Overview - Macromolecular Crystallography

1. Overexpression and crystallization2. Crystal characterization and data collection3. The diffraction experiment4. Phase problem

1. MIR (Multiple Isomorphous Replacement)- No homologous structure known- Incorporate heavy atoms- Locate bound heavy atoms- Phase refinement

2. MAD (Multiwavelength Anomalous Diffraction)- No homologous structure known- Incorporate anomalous scatterers- Locate anomalous scatterers- Phase refinement

3. MR (Molecular Replacement)- Requires homologous structure- Rotation function (determine orientation of known molecule in unknown crystal)- Translation function (determine position of search molecule in unknown crystal)- Rigid body refinement

5. Phase improvement6. Model building7. Refinement

2a. Crystal Characterization2b. Data Collection

Crystal Mounting

Capillary-mountedcrystal for room

temperature datacollection

Loop-mounted crystalfor cryogenic data

collection

Cold N2 Gas

Diffraction Photograph

Visualization of the reciprocal lattice

Asymmetric Unit, Unit Cell and Crystal

Asymmetric Units

2-fold Axisof Symmetry

N-fold Axis of Rotation

2422

Icosahedral symmetry

Molecular Symmetry I

C2 or 2 C3 or 3

Molecular Symmetry II

D2 / 222 D3 / 32

2-Fold and 21 Screw Axis

Mirror Planes and Inversion Centers

Mirror plane Inversion center

Point Groups I

Molecular symmetries are described by point groups

There are 32 point groups, which are relevant forcrystallography (those containing 2-, 3-, 4- and 6-fold axes ofsymmetry)

Additional point groups are possible (e.g. 52 or D5), but notas crystallographic symmetry operations

This does not mean that a pentamer with 5-fold symmetrycannot be crystallized, it only means that the five-fold axis ofrotation cannot be a crystallographic symmetry element

Point Groups II

n

n

n/m

nm

nm

n2

n/mm

n-fold axis of rotation

n-fold axis of rotation and inversion center

n-fold axis of rotation and perpendicularmirror plane

n-fold axis of rotation and mirror planecontaining n-fold axis

n-fold axis of rotation, mirror plane containing then-fold axis and inversion center

n-fold axis of rotation and perpendicular 2-fold

n-fold axis of rotation, perpendicular mirror planeand mirror plane containing n-fold axis

Cubic System

432

23

Simple Lattices I

2 1 3

A 2-fold axis (1 2) and alattice translation (2→3)generate a secondary 2-foldaxis (1 3)

Crystallographicsymmetry operations arevalid throughout thecrystal (global symmetry)

:-{ :-{

:-{ :-{

:-{ :-{ :-{

:-{ :-{ :-{

:-{ :-{ :-{

:-{ :-{ :-{

:-{ :-{ :-{

:-{ :-{ :-{

Simple Lattices II

A two-dimensional latticewith 2-fold symmetry axes(indicated by )perpendicular to the planeof the figure and 2-foldscrew axes (indicated bythe half-arrows) in theplane.

Crystal Systems

a = b = c,α = β = γ = 90°

Four 3-foldsP, I, FCubic

a = b = c,α = β = γ ≠ 90°

3-foldRRhombohedral

a = b ≠ c,α = β = 90°, γ = 120°

3-fold or 6-foldPTrigonal/Hexagonal

a = b ≠ c,α = β = γ = 90°

One 4-foldP, ITetragonal

α = β = γ = 90°Three perpendicular2-fold axes

P, C, I, FOrthorhombic

β ≠ 90°One 2-fold axisP, CMonoclinic

a ≠ b ≠ c,α ≠ β ≠ γ

NonePTriclinic

PropertiesMinimal SymmetryBravais TypeCrystal System

Primitive and Centered Unit Cells

a b

c

PrimitiveP

B-centeredB

(could also be A- or C-centered)

I-centeredI

F-centeredF

Bravais Lattices14 Bravais lattices

Some crystal systems haveonly one lattice type (triclinic,trigonal/hexagonal andrhombohedral), some havetwo (monoclinic andtetragonal), cubic has threeand orthorhombic four

Rhombohedral R can be setup either hexagonal orrhombohedral

Rhombohedral

Space Groups - Introduction

• Combination of 32 point groups with translationalsymmetry elements yields the 230 space groups

• All space groups are tabulated in the InternationalTables of Crystallography

• For chiral molecules (protein, DNA, RNA, etc.) only 65space groups are possible, i.e. those space groups whichdo not contain mirror planes, glide planes or inversioncenters

• Space group P212121 is the most common space groupfound with proteins and will be discussed in more detail

Space groups - P212121P212121 - Hermann-Mauguin symbolD2222 - Point groupOrthorhombic - Crystal system19 - Space group number

Space group diagrams -Projections onto the ab-, bc- and ac-

planes showing symmetryelements

Cartoon diagram illustrating thesymmetry operation acting on amolecule represented by the opencircle(+, above plane/ -, belowplane)

Origin - Location of origin with respect tothe symmetry elements

Asymmetric unit - One choice of theasymmetric unit

Symmetry operation -(1) Identity operation(2) 21 screw axis along z at a = 1/4

and b = 0

Space Group - P212121Symmetry operations:(1) x,y,z(2) -x+1/2, -y, z+1/2(3) -x, y+1/2, -z+1/2(4) x+1/2,-y+1/2,-z

Reflection conditions:For reflections of type h00only those with even h areobserved. Reflections withodd h are forbidden bysymmetry.Same for 0k0 and 00l

Asymmetric Unit, Unit Cell and Crystal

A Real Crystal

Space group P212121

4 asymmetric units

Molecularboundaries

Bovine pancreatictrypsin inhibitor(BPTI)

Empty spaces between molecules: Macromolecular crystals consist of protein andsolvent (30% - >80%). The Matthew’s coefficient describes crystal packing.

Packing Analysis I

Calculate unit cell volume: V = a • ( b x c) = abc (1 - cos2α - cos2β - cos2γ + 2 cosα cosβ cosγ)1/2

(if all angles = 90°: V = abc, if one angle (β) ≠90°: V= abc sin β)

Calculate volume of asymmetric unit:VAU = V / nAU

Calculate Matthew’s coefficient:VM = VAU / MrThe Matthew’s coefficient is between 1.7 - 4.0 Å3/Da

Calculate solvent content percentage:ρ = 1 - 1.23 / VM

Packing Analysis II

The MogA protein (molecular mass 21,500 Da) crystallizes inthe hexagonal space group P63 (six asymmetric units) with unitcell dimensions of a=b=66 Å, c=65 Å and γ=120°.

Calculate volume of unit cell: V= abc sin γ = 245206 Å3

Calculate volume of asymmetric unit: VAU = V / 6 = 40868 Å3

Calculate Matthew’s coefficient: VM = VAU / Mr = 1.9 Å3/Da

Calculate solvent content percentage: ρ = 1 - 1.23 / VM = 35%

This is a tightly packed crystal

Packing Analysis IIIThe MogA protein also crystallizes in orthorhombic space groupP212121 (4 asymmetric units) with unit cell dimensions of a=55Å, b=71 Å and c=165 Å. The protein is present as a trimer insolution.Calculate volume of unit cell: V = abc = 644325 Å3

Calculate volume of asymmetric unit: VAU = V / 4 = 161081 Å3

Calculate Matthew’s coefficient:1 monomer: VM = 7.5 Å3/Da2 monomers: VM = 3.75 Å3/Da3 monomers: VM = 2.5 Å3/Da4 monomers: VM = 1.9 Å3/Da5 monomers: VM = 1.5 Å3/Da

This crystal form contains a trimer in the asymmetric unit.

Packing Analysis IVHow to reconcile the results? The protein is a trimer in solution and also in both crystal forms

Hexagonal crystals: A monomer is present in the asymmetric unit A crystallographic 3-fold axis of symmetry generates the trimer Check the International Tables to find the 3-fold axis

Orthorhombic crystals: A trimer is present in the asymmetric unit There is a local (non-crystallographic) 3-fold axis of symmetry The monomers within the trimer do not have to be identical The monomers within the trimer will be similar Non-crystallographic symmetry elements can be detected

2a. Crystal Characterization2b. Data Collection

X-rays• High-energy electromagnetic radiation (λ ~ 1 Å)• Produced by x-ray generator or synchrotron• In-house: rotating anode x-ray generator with copper anode (λ = 1.5418 Å)• Synchrotron: Large scale research facilities

• DESY (Hamburg)• BESSY (Berlin)• ESRF (Grenoble)• SLS (Villigen, Switzerland)

• Interatomic distances are on the order of 1 ÅInterference effects