# Solving Crystal Structures From Two-wavelength X-ray Powder .._1_

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```					                                   Phasing
Today’s goal is to calculate phases (ap) for proteinase K using
MIRAS method (PCMBS and GdCl3).
What experimental data do we need?
1) from native crystal we measured |Fp| for native crystal
2) from PCMBS derivative we measured |FPH1(h k l)|
3) also from PCMBS derivative we measured |FPH1(-h-k-l)|
4) FH1 for Hg atom of PCMBS
5) from GdCl3 derivative we measured |FPH2(h k l)|
6) also from GdCl3 derivative we measured |FPH2(-h-k-l)|
7) FH2 for Gd atom of GdCl3

How many phasing triangles will we have for each structure factor if
we use all the data sets ?
FP ( h k l) = FPH ( h k l) - FH ( h k l) is one type of phase triangle.
FP ( h k l) = FPH(-h-k-l) * - FH (-h-k-l)* is another type of phase triangle.
Four Phase Relationships
PCMBS
FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l) isomorphous differences
FP ( h k l) = FPH1(-h-k-l)* - FH1 (-h-k-l)*   from Friedel mates (anomalous)

GdCl3
FP ( h k l) = FPH2( h k l) - FH2 ( h k l)     isomorphous differences

FP ( h k l) = FPH2 (-h-k-l)* - FH2 (-h-k-l)* from Friedel mates (anomalous)
FP ( h k l) = FPH1 ( h k l) - FH1 (h k l)
Imaginary axis
|Fp ( h k l) | = 1.8
2

1

Real axis

-2     -1                    1   2

-1

-2

Harker construction

1) |FP ( h k l) | –native measurement
FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l)
Imaginary axis
|Fp ( h k l) | = 1.8
|FH ( h k l) | = 1.4
aH (hkl)       = 45°                              |Fp |

Real axis
FH

Harker construction

1) |FP ( h k l) | –native measurement
2) FH (h k l) calculated from heavy
atom position.
FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l)
Imaginary axis
|FP ( hkl) | = 1.8
|FH1 ( hkl) | = 1.4
aH1 (hkl) = 45°                                           |Fp |
|FPH1 (hkl) | = 2.8
|FPH|
Real axis

FH

|FPH|

Harker construction
Let’s look at the quality                                        1) |FP ( h k l) | –native measurement
2) FH1 (h k l) calculated from heavy
of the phasing statistics                                        atom position.
3) |FP1(hkl)|–measured from
up to this point.                                           derivative.
Which of the following
SIR                                                graphs best represents
the phase probability
Imaginary axis                                    distribution, P(a)?

|Fp |
a)
0       90    180   270   360
|FPH|
Real axis

FH                               b)
|FPH|                            0       90   180    270   360

c)

0       90   180    270   360
The phase probability
SIR                                       distribution, P(a) is sometimes
shown as being wrapped
Imaginary axis                          around the phasing circle.

|Fp |

0     90    180   270   360
|FPH|
Real axis

FH

|FPH|                                      90

180                0

270
Which of the following is
SIR                                       the best choice of Fp?

Imaginary axis
90

|Fp |                   a)       180          0

90                                                   270

|FPH|
180                       0       Real axis
90

FH                                   b)       180          0

|FPH|                                          270
270
90
c)
180          0

270

Radius of circle is approximately |Fp|
SIR
Imaginary axis

|Fp |
best F =         |Fp|eia•P(a)da
90                                   a

|FPH|
180                         0   Real axis

Sum of
FH
probability
|FPH|
270                                  weighted
vectors Fp
Usually shorter than Fp
SIR
best F =       |Fp|eia•P(a)da
90                  a

180         0
Sum of
probability
270               weighted
vectors Fp

Best phase
SIR        Which of the following is
the best approximation
to the Figure Of Merit
(FOM) for this reflection?

90
a)    1.00
b)    2.00
180         0
c)    0.50
d)    -0.10
270

FOM=|Fbest|/|FP|
Radius of circle is approximately |Fp|
Which phase probability distribution would
yield the most desirable Figure of Merit?

a)                      b)
0
90

0
270
+
+     90

180                             180

270                        0
c)
270

+
+        90

180
SIR                                          Which of the following is
the best approximation
Imaginary axis
to the phasing power for
this reflection?

|Fp |
a)   2.50
|FPH|                 Fbest                           b)   1.00
Real axis

|Fp |
c)   0.50
FH
d)   -0.50
|FPH|
Phasing Power =               |FH|
Lack of closure

|FH ( h k l) | = 1.4
Lack of closure = |FPH|-|FP+FH| = 0.5
(at the aP of Fbest)
SIR                                        Which of the
Imaginary axis                    following is the most
desirable phasing
power?

|Fp |
a)   2.50
|FPH|                 Fbest                   b)   1.00
Real axis

|Fp |
c)   0.50
FH
d)   -0.50
|FPH|
Phasing Power =             |FH|
Lack of closure

What Phasing Power is sufficient to solve the structure?        >1
|FP ( hkl) | = 1.8
SIR                        Which of the following is
Imaginary axis
|FPHg (hkl) | = 2.8
the RCullis for this
reflection?
|Fp |
a)   -0.5
|FPH|                 Fbest                   b)   0.5
Real axis

|Fp |
c)   1.30
FH
d)   2.00
|FPH|
RCullis =   Lack of closure

isomorphous difference

From previous page, LoC=0.5
Isomorphous difference= |FPH| - |FP|
1.0 = 2.8-1.8
SIRAS           Isomorphous differences
Anomalous differences
Imaginary axis

FP ( h k l) = FPH(-h-k-l)* - FH (-h-k-l)*

Real axis

We will calculate SIRAS phases
using the PCMBS Hg site.

FP –native measurement
FH (hkl) and FH(-h-k-l) calculated
from heavy atom position.
FPH(hkl) and FPH(-h-k-l) –measured
from derivative. Point to these on
graph.

To Resolve the phase ambiguity
SIRAS                          Isomorphous differences
Anomalous differences

Imaginary axis

Which P(a) corresponds to SIR?
Real axis
Which P(a) corresponds to SIRAS?

0     90   180   270   360
SIRAS                          Isomorphous differences
Anomalous differences

Imaginary axis

Which graph represents FOM of SIRAS phase?
Which represents FOM of SIR phase?
Which is better? Why?

Real axis

90                90

180         0     180         0

270               270
Calculate phases and refine
parameters (MLPhaRe)
A Tale of Two Ambiguities
• We can solve both ambiguities in one
experiment.
Center of inversion ambiguity
Remember, because the position of
Hg was determined using a Patterson
map there is an ambiguity in
handedness.
The Patterson map has an additional
center of symmetry not present in the
real crystal. Therefore, both the site
x,y,z and -x,-y,-z are equally
consistent with Patterson peaks.
Handedness can be resolved by
calculating both electron density maps
and choosing the map which contains
structural features of real proteins (L-
amino acids, right handed a-helices).
If anomalous data is included, then
one map will appear significantly better
than the other.

Note: Inversion of the space group symmetry
(P43212 →P41212) accompanies inversion of the
Patterson map        coordinates (x,y,z→ -x,-y,-z)
Choice of origin ambiguity
• I want to include the Gd data (derivative 2)
in phase calculation.
• I can determine the Gd site x,y,z coordinates
using a difference Patterson map.
• But, how can I guarantee the set of
coordinates I obtain are referred to the same
origin as Hg (derivative 1)?
• Do I have to try all 48 possibilities?
Use a Cross difference Fourier to resolve
the handedness ambiguity
With newly calculated protein phases, fP, a protein
electron density map could be calculated.
The amplitudes would be |FP|, the phases would be fP.
r(x)=1/V*S|FP|e-2pi(hx+ky+lz-fP)

Answer: If we replace the coefficients with
|FPH2-FP|, the result is an electron density map
corresponding to this structural feature.
r(x)=1/V*S|FPH2-FP|e-2pi(hx-fP)

What is the second heavy atom, Alex.
When the difference FPH2-FP is taken, the protein
component is removed and we are left with only the
contribution from the second heavy atom.

This cross difference Fourier will help us in two ways:
1) It will resolve the handedness ambiguity by
producing a very high peak when phases are
calculated in the correct hand, but only noise when
phases are calculated in the incorrect hand.
2) It will allow us to find the position of the second
heavy atom and combine this data set into our
phasing. Thus improving our phases.
Phasing Procedures

1) Calculate phases for site x,y,z of Hg and run
cross difference Fourier to find the Gd site. Note
the height of the peak and Gd coordinates.
2) Negate x,y,z of Hg and invert the space group
from P43212 to P41212. Calculate a second set of
phases and run a second cross difference
Fourier to find the Gd site. Compare the height
of the peak with step 1.
3) Chose the handedness which produces the
highest peak for Gd. Use the corresponding
hand of space group and Hg, and Gd
coordinates to make a combined set of phases.
FH1(hkl)
FPH1(hkl)
FH1(-h-k-l)*
FPH1(-h-k-l)*
MIRAS
Imaginary axis

FH2(hkl)                                        Isomorphous
Deriv 2
FPH2(hkl)

Anomalous              Fp (hkl)
Deriv 1                                     Real axis

Isomorphous
Deriv 1

GdCl3      FPH = FP+FH            for isomorphous differences
FH1(hkl)
FPH1(hkl)             GdCl3                FP ( h k l) = FPH1 (-h-k-l)* - FH1 (-h-k-l)*
FH1(-h-k-l)*
FPH1(-h-k-l)*
Imaginary axis

FH2(hkl)                                                   Isomorphous
Deriv 2
FPH2(hkl)
FH2 (-h-k-l)*
FPH2(-h-k-l)*
Anomalous                Fp (hkl)
Deriv 1                                                 Real axis

Isomorphous                    Anomalous
Deriv 1                       Deriv 2

Harker Construction for MIRAS phasing (Multiple Isomorphous Replacement with Anomalous Scattering)
FH1(hkl)
FPH1(hkl)                                                   For opposite hand
FH1(-h-k-l)*
FPH1(-h-k-l)*                    Isomorphous
Deriv 1       Imaginary axis

FH2(hkl)
FPH2(hkl)                                                                    Anomalous
Deriv 2
FH2 (-h-k-l)*
FPH2(-h-k-l)*
Fp (hkl)
Real axis

Anomalous
Deriv 1
Isomorphous
Deriv 2

Note: in the presence of anomalous scattering, FH(HKL) ≠ FH(-H-K-L)* so if this opposite hand is the incorrect
hand, the phase angle (aopposite) will be significantly different than (-aoriginal). The electron density would not
just be inverted, but inverted and more noisy.
Density modification

A) Solvent flattening.
• Calculate an electron density map.
• If r<threshold, -> solvent
• If r>threshold -> protein
• Set density value in solvent region
to a constant (low).
• Transform flattened map to structure
factors
• Combine modified phases with
original phases.
• Iterate
MIR phased map +
MIRAS phased map   Solvent Flattening +
Histogram Matching
MIR phased map +
MIRAS phased map   Solvent Flattening +
Histogram Matching
Density modification
B) Histogram matching.
• Calculate an electron density
map.
• Calculate the electron density
distribution. It’s a histogram.
How many grid points on map
have an electron density
falling between 0.2 and 0.3
etc?
• Compare this histogram with
ideal protein electron density
map.
• Modify electron density to
resemble an ideal distribution.
Number of times a particular electron density value is observed.

Electron density value
Each grid point has a value r(x,y,z)

Y

X
Z
represent r(x,y,z) in shades of gray

Y

X
Z                    How many grid points have density 0?
How many grid points have density 1?
How many grid points have density x?
ntotal=130 pixels total

r=0 n=60    r=1 n=6     r=2 n=17   r=3 n=5   r=4 n=1

r=5 n=2     r=6 n=12    r=7 n=2    r=8 n=4   r=9 n=1

r=10 n=25
Histogram of electron density
distribution P(r).

Number of
pixels with
density (r)

0 1 2 3 4 5 6 7 8 9 10
Density (r)
Cumulative density distribution
function N(r).

Number of pixels with density < r
Number of pixels with
density (r)

0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
Density (r)
Cumulative density
Histogram matching density
distribution function N(r).

Number of pixels with density < r
Number of pixels with density < r

Cumulative density                                                                Cumulative density
observed                                                                             ideal
Pixels with r=4 in original map will be modified to r= 2.5 in the histogram matched map.
HOMEWORK
Barriers to combining phase
information from 2 derivatives
1) Initial Phasing with PCMBS
1) Calculate phases using coordinates you determined.
2) Refine heavy atom coordinates
2) Find Gd site using Cross Difference Fourier map.
1) Easier than Patterson methods.
2) Want to combine PCMBS and Gd to make MIRAS phases.
3) Determine handedness (P43212 or P41212 ?)
1) Repeat calculation above, but in P41212.
2) Compare map features with P43212 map to determine
handedness.
4) Combine PCMBS and Gd sites (use correct hand of
space group) for improved phases.
5) Density modification (solvent flattening & histogram
matching)
1) Improves Phases
6) View electron density map

```
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 views: 4 posted: 9/2/2012 language: English pages: 40