How to calculate average BLOSUM62 scores?

Question

I can understand the motive behind the BLOSUM62 matrix, this being a pairwise mutation matrix describing aggregate mutations between the 20 amino acids.

However how would you calculate the average BLOSUM62 score within a hydrophobic amino acids (K, D, E) vs hydrophilic amino acids (I, L, V)?

How about between those two groups?

Answer 1

I've not done this precise calculation. I would use point 3, but its complicated to implement.

Just for information BLOSUM62 is a static matrix 20x20 representing relative amino acid mutations. However, I am not exactly sure how the matrix was calculated. The algorithm looks quite simple, then a BLOSUM62 score can be obtained by comparing the observed matrix to the static matrix, this would provide a score. The paper is here.

Maximum likelihood approaches require a phylogeny.

There are three possibilities I can think of, the first one wouldn't work under a maximum likelihood approach:

1. Remove all hydrophobic residues from the alignment, check the BLOSUM62 score, remove all hydrophilic check the score etc... again a maximum likelihood approach is not meaningful in this example, but a BLOSUM62 score might be okay;
2. Recode the amino acids to generate 6-state amino acid data, or recode hydrophobics and in a separate data set hydrophilics and leave the rest uncoded. I've never done it but at a guess I, L, V could be represented by just one amino acid, e.g. L. If the observed BLOSUM62 matrix algorithm was calculated on the observed data the new amino acid e.g. L would have an average score.
3. Alternatively, manipulate the BLOSUM62 matrix, however its no longer a BLOSUM62 matrix, but it is highly amenable to maximum likelihood analysis. Thus I <-> V and I <-> L is set to zero (or massively negatively weight them) check likelihood, then hydrophilics are set to zero check likelihood. Do the same with the observed versus BLOSUM62 matrix as the control.

Point 3 what a maximum likelihood approach would do is say how close a given matrix is to observed data set for a given tree under a likelihood probability, the smallest is the best. So if you've got loads of matrices its a really cool approach. There is a membrane protein matrix - which is likely heavily skewed towards hydrophilicity. Basically, I'd test all the matrices - but not recode or delete the data set in any way. The matrix that is closer to the data set using likelihood is the best representation of its hydrophilicity vs. hydrophobicity.

Note Just finally mention you can simply calculate the observed amino acid matrix for that data set (via a program) and then average the amino acids. That might be easier and using parsimony its very quick but you'd need a tree. The underlying metric (or metric behind the metric) in all cases is mutations.