Introduction to Computational Biology Homework 2 Solution

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1 Itroductio to Computatioal Biology Homework 2 Solutio Problem 1: Cocave gap pealty fuctio Let γ be a gap pealty fuctio defied over o-egative itegers. The fuctio γ is called sub-additive iff it satisfies the followig: γ(k 1 + k k ) γ(k 1 ) + γ(k 2 ) γ(k ). (a) Show that a cocave γ, i.e. oe that satisfies γ(x+1) γ(x) γ(x) γ(x 1), is sub-additive if γ(0) 0. Hit: it is sufficiet to show that γ(k 1 + k 2 ) γ(k 1 ) + γ(k 2 ) ad express γ(k 1 + k 2 ) as γ(k 1 ) plus the icremets up to k 2. Solutio: We use the fact that γ is a gap fuctio, ad hece it is defied o itergers 0 oly. I fact, the statemet above is ot true otherwise. Before we prove the statemet, we give some examples whe the statemet is ot true to justify the eed for the coditio above. First, cosider the fuctio defied as follows: γ( 1) = 1, γ(0) = 0, γ(1) = 1, γ(2) = 1. This satisfies the cocavity coditio. Now γ(2 + ( 1)) = γ(1) = 1 > γ(2) + γ( 1) = 1 1 = 0. Therefore, we caot have γ defied o egative itegers. Secod, cosider the periodic fuctio that start at zero ad remais zero util x = 1/3, the goes up liearly from zero to 1 at x = 2/3, the goes dow lieraly to zero at x = 1. This is repeated for every iterval of legth 1. This fuctio satisfies the cocavity coditio because γ(x) = γ(x + 1). Now γ(1/3 + ǫ) is strictly positive. But γ(1/3) = γ(ǫ) = 0. Therefore, γ(1/3 + ǫ) > γ(1/3) + γ(ǫ). Hece, γ caot be defied o o-itegers. Now we prove the result for γ defied o itegers 0. We prove the statemet for γ(k 1 + k 2 ). The geeral case ca be easily obtaied by iductio. γ(k 1 + k 2 ) = γ(k 1 ) + k 2 i=1 γ(k 1 + i) γ(k 1 + i 1) γ(k 1 ) + k 2 i=1 γ(i) γ(i 1) by cocavity coditio = γ(k 1 ) + γ(1) γ(0) + γ(2) γ(1) γ(k 2 1) γ(k 2 2) + γ(k 2 ) γ(k 2 1) = γ(k 1 ) + γ(k 2 ) γ(0) γ(k 1 ) + γ(k 2 ) if γ(0) 0 Note that the first equality ad secod iequality are valid oly if both k 1 ad k 2 are 0. The ext set of questios are iteded to help you uderstad why the DP algorithm we saw i class requires γ to be cocave. Here s the algorithm agai: A(i 1,j 1) + s(i,j) (1) A(i,j) = max A(i,j k) γ(k), k = 1...j (2) A(i k,j) γ(k), k = 1...i (3) 1

2 Without loss of geerality, we restrict our attetio to aligmets that ed with a gap i x. Call such a aligmet of x 1...x i ad y 1...y j good if it eds with a gap of legth k i x for some k > 0 ad optimally aligs x 1...x i to y 1...y j k. (b) Show that a good aligmet of x 1...x i ad y 1...y j is ot ecessarily optimal. It is eough to give a couter example. Solutio: Cosider the followig aligmet: -A- AAA This algimet is good because -A aliged with AA is optimal (A- with AA is the oly other way ad is the same). However, the aligmet (-A-, AAA) is ot ecessarily optimal, depedig o γ. (c) Show that if γ is cocave, the a optimal aligmet (that eds with a gap i x) of x 1...x i ad y 1...y j is a good aligmet. Solutio: Cosider a optimal aligmet that eds with a gap of legth l 1 > 0 i x. The score of this aligmet is equal to S = S 1 γ(l 1 ), where S 1 is the score of the portio of the aligmet obtaied by excludig the gap of legth l 1. Assume that the aligmet is ot good, i.e. S 1 is ot optimal. The there must be aother aligmet A for that portio with score S 2 > S 1. This aligmet eds i a gap of legth l 2 0 i x. Therefore, S 2 = S 3 γ(l 2 ) (assume the covetio γ(0) = 0), where S 3 is the score of the portio of the aligmet A obtaied by excludig the gap of legth l 2. We will costruct a ew aligmet by cocateatig A with the gap of legth l 1 i x. Let s fid the core of this aligmet. The score is S = S 3 γ(l 2 + l 1 ). Usig part (a), S = S 3 γ(l 2 + l 1 ) S 3 γ(l 2 ) γ(l 1 ) = S 2 γ(l 1 ) > S 1 γ(l 1 ) = S. Hece S > S cotradictig that our aligmet was optimal. Therefore, it must have bee a good aligmet too. (d) Show that if γ is cocave, the for ay give aligmet of x 1...x i ad y 1...y j with score S, if we split the aligmet i two parts with scores S 1 ad S 2, the S 1 + S 2 S. Solutio: Assume the first part eds with a gap of legth l 1 0 ad the secod part start with a gap l 2 0. The S 1 = A γ(l 1 ) ad S 2 = γ(l 2 )+B. By the sub-additive property, S = A γ(l 1 +l 2 )+B A γ(l 1 ) γ(l 2 )+B = S 1 +S 2. (e) Put parts (b) ad (c) ad (d) together to argue that step (2) must check all k = 1...j, ad it computes the optimal aligmet of x 1...x i ad y 1...y j that eds with a gap i x. 2

3 Solutio: ecessity: Step (2) should fid the optimal aligmet that eds i a gap i x. This gap may have ay legth k. Moreover, the aligmet must be good (part c). Therefore, the score must be the sum of the score of some optimal aligmet ad a gap of legth k. But if such a good aligmet is foud, it may ot be optimal (part b); therefore, all good aligmet must be checked. correctess: The optimal aligmet that eds i a gap i x must exist ad it is good (part c). Therefore, step (2) will fid a k such that A(i,j k) is optimal ad correspods to a aligmet that eds with o gaps, ad A(i,j k) γ(k) is the score of the optimal aligmet that eds with a gap i x (of legth k). Sice S 1 + S 2 S for ay split of a aligmet (part d), step (2) is guarateed ot to compute ay score greater tha A(i,j k) γ(k). (f) Costruct a istace of a optimal aligmet (that eds with a gap i x) that is ot good. (you caot use a cocave gap fuctio accordig to part (c)). Argue that the algorithm above will ot work properly if such a istace ca be costructed. Solutio: Assume γ(0) = 0, γ(1) = 1, γ(2) = 5. Also assume that the score of a match is +1 ad the score of a mismatch is -1. Cosider the followig optimal aligmet with score = 3. -A- ABA Ay other aligmet will have a score of +1 5 = 4. I the aligmet above, the aligmet (-A, AB) is ot optimal. (-A, AB) has a score of 1 1 = 2. The aligmet (A-,AB) has a score of 1 1 = 0. Therefore, we have a optimal aligmet that is ot good. Assume that the optimal aligmet of x 1...x i ad y 1...y j eds with a gap of legth k i x ad it is good. This optimal aligmet correspods to aligig y 1...y j k with x 1...x i (o-optimally) ad aligig y j k+1...y j with a gap of legth k. If the algorithm does ot compute the correct value for A(i,j k), the we are doe because it will ot perform correctly o the istace x 1...x j ad y 1...y j k. If the algorithm computes the correct value for A(i,j k), the step (2) will compute at some poit i time A(i,j) = A(i,j k) γ(k) (or a larger value) which is greater tha the score of the optimal aligmet. Sice A(i,j) ever decreases, it will hold the wrog score forever, ad thus the algorithm fails o the istace x 1...x i ad y 1...y j. 3

4 Problem 2: Radom star aligmet Let M = S i D(S c,s i ) as defied i the star aligmet algorithm. Suppose that istead of S c, we choose a strig at radom to be the ceter of the star. Let M R be defied i a aalogous way whe S c is replaced by a radom strig. (a) Show that E[M R ] 2M ad argue that the expected score of the aligmet is at most a factor of 4 of optimal. Solutio: E[M R ] = 1 D(S i,s j ) 1 [D(S i,s c ) + D(S c,s j )] s j s i s j = 1 D(S i,s c ) + 1 D(S c,s j ) s j s i s i s j = 1 M + 1 M = 2M (b) Next, show that the media of M R is at most 3M. s j s i s i Solutio: If the media is strictly greater tha 3M, the the expected value is: a cotradictio. E[M R ] > 1 2 3M M = 2M (c) Fially, argue that the value of the multiple aligmet is at most a factor of 6 of optimal with probability at least 1/2. Solutio: evidet, see proof of Star aligmet. Problem 3: Cosesus strig Give a set of strigs S, let S c be the ceter strig as defied for the star aligmet of S, ad assume that the scorig scheme (distace) satisfies the triagular iequality. I additio, for ay strig T, defie the cosesus error E(T) = S i S D(T,S i). (a) Let S, ot ecessarily i S, be the strig that miimizes E(S ). Show that for ay strig T i S: Solutio: E(T) = E(T) ( S 2)D(T,S ) + E(S ) S i S {T } D(T,S i ) s i S {T } [D(T,S ) + D(S,S i )] 4

5 < S i S {T } D(T,S ) + E(S ) D(T,S ) = ( S 2)D(T,S ) + E(S ) (b) Show that if T S is the closest to S, the: E(T) E(S ) < 2 Solutio: Sice D(T,S ) D(S i,s ), the S D(T,S ) S i S D(S i,s ) = E(S ). Therefore, E(T) < 2E(S ) (c) Show that E(S c )/E(S ) < 2. Solutio: By defiitio, E(S c ) E(T). Problem 4: Example substitutio matrix Let s say we would like to build a DNA substitutio matrix (4x4 matrix) optimized for fidig 88% idetity aligmets. assume the backgroud frequecies are idetical, i.e. p i = 0.25 for each ucleotide i assume that all matches are equally probable assume that all mismatches are equally probable (a) Compute q ij for all i ad j. (b) Costruct the matrix usig the log-likelihood ratio. (c) Choose a scalig factor λ to make the substitutio matrix close to a iteger matrix. Solutio: Sice the idetity is 88%, ad all matches are equally probable, each match occurs with probability Therefore, q AA = q GG = q CC = q TT = Sice all mismatches (12 possibilities) are equally probable, we have q ij = (1 0.88)/12 = 0.12/12 = 0.01, where both i j {A,G,C,T }. Now p A = p G = p C = p T = 0.25, therefore: log q ii p i p i = 0.22/( ) = 1.26 log q ij p i p j = 0.01/( ) = 1.83 Scalig by 1/λ, where λ = 0.18, ad roudig we get a scorig scheme of +7 for a match, ad -10 for a mismatch. The matrix is: 5

6 Problem 5: Urevealig BLOSUM62 (a) Fid the 20x20 BLOSUM62 substitutio matrix olie. BLOSUM62 has the property that the backgroud probabilities ad the observed probabilities are cosistet, i.e. p i = j q ij. Solutio: A R N D C Q E G H I L K M F P S T W Y V A R N D C Q E G H I L K M F P S T W Y V (b) Give a symmetric ad cosistet substitutio matrix S, with a scalig factor λ, let M be the matrix e λs. Note M ij = qij p ip j. Let Y be the iverse of M (assumig M is ivertible). Show that the sum of the i th colum (or row) of Y must be equal to the backgroud probability p i. Hit: Cosider the vector p = [p 1,...,p ]. Show that pm = [1,...,1]. Use this result to compute pmy i two ways. Solutio: The i th etry of pm is p 1 M 1i + p 2 M 2i +...p M i = p 1 q 1i p 1 p i + p 2 q 2i p 2 p i +...p q i p p i 6

7 = 1 p i (q 1i + q 2i +...q i ) = 1 p i p i = 1 The last equality follows because the matrix is symmetric ad cosistet. Therefore, pm = [1,...,1]. So p = pmy = [1,...,1]Y. This shows that the sum of the i th colum of Y is equal to p i. (c) Usig the above strategy, ad kowig that λ = for BLOSUM62, fid the backgroud probabilities p i for BLOSUM62. (d) Usig part (c), fid the observed set of probabilities q ij. 7