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Interpreting difference Patterson Maps in Lab this week!

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Hurray!!!! SYMMETRY OPERATORS. FOR PLANE GROUP P2. 1) x,y. 2) -x,-y. x y -(-x y) 2x 2y ... HURRAY! we got back the coordinates. of our smiley faces!!!! (0,0) a ... – PowerPoint PPT presentation

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Title: Interpreting difference Patterson Maps in Lab this week!


1
Interpreting difference Patterson Maps in Lab
this week!
  • Calculate an isomorphous difference Patterson Map
    (native-heavy atom) for each derivative data set.
    We collected12 derivative data sets in lab
    (different heavy atoms at different
    concentrations)
  • HgCl2
  • PCMBS
  • Hg(Acetate)2
  • EuCl3
  • GdCl3
  • SmCl3
  • How many heavy atom sites per asymmetric unit, if
    any?
  • What are the positions of the heavy atom sites?
  • Lets review how heavy atom positions can be
    calculated from difference Patterson peaks.

2
Patterson Review
  • A Patterson synthesis is like a Fourier synthesis
    except for what two variables?

Fourier synthesis r(xyz)S Fhkl cos2p(hxkylz
-ahkl) hkl

Patterson synthesis P(uvw)S ?hkl cos2p(hukvlw
-?) hkl

Patterson synthesis P(uvw)S Ihkl cos2p(hukvlw
-0) hkl

3
Hence, Patterson density map electron density
map convoluted with its inverted image.
  • Patterson synthesis
  • P(uvw)S Ihkl cos2p(hukvlw)
  • Remembering IhklFhklFhkl
  • And Friedels law Fhkl F-h-k-l
    P(uvw)FourierTransform(FhklF-h-k-l)
  • P(uvw)r(uvw) r (-u-v-w)

4
Electron Density vs. Patterson Density
Lay down n copies of the unit cell at the origin,
where nnumber of atoms in unit cell.
For copy n, atom n is placed at the origin. A
Patterson peak appears under each atom.
Patterson Density Map single water molecule
convoluted with its inverted image.
Electron Density Map single water molecule in the
unit cell
5
Every Patterson peak corresponds to an
inter-atomic vector
  • 3 sets of peaks
  • Length O-H
  • Where?
  • Length H-H
  • Where?
  • Length zero
  • Where?
  • How many peaks superimposed at origin?
  • How many non-origin peaks?

Patterson Density Map single water molecule
convoluted with its inverted image.
Electron Density Map single water molecule in the
unit cell
6
Patterson maps are more complicated than electron
density maps. Imagine the complexity of a
Patterson map of a protein
Unit cell repeats fill out rest of cell with peaks
Patterson Density Map single water molecule
convoluted with its inverted image.
Electron Density Map single water molecule in the
unit cell
7
Patterson maps have an additional center of
symmetry
Patterson Density Map single water molecule
convoluted with its inverted image.
Electron Density Map single water molecule in the
unit cell
8
Calculating X,Y,Z coordinates from Patterson peak
positions (U,V,W)Three Examples
  1. Exceedingly simple 2D example
  2. Straightforward-3D example, Pt derivative of
    polymerase b in space group P21212
  3. Advanced 3D example, Hg derivative of proteinase
    K in space group P43212.

9
What Plane group is this?
10
Lets consider only oxygen atoms
Analogous to a difference Patterson map where we
subtract out the contribution of the protein
atoms, leaving only the heavy atom
contribution. Leaves us with a Patterson
containing only self vectors (vectors between
equivalent atoms related by crystal symmetry).
Unlike previous example.
11
How many faces?
  • In unit cell?
  • In asymmetric unit?
  • How many peaks will be in the Patterson map?
  • How many peaks at the origin?
  • How many non-origin peaks?

12
Symmetry operators in plane group p2
Coordinates of one smiley face are given as 0.2,
0.3. Coordinates of other smiley faces are
related by symmetry operators for p2 plane
group. For example, symmetry operators of plane
group p2 tell us that if there is an atom at
(0.2, 0.3), there is a symmetry related atom at
(-x,-y) (-0.2, -0.3). But, are these really
the coordinates of the second face in the unit
cell?
(-0.2,-0.3)
a
(0,0)
b
(0.2,0.3)
SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)
-x,-y
Yes! Equivalent by unit cell translation. (-0.21
.0, -0.31.0)(0.8, 0.7)
13
Patterson in plane group p2Lay down two copies
of unit cell at origin, first copy with smile 1
at origin, second copy with simile 2 at origin.
2D CRYSTAL
PATTERSON MAP
(-0.2,-0.3)
a
(0,0)
a
(0,0)
b
b
(0.2,0.3)
What are the coordinates of this Patterson self
peak? (a peak between atoms related by xtal
sym) What is the length of the vector between
faces?
SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)
-x,-y
  • Patterson coordinates (U,V) are simply
  • symop1-symop2. Remember this bridge!
  • symop1 X , Y 0.2, 0.3
  • symop2 -(-X,-Y) 0.2, 0.3
  • 2X, 2Y 0.4, 0.6 u, v

14
Patterson in plane group p2
(-0.4, -0.6)
(-0.2,-0.3)
a
(0,0)
b
(0.2,0.3)
(0.6, 0.4)
(0.4, 0.6)
SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)
-x,-y
PATTERSON MAP
2D CRYSTAL
15
Patterson in plane group p2
a
(0,0)
b
(0.6, 0.4)
(0.4, 0.6)
If you collected data on this crystal and
calculated a Patterson map it would look like
this.
PATTERSON MAP
16
Now Im stuck in Patterson space. How do I get
back to x,y coordinates?
Remember the Patterson Peak positions (U,V)
correspond to vectors between symmetry related
smiley faces in the unit cell. That is,
differences betrween our friends the space group
operators.
a
(0,0)
b
(0.6, 0.4)
(0.4, 0.6)
SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)
-x,-y
PATTERSON MAP
x , y -(-x, y) 2x , 2y
u2x, v2y
symop 1 symop 2
plug in Patterson values for u and v to get x and
y.
17
Now Im stuck in Patterson space. How do I get
back to x,y, coordinates?
SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)
-x,-y
a
(0,0)
b
x y -(-x y) 2x 2y
symop 1 symop 2
(0.4, 0.6)
set u2x v2y plug in Patterson values for u
and v to get x and y.
v2y 0.62y 0.3y
u2x 0.42x 0.2x
PATTERSON MAP
18
Hurray!!!!
SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)
-x,-y
a
(0,0)
b
x y -(-x y) 2x 2y
symop 1 symop 2
(0.2,0.3)
set u2x v2y plug in Patterson values for u
and v to get x and y.
2D CRYSTAL
v2y 0.62y 0.3y
u2x 0.42x 0.2x
HURRAY! we got back the coordinates of our
smiley faces!!!!
19
Devils advocate What if I chose u,v (0.6,0.4)
instead of (0.4,0.6) to solve for smiley face
(x,y)?
a
(0,0)
using Patterson (u,v) values 0.4, 0.6 to get x
and y.
b
v2y 0.62y 0.3y
u2x 0.42x 0.2x
(0.6, 0.4)
(0.4, 0.6)
using Patterson (u,v) values 0.6, 0.4 to get x
and y.
v2y 0.42y 0.2y
PATTERSON MAP
u2x 0.62x 0.3x
These two solutions do not give the same x,y?
What is going on??????
20
Arbitrary choice of origin
(-0.2,-0.3)
a
(0,0)
b
(-0.3,-0.2)
(0.2,0.3)
(0.3,0.2)
(0.8,0.7)
(0.7,0.8)
  • Original origin choice
  • Coordinates x0.2, y0.3.
  • New origin choice
  • Coordinates x0.3, y0.2.

Related by 0.5, 0.5 (x,y) shift
21
Recap
  • Patterson maps are the convolution of the
    electron density of the unit cell with its
    inverted image.
  • The location of each peak in a Patterson map
    corresponds to the head of an inter-atomic vector
    with its tail at the origin.
  • There will be n2 Patterson peaks total, n peaks
    at the origin, n2-n peaks off the origin.
  • Peaks produced by atoms related by
    crystallographic symmetry operations are called
    self peaks.
  • There will be one self peak for every pairwise
    difference between symmetry operators in the
    crystal.
  • Written as equations, these differences relate
    the Patterson coordinates u,v,w to atom
    positions, x,y,z.
  • Different crystallographers may arrive at
    different, but equally valid values of x,y,z
    that are related by an arbitrary choice of origin
    or unit cell translation.

22
Polymerase b example, P21212
  • Difference Patterson map, native-Pt derivative.
  • Where do we expect to find self peaks?
  • Self peaks are produced by vectors between atoms
    related by crystallographic symmetry.
  • From international tables of crystallography, we
    find the following symmetry operators.
  • X, Y, Z
  • -X, -Y, Z
  • 1/2-X,1/2Y,-Z
  • 1/2X,1/2-Y,-Z
  • Everyone, write the equation for the location of
    the self peaks. 1-2, 1-3, and 1-4 Now!

23
Self Vectors
  • X, Y, Z
  • -X, -Y, Z
  • 1/2-X,1/2Y,-Z
  • 1/2X,1/2-Y,-Z
  • X, Y, Z
  • -X, -Y, Z
  • u2x, v2y, w0

1. X, Y, Z 3. ½-X,½Y,-Z u2x-½,v-½,w2z
1. X, Y, Z 4. ½X,½-Y,-Z u-½,v2y-½,w2z
Harker sections, w0, v1/2, u1/2
24
Isomorphous difference Patterson map (Pt
derivative)
  • X, Y, Z
  • -X, -Y, Z
  • u2x, v2y, w0

1. X, Y, Z 3. ½-X,½Y,-Z u2x-½,v-½,w2z
1. X, Y, Z 4. ½X,½-Y,-Z u-½,v2y-½,w2z
Solve for x, y using w0 Harker sect.
25
Harker section w0
W0
  • X, Y, Z
  • -X, -Y, Z
  • u2x, v2y, w0

0.1682x 0.084x
0.2662y 0.133y
Does z0? No!
Solve for x, z using v1/2 Harker sect.
26
Harker Section v1/2
1. X, Y, Z 3. ½-X,½Y,-Z u2x-½,v-½,w2z
V1/2
0.3332x-1/2 0.8332x 0.416x
0.1502z 0.075z
27
Resolving ambiguity in x,y,z
  • From w0 Harker section x10.084, y10.133
  • From v1/2 Harker section, x20.416, z20.075
  • Why doesnt x agree between solutions? Arbitrary
    origin choice can bring them into agreement.
  • What are the rules for origin shifts? Can apply
    any of the Cheshire symmetry operators to convert
    from one origin choice to another.

28
Cheshire symmetry
From w0 Harker section xorig10.084,
yorig10.133 From v1/2 Harker section,
xorig20.416, zorig20.075
  1. X, Y, Z
  2. -X, -Y, Z
  3. -X, Y, -Z
  4. X, -Y, -Z
  5. -X, -Y, -Z
  6. X, Y, -Z
  7. X, -Y, Z
  8. -X, Y, Z
  9. 1/2X, Y, Z
  10. 1/2-X, -Y, Z
  11. 1/2-X, Y, -Z
  12. 1/2X, -Y, -Z
  13. 1/2-X, -Y, -Z
  14. 1/2X, Y, -Z
  15. 1/2X, -Y, Z
  16. 1/2-X, Y, Z
  17. X,1/2Y, Z
  1. 1/2X,1/2Y, Z
  2. 1/2-x,1/2-Y, Z
  3. 1/2-X,1/2Y, -Z
  4. 1/2X,1/2-Y, -Z
  5. 1/2-X,1/2-Y, -Z
  6. 1/2X,1/2Y, -Z
  7. 1/2X,1/2-Y, Z
  8. 1/2-X,1/2Y, Z
  9. 1/2X, Y,1/2Z
  10. 1/2-X, -Y,1/2Z
  11. 1/2-X, Y,1/2-Z
  12. 1/2X, -Y,1/2-Z
  13. 1/2-X, -Y,1/2-Z
  14. 1/2X, Y,1/2-Z
  15. 1/2X, -Y,1/2Z
  16. 1/2-X, Y,1/2Z
  17. X,1/2Y,1/2Z

Apply Cheshire symmetry operator 10 To x1 and y1
Xorig10.084 ½-xorig10.5-0.084 ½-xorig10.416
xorig2 yorig10.133 -yorig1-0.133yorig2
Hence, Xorig20.416, yorig2-0.133, zorig20.075
29
Advanced case,Proteinase K in space group P43212
  • Where are Harker sections?

30
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Symmetry operator 2-Symmetry operator 4
-x - y ½z - ( ½y
½-x ¼z) -½-x-y -½x-y ¼
33
Symmetry operator 2-Symmetry operator 4
-x - y ½z - ( ½y
½-x ¼z) -½-x-y -½x-y ¼
Plug in u. u-½-x-y 0.18-½-x-y 0.68-x-
y Plug in v.
v-½x-y 0.22-½x-y
0.72x-y Add two equations and solve for y.
0.68-x-y (0.72 x-y) 1.40-2y
-0.70y Plug y into first equation and solve for
x. 0.68-x-y 0.68-x-(-0.70) 0.02x
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Symmetry operator 3 -Symmetry operator 6
½-y ½x ¾z - ( -y -x
½-z) ½ ½2x ¼2z
Plug in v. v ½2x 0.48
½2x -0.022x -0.01x Plug
in w. w ¼2z 0.24
¼2z -0.012z -0.005z
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41
From step 3 Xstep3 0.02 ystep3-0.70
zstep3?.???
From step 4 Xstep4-0.02 ystep4 ?.??
zstep4-0.005
Clearly, Xstep3 does not equal Xstep4 . Use a
Cheshire symmetry operator that transforms
xstep3 0.02 into xstep4- 0.02. For example,
lets use -x, -y, z And
apply it to all coordinates in step
3. xstep3-transformed - (0.02)
-0.02 ystep3-transformed - (- 0.70) 0.70
Now xstep3-transformed xstep4 And ystep3 has
been transformed to a reference frame consistent
with x and z from step 4. So we arrive at the
following self-consistent x,y,z Xstep4-0.02,
ystep3-transformed0.70, zstep4-0.005 Or
simply, x-0.02, y0.70, z-0.005
The x, y coordinate in step 3 describes one of
the heavy atom positions in the unit cell. The x,
z coordinate in step 4 describes a symmetry
related copy. We cant combine these coordinates
directly. They dont describe the same atom.
Perhaps they even referred to different
origins. How can we transform x, y from step 3
so it describes the same atom as x and z in step
4?
42
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45
Use x,y,z to predict the position of a non-Harker
Patterson peak
  • x,y,z vs. x,y,z ambiguity remains
  • In other words x-0.02, y0.70, z-0.005 or
  • x0.02, y0.70, z-0.005
    could be correct.
  • Both satisfy the difference vector equations for
    Harker sections
  • Only one is correct. 50/50 chance
  • Predict the position of a non Harker peak.
  • Use symop1-symop5
  • Plug in x,y,z solve for u,v,w.
  • Plug in x,y,z solve for u,v,w
  • I have a non-Harker peak at u0.28 v0.28, w0.0
  • The position of the non-Harker peak will be
    predicted by the correct heavy atom coordinate.

46
x y z -( y x -z) x-y
-xy 2z
symmetry operator 1 -symmetry operator 5 u
v w
First, plug in x-0.02, y0.70, z-0.005 ux-y
-0.02-0.70 -0.72 v-xy 0.020.70
0.72 w2z2(-0.005)-0.01 The numerical value
of these co-ordinates falls outside the section
we have drawn. Lets transform this uvw by
Patterson symmetry u,-v,-w. -0.72,0.72,-0.01
becomes -0.72,-0.72,0.01 then add 1 to u and
v 0.28, 0.28, 0.01 This corresponds to the peak
shown u0.28, v0.28, w0.01 Thus, x-0.02,
y0.70, z-0.005 is correct. Hurray! We are
finished! In the case that the above test
failed, we would change the sign of x.
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