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--2-fold symmetry axis
Chose a 2-fold axis as the
origin.
o
b
a
On the crystal,
draw a unit cell
having this size
and shape.
How many molecules in
the unit cell? asymmetric
unit?
Draw an “o” at the upper
left corner of the cell and
make it the origin.
Estimate (x,y) coordinates
of oxygen atom in fractions
of a unit cell. Use the ruler
provided.
4 distinct two-fold axes
a
b
a
Unit Cell
b
Choice 1
Choice 2
Choice 3
Choice 4
4 Choices of origin
b
a
Choice 1
Choice 3
Choice 2
Choice 4
Which plane group?
a
b
What are the coordinates of the oxygen using
origin choice 1?
a
b
1
3
2
X=0.2
Y=0.2
What are the coordinates of the other oxygen in
the unit cell?
X1=0.20
Y1=0.20
Symmetry
operators in plane
group p2
X, Y
-X,-Y
a
b
X2=0.80
Y2=0.80
1
2
3
Always allowed to add or subtract multiples of 1.0
X=0.2
Y=-0.8
X=-0.8
Y=-0.8
b
X=0.2
Y=0.2
X=-.0.2
Y=0.8
X=1.2
Y=0.2
X=-.0.2
Y=1.8
X=1.8
Y=0.8
X=0.8
Y=0.8
X=0.2
Y=1.2
X=-0.8
Y=1.2
X=1.8
Y=-0.2
X=0.8
Y=-0.2
X=-.0.2
Y=-0.2a
X=-0.8
Y=0.2
X=1.2
Y=-0.8
X=1.2
Y=1.2
X=0.8
Y=1.8
X=1.8
Y=1.8
What would be the oxygen coordinates if we had
drawn the unit cell with origin choice 2?
a
b
1
2
3
X=0.70
Y=0.20
Oxygen coordinates with origin choice 3?
a
b
1
2
3
X=0.30
Y=0.30
Oxygen coordinates with origin choice 4?
a
b
1
2
3
4
X=0.80
Y=0.30
Cheshire operators
•
•
•
•
X ,Y
X+.5, Y
X+.5, Y+.5
X , Y+.5
X=0.20
Y=0.20
Choice 1
X=0.70
Y=0.20
Choice 2
X=0.30
Y=0.30
Choice 3
X=0.80
Y=0.30
Choice 4
The 4 choices of origin are equally valid but once a
choice is made, you must remain consistent.
b
a
Choice 1
Choice 2
X1=0.20
Y1=0.20
X1=0.70
Y1=0.20
X2=0.80
Y2=0.80
Choice 3
X2=0.30
Y2=0.80
Choice 4
X1=0.30
Y1=0.30
X2=0.70
Y2=0.70
X1=0.80
Y1=0.30
X2=0.20
Y2=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.
• If a structure is solved independently by two
crystallographers using different choices of
origin, their coordinates will be related by a
Cheshire opertor.
• 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)
– Mersalyl, PCMBS
– GdCl3
• Did a heavy atom bind? How many?
• 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)
Significance?
P(uvw)=r(uvw)
r (-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.
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
to interpret 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.
Plane group p2
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) x
y
(0.2,0.3)
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).
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) x
(0,0) u
y
v
(0.2,0.3)
Patterson coordinates (U,V) are simply
symop1-symop2.
SYMMETRY OPERATORS
FOR PLANE GROUP P2
1) x,y
2) -x,-y
Remember this bridge!
symop1= X , Y
-symop2= -(-X,-Y)
u=2X
v=2Y
u=2(0.2) v=2(0.3)
u=0.4, v=0.6
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 in plane group p2
(-0.4, -0.6)
(-0.2,-0.3)
(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
2D CRYSTAL
PATTERSON MAP
Patterson in plane group p2
(0,0) u
v
(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 between 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) u
v
(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) u
v
(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) u
v
(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) u
v
(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) x
y
(0.2,0.3)
(-0.3,-0.2)
(0,0) u
v
(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 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=0.01
Thus, x=-0.02, y=0.70, z=-0.005 is correct.
Hurray! We are finished!
(1) U,
(5)-U,
(9)-V,
(13) V,
V,
V,
U,
U,
W (2)-U,-V,
W (6) U,-V,
W (10) V,-U,
W (14)-V,-U,
W (3) U, V,-W (4)-U,-V,-W
W (7)-U, V,-W (8) U,-V,-W
W (11)-V, U,-W (12) V,-U,-W
W (15) V, U,-W (16)-V,-U,-W
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
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.
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_2011_scaled.mtz
Intensity measurements (IHKL)
native-yen-b3.sca
prok_native_mucheng.sca
prok-gdcl3-matthew.sca
prok-gdcl3-maya.sca
prok_gdcl3_sean.sca
prok_iodide_jungreem.sca
prok_mersalyl_dan.sca
prok_mersalyl_dan_xds.sca
prok-pcmbs-1-5-stephen.sca
prok-pcmbs-andrew.sca
prok_pcmbs_cuiwen.sca
prok_pcmbs_pmsf_jose.sca
prok-pcmbs-pmsf-yazan.sca
prok_pcmbs_riki.sca
prok_pcmbs_sabai.sca
prok_pcmbs_smriti.sca
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 a
reference native data set with the
best statistics: prok-native-yen-b3
-using the CCP4 program SCALEIT.
Scale intensities by a constant (k) and
resolution dependent exponential (B)
prok-native-yen
H
1
1
1
1
1
1
1
K
0
0
0
0
0
0
0
prok-gdcl3-matthew
L intensity sigma
10 106894.0 1698.0
11 41331.5
702.3
12 76203.2 1339.0
13 28113.5
513.6
14
6418.2
238.7
15 45946.4
882.7
16 26543.8
555.6
H
1
1
1
1
1
1
1
K
0
0
0
0
0
0
0
L
10
11
12
13
14
15
16
intensity sigma
40258.7 1222.9
25033.2
799.8
24803.6
771.5
11486.3
423.9
9180.5
353.6
25038.8
783.0
21334.6
686.4
comparison
106894.0
41331.5
76203.2
28113.5
6418.2
45946.4
26543.8
/
/
/
/
/
/
/
40258.7
25033.2
24803.6
11486.3
9180.5
25038.8
21334.6
=
=
=
=
=
=
=
2.65
1.65
3.07
2.45
0.70
1.83
1.24
-Probably first crystal
is larger than the
second.
-Multiply Saken’s
data by k and B to
put the data on the
same scale.
2q/l2
-B*sin
e
Initial intensity profile
25000
intensity
20000
15000
native-yen
pcmbs-andrew
10000
5000
0
803.6
3.66- 2.91- 2.54- 2.31- 2.14- 2.02- 1.91- 1.83- 1.762.91 2.54 2.31 2.14 2.02 1.91 1.83 1.76 1.70
low resolution
high resolution.
Scaling constant K=0.883
25000
intensity
20000
15000
native-yen
pcmbs-andrew
10000
5000
0
803.6
3.66- 2.91- 2.54- 2.31- 2.14- 2.02- 1.91- 1.83- 1.762.91 2.54 2.31 2.14 2.02 1.91 1.83 1.76 1.70
low resolution
high resolution.
after K=1.1329 and B=-5.75
25000
intensity
20000
15000
native-yen
pcmbs-andrew
10000
5000
0
803.6
3.66- 2.91- 2.54- 2.31- 2.14- 2.02- 1.91- 1.83- 1.762.91 2.54 2.31 2.14 2.02 1.91 1.83 1.76 1.70
low resolution
high resolution.
29 BOTH
0.0
Col Sort
Min
Max
num order
1 ASC
Num
%
Mean
Mean Resolution Type Column
Missing complete
0
39
0 100.00
abs. Low
20.9
High
label
20.9 56.46 1.70 H H
2 NONE
0
28
0 100.00
8.6
8.6 56.46 1.70 H K
3 NONE
0
59
0 100.00
22.3
22.3 56.46 1.70 H L
4 NONE
0.0
5 NONE
0.6 1060.8 885 96.70 291.57 291.57 56.46 1.70 F FP_native-yen
6 NONE
7 NONE
0.5
19.0
0 100.00
37.5 885 96.70
9.47
3.36
9.47 56.46 1.70 I FreeR_flag
3.36 56.46 1.70 Q SIGFP_native-yen
1.0 941.8 5734 78.65 286.40 286.40 56.46 1.79 F FP_native-mucheng
8 NONE
0.6
9 NONE
9.8 1454.3 4366 83.74 305.42 305.42 56.46 1.80 F FP_gdcl3-matthew
10 NONE
11 NONE
12 NONE
1.5 105.6 4366 83.74
0.9 1500.2
0.6
13 NONE -95.4
14 NONE
87.0 5734 78.65
0.0
42.9
17.38
15.07
17.38 56.46 1.79 Q SIGFP_native-mucheng
15.07 56.46 1.80 Q SIGFP_gdcl3-matthew
62 99.77 292.26 292.26 56.46 1.70 F FP_gdcl3-maya
62 99.77
97.7 173 99.36
52.7 173 99.36
4.95
0.12
7.98
4.95 56.46 1.70 Q SIGFP_gdcl3-maya
16.32 56.46 1.70 D D_gdcl3-maya
7.98 56.46 1.70 Q SIGD_gdcl3-maya
15 NONE
5.1 1352.8 986 96.33 293.14 293.14 48.02 1.70 F FP_gdcl3_sean
16 NONE
0.2
17 NONE -40.7
18 NONE
0.0
42.2 986 96.33
51.1 1341 95.01
49.3 1341 95.01
3.55
-0.06
5.44
3.55 48.02 1.70 Q SIGFP_gdcl3_sean
4.22 48.02 1.70 D D_gdcl3_sean
5.44 48.02 1.70 Q SIGD_gdcl3_sean
19 NONE 12.8 1623.3 491 98.17 291.95 291.95 43.41 1.70 F FP_iodide_jungreem
20 NONE
0.9 126.6 491 98.17
21 NONE -152.8 101.7 533 98.02
8.78
0.79
8.40 43.41 1.70 D D_iodide_jungreem
22 NONE
0.0 135.1 533 98.02
23 NONE
4.5 1958.6 1325 95.07 308.60 308.60 19.78 1.72 F FP_mersalyl_dan
24 NONE
0.3
25 NONE -26.0
26 NONE
0.0
19.5 1325 95.07
27.7 1360 94.94
29.1 1360 94.94
13.98
8.78 43.41 1.70 Q SIGFP_iodide_jungreem
1.60
0.01
2.65
13.98 43.41 1.70 Q SIGD_iodide_jungreem
1.60 19.78 1.72 Q SIGFP_mersalyl_dan
3.03 19.78 1.72 D D_mersalyl_dan
2.65 19.78 1.72 Q SIGD_mersalyl_dan
27 NONE 12.2 1488.7 2170 91.92 290.41 290.41 56.46 1.70 F FP_pcmbs-stephen
28 NONE
1.0 119.8 2170 91.92
14.94
14.94 56.46 1.70 Q SIGFP_pcmbs-stephen
0.0 23592 12.15
0.00 56.46 1.70 D D_pcmbs-stephen
31 NONE
1.1 1428.1 230 99.14 292.43 292.43 56.46 1.70 F FP_pcmbs-andrew
32 NONE
0.7
0.00
65.0 230 99.14
0.00 56.46 1.70 Q SIGD_pcmbs-stephen
6.64
91.4 369 98.63
6.64 56.46 1.70 Q SIGFP_pcmbs-andrew
-0.57
73.5 369 98.63
10.26 56.46 1.70 D D_pcmbs-andrew
34 NONE
0.0
35 NONE
1.0 1921.9 309 98.85 287.65 287.65 56.46 1.70 F FP_pcmbs_cuiwen
36 NONE
0.7
37 NONE -91.0
=======================
0.00
0.0
33 NONE -94.7
OVERALL FILE STATISTICS for resolution range 0.000 - 0.351
0.0 23592 12.15
30 BOTH
10.57
61.3 309 98.85
10.57 56.46 1.70 Q SIGD_pcmbs-andrew
7.28
89.9 358 98.67
7.28 56.46 1.70 Q SIGFP_pcmbs_cuiwen
0.99
93.4 358 98.67
10.27 56.46 1.70 D D_pcmbs_cuiwen
38 NONE
0.0
39 NONE
7.2 1534.3
89 99.67 285.42 285.42 40.68 1.70 F FP_pcmbs_pmsf_jose
11.59
11.59 56.46 1.70 Q SIGD_pcmbs_cuiwen
40 NONE
1.2 139.5
89 99.67
10.39
41 NONE -153.0 111.3 174 99.35
10.39 40.68 1.70 Q SIGFP_pcmbs_pmsf_jose
-0.68
13.99 40.68 1.70 D D_pcmbs_pmsf_jose
42 NONE
0.0 146.2 174 99.35
43 NONE
0.8 1290.2 424 98.42 287.16 287.16 48.02 1.70 F FP_pcmbs-pmsf-yazan
44 NONE
0.6
45 NONE -115.5
16.49
47.9 424 98.42
16.49 40.68 1.70 Q SIGD_pcmbs_pmsf_jose
6.93
95.1 765 97.15
6.93 48.02 1.70 Q SIGFP_pcmbs-pmsf-yazan
-1.24
55.5 765 97.15
9.48 48.02 1.70 D D_pcmbs-pmsf-yazan
46 NONE
0.0
10.98
47 NONE
6.1 1563.7 10388 61.32 332.65 332.65 43.41 2.00 F FP_pcmbs_riki
48 NONE
0.8
86.0 10388 61.32
10.98 48.02 1.70 Q SIGD_pcmbs-pmsf-yazan
7.75
49 NONE -128.4 124.3 10427 61.17
7.75 43.41 2.00 Q SIGFP_pcmbs_riki
0.24
88.8 10427 61.17
8.85 43.41 2.00 D D_pcmbs_riki
50 NONE
0.0
11.72
51 NONE
9.0 1270.0 543 97.98 287.74 287.74 48.02 1.70 F FP_pcmbs_sabai
52 NONE
0.6
52.9 543 97.98
6.08
53 NONE -83.4 102.5 808 96.99
11.72 43.41 2.00 Q SIGD_pcmbs_riki
6.08 48.02 1.70 Q SIGFP_pcmbs_sabai
0.43
61.2 808 96.99
9.45
7.94 48.02 1.70 D D_pcmbs_sabai
54 NONE
0.0
55 NONE
8.5 1452.5 1240 95.38 288.38 288.38 48.02 1.70 F FP_pcmbs_smriti
56 NONE
0.6
81.1 1240 95.38
6.07
57 NONE -89.0 102.4 1732 93.55
58 NONE
0.0
9.45 48.02 1.70 Q SIGD_pcmbs_sabai
6.07 48.02 1.70 Q SIGFP_pcmbs_smriti
-0.06
94.3 1732 93.55
9.11
No. of reflections used in FILE STATISTICS
8.88 48.02 1.70 D D_pcmbs_smriti
9.11 48.02 1.70 Q SIGD_pcmbs_smriti
26855
LIST OF REFLECTIONS
===================
0 0 12
17.00
?
186.54
5.69
278.24
116.84
2.90
0.00
3.25
0.00
0.00
118.72
0.00
84.13
0.99
5.60
0.00
0.00
0.00
245.47
8.95
0.00
0.00
142.73
?
?
4.71
0.00
5.05
?
11.23
0.00
?
163.16
4.84
0.00
0.00
0.00
171.07
259.16
11.71
0.00
0.00
0.00
159.70
264.64
7.11
0.00
0.00
0.00
194.52
5.51
?
0.00