bp219 2011-4.13

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

http://www.msg.ucsf.edu/

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

http://www.pdb.org/pdb/home/home.do

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

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


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


2pr.S


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)

Wave representation

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?


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?


and why do we care?


average amplitude? 313 e

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


and why do we care?


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

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

34Å 1/34Å-1

3.4Å 1/3.4Å-1

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.

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

4.6Å

b = 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

b.s=k

a.s=h

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

l=2dhklsin(q)

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

• Ronchi ruling:

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





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





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) 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) 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

Measure I(hkl)



Phases f(h,k,l)



Crystal


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

the signal towards some F true will beIntegral 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 haveF(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 haveF(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))/Fie m|F(s)| f(best) =Int P.Fwhere 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)) - Fbest f(best))]2df is minimum.

is minimum when dQ/dFbest = 0so Fbest f(best) = f |F| P(hkl) (f(hkl)) exp (if(hkl)) df

Fbest f(best) = m|F| center of ‘mass’ of the Probability distribution

m = Int P(hkl) (f(hkl)) cos( f - f(best))consider errors from one reflection, and its complex conjugate<(Dr)2> = 2/V2

Then F= Int Fcos( f - f(best))/ FNoise = 1/V Int F sin ( f - f(best))/F = F(1-m2)

mF where


2pr.S

s0sss

s0

s1 S

radius = 1/l

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

F(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

Amplitudes F(h,k,l) Electron density r(x,y,z)

Phases f(h,k,l)



Crystal


Change one side Change the other


Phases f(h,k,l)



Crystal



Phases f(h,k,l)



Crystal


Fhkl

Compare the ‘Observations’, with the Calculation-that depends on assumptions. =Difference Mapping!

Where Bias creeps in: 1. No Assumptions = no changes


Phases f(h,k,l)



Crystal


Fhkl

Compare the ‘Observations’, with the Calculation-that depends on assumptions. =Difference Mapping!

Where Bias creeps in: 1. Errors of Overinterpretation

Solvent FlatteningAverage multiple copies

2pr1.S

f1=6 electrons

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) +….

A reminder: Each individual F(hkl) has its own Phase F(hkl)

F(S)

2pr1.S

f1=6 electrons

F(S)



cos(q)

sin(q)


i = √(-1)

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

In reality, maybe 3 atoms are missing.How to see what is missing?

Suppose we interpret 7 atoms; but 3 remain to be found in density

2pr1.S

f1=6 electrons

F(S)



cos(q)

sin(q)


i = √(-1)

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


phase F(S)

|F(S)|observed

Observed F(hkl) will contain a part of that information..

F(S)

In reality The missing component isThe 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)


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


phase F(S)

|F(S)|observed

F(S)

In reality The missing component isThe 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)


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

phase F(S)

|F(S)|observed

It contains a component of the truth DF cos(q), and an error component DF sin(q)

q DF

The BEST WE CAN DO…IS DF, F(S)

F(S)

So what if we transform the Difference we

know DF =||F(S)|obs- |F(S)|calc| and F(S)

On average DF = ftrue cos(q)so On average towards the truth isDF cos(q) = ftrue cos2(q)What is the average value of cos2(q)?= ½ ( cos 2 +1)q

phase F(S)

|F(S)|observed

It contains a component of the truth DF cos(q)

and an error component DF sin(q)

q DFftrue

= ½ 0[ -f sin2 + ]/q q= ½ fie transform DF, F(S) = ½ true density for 3 missing atoms

+ 11/2

What contribution to the ‘missing truth’?

p/2d / q p/2

p/2p/2

F(S)

So what if we transform the Difference we

know DF =||F(S)|obs- |F(S)|calc| and F(S)

On average DF = ftrue cos(q)so On average towards the truth isDF cos(q) = ftrue cos2(q)

Since the average value of cos2(q)

=

phase F(S)

|F(S)|observed

It contains a component of the truth DF cos(q)

and an error component DF sin(q)

q DFftrue

= ½ f

ie transform DF, F(S) = ½ true density for 3 missing atoms+ noise from error components DF sin(q)

F(S)

phase F(S)

|F(S)|observed

q DFftrue

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 containsmore of the ‘missing’ truth than the 7 atomsplus some (small) noise.

Unbiased by any assumptionof the three missing atoms.

F(S)

USES: 1. Finding the missing parts of a model.The Transform of |F(S)|calc F(S) = 7 atoms assumed2DF F(S) = density for 3 missing atoms + ‘noise’(|F(S)|calc + 2DF) F(S) = all 10 atoms

Since DF =||F(S)|obs- |F(S)|calc| this is

2|F(S)|obs- |F(S)|calccalled a ‘2F0-Fc map’

phase F(S)

|F(S)|observed

q DFftrue

It is unbiased as to where the missingcomponents are

F(S)

USES: 2. Add a substrate, Grow a new crystalMeasure New |F(S)|obs+substrate Compare with the apo-protein.

Transform DF =||F(S)|obs+substrate- |F(S)|obs| F(S)or[2|F(S)|obs+substrate- |F(S)|obs ] F(S)= a ‘2F0-Fo map’

phase F(S)

|F(S)|observed

q DFftrue

It is unbiased as to where the missingsubstrate is.

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

Difference Maps: Errors and Signal

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

Fo-Fc maps identify everything ordered that is 'missing'

mapmap

-Eliminate Bias-Half electron content-See electrons

04/15/2023

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 Fo Fc (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/bi9813983Copyright © 1998 American Chemical Society

The closer you get –the lower the noise. Can see single electrons!

|2Fo-Fc| ac map

The structure showing‘thermal ellipsoids’ at50% Probability

|Fo-Fc| ac map

Difference Mapnoise ~20% of noise in the parent protein-and only two peaks!

Seeing small shifts-down to ±0.1Å

04/15/2023

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/bi9813983Copyright © 1998 American Chemical Society

2Fo-Fc maps (aqua and gold)Fo-Fc map in yellow and purple show individual hydrogen atoms!

What happens if we transform what we do know and can measure, namely I(S) = |F(S)|observed

P(r) = S |F(S)|2 e(-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)|2 e(-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.

Molecular Replacement

Patterson map

Intramolecular vectors before rotation

Colour-coded Patterson map

Intramolecular vectors after rotation

30e

20e

a

b

All vectors brought to a common origin.

Consider a structure of 2 atomsone 20 electrons, one 30 electrons-in a monoclinic ‘space group’ie. a two fold symmetry axis is present, symbolized by

Problem Set:

Interpret a Patterson mapin terms of heavy mercury atompositions

Density Modification- and Density Averaging.



Phases f(h,k,l)



Crystal



Phases f(h,k,l)



Crystal


Density Modification: Solvent Flattening


Phases f(h,k,l)



Crystal


Density Modification: Place expected atoms


Phases f(h,k,l)



Crystal


Density Modification: Non-Crystal Symmetry

Addendum on Icosohedral virus structures

Triangulation numbers of icosohedral virusesThere are 60T copies of each protein (asymmetric unit) in the capsid.

T=h2 + k2 + hkwhere h, k are integers

T=1,3,4,7,9,12,13, 16, 21,25are all possible.

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)


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

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)


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

cos(q)

sin(q)


i = √(-1)

2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)



cos(q)

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)



cos(q)

sin(q)


i = √(-1)


2pr1.S

f1=6 electrons

f7=8 electrons

2pr7.S

F(S)




Phases f(h,k,l)



Crystal


Rosalind Franklin and DNA


34Å


34Å 1/34Å-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…

Appendix: Slides for recall



Phases f(h,k,l)



Crystal


Anomalous Diffraction

ie Wavelength < absorption edge

Anomalous diffraction

Friedel’s Law: F(hkl)=F(-h,-k,-l)

a(hkl)=-a F(-h,-k,-l)

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

a

mFomFo

for F(hkl)

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….

-a

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 geometryMinimize 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.

A simple ‘Least squares‘ approach assumes that all errors are ‘Gaussian’, and that errors are all in the Amplitides Fo. 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.

Maximum Likelihood Refinement is better than ‘Least Squares Refinement’

bp219 2011-4.13

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