Electron Density

1

• Electron Density
• Structure factor amplitude defined as
• Funit cell(S) ?r r (r) exp (2pi r S) dr
• Using the inverse Fourier Transform
• r (r) ?r F(S) exp (-2pi r S) dS
• In practice you make a discrete inverse Fourier
  Transform
• r (r) Shkl Fhkl exp (-2pi Fhkl)
• But measure the X-ray diffraction intensities
• Ihkl ? Fhlk2
• The fact that you cannot measure Fhkl directly
  is called
• THE PHASE PROBLEM

2

• Solutions to the Phase Problem
• Direct methods
• - Based upon systematic relations between certain
  reflections.
• Need high resolution data relatively small
  systems.
• Overwhelmingly most popular for small molecule
  structures.
• Molecular replacement
• Find a molecule of known structure which is
  close enough to your protein of interest to
  provide a good first guess.
• Becoming more popular as the spectrum of
  possible structures is filled up.
• Heavy atom methods
• Soak an atom which is a strong scatter (eg. Hg,
  Fe, Pb, I,Se ..) into your crystal.
• Replace the methionines in your protein with
  selenio-methionine derivatives.
• Use Multiple Isomorphous replacement.
• Use Multiple or Single Anomolous Diffraction.
• An old powerful method for finding phases.

3

• Heavy atom methods
• Must search around for a heavy atom which binds
  within your crystal doesnt destroy the crystal
  lattice.
• Can be extremely frustrating!
• Many soaking, freezing diffraction
  experiments.
• - The heavy atom must bind in an ordered way to
  the protein.
• Suppose you have a protein with structure factor
  FP.
• For every X-ray intensity measured the addition
  of the heavy atom adds a term FH to the
  scattering

FH
FP
FPH
4

• Heavy atom methods
• Assuming that FH has an angle fH, the structure
  factor amplitude will be perturbed by
• DFPH FPH FP FH cos fP - fH

5

• Heavy atom methods
• If you can find the location of you heavy atom
  rH, then you can calculate the heavy atom
  structure factor
• FH fH exp 2pi rH Shkl
• just like any other atom within the protein.
• If you have already measured FH and FPH you can
  recover a constraint on the phases for the
  protein

FP
FPH
FH
6

• Additional constraints
• A single heavy atoms derivative gives you a two
  complex phases which may be correct for every
  reflection (h,k,l).
• For the approach to work crystals must be
  isomorphous (ie. Same a, b, c, a, b, g space
  group).
• In order to determine which phase is correct you
  need to find additional derivatives.

• Called Multiple Isomorphous Replacement

7

• Additional constraints
• If you have a second constraint
• Better to draw diagram with FP centred at origin
  rather than FH FPH.
• This then represents FP FPH FH
• Should recover one place where three circles
  intersect.
• This is your solution for the phases.

Possible solution for light green light blue
measurements.
FP
Possible solution for dark green dark blue
measurements.
8

• Combining phase information
• In practice there are errors with respect to
  each heavy atom experiment.
• - Therefore you recover a probability
  distribution for a phase, rather than an absolute
  phase.
• Best to take a weighted sum of the probabilities
  to determine the experimental phases.
• m is the length of the probability weighted
  experimental phase is called the figure of
  merit.
• Ideally m 1, but in practice m lt 1.

m
9

• Finding Heavy atom Positions
• The previous treatment relied on finding the
  heavy atom positions.
• In practice this is usually solvable using the
  Patterson Function.
• P(u,v,w) Shkl Fhkl 2 exp 2pi uS
• In this case you can calculate it without any
  phase information, but directly from the measured
  intensities.
• Note that Fhkl 2 Fhlk Fhkl hence
• P(u,v,w) r(r) ? r(r) ? r(r) r(r u) dr
• where we use the convolution theorem for Fourier
  Transforms.
• r(r u) represents the inverse of the real
  electron density.
• eg. If you had two heavy atoms in a unit cell
  the Patterson function would look like

10

• Patterson Function
• Since
• P(u,v,w) r(r) ? r(r) ? r(r) r(ru) dr
• This means that the Patterson Map gives you
• A central peak for u 0 since the density sits
  upon itself for all atoms (including the
  protein).
• An additional peak whenever one heavy atom sits
  upon another.
• It rapidly becomes very complicated but can be
  solved (used to be done by inspection) if you
  have a small number of heavy atoms.

11

• Difference Patterson Function
• Since you are looking for the scattering from
  the heavy atoms the protein adds only
  background its more convenient to calculate the
  difference Patterson
• P(u,v,w) Shkl ( FPH 2 - FP 2) exp 2pi
  uS
• This difference Patterson map gives you the
  vectors between the heavy atoms directly.
• - Somewhat easier to interpret than the Patterson
  for FPH itself since it has less background from
  protein-protein distances.

12

• Solving the Difference Patterson
• If you are successful with recovering a
  derivative recover a Patterson function then
  you must solve it to proceed.
• - The Patterson function gives you a set of
  constraints in 3D which are the distances between
  heavy atoms.
• There exist algorithms for finding unique
  solutions which work for a reasonable number of
  heavy atoms
• usually ten or less although I think the record
  may be more than twenty.
• The number of non-origin peaks is N(N-1) for N
  heavy atoms.

? invert
13

• Magnitude of intensity changes with heavy atoms
• May ask how one (or a few) heavy atoms within a
  protein of eg. 50 kDa could be seen?
• - eg. Hg has 80 e-
• - A protein a sea of eg. 4,000 non-carbon atoms
  with faverage 7 e- (ie 28,000 e- in total).
• - How can you see 80 e-/28000 e- 0.3 ?
• On average
• Fprotein faverage vN
• (N is the number of non-hydrogen atom random walk
  result) hence
• Iprotein ? Fatom 2 (faverage)2 N
• But for the heavy atom
• IH ? FH2
• Now
• Fprotein FH faverage vN FH
• Hence
• IPH faverage vN FH 2

14

• IPH faverage vN FH 2 (faverage
  vN)2 2 (faverage vN) FH FH 2
• Therefore
• DIPH (faverage vN)2 2 (faverage vN)
  FH FH 2 - (faverage vN)2
• 2 (faverage vN) FH FH 2
• The first term can be evaluate for eg. 4000
  non-hydrogen atoms
• 2 (faverage vN) FH 2 7 v4000 80
  70,835
• The second term
• FH 2 802 6,400
• And
• Iprotein ? Fatom 2 (faverage)2 N
  196,000
• Hence in this case
• IPH/Iprotein 71/196 36

15
Average intensity change if one heavy atom is
bound to a protein
In practice changes of 14 is a good phasing
measurement.
16

• Lack of Closure
• In practice when you have found the phases
  experimentally there is some mis-match
• The mis-match is called the lack of closure is
  given the symbol e.

Ideal case
Reality
FH
FP
e
FPH
17

• Phasing power
• From the miss-match you can estimate the
  phasing power
• Phasing Power v( S FH2 / S e2)
• A phasing power of 4 is excellent is rare
• A value between 1 2 is acceptable means that
  the scattering of the heavy-atom is larger than
  the lack of closure.

18

• First Map.
• Once you have recovered the experimental phases
  you can make a Fourier transform.
• At that point if the electron density is good
  enough you can build a structural model into the
  density.
• If this goes well you can then make interative
  rounds of structural refinement, phase
  improvement, and more model building until you
  recover a satisfactory model.

19

• Anomalous scattering
• An assumption to date was that X-rays scattering
  from an atom, fj, was a real number
• - In physics this is equivalent to assuming that
  all electrons can be treated as scattering from
  free electrons therefore contribute a phase
  change of p for the scattered X-ray.
• When you go close to the atomic energy levels of
  certain atoms then you can no longer assume that
  this holds.
• X-rays scatter with an anomalous scattering term
  near an absorption edge.

A transition from K to L shell electrons.
A photoelectron ejected from the K shell.
20

• Atomic absorption coeffecient
• For example copper has a K-absorption edge at
  1.38 Å due to the photoelectric effect.
• There are also transitions from K to L at 1.43
  Å.
• If you have Cu in your protein you record near
  1.43 Å you will have an anomalous scattering from
  this atom.

21

• Anomalous scattering
• To describe the anomalous scattering (which is
  wavelength dependent) we modify the atomic
  scattering factor for the particular heavy atom
• fanom f Df if'' f f' f''
• Two different symbols are used for the second
  term depending on what book you look at.

22

• Consequence for Friedel paris
• Earlier we showed that the reflection (h,k,l)
  its opposite (-h,-k,-l) have the same intensity
  since
• F-h-k-l Sj fj(Shkl) exp (2pi rj S-h-k-l)
  (Fhkl)
• therefore Ihkl I-h-k-l
• - These are called Friedel pairs have the same
  intensity
• When you have anomalous scattering this no
  longer holds

FPH(hkl)
FH
FPH(hkl)
FH
Fhkl
Fhkl
F-h-k-l
F-h-k-l
FH
FPH(-h-k-l)
FPH(-h-k-l)
FH
No anomalous scattering
With anomalous scattering
23

• When you have anomalous scattering this no
  longer holds
• Frequently draw the picture with the FPH(-h-k-l)
  reflected in the real axis.
• When scaling data it is possible to measure
  these differences due to the anomalous signal.

FPH(hkl) FPH(-h-k-l)
24

• DFano Patterson map
• If we make a measurement and are careful to
  measure all the Friedel pairs we can define
• DFano FPH(hkl) FPH(-h-k-l) f/2f
• where the scaling factor f/2f is put in for
  technical normalisation reasons.
• If we then calculate the Patterson map using
• Pano(u) Shkl DFano2 exp 2pi uS
• This gives us the inter-distance constraints for
  the anomolous scatterers.
• - This is powerful since you only need one set of
  observations not two, so the noise level is low
  even if DFano is small.

25

• Multiple Anomalous Diffraction
• It is possible to accurately tune synchrotron
  radiation.
• - It is therefore possible to collect diffraction
  data from the same crystal at different
  wavelengths.
• Near an X-ray absorption edge the real
  imaginary components f f change rapidly

0
5
f (electrons)
f (electrons)
1
-10
12.4 12.5 12.6 12.7 12.8
12.4 12.5 12.6 12.7 12.8
X-ray Energy (keV)
X-ray Energy (keV)
26

• Multiple Anomalous Diffraction
• As such it is possible to record three
  diffraction data sets from a single crystal
• - At the peak where f has its maximum.
• - At the peak where f has its minimum (ie. Most
  negative).
• - Far removed from the above two wavelengths.
• This is called Multiple Anomalous diffraction
  (MAD) since you use three wavelengths to get your
  data.
• - You can solve the structure from a single
  crystal (ie. dont need additional derivatives or
  another native data set).
• No problems with crystal being isomorphous.
• The text book has a detailed description of the
  algebra of solving structures using MAD, but for
  this course you can assume its (more or less) the
  same as having one native plus two derivative
  data sets.

27

• SAD SIR
• In addition to MIR MAD there is Single
  Isomorphous Replacement (SIR) or Single Anomalous
  Diffraction (SAD).
• - The same as MIR or MAD but with one data set
  rather than several.
• In SIR if your data is good you assume that your
  phases from one heavy atom are the sum of the two
  possible phases with 50 probability.
• - Then go ahead calculate a map.
• - The wrong choice of phases adds noise.
• Relies on the data being good enough to overcome
  additional noise.
• SAD is like SIR but with the variation that you
  also use the anomalous signal to give additonal
  information.

28

• Molecular Replacement
• If you believe that your protein may have a
  structure similar to that of another protein of
  known structure you can use that information.
• - Solving the structure by making a good guess!
• The problem is to place the possible structure
  correctly within the unit cell then you
  calculate a first electron density map using
  phases calculated from the know structure.

• Also useful if you have sub-domains of known
  structure.

29

• The Rotation Function
• Once you choose a molecular replacement model
  you first need to determine its orientation.
• Done by comparing the experimental Patterson Map
  with one calculated from your candidate
  structure.
• The Patterson map is sensitive to the
  orientation of the molecule but not its position
  within the unit cell.

Rotation
30

• Cross Rotation Function
• An overlap function is defined as R of the
  experimental P(u) with the rotated version of the
  candidate model Patterson Pr(ur) is defined as
• R(a,b,g) ? P(u) Pr(ur) d(u)
• Where a, b g are normally the Euler angles
  describing rotation.
• R(a,b,g) is maximal when the rotational angle is
  correct is not affected by where in the unit
  cell the candidate structural model is placed.

• Poor correlation because the overlap is not
  excellent.
• If rotated can recover perfect overlap ie.
  R(a,b,g) is a maximum.

31

• Translation function
• Once you have found the optimal rotational
  orientation next need to find the correct
  translation.
• In this case the Patterson function is useless
  since its insensitive to translation.
• What one does is move the molecule around in the
  unit cell calculate the theoretical Fhklcalc
  values compare these with the experimental
  Fhklobs.

?
32

• R-factor Correlation Coeffecient
• Two numbers are optimised
• The R-factor
• R Shkl Fobs - kFcalc/ (Shkl Fobs)
• which should be minimised
• - ie. calculated structure factors are as close
  as possible to the experimental structure
  factors.
• The Standard Linear Correlation Coeffecient
• C Shkl (Fobs2 - ltFobs2gt) (Fcalc2 -
  ltFcalc2gt)
• Shkl (Fobs2 - ltFobs2gt)2 Shkl (Fcalc2 -
  ltFcalc2gt)2 -1/2
• which should be maximised.
• ie. When Fobs is much greater than the average,
  Fcalc should also be greater than the average
  etc.
• Useful values are C gt 30 Rlt 55 .

33

• Direct methods
• If you have a small number of atoms very good
  resolution you may recover a structure from a
  native data set.
• The concept is that there are phase-relations
  between different Bragg reflections
• f(h1) f(h2) f(-h1 h2) 0
• Geometrical arguments can be used to show that
  this holds when atoms sit on the lattice planes
  but not in-between lattice planes (chapter 11 of
  the textbook).
• In practice this assumption holds approximately
  for very strong reflections, which are strong
  since most of the atoms are scattering in phase
  therfore on the same lattice plane.
• - This assumption itself breaks down for large
  proteins.

34

• Shake Bake
• Direct methods is very successful for small
  molecule crystallography.
• - In practice you pick a few phases derive the
  rest from the triplet relations for a limited
  number of strong reflections.
• - For molecules with gt 150 non-hydrogen atoms the
  unit cell is so evenly filled with atoms that the
  phase triplet relation doesnt work.
• An algorithm has been written to extend up to
  about 1000 non-hydrogen atoms but requires about
  1.2 Å data.
• The principle is that the phase triplet is no
  longer set to zero but obeys a probability
  distribution.
• You then shake the phase angles in reciprocal
  space bake out the low density regions in real
  space.

35

• Protein Data Bank Entries
• This month the number of entries in the protein
  data bank will surpass 30,000.
• Current deposit rate is approaching 6,000 per
  year or 20 per day.
• - Not all are unique!
• X-ray crystallography is responsible for approx
  75
• - Phasing methods have been very successful!