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Conformational Space Conformational Space Conformation of a molecule: specification of the relative positions of all atoms in 3D-space, Typical parameterizations: List of coordinates of atom centers List of torsional angles (e.g., the f-y-c for a protein) Conformational space: Space of all conformations Conformational Space qj qi qN-1 q2 qN q1 Conformational Space q0 q1 qn q4 q3 Relation to Robotics/Graphics q0 q1 q2 qn t(t) q4 Configuration space q3 Need for a Metric Simulation and sampling techniques can produce millions of conformations Which conformations are similar? Which ones are close to the folded one? Do some conformations form small clusters (e.g. key intermediates while folding)? Metric in Conformational Space A metric over conformational space C is a function: d: c,c’ C d(c,c’) +{0} such that: d(c,c’) = 0 c = c’ d(c,c’) = d(c’,c) d(c,c’) + d(c’,c”) d(c,c”) (non-degeneracy) (symmetry) (triangle inequality) But not all metrics are “good” Euclidean metric: d(c,c’) = Si=1,...,n(|fi-fi’|2+ |yi-yi’|2) Metric in Conformational Space A “good” metric should measure how well the atoms in two conformations can be aligned Usual metrics: cRMSD, dRMSD RMSD Given two sets of n points in 3 A = {a1,…,an} and B = {b1,…,bn} The RMSD between A and B is: RMSD(A,B) = [(1/n)Si=1,…,n||ai-bi||2]1/2 where ||ai-bi|| denotes the Euclidean distance between ai and bi in 3 RMSD(A,B) = 0 iff ai = bi for all i cRMSD Molecule M with n atoms a1,…,an Two conformations c and c’ of M ai(c) is position of ai when M is at c cRMSD(c,c’) is the minimized RMSD between the two sets of atom centers: minT[(1/n)Si=1,…,n||ai(c) – T(ai(c’))||2]1/2 where the minimization is over all possible rigid-body transform T cRMSD cRMSD verifies triangle inequality cRMSD takes linear time to compute Often, cRMSD is restricted to a subset of atoms, e.g., the Ca atoms on a protein’s backbone Representation Restricted to Ca Atoms Protein 1tph - The positions of AA residue centers (Cα atoms) mainly determine the structure of a protein. - In structural comparison, people usually work only on the backbone of Cα atoms, and neglect the other atoms. Possible project: Design a method for efficiently finding nearest neighbors in a sampled conformation space of a protein, using the cRMSD metric. dRMSD Molecule M with n atoms a1,…,an Two conformations c and c’ of M {dij(c)}: nn symmetrical intra-molecular distance matrix in M at c dRMD(c, c’) is : [(1/n(n-1))Si=1,…,n-1Sj=i+1,…,n(dij(c) – dij(c’))2]1/2 {dij} is usually restricted to a subset of atoms, e.g., the Ca atoms on a protein’s backbone Intra-Molecular Distance Matrix Distances between Ca pairs of a protein with 142 residues. Darker squares represent shorter distances. Intra-Molecular Distance Matrix 45 40 85 1 Distances between Ca pairs of a protein with 142 residues. Darker squares represent shorter distances. Intra-Molecular Distance Matrix dRMSD Molecule M with n atoms a1,…,an Two conformations c and c’ of M {dij(c)}: nn symmetrical intra-molecular distance matrix in M at c dRMSD(c, c’) = [(2/n(n-1))Si=1,…,n-1Sj=i+1,…,n(dij(c) – dij(c’))2]1/2 {dij} is usually restricted to a subset of atoms, e.g., the Ca atoms on a protein’s backbone dRMSD Molecule M with n atoms a1,…,an Two conformations c and c’ of M {dij(c)}: nn symmetrical intra-molecular distance matrix in M at c dRMSD(c, c’) = [(2/n(n-1))Si=1,…,n-1Sj=i+1,…,n(dij(c) – dij(c’))2]1/2 {dij} is usually restricted to a subset of atoms, e.g., the Ca atoms on a protein’s backbone Advantage: No aligning transform Drawback: Takes quadratic time to compute Is dRMSD a metric? dRMSD(c, c’) = [(2/n(n-1))Si=1,…,n-1Sj=i+1,…,n(dij(c) – dij(c’))2]1/2 is a metric in the n(n-1)/2-dimensional space, where a conformation c is represented by {dij(c)} But, in this representation, the same point represents both a conformation and its mirror image k-Nearest-Neighbors Problem Given a set S of conformations of a protein and a query conformation c, find the k conformations in S most similar to c (w.r.t. cRMSD, dRMSD, other metric) Can be done in time O(N(log k + L)) where: - N = size of S - L = time to compare two conformations k-Nearest-Neighbors Problem The total time needed to compute the k nearest neighbors of every conformation in S is O(N2(log k + L)) Much too long for large datasets where N ranges from 10,000’s to millions!!! Can be improved by: 1. Reducing L 2. More efficient algorithm (e.g., kd-tree) kd-Tree In a d-dimensional space, where d>2, range searching for a point takes O(dn1-1/d) k-Nearest-Neighbors Problem Idea: simplify protein’s description Assume that each conformation is described by the coordinates of the n Ca atoms cRMSD O(n) time dRMSD O(n2) time This representation is highly redundant Proximity along the chain entails spatial proximity d 3l Atoms can’t bunch up, hence far away atoms along the chain are on average spatially distant ci cj m-Averaged Approximation Cut the backbone into fragments of m Ca atoms Replace each fragment by the centroid of the m Ca atoms Simplified cRMSD and dRMSD 3n coordinates 3n/m coordinates Evaluation: Test Sets [Lotan and Schwarzer, 2003] 8 diverse proteins (54 -76 residues) Decoy sets of N =10,000 conformations from the Park-Levitt set [Park et al, 1997] Correlation: m cRMSD dRMSD 3 0.99 0.96-0.98 4 0.98-0.99 0.94-0.97 6 0.92-0.99 0.78-0.93 9 0.81-0.98 0.65-0.96 12 0.54-0.92 0.52-0.69 Higher correlation for random sets ( greater savings) Running Times Further Reduction for dRMSD 1) Stack m-averaged distance matrices as vectors of a matrix A N r 1 n n r 1 2 m m A Vector ai of elements of distance matrix of ith conformation (i = 1 to N) 1 dRMSDm (c,c ai -aj i j )= r 2 Further Reduction for dRMSD 1) Stack m-averaged distance matrices as vectors of a matrix A 2) Compute the SVD A = UDVT SVD Decomposition N r A (rxN) = Vector aj of elements of distance matrix of jth conformation (j = 1 to N) U (rxr) D (rxr) Diagonal matrix Orthonormal (rotation) matrix VT (rxN) SVD Decomposition N r A (rxN) = Vector aj of elements of distance matrix of jth conformation (j = 1 to N) U (rxr) s1 s2 0 0 sr VT (rxN) Diagonal matrix s1 s2 ... sr 0 (singular values) Orthonormal (rotation) matrix SVD Decomposition N r A (rxN) = Vector aj of elements of distance matrix of jth conformation (j = 1 to N) U (rxr) VT D (rxr) (rxN) vjT vkT Diagonal matrix Orthonormal (rotation) matrix Matrix with orthonormal rows vi and vj are orthogonal unit Nx1 vectors SVD Decomposition N A (rxN) r = U (rxr) D (rxr) y X VT (rxN) Representation of A in space (X,Y) 1 dRMSDm (c,c )= ai -aj i j r Y does not depend on the coordinate system! r-dimensional space r 1 n n 1 2 m m x 2 SVD Decomposition N r s1 s2 A (rxN) = U (rxr) v1 T v2 T s3 sr D (rxr) VT (rxN) ||s1v1|| ||s2v2|| ... SVD Decomposition N r s1 s2 A (rxN) = U (rxr) v1 T v2 T s3 sr vpT D (rxr) VT (rxN) p principal components SVD Decomposition N r s1 A (rxN) = U (rxr) v1 T v2 T s2 sp vpT 0 D (rxr) VT (rxN) p principal components Further Reduction for dRMSD 1) Stack m-averaged distance matrices as vectors of a matrix A 2) Compute the SVD A = UDVT 3) Project onto p principal components Correlation between dRMSD and dRMSD4PC dRMSD is PC 4 reduced to summing up 12 to 20 terms (instead of ~ 80 to 200, since the proteins have 54 to 76 amino acids) Complexity of SVD SVD of rxN matrix, where N > r, takes O(r2N) time Here r ~ (n/m)2 So, time complexity is O(n4N) Would be too costly without m-averaging Evaluation for 1CTF Decoy Sets [Lotan and Schwarzer, 2003] N = 100,000, k = 100, 4-averaging, 16 PCs 70% correct, with furthest NN off by 20% Brute-force: 84 h Brute-force + m-averaging: 4.8 h Brute-force + m-averaging + PC: 41 min kD-tree + m-averaging + PC: 19 min Speedup greater than x200 6k approximate NNs contain all true k NNs Use m-averaging and PC reduction as fast filters