Crystallography -- lecture 21

1. Crystallography -- lecture 21 Sidechain chi angles Rotamers Dead End Elimination Theorem

2. Sidechain space is discrete, almost A random sampling of Phenylalanine sidechains, when superimposed, fall into three classes: rotamers. This simplifies the problem of sidechain modeling.All we have to do is select the right rotamers and we're close to the right answer.

3. Rotamer

4. N N N CG CG CG CA CA CA H H H CB CB CB H H H O=C O=C O=C H H H What determines rotamers 3-bond or 1-4 interactions define the preferred angles, but these may differ greatly in energy depending on the atom groups involved. "m" "p" "t" -60° gauche 180° anti/trans +60° gauche

5. Measure a rotamer

6. Rotamer Libraries Rotamer libraries have been compiled by clustering the sidechains of each amino acid over the whole database. Each cluster is a representative conformation (or rotamer), and is represented in the library by the best sidechain angles (chi angles), the "centroid" angles, for that cluster. Two commonly used rotamer libraries: *Jane & David Richardson: http://kinemage.biochem.duke.edu/databases/rotamer.php Roland Dunbrack: http://dunbrack.fccc.edu/bbdep/index.php

7. Protein cores are tightly packed The environment of a buried leucine in 1A07. The interior of a protein is tightly packed. Bad packing produces voids or collisions.

8. Sidechain modeling Given a backbone conformation and the sequence, can we predict the sidechain conformations? ≠ Energy calculations are sensitive to small changes. So the wrong sidechain conformation will give the wrong energy.

9. Goal of sidechain modeling Given the sequence and only the backbone atom coordinates, accurately model the positions of the sidechains. fine lines = true structure think lines = sidechain predictions using the method of Desmet et al. Desmet et al, Nature v.356, pp339-342 (1992)

10. Dead end elimination theorem • There is a global minimum energy conformation (GMEC), where each residue has a unique rotamer. • In other words: GMEC is the set of rotamers that has the lowest energy. • Energy is a pairwise thing. Total energy can be broken down into pairwise interactions. Each atom is either fixed (backbone) or movable (sidechain). fixed-fixed fixed-movable movable-movable E is a constant, =Etemplate E depends on rotamer, but independent of other rotamers E depends on rotamer, and depends on surrounding rotamers

11. Theoretical complexity of sidechain modeling The Global Minimum Energy Configuration (GMEC) is one, unique set of rotamers. How many possible sets of rotamers are there? n1 n2 n3 n4 n5 … nL where n1 is the number of rotamers for residue 1, and so on. Estimated complexity for a protein of 100 residue, with an average of 5 rotamers per position: 5100 = 8*1069 DEE reduces the complexity of the problem from 5L to approximately (5L)2

12. Dead end elimination theorem • Each residue is numbered (i or j) and each residue has a set of rotamers (r, s or t). So, the notation ir means "choose rotamer r for position i". • The total energy is the sum of the three components: fixed-movable fixed-fixed movable-movable Eglobal = Etemplate + iE(ir) + ijE(ir,js) where r and s are any choice of rotamers. NOTE: Eglobal ≥ EGMEC for any choice of rotamers.

13. Dead end elimination theorem • If ig is in the GMEC and it is not, then we can separate the terms that contain ig or it and re-write the inequality. EGMEC = Etemplate + E(ig) + jE(ig,jg) + jE(jg) + jkE(jg,kg) ...is less than... EnotGMEC = Etemplate + E(it) + jE(it,jg) + jE(jg) + jkE(jg,kg) Canceling all terms in black, we get: E(ir) + j E(irjs) > E(ig) + j E(ig,js) So, if we find two rotamers ir and it, and: E(ir) + j mins E(irjs) > E(it) + j maxs E(it,js) Then ir cannot possibly be in the GMEC.

14. Dead end elimination theorem E(ir) + j mins E(irjs) > E(it) + j maxs E(it,js) This can be translated into plain English as follows: If the "worst case scenario" forrotamer t is better than the "best case scenario" for rotamer r, then you can eliminate r.

15. Exercise: Dead End Elimination Using the DEE worksheet: (1) Find a rotamer that satisfies the DEE theorem. (2) Eliminate it. (3) Repeat until each residue has only one rotamer. What is the final GMEC energy?

16. DEE exercise Three sidechains. Each with three rotamers. Therefore, there are 3x3x3=27 ways to arrange the sidechains. • Each rotamer has an energy E(r), which is the non-bonded energy between sidechain and template. • Each pair of rotamers has an interaction energy E(r1,r2), which is the non-bonded energy between sidechains. 3 a b c 1 2

17. r1 1 2 3 DEE exercise a b c a b c a b c -1 1 1 3 5 1 5 5 -1 -2 2 5 0 5 -1 0 0 0 0 0 5 0 0 0 0 0 12 E(r2) a b c 1 -1 3 5 1 5 5 1 1 -1 0 0 1 12 5 0 4 3 0 a b c r2 2 E(r1,r2) -2 0 0 2 5 0 5 -1 0 0 12 4 0 5 3 1 0 0 a b c 3 0 0 5 0 0 0 0 0 12 E(r1)

18. DEE exercise: instructions If the “best case scenario” for r1 is worse than the “worst case scenario” for r2 you can eliminate r1. (1) The best (worst) energies are found using the worksheet: Add E(r1) to the sum of the lowest (highest) E(r1,r2) that have not been previously eliminated. (2) There are 9 possible DEE comparisons to make: 1a versus 1b, 1a versus 1c, 1b versus 1c, 2a versus 2b, etc. etc. For each comparison, find the minimum and maximum energy choices of the other rotamers. If the maximum energy ofr1 is less than the minimum energy ofr2, eliminate r2. (3) Scratch out the eliminated rotamer and repeat until one rotamer per position remains.