hw4_pquestion5 - 3 with S c S c → S ∗ c S 3 → S ∗ 3...

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[Answer] The order that strings are aligned to S c doesn’t matter, because introducing spaces in any aligned pair of strings will not change the score of the final alignment. In the center-star method, the multiple alignment is constructed by successively aligning each new string to the center string S c . Consider align- ing 3 strings S 1 , S 2 , and S 3 . Let S 1 be the center string S c . If we align S 2 and S c first, we have: step 1: align S c with S 2 S c → S c S 2 → S 2 . step 2: align S c with S 3 S c → S ′′ c S 3 → S ′′ 3 S 2 → S ′′ 2 If we align S 3 and S c first, we have: step 1: align S
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Unformatted text preview: 3 with S c S c → S ∗ c S 3 → S ∗ 3 . step 2: align S 2 with S ∗ c S ∗ c → S ∗∗ c S 2 → S ∗∗ 2 S ∗ 3 → S ∗∗ 3 Each time we only align a string not yet included in the alignment, we need to add spaces into all aligned strings to pad them up to the same length as the new alignment. The padding rule is that two opposing spaces have zero scores as we consider them a match. That is, the space-padding step does not affect the multiple alignment score. Therefore, the order does not affect the final alignment....
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This note was uploaded on 10/01/2009 for the course CS BCB/Co taught by Professor Olivereulenstein during the Fall '06 term at Iowa State.

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