B1 219 ROBERT M. STROUD

• Understanding crystallography and structures• Interactive – Bring conundra –• Laboratory course:

– Crystallize a protein– Determine structure– Visit ‘Advanced Light Source’ (ALS) for data

Determining Atomic Structure• X-ray crystallography = optics l ~ 1.5Å (no lenses)• Bond lengths ~1.4Å• Electrons scatter X-rays; ERGO X-rays ‘see electrons’• Resolution –Best is l/2 Typical is 1 to 3 Å• Accuracy of atom center positions ±1/10 Resolution

2q

Where do X-rays come from?=accelerating or decelerating electrons

Resources:

Crystallography accessible to no prior knowledge of the field or its mathematical basis. The most comprehensive and concise reference Rhodes' uses visual and geometric models to help readers understand the basis of x-ray crystallography.

http://www.msg.ucsf.eduComputingCalculation software-all you will ever needOn line course for some items:http://www-structmed.cimr.cam.ac.uk/course.html

http://bl831.als.lbl.gov/~jamesh/movies/

Dr Chris Waddling msg.ucsf.edu

Dr James Holton UCSF/LBNL

Optical image formation, - without lenses

TopicsSummmary: Resources1 Crystal lattice optical analogues photons as waves/particles2 Wave addition complex exponential3. Argand diagram4. Repetition ==sampling fringe function5. Molecular Fourier Transform Fourier Inversion theorem sampling the transform as a product6. Geometry of diffraction7. The Phase problem heavy atom Multiple Isomorphous Replacement) MIR Anomalous Dispersion Multi wavelength Anomalous Diffraction MAD/SAD8. Difference Maps and Errors9. Structure (= phase) Refinement Thermal factors Least squares Maximum likelihood methods10. Symmetry –basis, consequences

Topics11. X-ray sources: Storage rings, Free Electron Laser (FELS)12. Detector systems

13. Errors, and BIG ERRORS! –RETRACTIONS

14. Sources of disorder15. X-ray sources: Storage rings, Free Electron Laser (FELS)

If automated- why are there errors? What do I trust? Examples of errors trace sequence backwards, mis assignment of helices etc

The UCSF beamline 8.3.1

UCSF mission bay

NH3 sites and the role of D160 at 1.35Å Resolution

Data/Parameter ratio is the same for all molecular sizes at the same resolution dmin

ie. quality is the same!

30S 50S

Growth in number and complexity of structures versus time

Macromolecular Structures

The universe of protein structures: Our knowledge about protein structures is increasing..

• 65,271 protein structures are deposited in PDB (2/15/2010).

• This number is growing by > ~7000 a year • Growing input from Structural Genomics HT structure

determination (>1000 structures a year)

X-Ray Crystallography for Structure Determination

Goals: 1. How does it work2. Understand how to judge where errors may lurk3. Understand what is implied, contained in the Protein Data Bank PDB http://www.pdb.org/pdb/home/home.do

Resolution: - suspect at resolutions >3 ÅR factor, and Rfree : statistical ‘holdout test’Wavelength ~ atom sizeScattering from electrons = electron densityAdding atoms? howObservations Intensity I(h,k,l) = F(hkl).F*(hkl)Determine phases y(hkl)Inverse Fourier Transform === electron densityJudging electron density- How to interpret?Accuracy versus Reliability

A Typical X-ray diffraction pattern

~100 microns

qmax=22.5° if l=1ÅResolution 1.35Å2qmax=45°

The Process is re-iterative, and should converge-but only so far!

Intensities I(h,k,l) Electron density r(x,y,z)

Phases f(h,k,l)

Atom positions (x,y,z)

Known:Amino acid sequenceLigandsBond lengths anglesConstraints on geometry

Crystal

Experimentalheavy atom labelsselenium for sulfurTrial & error similar structure

calculated I(h,k,l)

Resolution dmin = l /2 sin (qmax)differs from Rayleigh criterion

dmin = l /2 sin (qmax) is the wavelength of the shortest wave used to construct the density map

The Rayleigh Criterion• The Rayleigh criterion is the generally accepted criterion for

the minimum resolvable detail - the imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.

compared withsin (qmax) = l /2 dmin

How do we judge the Quality of structure?

2. Overall quality criteria: agreement of observations with diffraction calculated from the interpreted structure.

3. Since we refine the structureTo match the Ihkl overfitting ?

Define Rfree for a ‘hold-out ‘ set of observations.

4. OK? R < 20%, R free< 25%

5. But the experimental errors in measuring Fo are ~ 3%.inadequate models of solvent, atom motion, anharmonicisity

6 Accuracy ~ 0.5*res*R

Crystal lattice is made up of many ‘Unit Cells’Unit cell dimensions are 3 distances a,b,c and angles between them , ,a b g

Repetition in ‘Real space’

Causes Sampling in ‘scattering space’

A ‘section’ throughScattering patternof a crystal l=0Note symmetry,Absences for h=even k=even

h

k

Crystal lattice is made up of many ‘Unit Cells’Unit cell dimensions are 3 distances a,b,c and angles between them , ,a b g

Repetition in ‘Real space’

Causes Sampling in ‘scattering space’

Vabc = |a.(bxc)|Vabc = |b.(cxa)|Vabc = |c.(axb)|

Scattering

Adding up the scattering of Atoms:‘interference’ of waves

Waves add out of phaseby 2p[extra path/l]

In general they add up to somethingamplitude In between -2f and +2f.For n atoms

Just 2 atoms…

Many atoms add by the same rules.

Different in every direction.

Summary of the Process from beginningto structure..

Optical Equivalent: eg slide projector; leave out the lens..Optical diffraction = X-ray diffraction

Image of the object

Remove the lens= observe scatteringpattern

object

film

object

object

object

Scattering

Adding up the scattering of Atoms:‘interference’ of waves

vectors revisited…• Vectors have magnitude and direction• Position in a unit cell

– r = xa + yb + zc where a, b, c are vectors, x,y,z are scalars 0<x<1– a.b = a b cos (q) projection of a onto b -called ‘dot product’– axb = a b sin (q) a vector perpendicular to a, and b proportional to area in magnitude

-- called cross product– volume of unit cell = (axb).c = (bxc).a = (cxa).b = -(bxa).c = -(cxb).a – additivity: a + b = b + a– if r = xa + yb + zc and s = ha* + kb* + lc* then

• r.s = (xa + yb + zc ).(ha* + kb* + lc* ) • = xh a.a* + xka.b* + xla.c* +yhb.a* + ykb.b* + ylb.c* + zhc.a* +zhc.a* + zkc.b* + zlc.c*

– as we will see, the components of the reciprocal lattive can be represented in terms of a* + b* + c*, where a.a* = 1, b.b*=1, c.c*=1

– and a.b*=0, a.c*=0 etc.. r.s = (xa + yb + zc ).(ha* + kb* + lc* ) • = xh a.a* + xka.b* + xla.c* +yhb.a* + ykb.b* + ylb.c* + zhc.a* +zhc.a* + zkc.b* + zlc.c*• = xh + yk + zl

Adding up the scattering of Atoms:‘interference’ of waves

2pr.S

Adding up the scattering of Atoms:‘interference’ of waves

2pr.SF(S)

Revisit..

Revision notes on McClaurin’stheorem.

It allows any function f(x) to be definedin terms of its value at some x=a valueie f(a), and derivatives of f(x) at x=a,namely f’(a), f’’(a), f’’’(a) etc

Extra Notes on Complex Numbers(p25-28)

2pr.SF(S)Argand Diagram.. F(S) = |F(s)| eiq

Intensity = |F(s)|2

How to represent I(s)?

I(s) = |F(s)|2 = F(S) .F*(S)

proof?

Where F*(S) is defined to be the ‘complex conjugate’ of F(S) = |F(s)| e-iq

so |F(s)|2 = |F(s)|[cos(q) + isin(q)].|F(s)|[cos(q) - isin(q)] = |F(s)|2 [cos2 (q) + sin2 (q)] = |F(s)|2 R.T.P.

q

-2pr.S

F*(S)

-q

F*(S) is the complex conjugate of F(S), = |F(s)| e-iq

(c+is)(c-is)=cos2 q - c q is q + c q is q + sin2q

so |F(s)|2 = F(S) .F*(S)

Origin Position is arbitrary..proof..

So the origin is chosen by choice of: a) conventional choice in each space group-eg Often on a major symmetry axis- BUT for strong reasons—see ‘symmetry section’.

Even so there are typically 4 equivalent major symmetry axes per unit cell..

b) chosen when we fix the first heavy metal (or Selenium) atom position, -all becomes relative to that.

c) chosen when we place a similar moleculefor ‘molecular replacement’ = trial and errorsolution assuming similarity in structure.

Adding waves from j atoms…F(S)= G(s) =

and why do we care?

How much difference will it make to the average intensity? average amplitude?

if we add a single Hg atom?

The ‘Random Walk’ problem? (p33.1-33.3)

What is the average sum of n steps in random directions?

(What is the average amplitude<|F(s)|> from an n atom structure?)

-AND why do we care?!........

How much difference from adding amercury atom (f=80).

The average intensity for ann atom structure, each of f electronsis <I>= nf2

The average amplitude is Square rootof n, times f

and why do we care?

How much difference will 10 electrons make to the average intensity? average amplitude?

average difference in amplitude?average difference in intensity?

if we add a single Hg atom?

and why do we care?

How much difference will 10 electrons make to the average intensity? 98,000 e2

average amplitude? 313

average difference in amplitude?average difference in intensity?

if we add a single Hg atom?

and why do we care?

How much difference will 10 electrons make to the average intensity? 98,000 e2

average amplitude? 313 e

average difference in amplitude?2.2% of each amplitude! average difference in intensity?98,100-98000=100 (1%)

if we add a single Hg atom?

and why do we care?

How much difference will 80 electrons make to the average intensity? 98,000 e2

average amplitude? 313 e

average difference in amplitude?18 % of each amplitude! average difference in intensity?104,400-98000=6400 (6.5%)

or Hg atom n=80e

WILSON STATISTICS

What is the expectedintensity of scattering versus the observedfor proteins of i atoms,?

on average versus resolution |s|?

Bottom Lines:

This plot should provide The overall scale factor(to intercept at y=1)

The overall B factor

In practice for proteins it has bumpsin it, they correspond to predominantor strong repeat distances in the protein.For proteins these are at 6Å (helices)3Å (sheets), and 1.4Å (bonded atoms)

Topic: Building up a Crystal

1 Dimension

Scattering from an array of points, is the same as scattering from one point,SAMPLED at distances ‘inverse’ to the repeat distance in the object

The fringe function

Scattering from an array of objects, is the same as scattering from one object,SAMPLED at distances ‘inverse’ to the repeat distance in the object eg DNA

Scattering from a molecule is described by

F(s)= Si fi e(2pir.s)

Consequences of being a crystal?• Repetition = sampling of F(S)

34Å 1/34Å-1

Object repeated 1/l2/l

Transform of two hoizontal lines defined y= ± y1

F(s) = Int [x=0-inf exp{2pi(xa+y1b).s} + exp{2pi(xa-y1b).s}dVr ]=2 cos (2py1b).s * Int [x=0-inf exp{2pi(xa).s dVr]

for a.s=0 the int[x=0-inf exp{2pi(xa).s dVr] = total e content of the linefor a.s≠0 int[x=0-inf exp{2pi(xa).s dVr] = 0

hence r(r) is a line at a.s = 0 parallel to b, with F(s) = 2 cos (2py1b).s -along a vertical line perpendicular to the horizontal lines.

Transform of a bilayer..

b=53Å

4.6Å

53Å Repeat distance 40Å inter bilayer spacing

4.6Å

53Å Repeat distance

40Å inter bilayer spacing

The Transform of a Molecule

How to calculate electron density?

Proof of the Inverse Fourier Transform:

Definitions:

Build up a crystal from Molecules…

First 1 dimension,a direction

a.s=h

b.s=k

hkl=(11,5,0)

Measure I(hkl)F(hkl)= √I(hkl)Determine f(s) = f(hkl)

Calculate electron density map

The density map is made up of interfering density waves through the entire unit cell…

Geometry of Diffraction

A Typical X-ray diffraction pattern

~100 microns

A Typical X-ray diffraction pattern

~100 microns

l=2dhklsin(q)

What if ‘white radiation’? representationLaue picture.

Compute a Transform of a series of parallel lines

Adding up the scattering of Atoms:‘interference’ of waves

2pr.S

s0sss

s0

s1 S

radius = 1/l

I(S)= F(S).F(S)*

F(S)

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)

Sum of 7 atoms scattering

Result is a wave of amplitude F(S)phase f(s)

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)

Sum of 7 atoms scattering

Result is a wave of amplitude F(S)phase F(S)

cos(q)

sin(q)

e(iq) = cos(q) + i sin(q)

i = √(-1)

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)

Sum of 7 atoms scattering

Result is a wave of amplitude |F(S)|phase F(S)

cos(q)

sin(q)

e(iq) = cos(q) + i sin(q)

i = √(-1)

F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)

Sum of 7 atoms scattering

Result is a wave of amplitude |F(S)|phase F(S)

cos(q)

sin(q)

e(iq) = cos(q) + i sin(q)

i = √(-1)

F(S) = Sj fj e(2pirj.S)

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)

Sum of 7 atoms scattering

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)

F(S) = Sj fj e(2pirj.S)

e(iq) = cos(q) + i sin(q)

The Process is re-iterative, and should converge-but only so far!

Intensities I(h,k,l) Electron density r(x,y,z)

Phases f(h,k,l)

Atom positions (x,y,z)

Known:Amino acid sequenceLigandsBond lengths anglesConstraints on geometry

Crystal

Experimentalheavy atom labelsselenium for sulfurTrial & error similar structure

F(S) = Sj fj e(2pirj.S)

Scattering pattern is the Fourier transform of the structure

Structure is the ‘inverse’Fourier transform of the Scattering pattern

r(r) = S F(S) e(-2pir.S)

Rosalind Franklin and DNA

F(S) = Sj fj e(2pirj.S)

Scattering pattern is the Fourier transform of the structure

Structure is the ‘inverse’Fourier transform of the Scattering pattern

r(r) = S F(S) e(-2pir.S)

FT

FT-1

FT-1

FT

This is all there is? YES!!

a

b

1/a

1/b

F(S) = Sj fj e(2pirj.S)

Scattering pattern is the Fourier transform of the structure

Structure is the ‘inverse’Fourier transform of the Scattering pattern

r(r) = S F(S) e(-2pir.S)

FT

FT-1

FT-1

FT

But we observe |F(S)|2 and there are ‘Phases’

a

b

1/a

1/bWhere F(S) has phaseAnd amplitude

F(S) = Sj fj e(2pirj.S)

Scattering pattern is the Fourier transform of the structure

Structure is the ‘inverse’Fourier transform of the Scattering pattern

r(r) = S F(S) e(-2pir.S)

FT

FT-1

FT-1

FT

This is all there is?

a

b

1/a

1/bF(h,k,l) = Sj fj e(2pi (hx+ky+lz))

r(x,y,z) = S F(h,k,l) e(-2pir.S)

S

Sh=15, k=3,

Relative Information in Intensities versus phases

r(r) = S F(S) e(-2pir.S)

r(r) duck r(r) cat

F(S) = Sj fj e(2pirj.S)F(S) duck |F(S)|

f(s)

F(S) cat

|F(S)|duck

f(s) cat

Looks like a …..

Relative Information in Intensities versus phases

r(r) = S F(S) e(-2pir.S)

r(r) duck r(r) cat

F(S) = Sj fj e(2pirj.S)F(S) duck |F(S)|

f(s)

F(S) cat

|F(S)|duck

f(s) cat

Looks like a CATPHASES DOMINATE:-Incorrect phases = incorrect structure-incorrect model = incorrect structure-incorrect assumption = incorrect structure

The Process is re-iterative, and should converge-but only so far!

Intensities I(h,k,l) Electron density r(x,y,z)

Phases f(h,k,l)

Atom positions (x,y,z)

Known:Amino acid sequenceLigandsBond lengths anglesConstraints on geometry

Crystal

Experimentalheavy atom labelsselenium for sulfurTrial & error similar structure

Consequences of being a crystal?• Repetition = sampling of F(S)

34Å

Consequences of being a crystal?• Repetition = sampling of F(S)

34Å 1/34Å-1

Consequences of being a crystal?• Sampling DNA = repeating of F(S)

34Å 1/34Å-1

3.4Å 1/3.4Å-1

Object repeated 1/l2/l

Object ScatteringBuild a crystal

Early Definitions:

a.s=h

b.s=k

hkl=(11,5,0)

Measure I(hkl)F(hkl)= √I(hkl)Determine f(s) = f(hkl)

Calculate electron density map

The density map is made up of interfering density waves through the entire unit cell…

Measure I(hkl)

The Process is re-iterative, and should converge-but only so far!

Intensities I(h,k,l) Electron density r(x,y,z)

Phases f(h,k,l)

Atom positions (x,y,z)

Known:Amino acid sequenceLigandsBond lengths anglesConstraints on geometry

Crystal

Experimentalheavy atom labelsselenium for sulfurTrial & error similar structure

Anomalous diffraction