walter lecture 3 phasing final

8/4/2019 Walter Lecture 3 Phasing FINAL

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9/25/20

The problem………

h, k , l, I and sigma I have been collected in an X‐ray diffraction experiment

We have converted I to F, F=SQRT(I). How do we get coordinate positions for the macromolecular structure?

experimental data

Structure – PDB file

File of structure factor amplitudes

Electron density equation (Density rho (e‐/Ǻ3) at x,y,z in the unit cell)

|F(hkl)| structure factor amplitude (scalar) measured in the diffraction experiment.

“ ”α p ase n ra ans. e p ase o a re ec on. os ur ng e rac on experiment

Thus, F(hkl) is a vector with amplitude |F(hkl)| and phase α(hkl).

h k l are the miller indices!!!

x y z fractional coordinates of the unit cell. Sampling from 0‐1 in each direction.

electron density (ρ(xyz)) is calculated by summing all structure factors at each xyz (sampling grid) position in the unit cell. Thus, ALL reflections contribute to the electron density at each place in the cell.

Reduce the equation to 1‐D example to learn more about the equation. Homework

elec‐dens.xls.



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Electron density waves are described by vectors. 2 components amplitude and phase

α

|Fhkl|= amplitude of the wave

= relative phase of the wave compared

°α . . .

If α = 0, density wave will always have

maximum at x=0X=0

wave freq.

One rotation around circle = 1 unit dist in x.

Shift position of wave, relative to origin.

= if h=1 vector rotates 360 in 1 x trans.

if h=2 vector rotation 720 in 1 unh

X=1

The electron density in the unit cells of a crystal is a periodic wave function. It may be decomposed into its component waves .These component waves correspond to

the family of planes described by

miller indices h,k,l

h F α

frequency of each electron density wave.

In the diffraction experiment, we

measure intensity of reflections at different h k l. However, we have lost the

relative phase angles. In this figure,

A (F) is measured at different n (.e.g h). We do not know ψ (α). This is the phase

108°

problem in X‐ray crystallography.

If we know F and α we can calculate

electron density.

Since we know both lets calculate the

periodic wave. .xls file

144

144°

162°

234°



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h= 1, k=0, α = 0

h= 2,k=0,α = 0

Relationship between X‐ray diffraction and electron density waves

Electron density waves

1. Frequencies and directions determined by the miller indices.

2. Amplitudes and phases determined by the positions and types of atoms in the unit cell

X‐ray waves

1. Monochromatic (constant λ and constant freq.

2. Positions of waves on the detector determined by the crystal system , unit cell constants

and must satisfy Bragg’s law.

3. Diffracted beams exhibit different phases relative to one another (hit detector at different

times).

. or a g ven exper men a se up, e rac e n ens y epen en on e pos on an type of atoms in the unit cell.

(See the structure factor equation).



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FURTHER BREAK DOWN OF THE STRUCTURE FACTOR OR ANY VECTOR into components A, B.

A=|F(hkl)|cosα

B=|F(hkl)|sinα

α=tan‐1 (B/A)

|Fhkl| = SQRT(A2 + B2)

I = A2 + B2

Aα

A=1, B=0 A=0.71, B=0.71

If |F| = 1

f1, f2, and f3 are the scattering contribution of 3 different atoms in the unit cell of a crystal.

F (hkl) corresponds to a waves scattered

from all atoms in the crystal unit cell for a given set of planes h, k , l.

STRUCTURE FACTOR summation of scattering vectors of individual atoms in unit cell

Note all atoms contribute to the

measured intensity. Thus, to each structure factor.

In this figure, atom 1 scattering is greater 2 is greater than 3.

Scattering for each atom

ro en own into A=f1cos φ and B= f 1sin φ terms

The A and B components are Summed to generate the structure factor F = SQRT(A2+B2) and with phase =Atan2(B,A), or tan‐1 B/A



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Structure Factor equation (Summation over all atoms in the unit cell for each h, k, l.)

The structure factor F(hkl) (vector) may be calculated if the x,yz positions of atoms are known.

From x, y, z (fractional coordinates), α (phase) in radians can be calculated.

The amplitude of the structure factor |F(hkl)| is defined by the atomic scattering

factors ( f ) of atoms in the crystal

h k l are the miller indices!!!

Each F(hkl) consists of scattering contributions of all atoms in the unit cell.

Conclusion: To determine the phases of reflections, α, you must know at least one coordinate position in the crystal.

The Isotropic Atomic Scattering Factor

B=8π2<u>2

1. Resolution dependent

2. Reduced by movement of the atom

Scattering factor (f) plot for oxygen vs. resolution at B=0A2 and B=10A2

B‐factor found in PDB files

Isotropic scattering factors in electrons found in International tables Vol. 4. (Table 2.2)

Your B‐factor of 10 corresponds to a mean atom displacement of 0.36 A



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H =1e at 0.0

5A

S=16e at 0.0

1.7A

2.5A

UNIT CELL AND DIFFRACTION PATTERN OF A 1D CRYSTAL LEARN2

B=10A2 for each atom



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DIFFRACTION DATA FOR THE 1D CRYSTAL LEARN2

Role of the scattering factor in Altering the structure factor calculation

Isotropic f : 6 e‐ 11.39 ‐4.62 ‐10.41 ‐113.93

Resolution correction 7.52 ‐3.45 ‐6.69 ‐117.26

Resolution + B factor correction 5.04 ‐2.38 ‐4.48 ‐117.26

resolution

Thermal motion of atoms



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Structure factor calculation for the h=8 reflection with no resolution of B factor correction:

Note: green e‐density wave is across entire crystal!

2 atoms 6 elec. each

1) x=0.1, y=0.1 6elec.

2) x=0.6, y=0.45 6 elec

Refer to scat fact table.

Calculation of structure factors from atoms

Calculate the structure

factor of the 2, 0 reflection.

Do not apply resolution or B corrections to the scattering factor f



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h=2, k=0, l=0 reflection Calculation of a structure factor for the

f *

Answer=

Where is the 3,4 on the

Reciprocal lattice?

Calculate F and phase for the h=3, k=4 reflection.

. Phi=‐126



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1, 0 0 reflection: Measured but absent reflection ‐ SG determination

2 atoms 6 elec. each

1) x=0.1, y=0.1 6elec.

2) x=0.6, y=0.45 6 elec

The distance between atom 1 and atom 2 in the

X direction is 0.5. thus, the 1 0 0 wave cannot

represent + dens for both atoms. This is the

basis for systemic absences in diffraction data

which are used to determine the space groupLook back at the 2 0 0

Compared to 100 or 300, large F for h=4, k=0 reflection.



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HOW do you find x, y z position of atoms in crystal from just |F(hkl)| data?

No appropriate model structure or structural model too different

Often heavy atoms don’t bind OR don’t bind specifically

Heavy‐atoms change the unit cell – non‐isomorphous

Generally, requires labeling of the protein (SeMet). Great if molecule can be produced in e. coli. Need anomolous scatter e.g Fe protein



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Fph Fp Fh

FPH = FP + FHFor SIR or MIR Phasing Method

Dataset of

Native

crystal

Dataset of Native

Crystals derivatized

With heavy‐atom

Hg, or U, or Pb, or Pt

(look at scattering

factor table)

Estimate of

Heavy‐atom

contribution

easure e n ens y amp u e erences e ween eavy‐a om a ase p an Native dataset (Fp).

Subtraction of Fph – Fp is an estimate of the contribution of the heavy‐atom to the diffraction

pattern.

Find the phase of the heavy atom (Fh). Need to find the xyz position of the atom: Patterson Map.

Native crystal derivative crystal

Frames of data showing INTENSITY DIFFERENCEs between Native and derivatized crystal

Weak‐strong‐weak Weak‐weak‐weakSpot intensities



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Patterson map – allows the identification of x, y, z positions of atoms without phases

Vector map is zero everywhere except when the ends of the vector u, v, w, are atoms x, y, z.

atoms. 2 carbons peak should be 6x6=36e

The map contains all “self vectors” (e.g. atom1xyz‐atom1xyz=0,0,0). Thus, the origin is the sum of the squares of the atomic numbers of all atoms in the unit cell.

example, atoms at x1 and x2 in the crystal will be located at “u” positions x1‐x2 and x2‐x1 in the Patterson Map. As well as x1‐x1 and x2‐x2.

If N atoms in a unit cell The Patterson map will contain N2 atoms

Bottom line: Patterson Maps provide a way to determine x, y, z positions

without phases

Note differences between Patterson equation and



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u = x2‐x1, v=y2‐y1, w=z2‐z1 u = x1‐x1, v= y1‐y1, w=z1‐z1

u = x2‐x2, v=y2‐y2, w=z2‐z2

or u=0,0,0

Features of the Patterson map

u = x1‐x2, v=y1‐y2, w=z1‐z2 4 patterson peaks

1. Always centrosymmetric

2. Has the same lattice as the original space group

3. Remove all translational components of the original space group add the center…

4. Total of N*N peaks or N‐1*N to determine non origin peaks.

ymme ry

If Real cell SG = P21 x, y, z ‐x, y+1/2, ‐z

Then Patterson symmetry P2/m x, y, z ‐x, y, ‐z

‐x,‐y,‐z x, ‐y, z

Space group symmetry helps locate atoms in Patterson Maps – HARKER VECTORS

SG symmetry P2

Equiv positions

1. x, y, z2. ‐x, y, ‐z

HARKER VECTORS

u=x‐ ‐x, v= y‐y, w=z‐ ‐ z

= u=2x, v=0, w=2z

u= ‐x‐x, v=y‐y, w=‐z‐z

u= ‐2x, v=0, w=‐2z

OR x=u/2 and z=w/2 y



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1D Patterson Map for C=0.1 O=0.7 FROM LEARN2 data

400

600

800

1000

1200

Series1

-200

0

200

0 0.2 0.4 0.6 0.8 1

1. If you define the location of the heavy

FPH = FP + FH

Remember these are each VECTORS….we only have amplitudes

atom, you can calculate the vector FH. With FH you can define the phase of the

protein (αP).



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SIR

FPHFP

Two possible phases possible

with a single derivative

FPH

FH

FP

The position of a heavy‐atom has been determined. Thus, FH is a vector.

We want to determine the phase of Fp

SIR MIR

Using 2 derivatives resolves the ambiguity in the phasing

FH2

3 intensity data sets,

1. FP – native (no heavy atom)2. FPH1 – heavy atom 1 3. FPH2 – heavy atom 2

FH1

SIR methods some times work. Correct phases are selected using a method called solvent

leveling. E.g. Phases must be consistent with the physical rules of protein structures and crystals.

Proteins density is positive and continuous and solvent channels are constant and low. Starting

estimates of % solvent is taken from Vm



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80

100

120

140

40

60

80

10 0

12 0

14 0

AB

1D electron density maps From LEARN2 DATA SET

-20

0

20

40

0 0. 2 0.4 0.6 0.8 1

-20

0

0 0.2 0.4 0.6 0.8 1 1.

C

A= Electron density based on SFs calculated with atoms C=0.1, O=0.7

B= Electron density based on SFs calculated only with atom O=0.7

-20

0

20

40

60

80

10 0

12 0

14 0

16 0

0 0.2 0.4 0.6 0.8 1

C = Phases from B calculation , with structure

factors from A.

Partial and inaccurate phases, based only on the oxygen atom, reveal the location of the carbon in the e‐ density map.

Mo

d

c

c

c

c

#1 #2

Collect an X‐ray diffraction data set.

Calculate structure factor amplitudes for Known protein structure (Pmodel)

model

c

c

c

c

However, on t now w ere t s mo e s n t e un t ce …

Molecular Replacement

1. Finds the correct rotation of Pmodel relative to Pcrystal (Pc). 2. Finds the correct translation of Pmodel in Pcrystal.

3. This is effective when Pmodel is structurally similar to Pcrystal. The breakdown

occurs somewhere around ~35% or less sequence identity.



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Fph Fp Fh

FPH = FP + FH

Difference between SIR/MIR (heavy‐atom) and SAD/MAD anomalous phasing methods

SIR/MIR

1. Intensity differences between Fph and Fp

2. Requires data collection on 2 crystals

SAD/MAD |Fp(hkl)| α(hkl) = |Fp(‐h‐k‐l)|, ‐α(‐h‐k‐l)

Friedels Law = I of 111 = I of ‐1‐1‐1

Breakdown in Friedel’s Law for

Anomalous

scatterer

1. Collect Friedel pairs of reflections, which exhibit Intensity differences

Fph+ and Fph‐ ( e.g. the 111 and the ‐1‐1‐1. Related by 180° rotation of crystal

2. For SAD, requires data collection on 1 crystal.. Perfect isomorphism.

3. This is generally Selenomethionine

Anomalous scattering atoms

Anomalous Scattering http://skuld.bmsc.washington.edu/scatter/



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Anomalous scattering (AS) causes a 90° phase shift in the scattering factors. The

size of the AS factor components ( f’ (delta f) and f’’) are dependent on the atom.

As a result of AS, Freidel’s law breaks down for the anomalous scatter. This results in Intensity differences between F hkl and F(‐h‐k‐l), which can be used to locate

Anomalous scatters (using patterson functions) and calculating FH for these atomsResulting in phases for FP

Vector diagram showing the change inthe scattering factor, which causes

intensity differences that can be used

For phasing.

f’ and f’’ plot for Selenium

Collect the SAD data at the absorbtion edge of the anomalous scatter



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Friedel pairs

h k l plane

Breakdown in Friedel’s law

FPH (hkl)≠FPH (‐h‐k‐l) when an anomalous

scattering atom is

present

At the appropriate X‐ray energy, the white atom Scatters anomalousl and Friedel’s law breaks down. This is

‐h ‐k ‐l plane

FPH

FPH

because the phase of the anomalous scatterer is advanced by 90 degrees (FIXED) relative to the other atoms. This results

in different intensities at I h,k,l and I –h,‐k, ‐l. As a result,FPH+ and FPH‐ are different. With no anomalous scattering, the location of FPH+

And FPH‐ ( labeled in green) are identical and would have identical intensities.

These differences can be used to identify the anomalous scatterer and generate phases for the protein just as in SIR/MIR methods.

walter lecture 3 phasing final

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