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## Patterson Space and Heavy Atom Isomorphous Replacement

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### Patterson map represents all inter-atomic vectors ... http://www.cryst.bbk.ac.uk/xtal/mir/patt2.htm. R ... Protein Patterson maps are a mess ... – PowerPoint PPT presentation

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Title: Patterson Space and Heavy Atom Isomorphous Replacement

1
Patterson Space and Heavy Atom Isomorphous
Replacement
MIR Multiple heavy atom Isomorphous Replacement
phasing
SIR Single heavy atom Isomorphous Replacement
phasing
2
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.
3
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
4
Phase zero Fourier transform
Whatever the true phases of the Fs, 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
5
Patterson maps
phase
Reverse transform with phases
Reverse transform without phases
This is the Patterson function
it uses the measured amplitude, no phase
6
The Patterson map is the centrosymmetric
projection
Patterson space
real space
Patterson space
r
phase 0
phasea
Using observed amplitudes, but setting all phases
to 0 creates a centro-symmetric image of the
molecule.
7
Patterson map represents all inter-atomic vectors
To generate a centrosymmetric projection in 2D,
draw all inter-atomicvectors, then move the
tails to the origin. The heads are wherepeaks
would be.
Move each vector to the origin
For example, take glycine, 5 atoms (not counting
Hs)
8
Patteron map for Gly in P1
unit cell vector peak
translational symmetry peaks
Can you reassemble glycine from this?
9
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
10
Patterson peaks generated by symmetry operations
are on Harker sections
P21
c
a
b
R
R
R
Patterson space
Real space
Harker section z0.5
11
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)
12
Non-Harker sections tell us inter-atomic 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.
13
Simple case 2 atoms, P21
In a Harker section
y
The xy-position is found relative to the 2-fold
axis, for each atom
x
z
In a non-Harker section
y
The relative Z position is found for one atom
relative to the other.
x
14
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.
15
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.
16
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.
17
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
18
Subtracting Fourier transforms
We cant do FPH FP FH i directly because we
dont know the phases. Assume, for the moment,
that FH ltlt FP , FPH . Then the phases of FP
and FPH are approximately the same. So,
FPHFP FPH FP
FP
FH
FPH
In that case, the Patterson map based on
FPHFP will be approximately equal to
Patterson map based on the true FH .
19
A difference Patterson is dominated by the heavy
atoms
cross-peaks
self-peaks
20
In class exercise solving a simple Patterson
Patterson peak (0.5, 0.1, 0.333) 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.
a
b
21
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.
22
Calculating FH from the heavy atom coordinates
gall heavy atoms
Once we know the heavy atom coordinates, we can
calculate amplitude and phase for all reflections
FH.
23
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.
24
FP FH FPH
Harker diagram method for discovering phase from
amplitudes.
We know only amplitude
We know amplitude and phase
FPH
One heavy atom (SIR) There are two ways to make
FH
FP
25
Two heavy atom derivatives (MIR), unambigous
phases
FPH
FH
FP FH1 FPH1 FP FH2
FPH2 Or FPFPH1FH1 FPH2FH2
FP
26
In class exercise Solve the phase problem for
one F using two FHs
Draw three circles with the three diameters
(scale doesnt matter) Offset the PH1 circle from
the P circle by -FH1 Offset the PH2 circle from
the P circle by -FH2 Find the intersection of the
circles
FP 29.0 FPH1 26.0 FPH2 32.0
FH1 7.8 aH1 155 FH2 11.0 aH1 9
27
Crystal packing, solvent content, Mathews number,
reciprocal space symmetry, amplitude error and
phase error.
28
Crystal packing
Protein crystal packing interactions are
salt-bridges and H-bonds 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.
29
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,y1/2,z1/2)(x1/2,y1/2,z)(x1
/2,y,z1/2).
So the asymmetric unit will be 1/4 of the unit
cell.
asu
30
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 (VasuVcell/Z) (3)Calculate the
approximate volume of your molecule using the
molecular weight and the density of protein,
Å3/ml (4)Assume the number of molecules in the
asu is n. Calculate percent solvent in the
crystal Psol100(Vasu-n Vmol)/ Vasu
31
Quicker and easier Matthews number
Matthews number is VM Vcell/MWZ 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.
32
In class exercise how many molecules are in the
asymmetric unit?
dimensions are 40x50x60Å, orthorhombic. P212121
(Z4). Use Matthews number to get n.
VM Vcell/(nMWZ)
Get molecular volume.
Vmol((MW/(1.35g/ml))/Nav)1024Å3/ml 1.23MW
Percent solvent?
Psol100(Vasu-n Vmol)/ Vasu
33
Molecular weight is roughly 62110
6820g/mol Vmol 1.23 6820 8390Å3 Vasu
(405060)/4 30000Å3 if n1, VM 120000/(4
6820) 4.39 too high!! if n2, VM 120000/(4
26820) 2.2 just right so n2 Percent solvent
(30000 - 28390)/30000 44
34
Oh no. We cant measure phases!
The Phase Problem
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 Aeia 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. Thats why we say we solved the
crystal structure, instead of measured or
determined it.
35
Phase is more important than amplitude
colorphase angle darknessamplitude
36
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) gt (-h,-k,l) gt (h,k,-l)
mirror
2-fold
Friedel
(h,k,-l)
37
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.
38
Transposing the matrix is the same as reversing
the order of multiplication
Flip the matrix and change the order, and you get
the same thing (written as a column instead of a
row)
39
The Unique set is the asymmetric unit of
reciprocal space
Unique data
40
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 a.
width of circle is 1s deviation, derived from
data collection statistics.