15.12 Applications of Suffix Trees

Transcription

1 248 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, Appliations of Suffix Trees 1. Searhing for exat patterns 2. Minimal unique substrings 3. Maximum unique mathes 4. Maximum repeats 5. Approximate repeats Searhing for exat patterns To determine whether a string q ours in a string t, follow the path from the root of the suffix tree ST (t) as direted by the haraters of q. If at some point you annot proeed, then q does not our in t, otherwise it does. b a b Example: Text abab. a b b a 5 The query abb is not ontained in abab (Navigating into the tree and mathing the first two letters ab, we arrive at the node ab, however there is no b-edge leaving from there.) The query aba is ontained in abab. Determining whether q ours in t requires O( q ) time Finding all ourrenes To find all positions where the query q is ontained in t, annotate eah leaf s i of the suffix tree with the position i at whih the suffix i starts in t. Then, after mathing q to a path in the tree, visit all nodes below the path and return the annotated values. This works beause any ourrene of q in t is the prefix of one of these suffixes. The number of nodes below the path is at most twie the number of hits and thus finding and olleting all hits takes time O( q + k), where k is the number of ourrenes. (Note that in the disussed lazy suffix tree implementation we do not use this leaf annotation but rather ompute the positions from the lp values, to save spae...) Minimal Unique Substrings Definition Given a sequene s Σ, and a number L > 0. The minimal unique substrings problem onsists of enumerating all substrings u Σ of s satisfying the following properties: u ours exatly one in s, u L, and all proper prefixes of u our at least twie in s.

2 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, Example: Let s = abab and L = 2. Then the minimal unique substrings are aba and ba. Note, that bab is not a minimal unique substring, sine the proper prefix ba of bab is already unique, i.e. the minimality ondition does not hold. Question: How an suffix trees be used to solve the minimal unique substring problem? Appliations of this in primer design Appliation: Maximum Unique Mathes Modern sequening and omputational tehnologies and advanes in bioinformatis has made whole genome sequening possible. One resulting hallenge is the fast alignment of whole genomes. Dynami programming is too slow for aligning two large genomes. Heuristis suh as BLAST or FASTA are not designed to perform pairwise alignments of two very long sequenes. One very suessful approah is based on identifying maximum unique mathes, whih is based on the assumption that one expets to substrings ourring in two similar genomes. Maximum unique mathes (MUMs) are almost surely part of a good alignment of the two sequenes and so the alignment problem an be redued to aligning the sequene in the gaps between the MUMs. Definition Given two sequenes s, t Σ, and a number L > 0. mathes problem (MUM-problem) is to find all sequenes u Σ with: The maximum unique u L, u ours exatly one in s and one in t, and for any harater a Σ neither ua nor au ours both in s and t. In other words, a MUM is a sequene m that ours preisely one in s and one in t, and is both right maximum and left maximum with this property (meaning that ma and am both do not have the uniqueness property, for any letter a). For example: if s =mostbeautifulandwildorsia t =greenleanandnulearfree, then there is only one MUM of length 3, namely This problem an be solved in O( s + t ) time using a so-alled generalized suffix tree: and To find all MUMs, generate the generalized suffix tree T for s%t, where % is a separator with % / s and % / t. Any path in T from the root to some node u that has preisely two hildren, one in s and one in t, orresponds to a right maximum unique math. To determine whether u is left maximum, too, simply hek whether both preeding letters in s and t differ. Example For s =agga and t =agagga, onstrut the suffix tree for s%t:

3 250 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, 2007 The node ga has preisely two leaves as hildern, one representing a suffix that starts in s and the other one that starts in t. Thus, the word ga ours preisely one in both s and t, and is right maximum. The word is also left-maximum as the preeding haraters in s (=g) and t (=) differ. Note, however, that the string ag is not a MUM, sine the node ag has three hildren. MUMs are apparent in dotplots: Example

4 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, Example Appliations Appliations of MUMs in sequene analysis: Detetion of large-sale inversions in baterial genomes Detetion of whole-genome dupliations, for example in the genome of Arabidopsis thaliana. Comparison of different assemblies of the same genome at different stages of sequening and finishing. Comparison of assemblies of the same data using different assembly algorithms Deteting Repeats A puzzling observation in the early days of moleular biology was that genome size does not orrelate well with organismal omplexity. For example, Homo sapiens has a genome that is 200 times as large as that of the yeast S. erevisiae, but 100 times as small as that of Amoeba dubia. This so-alled C-value paradox was largely resolved with the reognition that many genomes onsist of a large amount of repetitive sequene.

5 252 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, 2007 Repeats in human (Nature, vol. 409, pg. 880, 15. Feb 2000) Repeats A PubMed query for (repeat OR repetitive) AND (protein OR nuleotide sequene) on May 13, 2006 produed the following results: hits without limits to speifi fields hits when restriting to title/abstrat 1170 hits when restriting to title Keywords assoiated with the hits: imprinting / RNA editing / diseases / repair / genome organization / viruses / protein families One an distinguish between loal, small-sale repeats with known funtion or origin, simple repeats, loal and interspersed, with less known funtion, and omplex interspersed repeats with unknown funtion. Examples of loal repeats are: palindromi sequenes (regulation of DNA transription), eg. sequenes, restrition enzyme reognition inverted repeats flanking transposons (orientation), repeats in viruses,

6 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, non-oding RNAs suh as rrnas and trnas (must be produed in large opy numbers), protein family members (eg. globins), lustered genes (eg. histones), and motifs and domains of proteins. Examples of simple repeats are: often seen as tandem repeats: eg. hromosome, TTAGGG up to several opies at the end of every human simple tandem repeats of length 3 nuleotides are linked to ertain diseases (Huntington, et.), and satellite DNA (forensi appliations). Examples of interspersed repeats are: SINEs - Short Interspersed Nulear sequenes (Alu repeats, MIRs), LINEs, and LTR elements Maximum Repeats Definition Given a sequene t = t 1 t 2... t n. Let a substring t[i, j] := t i... t j be represented by the pair (i, j). A pair R = (l, r) of different substrings l = (i, j) and r = (i, j ) of t is alled a repeat, if i < i and t i... t j = t i... t j. We all l and r the left and right instane of the repeat R, respetively. t i j i j t... t j i = t... t i j Definition A repeat R = ((i, j), (i, j )) is alled left maximum, if i = 1 or t i 1 t i 1, and right maximum, if j = n or t j+1 t j +1, and maximum, if it is both left and right maximum. t i j i j = at... t i j b t i... tj d Note that the definition allows for overlapping repeats. maximum a and b d Example The string g a g t g a g ontains the following repeats of length 2: ((1, 4), (7, 10)) gag maximum ((1, 3), (7, 9)) gag left maximum ((2, 4), (8, 10)) ag right maximum ((1, 2), (7, 8)) ga left maximum ((2, 3), (8, 9)) ag ((3, 4), (9, 10)) g right maximum

7 254 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, 2007 Computing all Maximum Repeats We will disuss how to ompute all maximum repeats for a string t of length n. Generally we also want to permit a prefix or a suffix of t to be part of a maximum repeat. To model this we simply add a harater to the start of t and one to the last of t that do no our elsewhere in t, e.g.: t = x g g g y g g z Let ST (t) be the suffix tree for t. A onnetion between suffix trees and maximum repeats is stated in the following Lemma Let ST (t) be a suffix tree for string t. If a string r is a maximum repeat in t, then r is the label of an internal node r in ST (t). The following is maybe a bit surprising Theorem There are at most n maximum repeats in any string of length n. Proof: Sine ST (t) has n leaves, it has at most n internal nodes. Thus the theorem follows immediately from the preeding lemma. Thus beause of the theorem above we will need only to look at internal nodes, but whih speifi internal nodes orrespond to maximum repeats? The following algorithm will answer this question. From now on we an ignore all leaf edges from the root (why?). The algorithm proeeds in two phases: In the first phase, every leaf node v of ST (t) is annotated by (a, i), where i denotes the position of the suffix v = t i... t n assoiated with v and a = t i 1 is the letter that ours immediately before the suffix. Example: Partial suffix tree for t = x g g g y g g z : g z g ygg ggygg ygg ygg g ygg

8 Algorithms in Bioinformatis II, SoSe 07, ZBIT, D. Huson, May 14, With leaf annotations: g 12 z g 11 ygg g g 6 ggygg ygg ygg g g 9 g ygg y 8 g 3 x 2 Note that a substring r an only be a maximum repeat iff its branhing node r has at least two leaves with different letters in their annotations. Thus in order to find and print those we need to ombine the leaf annotations appropriately. For every leaf node v and Σ we set: { {i}, if = ti 1, and A(v, ) =, else, where i is the start position of the orresponding suffix v. In the seond phase of the algorithm, we extend this annotation to all branhing nodes, bottom-up: Let w be a branhing node with hildren v 1,..., v h and assume we have omputed A(v j, ) for all j {1,..., h} and all Σ. For eah letter Σ set: A(w, ) := h A(v j, ). j=1 Note that this is a disjoint union and A(w, ) is the set of all start positions of w in t for whih t i 1 =. 12 z g 11 g ygg g g 6 ggygg ygg ygg g g 9 g ygg y 8 g 3 x 2 ((8,9),(10,11)) ((3,4),(10,11) ((5,6),(8,9)) ((3,4),(5,6)) g 3 5,10 y 8 g ygg g 5 10 ygg y 8 g 3 g 3 5,10 x 2 y 8 ggygg g 3 y 8 x 2 ((3,6),(8,11)) Reporting All Maximum Repeats In a bottom-up traversal, for eah branhing node w we first determine A(w, ) for all Σ and then report all maximum repeats of the substring w that have length L: Let q be the depth of node w, i.e., q is equal to the length of w. Then there is a maximum repeat for w, if w has a pair of hildren v f and v g suh that there is a letter and a letter d, with d and A(v f, ) and A(v g, d). In this ase we output R((i, i + q 1), (j, j + q 1)) for all i that are in A(v f, ) and j that are in A(v g, d). The result of the algorithm for maximum repeats of length L 2 for the example is: