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1182: 55 Residue Asymmetric Dimer Design: HBond Networks

Closed since over 6 years ago

Intermediate Overall Design


January 14, 2016
Max points

This puzzle has two 55 residue chains, but they are not symmetric; you have 110 independent residues to work with! The Hbond Network Filter encourages players to bury satisfied Hbond networks at the dimer interface, but note that no more than four inter-molecular Hbonds may contribute to the filter! There are several other filters in effect; see the puzzle comments for details. The Baker Lab will run folding predictions on your solutions for this puzzle, and those that perform well will be synthesized in the lab. Remember, you can use the Upload for Scientists button for up to 5 designs that you want us to look at, even if they are not the best-scoring solutions!

This puzzle will be online for two weeks.

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bkoep Staff Lv 1

Residue IE Score: Monitors that all PHE, TYR, and TRP residues are scoring well.

HBond Network: Rewards networks comprising at least three hydrogen bonds. Up to 4 hydrogen bonds may span the dimer interface. In addition, at least 75% of all polar atoms in a network must make hydrogen bonds.

Core Existence: Complex: Ensures that at least 34 residues are buried in the core of the entire complex (including the interface between monomer units).

jeff101 Lv 1

Having an asymmetric dimer made from two different 55-residue monomers is
like having a puzzle with 110 residues. Having a symmetric dimer made from
two identical 55-residue monomers is more like having a puzzle with only
55 residues. Maybe this is why they have given us two weeks for this puzzle.

retiredmichael Lv 1

will be interested to see if anyone was able to get an hbond on the interface, all of mine were on the outside, wondering if the filter was to tight to allow for the bigger side chains to fit