BLAST, using Smith-Waterman in the last step and S-score

Question

In my bioinformatics course we studied the BLAST agorithm. After finding the HSP's and joining the HSPs together that are close enough and in a correct diagonal trend, we perform a next step: namely a Smith-Waterman alignment. Why do we do this? Didn't BLAST already find the alignment by joinging the HSPs?

I also have a second question about BLAST. When calculating the E-value with the formula:

$E=K \times m \times n \times e ^{\lambda S}$

Which score is this S exactly? Is it the score found for the HSP with the used substitution matrix? What is the score if we have multiple HSPs in one database entry? Is the E-value, and thus the S-score, calculated seperatly for every HSP or is it a value for the entry in total?

Answer 1

The BLAST algorithm is a heuristic to find which the path through the matrices formed between a query sequence and a target sequences from the database is ‘best’. Such paths represent an alignment which is has a completely different basis to that of the Smith-Waterman dynamic programming algorithm. Not only is it ungapped, but it based on comparison of tuples with matches scored according to frequency, and then proceeds by trying to join those with certain relative scores.

The speed of the program is partly the result of rapidly discarding unpromising paths, storing and continually refining only the current best candidates. The alignments will not be Smith-Waterman alignments, but there is no point in performing a Smith-Waterman alignment on each high-scoring sequence pair as this is not the basis of inclusion in the pool of best candidates. Performing such an alignment would be a waste of time for those candidates that will ultimately be discarded.

Only when the run is complete is it appropriate to perform the Smith–Waterman alignment on the final top candidates to produce a gapped alignment from which statistics such as the E-value are calculated (using S, the raw score from such Smith–Waterman alignments).