Canal-U

4.7. Alignment costs

We have seen how we can compute the cost of the path ending on the last node of our grid if we know the cost of the sub-path ending on the three adjacent nodes. It is time now to see more deeply why these costs are used to compute the cost in the last node. So again, we saw how we can compute the cost here of the path ending on that node if we know the cost of the sub-path ending on these three red nodes. Indeed, if we come from that node, the cost on that node will be the cost of that node plus the cost of the gap that is an insertion cost. If we come from that node, the situation is similar, the cost of that node will be the cost of thatone plus the same insertion cost. If we come from that node, thecost here will be the sum of the cost of that node plus the code of the substitution of the letter N of the sequence by letter M of the second sequence. And this cost is known withinthe substitution matrix, here, which tells for each pair of characters, here in the four letter alphabet, the cost of substituting oneletter, one nucleotide by another one. So, now since we want to havehere the minimal cost what we do is we take, we return as the value of the cost here the minimum of these three expressions.