Solutions to Exercise 4: Atomistic/Smooth Molecular Models

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1 Solutions to Exercise 4: Atomistic/Smooth Molecular Models CS384R, CAM 395T, BME 383J: Fall 2010 November 19, 2010 Question 1 Consider the atomic representations (as per lectures, and the molecular modeling book), of the twenty protein residues side-chains SC. (a) Construct one to three spherical ball (quasi-atoms) topology preserving representations SCcoarse of each of the following five protein sidechains R = leu, pro, lys, tyr, trp. (b) Give the location and radii of the quasi-atoms such that any Hausdorff distance between R and Rcoarse is minimal. [Note, the Hausdorff distance is defined in the mol. modeling book chap. 2] Figure 1: List of twenty amino acids Soln. (a) Though the twenty amino acides (shown in Figure 1) differ in size and topology, there are certain similarities. We can categorize the side chains (or parts of them) into four distinct classes based on their topology. These classes are- 1

2 Chain: This is the most basic topology formed by a series of alkyles. For example, lys, glu, ile etc. are formed of a chain. Brach: Branch is present when a chain splits into two chains [or, three chains meet]. For example, leu, val, gln etc. Phenol: Phenol are the single loops present in tyr, phe, his. indol: Indol is a double loop present in trp. The five cases mentioned in the question were taken from different classes. Hence the techniques we describe here can be generalized to cover all twenty amino acids. (a) chain:(lys) (b) branch:(leu) (c) one ring:(tyr) (d) two rings:(trp) Figure 2: (a) Chain structure of butyl which has four alkyls. (b) Branch structure of isobutyl which has two carbon alkyls on the main branch and one alkyl on each of the two minor branches. (c) Ring structure of phenol. (d) Double ring structure of indole. For each structure (a-d), from left to right, we show how the structure is hierarchically grouped into CG beads. The purple rectangles represent the CG beads. 2

3 Figure 2 shows how each of these are clustered in several stages of coarse representations. The first stage of grouping is shown in the rightmost panel of each figure and gradually groups larger and larger clusters until it reaces a one bead model (lefttmost). The choice of the pairs to be clustered in each step have been made to preserve the topology as much as possible. Note that, for the case of lys (Figure 2(a)), even a 1 bead model is topology preserving, and 2 beads are more than sufficient. Though pro is quite similar in structure to lys (as both are chains), but pro has two connections to the backbone (see Figure 1). Hence, the clustering depends on the clustering of the backbone. If the C and N on the backbone is already clustered, then a 1 bead model for pro is not topology preserving. On the other hand, if the C and N on the backbone are not clustered, then a 1 bead model suffices. For branches like leu (Figure 2(b)), at least 3 beads are needed to preserve the topology. For tyr, at least 3 beads are neccessary to preserve the ring topology. The best topology preserving option is to use 5 beads altogether. If we want to use 3 beads, then joining the top and bottom beads to the nearest bead on the ring works. But in general it is found that the rings are very stable in nature and is hence considered rigid and so clustering the beads on the ring into a single bead is more biophysically justified option. The case of trp is similar. (b) Choosing the center and radii of the beads of the final coarse-grained model is a high degree optimization problem. Let, the center and radii of the final beads are {C i, R i } and the centers and radii of the initial atoms are {c j, r j }. Let, the Hausdorff distance between each pair of initial atoms and final coarse beads be defined as H ij s. Hence our optimization problem is argmin Ci,R i (max i,j H ij ) (1) However, each H ij is a quartic (Refer to the definition of Hausdorff distance from the mol. modeling book chap. 2). Hence, finding globally optimal solutions is often impossible. Hence, we can reduce the clustering problem into a series of two way merges and locally minimize the Hausdorff distance in each step. This can be repeatedly applied according to the steps outlined in part (a) to produce the center and radii of the final 2 or 3 beads. In the 2D example in Figure 3, h 1 and h 2 are one-way Hausdorff distances from the two quasiatoms to the bead representing the cluster, and h 3 and h 4 are the one-way Hausdorff distances in the other direction. Our goal to select a center C and radius r for the cluster such that max 1 i 4 h i is minimized. Without loss of generality, assume that the centers C 1 and C 2 of the two circles lie on the X axis. The dark curve which goes through the intersections of the two circles and intersects the cluster at point P, is the power diagram bisector of the two circles. Let, it intersects the X axis at the origin. Hence, assuming that the distance from the bisector to the circles along the X axis is l, the centers are at ( r 1 + l, 0) and (r 2 l, 0) where r 1 and r 2 are their radii. Clearly, to minimize both h 1 and h 2 at the same time, we should place the center of the cluster precisely at (r 2 r 1, 0). It remains to compute the radius r. By fixing the center, we can now express h 1 and h 2 as functions of r as follows h 1 (r) = h 2 (r) = r 1 + r 2 l r. Note that, h 3 and h 4 can be computed as the distance from the point P to the two circles [see mol. modeling book chap. 2 for proof]. Location of the point P can be computed by solving the following two equations- (x (r 2 r 1 )) 2 + y 2 r 2 = 0 (2) (x ( r1 + l)) 2 + y 2 r 1 = (x (r 2 l)) 2 + y 2 r 2 (3) 3

4 Figure 3: Computing the Hausdorff distance which can be solved to get the X and Y coordinates of P as functions of r as follows- (r l) 2 x 2 + 2(r 1 r 2 )[(r 1 l) 2 + (r 2 l) 2 ]x (2r 1 l)(2r 2 l)(r 1 r 2 ) 2 = 0 (4) y = ±[ x 2 + 2(r 2 r 1 )x + r 2 (r 1 r 2 ) 2 ] 1/2 (5) Now the distances h 3 = h 4 can be represented as functions of r as well and we can find the value of r by solving h 1 (r) = h 3 (r). This technique need to be repeated for each pair of (quasi)atoms clustered. Note that, during the first step of clustering, r 1, r 2 and l are fixed by looking up the bond length and radii given in Table A.4 and Table A.2 of the appendix of the mol-model book. In the next steps, the center and radii computed in previous steps are used. Bond angles given in Table A.5 is used in all but the first step to find the transformation that maps the atom centers to the X axis as required. Question 2 Consider the atomic representations of protein secondary structures (namely, α-helices and β-sheets). Construct a coarsest possible topology preserving, geometric shape representation of an average α-helix and an average β-sheet Soln. α-helix The amino acids in an average α-helix are arranged in a right-handed helical structure with each amino acid corresponding to a 100 turn in the helix and a 1.5 Å translation along the helical axis. Thus there are 13 atoms and 3.6 amino acid residues per turn, and each turn is 5.4 Å wide (see Figure 4(a)). Hence, the coarsest topology preserving shape representation of an average α-helix is thus a tubular solid. One way to represent such a surface is described here- 4

5 Figure 4: (a) Topology of an α-helix (b) A topology preserving shape representation 1. Take a curve representing the medial axis of the tube. 2. Sample points on the curve and define planes normal to the curve at those points. 3. Define circles of radius 2.7 centered at the sampled points and lying the normal planes. 4. Compute a smooth canal surface joining the circles (similar to problem 1 in the previous exercise). This representation, however, does not maintain the hollow inside the tube. To represent the hollow, we need to erect to concentric surfaces and the solid will be the spaces between these two surfaces. The inner surface is defined by sampling circles of radius equal to half of the width of an average hollow inside an α-helix, and joining them (see Figure 4(b)). So, an average helix is parameterized by the medial curve and two cross sectional radii. β-sheet The β-sheet is formed by hydrogen bonds connecting multiple parallel or anti-parallel β-strands. Each β-strand can be viewed as a helical structure with two residues per turn. The distance between two such consecutive residues is 3.47 Å in anti-parallel β-sheets and 3.25 Å in parallel β-sheets (see Figure 5). We propose to represent each β-sheet as a shell in a quad prism element (see Figure 5). To construct the prism element, we first align the medial plane of the β-sheet to the XY plane and compute a bounding box round the β-sheet. Then, the normals at the four corners are computed. These normals are extended in both direction to produce the prism scaffold. The shell surface inside is parameterized using a tensor of bi-cubic or bi-quartic with a linear kernel. See chapter 3 of the finite elements book for details on computing the coefficients of a prism element. Question 3 Consider a scalar function F defined on a two dimensional bounded domain D, and the level set curve family Q : F = c for various c R 1. Additionally, consider the region within domain D which 5

6 Figure 5: Topology of a β-sheet Figure 6: Shell representation of a β-sheet contains the gradient F evaluated at Q, as exterior to the level set Q. For the twin cases where F is a polynomial of degree d = 1, 2, and D is a triangle, derive the length of Q as a function of c. [Extra credit: For the twin cases of F over triangle D, derive the area of the interior region in D bounded by Q as a function of c.] Soln. Degree 1 For the case where F is a polynomial of degree d = 1 (i.e. a plane), let us assume that the values of F at the three endpoints of the triangle have the following relationship F (v 1 ) < F (v 3 ) < F (v 2 ). So, the level set family has length zero for the ranges c < F (v 1 ) and c > F (v 3 ). Also, the length L(Q) increases monotonically between F (v 1 ) c F (v 3 ) and decreases monotonically between 6

7 Figure 7: Length of level set family of a linear function F (v 3 ) c F (v 2 ). See Figure 7 for an example. Hence the length of the level set family Q is a piecewise linear C 0 B-spline. [Extra Credit] Figure 8: Area of level set family of a linear function Soln. The area as a function of c is simply an integral of the length within the range [F (v 1 ), c]. For this 7

8 Degree 1 particular case, the area is a cubic (see Figure 8) Figure 9: Length of level set family of a quadratic function. Blue curves on the left show the surface defined by F, red line segments are used to show the control polygon defining the surface. Green curves on the surface trace out some example level sets. The green curve on the right displays the length of the level set as a function of c. For the case where F is a polynomial of degree d = 2, we can consider the F as a B-spline surface with six knots at the endpoints and the midpoints of the edges of the triangular domain D (see Figure 9). Similar to before, we can impose some ordering on the values of F evaluated at the knots. The length of the level set curves (each curve is now quadratic), would be zero at the highest and lowest valued knots. Each level set curve is an intersection of a quadratic surface with a plane and is hence quadratic. The lengths of these curves increase/decrease between each pair of control points and is upper-bounded by the sum of the lengths of the line segments on the control polygon for the same c value. So, these are defined as piecewise functions defined within ranges specified by the values at the control points. [Extra Credit] Again, the area as a function of c is simply an integral of the length within the range [F (v 1 ), c]. Note that, within this range, there can be mutiple subranges, based on the value of c. The intergal is hence decomposed and expressed as a sum of integrals. Question 4 Consider an arrangement of n 2 charged circular disks (2D atoms) of radius r with centers on a uniform rectilinear 2D nxn grid G with grid step size l. If q is the charge density per unit area of each disk, what is the total charge density of the arrangement as a function of r? Note, that the topology/geometry of the union of the disks varies for discrete ranges of r for fixed l. Soln. The topology of the union has three distinct stages with varying r. We discuss each of the topologies below and compute the charge density, d = na0 A U where A 0 is the area of a single circle and A U is the area of the union. 8

9 Case 1: (0 < r l/2) Figure 10: 0 < r l/2 In this case (see Figure 10), the topology is a collection of disjoint circles. Hence A U = na 0 and d = 1. Case 1: (l/2 < r l/ 2) Figure 11: l/2 < r l/ 2 In this case (see Figure 11), the topology has a single connected component, but contains holes. Each of the 4 circles centered on the corners of the grid intersects 2 other circles. Each of the 4(n 2) circles having centers on the edges of the grid, but not on the corners, intersects with 3 other circles. Each of the rest ((n 2) 2 ) of the circles intersect with 4 other circles. Hence, total number of pairwise intersections = 4(n 2) (n 2) + 8 = 4n 2 4n. So, there are 2n 2 2n overlap areas. Let, the area of any of these overlap areas be A i. The charge density can now be computed as d = Case 1: (r l/ 2) na 0 na 0 2n(n 1)A i. 9

10 Figure 12: r l/ 2 In this case (see Figure 12), the topology has a single connected component, without holes. Now, the entire area inside the grid is covered and belongs to the union. With carying r, the number of intersections would vary a lot. So, instead of counting all intersections we can compute the area by subtracting the area of the shaded regions (see Figure 12 left) from the larger square with sides L = 2r + (n 1)l. Let, the area of each of the shaded regions at the corners are A j and the area of each of the smaller shaded regions be A k. Hence, A U = L 2 4A j 4(n 1)A k. The charge density can now be computed. Now we discuss the computations of A i, A j and A k. Figure 13: (a) Computing A i. (b) Computing A j. (c) Computing A k Computing A i : See Figure 13(a). c 1 A = c 1 B = c 2 A = c 2 B = r and c 1 c 2 = l. Hence, Ac 1 E = cos 1 l 2r. Let, θ = 2 Ac 1 E Hence, A i = 2 (Area of the circle under arc AB - Area of the triangle c 1 AB) = 1 2 r2 (θ sin θ). Computing A j : See Figure 13(b). A j = r 2 (1 π 4 ) Computing A k : See Figure 13(c). AB = l and AC = AE = BE = BD = r. Let θ = EAC = sin 1 AB/2 AE = sin 1 l 2r. Hence, A k = (Area of ABCD - Area of ABE - 2*Area of the circle under arc CE) = lr r 2 sinθ r 2 θ 10