macromolecular crystallography part ishelx.uni-ac.gwdg.de/.../pdfs/gruene_day1-lecture.pdf ·...

Molecular Biology Course 2009

Macromolecular Crystallography Part ITim Grüne

University of Göttingen

Dept. of Structural Chemistry

November 2009

http://shelx.uni-ac.gwdg.de

[email protected]

http://shelx.uni-ac.gwdg.de

Molecular Biology Course 2009 Macromolecular Crystallography I Tim Grüne

Techniques for Structural Biology

Nuclear Magnetic Resonance Electron Microscopy

X-ray Crystallography Neutron Diffraction

X-Ray Crystallography 1/61


Macromolecular Crystallography

Crystal

Final ModelElectron

Density Map

X-ray

Diffraction

Introduction 2/61


X-Ray Crystallography

X-Rays 3/61


Spectrum of Electromagnetic Waves

Name wavelength λ Name wavelength λ

Radio waves 30cm–10km Microwaves 1mm–30cm

Infrared 800nm – 1mm Visible Light 400nm – 800nm

Ultra violet 1nm – 400nm X-rays 10pm – 1nm

Gamma <1pm (1nm = 10 Å, 10pm = 0.01 Å)

X-Rays 4/61


Why X–Rays?

• Rule of thumb in Optics: Objects smaller than λ/2 invisible

• Typical bond lengths: around 1.5 Å(C-C = 1.54 Å)

• Crystallography: typical wavelengths between 0.8 Å and 2 Å

X–Rays 5/61


Generating X–Rays 1/2: Rotating Anode

Fixed wavelength at 1.54 Å (Cu) or 0.7 Å (Mo).

X–Rays 6/61


Generating X–Rays 1/2: Synchrotron

Variable wavelength, high intensity.

X–Rays 7/61


X-Ray Crystallography

Crystals 8/61


Example of crystals

Ice Crystals Salt

CuSO4 Cueva de los Cristales, Mexico

Inorganic crystals and es-

pecially ionic crystals are

quite stable w.r.t. tempera-

ture, pressure,. . .

Crystals 9/61


Macromolecular Crystals

• Crystals of (macro-) molecular compounds grow from

and in solution

• Very fragile

• Level of humidity must be kept constant

• sensitive of X–ray damage

Crystals 10/61


Inside Crystals

The regular packing and repetition of one compound (molecule, dimer, . . . ) defines the term

crystal. The regularity is the reason why we can use X–rays to determine the structure of the

compound inside.

Crystals 11/61


Diffraction Experiment

X–Ray Diffraction 12/61


The Disadvantage of X–rays

��

��

��

��

��

��

Screen

visible light

object (focussing) lense image

Light is scattered from an object in all directions. A lense (e.g., eye lense, camera, micro-

scope, or telescope) collects some of the scattered light and focusses it on a screen (and

eventually the retina). This creates an image of the object.



The Disadvantage of X–rays

��

��

��

��

Screen

object

X-rays

no image - "blur"object

X–rays are too energetic to be bent like light. They run straight through any lense, and the

image cannot be reconstructed.



Crystal in X–Ray Beam

��

��

SpotsCrystalX-ray source



Principle of an X–ray experiment

X-ray beam

λ ≈ 1Å(0.1nm)

crystal ≈ (0.2mm)3

(T. Schneider)

Diffraction patternon detector

The crystal diffracts in all directions, even backwards. Since the detector cannot cover the

full sphere around the crystal, the crystal is rotated during an experiment.

One typical data set consists of hundreds to thousands of single images (frames).X–Ray Diffraction 16/61


Crystals and X–Rays



Path Lengths

��

��

CrystalX-ray source

Each atom scatters in each direction. Depending on the position on the screen, two rays from

corresponding atoms have different path lengths. This causes constructive or destructive

interference on the screen.



Interference

Because of the regularity of crys-tals, there are places on thescreen where δ = path difference

wavelength isa multiple of 2π. That’s where thediffraction image has spots.

At all other places there is only

noise.

−30

−20

−10

0

10

20

30

−3 −2 −1 0 1 2 3

δ = 0 (30 rays)

−6

−4

−2

0

2

4

6

−3 −2 −1 0 1 2 3

δ = π (180°) (6 rays)



Interference

Because of the regularity of crys-tals, there are places on thescreen where δ = path difference

wavelength isa multiple of 2π. That’s where thediffraction image has spots.

At all other places there is only

noise.

−30

−20

−10

0

10

20

30

−3 −2 −1 0 1 2 3

δ = 0 (30 rays)

−6

−4

−2

0

2

4

6

−3 −2 −1 0 1 2 3

δ = π (180°) (6 rays)

−6

−4

−2

0

2

4

6

−3 −2 −1 0 1 2 3

δ = 0.10*π (18°) (6 rays)

−6

−4

−2

0

2

4

6

−3 −2 −1 0 1 2 3

δ = 0.10*π (18°) (30 rays)



Crystallography — Understanding the Experiment

Crystallography 21/61


The Unit Cell

A crystal consists of imaginary unit cells which are “piled up” like bricks to compose the whole

crystal.

→ → c

ab

γ

αβ

The unit cell is characterised by the three side lengths, a, b, c and angles α, β, γ.

α: angle between b and c

β: angle between c and a

γ: angle between a and b



Spot Prediction

• The position of the spots depends only on the unit cell, not on the molecule that makes

the crystal.

• For data processing it is important to know the expected spot positions on the detector.

• The positions can be predicted with the concept of virtual planes in the crystal lattice.



Lattice Planes

• a, b: Sides of the (2D) Unit cell

• Dots: Corners of other unit cells

• Planes: Connect any two dots and

shift parallel (in 3D: any three dots not

on a line)

• Purple lines divide a and b into 1 part

each: call it the (1,1)-plane

• Green lines divides a into 2 parts and

b into 5 parts: call it the (2,5)-plane.

In a 3-dimensional crystal one needs 3 corners for one plane and three intersections of the

unit cell. The resulting three integer numbers are called the Miller-Index of the plane.



Bragg’s Law

Imagine the X-rays are reflected at the crystal planes. Like in normal optics we can assume

that the input angle equals θ the output angle.

��

��

e.g. (2 5) plane

dX-rays θ

θ

When the wavelength

λ is fixed, there is

only constructive inter-

ference for the X-rays

reflected from

one lattice plane, when nλ2d = sin θ (for any positive integer n) (Bragg’s Law).

In practice, only n = 1 needs to be considered because for larger n the resulting spot is too week to be detected.



Diffraction

Diffraction 26/61


Diffraction Pattern

Every lattice plane creates exactly one spot on the detector. It is not a “streak” (or similar) but

really a spot because the crystal is very small compared to the distance to the detector.

beam

crystal

image

X-ray source

recorded

Diffraction 27/61


Number of Reflections

The size of a typical protein crystal is about 150µm×150µm×150µm. A unit cell is about

50Å×50Å×50Å= 0.005µm× 0.005µm× 0.005µm.

I.e., in each direction there are 1500.005 = 30,000 lattice points, i.e. 30,000 × 30,000 ×

30,000 = 27,000,000,000,000 possible Miller-Indices, i.e., 27,000,000,000,000 pos-

sible reflections.

A typical data collection would probably detect a few hundred thousand or maybe million of

them — still quite a few.

Diffraction 28/61


Information Content of Reflections

• The position of each reflection depends only on

1. Unit Cell (a, b, c, α, β, γ)

2. Crystal orientation w.r.t. the incident beam direction

3. Wavelengthλ

but not the atoms inside.

• The intensity depends on the molecule(s) (atoms) inside the crystal.

The goal of a diffraction experiment therefore is to determine the unit cell parameters (a, b, c,

α, β, γ) and the reflection intensities as accurate as possible.

Diffraction 29/61


Resolution of a Data Set

A perfect, infinitely large crystal would diffract all around, also backwards. A real crystal has

a maximal angle θmax beyond which no reflections are recorded.

θ

θ

beam 2θ

crystal

recordedimage

The distance d = λ2 sin θ is called the resolution of the reflection, and the smallest distance

dmin = λ2 sin θmax

is the resolution of the data set

Diffraction 30/61


Examples for the Resolution of Electron Density Maps

The images show three times the same region of a protein map at different resolutions.

N.B.: Crystallographers usually speak of “high resolution” when the number is small (e.g.

1.2Å) and of “low resolution” when the number is large (e.g. 3Å).

Diffraction 31/61


Crystal Symmetry

Symmetry 32/61


What is Symmetry?

We say something is “symmetric” when it looks identical after moving it:

This applies to the single unit cell as well as the whole crystal lattice.Symmetry 33/61


Why Symmetry?

The main benefit from taking crystal symmetry into account:

higher data accuracy

Crystal symmetry forces the intensities of some reflections to be identical, even though they

are most likely measured with different intensities due to experimental errors. The average of

such reflections will come closer to the real value and also estimate the error.

Symmetry 34/61


Symmetry: Classification of the unit cell

Seven different categories of “boxes” or unit cells can be distinguished.

a 6= b 6= c

α 6= β 6= γ

triclinic

a 6= b 6= c

α = γ = 90◦ 6= β

a = b 6= c

tetragonal

a 6= b 6= c

orthorhombic

a = b = c

α = β = 90◦

γ = 120◦

hexagonal monoclinic

α=β

=γ

=90◦

cubic

trigonal

a = b 6= c

α = β = 90◦

γ = 120◦

b

a

a = b = c

γ

βcα

Triclinic is the most general one (six numbers required for characterisation), whereas cubic is

the most special one (only one number).

Symmetry 35/61


Building Crystal with Symmetry

The total lattice can have more symmetry than the unit cell by itself:

a = b

γ = 120◦

The hexagonal crystal lattice has a 6-fold rotation axis while the unit cell does not.Symmetry 36/61


Possible Symmetry Elements

The following symmetry elements can occur in a crystal:

rotations ( only 2-, 3-, 4- and

6-fold axes possible)

3-fold 4-fold 6-fold2-fold

mirrors and inversion cen-

tres (only small molecules!)

centre of inversionmirror plane

Symmetry 37/61


No 5-fold axis

There is no 5-fold rotation axis for crystals:

The pentagon does not cover the plane with-

out gaps.

Symmetry 38/61


Space Groups

• One can combine symmetry elements.

• One possible combination of one or more elements is called a space group.

• There are 230 space groups.

• Macromolecules do not allow inversions and mirrors

• There are only 65 space groups possible for macromolecules.

Symmetry 39/61


Example: Spacegroup P6

The unit cell is only a concept in order to understand the experiment. These pictures show

an example of a DNA-oligo. The spacegroup is P6, i.e., there is a six-fold rotation axis along

one axis.Symmetry 40/61


Data Collection, Processing, and Scaling

Data Collection, Processing, and Scaling 41/61


Data in Frames

Detectors are planar and also limited in size. To capture all reflections, the crystal is therefore

rotated during data collection. The data are recorded during a rotation of typically 0.5 − 1◦

per frame.

60◦ − 61◦

X-ray beam

φ rotation

0◦ − 1◦

X-ray beam

φ rotation

Between the recording of two frames the detector is read out and erased.



Data Processing

Only the reflections are interesting: their Miller-Indices, Intensities, and error estimates.

The determination of these quantities is called data processing or data integration. It results

in a simple list of reflections:

det.–coord’sH K L Intensity error x y z[◦]-3 0 -3 4,162 153.7 1181.5 1235.6 107.4-3 -3 0 2,747 107.5 1110.9 1205.1 76.0-3 0 3 3,946 145.1 1156.2 1233.4 18.31 1 -4 5,933 213.9 1215.0 1226.7 165.04 1 -1 5,640 206.4 1209.5 1074.0 57.3

Even the coordinates of the reflection (x,y,z) are discarded after the next step, the data scal-

ing.



Data Scaling

Crystallographic Theory assumes a perfect crystal — quite unrealistic. To better match ex-

periment and perfection, the data must be scaled:

1. Scaling of reflections within one dataset — this must always be done:

• The intensity of the beam might change from frame to frame — especially at syn-

chrotron sources

• The crystal might be larger in one direction than another — spot intensity increases

with the volume the beam traversed.

• Different regions of the detector have different sensitivity

2. Scaling of reflections of different datasets, e.g. from two different crystals. This is only

necessary if two or more datasets exist.



Friedel’s Law

Even in the simplest space group (P1) with no symmetries, scaling can be carried out be-

cause of Friedel’s law: Reflections with negated indices, i.e., (h, k, l) and (−h,−k,−l) have

the same intensity. They arise from reflection at the same set of planes, but on opposite

sides.

i.e. rotation of latticeAfter rotation of crystal,

X-rays

“bottom”

“top”

leads to reflection (−h − k − l)

Before rotation of crystal,i.e. rotation of lattice

X-rays“top”

“bottom”

leads to reflection (h k l)



The Phase Problem

The Phase Problem 46/61


Intensities and Atoms

Spots and intensities do not tell us much about the molecule inside the crystal. Where is the

connection?

Physically the spots are the result from the interaction of the electrons in the crystal and the

X-rays. At every point (x, y, z) in the crystal these electrons give rise to an effective electron

density ρ(x, y, z).

If we knew the electron density ρ(x, y, z) inside the crystal, the intensity I(hkl) of the reflec-

tion with Miller-Indices (hkl) could be calculated as

I(hkl) =∣∣∣∣const ∗ ∫

unit cellρ(x, y, z)e2πi(hx+ky+lz)

∣∣∣∣2 (1)



The Structure Factor

The term const ∗ ∫unit cell ρ(x, y, z)e(2πihx+ky+lz) is a complex number. It is called the

structure factor of the reflection (hkl).

As mathematical formula one writes

|F (hkl)|eiφ(hkl) = const ∗∫unit cell

ρ(x, y, z)e2πi(hx+ky+lz) (2)

|F (hkl)| structure factor amplitude of reflection (hkl). Important: I(hkl) = |F (hkl)|2

φ(hkl) phase of reflection (hkl)



The Structure Factor

The equation (1), I(hkl) =∣∣const ∗ ∫

unit cellρ(x, y, z)e2πi(hx+ky+lz)

∣∣2 cannot be reversed, even though

that’s what we need: calculate the electron density ρ(x, y, z) from the measured intensities

I(hkl), not vice versa.

The equation (2), |F (hkl)|eiφ(hkl) = const ∗∫unit cell

ρ(x, y, z)e2πi(hx+ky+lz) is a so-called Fourier transforma-

tion and can be reversed:

ρ(x, y, z) =1

const

∑h,k,l|F (hkl)| eiφ(hkl)e−2πi(hx+ky+lz)



The Phase Problem

The previous equation for the electron density requires both structure factor amplitude

|F (hkl)| and phase φ(hkl) as input.

Only the amplitudes can be derived directly from the measured intensities I(hkl), but not the

phase.

This fact is called the phase problem.

yields |F (h, k, l)|, but not φ(h, k, l)

The loss of the phase can

be compared with a projec-

tion on a plane wall: The

eye may see a three dimen-

sional object — but which

face points forward?



The Importance of Phases

The phase of the structure factor contains the main information about the shape of the

molecule. (http://www.ysbl.york.ac.uk/~cowtan/fourier/fourier.html )

inverse FT

inverse FT

FT

|F (h, k, l)|

φ(h, k, l)

The phase φ of the duckdetermines the picture

|F (h, k, l)|, φ(h, k, l)

|F (h, k, l)|, φ(h, k, l)


http://www.ysbl.york.ac.uk/~cowtan/fourier/fourier.html


Phasing

Phasing 52/61


Techniques for Retrieving Phases — Overview

The phase problem makes it necessary to recover the phase information by indirect means.

It is one of the major efforts of macromolecular crystallography to determine good phases.

The main methods used for macromolecular crystals are:

1. molecular replacement

2. isomorphous replacement

3. anomalous dispersion

Phasing 53/61


Method 1: Molecular Replacement

The idea: Compare structure with a similar, already known structure

Background:

• By November 2009, the Protein Data Base (PDB, www.pdb.org) contained more than

60,000 NMR- and X–ray-structures of macromolecules.

• Only very few new structures have a new, unknown fold.

• Sequence homology of as little as 30% can cause a fold similar enough for Molecular

Replacement to succeed.

Phasing 54/61

http://www.pdb.org



The Technique

1. Calculate the expected data from a “crystal” with the know structure and the new unit cell.

2. Compare calculated and measured data.

3. When they match, combine the measured intensities with the calculated phases (and

hope that these phases are similar enough to the real ones)

Phasing 55/61



Problem: The measured data reveal the unit cell, but not the placement and orientation of

the molecule within the unit cell

Solution: Try all possible locations. This requires a lot of computing power and therefore is

done in two steps:

Rotational Search The so-called Patterson Function can be calculated both from measured

intensities and from the known model. It happens to depend only on the orientation but

not the location of the model.

Translational Search once the correct orientation is found, place the known model at many

different locations in the unit cell. At every position, the calculated data and measured

data are compared for best fit.

Phasing 56/61


Method 2: Isomorphous Replacement

Isomorphous Replacement is the oldest method of phasing for protein structures. It is based

on the idea that introduction of a small molecule into the macromolecular crystal does not

change the structure of the macromolecule. On the other hand, a few heavy metal atoms can

contribute detectably to the structure factors and hence introduce changes in the reflection

intensities.

Common heavy metals are Hg (80e−),

Pb (82e−), Au (79e−), Pt (78e−), or U

(92e−). They can be incorporated by co-

crystallisation or by soaking after the crys-

tals have grown.

The first protein structures like myoglobin

or hemoglobin were solved by isomor-

phous replacement, making use of the

iron in the heme–cluster. G. Sheldrick

Phasing 57/61


Method 2: Isomorphous Replacement

In order to use the extra information, one needs at least two data sets: a native one (no heavy

metal) and a derivative (with heavy metal).

|FH | , φH

|FT | , φT

construction

Harker-co-ordinates

subtract

derivative: |FT |

native: |FP |

Phasing 58/61


Method 3: Anomalous Dispersion

• For a normal diffraction experiment, Friedel’s law is valid (intensities of the reflections

(h, k, l) and (−h,−k,−l) are equal).

• A wavelength at the transition energy of a special atom in the crystal (e.g. Se,Au, U )

causes so-called anomalous dispersion, brought about by a shell transition of an inner

shell electron.

• In the presence of anomalous dispersion, Friedel’s Law breaks down, i.e. I(h, k, l) 6=

I(−h,−k,−l). The theoretical treatment is very similar to that of isomorphous replace-

ment.

Phasing 59/61


Method 3: Anomalous Dispersion — MAD and SAD

The two main applications of anomalous dispersion are MAD and SAD:

• MAD or Multi-wavelength Anomalous Dispersion uses data-sets collected at two or more

different wavelength. Up to experimental error this allow to exactly determine the phase

φ(hkl) for every reflection.

• SAD or Single-wavelength Anomalous Dispersion uses only one wavelength. Every phase

is determined up to a two-fold ambiguity. Taking the mean value often suffices forsuccess-

ful phasing.

Phasing 60/61


Examples of Phase Quality

Mean phases from SAD Resolved twofold ambiguity

Final (refined) phases

G.M. Sheldrick

Phasing 61/61

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