Understanding Crystallography and Protein Structures

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: http://www.msg.ucsf.edu Computing Calculation software-all you will ever need On line course for some items: http://www-structmed.cimr.cam.ac.uk/course.html Dr Chris Waddling msg.ucsf.edu Dr James Holton UCSF/LBNL 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://bl831.als.lbl.gov/~jamesh/movies/

Optical image formation, - without lenses

Topics Summmary: Resources 1 Crystal lattice optical analogues photons as waves/particles 2 Wave addition complex exponential 3. Argand diagram Repetition ==sampling fringe function 5. Molecular Fourier Transform Fourier Inversion theorem sampling the transform as a product 6. Geometry of diffraction 7. The Phase problem heavy atom Multiple Isomorphous Replacement) MIR Anomalous Dispersion Multi wavelength Anomalous Diffraction MAD/SAD 8. Difference Maps and Errors 9. Structure (= phase) Refinement Thermal factors Least squares Maximum likelihood methods 10. Symmetry –basis, consequences

Topics 11. X-ray sources: Storage rings, Free Electron Laser (FELS) 12. Detector systems 13. Errors, and BIG ERRORS! –RETRACTIONS 14. Sources of disorder 15. X-ray sources: Storage rings, Free Electron Laser (FELS)

The UCSF beamline 8.3.1 UCSF mission bay If automated- why are there errors? What do I trust? Examples of errors trace sequence backwards, mis assignment of helices etc

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 dminie. quality is the same! 30S 50S

Macromolecular Structures Growth in number and complexity of structures versus time

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 work 2. Understand how to judge where errors may lurk 3. Understand what is implied, contained in the Protein Data Bank PDB http://www.pdb.org/pdb/home/home.do Resolution: - suspect at resolutions >3 Å R factor, and Rfree: statistical ‘holdout test’ Wavelength ~ atom size Scattering from electrons = electron density Adding atoms? how Observations Intensity I(h,k,l) = F(hkl).F*(hkl) Determine phases y(hkl) Inverse Fourier Transform === electron density Judging electron density- How to interpret? Accuracy versus Reliability

A Typical X-ray diffraction pattern ~100 microns

qmax=22.5° if l=1Å Resolution1.35Å 2qmax=45°

The Process is re-iterative, and should converge-but only so far! Crystal Intensities I(h,k,l) Electron density r(x,y,z) calculated I(h,k,l) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

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 with sin (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 structure To match the Ihkloverfitting ? 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 h A ‘section’ through Scattering pattern of a crystal l=0 Note symmetry, Absences for h=even k=even k Causes Sampling in ‘scattering space’ Repetition in ‘Real space’

Scattering Adding up the scattering of Atoms: ‘interference’ of waves Waves add out of phase by 2p[extra path/l]

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

Many atoms add by the same rules. Different in every direction.

Summary of the Process from beginning to structure..

Optical Equivalent: eg slide projector; leave out the lens..Optical diffraction = X-ray diffraction object Image of the object object Remove the lens= observe scattering pattern film 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 + zcwhere 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* ) = xha.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* ) = xha.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 F(S) 2pr.S

Revision notes on McClaurin’s theorem. It allows any function f(x) to be defined in terms of its value at some x=a value ie 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)

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. F(S) 2pr.S q

F*(S) is the complex conjugate of F(S), = |F(s)| e-iq (c+is)(c-is)=cos2q -cqisq+ cqisq+ sin2q so |F(s)|2 =F(S) .F*(S) -2pr.S -q 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 molecule for ‘molecular replacement’ = trial and error solution 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 a mercury atom (f=80).

The average intensity for an n atom structure, each of f electrons is <I>= nf2 The average amplitude is Square root of 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? or Hg atom n=80e 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%)

WILSON STATISTICS What is the expected intensity of scattering versus the observed for 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 bumps in it, they correspond to predominant or 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

Consequences of being a crystal? Sampling DNA = repeating of F(S) 1/3.4Å-1 3.4Å 34Å 1/34Å-1

Transform of two hoizontal lines defined y= ± y1 F(s) = Int [x=0-infexp{2pi(xa+y1b).s} + exp{2pi(xa-y1b).s}dVr ] =2 cos (2py1b).s * Int [x=0-infexp{2pi(xa).s dVr] for a.s=0 the int[x=0-infexp{2pi(xa).s dVr] = total e content of the line for a.s≠0 int[x=0-infexp{2pi(xa).s dVr] = 0 hence r(r)isa line at a.s= 0 parallel to b, with F(s) = 2 cos (2py1b).s -along a vertical line perpendicular to the horizontal lines. This structure repeats along the b direction thus peaks occur at b.s= k (where k is integer only) Transform of a bilayer.. b=53Å

b = 53Å Repeat distance 40Å inter bilayer spacing 4.6Å

53Å Repeat distance 40Å inter bilayer spacing 4.6Å

How to calculate electron density? Proof of the Inverse Fourier Transform:

Build up a crystal from Molecules… First 1 dimension, a direction

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…

Compute a Transform of a series of 50%b/50%w parallel lines Ronchi ruling:

This is all there is? YES!! Scattering pattern is the Fourier transform of the structure FT F(S) = Sjfj e(2pirj.S) FT-1 Structure is the ‘inverse’ Fourier transform of the Scattering pattern 1/a r(r) = SF(S) e(-2pir.S) a FT b 1/b FT-1

But we observe |F(S)|2 and there are ‘Phases’ Scattering pattern is the Fourier transform of the structure FT F(S) = Sjfj e(2pirj.S) FT-1 Structure is the ‘inverse’ Fourier transform of the Scattering pattern 1/a r(r) = SF(S) e(-2pir.S) a FT Where F(S) has phase And amplitude b 1/b FT-1

This is all there is? Scattering pattern is the Fourier transform of the structure FT F(S) = Sjfj e(2pirj.S) S FT-1 Structure is the ‘inverse’ Fourier transform of the Scattering pattern 1/a r(r) = SF(S) e(-2pir.S) a FT b F(h,k,l) = Sjfj e(2pi(hx+ky+lz)) S 1/b h=15, k=3, r(x,y,z) = SF(h,k,l) e(-2pir.S) FT-1

Relative Information in Intensities versus phases r(r) duck r(r) cat |F(S)| F(S)= Sjfj e(2pirj.S) F(S) duck F(S) cat f(s) |F(S)|duck r(r) = SF(S) e(-2pir.S) f(s) cat Looks like a …..

Relative Information in Intensities versus phases r(r) duck r(r) cat |F(S)| F(S)= Sjfj e(2pirj.S) F(S) duck F(S) cat f(s) |F(S)|duck r(r) = SF(S) e(-2pir.S) f(s) cat Looks like a CAT PHASES 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! Crystal Intensities I(h,k,l) Electron density r(x,y,z) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Phase determination by any means, ends up as a probabilty distribution. So Fh,k,l, f(h,k,l) Then what to use for the best map? r(r) = SF(S) e(-2pir.S) ? the signal towards some F true will be Integral P(F(S)) cos (f(h,k,l) –ftrue) and the ‘noise ‘ will be Integral P(F(S)) sin (f(h,k,l) –ftrue) The map with the least noise will have F(s) = center of mass of P(F(S))

Phase determination by any means, ends up as a probabilty distribution. So Fh,k,l, f(h,k,l) Then what to use for the best map? r(r)best= F(S) e(-2pir.S) ? The map with the least noise will have F(s) = center of mass of P(F(S))= Int P F sin(f-f(best)) is a minimum. Then m = Int P F cos(f-f(best))/F iem|F(s)| f(best) =Int P.F where m = figure of merit = Int P(F) F(s) noise = Int F(s) sin

If a map is produced with some f(hkl) The probability of it being correct is P(hkl)P(hkl) (f(hkl)) Maximum value of P(hkl) (f(hkl)) gives the ‘Most probable’ map Map with the least mean square error, is when noise is minimum, Int find f(best) such that Q= Intf [|F| P(hkl) (f(hkl))exp (if(hkl)) - Fbestf(best))]2df is minimum. is minimum when dQ/dFbest= 0 so Fbestf(best) = f |F| P(hkl) (f(hkl))exp (if(hkl))df Fbestf(best) = m|F| center of ‘mass’ of the Probability distribution m = IntP(hkl) (f(hkl))cos(f - f(best)) consider errors from one reflection, and its complex conjugate <(Dr)2> = 2/V2 Then F= IntFcos(f - f(best))/ F Noise = 1/V Int F sin (f - f(best))/F = F(1-m2) mF where

s0sss Adding up the scattering of Atoms: ‘interference’ of waves I(S)= F(S).F(S)* s1 S s0 radius = 1/l F(S) 2pr.S

How to be ‘UNBIASED’ The Observations are constant F(hkl) The phases change according to presumptions.. Remove all assumptions about any questionable region as early as possible Re-enforce the observations at each cycle and try to improve the phases without bias It is an objective method (prone to human aggressive enthusiasm.

AXIOM: Forward FT Back FT-1 are Truly Inverse Crystal Amplitudes F(h,k,l) Electron density r(x,y,z) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Change one side Change the other Crystal Amplitudes F(h,k,l) Electron density r(x,y,z) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Where Bias creeps in: 1. No Assumptions = no changes Crystal Compare the ‘Observations’, with the Calculation -that depends on assumptions. =Difference Mapping! Amplitudes F(h,k,l) Electron density r(x,y,z) Fhkl Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Where Bias creeps in: 1. Errors of Overinterpretation Crystal Compare the ‘Observations’, with the Calculation -that depends on assumptions. =Difference Mapping! Amplitudes F(h,k,l) Electron density r(x,y,z) Solvent Flattening Average multiple copies Fhkl Phases f(h,k,l) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

A reminder: Each individual F(hkl) has its own Phase F(hkl) Result is a wave of amplitude |F(S)| phase F(S) Sum of 7 atoms scattering 2pr7.S i = √(-1) sin(q) F(S) cos(q) F(S) 2pr1.S f1=6 electrons e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

Suppose we interpret 7 atoms; but 3 remain to be found in density Result is a wave of amplitude |F(S)| phase F(S) In reality, maybe 3 atoms are missing. How to see what is missing? 2pr7.S i = √(-1) Sum of 7 atoms scattering sin(q) F(S) cos(q) 2pr1.S f1=6 electrons e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

Observed F(hkl) will contain a part of that information.. Sum of 7 atoms scattering Result is a wave of amplitude |F(S)| phase F(S) In reality, maybe 3 atoms are missing. How to see what is missing? |F(S)|observed 2pr7.S i = √(-1) sin(q) F(S) cos(q) phase F(S) 2pr1.S f1=6 electrons e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

In reality The missing component is The Difference F(S) obs- F(S)calc It has all the missing information. Transform it= all missing parts. But all we know is |F(S)|observed|F(S)|calc and current (biased) phase F(S) So what if we take the Difference we know||F(S)|obs- |F(S)|calc| and F(S) In reality, maybe 3 atoms are missing. How to see what is missing? |F(S)|observed 2pr7.S F(S) 2pr1.S phase F(S) e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

The BEST WE CAN DO…IS DF, F(S) It contains a component of the truth DF cos(q), and an error component DF sin(q) In reality The missing component is The Difference F(S) obs- F(S)calc It has all the missing information. Transform it= all missing parts. But all we know is |F(S)|observed|F(S)|calc and current (biased) phase F(S) So what if we take the Difference we know||F(S)|obs- |F(S)|calc| and F(S) = DF and F(S) cos(q) |F(S)|observed 2pr7.S DF q F(S) 2pr1.S phase F(S) e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

What contribution to the ‘missing truth’? It contains a component of the truth DF cos(q) and an error component DF sin(q) So what if we transform the Difference we knowDF =||F(S)|obs- |F(S)|calc| and F(S) On average DF = ftruecos(q) so On average towards the truth is DF cos(q) = ftruecos2(q) What is the average value of cos2(q)? = ½ ( cos 2q +1) cos(q) |F(S)|observed 2pr7.S DF q ftrue F(S) p/2 + 1 dq/ 1/2 p/2 2pr1.S p/2 = ½ 0[ -f sin2q + q]/ = ½ f ie transform DF, F(S) = ½ true density for 3 missing atoms phase F(S) p/2 e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

It contains a component of the truth DF cos(q) and an error component DF sin(q) So what if we transform the Difference we knowDF =||F(S)|obs- |F(S)|calc| and F(S) On average DF = ftruecos(q) so On average towards the truth is DF cos(q) = ftruecos2(q) Since the average value of cos2(q) = cos(q) |F(S)|observed 2pr7.S DF q ftrue F(S) 2pr1.S = ½ f ie transform DF, F(S) = ½ true density for 3 missing atoms + noise from error components DF sin(q) phase F(S) e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

ie transform DF, F(S) = ½ true density for 3 missing atoms + noise from error components DF sin(q) So transform 2DF, F(S) = true density for 3 missing atoms + 2*noise from error components DF sin(q) This Difference map contains more of the ‘missing’ truth than the 7 atoms plus some (small) noise. Unbiased by any assumption of the three missing atoms. cos(q) |F(S)|observed 2pr7.S DF q ftrue F(S) 2pr1.S phase F(S) e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

USES: 1. Finding the missing parts of a model. The Transform of |F(S)|calcF(S) = 7 atoms assumed 2DF F(S) = density for 3 missing atoms + ‘noise’ (|F(S)|calc + 2DF) F(S) = all 10 atoms SinceDF =||F(S)|obs- |F(S)|calc| this is 2|F(S)|obs- |F(S)|calccalled a ‘2F0-Fc map’ cos(q) |F(S)|observed 2pr7.S DF q ftrue F(S) It is unbiased as to where the missing components are 2pr1.S phase F(S) e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

USES: 2. Add a substrate, Grow a new crystal Measure New |F(S)|obs+substrateCompare with the apo-protein. TransformDF =||F(S)|obs+substrate- |F(S)|obs|F(S) or [2|F(S)|obs+substrate- |F(S)|obs ] F(S) = a ‘2F0-Fo map’ cos(q) |F(S)|observed 2pr7.S DF q ftrue F(S) It is unbiased as to where the missing substrate is. 2pr1.S phase F(S) e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

Difference Maps: Errors and Signal

Fo-Fc maps identify everything ordered that is 'missing' mapmap -Eliminate Bias -Half electron content -See electrons

Probability, Figure of merit, and the Heavy atom method of Phase Determination.

http://www.ruppweb.org/Xray/101index.html

The Best Fourier: Minimizing errors in the density map.

A Dfference map shows 1/3 occupied NH3 sites and the role of D160 at 1.35Å Resolution. Here are 0.3 NH3 peaks! Khademi..Stoud 2003

The closer you get –the lower the noise. Can see single electrons! |2Fo-Fc| ac map The structure showing ‘thermal ellipsoids’ at 50% Probability |Fo-Fc| ac map Figure 3 The catalytic triad. (A) Stereoview displaying Model H superimposed on the 2Fo Fc (model H phases) at 1 (aqua) and 4 (gold). The densities for C and N in His 64 are weaker than in Asp 32. The Asp 32 CO2 bond at 4 is continuous, while the density for the C and O1 are resolved. (B) Schematic of the catalytic residues and hydrogen bonded neighbors with thermal ellipsoid representation countered at 50% probability (29). Catalytic triad residues Ser 221 and His 64 show larger thermal motion than the Asp 32. Solvent O1059 appears to be a relatively rigid and integral part of the enzyme structure. (C) Catalytic hydrogen bond (CHB). A FoFc (model H phases) difference map contoured at +2.5 (yellow) and 2.5 (red) and a 2Fo Fc (model H phases) electron density map contoured at 4 (gold). The position of the short hydrogen atom (labeled HCHB) in the CHB is positioned in the positive electron density present between His 64 N1 and Asp 32 O2. Published in: Peter Kuhn; Mark Knapp; S. Michael Soltis; Grant Ganshaw; Michael Thoene; Richard Bott; Biochemistry 1998, 37, 13446-13452. DOI: 10.1021/bi9813983 Copyright © 1998 American Chemical Society 4/11/11

Seeing small shifts -down to ±0.1Å Difference Map noise ~20% of noise in the parent protein -and only two peaks!

2Fo-Fc maps (aqua and gold) Fo-Fc map in yellow and purple show individual hydrogen atoms! Figure 2 (A) Electron density for Asp 60. Fo Fc difference electron density map, contoured at +2.5 (yellow) and 2.5 (purple), is superimposed on a 2Fo Fc electron density map contoured at 1 (aqua) and 4 (gold), based on model H (Table 2). No hydrogen atoms were included for residues Asp 60, Gly 63, Thr 66, and solvent. At the 4 contour, the electron density between C and both O1 and O2 are equivalent as expected. (B) H2O hydrogen bonding in an internal water channel. Electron density; 2Fo Fc (model H phases) contoured at 4 (gold) level is superposed on Fo Fc (model H phases) difference electron density contoured at +2.5 (yellow) and 2.5 (purple) level. Peaks for both hydrogen atoms on solvent O1024 are seen along with hydrogen atoms of neighboring solvent and Thr 71 O1. A zigzag pattern of alternating proton donors and acceptors can be seen extending from Thr 71 O1 in the interior, through a chain of four solvent molecules ending at the surface of the enzyme (Figure 1). Published in: Peter Kuhn; Mark Knapp; S. Michael Soltis; Grant Ganshaw; Michael Thoene; Richard Bott; Biochemistry 1998, 37, 13446-13452. DOI: 10.1021/bi9813983 Copyright © 1998 American Chemical Society 4/11/11

What happens if we transform what we do know and can measure, namely I(S) = |F(S)|observed P(r) = S|F(S)|2e(-2pir.S)

Transform of I(hkl) = Transform of F2(hkl)… What happens if I transform what I do know and can measure, namely I(S) = |F(S)|observed ie P(r) = S|F(S)|2e(-2pir.S) = S [F(S).F*(S)]e(-2pir.S) ?

Arthur Lindo Patterson (1902-1966), innovative crystallographer who devised the well-known Patterson method (Patterson function) used in crystal-structure determination.

Intramolecular vectors before rotation

Intramolecular vectors after rotation

All vectors brought to a common origin. 20e Consider a structure of 2 atoms one 20 electrons, one 30 electrons -in a monoclinic ‘space group’ ie. a two fold symmetry axis is present, symbolized by 30e b a

Problem Set: Interpret a Patterson map in terms of heavy mercury atom positions

Density Modification- and Density Averaging.

Density Modification: Solvent Flattening Crystal Intensities I(h,k,l) Electron density r(x,y,z) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Density Modification: Place expected atoms Crystal Intensities I(h,k,l) Electron density r(x,y,z) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Density Modification: Non-Crystal Symmetry Crystal Intensities I(h,k,l) Electron density r(x,y,z) Known: Amino acid sequence Ligands Bond lengths angles Constraints on geometry Phases f(h,k,l) Experimental heavy atom labels selenium for sulfur Trial & error similar structure Atom positions (x,y,z)

Addendum on Icosohedral virus structures

Triangulation numbers of icosohedral viruses There are 60T copies of each protein (asymmetric unit) in the capsid. T=h2 + k2 + hk where h, k are integers T=1,3,4,7,9,12,13, 16, 21,25 are all possible.

Sum of 7 atoms scattering Result is a wave of amplitude F(S) phase f(s) f7=8 electrons 2pr7.S F(S) 2pr1.S f1=6 electrons

Sum of 7 atoms scattering Result is a wave of amplitude F(S) phase F(S) f7=8 electrons 2pr7.S i = √(-1) sin(q) F(S) cos(q) 2pr1.S f1=6 electrons e(iq) = cos(q) + isin(q)

Sum of 7 atoms scattering Result is a wave of amplitude |F(S)| phase F(S) f7=8 electrons 2pr7.S i = √(-1) sin(q) F(S) cos(q) 2pr1.S f1=6 electrons e(iq) = cos(q) + isin(q) F(S) = f1 e(2pir1.S) + f2 e(2pir2.S) +….

Sum of 7 atoms scattering Result is a wave of amplitude |F(S)| phase F(S) f7=8 electrons 2pr7.S i = √(-1) sin(q) F(S) cos(q) 2pr1.S f1=6 electrons e(iq) = cos(q) + isin(q) F(S) = Sjfj e(2pirj.S)

Sum of 7 atoms scattering f7=8 electrons 2pr7.S F(S) 2pr1.S f1=6 electrons

f7=8 electrons 2pr7.S F(S) 2pr1.S f1=6 electrons

e(iq) = cos(q) + isin(q) F(S) = Sjfj e(2pirj.S)

Scattering pattern is the Fourier transform of the structure F(S) = Sjfj e(2pirj.S) Structure is the ‘inverse’ Fourier transform of the Scattering pattern r(r) = SF(S) e(-2pir.S)

Consequences of being a crystal? Repetition = sampling of F(S) 34Å

Build a crystal Object Scattering

ie Wavelength < absorption edge

Friedel’s Law: F(hkl)=F(-h,-k,-l) a(hkl)=-a F(-h,-k,-l)

http://www.ruppweb.org/Xray/101index.html for any/all elements..

Anomalous scattering goes against Friedels Law. –Hence it can be used to break the ambiguity in phase from a single heavy metal derivative, --or to determine the phase of F(hkl)

The Phase probabilty diagram now chooses between the two choices for F(hkl) and F(-h,-k,-l) Pa a

The Phase probabilty diagram now chooses between the two choices for F(hkl) and F(-h,-k,-l) Pa for F(hkl) a mFo mFo

And now for F(-h,-k,-l) Pa a -a mFo mFo Phase Combination Pa (hkl) = P(hkl)ano. P(hkl)heavy atom 1. P(hkl)heavy atom 2….

And now for F(-h,-k,-l) Pa a mFo mFo Phase Combination Pa (hkl) = P(hkl)ano. P(hkl)heavy atom 1. P(hkl)heavy atom 2….

Refinement Least Squares Refinement is common when errors in observations are presumed to be random errors that obey Gaussian statistics. Refine xi,yi,zi, Bi with respect to the Fo Minimize E = Shkl1/s2(k|Fobs|-|Fcalc|)2 with respect to (xyzB)i of all atoms. To include an energy term, that constrains the structure toward acceptable geometry Minimize E = (1-w) Energy + w Shkl1/s2(k|Fobs|-|Fcalc|)2 where w is the fractional weighting on geometry versus X-ray terms. Energy has vdW, torsional restraints, bond length and dihedral angles. Maximum Likelihood refinement seeks the most probable solution most consistent with all observations. ie Least squares refinement alone minimizes the difference between |Fo| and |Fc|, however

Principle of maximum likelihood The basic idea of maximum likelihood is quite simple: the best model is most consistent with the observations. Consistency is measured statistically, by the probability that the observations should have been made. If the model is changed to make the observations more probable, the likelihood goes up, indicating that the model is better. (You could also say that the model agrees better with the data, but bringing in the idea of probability defines "agreement" more precisely.) An example of likelihood The behaviour of a likelihood function will probably be easier to understand with an explicit example. To generate the data for this example, I generated a series of twenty numbers from a Gaussian probability distribution, with a mean of 5 and a standard deviation of 1. We can use these data to deduce the maximum likelihood estimates of the mean and standard deviation.

In the following figure, the Gaussian probability distribution is plotted for a mean of 5 and standard deviation of 1. The twenty vertical bars correspond to the twenty data points; the height of each bar represents the probability of that measurement, given the assumed mean and standard deviation. The likelihood function is the product of all those probabilities (heights). As you can see, none of the probabilities is particularly low, and they are highest in the centre of the distribution, which is most heavily populated by the data.

What happens if we change the assumed mean of the distribution? The following figure shows that, if we change the mean to 4, the data points on the high end of the distribution become very improbable, which will reduce the likelihood function significantly. Also, fewer of the data points are now in the peak region.

What happens if we have the correct mean but the wrong standard deviation? In the following figure, a value of 0.6 is used for the standard deviation. In the heavily populated centre, the probability values go up, but the values in the two tails go down even more, so that the overall value of the likelihood is reduced.

Similarly, if we choose too high a value for the standard deviation (two in the following figure), the probabilities in the tails go up, but the decrease in the heavily-populated centre means that the overall likelihood goes down. In fact, if we have the correct mean the likelihood function will generally balance out the influence of the sparsely-populated tails and the heavily-populated centre to give us the correct standard deviation.

We can carry out a likelihood calculation for all possible pairs of mean and standard deviation, shown in the following contour plot. As you see, the peak in this distribution is close to the mean and standard deviation that were used in generating the data from a Gaussian distribution.

Maximum Likelihood Refinement is better than ‘Least Squares Refinement’ A simple ‘Least squares‘ approach assumes that all errors are ‘Gaussian’, and that errors are all in the AmplitidesFo. But they are not.. To use the structure factor distributions to predict amplitudes, we have to average over all possible values of the phase, and this changes the nature of the error distributions. The rms error in the structure factors involves a vector difference, but we can turn our amplitude difference into a vector difference by assigning both amplitudes the calculated phase.

Understanding Crystallography and Protein Structures

Understanding Crystallography and Protein Structures

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