Lecture 6: Protein Structure
comparison
Computational Aspects of Molecular
Structure
Instructor: Teresa Przytycka, PhD

• In evolution, structure is better preserved than
sequence
• Structure comparison gives a powerful method for
searching for homologous proteins .
• Structure comparison allow to study protein
evolution
• To classify structures
Why compare structures?

Superposition of two structures

Structural similarity between Acetylcholinesterase
and Calmodulin
(Tsigelny et al, Prot Sci, 2000, 9:180

Estimating Quality of the alignment:
Root Mean Square Distance (RMSD)
∑=
=
N
i
ii bad
N
BARMS
1
2
' ),(
1
),(
A= a1 … an ; B= b1 … bm ;
Assume that ai is aligned with bi’ ;
d(ai,bi) is the Euclidian distance between ai and bi.

Problems with RMSD
A small local alignment error can propagate and the
quality of alignment nay be underestimated

Finding maximum common
substructure is NP hard
Goal: Find the maximum subset of dots that are in both sets in the
same relative position
We can superimpose 6 points
NP – hard: only
exponential time
algorithms are known

Methods
• Dynamic programming similar to sequence
alignment (we will discuss potential problems)
• Identify pairs of fragments (usually secondary
structures) that are similar and try to glue them
together into consistent alignment
• Presenting it as an optimization problem and using
algorithms as simulated annealing, brunch and
bound etc.
• Fast screening methods to that filter structure pairs
to be compared by more elaborate algorithms

Dynamic programming and it’s
limitations
• This is not a clean dynamic programming type of
problem but some program (e.g. SSAP) use DP as
heuristic approach.
• Idea: score for a pair of two aligned residues is
computed based on whether they are in the same
context with respect to their (3D) environment.
• Environment is defined by the proximity to other close
residues
A, B have similar environments thus we
aligned them
…but…
After we remove x similarity is lost
We don’t know what are your neighbors
until you do whole alignment!
A
B
x
y z

Example: SSAP
A, B- two fragments of protein
structure
views from i and k can be
compared by calculating the
difference between the
corresponding vectors (that is
vectors from i to all other nodes
and from k to all other nodes.

Double dynamic programming used
by SSAP
• Using DP find optimal path
Score between two vectors
are a/(b+δ) where a,b constant
learned on PDB
• Sum all optimal paths in the
summary matrix (top)
• Other scores added: solvent
accessibility, torsion angle,
volume
• Relative weight of these
contribution optimized based on
some pdb structures
• Do second dynamic
programming step on the
summary matrix.

Going around the above problems
and still using DP
• Method one: double dynamic programming
• Method two: “iterative” dynamic programming
1. Let the current alignment be any alignment.
2. For every residuum compute vector describing its
environment using current alignment
3. Find best alignment using dynamic programming
4. Iteratate 2, 3 using the computed alignment as current

SHEBA
J.Joung, B.Lee (2000) Protein Engineering 535-543
STEP 1: Initial alignment. Scoring function for the first iteration of
DP is as follows
a i i’ = score_ for_anino_acid_similarity+
score_for_similarity_of_secodry_structures_it_belongs_to +
similarity_in_watter_accesibility
Iterative improvement:
STEP 2: Superimpose the structures so that the distances between
aligned residues are minimized.
STEP 3: Using DP find max. number of aligned pairs whose
distance is <3.5 A.
Iterate 2 and 3
REPEAT WHOLE PRCEDURE WITH A DIFFERENT INITIAL
ALIGNMENT (change first scoring function).

DALI
• Dali is based on the comparison of intra-molecular
distance matrices.
• The original Dali (Holm L and Sander C 1993,
Protein structure comparison by alignment of
distance matrices, J. Mol. Biol. 233:123-38) used
a simulated annealing algorithm.
• A recent implementation, called DaliLite (Holm L
and Park J, DaliLite workbench for protein
structure comparison, (Bioinformatics 16:566-7),
used a branch-and-bound strategy.

Contact matrix
d(i,j) = distance( cα i , cα j )
Idea: Similar structures have similar contact matrices
Contact matrix n x n matrix where n = #residues
Below, pairs with d(i,j) below a certain treschold are gray and the rest
is white

Combinatorial Extension algorithm (CE)
Shindyalov & Bourne, Proot. Eng. 1998 739-747
• Identify all pairs of fragments that can be
reasonably aligned without gaps: AFP –
aligned fragment pairs (length <=8)for
example using Contact Map similarity (see
next slides)
• Extend the fragments using a heuristic (no
global optimization)

Dali solves optimization problem
),()),((
),(
B
ij
A
ij
B
iji j
A
ij ddwdd
BAS
∑ ∑ Δ−
=
θ
i,j pairs of residues from “core” = aligned part
D deviation of intramolecular C_alpha distance relative to their
arithmetic mean
θ – threshold similarity set empirically to 0.2 (20%)
ω – exp(-d2/r2) r = 20A – down weight contribution from distant
pairs
Find set of aligned residues pairs (i,j) that maximize the function

Finding All fragments
• Consider all possible pairs of 8x8 submatrices of the
contact matrices. Such matrices are small enough that
the problem can be solved optimally.
• Put the fragments together using a Monte
Carlo algorithm (slow process) –older version
• New version brunch and bound

Remarks
• Another method: Combinatorial Extension
(CE) also starts identifying such short
fragments but puts them together using a
variant of dynamic programming

Methods based on Secondary
Structure Alignments

Reducing the size of
representation of protein fold
All atom
Back bone atoms
Polygonal chain Cα-atoms

Reducing the size of
representation of protein fold
Secondary structure vectors

Approach based on comparing
secondary structure arrangement
Motivation:
1. Folds are often
defined as
arrangement of
secondary structure
elements (sse).
2. Why not to compare
arrangement of sse
rather than going
down to atomic
level?
1EJ9: Human topoisomerase I

VAST- graph theoretical approach
• http://www2.ncbi.nlm.nih.gov/Structure/VAST/vast.shtml
• Treat each secondary structure as a vector of direction and
length corresponding to the direction and length of the
secondary structure. Attributes of such vector include the
type of secondary structure, number of residues, etc.
• For two secondary structure provide a way of describing the
relative spatial position of secondary structures – distance,
angle, etc.
• VAST finds maximal subset of secondary structures that are
in the same relative positions in compared protein structures
and in the same order within the structure.

Step 1: represent secondary
structures as vectors

VAST: Calculate (rik, zik)
3
1
z
For both the query and
target structures,
For each SSE k,
set the origin at the
midpoint of k.
Then calculate rik and
zik for the endpoints of
SSEs i ≠ k.
Vector position relative to the xy plane
xyz13
r13

VAST: Create Comparison Graph
IL-4
IL-6
3 1
4
6
1
2
3
5
1 2 3 4 5 6
1
2
3
4
5
4
2
5
Nodes: r13<>r12
z13<>z12
Arcs: φ16<>φ15
must follow
sequence order
Select path with highest “weights”
N
N
C
C

VAST: Refinement
Aligned residues
are red
Alignment extended to the
end of this strand
Aligned SSEs guide the
alignment of the Cα
atoms
Alignments are allowed to
extend beyond SSE
boundaries
Refined alignment is
computed via a Gibbs
sampling algorithm (i.e.
Monte Carlo)

VAST credits & www
http://www.ncbi.nlm.nih.gov/Structure/
• Steve Bryant
• John Spouge
• Jean-Francois Gibrat
• Paul Thiessen
• Tom Madej
• Eric Sayers

Missed
similarities:
circular
permutations
From:
J.Jung, BK Lee
Protein Science 2001
1881-1885

Geometric Hashing
(Indexing / Fold invariants)

Hashing
Hash table
Hash function: assigns indexes in hash table to the
objects.
Hash function
list of all the words
with given hash
value
Set of objects

Choosing hash function for protein
structures
?
?
? ?
?
?
• Ideally: Different
folds: different hash
and same fold same
hash values.
• Problem – “same”
fold does not mean
identical structure.l
• Modified goal: Same
fold – “similar” hash
values.

Hashing for protein structures
• Given is a query structure and a data base of
structures
• Find fast way of searching similar structure in a
data base.
• Idea: assign to each protein a list of features.
• Identify protein that have the same (or similar
features)
• Example: feature: number of helices and strands
in the structure. Proteins that have very different
secondary structure composition than the query
protein are filtered out and in a subsequent phase
only proteins with similar secondary structure
composition are compared.

Hashing function (key function)
• In general it is many to one - we accept the
fact that different folds may lead to the
same result but we want to minimize such
overlaps.

Key function that describes relative
position of secondary structure vectors

Assume d = 2
Hopefully all related
triples are hashed in the
neighborhood of the key of the
query, in practice there may be
some false positive /negative

Practical considerations
• Dimension d cannot be to large, or else finding all
neighbors is becoming costly
• There are data structures that are designed for
searching for neighbors is d-dimensional structure
• Examples of good hash (key) function
– Angles between vectors
– Distance between midpoints
• Agreement of the key function on three vectors is
usually not enough to declare possible similarity.
We have to require a larger number of matches,
how large – depends on the size of structure.

Geometric hashing
• R. Nussinov and H.J. Wolfson. Efficient
detection of three-dimensional structural motifs
in biological macromolecules by computer
vision techniques. Proc. Natl. Acad. Sci. USA.,
88:10495-10499, 1991.
• L. Holm and C. Sander. 3-d lookup: Fast
protein structure database searches at 90 %
reliability. Proceedings of the Third
International Conference on Intelligent
Systems for Molecular Biology 179-187.

Projection to Rn
• Encode the structure as a n-dimensional real
vector
• Reduce the problem of comparing structure
to computing Euclidean distance between the
vectors
Problem:
How to find a good encoding?

Idea
Same ?
S S’
I I
I(S) I(S’)
Easier comparison

Properties of an invariant
• S = S’ ! I(S) = I(S’)
• I(S) = I(S’) !S=S’ (that is not always)
• “Strength” of an invariant: how likely two
different object receive the same invariant

“Shape descriptors” for polygonal lines
• Motivated by Vasiliev knot invariants
• Introduced by Rogen and Bohr, (Math.
Biosciences, 2002).
• Rogen and Fain (PANS 2002)

Main Idea
• Consider a polygonal line embedded in 3D
• Consider a projection of such line on a plane and count the
crossings (with or without sign)
W(i1,i2)=The number of crossing depends on projection, but
the average number of crossing over all possible projection
is an invariant of an embedded line: CAN BE EASILY
COMPUTED
+ _

The average over all projection of the “diagram”
crossing number
Wr(γ) = 1/(4π) w(t1,t2) dt1 dt2
For polygonal lines one can replace it with summations
over all pairs of intervals
Wr(γ) = Σ i1,i2 W(i1,i2)
Where W is the integral as above but restricted to the
two intervals.
Writhe
γ x γ - diagonal

Can count both the average of signed
crossings and unsigned crossings
• Singed:
I (1,2) = Wr(γ) = Σ i1,i2 W(i1,i2)
• Unsigned:
I |1,2| = Wr(γ) = Σ i1,i2 |W(i1,i2)| (same as
above but the crossings are unsigned)
Crossings between projections of two
intervals averaged over all projections

Towards stronger invariants
• I (1,2) :
– look at a pair of segments,
– compute average crossing number for the pair,
– sum over all pairs
• Extending the concept (following Vasiliev
knot invariant) to I (1,2) (3,4) :
– consider 2 pairs of intervals at a time
– compute product W(i1,i2) W(i2,i4) for the two pairs
– sum over all possible pairs of pairs
– … you can also consider triplets and so on.

PRIDE
• Carugo, Pongor 2002
Consider set of distances C .
Build histogram of these distances and
compare them.
(for several different values of n)

LFF (Local Feature Frequency)
Choi, Kwon, Kim 2003
• For each structure, subdivide the Ca-Ca
distance matrix into submatirces
corresponding to overlapping fragments
• Select 100 such submatrices to be
representative “models”
• For every protein compute the distribution of
these selected patterns in the protein structure
• To compare protein structure – compare these
distributions.

models
Count the number of occurrences
Of each model (here first) in the
structure and report it on the
corresponding position (here first)
on the 100-long vector.
Comparing structures is reduced to
comparing vectors