Transcript Document 7731751

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a
b
Draw location of 2-fold
symmetry axis.
Draw unit cell.
Estimate (x,y)
coordinates of oxygen
atom in fractions of a
unit cell.
a
b
Two-folds
a
b
Which plane group?
Choice of origin
a
a
b
b
a
a
b
b
a
Origin choice 1
b
Draw location of 2-fold
symmetry axis.
Draw unit cell.
Estimate (x,y)
coordinates of oxygen
atom in fractions of a
unit cell.
What are the coordinates of the oxygen in the
asymmetric unit?
a
b
+
1
2
3
4
X=0.2
What are the coordinates of the oxygen in the
asymmetric unit?
a
b
+
X=0.20
Y=0.20
What are the coordinates of the oxygen in the
asymmetric unit?
a
b
X=0.20
Y=0.20
+
1
2
3
4
X=0.80
Y=0.80
Symmetry operators
in plane group p2
X, Y
-X,-Y
Always allowed to add or subtract multiples of 1.0
a
b
X=0.2
Y=0.2
X=1.2
Y=0.2
X=2.2
Y=0.2
+
X=0.8
Y=0.8
X=1.8
Y=0.8
X=2.8
Y=0.8
How many unique 2-fold axes in
a
unit cell?
b
New coordinates for oxygen?
a
b
1
2
3
X=0.70
Y=0.20
New coordinates for oxygen?
a
b
1
2
3
4
X=0.80
Y=0.30
New coordinates for oxygen?
a
b
1
2
3
X=0.30
Y=0.30
Oxygen coordinates related by
alternate
choices
of
origin
a
b
X=0.20
Y=0.20
X=0.70
Y=0.20
X=0.80
Y=0.30
X=0.30
Y=0.30
Cheshire operators
•
•
•
•
X ,Y
X+.5, Y
X+.5, Y+.5
X , Y+.5
X=0.20
Y=0.20
X=0.70
Y=0.20
X=0.80
Y=0.30
X=0.30
Y=0.30
Interpreting difference Patterson
Maps in Lab this week!
• Calculate an isomorphous difference Patterson Map
(native-heavy atom) for each derivative data set. We
collected 5 derivative data sets in lab (different heavy atoms
at different concentrations)
–
–
–
–
–
HgCl2
PCMBS
Hg(Acetate)2
EuCl3
GdCl3
• 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 Review
A Patterson synthesis is like a Fourier synthesis
except for what two variables?
Fourier synthesis
r(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl)
hkl
Patterson synthesis
P(uvw)=S I?hkl
-?)
hkl cos2p(hu+kv+lw -0)
hkl
hkl
Hence, Patterson density map=
electron density map convoluted
with its inverted image.
Patterson synthesis
P(uvw)=S Ihkl cos2p(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)=r(uvw) r (-u-v-w)
Electron Density vs. Patterson
Density
a
a
b
1
3H
H2
Electron Density Map
single water molecule in
the unit cell
Lay down n
copies of the
unit cell at
the origin,
where
n=number of
atoms in unit
cell.
For copy n,
atom n is
placed at
the origin.
A Patterson peak
appears under
each atom.
b
H
H
H
H
H
Patterson Density Map
single water molecule
convoluted with its
inverted image.
Every Patterson peak corresponds
to an inter-atomic vector
3 sets of peaks:
a
b
1
3H
H2
Electron Density Map
single water molecule in
the unit cell
Length O-H
Where?
Length H-H
Where?
Length zero
Where?
How many
peaks
superimposed
at origin?
How many nonorigin peaks?
H
H
H
H
H
Patterson Density Map
single water molecule
convoluted with its
inverted image.
Patterson maps are more complicated
than electron density maps.
Imagine the complexity of a Patterson map of a protein
a
b
H
H
Electron Density Map
single water molecule in
the unit cell
Unit cell repeats fill
out rest of cell with
peaks
H
H
Patterson Density Map
single water molecule
convoluted with its
inverted image.
Patterson maps have an additional
center of symmetry
a
b
H
H
H
H
H
H
H
H
Electron Density Map
single water molecule in
the unit cell
plane group pm
Patterson Density Map
single water molecule
convoluted with its
inverted image.
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 b in space group P21212
3. Advanced 3D example, Hg derivative of
proteinase K in space group P43212.
What Plane group is this?
b
Let’s consider only oxygen atoms
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?
(0,0) a
b
•In unit cell?
•In asymmetric
unit?
•How many peaks
will be in the
Patterson map?
•How many peaks
at the origin?
•How many nonorigin peaks?
Symmetry operators in
plane group p2
(-0.2,-0.3)
(0,0) a
b
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.
(0.2,0.3)
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).
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
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 p2
Lay down two copies of unit cell at origin, first copy with smile 1 at origin, second copy with simile 2 at origin.
2D CRYSTAL
(-0.2,-0.3)
PATTERSON MAP
(0,0) a
b
(0,0) a
b
(0.2,0.3)
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
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
(-0.4, -0.6)
(-0.2,-0.3)
(0,0) a
b
(0,0) a
b
(0.2,0.3)
(0.6, 0.4)
(0.4, 0.6)
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
2D CRYSTAL
PATTERSON MAP
Patterson in plane group p2
(0,0) a
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
Now I’m 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.
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
symop #1
symop #2
x, y
-(-x, –y)
2x , 2y
u=2x, v=2y
(0,0) a
b
(0.6, 0.4)
(0.4, 0.6)
PATTERSON MAP
plug in Patterson values
for u and v to get x and y.
Now I’m stuck in Patterson space. How
do I get back to x,y, coordinates?
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
symop #1
symop #2
x y
-(-x –y)
2x 2y
(0,0) a
b
(0.4, 0.6)
set u=2x v=2y plug in Patterson values
for u and v to get x and y.
u=2x
0.4=2x
0.2=x
v=2y
0.6=2y
0.3=y
PATTERSON MAP
Hurray!!!!
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
symop #1
symop #2
x y
-(-x –y)
2x 2y
(0,0) a
b
(0.2,0.3)
set u=2x v=2y plug in Patterson values
for u and v to get x and y.
u=2x
0.4=2x
0.2=x
v=2y
0.6=2y
0.3=y
HURRAY! we got back the coordinates
of our smiley faces!!!!
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)?
using Patterson (u,v) values 0.4, 0.6
to get x and y.
v=2y
u=2x
0.6=2y
0.4=2x
0.3=y
0.2=x
(0,0) a
b
(0.6, 0.4)
(0.4, 0.6)
using Patterson (u,v) values 0.6, 0.4
to get x and y.
u=2x
0.6=2x
0.3=x
v=2y
0.4=2y
0.2=y
PATTERSON MAP
These two solutions do not give the same x,y? What is going on??????
Arbitrary choice of origin
(-0.2,-0.3)
(0,0) a
b
(0.2,0.3)
(-0.3,-0.2)
(0,0) a
b
(0.8,0.7)
(0.3,0.2)
(0.7,0.8)
• Original origin choice
• Coordinates x=0.2, y=0.3.
• 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 b example, P21212
•
•
•
•
1.
2.
3.
4.
•
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,1/2+Y,-Z
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.
2.
3.
4.
X,
Y, Z
-X,
-Y, Z
1/2-X,1/2+Y,-Z
1/2+X,1/2-Y,-Z
1. X,
Y, Z
2.-X, -Y, Z
u=2x, v=2y, w=0
1.
X, Y, Z
3. ½-X,½+Y,-Z
u=2x-½,v=-½,w=2z
Harker sections, w=0, v=1/2, u=1/2
1.
X, Y, Z
4. ½+X,½-Y,-Z
u=-½,v=2y-½,w=2z
Isomorphous difference Patterson
map (Pt derivative)
W=0
V=1/2
U=1/2
1. X,
Y, Z
2.-X, -Y, Z
u=2x, v=2y, w=0
1.
X, Y, Z
3. ½-X,½+Y,-Z
u=2x-½,v=-½,w=2z
1.
X, Y, Z
4. ½+X,½-Y,-Z
u=-½,v=2y-½,w=2z
Solve for x, y using w=0 Harker sect.
Harker section w=0
W=0
1. X,
Y, Z
2.-X, -Y, Z
u=2x, v=2y, w=0
0.168=2x
0.084=x
0.266=2y
0.133=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, Z
3. ½-X,½+Y,-Z
u=2x-½,v=-½,w=2z
0.333=2x-1/2
0.833=2x
0.416=x
0.150=2z
0.075=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
From w=0 Harker section xorig1=0.084, yorig1=0.133
From v=1/2 Harker section, xorig2=0.416, zorig2=0.075
1.
2.
3.
4.
5.
6.
7.
8.
X,
-X,
-X,
X,
-X,
X,
X,
-X,
9.
10.
11.
12.
13.
14.
15.
16.
1/2+X,
1/2-X,
1/2-X,
1/2+X,
1/2-X,
1/2+X,
1/2+X,
1/2-X,
Y,
-Y,
Y,
-Y,
-Y,
Y,
-Y,
Y,
Z
Z
-Z
-Z
-Z
-Z
Z
Z
33.
34.
35.
36.
37.
38.
39.
40.
1/2+X,1/2+Y,
1/2-x,1/2-Y,
1/2-X,1/2+Y,
1/2+X,1/2-Y,
1/2-X,1/2-Y,
1/2+X,1/2+Y,
1/2+X,1/2-Y,
1/2-X,1/2+Y,
1/2+X,
1/2-X,
1/2-X,
1/2+X,
1/2-X,
1/2+X,
1/2+X,
1/2-X,
Z
Z
-Z
-Z
-Z
-Z
Z
Z
Y,
-Y,
Y,
-Y,
-Y,
Y,
-Y,
Y,
Z
Z
-Z
-Z
-Z
-Z
Z
Z
41.
42.
43.
44.
45.
46.
47.
48.
Y,1/2+Z
-Y,1/2+Z
Y,1/2-Z
-Y,1/2-Z
-Y,1/2-Z
Y,1/2-Z
-Y,1/2+Z
Y,1/2+Z
17.
18.
19.
20.
21.
22.
23.
24.
X,1/2+Y,
-X,1/2-Y,
-X,1/2+Y,
X,1/2-Y,
-X,1/2-Y,
X,1/2+Y,
X,1/2-Y,
-X,1/2+Y,
Z
Z
-Z
-Z
-Z
-Z
Z
Z
49.
50.
51.
52.
53.
54.
55.
56.
X,1/2+Y,1/2+Z
-X,1/2-Y,1/2+Z
-X,1/2+Y,1/2-Z
X,1/2-Y,1/2-Z
-X,1/2-Y,1/2-Z
X,1/2+Y,1/2-Z
X,1/2-Y,1/2+Z
-X,1/2+Y,1/2+Z
25.
26.
27.
28.
29.
30.
31.
32.
X,
-X,
-X,
X,
-X,
X,
X,
-X,
Y,1/2+Z
-Y,1/2+Z
Y,1/2-Z
-Y,1/2-Z
-Y,1/2-Z
Y,1/2-Z
-Y,1/2+Z
Y,1/2+Z
57.
58.
59.
60.
61.
62.
63.
64.
1/2+X,1/2+Y,1/2+Z
1/2-X,1/2-Y,1/2+Z
1/2-X,1/2+Y,1/2-Z
1/2+X,1/2-Y,1/2-Z
1/2-X,1/2-Y,1/2-Z
1/2+X,1/2+Y,1/2-Z
1/2+X,1/2-Y,1/2+Z
1/2-X,1/2+Y,1/2+Z
Apply Cheshire symmetry operator #10
To x1 and y1
Xorig1=0.084
½-xorig1=0.5-0.084
½-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+y
½-x
-½-x-y -½+x-y
½+z
¼+z)
¼
Symmetry operator 2
-Symmetry operator 4
-x
-y
- ( ½+y
½-x
-½-x-y -½+x-y
½+z
¼+z)
¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-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+y
½-x
-½-x-y -½+x-y
½+z
¼+z)
¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-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+y
½-x
-½-x-y -½+x-y
½+z
¼+z)
¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-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+y
½-x
-½-x-y -½+x-y
½+z
¼+z)
¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-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 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.
The x, y coordinate
in step
3 describes
one of the heavy atom
For
example,
positions
in thelet’s
unit use:
cell. The x, z coordinate in step 4 describes a
-x, -y,copy.
z
symmetry related
We can’t combine these coordinates
And
apply
it todon’t
all coordinates
stepatom.
3.
directly.
They
describe the in
same
Perhaps they even
xreferred
= - (+0.02)
to different
origins.= -0.02
step3-transformed
ystep3-transformed = - (- 0.70) = +0.70
Now
xstep3-transformed
= xx,
How can
we transform
y from step 3 so it describes
step4
And
ystep3atom
has been
transformed
to a
the same
as x and
z in step 4?
reference frame consistent with x and z from
step 4. So we arrive at the following selfconsistent x,y,z:
Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
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, let’s 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 selfconsistent 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,
-X,
-Y,
Y,
Y,
-Y,
X,
-X,
Y, Z
-Y, Z
X, 1/4+Z
-X, 1/4+Z
X, -Z
-X, -Z
-Y, 1/4-Z
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,
-X,
-Y,
Y,
Y,
-Y,
X,
-X,
Y, 1/2+Z
-Y, 1/2+Z
X, 3/4+Z
-X, 3/4+Z
X, 1/2-Z
-X, 1/2-Z
-Y, 3/4-Z
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 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, let’s 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 selfconsistent 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,
-X,
-Y,
Y,
Y,
-Y,
X,
-X,
Y, Z
-Y, Z
X, 1/4+Z
-X, 1/4+Z
X, -Z
-X, -Z
-Y, 1/4-Z
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,
-X,
-Y,
Y,
Y,
-Y,
X,
-X,
Y, 1/2+Z
-Y, 1/2+Z
X, 3/4+Z
-X, 3/4+Z
X, 1/2-Z
-X, 1/2-Z
-Y, 3/4-Z
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 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, let’s 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 selfconsistent 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,
-X,
-Y,
Y,
Y,
-Y,
X,
-X,
Y, Z
-Y, Z
X, 1/4+Z
-X, 1/4+Z
X, -Z
-X, -Z
-Y, 1/4-Z
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,
-X,
-Y,
Y,
Y,
-Y,
X,
-X,
Y, 1/2+Z
-Y, 1/2+Z
X, 3/4+Z
-X, 3/4+Z
X, 1/2-Z
-X, 1/2-Z
-Y, 3/4-Z
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 remains
In 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
symmetry operator 1
-symmetry operator 5 -( y
x -z)
u
v
w
x-y -x+y 2z
First, plug in x=-0.02, y=0.70, z=-0.005
u=x-y = -0.02-0.70 =-0.72
v=-x+y= +0.02+0.70= 0.72
w=2z=2*(-0.005)=-0.01
The numerical value of these coordinates 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=0.01
Thus, 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.
Symmetry Operators are the Bridge between
Atomic Coordinates and Patterson Peaks
symop #1
symop #2
(-0.2,-0.3)
x, y
-(-x, –y)
2x , 2y
u=2x, v=2y
(0,0) x
y
(0,0) u
v
(0.2,0.3)
(0.6, 0.4)
(0.4, 0.6)
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
PATTERSON MAP
Patterson space
U=0.5
Crystal space
P43212 Symmetry operator difference 3-6
Calculate Y and Z
Calculate X and Y
Cheshire
operator
applied to
Y and Z if
two values
of Y do not
match
X,Y,Z referred to a
common origin.
u,v,w
If prediction lies outside Patterson asymmetric unit (0→0.5, 0→0.5,0→ 0.5) use
Patterson symmetry operators to find the symmetry equivalent peak in the
asymmetric unit. If the predicted peak is absent, then negate x value and recalculate u,v,w. Predicted peak should be present if algebra is correct.
m230d_2009_scaled.mtz
1) prok-native-christina-a1.sca
2) prok-native-rob-mike.sca
3) prok-eucl3-1-5-andre-nikkole.sca
4) prok-gdcl3-1-1-hui.sca
5) prok-gdcl3-1-5-wei.sca
6) prok-mersalyl-1-5-domi.sca
7) prok-pmsf-1-a-nikkole-andre2.sca
8) prok-smcl3-1-1-mike-rob.sca
9) prok-pcmbs-1-1-zheng.sca
Ihkl
Ihkl
Ihkl
Ihkl
Ihkl
Ihkl
Ihkl
Ihkl
Convert intensities to
structure factor amplitudes
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Fhkl
Scale all data sets to a
reference native data set
Which Derivatives will produce
the best phases?
1) Andre= FP_native-christina minus FP_Eu-1-5-andre-nikkole
2) Domi= FP_native-christina minus FP_mersalyl-1-5-domi
3) Nikkole= FP_native-rob-mike minus FP_Eu-1-5-andre-nikkole
4) Zheng= FP_native-rob-mike minus FP-pcmbs-zheng
5) Christina= FP_native-christina minus FP-pcmbs-zheng
6) Hui=
FP_native-christina minus FP_Gd-1-1-hui
7) Michael= FP_native-rob-mike minus FP_Sm-1-1-rob-mike
8) Rob=
FP_native-christina minus FP_Sm-1-1-rob-mike
9) Wei=
FP_native-christina minus FP_Gd-1-5-wei