1. Short fast history of protein design Site-directed mutagenesis -- protein engineering (J. Wells, 1980's) Coiled coils, helix bundles (W. DeGrado, 1980's-90's) Extreme protein stabilization (S. Mayo, 1990's) Binding pocket design (H. Hellinga, 2000) New fold design (B. Kuhlman, 2002-4) Protein-protein interface design (J. Gray, 2004) Experimental (non-computational) approaches: in vitro evolution phage display **Other names in protein design: Hill, Vriend, Regan, D. Baker, Richardson, Dunbrack, Choma, several more.

2. The goal of sequence design Given a desired structure, find an amino acid sequence that folds to that structure. MIKYGTKIYRINSDNSG KJHGCKAHNEEEGHA design folding To do this, we must assign an energy to each possible sequence.

3. Theoretical complexity of sequence design To design THE OPTIMAL sequence, we need the best amino acid, and its best rotamer at every position. We can treat each position as one of 193 possible rotamers. That's 191 rotamers in the Richardson library, plus Gly and Ala (which have no rotamers) How many possible sets of rotamers are there for a protein of length 100? 193 100 = 3.6*10 228 DEE reduces the complexity of sequence design to about (193L) 2 = 3.6*10 8

4. Sequence space maps to structure space..as many-to-one. This means that there is a lot of potential for "slop" in a sequence design. Moderately big sequence changes are possible, and the sequence can still fold to the same general structure. sequence families fold Good news for protein designers

5. Dead end elimination theorem E(i r ) +  j min s E(i r j s ) > E(i t ) +  j max s E(i t,j s ) This can be translated into plain English as follows: If the "worst case scenario" for t is better than the "best case scenario" for r, then you always choose t. reminder

6. DEE algorithm E(r 1 ) Find two columns (rotamers) within the same residue, where one is always better than the other. Eliminate the rotamer that can always be beat. (repeat until only 1 rotamer per residue)

7. DEE with alternative sequences abcabc 1 2 3 -111 351 55-1 -2 1 5 2 0 5 -1 2 0 0 0 3 0 0 1 1 12 5 0 -3 4 3 0 1 -135 155 11-1 -200 250 5-10 223 0124 053 100 1-31 r1r1 r2r2 E(r 1,r 2 ) 1 2 3 21 3 abcabc abcabc abababab abcabcabcabcab a b 00500 000122 0 5 0 12 2 E(r 2 ) E(r 1 ) Asp Leu “Rotamers” within the DEE framework can have different atoms. i.e. they can be different amino acids. Using DEE, we choose the best set of rotamers. Now we have the sequence of the lowest energy structure. In the example, we have D or L at position 3.

8. Sequence design using DEE Selected residues (or all) are chosen for mutating. Selected (or all) amino acids are allowed at those positions. For the selected amino acids, all rotamers are considered. Now "rotamer" comes to mean the amino acid identity and its conformation. Since there are as many as 193 rotamers in the rotamer library for all amino acids, each selected position can have as many as 193 "rotamers." If "fine grained" rotamers are used, this number may be much larger.

9. DEE with alternative sequences and ligands abcabc L 2 3 -111 351 55-1 -2 1 5 2 0 5 -1 2 0 0 0 3 0 0 1 1 12 5 0 -3 4 3 0 1 -135 155 11-1 -200 250 5-10 223 0124 053 100 1-31 r2r2 E(r 1,r 2 ) L 2 3 21 3 abcabc abcabc abababab a bc a b 00500 000122 0 5 0 12 2 E(r 2 ) E(r 1 ) Asp Leu Ligands can have multiple conformations and locations within the active site. In DEE, each position of the ligand is another “rotamer”, i.e. another row and column in the DEE matrix. Ligand conformers. r1r1

10. 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. ≠

11. Goal of sidechain modeling Desmet et al, Nature v.356, pp339-342 (1992) 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.

12. 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.

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

14. 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 *rotamers of W on the previous page are from the Richardson library.

15. 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-movablemovable-movablefixed-fixed E is a constant, =E template E depends on rotamer, but independent of other rotamers E depends on rotamer, and depends on surrounding rotamers

16. 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? n 1 n 2 n 3 n 4 n 5 … n L where n 1 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: 5 100 = 8*10 69 DEE reduces the complexity of the problem from 5 L to approximately (5L) 2

17. 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 i r means "choose rotamer r for position i". The total energy is the sum of the three components: NOTE: E global ≥ E GMEC for any choice of rotamers. E global = E template +  i E(i r ) +  i  j E(i r,j s ) where r and s are any choice of rotamers. fixed-fixed fixed-movable movable-movable

18. Dead end elimination theorem If i g is in the GMEC and i t is not, then we can separate the terms that contain i g or i t and re-write the inequality. E(i r ) +  j min s E(i r j s ) > E(i t ) +  j max s E(i t,j s ) E GMEC = E template + E(i g ) +  j E(i g,j g ) +  j E(j g ) +  j  k E(j g,k g ) E notGMEC = E template + E(i t ) +  j E(i t,j g ) +  j E(j g ) +  j  k E(j g,k g )...is less than... E(i r ) +  j E(i r j s ) > E(i g ) +  j E(i g,j s ) Canceling all terms in black, we get: So, if we find two rotamers i r and i t, and: Then i r cannot possibly be in the GMEC.

19. Dead end elimination theorem E(i r ) +  j min s E(i r j s ) > E(i t ) +  j max s E(i t,j s ) If the "worst case scenario" for rotamer t is better than the "best case scenario" for rotamer r, then you can eliminate r. This can be translated into plain English as follows:

20. 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?

21. DEE exercise abcabc 1 2 3 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(r 1, r 2 ), which is the non-bonded energy between sidechains.

22. DEE exercise -111 351 55-1 -225 05-1 000 001 1250 430 -135 155 11-1 -200 250 5-10 0124 053 100 r1r1 r2r2 E(r 1,r 2 ) 1 2 3 213 abcabc abcabc abcabc abcabcabcabcabcabc 0050000012 0 5 0 10 E(r 2 ) E(r 1 )

23. DEE exercise: instructions (1) The best (worst) energies are found using the worksheet: Add E(r 1 ) to the sum of the lowest (highest) E(r 1,r 2 ) 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 of r 1 is less than the minimum energy of r 2, eliminate r 2. (3) Scratch out the eliminated rotamer and repeat until one rotamer per position remains. If the “best case scenario” for r 1 is worse than the “worst case scenario” for r 2 you can eliminate r 1.