Molecular Crystallography
1. Molecular crystallography
2. Snow crystals
3. Crystallographic scaling
4. Axial-symmetric proteins
5. Integral lattices
6. Perspectives
Nijmegen, 21.04.08
A. Janner
Overview
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Molecular Crystallography
Quanosine 5’-phosphate tetramer
(Zimmerman, JMB 106 (1976) 663-677)
[0 6 2]
Cubic Form lattice
Sugar-Phosphate
[-3 3 2]
[3 3 2]
Bases:
Guanine
[-6 0 2]
[6 0 2]
Central
Hole
[-3 -3 2]
[3 -3 2]
Envelope
[0 -6 2]
molecular crystallography
[-6 0 2]
[6 0 2]
[-6 0 0]
[6 0 0]
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Dendritic Snow Crystal with Growth Lattice
Bentely & Humphreys, Snow Crystals, Dover, 1962 (167.8)
bh167.8-84 (fig5)
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Dendritic Snow Crystal with Growth Lattice
Branching sites at points of the growth lattice
BH 167.8
bh167.8-84 (fig5)
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Facet-like Snow Crystal with Growth Lattice
Bentely & Humphreys, Snow Crystals, Dover, 1962 (114.8)
bh114.8-83 (fig4b)
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Facet-like Snow Crystal with Growth Lattice
Regular hexagons with center and vertices at points of the growth lattice
BH 114.8
bh114.8-83 (fig4b)
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Indexed Snow Crystal
Bentely & Humphreys, Snow Crystals, Dover, 1962 (53.1)
BH 53.1
bh167.8-84 (fig5)
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Indexed Snow Crystal
Hexagrammal scaled form
04
44
01
10
40
-4 0
-4 -4
0 -4
(bh53.1-86)
bh167.8-84 (fig5)
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Crystallographic Scaling
Scaling
with scaling factor λ
1D (linear)
2D (planar)
3D (isotropic)
Higher dimensional
Xλ (x, y, z) = (λx, y, z)
Pλ (x, y, z) = (λx, λy, z)
Iλ (x, y, z) = (λx, λy, λz)
.......
Cryst.Scal.
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Crystallographic Scaling
Scaling
with scaling factor λ
1D (linear)
2D (planar)
3D (isotropic)
Higher dimensional
Crystallographic
in general:
Cryst.Scal.
Xλ (x, y, z) = (λx, y, z)
Pλ (x, y, z) = (λx, λy, z)
Iλ (x, y, z) = (λx, λy, λz)
.......
transforming a lattice into a lattice
Sλ Λ = Λ
Sλ integral invertible
Sλ Λ = Σ
Σ ⊆ Λ or Λ ⊆ Σ
Sλ rational invertible
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Hexagrammal Scaling
mid-edge
vertex
02
01
-1 1
10

S=
2 0
0 2


λ = 2, ϕ = 0
01
21
20
10

S=
λ=
√
2 −1
1
1


3, ϕ = 30o
mid-edge vertex
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Pentagonal Case
Polygrammal Scaling
Star Pentagon:
1000
Schäfli Symbol {5/2}
0100
Scaling matrix: (planar scaling)
-1 -1 0 -2

0 1 -1 1
-1 -1 -1 -1
2112
1 -1 1 0
2̄

 0


 1̄

1̄

1 0
1̄
1̄ 1

1̄ 


0 

2̄
1 1̄
0 1
-2 0 -1 -1
Scaling factor:
-1/τ 2 = −0.3820...
0010
0001
Pentagram
(τ =
√
1+ 5
2
= 1.618...)
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Hexagrammal Scaling Symmetry of Snow Crystals
Facet-like snow flake
Dendritic-like snow flake
(Sci.Am. 2)
Sa1+Sa2[SAb]
(Sci.Am. 1)
Scientific American (1961)
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Hexagrammal Scaling Symmetry of Snow Crystals
Mid-edge star hexagons: λM E = 1/2
(Sci.Am. 2)
√
Vertex star hexagons: λV E = 1/ 3
(Sci.Am. 1)
Sa1+Sa2[SAb]
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R-phycoerythrin (trigonal hexamer)
Hexagonal form lattice
Hexagrammal mid-edge scaling
y
04
44
01
-4 0
-4 -4
R-phycoerythrin 1-hex.
re x
40
r
° 10
0 -4
Chang et al., J.Mol.Biol 262 (1996) 721-731 (PDB 1lia)
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Mitochondrial creatine kinase (tetragonal octamer)
Square form lattice with scaling relations
07
-6 6
-1 6
16
66
Gly365
-6 1
-7 0
-6 -1
-1 1
11
61
70
-1 -1
1 -1
6 -1
Thr1
-6 -6
creatine kin. square
-1 -6 1 -6
0 -7
6 -6
Fritz-Wolf, Schnyder, Wallimann & Kabsch, Nature (1996) 341-345 (1crk)
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Cyclophilin A (Decamer)
Ke and Mayrose (PDB 2rma)
τ = 1.61803... the Golden Ratio
τ
1
τ
GLY(14)
Cycloph.decam.pentam.
– p. 12/??
Cyclophilin A (Pentamer)
Pentamer: Pentagrammal scaled form
τ
1
τ
GLY(14)
Cycloph.decam.pentam.
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Mitochondrial creatine kinase (tetragonal octamer)
y
[-6 6 4]
[-1 6 4] [1 6 4]
[6 6 4]
Gly365
[6 1 4]
[-6 1 4]
x
[-6 -1 4]
[-6 -6 4]
Cubic indexed form
[-6 -6 4]
[6 -1 4]
[-1 -6 4] [1 -6 4]
[6 -6 4]
z
[-1 -6 4] [1 -6 4]
[6 -6 4]
Gln366
x
[-6 -6 -4]
Creatine Kinase cubic
[-1 -6 -4][1 -6 -4]
[6 -6 -4]
Fritz-Wolf, Schnyder, Wallimann & Kabsch, Nature (1996) 341-345 (1crk)
– p. 13/??
Cyclophilin: Isometric Decagonal Lattice
y
Glu15
[1 0 0 0, 2]
[-2 1 -2 -1, 2]
[3 0 2 1, 2]
[0 1 0 0, 2]
[-3 -1 -2 -3, 2]
C
τ
[3 0 1 2, 2]
[1 -1 2 -1, 2]
r
°
P
1
A
[-1 -1 -1 -1, 2]
x
[3 1 1 3, 2]
τr
°
3
Q
[-1 2 -1 1, 2]
τ
[2 1 0 3, 2]
D
[-3 -2 -1 -3, 2]
[0 0 1 0, 2]
r0 = a = c
[1 2 0 3, 2]
[-1 -1 1 -2, 2]
[0 0 0 1, 2]
z
[-1 2 -1 1, 2]
[2 1 0 3, 2]
Glu15
2r
°
x
4r
°
Glu15
[-1 2 -1 1,-2]
Cyclo. Iso-pentagonal
[2 1 0 3,-2]
Ke et al., Current Biology Structure, 2 (1994) 33-44
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R-phycoerythrin: Isometric hexagonal
y
[0 4 4]
[-4 0 4]
a = 4r
°
[4 4 4]
x
r
°
[4 0 4]
[-4 -4 4]
[0 -4 4]
z
[-4 -4 4]
[0 -4 4]
[4 0 4]
4r°
R-phycoerythrin 1-hex.
x
[-4 -4 -4]
[0 -4 -4]
[4 0 -4]
– p. 15/??
Distribution of Hexagonal Inorganic Crystals
1/√2
1
√2
√(8/3)
√6
1000
964 Hexagonal isometric lattices
as for the molecular forms of:
800
- Hexameric R-Phycoerythrin
- Trimeric Outer Membrane Protein F
600
400
200
0
0
1
2
3
4
5
c/a
Inorganic Crystal Structure Database (ICSD)
12’000 hexagonal entries
hexag. inorg.
(collaboration R. de Gelder)
– p. 16/??
Distribution of Pseudo-Tetragonal Orthorhombic Crystals
1
√2
√6
3
900
800
700
600
500
400
300
200
100
0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
c/a = c/b
Inorganic Crystal Structure Database (ICSD)
4095 pseudo-tetragonal entries
pseudo-tetr. inorg.
(collaboration R. de Gelder)
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From Structural Relations to Symmetry
System
Axial proteins, Viral capsid, Holoenzymes (Ferritin, SOR)
Property
External envelope - Central hole
Same form lattice
Integral lattice
Internal - External polygons & polyhedra Indexed vertices
Relation
Crystallographic scaling
Integral lattice
Star polygons, scaled capsid
Rational (c/a)2
Problems
Infinite order point group
Finite structures
Crystallographic scaling is not a molecular symmetry
Axial ratio: not determined by crystallographic laws
Solution?
Finite Higher-Dimensional crystallographic point groups
relation - symmetry
– p. 18/??
4D Symmetry of pentagonal and decagonal star polygons
I4 = {A5 = B 4 = 1, [A, B] = A2 } ∈ G`(4, Z)
I4 orbits
{5/1} = {I4 | [1000]}
Pentagon
{10/3} = {I4 |[1-100]}
Decagram {10/3}
{5/2} = {I4 | [0-1-10]}
Pentagram {5/2}
{5/2}*{5/2} = {I4 | [1221]}
Squared Pentagram
BBNWZ, Crystallographic groups of four-dimensional space (isom 20.5, p.242)
– p. 19/??
Frank’s cubic hexagonal lattice
Hexagonal four-indices
1922: Weber
0001
Kr = hm + kn + lp = hu + kv + it + lw
i=-k-l
0100
u=
w=p
2m−n
3
v=
−m+2n
3
t=
−m−n
3
1000
0010
K = (hkl) = (hkil)
r = [mnp] = [uvtw]
4D cubic
1965: Frank
[001]
[0100]
[0001]
c2
[1000]
[100]
a2
[0010]
ϕ
weber-frank
=
-[2110]/3
c2
=
1
[21̄1̄0]
3
6 2
c
9
c
a
=
q
3
2
cos ϕ = √(2/3)
– p. 20/??
Thanking for your attention!
thanks
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