdb
Poor phases yield self-fulfilling prophesies
Amplitudes: Karlé
Amplitudes: Hauptmann
Phases:
Karlé
Phases:
Phases:
Karlé
Amplitudes: Hauptmann
If Karlé phases Hauptmann,
Hauptmann is Karléd!
Hauptmann
Heavy atom methods
?
Can we do X-ray holography?
Can we do holography with crystals?
In principle yes, but the limited coherence
length requires a local reference scatterer.
For a particular h,k,l
FH2
FPH1
FP
FH1
FPH1
we can collect all knowledge about
amplitudes and phases in a diagram
(the so-called Harker diagram)
• 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.
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.
Which elements are useful for MAD data collection?
0.5 Å
25 keV
LIII
64-
7 keV
K
26-46
1.8 Å
The MAD periodic table
H
He
Li Be
B
N
O
F
Na Mg
Al Si P
S
Cl Ar
K
Ca Sc Ti V
Rb Sr Y
C
Ne
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I
Cs Ba La Hf Ta W
Xe
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
All phasing can be done on one crystal.
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.
F
F
2i hxky lz 
h ,k ,l
  x, y, z e
2i  hxky lz 
h ,  k ,  l
  x, y, z e
assuming
F
dxdydz
  x, y , z     x, y , z 
2i  hx ky  lz 
h,  k , l
    x, y , z  e
with absorption:
dxdydz
dxdydz  F h ,k ,l
  x, y , z     x, y , z 
Direct methods
?
Atomic resolution data
the best approach for small molecules
If atoms can be treated as point-scatterers, then
if all involved structure factors are strong
100 atoms in the unit cell
a small protein
The method is blunt for lower resolution or too many atoms.
Three-beam phasing
?
very low mosaicity data
an exciting, but not yet practical way
An example from our work
(solved by a combination of MAD and MR)
Metal ions
Can we tell from the fluorescence scans?
Compton
Zn
Cu
Fe
Ni
Co
Normally yes, but not in this case!
Can we tell from the anomalous signal?
order in the periodic table: Fe, Co, Ni, Cu, Zn
Here´s the maps!
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. Sabała
Thanks to Gleb Bourenkov, MPI Hamburg c/o DESY