/* Algorithm:
* 1. Calculate all of the interatomic distances between the proteins in one structure
* and lable these Dij, Eij for the other
* 2. Wij = 1 - ((min(Dij, Eij) / Doff)^2) if min(Dij, Eij) < Doff
* Wij = 0 if min(Dij, Eij) >= Doff
* 3. Score function:
* Q = Sum(Wij * exp(-k|Dij - Eij|)) / sum(Wij) (k = 0.5)
* Q is close to 1 if two structures are similar.
*/
/// <summary>
/// weighted Q function
/// Input: two structures with exactly same sequences
/// but may exist missing residues
/// Output: weighted Q score
/// </summary>
/// <param name="interChain1"></param>
/// <param name="interChain2"></param>
public double WeightQFunc(ChainContact interfaceChain1, ChainContact interfaceChain2)
{
// match residue sequence numbers
// in case there are missing residues in coordinates in a protein
double weightDistSum = 0.0;
double weightSum = 0.0;
Dictionary <string, AtomPair> atomContactHash1 = interfaceChain1.ChainContactInfo.atomContactHash;
Dictionary <string, AtomPair> atomContactHash2 = interfaceChain2.ChainContactInfo.atomContactHash;
foreach (string conKey in atomContactHash1.Keys)
{
AtomPair atomPair = (AtomPair)atomContactHash1[conKey];
double dist1 = atomPair.distance;
if (atomContactHash2.ContainsKey(conKey))
{
double dist2 = ((AtomPair)atomContactHash2[conKey]).distance;
double weightTemp = Math.Pow((Math.Min(dist1, dist2) / AppSettings.parameters.contactParams.cutoffResidueDist), 2);
double weight = Math.Pow(1 - weightTemp, 2);
weightSum += weight;
weightDistSum += weight * Math.Exp(Math.Abs(dist1 - dist2) * -0.5);
}
}
if (weightSum == 0)
{
return(-1.0);
}
return(weightDistSum / weightSum);
}
/// <summary>
/// get the interfaces in an ASU
/// </summary>
/// <param name="asuChainsHash"></param>
public InterfaceChains[] GetInterfacesInAsu(string pdbId, Dictionary <string, AtomInfo[]> asymUnit)
{
// build trees for the biological unit
Dictionary <string, BVTree> asuChainTreesHash = BuildBVtreesForAsu(asymUnit);
// calculate interfaces
List <InterfaceChains> interfaceList = new List <InterfaceChains> ();
List <string> chainAndSymOpList = new List <string> (asuChainTreesHash.Keys);
chainAndSymOpList.Sort();
int interChainId = 0;
for (int i = 0; i < chainAndSymOpList.Count - 1; i++)
{
for (int j = i + 1; j < chainAndSymOpList.Count; j++)
{
ChainContact chainContact = new ChainContact(chainAndSymOpList[i].ToString(), chainAndSymOpList[j].ToString());
ChainContactInfo contactInfo = chainContact.GetChainContactInfo((BVTree)asuChainTreesHash[chainAndSymOpList[i]],
(BVTree)asuChainTreesHash[chainAndSymOpList[j]]);
if (contactInfo != null)
{
interChainId++;
InterfaceChains interfaceChains = new InterfaceChains(chainAndSymOpList[i].ToString(), chainAndSymOpList[j].ToString());
// no need to change the tree node data
// only assign the refereces
interfaceChains.chain1 = ((BVTree)asuChainTreesHash[chainAndSymOpList[i]]).Root.CalphaCbetaAtoms();
interfaceChains.chain2 = ((BVTree)asuChainTreesHash[chainAndSymOpList[j]]).Root.CalphaCbetaAtoms();
interfaceChains.interfaceId = interChainId;
interfaceChains.firstSymOpString = (string)chainAndSymOpList[i];
interfaceChains.secondSymOpString = (string)chainAndSymOpList[j];
interfaceChains.seqDistHash = chainContact.ChainContactInfo.GetBbDistHash();
interfaceList.Add(interfaceChains);
//chainContact = null;
}
}
}
// return interface chains
return(interfaceList.ToArray());
}
/// <summary>
///
/// </summary>
/// <param name="interChain1"></param>
/// <param name="interChain2"></param>
/// <returns></returns>
public double CountQFunc(ChainContact contactInfo1, ChainContact contactInfo2)
{
int matchPairNum = 0;
Dictionary <string, AtomPair> contactHash1 = contactInfo1.ChainContactInfo.atomContactHash;
Dictionary <string, AtomPair> contactHash2 = contactInfo2.ChainContactInfo.atomContactHash;
int totalPairNum = contactHash1.Count + contactHash2.Count;
foreach (string seqIdPair in contactHash1.Keys)
{
if (contactHash2.ContainsKey(seqIdPair))
{
matchPairNum++;
// for those pairs in both structures
// only count once
totalPairNum--;
}
}
if (totalPairNum == 0)
{
return(-1.0);
}
return((double)matchPairNum / (double)totalPairNum);
}
