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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 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

Patterson maps

phase

Reverse transform with phases

Reverse transform without phases

This is the Patterson function

it uses the measured amplitude, no phase

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.

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)

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

R

R

R

Patterson space

Real space

Harker section z0.5

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 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.

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

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 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 .

A difference Patterson is dominated by the heavy

atoms

cross-peaks

self-peaks

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

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

gall 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.

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

the vector sums add up.

FH

FP

Two heavy atom derivatives (MIR), unambigous

phases

FPH

FH

FP FH1 FPH1 FP FH2

FPH2 Or FPFPH1FH1 FPH2FH2

FP

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

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 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.

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

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,

about 1.35 g/ml.Vmol ((MW/(1.35g/ml))/Nav)1024

Å3/ml (4)Assume the number of molecules in the

asu is n. Calculate percent solvent in the

crystal Psol100(Vasu-n Vmol)/ Vasu

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.

In class exercise how many molecules are in the

asymmetric unit?

Your molecule has 62 residues. Your crystal cell

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

answer

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

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.

Phase is more important than amplitude

colorphase angle darknessamplitude

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)

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.

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)

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 a.

width of circle is 1s deviation, derived from

data collection statistics.