Grafting – preparing the co-complex

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  1. The carbohydrate residues must be in the HETATM section. This is the default for files taken from the PDB.
  2. The carbohydrate molecule must contain “whole” residues. If a residue is missing the reducing oxygen atom (the O1) grafting will attempt to graft with the beta version, and provide the alpha version for you to test. However, grafting will not continue if a residue is missing any of the other atoms necessary for determining the monosaccharide. For example, a GlcNAcβ that is missing the O6 atom. These types of structures can occur in the PDB for a variety of reasons. If you know what the residue should be, in most cases it will be possible for you to build a reasonable guesstimate of the missing atoms’ position. Example structures can be found in the Cambridge structural database, the PDB, or by using the carbohydrate builder on

Known Issues:

  1. Carbohydrate residue numbers must be unique. We plan to fix this, but for now please manually renumber the carbohydrate residues so that each residue has a unique id.

Further cautions:

  1. The carbohydrate must be a minimal binding determinant. This is a key concept! If you have a co-complex with a large N-glycan, but really the protein has affinity for just a disaccharide, the grafting predictions will fail to reproduce the experiment. It will just graft glycans that contain the large N-glycan structure. Similarly, having something that is just a part of the MBD will cause a mismatch between the predictions and experiment. Interestingly, you can repeat grafting with structures of lots of putative MBDs, and use agreement with experiment as the criteria for selecting MBD(s).
  2. Some PDB files contain multiple copies of the same complex: Screenshot from 2016-01-22 12:30:02Grafting will still be performed, but neighboring molecules may prevent binders from fitting in the binding site. This is common in crystal structures as the unit cell can contain the same molecule in a different orientation. Just delete down to one molecule.
  3. If multiple carbohydrate molecules are present, only one will be grafted to. In the example below, only one of the red carbohydrate molecules would be grafted to.Screenshot from 2016-01-22 12:33:40
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