Patterson Space and Heavy Atom
Isomorphous Replacement
MIR = Multiple heavy atom Isomorphous
Replacement phasing
SIR = Single heavy atom Isomorphous
Replacement phasing
Amplitudes for a 2 atom crystal
The amplitude of a wave F(h k l) scattered by just 2
atoms depends on the distance distance between them in
the (h k l) direction. Ignoring phase, what happens to the
amplitude as the two atoms slide perpendicular to the
Bragg planes? It stays the same. Only the phase changes.
Largest amplitudes are when
atoms are in-phase
If two atoms are in-phase, they have the largest
amplitude.
If one of the atoms (either one) is at the origin, then the
phase is 0°
Phase zero Fourier transform
Whatever the true phases of the F’s, if they are set to zero,
then the density goes to the points of intecsection, which
are the interatom vectors.
The phase zero inverse Fourier is called a
Patterson Map
Patterson maps
phase
Reverse transform
with phases
Reverse transform
without phases

(r )   F h e
i 2 h  r
h
.
h
.

P(r )   F h e
This is the Patterson function

F h e
 i2 h  r
h
.
it uses the measured
amplitude, no phase
 i2 h  r 
“Patterson space”
The Patterson map is the
centrosymmetric projection
real space
Patterson space
r
phase=
phase = 0
Using observed amplitudes, but setting all phases to 0 creates
a centro-symmetric image of the molecule.
Patterson map represents all
inter-atomic vectors
To generate a centrosymmetric projection in 2D, draw all inter-atomic
vectors, then move the tails to the origin. The heads are where
peaks would be.
For example, take glycine, 5
atoms (not counting H’s)
Move each vector to the origin
Patteron map for Gly in P1
unit cell vector peak
translational
symmetry
peaks
Can you reassemble glycine from this?
For small molecules,
vector/geometry problem can be
solved
...if you know the geometry (bond lengths, angles) of the molecule
http://www.cryst.bbk.ac.uk/xtal/mir/patt2.htm
Patterson peaks generated by
symmetry operations are on
Harker sections
P21
c
a
b
Real space
Patterson space
Harker section z=0.5
Harker sections tell us the
location of atoms relative to the
cell axes
y
x
Divide this vector
by two to get the
XY coordinates.
(The Z position
is found on
other sections)
Non-Harker sections tell us interatomic vectors not related by
symmetry
If there is more than one atom in the asu, you can get the
vector between them by searching for peaks in non-Harker
sections of the Patterson. (like the glycine example)
Then, combining knowledge from Harker sections (giving
absolute positions) and non-Harker sections (giving relative
positions) we can get the atomic coordinates.
Simple case: 2 atoms, P21
In a Harker section
y
x
z
y
x
The xy-position
is found
relative to the
2-fold axis, for
each atom
In a non-Harker section
The relative Z
position is found
for one atom
relative to the
other.
Protein Patterson maps are a
mess
Even if atomic resolution data is available, no individual
atom-atom peaks can be identified because of overlap.
The number of atom-atom peaks goes up as the square of
the number of atoms.
In class exercise: make a
Patterson map
Using the Escher Web Sketch:
Set the space group to P4mm
Place one atom at (0.4, 0.25)
Draw the Patterson vectors.
Place a second atom (different color) at (0.2, 0.1)
Draw the Patterson vectors.
Multiple isomorphous replacement
=Turning proteins into small molecules by soaking in heavy atoms
=
+
The Fourier transform (i.e. diffraction pattern) of a heavy
atom derivitive is the vector sum of the transforms of the
protein and the heavy atoms.
NOTE: protein and protein-heavyatom crystals must be isomorphous.
Subtracting amplitudes is like
subtracting density, almost
=
–
FH = FPH –FP
FH, the Fourier transform of the heavy atoms is the vector
difference of FPH and FH.
|FH| ≈ |FPH| – |FP|
Subtracting Fourier transforms
We can’t do FPH – FP = FH i directly because we don’t know the
phases.
Assume, for the moment, that |FH| << |FP| , |FPH| . Then the phases
of F and F are approximately the same. So, |FPH–FP| ≈ |FPH| – |FP|
P
PH
FP
FH
FPH
In that case, the Patterson map based on |FPH|–|FP| will be
approximately equal to Patterson map based on the true |FH| .
A difference Patterson is
dominated by the heavy atoms
self-peaks
cross-peaks
In class exercise: solving a
simple Patterson
Patterson peak (0.5, 0.1, 0.333)
Space group is P31
a
Where are the 3 heavy atoms?
(1) Draw a trigonal unit cell
(2) Draw an equilateral triangle around the cell
origin such that XY from these two vectors are
the sides.
b
(3) Estimate the coordinates of the triangle
vectices.
In class exercise: solving a
Patterson
Patterson peaks at (0.4, 0.2, 0.333),(-0.1, 0.5,
0.333), (0.5,-0.3,333)
Space group is P31
Where are the 3 heavy atoms?
(1) Draw a trigonal unit cell
(2) Draw an equilateral triangle around the cell
origin such that XY from these two vectors are
the sides.
(3) Estimate the coordinates of the triangle
vectices.
Calculating FH from the heavy
atom coordinates
FH (h )   f ge
i2 h  rg
g=all
.
heavy
atoms
Once we know the heavy atom coordinates, we
can calculate amplitude and phase for all
reflections FH.
We can represent a structure factor
of unknown phase as a circle
i
R
Radius of the circle is the amplitude. The true F
lies somewhere on the circle.
Harker diagram method for discovering
phase from amplitudes.
We know
amplitude
and phase
|FP + FH| = |FPH|
FPH
FH
FP
We know only
amplitude
One heavy
atom (SIR):
There are two
ways to make
the vector sums
add up.
Two heavy atom derivatives
(MIR), unambigous phases
FPH
FH
|FP + FH1| = |FPH1|
|FP + FH2| = |FPH2|
Or
|FP|=|FPH1–FH1|= |FPH2–FH2|
FP
In class exercise: Solve the phase
problem for one F using two FH’s
|FP| = 29.0
Draw three circles with the three diameters (scale
doesn’t matter)
|FPH1| = 26.0
Offset the PH1 circle from the P circle by -FH1
|FPH2| = 32.0
Offset the PH2 circle from the P circle by -FH2
FH1 = 7.8 H1 =155°
FH2 = 11.0 H1 =9°
Find the intersection of the circles
Additional topics
Crystal packing, solvent content, Mathews number, reciprocal
space symmetry, amplitude error and phase error.
Crystal packing
Protein crystal packing interactions are salt-bridges and Hbonds mostly. These are much weaker than the
hydrophobic interactions that hold proteins together. This
means that (1) protein crystals are fragile, and (2) proteins
in crystals are probably not significantly distorted from
their native conformations.
Reminder: The asymmetric unit
= the smallest region of the unit cell that is sufficient to
generate the whole unit cell by symmetry.
For space group P1, the whole cell is the asymmetric unit.
For other space groups, it is the fraction of the unit cell
corresponding to 1/Z where
Z = the number of equivalent positions.
For P212121 Z = 4
(x,y,z)
(–x,–y+1/2,–z+1/2)
(–x+1/2,y+1/2,–z)
(x+1/2,–y,z+1/2).
So the asymmetric unit
will be 1/4 of the unit
cell.
asu
How many molecules are in the asu?
(1)First find the total volume of the unit cell, Vcell.
(2)Divide by the number of equivalent positions, Z, this is the volume of
the asymmetric unit (Vasu=Vcell/Z)
(3)Calculate the approximate volume of your molecule using the
molecular weight and the density of protein, about 1.35 g/ml.
Vmol = ((MW/(1.35g/ml))/Nav)•1024Å3/ml
(4)Assume the number of molecules in the asu is n. Calculate percent
solvent in the crystal:
Psol=100%•(Vasu-n Vmol)/ Vasu
Quicker and easier: Matthews’ number
Matthews’ number is VM= Vcell/MW•Z
Molecular weight is roughly the number of residues in the
sequence x 110.
Z is the number of equivalent positions.
Matthews’ number for proteins varies from 1.66 to 4.0 for
crystals with 30% to 75% solvent. If VM is too high, there
must be more molecules in the asu than you thought.
In class exercise: how many molecules
are in the asymmetric unit?
Your molecule has 62 residues. Your crystal cell dimensions
are 40x50x60Å, orthorhombic. P212121 (Z=4). Use
Matthews’ number to get n.
VM= Vcell/(nMW•Z)
Get molecular volume.
Vmol=((MW/(1.35g/ml))/Nav)•1024Å3/ml ≈ 1.23MW
Percent solvent?
Psol=100%•(Vasu-n Vmol)/ Vasu
answer
Molecular weight is roughly 62*110 = 6820g/mol
Vmol = 1.23 * 6820 = 8390Å3
Vasu = (40*50*60)/4 = 30000Å3
if n=1, VM = 120000/(4* 6820) = 4.39 too high!!
if n=2, VM = 120000/(4* 2*6820) = 2.2 =just right
so n=2
Percent solvent = (30000 - 2*8390)/30000 = 44%
The Phase Problem
Oh no. We can’t measure phases!
X-ray detectors (film, photomultiplier tubes, CCDs, etc)
can measure only the intensity of the X-rays (which is
the amplitude squared), but we need the full wave
equations Aei for each reflection to do the reverse Fourier
transform.
And because it is called the phase “problem”, the process
of getting the phases is called a “solution”. That’s why we
say we “solved” the crystal structure, instead of
“measured” or “determined” it.
Phase is more
important than
amplitude
color=phase angle
darkness=amplitude
Reciprocal space symmetry
Rotating the crystal rotates the reciprocal lattice by the same
amount. Applying a mirror or inversion does the same to the
reciprocal lattice.
So the symmetry in the reciprocal lattice is the same as the
symmetry in the crystal, with the addition of Friedel
symmetry, but without translational symmetry.
The addition of Friedel symmetry causes mirror planes to
appear in reciprocal space when there is 2-fold rotational
symmetry in the crystal.
(h,k,l)
(h,k,l) ===> (-h,-k,l) ===> (h,k,-l)
2-fold
Friedel
mirror
(h,k,-l)
Reciprocal space symmetry
operators are the transpose
...of the real space symmetry operators.
The forward transform is
summed over all atoms in
the asu, and all symmetry
operators Z.
So the reverse transform
is summed over the
unique set, times Z.
F(h )    f g e
i 2 h  Zrg
Z
g
.

(r )    F h e
Z
h
.
 
 i2  h Z r
Transposing the matrix is the same as
reversing the order of multiplication
a b
x y z 
d e
g h
c 
f 
  xa  yd  zg xb  ye  zh xc  yf  zi
i 
Flip the matrix and change the order, and you get the same thing
(written as a column instead of a row)
a d gx xa  yd  zg
b e hy  xb  ye  zh 

  

c f i z  xc  yf  zi 
The Unique set is the
“asymmetric unit” of reciprocal
space
Unique
data
Initial phases
Phases are not measured
exactly because amplitudes
are not measured exactly.
Error bars on FP and FPH
create a distribution of
possible phase values .
width of circle is 1s
deviation, derived
from data collection
statistics.