Today: compute the
experimental electron density
map of proteinase K
Fourier synthesis
r(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl)
hkl
3 Crystals
5 Measured Quantities
Native
H
1
1
1
1
1
1
1
1
1
K
1
1
1
1
1
1
1
1
2
L
1
2
3
4
5
6
7
8
0
PCMBS (Hg)
|FP|
|FPH (+)| |FPH (-)|
681.4 725.8 722.4
752.8 733.6 695.3
332.1 444.5 456.2
526.9 575.8 564.7
719.2 827.8 805.4
358.4 349.8 354.2
273.3 359.4 390.8
400.7 362.5 411.2
162.5
73.8 132.3
EuCl3 (Eu)
|FPH (+)| |FPH (-)|)
730.8 707.6
813.9 805.3
296.1 312.5
527.4 518.3
759.6 766.3
375.6 358.9
300.5 286.6
396.7 411.5
149.8 159.8
Measureable differences
1) Isomorphous differences
(between native and derivative)
2) Anomalous Differences
(i.e. differences between Friedel pairs):
3) Dispersive differences
(differences due to changing the
wavelength):
FP(hkl)=FPH (hkl) - fH(hkl)
FP(hkl)=FPH(-h-k-l)* - fH(-h-k-l)*
FP(hkl)=FPH (hkl)
ln
- fH(hkl) l
n
3 Crystals
5 Measured Quantities
Native
H
1
1
1
1
1
1
1
1
1
K
1
1
1
1
1
1
1
1
2
L
1
2
3
4
5
6
7
8
0
PCMBS (Hg)
|FP|
|FPH (+)| |FPH (-)|
681.4 725.8 722.4
752.8 733.6 695.3
332.1 444.5 456.2
526.9 575.8 564.7
719.2 827.8 805.4
358.4 349.8 354.2
273.3 359.4 390.8
400.7 362.5 411.2
162.5
73.8 132.3
EuCl3 (Eu)
|FPH (+)| |FPH (-)|
730.8 707.6
813.9 805.3
296.1 312.5
527.4 518.3
759.6 766.3
375.6 358.9
300.5 286.6
396.7 411.5
149.8 159.8
For EuCl3 (Eu)
PCMBS (Hg)
Vector equations for this
experiment
Isomorphous and
Anomalous Differences
Isomorphous and
Anomalous Differences
FP(hkl)=FPHg (hkl) - fHg(hkl)
FP(hkl)=FPHg(-h-k-l)* - fHg(-h-k-l)*
FP(hkl)=FPEu (hkl) - fEu(hkl)
FP(hkl)=FPEu(-h-k-l)* - fEu(-h-k-l)*
Vector equations for this
experiment
Isomorphous Differences
FP(hkl)=FPHg (hkl) - fHg(hkl)
We have collecting data on the native and
derivative crystals.
We know the coordinates of Hg.
How many unknown quantities remain?
SIR Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(+) - fHg(+)
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
|FP |
250
Real axis
-500
-250
250
-250
-500
500
SIR Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(+) - fHg(+)
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
|FP |
+f’+if[e2pi*(h(x)+k(y)+l(z))+
fHg=fHg
e2pi*(h(-x)+k(-y)+l(½+z))+
e2pi*(h(½-y)+k(½+x)+l(¾+z)+
e2pi*(h(½+y)+k(½-x)+l(¼+z)+
e2pi*(h(½-x)+k(½+y)+l(¾-z)+
e2pi*(h(½+x)+k(½-y)+l(¼-z)+
e2pi*(h(y)+k(x)+l(-z)+
e2pi*(h(-y)+k(-x)+l(½-z)]
250
Real axis
-500
-250
x,y,z are Hg coordinates from Patterson map (0.197, 0.755, 0.935)
f’ and f” are anomalous scattering corrections specific for wavelength used.
fHg is a real number proportional to the number of e- in Hg
250
-250
-500
500
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)|
9 2 1 486 586 611
fHg ≠ |FP|-|FPH (+)|
FP=FP·Hg(+) - fHg(+)
|FPH (+)| |FPH (-)|
536 499
Imaginary axis
500
|FP |
No!
[e2pi*(h(x)+k(y)+l(z))+
fHg=fHg
e2pi*(h(-x)+k(-y)+l(½+z))+
e2pi*(h(½-y)+k(½+x)+l(¾+z)+
e2pi*(h(½+y)+k(½-x)+l(¼+z)+
e2pi*(h(½-x)+k(½+y)+l(¾-z)+
e2pi*(h(½+x)+k(½-y)+l(¼-z)+
e2pi*(h(y)+k(x)+l(-z)+
e2pi*(h(-y)+k(-x)+l(½-z)]+f’+if”
250
Real axis
-500
-250
x,y,z are Hg coordinates from Patterson map (0.197, 0.755, 0.935)
f’ and f” are anomalous scattering corrections specific for wavelength used.
fHg is a real number proportional to the number of e- in Hg
250
-250
-500
500
SIR Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(+) - fHg(+)
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
|FP |
|fHg|=282
aHg=58°
250
Real axis
58°
fHg(+)
-500
-250
250
-250
-500
500
SIR Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(+) - fHg(+)
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
|FP |
250
Real axis
-500
-250
250
-fHg(+)
-250
Let’s look at the quality
of the phasing statistics
P·Hg(+)|
up to this |F
point.
-500
500
SIR Phasing
H K L |FP|
9 2 1 486
Which of the following
graphs best represents
the phase probability
distribution, P(a)?
|FPH (+)| |FPH (-)| |FPH (+)|
586 611
536 4
Imaginary axis
500
|FP |
a)
250
0
90
180
270
360
Real axis
-500
-250
250
500
-fHg(+)
b)
-250
0
90
180
270
360
-500
c)
|FP·Hg(+)|
0
90
180
270
360
SIR Phasing
H K L |FP|
9 2 1 486
The phase probability
distribution, P(a) is sometimes
shown as being wrapped
around the phasing circle.
|FPH (+)| |FPH (-)| |FPH (+)|
586 611
536 4
Imaginary axis
500
|FP |
250
0
90
180
270
360
Real axis
-500
-250
250
500
-fHg(+)
90
-250
-500
180
0
|FP·Hg(+)|
270
SIR Phasing
Which of the following is
the best choice of Fp?
H K L |FP|
9 2 1 486
|FPH (+)| |FPH (-)| |FPH (+)|
586 611
536 4
Imaginary axis
90
a)
180
500
90
0
|FP |
270
180
-500
90
b)
180
0500Real axis
0
270
270
-500
90
c)
180
0
270
Radius of circle is approximately |Fp|
|FP·Hg(+)|
SIR Phasing
H K L |FP|
9 2 1 486
|FPH (+)| |FPH (-)| |FPH (+)|
586 611
536 4
Imaginary axis
500
|Fp|eia•P(a)da
best FP =
90
|FP |
a
Sum of
probability
weighted
vectors Fp
Usually shorter than Fp
180
-500
0500Real axis
270
-500
|FP·Hg(+)|
SIR Phasing
H K L |FP|
9 2 1 486
|FPH (+)| |FPH (-)| |FPH (+)|
586 611
536 4
Imaginary axis
500
|Fp|eia•P(a)da
best FP =
90
a
Sum of
probability
weighted
vectors Fp
Usually shorter than Fp
180
-500
|Fbest |
Real axis
0500
270
-500
|FP·Hg(+)|
SIR
|Fp|eia•P(a)da
best F =
90
a
0
180
270
Best phase
Sum of
probability
weighted
vectors Fp
SIR
Which of the following is
the best approximation
to the Figure Of Merit
(FOM) for this reflection?
90
0
180
a)
b)
c)
d)
1.00
2.00
0.50
-0.10
270
FOM=|Fbest|/|FP|
Radius of circle is approximately |Fp|
Which phase probability distribution would
yield the most desirable Figure of Merit?
b)
a)
0
90
270
0
180
270
+
+
90
180
c)
0
+
+
270
180
90
SIR
Which of the following is
the best approximation
to the phasing power for
this reflection?
Imaginary axis
|Fp |
Fbest
|FPH|
Real axis
fH
|Fp |
a)
b)
c)
d)
2.50
1.00
0.50
-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
following is the most
desirable phasing
power?
Imaginary axis
|Fp |
Fbest
|FPH|
Real axis
fH
|Fp |
a)
b)
c)
d)
2.50
1.00
0.50
-0.50
|FPH|
Phasing Power =
|fH|
Lack of closure
What Phasing Power is sufficient to solve the structure?
>1
|FP ( hkl) | = 1.8
|FP•Hg (hkl) | = 2.8
SIR
Imaginary axis
Which of the following is
the RCullis for this
reflection?
a)
b)
c)
d)
|Fp |
Fbest
|FPH|
Real axis
fH
|Fp |
-0.5
0.5
1.30
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 Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(-)* - fHg(-)*
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
|FP |
|fHg|=282
aHg*=48°
250
Real axis
fHg(-)*
-500
-250
250
-fHg(+)
48°
-250
|FP·Hg(+)|
-500
500
SIRAS Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(-)* - fHg(-)*
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
|FP |
250
Real axis
-500
-250
-fHg(-)*
250
-fHg(+)
-250
|FP·Hg(-)|
|FP·Hg(+)|
-500
500
SIRAS
Isomorphous differences
Anomalous differences
Which P(a) corresponds to SIR?
Which P(a) corresponds to SIRAS?
0
90
180
270
360
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 (Lamino acids, right handed a-helices).
If anomalous data is included, then
one map will appear significantly better
than the other.
Patterson map
Note: Inversion of the space group symmetry
(P43212 →P41212) accompanies inversion of the
coordinates (x,y,z→ -x,-y,-z)
Choice of origin ambiguity
• I want to include the Eu data (derivative 2)
in phase calculation.
• I can determine the Eu 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, aP, a protein
electron density map could be calculated.
The amplitudes would be |FP|, the phases would be aP.
r(xyz)=1/V*S|FP|e-2pi(hx+ky+lz-aP)
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-aP)
What is the second heavy atom, Alex.
When the difference FPH2-FP is taken, the protein contribution
is cancelled 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
-high peak in difference map calculated with aP in correct hand
-only noise in difference map calculated with aP in incorrect hand.
2) It will improve our electron density map of the protein
-identify the position of the 2nd heavy atom
-include 2 new vector equations for Eu (more accurate aP)
Phasing Procedures
1) Calculate phases for site x,y,z of Hg and run
cross difference Fourier to find the Eu site.
-Note the height of the peak and Eu 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 Eu site.
-Compare the height of the peak with step 1.
3) Chose the handedness which produces the
highest peak for Eu. Use the corresponding
hand of space group and Hg, and Eu
coordinates to make a combined set of phases.
MIRAS Phasing
H K L |FP|
9 2 1 486
FP=FP·Hg(+) - fHg(+)
|FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
586 611
536 499
Imaginary axis
500
FP=FP·Hg(-)* - fHg(-)*
250
Real axis
-500
fH+f’ f”(-) aH
281
27
53°
250
-250
|FP | = 486 ± 2
H K L
9 2 1
-250
fH+f’ f” aH
100
24 -114°
-500
500
SIR
SIRAS
MIRAS
SIR
SIRAS
MIRAS
Density modification
A) Solvent flattening.
• Calculate an electron density map.
• If r<threshold, -> solvent
• If r>threshold -> protein
• Build a mask
• Set density value in solvent region
to a constant (low).
• Transform flattened map to structure
factors
• Combine modified phases with
original phases.
• Iterate
MIRAS phased map
MIR phased map +
Solvent Flattening +
Histogram Matching
MIRAS phased map
MIR 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
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 Eu site using Cross Difference Fourier map.
1) Easier than Patterson methods.
2) Want to combine PCMBS and Eu 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 Eu 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