Multiple Sequence Alignment Mult-Seq-Align allows to detect similarities which cannot be detected with Pairwise-Seq-Align methods. Detection of family characteristics. Three questions: 1.Scoring 2.Computation of Mult-Seq-Align. 3.Family representation.

Multiple Sequence Alignment

Example of MSA (Multiple Sequence Alignment)

Scoring: SP (sum of pairs) SP – the sum of pairwise scores of all pairs of symbols in the column. ρ 3 (-,A,A) = (-,A)+(-,A)+(A,A) SP Total Score = Σ ρ i Here, we will assume that: (-,-) = 0

Induced pairwise alignment Induced pairwise alignment or projection of a multiple alignment. a(S 1, S 2 ) a(S 2, S 3 ) a(S 1, S 3 ) (-,-) = 0 SP Total Score = Σ i<j score[ a(S i, S j ) ]

Dyn.Prog. Solution

Dynamic Programming Solution The best multiple alignment of r sequences is calculated using an r- dimensional hyper-cube The size of the hyper-cube is O( Πn i ) Time complexity O(2 r n r ) * O( computation of the ρ function ). Exact problem is NP-Hard (metrics: sum-of-pairs or evolutionary tree). more efficient solution is needed

Multiple Alignment from Pairwise Alignments ? Problem: The best pairwise alignment does not necessary lead to the best multiple alignment.

Pattern-APattern-X Pattern-APattern-X Pattern-B Pattern-XPattern-B Pattern-D S1 S3 S2 S1S2S1S3S2S3 Pattern-APattern-BPattern-D Empty Correct Solution S1S2S3 Pattern-X

Center Star Alignment S1S1 S2S2 S3S3 SkSk ScSc S k-1 S k-2 (a)Scoring scheme – distance. (b)Scoring scheme satisfies the triangle inequality: for any character a,b,c dist(a,c) ≤ dist(a,b) + dist(b,c) (in practice not all scoring matrices satisfy the triangle inequality) (c) D(S i, S j ) – score of the optimal pairwise alignment. (d) D(M) = Σ i<j a M (S i, S j ) – score of the multiple alignment M. (e) a M (S i, S j ) – pairwise alignment/score induced by M.

S1S1 S2S2 S3S3 SkSk ScSc S k-1 S k-2 The Center Star Algorithm: (a) Find S c minimizing Σ i  c D(S c, S i ). (b) Iteratively construct the multiple alignment M c : 1. M c ={S c } 2. Add the sequences in S\{S c } to M c one by one so that the induced alignment a Mc (S c, S i ) of every newly added sequence S i with S c is optimal. Add spaces, when needed, to all pre-aligned sequences. Running time: * O(n 2 ). AC-BC DCABC AC--BC DCAAB C AC--BC DCA-BC DCAAB C

D(M c ) is at most twice the score of the D(M opt ) D (M c ) / D (M opt ) ≤ 2(k-1)/k ( < 2 ) Proof: (a) a(S i, S j ) ≥ D (S i, S j ) (any induced align. is not better than optimal align.) a Mc (S c, S j ) = D (S c, S j ) (b) a Mc (S i, S j ) ≤ a Mc (S i, S c ) + a Mc (S c, S j ) = D (S i, S c ) + D (S c, S j ) (follows from the triangle inequality) (c) 2 D(M c ) = Σ i=1..k Σ j=1..k,j  i a Mc (S i, S j ) ≤ Σ i=1..k Σ j=1..k,j  i ( a Mc (S i, S c ) + a Mc (S c, S j ) )= 2(k-1) Σ j  c a Mc (S c, S j ) = 2(k-1) Σ j  c D(S c, S j )

(d) k Σ j=1..k,j  c D(S c, S j ) = Σ i=1..k Σ j=1..k,j  c D(S c, S j ) ≤ Σ i=1..k Σ j=1..k,j  i D(S i, S j ) ≤ Σ i=1..k Σ j=1..k,j  i a Mopt (S i, S j ) = 2 D(M opt ) (e) → 2 D(M c ) ≤ 2(k-1) Σ j  c D(S c, S j ) k Σ j  c D(S c, S j ) ≤ 2 D(M opt ) → D(M c )/(k-1) ≤ Σ j  c D(S c, S i ) Σ j  c D(S c, S i ) ≤ 2 D(M opt )/k → D (M c ) / D (M opt ) ≤ 2(k-1)/k