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# Geometric and Kinematic Models of Proteins - PowerPoint PPT Presentation

Geometric and Kinematic Models of Proteins. Study of movement independent of the forces that cause them. What is Kinematics?. Protein. Long sequence of amino-acids (dozens to thousands), also called residues from a dictionary of 20 amino-acids. Role of Geometric and Kinematic Models.

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Geometric and Kinematic Models of Proteins

What is Kinematics?

• Long sequence of amino-acids (dozens to thousands), also called residues from a dictionary of 20 amino-acids

• Represent the possible shapes of a protein (compare/classify shapes, find motifs)

• Answer proximity queries: Which atoms are close to a given atom? (computation of energy)

• Compute surface area (interaction with solvent)

• Find shape features, e.g., cavities (ligand-protein interaction)

• Large number of atoms Combinatorial problems

• Large number of degrees of freedom  Large-dimensional conformation space

• Need to efficiently update information during simulation (surface area, proximity among atoms):

• What is the position of every atom in some given coordinate system?

• Which atoms intersect a given atom?

• What atoms are within some distance range from another one?

• Complex metric in conformational space

• Many shape matching issues

Geometric Models of Bio-Molecules

• Hard-sphere model (van der Waals radii)

• Van der Waals surface

Van der Waals interactions between twoatoms result from induced polarization

effect (formation of electric dipoles). Theyare weak, except at close range.

The van der Waals force is the force to which the gecko's unique ability to cling to smooth surfaces is attributed!

12-6 Lennard-Jones potential

Geometric Models of Bio-Molecules

• Hard-sphere model (van der Waals radii)

• Van der Waals surface

Van der Waals radii in Å

Geometric Models of Bio-Molecules

• Hard-sphere model (van der Waals radii)

• Van der Waals surface

• Solvent- accessible surface

• Molecular surface

Probe of 1.4Å

Probe of 5Å

Computation of Hard-Sphere Surface (Grid method [Halperin and Shelton, 97])

• Each sphere intersects O(1) spheres

• Computing each atom’s contribution to molecular surface takes O(1) time

• Computation of molecular surface takes Θ(n) time

Why?

Computation of Hard-Sphere Surface (Grid method [Halperin and Shelton, 97])

• Each sphere intersects O(1) spheres

• Computing each atom’s contribution to molecular surface takes O(1) time

• Computation of molecular surface takes Θ(n) time

Why?

D. Halperin and M.H. Overmars Spheres, molecules, and hidden surface removalComputational Geometry: Theory and Applications 11 (2), 1998, 83-102.

D. Halperin and C.R. Shelton A perturbation scheme for spherical arrangements with application to molecular modelingComputational Geometry: Theory and Applications 10 (4), 1998, 273-288.

Possible project: Design software to update surface area during molecule motion

Other approach: Alpha shapes http://biogeometry.duke.edu/software/alphashapes/pubs.html

• United-atom model: non-polar H atoms are incorporated into the heavy atoms to which they are bonded

• Lollipop model: the side-chains are approximated as single spheres with varying radii

• Bead model: Each residue is modeled as a single sphere

• Stick (bond) model

• Stick (bond) model

• Small-sphere model

(x4,y4,z4)

(x5,y5,z5)

(x6,y6,z6)

(x8,y8,z8)

(x7,y7,z7)

(x1,y1,z1)

Kinematic Models of Bio-Molecules

• Atomistic model: The position of each atom is defined by its coordinates in 3-D space

(x3,y3,z3)

(x2,y2,z2)

p atoms  3p parameters

Drawback: The bond structure is not taken into account

The atomistic model does not encode this kinematic structure( algorithms must maintain appropriate bond lengths)

• Atomistic model: The position of each atom is defined by its coordinates in 3-D space

• Linkage model:The kinematics is defined byinternalcoordinates (bond lengths and angles, and torsional angles around bonds)

T?

• Update the position of each atom in world coordinate system

• Determine which pairs of atoms are within some given distance(topological proximity along chain  spatial proximitybut the reverse is not true)

T(x)

y

T

x

x

Rigid-Body Transform

x

2-D Case

y

x

x

2-D Case

y

x

x

2-D Case

y

x

x

2-D Case

y

x

x

2-D Case

y

x

x

2-D Case

y

Rotation matrix:

cos q -sin qsin qcos q

j

i

q

ty

tx

x

x

2-D Case

y

Rotation matrix:

i1 j1i2j2

j

i

q

ty

tx

x

x

2-D Case

y

Rotation matrix:

a

i1 j1i2j2

a

b

j

i

=

b’

q

ty

a’

b’

b

q

a

tx

a

a’

x

x

2-D Case

v

Transform of a point?

y

y’

q

y

ty

x’cos q -sin qtxx tx + x cosq – y sin q

y’ = sin q cos qtyy = ty + x sin q + y cos q

1 0 0 1 1 1

x

x’

tx

x

x

Homogeneous Coordinate Matrix

i1 j1txi2 j2ty 0 0 1

• T = (t,R)

• T(x) = t + Rx

q2

q1

3-D Case

z

y

x

y

i

z

j

k

x

Homogeneous Coordinate Matrix in 3-D

i1 j1 k1txi2 j2 k2tyi3 j3 k3tz 0 0 0 1

with:

• i12 + i22 + i32 = 1

• i1j1 + i2j2 + i3j3 = 0

• det(R) = +1

• R-1 = RT

y

x

Example

cos q 0 sinq tx

0 1 0 ty

-sin q 0 cos q tz

0 0 0 1

q

q

Rotation Matrix

R(k,q)=

kxkxvq+ cqkxkyvq- kzsqkxkzvq+ kysq

kxkyvq+ kzsqkykyvq+ cqkykzvq- kxsq

kxkzvq- kysqkykzvq+ kxsqkzkzvq+ cq

where:

• k = (kx ky kz)T

• sq = sinq

• cq = cosq

• vq = 1-cosq

y

x

y

i

z

j

k

x

x’ i1 j1 k1 txx

y’ i2 j2 k2 tyy

z’ i3 j3 k3 tzz

1 0 0 0 1 1

=

Homogeneous Coordinate Matrix in 3-D

(x,y,z)

(x’,y’,z’)

Composition of two transforms represented by matrices T1 and T2 : T2T1

What is the potential problem with homogeneous coordinate matrix?

• Rigid bodies are:

• atoms (spheres), or

• groups of atoms

• Build the assembly of the first 3 atoms:

• Place 1st atom anywhere in space

• Place 2nd atom anywhere at bond length

• Build the assembly of the first 3 atoms:

• Place 1st atom anywhere in space

• Place 2nd atom anywhere at bond length

• Place 3rd atom anywhere at bond length with bond angle

x

y

Coordinate Frame

• Build the assembly of the first 3 atoms:

• Place 1st atom anywhere in space

• Place 2nd atom anywhere at bond length

• Place 3rd atom anywhere at bond length with bond angle

• Introduce each additional atom in the sequence one at a time

x

y

1 0 0 0cb -sb0 0 1 0 0 d

0 ct -st 0 sbcb 0 0 0 100

0 st ct 0 0 0 1 0 0 010

0 0 0 1 0 0 0 1 0 0 0 1

Ti+1 =

Bond Length

x

y

1 0 0 0cb -sb0 0 1 0 0 d

0 ct -st 0 sbcb 0 0 0 100

0 st ct 0 0 0 1 0 0 010

0 0 0 1 0 0 0 1 0 0 0 1

Ti+1 =

Bond angle

x

y

1 0 0 0cb -sb0 0 1 0 0 d

0 ct -st 0 sbcb 0 0 0 100

0 st ct 0 0 0 1 0 0 010

0 0 0 1 0 0 0 1 0 0 0 1

Ti+1 =

Torsional (Dihedral) angle

x

y

1 0 0 0cb -sb0 0 1 0 0 d

0 ct -st 0 sbcb 0 0 0 100

0 st ct 0 0 0 1 0 0 010

0 0 0 1 0 0 0 1 0 0 0 1

Ti+1 =

Transform Ti+1

y

i+1

Ti+1

z

x

t

i-1

d

i

b

i-2

x

y

1 0 0 0cb -sb0 0 1 0 0 d

0 ct -st 0 sbcb 0 0 0 100

0 st ct 0 0 0 1 0 0 010

0 0 0 1 0 0 0 1 0 0 0 1

Ti+1 =

Transform Ti+1

y

i+1

J.J. Craig. Introduction to Robotics. Addison Wesley, reading, MA, 1989.

Zhang, M. and Kavraki, L. E.. A New Method for Fast and Accurate Derivation of Molecular Conformations. Journal of Chemical Information and Computer Sciences, 42(1):64–70, 2002.http://www.cs.rice.edu/CS/Robotics/papers/zhang2002fast-comp-mole-conform.pdf

Ti+1

z

x

t

i-1

d

i

b

i-2

T1

0

1

T2

-1

-2

Ti+2

k-1

Ti+1

i+1

Tk

k

i

Tk(i) = Tk…Ti+2 Ti+1 position of atom k in frame of atom i

• Tk(i) = Tk…Ti+2 Ti+1

• Atom j between i and k

• Tk(i) = Tj(i)Tj+1Tk(j+1)

• A parameter between j and j+1 is changed

• Tj+1 Tj+1

• Tk(i)  Tk(i) = Tj(i)Tj+1 Tk(j+1)

Root group of 3 atoms

p atoms  3p -6 parameters

T0

world coordinate system

Root group of 3 atoms

p atoms  3p -6 parameters

In physiological conditions:

• Bond lengths are assumed constant [depend on “type” of bond, e.g., single: C-C or double C=C; vary from 1.0 Å (C-H) to 1.5 Å (C-C)]

• Bond angles are assumed constant[~120dg]

• Only some torsional (dihedral) angles may vary

• Fewer parameters: 3p-6   p-3

C

C

N

Ca

3.8Å

Bond Lengths and Angles in a Protein

w: Ca Ca

f: C  C

y: N  N

w = p

w

side-chain group

C

N

Ca

f=0

Convention for f-y Angles

• fis defined as the dihedral angle composed of atoms Ci-1–Ni–Cai–Ci

• If all atoms are coplanar:

• Sign of f: Use right-hand rule. With right thumb pointing along central bond (N-Ca), a rotation along curled fingers is positive

• Same convention fory

C

Ca

N

C

f=p

They assign probabilities to φ-ψ pairs based on frequencies in known folded structures

ψ

φ

The sequence of N-Ca-C-… atoms is the backbone (or main chain)

Rotatable bonds along the backbone define the f-y torsional degrees of freedom

Small side-chains with c degree of freedom

c

c

c

c

c

Cb

Ca

f-y-c Linkage Model of Protein

0 to 4 c angles: c1, ..., c4

Kinematic Models of Bio-Molecules

• Atomistic model:The position of each atom is defined by its coordinates in 3-D spaceDrawback: Fixed bond lengths/angles are encoded as additional constraints. More parameters

• Linkage model:The kinematics is defined byinternal parameters (bond lengths and angles, and torsional angles around bonds)Drawback: Small local changes may have big global effects. Errors accumulate. Forces are more difficult to express

• Simplified (f-y-c) linkage model: Fixed bond lengths, bond angles and torsional angles are directly embedded in the representation.Drawback: Fine tuning is difficult

 Computational errors may accumulate

x’ effect i1 j1 k1 tx x

y’ i2 j2 k2 ty y

z’ i3 j3 k3 tz z

1 0 0 0 1 1

=

Drawback of Homogeneous Coordinate Matrix

• Too many rotation parameters

• Accumulation of computing errors along a protein backbone and repeated computation

• Non-redundant 3-parameter representations of rotations have many problems: singularities, no simple algebra

• A useful, less redundant representation of rotation is the unitquaternion

R( effectr,q)

R(r,q+2p)

Unit Quaternion

R(r,q) = (cosq/2, r1sin q/2, r2sinq/2, r3sinq/2)

=cosq/2 + rsin q/2

Space of unit quaternions: Unit 3-sphere in 4-D space with antipodal points identified

P =p0+ p

Q =q0+ q

Product R = r0 + r = PQ

r0 = p0q0 – p.q(“.” denotes inner product)

r = p0q + q0p + pq (“” denotes outer product)

Conjugate of P:P* = p0-p

Point x = (x,y,z)  quaternion 0 + x

Transform of translation t = (tx,ty,tz) and rotation (n,q)

Transform of x is x’

0 + x’ = R(n,q)(0 + x) R*(n,q)+ (0 + t)