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Bioinformatics 2 -- Lecture 3

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Bioinformatics 2 -- Lecture 3 Bioinformatics 2 -- lecture 3 Rotation and translation Superposition, structure comparison Structure classification Rotation is angular addition y atom starts at (x=|r|cosα, y=|r|sinα) axis of (x’,y’) rotation = Cartesian β (x,y) origin r α x ..rotates to... (x'=|r|cos(α+β), y'=|r|sin(α+β)) Convention: angles are measured counter-clockwise. 1 Sum of angles formula cos (α+β) = cos α cos β − sin α sin β sin (α+β) = sin α cos β + sin β cos α A rotation matrix y α x = |r|cos (x’,y’) y = |r|sin α β (x,y) r α x x' = |r| cos (α+β) y' = |r| sin (α+β) = |r|(cos α cos β − sin α sin β) = |r|(sin α cos β + sin β cos α) = (|r| cos α) cos β − (|r| sin α)sin β = (|r| sin α) cos β + (|r| cos α) sin β = x cos β − y sin β = y cos β + x sin β x' cosββ− sin r cosα cosββ− sin x = = y' sinββ cos rsinα sinββ cos y rotation matrix is the same for any r, any α. 2 A rotation around a principle axis The Z coordinate stays the same. X and Y change. cosββ− sin 0 R = sinββ cos 0 z 001 The Y coordinate stays the same. X and Z change. cosγγ0 sin R = 010 y −sinγγ0 cos The X coordinate stays the same. Y and Z change. 10 0 0 cosαα− sin Rx = 0 sinαα cos A 3D rotation matrix Is the product of 2D rotation matrices. cosββ− sin 0cosγγ0 − sin cosβγ cos− sin β cos β sinββ cos 0 01 0 = sinβγ cos cos β− sin βγ sin 001 sinγγ00 cos sinγγ cos Rotatio 3D rotati n around z n around y- on -axis axis 3 What happens when you move the mouse: 1. Mouse sends mouse coordinates (∆x,∆y) to the running program y 2. Rotation angles are calculated: x θ ∆ θ ∆ x = x*scale, y = y*scale 3. Rotation matrices are calculated: 10 0 θθ− cosyy0 sin =− θθ R 0 cosxx sin = 01 0 x Ry 0 sinθθ cos θθ xx sinyy0 cos What happens when you move the mouse (cont'd): 4. New atomvv coordinates are calculated rr' = RR yx 5. The scene is rendered using the new coordinates. 4 When multiplying matrices, the order matters. θθ− 10 0 cosyy0 sin = 01 0 0 cosθθ− sin RRyx xx θθ θθ sinyy0 cos 0 sinxx cos θθθθθ−− cosyxyyx sin sin sin cos = θθ− 0 cosxx sin θθθθθ sinyxyxy sin cos cos cos This is the matrix if the X-rotation is first, then the Y-rotation. Rotating in opposite order gives a different matrix θθ− 10 0 cosyy0 sin =−0 cosθθ sin 01 0 RRxy xx θθ θθ 0 sinxx cos sinyy0 cos θθ− cosyy0 sin = −−θθ θ θ θ sinxy sin cos x sin x cos y θθ θ θθ sinyx cos sin x cos xy cos 5 Reversing the rotation For the opposite rotation, flip the matrix. AB T AC This is the “transpose” = CD BD x' cosββ sin x = y' −sinββ cos y The inverse matrix = The transposed matrix. cosββ sin cosββ− sin 10 = −sinββ cos sinββ cos 01 NOTE: cosβ cosβ + sinβ sinβ = 1 Example: rotation in 2 steps Rotate the vector v=(1.,2.,3.) around Z by 60°, then around Y by -60° °− ° −+− cos60 sin 60 01 1(.) 0 5 2 (. 0 866 ) 3 () 0 1. 232 °° = ++ = sin60 cos 60 0 2 1(. 0 866 ) 2 (.) 0 5 3 () 0 1. 866 001 3 0031++() 3 cos60°−° 0 sin 60 −1. 232 −+−1.(.).()(.) 232 0 5 1 866 0 3 0 866 −3. 214 01 0 1. 866 = −++1.().()() 232 0 1 866 1 3 0 = 1. 866 sin60°° 0 cos 60 3 −++1.(.).()(.) 232 0 866 1 866 0 3 0 5 0. 433 6 Right-handed 90° rotations: z 90° rotation around 10 0 00− 1 X y 01 0 x x 001 010 Y z − 100 y 010− y Z 100 x 001 z Helpful hint: For a R-handed rotation, the -sine is up and to the right of the +sine. In class exercise: rotate a point (x,y,z) = (1., 4., 7.) Rotate this point by 90° around the Z-axis Then... Rotate the new point by 90° around the Y-axis. What are the new coordinates? 7 3D rotation conventions: α β γ axis of Euler angles, z’’ x’ z rotation: cosγγ− sin 010 0cosαα− sin 0 sinγγ cos00 cos ββ− sin sinαα cos 0 ββ Order of 0010 sin cos 001 rotations: 3 2 1 Each rotation is around a principle axis. Polar angles, φψκ z’’’’ y’’’ z’’-y’ -z cosφφ− sin 0cosϕϕ0 − sin cosκκ− sin 0 cosϕϕ0 sin cosφφ sin 0 sinφφ cos 0 01 0 sinκκ cos 0 010−sinφφ cos 0 001 sinϕϕ0 cos 001−sinϕϕ0 cos 001 5 4 3 2 1 Net rotation = κ, around an axis axis defined by φ and ψ Polar angles z = north pole ψ κ y φ x = prime meridean @ equator Rotation of κ degrees around an axis axis located at φ degrees longitude and ψ degrees latitude 8 Special properties of rotation matrices •They are square, 2x2 or 3x3(higher dimensions in principle) •The product of any two rotation matrices is a rotation matrix. •The inverse equals the transpose, R-1 = RT •Every row/column is a unit vector. •Any two rows/columns are orthogonal vectors. •The cross-product of any two rows equals the third. •|x| = |Rx|, where R is a rotation matrix. Read more about rotation matrices at: http://mathworld.wolfram.com/RotationMatrix.html RMSD Root Mean Square Deviation in superimposed coordinates is the standard measure of structural difference. Similar to “standard deviation” which is the square-root of the variance. Orengo p.88 r r ()− 2 ∑ xyii iN=1, N Where xi are the coordinates from molecule 1 and yi are the equivalent* coordinates from molecule 2. *Which atoms are equivalent is based on an alignment. 9 pseudo-pseudocode program for computing rmsd sum = 0 N = 0 while <there is data to read> { /** Read coordinates for two aligned positions. read (x1,y1,z1) , (x2,y2,z2) /** Compute the distance2 2 2 2 d =(x1-x2) +(y1-y2) +(z1-z2) /** sum it. sum = sum + d /** keep track of how many pairs there are. N = N + 1 } /** Average and take square root. rmsd = sqrt(sum/N) Least squares superposition Problem: find the rotation matrix, M, and a vector, v, that minimize the following quantity: +−2 ∑ Mxii v y i Where xi are the coordinates from one molecule and yi are the equivalent* coordinates from another molecule. *equivalent based on alignment 10 Mapping structural equivalence = aligning the sequence Any position that is aligned is included in the sum of squares. 4DFR:A ISLIAALAVDRVIGMENAMPWNLPADLAWFKRNTLDKPVIMGRHTWESIG-RPLPGRKNI 1DFR:_ TAFLWAQNRNGLIGKDGHLPWHLPDDLHYFRAQTVGKIMVVGRRTYESFPKRPLPERTNV 4DFR:A ILSSQ-PGTDDRVTWVKSVDEAIAAC--GDVPEIMVIGGGRVYEQFLPKAQKLYLTHIDA 1DFR:_ VLTHQEDYQAQGAVVVHDVAAVFAYAKQHLDQELVIAGGAQIFTAFKDDVDTLLVTRLAG 4DFR:A EVEGDTHFPDYEPDDWESVFSEFHDADAQNS--HSYCFKILERR 1DFR:_ SFEGDTKMIPLNWDDFTKVSSRTVEDT---NPALTHTYEVWQKK Unaligned positions are not. Least squares At the position of best superposition, +≅ we have an approximate equality: Mxii v y First we eliminate v by translating the center of mass of both molecules to the origin.Now Mx''≅ y we have: ii We have one equation (i) for each atom, M has 9 unknowns. If there are more equations than unknowns, there is a unique solution. What is it? See the next two slides. 11 Least squares Least squares solves a set of a linear equations in the form : x11a1+x12a2+ … x1NaN = y1 M x x21a1+x22a2+ … x2NaN = y2 • • • xM1a1+xM2a2+ … xMNaN = yM s (M > N) known coefficientsknown N value This is 'shorthand' unknowns (N) notation for the a equations. x = y matrix vector vector Least squares, continued a Green elements are known. x = y Orange are unknown. Fat Rectangles are matrices. Thin rectangles are vectors. Multiply both sides by xT a = xT x y transpose of x. "Squaring" "Squared" matrix is now, uh, T y square. Which means we can x x a T = x invert it. (We can use the "LU decomposition" method.) Multiplying both sides by the y T -1 T a = (x x) x inverse of "squared" matrix solves for a. Summary: a = (xTx)-1 xTy 12 least-squares superimposed molecules In class exercise: Superimpose 2 molecules by hand •Open insightII •Open two PDB files: 3shd.pdb, 1jzm.pdb •Display first chain ("A" usually) as "trace only" •Choose a better color if necessary. •Adjust the depth queuing if necessary. •Superimpose the two molecules Use the reference pages that follow. Do these pairs: 3sdh.pdb 1jzm.pdb (hard) 3sdh.pdb 1h97.pdb (harder) 3sdh.pdb 1phn.pdb (hardest) 13 InsightII reference: getting a molecule: Molecule-->Get--> Getfile type=PDB Filename= Trace of chain A only: Molecule-->Display--> Display operation=Only Molec. attrib.=Atoms Atom set=Trace Molec. Spec= :A* left-click on a molecule to see the object name chain ID=A atom spec. Look for chain ID, if any. InsightII reference: invis. dark change depth queuing: slab hit F12 bright click here slab thick invis. viewer Color molecule: Molecule-->Color--> Attribute=Atoms Molec. Spec= Atom set=specified Color=<select> object name Delete objects: type: Delete Object or Delete Object * 14 InsightII reference: How to superimpose two molecules: •Select first molecule, hit F10 and click on it. Only first molecule will move. Move it to superimpose on 2nd molecule. •Select "world" ( hit F11 ) Rotate both together to get a different view. •Select either molecule (F10 + click) Refine the superposition. Repeat until converged. Note: you can connect to a molecule by typing: connect XXX where XXX is the "object name" for your molecule.
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