<p>Given the PDB or mmCIF information of a homomeric protein assembly whose global symmetry is unknown, how might one design an algorithm to find which point group symmetry class it belongs to (cyclic <span class="math-container">$C_n$</span>, dihedral <span class="math-container">$D_n$</span>, or neither) without requiring manual inspection?</p>
<p>One idea I had for detecting cyclic symmetry was to calculate the centroid of each subunit in the homomer. If there indeed is some cyclic symmetry, the axis must pass through the global centroid, and it must be normal to the plane formed by the individual subunit centroids. To test if this axis is indeed valid, one can rotate the entire protein about the axis by 2π / n, and check that the atomic coordinates align closely with those of the original (using RMSD for example).</p>
<p>But I'm having trouble thinking about dihedral symmetry, specifically for simple homomers. Any help on this would be appreciated.</p>

CC BY-SA 4.0
2024-03-20T21:20:56.917Z
https://core.ac.uk/download/pdf/190349304.pdf PROSHADE for a coordinate input file , calculates the theoretical map neverthless; there should bee a reason for that approach ...
pippo1980
CC BY-SA 4.0
2024-03-20T21:26:08.330Z
what if you have cyclic symmetry and only two subunits :  it must be normal to the plane formed by the individual subunit centroids , my geometry is really bad but how many plane passes for 2 points ?
pippo1980
CC BY-SA 4.0
2024-03-20T21:29:53.627Z
copying from PROSHADE : Following from the definition of the chiral dihedral symmetry, a simple al-
gorithm can be devised to detect them from the already available list of detected
cyclic symmetries. Specifically, the list of the cyclic symmetries detected by the
algorithm described in the section 4.3 can be searched for any pairs of symmetries
with symmetry axes perpendicular to each other. The resulting list of symme-
try pairs then, by definition, is the list of all detected chiral dihedral symmetries
present in the structure.
pippo1980
<protein-structure><pdb>
Protein symmetry classification from PDB data
# Protein symmetry classification from PDB data

(Asked by: username_0 on 2024-03-20)

Given the PDB or mmCIF information of a homomeric protein assembly whose global symmetry is unknown, how might one design an algorithm to find which point group symmetry class it belongs to (cyclic $C\_n$, dihedral $D\_n$, or neither) without requiring manual inspection?

One idea I had for detecting cyclic symmetry was to calculate the centroid of each subunit in the homomer. If there indeed is some cyclic symmetry, the axis must pass through the global centroid, and it must be normal to the plane formed by the individual subunit centroids. To test if this axis is indeed valid, one can rotate the entire protein about the axis by 2π / n, and check that the atomic coordinates align closely with those of the original (using RMSD for example).

But I'm having trouble thinking about dihedral symmetry, specifically for simple homomers. Any help on this would be appreciated.

**Tags:** protein-structure, pdb
> **Comments:**
> **username_1**: https://core.ac.uk/download/pdf/190349304.pdf PROSHADE for a coordinate input file , calculates the theoretical map neverthless; there should bee a reason for that approach ...  
> **username_1**: what if you have cyclic symmetry and only two subunits :  it must be normal to the plane formed by the individual subunit centroids , my geometry is really bad but how many plane passes for 2 points ?  
> **username_1**: copying from PROSHADE : Following from the definition of the chiral dihedral symmetry, a simple al- gorithm can be devised to detect them from the already available list of detected cyclic symmetries. Specifically, the list of the cyclic symmetries detected by the algorithm described in the section 4.3 can be searched for any pairs of symmetries with symmetry axes perpendicular to each other. The resulting list of symme- try pairs then, by definition, is the list of all detected chiral dihedral symmetries present in the structure.  

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## 0 answers

