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

ρ(r r ) = F

r h ( )r

h .

∑ e−i2πr h •r r = F

r h ( )r

h .

∑ e−i2πr h •r r +αReverse transform

with phases

P(r r )= F

r h ( )r

h .

∑ e−i2πr h •r r Reverse transform

without phases

it uses the measured amplitude, no phase

This is the Patterson function

phase

real space Patterson space

The Patterson map is the centrosymmetric projection

phase= phase = 0

“Patterson space”

r

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.

For example, take glycine, 5 atoms (not counting H’s)

Move each vector to the origin

Patteron map for Gly in P1

Can you reassemble glycine from this?

translational symmetry peaks

unit cell vector peak

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 sectionsP21

Real spacePatterson space

c

Harker section z=0.5

a

b

Harker sections tell us the location of atoms relative to the

cell axes

Divide this vector by two to get the XY coordinates.

x

y

(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

x

y

The relative Z position is found for one atom relative to the other.

z

x

yIn a non-Harker section

In a Harker section

The xy-position is found relative to the 2-fold axis, for each atom

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

= +

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.

=Turning proteins into small molecules by soaking in heavy atoms

Subtracting amplitudes is like subtracting density, almost

= –

FH, the Fourier transform of the heavy atoms is the vector difference of FPH and FH.

|FH| ≈ |FPH| – |FP|

FH = FPH –FP

Subtracting Fourier transformsWe 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 FP and FPH are approximately the same. So,

FHFP

FPH

In that case, the Patterson map based on |FPH|–|FP| will be

approximately equal to Patterson map based on the true |FH| .

|FPH–FP| ≈ |FPH| – |FP|

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

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

FH(r h )= f g( )

.

∑ ei2π

r h •

r r g

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

R

i

Radius of the circle is the amplitude. The true F lies somewhere on the circle.

|FP + FH| = |FPH|We know amplitude and phase

We know only amplitude

FH

FPH

FP

One heavy atom (SIR): There are two ways to make the vector sums add up.

Harker diagram method for discovering phase from amplitudes.

|FP + FH1| = |FPH1||FP + FH2| = |FPH2|Or|FP|=|FPH1–FH1|= |FPH2–FH2|

Two heavy atom derivatives (MIR), unambigous phases

FH

FPH

FP

In class exercise: Solve the phase problem for one F using two FH’s

|FP| = 29.0

|FPH1| = 26.0

|FPH2| = 32.0

FH1 = 7.8 H1 =155°

FH2 = 11.0 H1 =9°

Draw three circles with the three diameters (scale doesn’t 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

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,–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?

Percent solvent?

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)

Vmol=((MW/(1.35g/ml))/Nav)•1024Å3/ml ≈ 1.23MW

Get molecular volume.

Psol=100%•(Vasu-n Vmol)/ Vasu

answerMolecular 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%

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.

The Phase Problem

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)2-fold Friedel

(h,k,l)

(h,k,-l)mirror

Reciprocal space symmetry operators are the transpose

...of the real space symmetry operators.

F(r h ) = f g( )

g.

∑ ei2π

r h •Z

r r g

Z∑The forward transform is

summed over all atoms in the asu, and all symmetry operators Z.

ρ(v r ) = F

r h ( )r

h .

∑ e−i2π

r h •Z( )

r r

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

x y z( )

a b c

d e f

g h i

⎛

⎝

⎜ ⎜

⎞

⎠ ⎟ ⎟ = xa+yd+zg xb+ye+zh xc+yf+zi( )

a d g

b e h

c f i

⎛

⎝

⎜ ⎜

⎞

⎠ ⎟ ⎟

x

y

z

⎛

⎝

⎜ ⎜

⎞

⎠ ⎟ ⎟ =

xa+yd+zg

xb+ye+zh

xc+yf+zi

⎛

⎝

⎜ ⎜

⎞

⎠ ⎟ ⎟

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 phasesPhases 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 1 deviation, derived from data collection statistics.