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1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
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1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
ba
Draw the location of all 2-fold symmetry axes using the ellipse symbol.
Draw a unit cell.
How many molecules in the unit cell? asymmetric unit?
Estimate (x,y) coordinates of oxygen atom in fractions of a unit cell. Use the ruler provided.
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li H
Li
H
Li
H
Li
H
LiH
Li
H
Li
H
Li
H
LiH
Li
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Two-folds
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
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Li
H
Li H
Li
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LiH
Li
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Li
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Li
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LiH
Li
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Unit Cell
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Li
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Li H
Li
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LiH
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Li
H
Li
H
LiH
Li
Choice 1
Choice 2
Choice 3
Choice 4
4 Choices of originH
H
Li
Li H
H
Li
Li H
H
Li
Li
H
H
Li
Li H
H
Li
Li H
H
Li
Li
H
H
Li
Li H
H
Li
Li H
H
Li
Li
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H
H
Li
Li H
H
Li
Li H
H
Li
Li
H
H
Li
Li H
H
Li
Li H
H
Li
Li
H
H
Li
Li H
H
Li
Li H
H
Li
Li
H
H
Li
Li H
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Li
Li H
H
Li
Li
H
H
Li
Li H
H
Li
Li H
H
Li
Li
H
H
Li
Li H
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Li
Li H
H
Li
Li
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H
Li
Li H
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Li
Li H
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Li
Li
H
H
Li
Li H
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Li
Li H
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Li
Li
H
H
Li
Li H
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Li
Li H
H
Li
Li
Choice 1 Choice 2
Choice 3 Choice 4
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Which plane group?
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Li H
Li
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LiH
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LiH
Li
X=0.2
What are the coordinates of the oxygen using origin choice 1?
ba
H
Li
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Li
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Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li H
Li
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H
Li
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LiH
Li
H
Li
H
Li
H
LiH
Li
X=0.20Y=0.20
1 2 3 41
2
3
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
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Li
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Li H
Li
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Li
H
Li
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LiH
Li
H
Li
H
Li
H
LiH
Li
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X=0.20Y=0.20
What are the coordinates of the other oxygen in the unit cell?
X=0.80Y=0.80
Symmetry operators in plane group p2X, Y-X,-Y
1
2
3
4
1 2 3 4
ba
Always allowed to add or subtract multiples of 1.0
H
Li
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Li
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Li
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Li
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Li
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Li
H
Li H
Li
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Li
H
Li
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LiH
Li
H
Li
H
Li
H
LiH
Li
X=0.2Y=0.2
X=1.8Y=0.8
X=1.2Y=0.2
X=0.8Y=0.8
X=2.2Y=0.2
X=2.8Y=0.8
H
Li
H
Li
H
Li
H
Li
H
Li
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Li
H
Li
H
Li
H
Li H
Li
H
Li
H
Li
H
LiH
Li
H
Li
H
Li
H
LiH
Li
What would be the oxygen coordinates if we had drawn the unit cell with origin choice 2?b
a
1 2 3
X=0.70Y=0.20
1
2
3
4
Oxygen coordinates with origin choice 3?
X=0.30Y=0.30
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li
H
Li H
Li
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LiH
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LiH
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1 2 31
2
3
H
Li
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Li
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Li
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Li
H
Li
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Li
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Li
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Li
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Li H
Li
H
Li
H
Li
H
LiH
Li
H
Li
H
Li
H
LiH
Li
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Oxygen coordinates with origin choice 4?
1 2 3 4
X=0.80Y=0.30
1
2
3
Cheshire operators
• X , Y
• X+.5, Y
• X+.5, Y+.5
• X , Y+.5
X=0.20Y=0.20
X=0.70Y=0.20
X=0.30Y=0.30
X=0.80Y=0.30
Choice 1
Choice 2
Choice 3
Choice 4
The 4 choices of origin are equally valid but once a choice is made, you must remain consistent.
H
H
Li
Li H
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Li H
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Li H
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Li H
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Li H
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Li H
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Li
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H
H
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Li H
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Li
Li H
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Li
Li
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Li H
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Li H
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H
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Li H
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Li
Li H
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Li H
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Li H
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Li
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Li
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Li H
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Li H
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Li H
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Li H
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Li
Li H
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Li
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Li H
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Li
Li H
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Li
Li
H
H
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Li H
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Li
Li H
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Li
Li
Choice 1 Choice 2
Choice 3 Choice 4
X1=0.20Y1=0.20
X1=0.70Y1=0.20
X1=0.80Y1=0.30
X1=0.30Y1=0.30
X2=0.80Y2=0.80
X2=0.30Y2=0.80
H
Li
H
Li
X2=0.70Y2=0.70
X2=0.20Y2=0.70
What did we learn?
• There are multiple valid choices of origin for a unit cell.
• The values of x,y,z for the atoms will depend on the choice of origin.
• To compare atomic coordinates between structures solved with different origins, a Cheshire operator must be applied.
• Adding 1 to x, y, or z, or any combination of x, y, and z is valid. It is just a unit cell translation.
Interpreting difference Patterson Maps in Lab this week!
• Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 16 derivative data sets in lab (different heavy atoms at different concentrations)– HgCl2 Hg(Acetate)2 PCMBS– PIP– EuCl3 GdCl3 SmCl3
• How many heavy atom sites per asymmetric unit, if any?
• What are the positions of the heavy atom sites?• Let’s review how heavy atom positions can be
calculated from difference Patterson peaks.
Patterson synthesis
P(uvw)=?hkl cos2(hu+kv+lw -)
hkl
Patterson synthesis
P(uvw)=Ihkl cos2(hu+kv+lw -)
hkl
Patterson Review
A Patterson synthesis is like a Fourier synthesis except for what two variables?
Fourier synthesis
(xyz)=|Fhkl| cos2(hx+ky+lz -hkl)
hkl
Hence, Patterson density map= electron density map convoluted
with its inverted image.
Patterson synthesis
P(uvw)=Ihkl cos2(hu+kv+lw)Remembering Ihkl=Fhkl•Fhkl*
And Friedel’s law Fhkl*= F-h-k-l P(uvw)=FourierTransform(Fhkl•F-h-k-l)
P(uvw)=(uvw) (-u-v-w)
Significance?
P(uvw)=(uvw) (-u-v-w)
The Patterson map contains a peak for every interatomic vector in the unit cell.
The peaks are located at the head of the interatomic vector when its tail is placed at the origin.
a
b
Electron Density vs. Patterson Density
H H
H H
H HH H
Electron Density Mapsingle water molecule in the unit cell
Patterson Density Mapsingle water molecule convoluted with its inverted image.
a
bLay down n copies of the unit cell at the origin, where n=number of atoms in unit cell.
1
23
For copy n, atom n is placed at the origin.A Patterson peak appears under each atom.
Every Patterson peak corresponds to an inter-atomic vector
H H
Electron Density Mapsingle water molecule in the unit cell
Patterson Density Mapsingle water molecule convoluted with its inverted image.
a
b
3 sets of peaks:
Length O-HWhere?
Length H-HWhere?
Length zero Where?
How many peaks superimposed at origin?
How many non-origin peaks?
1
23H H
H HH H
Patterson maps are more complicated to interpret than electron density maps.
Imagine the complexity of a Patterson map of a protein
Electron Density Mapsingle water molecule in the unit cell
Patterson Density Mapsingle water molecule convoluted with its inverted image.
a
b
Unit cell repeats fill out rest of cell with peaksH H
H H
Patterson maps have an additional center of symmetry
Electron Density Mapsingle water molecule in the unit cell
Patterson Density Mapsingle water molecule convoluted with its inverted image.
a
b
H H H H
H H H H
plane group pm plane group p2mm
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 in space group P21212
3. Advanced 3D example, Hg derivative of proteinase K in space group P43212.
Plane group p2
H
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H
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H H
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HH
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H
b
Let’s consider only oxygen atoms
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H H
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HH
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H
b
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.
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?
(0,0) ab
Symmetry operators in plane group p2
(0,0) ab
(0.2,0.3)
(-0.2,-0.3)
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -x,-y
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?
Yes! Equivalent by unit cell translation.(-0.2+1.0, -0.3+1.0)=(0.8, 0.7)
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.
(0,0) ab
a(0,0)b
(0.2,0.3)
(-0.2,-0.3)
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -x,-y
PATTERSON MAP2D CRYSTAL
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?
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
Patterson in plane group p2
(0,0) ab
a(0,0)b
(0.2,0.3)
(-0.2,-0.3)
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -x,-y
PATTERSON MAP2D CRYSTAL
(0.4, 0.6)
(-0.4, -0.6)
(0.6, 0.4)
Patterson in plane group p2
a(0,0)b
PATTERSON MAP
(0.4, 0.6)
If you collected data on this crystaland calculated a Patterson mapit would look like this.
(0.6, 0.4)
Now I’m stuck in Patterson space. How do I get back to x,y coordinates?
a(0,0)b
PATTERSON MAP
(0.4, 0.6)
Remember the Patterson Peak positions (U,V) correspond to vectors between symmetry related smiley faces in the unit cell. That is, differences between our friends the space group operators.
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -x,-y
x , y -(-x, –y) 2x , 2y u=2x, v=2y
symop #1symop #2
(0.6, 0.4)
plug in Patterson valuesfor u and v to get x and y.
Now I’m stuck in Patterson space. How do I get back to x,y, coordinates?
a(0,0)b
PATTERSON MAP
(0.4, 0.6)
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -x,-y
x y-(-x –y)2x 2y
symop #1symop #2
set u=2x v=2y plug in Patterson valuesfor u and v to get x and y.
u=2x0.4=2x0.2=x
v=2y0.6=2y0.3=y
Hurray!!!!
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -x,-y
x y-(-x –y)2x 2y
symop #1symop #2
set u=2x v=2y plug in Patterson valuesfor u and v to get x and y.
u=2x0.4=2x0.2=x
v=2y0.6=2y0.3=y
HURRAY! we got back the coordinatesof our smiley faces!!!!
(0,0) ab
(0.2,0.3)
2D CRYSTAL
Devil’s 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)b
PATTERSON MAP
(0.4, 0.6)
(0.6, 0.4)
using Patterson (u,v) values 0.4, 0.6to get x and y.
u=2x0.4=2x0.2=x
v=2y0.6=2y0.3=y
u=2x0.6=2x0.3=x
v=2y0.4=2y0.2=y
using Patterson (u,v) values 0.6, 0.4to get x and y.
These two solutions do not give the same x,y? What is going on??????
Arbitrary choice of origin
• Original origin choice• Coordinates x=0.2, y=0.3.
(0,0) ab
(0.2,0.3)
(-0.2,-0.3)
(0.8,0.7) (0.3,0.2)
(-0.3,-0.2)
(0.7,0.8)
(0,0) ab
• New origin choice• Coordinates x=0.3, y=0.2.
Related by 0.5, 0.5 (x,y) shift
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.
Polymerase 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.1. X, Y, Z2. -X, -Y, Z3. 1/2-X,1/2+Y,-Z4. 1/2+X,1/2-Y,-Z• Everyone, write the equation for the location of the self
peaks. 1-2, 1-3, and 1-4 Now!
Self Vectors
1. X, Y, Z2. -X, -Y, Z3. 1/2-X,1/2+Y,-Z4. 1/2+X,1/2-Y,-Z
1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0
1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z
1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z
Harker sections, w=0, v=1/2, u=1/2
Isomorphous difference Patterson map (Pt derivative)
W=0
V=1/2
U=1/2
1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0
1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z
1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z
Solve for x, y using w=0 Harker sect.
Harker section w=0
W=0 1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0
0.168=2x0.084=x
0.266=2y0.133=y
Does z=0? No!
Solve for x, z using v=1/2 Harker sect.
Harker Section v=1/2
V=1/2
1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z
0.333=2x-1/20.833=2x0.416=x
0.150=2z0.075=z
Resolving ambiguity in x,y,z
• From w=0 Harker section x1=0.084, y1=0.133• From v=1/2 Harker section, x2=0.416, z2=0.075• Why doesn’t x agree between solutions? They
differ by an origin shift. Choose the proper shift to bring them into agreement.
• What are the rules for origin shifts? Cheshire symmetry operators relate the different choices of origin. You can apply any of the Cheshire symmetry operators to convert from one origin choice to another.
Cheshire symmetry
1. X, Y, Z2. -X, -Y, Z3. -X, Y, -Z4. X, -Y, -Z5. -X, -Y, -Z6. X, Y, -Z7. X, -Y, Z8. -X, Y, Z
9. 1/2+X, Y, Z10. 1/2-X, -Y, Z11. 1/2-X, Y, -Z12. 1/2+X, -Y, -Z13. 1/2-X, -Y, -Z14. 1/2+X, Y, -Z15. 1/2+X, -Y, Z16. 1/2-X, Y, Z
17. X,1/2+Y, Z18. -X,1/2-Y, Z19. -X,1/2+Y, -Z20. X,1/2-Y, -Z21. -X,1/2-Y, -Z22. X,1/2+Y, -Z23. X,1/2-Y, Z24. -X,1/2+Y, Z
25. X, Y,1/2+Z26. -X, -Y,1/2+Z27. -X, Y,1/2-Z28. X, -Y,1/2-Z29. -X, -Y,1/2-Z30. X, Y,1/2-Z31. X, -Y,1/2+Z32. -X, Y,1/2+Z
33. 1/2+X,1/2+Y, Z34. 1/2-x,1/2-Y, Z 35. 1/2-X,1/2+Y, -Z36. 1/2+X,1/2-Y, -Z37. 1/2-X,1/2-Y, -Z38. 1/2+X,1/2+Y, -Z39. 1/2+X,1/2-Y, Z40. 1/2-X,1/2+Y, Z
41. 1/2+X, Y,1/2+Z42. 1/2-X, -Y,1/2+Z43. 1/2-X, Y,1/2-Z44. 1/2+X, -Y,1/2-Z45. 1/2-X, -Y,1/2-Z46. 1/2+X, Y,1/2-Z47. 1/2+X, -Y,1/2+Z48. 1/2-X, Y,1/2+Z
49. X,1/2+Y,1/2+Z50. -X,1/2-Y,1/2+Z51. -X,1/2+Y,1/2-Z52. X,1/2-Y,1/2-Z53. -X,1/2-Y,1/2-Z54. X,1/2+Y,1/2-Z55. X,1/2-Y,1/2+Z56. -X,1/2+Y,1/2+Z
57. 1/2+X,1/2+Y,1/2+Z58. 1/2-X,1/2-Y,1/2+Z59. 1/2-X,1/2+Y,1/2-Z60. 1/2+X,1/2-Y,1/2-Z61. 1/2-X,1/2-Y,1/2-Z62. 1/2+X,1/2+Y,1/2-Z63. 1/2+X,1/2-Y,1/2+Z64. 1/2-X,1/2+Y,1/2+Z
From w=0 Harker section xorig1=0.084, yorig1=0.133From v=1/2 Harker section, xorig2=0.416, zorig2=0.075
Apply Cheshire symmetry operator #10To x1 and y1 Xorig1=0.084½-xorig1=0.5-0.084½-xorig1=0.416 =xorig2
yorig1=0.133-yorig1=-0.133=yorig2
Hence,Xorig2=0.416, yorig2=-0.133, zorig2=0.075
Advanced case,Proteinase K in space group P43212
• Where are Harker sections?
Symmetry operator 2-Symmetry operator 4
-x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Symmetry operator 2-Symmetry operator 4
-x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Plug in u. u=-½-x-y0.18=-½-x-y0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y
Add two equations and solve for y.
0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y
Plug y into first equation and solve for x.
0.68=-x-y 0.68=-x-(-0.70) 0.02=x
Symmetry operator 2-Symmetry operator 4
-x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Plug in u. u=-½-x-y0.18=-½-x-y0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y
Add two equations and solve for y.
0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y
Plug y into first equation and solve for x.
0.68=-x-y 0.68=-x-(-0.70) 0.02=x
Symmetry operator 2-Symmetry operator 4
-x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Plug in u. u=-½-x-y0.18=-½-x-y0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y
Add two equations and solve for y.
0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y
Plug y into first equation and solve for x.
0.68=-x-y 0.68=-x-(-0.70) 0.02=x
Symmetry operator 2-Symmetry operator 4
-x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Plug in u. u=-½-x-y0.18=-½-x-y0.68=-x-y Plug in v. v=-½+x-y 0.22=-½+x-y 0.72=x-y
Add two equations and solve for y.
0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y
Plug y into first equation and solve for x.
0.68=-x-y 0.68=-x-(-0.70) 0.02=x
Symmetry operator 3-Symmetry operator 6
½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z
Plug in v. v= ½+2x 0.48= ½+2x-0.02=2x-0.01=x Plug in w. w= ¼+2z 0.24= ¼+2z -0.01=2z -0.005=z
Symmetry operator 3-Symmetry operator 6
½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z
Plug in v. v= ½+2x 0.46= ½+2x-0.04=2x-0.02=x Plug in w. w= ¼+2z 0.24= ¼+2z -0.01=2z -0.005=z
Symmetry operator 3-Symmetry operator 6
½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z
Plug in v. v= ½+2x 0.46= ½+2x-0.04=2x-0.02=x Plug in w. w= ¼+2z 0.24= ¼+2z -0.01=2z -0.005=z
From step 3Xstep3= 0.02 ystep3=-0.70 zstep3=?.???From step 4Xstep4=-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, let’s use: -x, -y, zAnd apply it to all coordinates in step 3.xstep3-transformed = - (+0.02) = -0.02ystep3-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-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, 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 can’t combine these coordinates directly. They don’t describe the same atom. Perhaps they evenreferred to different origins.
How can we transform x, y from step 3 so it describesthe same atom as x and z in step 4?
From step 3Xstep3= 0.02 ystep3=-0.70 zstep3=?.???From step 4Xstep4=-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, let’s use: -x, -y, zAnd apply it to all coordinates in step 3.xstep3-transformed = - (+0.02) = -0.02ystep3-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-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
Cheshire Symmetry Operators for space group P43212
X, Y, Z -X, -Y, Z -Y, X, 1/4+Z Y, -X, 1/4+Z Y, X, -Z -Y, -X, -Z X, -Y, 1/4-Z -X, Y, 1/4-Z
1/2+X, 1/2+Y, Z 1/2-X, 1/2-Y, Z 1/2-Y, 1/2+X, 1/4+Z 1/2+Y, 1/2-X, 1/4+Z 1/2+Y, 1/2+X, -Z 1/2-Y, 1/2-X, -Z 1/2+X, 1/2-Y, 1/4-Z 1/2-X, 1/2+Y, 1/4-Z
X, Y, 1/2+Z -X, -Y, 1/2+Z -Y, X, 3/4+Z Y, -X, 3/4+Z Y, X, 1/2-Z -Y, -X, 1/2-Z X, -Y, 3/4-Z -X, Y, 3/4-Z
1/2+X, 1/2+Y, 1/2+Z 1/2-X, 1/2-Y, 1/2+Z 1/2-Y, 1/2+X, 3/4+Z 1/2+Y, 1/2-X, 3/4+Z 1/2+Y, 1/2+X, 1/2-Z 1/2-Y, 1/2-X, 1/2-Z 1/2+X, 1/2-Y, 3/4-Z 1/2-X, 1/2+Y, 3/4-Z
From step 3Xstep3= 0.02 ystep3=-0.70 zstep3=?.???From step 4Xstep4=-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, let’s use: -x, -y, zAnd apply it to all coordinates in step 3.xstep3-transformed = - (+0.02) = -0.02ystep3-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-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
Cheshire Symmetry Operators for space group P43212
X, Y, Z -X, -Y, Z -Y, X, 1/4+Z Y, -X, 1/4+Z Y, X, -Z -Y, -X, -Z X, -Y, 1/4-Z -X, Y, 1/4-Z
1/2+X, 1/2+Y, Z 1/2-X, 1/2-Y, Z 1/2-Y, 1/2+X, 1/4+Z 1/2+Y, 1/2-X, 1/4+Z 1/2+Y, 1/2+X, -Z 1/2-Y, 1/2-X, -Z 1/2+X, 1/2-Y, 1/4-Z 1/2-X, 1/2+Y, 1/4-Z
X, Y, 1/2+Z -X, -Y, 1/2+Z -Y, X, 3/4+Z Y, -X, 3/4+Z Y, X, 1/2-Z -Y, -X, 1/2-Z X, -Y, 3/4-Z -X, Y, 3/4-Z
1/2+X, 1/2+Y, 1/2+Z 1/2-X, 1/2-Y, 1/2+Z 1/2-Y, 1/2+X, 3/4+Z 1/2+Y, 1/2-X, 3/4+Z 1/2+Y, 1/2+X, 1/2-Z 1/2-Y, 1/2-X, 1/2-Z 1/2+X, 1/2-Y, 3/4-Z 1/2-X, 1/2+Y, 3/4-Z
From step 3Xstep3= 0.02 ystep3=-0.70 zstep3=?.???From step 4Xstep4=-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, let’s use: -x, -y, zAnd apply it to all coordinates in step 3.xstep3-transformed = - (+0.02) = -0.02ystep3-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-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
Cheshire Symmetry Operators for space group P43212
X, Y, Z -X, -Y, Z -Y, X, 1/4+Z Y, -X, 1/4+Z Y, X, -Z -Y, -X, -Z X, -Y, 1/4-Z -X, Y, 1/4-Z
1/2+X, 1/2+Y, Z 1/2-X, 1/2-Y, Z 1/2-Y, 1/2+X, 1/4+Z 1/2+Y, 1/2-X, 1/4+Z 1/2+Y, 1/2+X, -Z 1/2-Y, 1/2-X, -Z 1/2+X, 1/2-Y, 1/4-Z 1/2-X, 1/2+Y, 1/4-Z
X, Y, 1/2+Z -X, -Y, 1/2+Z -Y, X, 3/4+Z Y, -X, 3/4+Z Y, X, 1/2-Z -Y, -X, 1/2-Z X, -Y, 3/4-Z -X, Y, 3/4-Z
1/2+X, 1/2+Y, 1/2+Z 1/2-X, 1/2-Y, 1/2+Z 1/2-Y, 1/2+X, 3/4+Z 1/2+Y, 1/2-X, 3/4+Z 1/2+Y, 1/2+X, 1/2-Z 1/2-Y, 1/2-X, 1/2-Z 1/2+X, 1/2-Y, 3/4-Z 1/2-X, 1/2+Y, 3/4-Z
Use x,y,z to predict the position of a non-Harker Patterson peak
• x,y,z vs. –x,y,z ambiguity remainsIn other words x=-0.02, y=0.70, z=-0.005 or x=+0.02, y=0.70, 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 u=0.28 v=0.28, w=0.0• The position of the non-Harker peak will be predicted by the correct
heavy atom coordinate.
x y z -( y x -z) x-y -x+y 2z
symmetry operator 1-symmetry operator 5u v w
First, plug in x=-0.02, y=0.70, z=-0.005
u=x-y = -0.02-0.70 =-0.72v=-x+y= +0.02+0.70= 0.72w=2z=2*(-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 u=0.28, v=0.28, w=0.01Thus, x=-0.02, y=0.70, z=-0.005 is correct. Hurray! We are finished!
In the case that the above test failed, we would change the sign of x.
(1) U, V, W (2)-U,-V, W (3) U, V,-W (4)-U,-V,-W (5)-U, V, W (6) U,-V, W (7)-U, V,-W (8) U,-V,-W (9)-V, U, W (10) V,-U, W (11)-V, U,-W (12) V,-U,-W(13) V, U, W (14)-V,-U, W (15) V, U,-W (16)-V,-U,-W
Symmetry Operators are the Bridge between Atomic Coordinates and Patterson Peaks
u(0,0)v
PATTERSON MAP
(0.4, 0.6)
x , y -(-x, –y) 2x , 2y u=2x, v=2y
symop #1symop #2
(0.6, 0.4)
(0,0) xy
(0.2,0.3)
(-0.2,-0.3)
SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y 2) -x,-y
Assignment
• Calculate an isomorphous difference Patterson map in lab.
• Solve the positions of the heavy atom (x,y,z) from the peaks in the map (u,v,w).– follow the procedures in the handout– write neatly– check your answer
• Next week hand in your calculation. • We will test the accuracy of your solution and
use it to calculate phases and electron density.
Calculate Y and Z
Calculate X and Y
X,Y,Z referred to a common origin.
If prediction lies outside Patterson asymmetric unit (0→0.5, 0→0.5,0→ 0.5) usePatterson symmetry operators to find the symmetry equivalent peak in theasymmetric unit. If the predicted peak is absent, then negate x value and re-calculate u,v,w. Predicted peak should be present if algebra is correct.
Crystal space
U=0.5
W=0.25 Cheshire operator applied to Y and Z if two values of Y do not match
P43212 Symmetry operator difference 1-5
x y z
-( y
x -z)
x-y -x+y 2z
u,v,w
Check answer for peak
off Harker section.
P43212 Symmetry operator difference 3-6
P43212 Symmetry operator difference 2-4
Patterson space
m230d_2010_scaled.mtzprok-eu-yanran.scaprok-gd-1-10-david.scaprok-gd-1-1-keith.scaprok-gd-1-1-shewit.scaprok-gd-1-4-wei.scaprok-hgac-1-4-ralph.scaprok-hgcl2-1-1-huimin.scaprok-hgcl2-travis.scaprok-native-amanda.scaprok-native-natalie.scaprok-native-saken.scaprok-pcmbs-1-1-alex.scaprok-pcmbs-1-1-toby.scaprok-pcmbs-alexsin.scaprok-pip-1-1-annie.scaprok-pip-1-1-lauren.scaprok-pip-1-4-christine.scaprok-pmsf-angelica.scaprok-pmsf-camellia.scaprok-pmsf-catherine.scaprok-pmsf-yuewei.scaprok-sm-1-1-bill.scaprok-sm-1-4-xinghong.sca
Intensity measurements (IHKL)
All data sets were appended into a spreadsheet. Each column contains IHKL of a different data set. Each row specifies a different HKL.-using the CCP4 program CAD.
Intensity measurements were converted to structure factor amplitudes (|FHKL|) -using the CCP4 program TRUNCATE.
All data sets were scaled to areference native data set with the best statistics: prok-native-natalie-using the CCP4 program SCALEIT.
Scale intensities by a constant (k) and resolution dependent exponential (B)
H K L intensity sigma1 0 10 106894.0 1698.01 0 11 41331.5 702.31 0 12 76203.2 1339.01 0 13 28113.5 513.61 0 14 6418.2 238.71 0 15 45946.4 882.7 1 0 16 26543.8 555.6
prok-native-natalie
106894.0 / 40258.7 = 2.65 41331.5 / 25033.2 = 1.65 76203.2 / 24803.6 = 3.07 28113.5 / 11486.3 = 2.45 6418.2 / 9180.5 = 0.70 45946.4 / 25038.8 = 1.83 26543.8 / 21334.6 = 1.24
prok-native-saken
H K L intensity sigma1 0 10 40258.7 1222.91 0 11 25033.2 799.81 0 12 24803.6 771.51 0 13 11486.3 423.91 0 14 9180.5 353.61 0 15 25038.8 783.01 0 16 21334.6 686.4
comparison-Probably Natalie used a larger crystal than Saken. -Multiply Saken’s data by k and B to put the data on the same scale.
OVERALL FILE STATISTICS for resolution range 0.000 - 0.399
=======================
Col Sort Min Max Num % Mean Mean Resolution Type Column
num order Missing complete abs. Low High label
1 ASC 0 42 0 100.00 22.2 22.2 56.53 1.60 H H
2 NONE 0 30 0 100.00 9.1 9.1 56.53 1.60 H K
3 NONE 0 63 0 100.00 23.8 23.8 56.53 1.60 H L
4 NONE 0.0 19.0 0 100.00 9.48 9.48 56.53 1.60 I FreeR_flag
5 NONE 6.2 2336.4 1623 94.96 232.23 232.23 40.76 1.60 F FP_native-natalie
6 NONE 0.8 34.0 1623 94.96 3.59 3.59 40.76 1.60 Q SIGFP_native-natalie
7 NONE 6.5 1420.2 5392 83.26 248.69 248.69 33.97 1.70 F FP_native-amanda
8 NONE 0.6 56.2 5392 83.26 3.51 3.51 33.97 1.70 Q SIGFP_native-amanda
9 NONE 1.6 2486.3 9128 71.66 278.06 278.06 48.05 1.79 F FP_native-saken
10 NONE 1.0 56.3 9128 71.66 5.84 5.84 48.05 1.79 Q SIGFP_native-saken
11 NONE 2.0 2428.1 5239 83.74 258.80 258.80 56.53 1.69 F FP_Eu-yanran
12 NONE 1.5 117.4 5239 83.74 11.53 11.53 56.53 1.69 Q SIGFP_Eu-yanran
13 NONE -109.8 175.0 5280 83.61 -2.52 15.74 56.53 1.69 D D_Eu-yanran
14 NONE 0.0 80.8 5280 83.61 18.37 18.37 56.53 1.69 Q SIGD_Eu-yanran
15 NONE 4.1 2334.3 5370 83.33 254.01 254.01 56.53 1.70 F FP_Gd-david
16 NONE 0.7 37.2 5370 83.33 3.57 3.57 56.53 1.70 Q SIGFP_Gd-david
17 NONE -52.0 47.8 5415 83.19 -0.99 8.20 56.53 1.70 D D_Gd-david
18 NONE 0.0 28.4 5415 83.19 5.75 5.75 56.53 1.70 Q SIGD_Gd-david
19 NONE 13.7 2166.6 14898 53.75 307.13 307.13 48.05 1.91 F FP_Gd-keith
20 NONE 2.3 178.1 14898 53.75 23.20 23.20 48.05 1.91 Q SIGFP_Gd-keith
21 NONE -234.6 266.1 15375 52.27 -0.08 22.57 48.05 1.91 D D_Gd-keith
22 NONE 0.0 287.9 15375 52.27 35.67 35.67 48.05 1.91 Q SIGD_Gd-keith
23 NONE 7.0 2361.1 5566 82.72 253.74 253.74 56.53 1.70 F FP_Gd-shewit
24 NONE 1.6 161.6 5566 82.72 11.67 11.67 56.53 1.70 Q SIGFP_Gd-shewit
25 NONE -128.6 103.2 5651 82.46 -1.60 17.22 56.53 1.70 D D_Gd-shewit
26 NONE 0.0 102.4 5651 82.46 18.70 18.70 56.53 1.70 Q SIGD_Gd-shewit
27 NONE 3.7 2404.7 480 98.51 233.25 233.25 56.53 1.60 F FP_Gd-wei
28 NONE 0.8 19.5 480 98.51 3.33 3.33 56.53 1.60 Q SIGFP_Gd-wei
29 NONE -77.1 64.6 601 98.13 0.31 12.27 56.53 1.60 D D_Gd-wei
30 NONE 0.0 32.2 601 98.13 5.45 5.45 56.53 1.60 Q SIGD_Gd-wei
31 NONE 7.0 2175.4 5797 82.00 256.78 256.78 56.53 1.71 F FP_HgAc-ralph
32 NONE 0.7 76.2 5797 82.00 6.24 6.24 56.53 1.71 Q SIGFP_HgAc-ralph
33 NONE -66.7 57.1 5912 81.65 0.35 6.39 56.53 1.71 D D_HgAc-ralph
34 NONE 0.0 55.2 5912 81.65 10.02 10.02 56.53 1.71 Q SIGD_HgAc-ralph
35 NONE 4.1 2025.3 5506 82.91 263.03 263.03 56.53 1.70 F FP_HgCl2-huimin
36 NONE 1.4 96.4 5506 82.91 7.62 7.62 56.53 1.70 Q SIGFP_HgCl2-huimin
37 NONE -96.9 125.9 6249 80.60 1.60 9.71 56.53 1.70 D D_HgCl2-huimin
38 NONE 0.0 98.4 6249 80.60 12.11 12.11 56.53 1.70 Q SIGD_HgCl2-huimin
39 NONE 9.1 2020.0 8928 72.28 254.72 254.72 40.76 1.69 F FP_HgCl2-travis
40 NONE 1.8 147.6 8928 72.28 11.15 11.15 40.76 1.69 Q SIGFP_HgCl2-travis
41 NONE -140.7 137.7 10384 67.76 -3.31 13.56 40.76 1.69 D D_HgCl2-travis
42 NONE 0.0 166.3 10384 67.76 17.25 17.25 40.76 1.69 Q SIGD_HgCl2-travis
43 NONE 6.1 2276.3 5412 83.20 254.27 254.27 56.53 1.70 F FP_PCMBS-alexj
44 NONE 0.8 49.5 5412 83.20 3.78 3.78 56.53 1.70 Q SIGFP_PCMBS-alexj
45 NONE -75.3 72.9 5413 83.20 0.23 7.72 56.53 1.70 D D_PCMBS-alexj
46 NONE 0.0 33.8 5413 83.20 5.98 5.98 56.53 1.70 Q SIGD_PCMBS-alexj
47 NONE 0.7 2275.5 5067 84.27 251.80 251.80 56.53 1.69 F FP_PCMBS-alexs
48 NONE 0.5 50.7 5067 84.27 3.90 3.90 56.53 1.69 Q SIGFP_PCMBS-alexs
49 NONE -71.6 78.1 5319 83.49 0.60 9.27 56.53 1.69 D D_PCMBS-alexs
50 NONE 0.0 34.1 5319 83.49 6.16 6.16 56.53 1.69 Q SIGD_PCMBS-alexs
51 NONE 2.2 2246.6 337 98.95 233.26 233.26 48.05 1.60 F FP_PCMBS-toby
52 NONE 1.1 31.4 337 98.95 3.04 3.04 48.05 1.60 Q SIGFP_PCMBS-toby
53 NONE -63.7 68.4 458 98.58 -0.68 7.52 48.05 1.60 D D_PCMBS-toby
54 NONE 0.0 27.6 458 98.58 4.86 4.86 48.05 1.60 Q SIGD_PCMBS-toby
55 NONE 6.2 2265.1 5371 83.33 247.65 247.65 48.05 1.70 F FP_PIP-annie
AND SO ON…………….
No. of reflections used in FILE STATISTICS 32212
LIST OF REFLECTIONS
===================
0 0 4 2.00 1816.34 26.07 ? ? 1882.86
56.33 1194.04 117.40 0.00 0.00 1499.12
26.90 0.00 0.00 ? ? ?
? 1188.25 43.78 0.00 0.00 1544.44
19.48 0.00 0.00 1445.92 74.58 0.00
0.00 1601.80 51.68 0.00 0.00 1733.87
88.65 0.00 0.00 1782.74 49.52 0.00
0.00 ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? 1549.52 63.20 0.00
0.00 ? ? ? ? ?
? 1954.73 56.31 ? ? ? ?
