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Constraints and Bioinformatics: Results and Challenges Agostino Dovier Dept. Mathematics and Computer Science, University of Udine, Italy Cork, Sept. 4, 2015 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept.

  1. Genomics: Haplotype Inference Definitions Haplotype Inference Each person inherits two haplotypes (from the mother and from the father) for most regions of the genome. G A T C T G T A C T G A G T · · · · · · G A T C T G T A C T G A A T · · · · · · ⇑ ⇑ ⇑ ∗ ⇑ ∗ In some typical positions, the bases are subject to mutations. In the most common case, there is a Single Nucleotide Polymorphism (SNP). Mutations are C ↔ T and A ↔ G Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 14 / 98

  2. Genomics: Haplotype Inference Definitions Haplotype Inference Single Nucleotide Polymorphism (SNP) Each person has two haplotypes (from the mother and from the father) for most regions of the genome: G A A T C T T C G T A C T G A G T G A A T C T T C G T A C T G A A T Let us focus on the SNPs: A C T G A C T A We encode SNPs according to: A �→ 0 C �→ 0 G �→ 1 T �→ 1 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 15 / 98

  3. Genomics: Haplotype Inference Definitions Haplotype Inference Single Nucleotide Polymorphism (SNP) Each person has two haplotypes (from the mother and from the father) for most regions of the genome: G A A T C T T C G T A C T G A G T G A A T C T T C G T A C T G A A T Let us focus on the SNPs: A C T G A C T A We encode SNPs according to: A �→ 0 C �→ 0 G �→ 1 T �→ 1 0 0 1 1 0 0 1 0 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 15 / 98

  4. Genomics: Haplotype Inference Definitions Haplotype Inference Single Nucleotide Polymorphism (SNP) Each person has two haplotypes (from the mother and from the father) for most regions of the genome: G A A T C T T C G T A C T G A G T G A A T C T T C G T A C T G A A T Let us focus on the SNPs: A C T G A C T A We encode SNPs according to: A �→ 0 C �→ 0 G �→ 1 T �→ 1 0 0 1 1 0 0 1 0 But this is the situation of complete knowledge. In practice, we can detect a mismatch but not its single components. 0 0 1 2 The genotype is set to 2 if there is a mismatch Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 15 / 98

  5. Genomics: Haplotype Inference Definitions Haplotype Inference Looking for an explanation Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 16 / 98

  6. Genomics: Haplotype Inference Definitions Haplotype Inference Looking for an explanation Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 16 / 98

  7. Genomics: Haplotype Inference Definitions Haplotype Inference Looking for an explanation Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 16 / 98

  8. Genomics: Haplotype Inference Definitions Haplotype Inference A string of { 0 , 1 } ∗ is called a haplotype A string of { 0 , 1 , 2 } ∗ is called a genotype Two equal length haplotypes generate a unique genotype The rules are 0 ⊕ 0 = 0, 1 ⊕ 1 = 1, 0 ⊕ 1 = 2 E.g., 0010 , 0101 ⇒ 0222 If we have a genotype, we can only conjecture (potentially exponentially many) haplotype s that generated it (observe that, e.g., 0110 , 0001 ⇒ 0222) Biological experiments allow us to know genotypes! Investigating sets of genotypes for a population, helps in understanding the relationships between SNPs and physical features as well as medical information Since genotypes are introduced in evolution, it is reasonable to find minimal sets of haplotypes explaining the known genotypes. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 17 / 98

  9. Genomics: Haplotype Inference Model Haplotype Inference Let H be the set of haplotypes (of given length n ) and G be a set of genotypes (of the same length n ) . Given h 1 , h 2 ∈ H and g ∈ G , { h 1 , h 2 } explains g if and only if | h 1 | = | h 2 | = | g | and ∀ i ∈ [ 1 .. n ] : g [ i ] ≤ 1 − → h 1 [ i ] = h 2 [ i ] = g [ i ] g [ i ] = 2 − → h 1 [ i ] � = h 2 [ i ] A set of haplotypes H explains a set of genotypes G if for all g ∈ G there are h 1 , h 2 ∈ H such that { h 1 , h 2 } explains g . Given a set of genotypes G and an integer k , the haplotype inference problem (HIP) by pure parsimony is the problem of finding a set H that explains G and such that | H | = k (decision version—NP complete). Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 18 / 98

  10. Genomics: Haplotype Inference Model Haplotype Inference CP encoding Let us focus on the decisional version: Is there an explanation for G with k haplotypes? Generate m = 2 | G | vectors of 0-1 FD variables H 1 , . . . , H m of length n Add a < -lexicographical constraint on each pair ( H 1 , H 2 ) , ( H 3 , H 4 ) , . . . , ( H m − 1 , H m ) (repetitions in different pairs are allowed!) Build a constraint of the form: ( ∀ G i ∈ G ) ( � H 2 i − 1 , H 2 i � explain G ) Namely: � G i [ j ] ≤ 1 → ( H 2 i 1 [ j ] = H i 2 [ j ] = G 2 i [ j ]) ∧ n � � G i [ j ] = 2 → ( H 2 i 1 [ j ] � = H 2 i [ j ]) j = 1 We need to state (using constraints!) that |{ H 1 , . . . , H m }| = k . Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 19 / 98

  11. Genomics: Haplotype Inference Model Haplotype Inference 2nd CP encoding For a , b ∈ [ 1 .. m ] we set F a , b ↔ � n i = 1 H a [ i ] = H b [ i ] . Namely F a , b is a Boolean variable that is true iff H a and H b will be equal in the solution Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 20 / 98

  12. Genomics: Haplotype Inference Model Haplotype Inference 2nd CP encoding For a , b ∈ [ 1 .. m ] we set F a , b ↔ � n i = 1 H a [ i ] = H b [ i ] . Namely F a , b is a Boolean variable that is true iff H a and H b will be equal in the solution Then define M a ↔ � m b = a + 1 F a , b M a is again a Boolean variable that is true if and only if there is another vector in H a + 1 , H a + 2 , . . . , H m equal to H a Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 20 / 98

  13. Genomics: Haplotype Inference Model Haplotype Inference 2nd CP encoding For a , b ∈ [ 1 .. m ] we set F a , b ↔ � n i = 1 H a [ i ] = H b [ i ] . Namely F a , b is a Boolean variable that is true iff H a and H b will be equal in the solution Then define M a ↔ � m b = a + 1 F a , b M a is again a Boolean variable that is true if and only if there is another vector in H a + 1 , H a + 2 , . . . , H m equal to H a The size of H can be therefore expressed as � n a = 1 ( 1 − M a ) (viewing Boolean truth values as 0/1) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 20 / 98

  14. Genomics: Haplotype Inference References Haplotype Inference Some References Gusfield and Orzack. Haplotype Inference (Survey, and ILP formulations) In CRC Handbook on Bioinformatics, 2006 Lancia, Pinotti, Rizzi. [LPR04] Haplotyping Populations by Pure Parsimony: Complexity of Exact and Approximation Algorithms. INFORMS Journal on Computing 16(4):348–359, 2004. Graça, Marques-Silva, Lynce, Oliveira. Several works on SAT-based and specialized 0-1 ILP for Haplotype Inference. (e.g. WCB 08, WCB 09) Di Gaspero, Roli. Stochastic local search for large-scale instances of the haplotype inference problem by pure parsimony. J. Algorithms 63(1-3): 55-69 (2008) (also in WCB 08). Erdem, Erdem, Türe. HAPLO-ASP: Haplotype Inference Using Answer Set Programming. LPNMR 2009: 573–578 James Cussens Maximum likelihood pedigree reconstruction using integer programming. WCB 10. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 21 / 98

  15. Genomics: Phylogenetic trees Phylogenetics Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 22 / 98

  16. Genomics: Phylogenetic trees Introduction Phylogenetic trees Basics A phylogeny describes evolutionary relationships among entities. Comparative biology: investigates similarities and differences More reliable than pattern matching Applied outside biology: e.g. Indo-European languages [Erdem03] Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 23 / 98

  17. Genomics: Phylogenetic trees Introduction Phylogenetic trees Basics The entities a set L of elementary taxonomic units, known as taxa (e.g., L = { English , German , French , Spanish , Italian } or L = { dog , cat , horse , chicken } ) A set C of characters is assigned to each element of L (e.g., characters “hand” and “father”, or characters “number of legs”, “length of the tail”, etc.) Characters are evaluated with FD values (e.g. {1 (hand), 2 (mano/main)} for “hand” and {1 (father/padre), 2 (vater/père)} for “father”) Each element in L is assigned a value for each character. Let us focus on Boolean characters Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 24 / 98

  18. Genomics: Phylogenetic trees Model Phylogenetic tree reconstruction A phylogeny ( V , E , L , C , D , f ) for a set L of taxa is a finite binary tree ( V , E ) with leaves L ⊆ V (taxa=leaves, with a slight abuse of notation) along with two finite sets C and D and a function f : L × C − → D . V \ L describes the ancestral units and E evolutionary relationships. C is the set of characters, and D contains their domain values (also knows are states) f labels every leaf v ∈ L by assigning a state for each character i ∈ C Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 25 / 98

  19. Genomics: Phylogenetic trees Model Phylogenetic trees Example (from Erdem11) A phylogeny ( V , E , L , C , D , f ) where L = { English , German , French , Spanish , Italian } (taxa) C = { Hand , Father } (characters), D = { 1 , 2 } (states). Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 26 / 98

  20. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees Example (from Erdem 2011) A character i ∈ C is compatible with a phylogeny if the taxa that present the same value for i are connected by a subtree. Character Hand is compatible with the above tree Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 27 / 98

  21. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees Example (from Erdem 2011) A character i ∈ C is compatible with a phylogeny if the taxa that present the same value for i are connected by a subtree. Character Hand is compatible with the above tree Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 27 / 98

  22. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees Example (from Erdem 2011) A character i ∈ C is compatible with a phylogeny if the taxa that present the same value for i are connected by a subtree. Character Hand is compatible with the above tree Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 27 / 98

  23. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees Example (from Erdem 2011) A character i ∈ C is compatible with a phylogeny if the taxa that present the same value for i are connected by a subtree. Otherwise it is incompatible Character Father is incompatible with the above tree Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 28 / 98

  24. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees Example (from Erdem 2011) A character i ∈ C is compatible with a phylogeny if the taxa that present the same value for i are connected by a subtree. Otherwise it is incompatible Character Father is incompatible with the above tree Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 28 / 98

  25. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees Example (from Erdem 2011) A character i ∈ C is compatible with a phylogeny if the taxa that present the same value for i are connected by a subtree. Otherwise it is incompatible Character Father is incompatible with the above tree Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 28 / 98

  26. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees k -incompatibility The above subtree requirement implicitly states that when a character changes (in the evolution) it never go back to the previous value (Camin-Sokal). Moreover, that the change occurs in a unique place (Dollo). Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 29 / 98

  27. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees k -incompatibility The above subtree requirement implicitly states that when a character changes (in the evolution) it never go back to the previous value (Camin-Sokal). Moreover, that the change occurs in a unique place (Dollo). k -INCOMPATIBILITY PROBLEM Given sets L (taxa/leaves), C (characters), and D (states), a function → D , and k ∈ N , decide the existence of a phylogeny f : L × C − ( V , E , L , C , D , f ) with at most k incompatible characters. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 29 / 98

  28. Genomics: Phylogenetic trees Character compatibility Phylogenetic trees k -incompatibility The above subtree requirement implicitly states that when a character changes (in the evolution) it never go back to the previous value (Camin-Sokal). Moreover, that the change occurs in a unique place (Dollo). k -INCOMPATIBILITY PROBLEM Given sets L (taxa/leaves), C (characters), and D (states), a function → D , and k ∈ N , decide the existence of a phylogeny f : L × C − ( V , E , L , C , D , f ) with at most k incompatible characters. This problem is NP-complete (Day, Sankoff 1986). The number of possible phylogenies is exponential in L NP-complete (Day, Sankoff 1986). Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 29 / 98

  29. Genomics: Phylogenetic trees CP Modeling Encoding Input Input vector L of n elements (taxa) each of them characterized by a m -tuple of (character) values. For simplicity, let us focus on Boolean encodings. E.g. m = 3 , n = 4: L = [[ 0 , 1 , 1 ] , [ 1 , 0 , 0 ] , [ 1 , 1 , 0 ] , [ 1 , 0 , 1 ]] (four elements/taxa with three characters) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 30 / 98

  30. Genomics: Phylogenetic trees CP Modeling Encoding: Binary tree The Tree can be represented by a FD vector of t = 2 n − 1 7 elements valued in ( n +) 1 , . . . , t + 1. 6 5 Tree [ i ] = j means that node i is a son of node j . For the root 1 2 4 3 r , Tree [ r ] = t + 1. The tree is binary: for 1 2 3 4 5 6 7 i = 1 , . . . , n : 5 5 6 6 7 7 8 count ( i , Tree , ≤ , 2 ) Symmetries: � Taxa are the leaves of the tree: nodes 1 . . . n � Tree [ 1 ] = n + 1 � Tree [ t ] = t + 1 ( t is the root) � For i , j ∈ { 1 , . . . , t } : i < j → Tree [ i ] ≤ Tree [ j ] Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 31 / 98

  31. Genomics: Phylogenetic trees CP Modeling Encoding Hypercube tree Each node of the tree is assigned a m -tuple of Boolean Values. This is stored in a vector Chars. Chars [ 1 ] –Chars [ n ] are assigned using the input L . Values for internal nodes must be computed. For i < j , if Tree [ i ] = j , the Hamming difference of the corresponding tuples is 1. Precisely: � m � � Tree [ i ] = j → | Chars [ i ][ ℓ ] − Chars [ j ][ ℓ ] | = 1 ℓ = 1 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 32 / 98

  32. Genomics: Phylogenetic trees CP Modeling Encoding Hypercube tree Each node of the tree is assigned a m -tuple of Boolean Values. This is stored in a vector Chars. Chars [ 1 ] –Chars [ n ] are assigned using the input L . Values for internal nodes must be computed. For i < j , if Tree [ i ] = j , the Hamming difference of the corresponding tuples is 1. Precisely: � m � � Tree [ i ] = j → | Chars [ i ][ ℓ ] − Chars [ j ][ ℓ ] | = 1 ℓ = 1 Actually, we can either relax the above constraint to ≤ 1 (see e.g. hand/father example, italian and spanish) or (alternatively) Add the redundant constraint AllDifferentTuples ( Chars ) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 32 / 98

  33. Genomics: Phylogenetic trees CP Modeling Encoding k -incompatibility We need to state that a character changes (actually, increases) in at most one node. This makes the tree compatible with that character. Let Comp be a vector of m elements (one per character). For i < j , let F i , j = 1 if Tree [ i ] = j , F i , j = 0 otherwise. Then, for ℓ = 1 , . . . , m (and i , j = 1 , . . . , n : � Comp [ ℓ ] = F i , j ( Chars [ i ][ ℓ ] − Chars [ j ][ ℓ ]) i < j Basically, after variable instantiation, Comp [ ℓ ] will contain the number of changes of character ℓ in the tree. The number of values different from 1 and 0 in Comp is forced to be less than or equal to k . Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 33 / 98

  34. Genomics: Phylogenetic trees References (Some) References Day, W.H.E., Johnson D.S., Sankoff, D. The Computational complexity of Inferring Rooted Phylogenies by Parsimony. Math. Biosciences 81:33–42, 1986. Day, W.H.E., Sankoff, D. Computational complexity of Inferring Phylogenies by Compatibility. Systematic Zoology 35(2):224–229, 1986. Erdem E., Lifschitz V., Nakhleh L., Ringe D. Reconstructing the Evolutionary History of Indo-European Languages Using Answer Set Programming. PADL 2003: 160-176. Thomas Schiex et al. Papers on complex pedigree reconstructions using weighted constraint satisfaction. In WCB 05, WCB 06, WCB 07. Erdem E. Applications of Answer Set Programming in Phylogenetic Systematics MG65, LNCS 6565, 2011. Moore N.C.A., and Prosser P . The Ultrametric Constraint and its Application to Phylogenetics. (Supertree construction). J. Artif. Intell. Res. 32:901–938, 2008 (also in WCB 06): ( x > y = z ) ∨ ( y > x = z ) ∨ ( z > x = y ) ∨ ( x = y = z ) Le Tiep, Nguyen Hieu, Pontelli Enrico, and Cao Son Tran. ASP at Work: An ASP Implementation of PhyloWS. ICLP 2012, LIPICS vol 17. (also in WCB 12) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 34 / 98

  35. RNA secondary structure prediction DNA and RNA RNA and Central Dogma C T A G G C C C G T A A DNA T C T G A C G G G transcription A T C A G C G C U A G C C U A mRNA translation S A S L Protein RNA is a sequence of nucleotides (A,C,G,U) that (often) is just an intermediary between DNA and proteins DNA strands are transcribed to mRNA, in order to exit the cell’s nucleus Nucleotides replacement: DNA T �→ RNA U. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 35 / 98

  36. RNA secondary structure prediction DNA and RNA RNA Secondary Structure Stem Loop A G CU U U U C U U G C U U G A G C G A U U UG C G U Mismatch RNA folds according to favorable matchings (A-U, C-G, ∼ U-G) The secondary structure is the set of its base pairings Secondary structure determines the 3D properties Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 36 / 98

  37. RNA secondary structure prediction DNA and RNA RNA Secondary Structure RNA folds according to favorable matchings (A-U, C-G, ∼ U-G) The secondary structure is the set of its base pairings Secondary structure determines the 3D properties Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 36 / 98

  38. RNA secondary structure prediction Definitions Mathematically A RNA sequence � s = s 1 s 2 · · · s n is a string in { A , C , G , U } ∗ A RNA secondary structure is a (partial) injective function P ⊆ { 1 , . . . , n } 2 such that ( i , j ) ∈ P ↔ ( j , i ) ∈ P ( i , j ) ∈ P only if ( s i , s j ) ∈ { ( A , U ) , ( U , A ) , ( C , G ) , ( G , C ) , ( U , G ) , ( G , U ) } 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 A C C U G G U A U C G A C A Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 37 / 98

  39. RNA secondary structure prediction Definitions Mathematically A RNA sequence � s = s 1 s 2 · · · s n is a string in { A , C , G , U } ∗ A RNA secondary structure is a (partial) injective function P ⊆ { 1 , . . . , n } 2 such that ( i , j ) ∈ P ↔ ( j , i ) ∈ P ( i , j ) ∈ P only if ( s i , s j ) ∈ { ( A , U ) , ( U , A ) , ( C , G ) , ( G , C ) , ( U , G ) , ( G , U ) } We are interested in a solution with maximal pairings (and/or minimizing a more complex energy function) 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 A C C U G G U A U C G A C A Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 37 / 98

  40. RNA secondary structure prediction Complexity Complexity The general problem is NP-complete [Lyngsø and Pedersen 2000]. A large sub-class has polynomial time complexity: the absence of pseudo-knots, e.g. (8,10). 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 A C C U G G U A U C G A C A Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 38 / 98

  41. RNA secondary structure prediction Complexity Pseudo-knots To avoid pseudo-knots, we impose a constraint: If i < ℓ < j and ( i , j ) ∈ P , and ( ( ℓ, k ) ∈ P or ( k , ℓ ) ∈ P ), then i < k < j . 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 A C C U G G U A U C G A C A Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 39 / 98

  42. RNA secondary structure prediction Modeling A simple CP encoding Input s 1 , . . . , s n ∈ { A , C , G , U } Variables Pairs = [ P 1 , . . . , P n ] with domain 0 .. n . Let S x = { i ∈ { 1 , . . . , n } | s i = x } . If s i = A , then dom ( P i ) = { 0 } ∪ S U . If s i = C , then dom ( P i ) = { 0 } ∪ S G . If s i = G , then dom ( P i ) = { 0 } ∪ S C ∪ S U . If s i = U , then dom ( P i ) = { 0 } ∪ S A ∪ S G . For i = 1 , . . . , n , if P i > 0 then P P i = I . If P i = 0 no constraint. In CLP(FD) we can state: element ( P + 1 , [ I | Pairs ] , I ) Pseudo-knots: If P i > 0 then ( P i + 1 ∈ [ i + 3 .. P P i − 1 ]) ∨ ( P i + 1 = 0 ) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 40 / 98

  43. RNA secondary structure prediction Modeling A simple CP encoding As cost function we want either to maximize contacts or (as done by Dahl-Bavarian, WCB 05), a solution close to the statistics, namely 35% for AU, 53% for CG, 12% for GU. Let NC = n − # contacts We minimize therefore a weighted sum of the form #( AU ) − . 35 ( n − NC ) #( CG ) − . 53 ( n − NC ) NC c 1 + c 2 + c 3 n n n ( c 1 , c 2 , c 3 constants that can be changed. The denominator n can be omitted for minimization) Other functions can be implemented, of course. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 41 / 98

  44. RNA secondary structure prediction Modeling (Some) References M. Zucker and P . Stiegler. Optimal computer folding of large RNA sequences using thermodynamics and auxiliary information. Nucleid Acid Research, 9(1):133–148, 2981. R.B. Lyngsø and C.N.S Pedersen. RNA Pseudoknot prediction in Energy-Based Models. J. of Computational Biology 7(3/4), 2000. G. Blin, G. Fertin, I. Rusu, and C. Sinoquet. Extending the hardness of RNA secondary structure comparison. LNCS 4614, pp. 140–151, 2007. M. Bauer, G.W. Klau, and K. Reinert. Accurate multiple sequence-structure alignment of RNA sequences using combinatorial optimization. BMC Bioinformatics, 8, 2007. M. Bavarian and V. Dahl. Constraint Based Methods for Biological Sequence Analysis. J. Universal Computer Science 12(11):1500–1520, 2006 (also in WCB 05). A. Dal Palù, M. Möhl, and S. Will. A Propagator for Maximum Weight String Alignment with Arbitrary Pairwise Dependencies. CP 2010: 167-175 (also in WCB 10) Alexander Bau, Johannes Waldmann and Sebastian Will: RNA Design Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 42 / 98

  45. Protein Structure Prediction Protein Structure Prediction Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 43 / 98

  46. Protein Structure Prediction Central Dogma and Proteins Proteins and Central Dogma T A C G G C C C G T A A DNA T C T G A C G G G transcription A C T A G C G C U A G C C U A mRNA translation Protein S A S L The translation phase starts from a mRNA sequence and associates a protein sequence Proteins are made of amino acids (20 common different types) Amino acids are defined by letters { A , . . . , Z } \ { B , J , O , U , X , Z } Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 44 / 98

  47. Protein Structure Prediction Central Dogma and Proteins Universal code G A C U G A C U G A C U G A C U G A C U G G G G G G G C A G G G C G C G G C C A G A A C A A A C G U A A G C A A U A U A A U C U C C C C C U C C U A U A U U U U U U U U ⊣ G E D A V R S K T M R Q H P L W ⊣ C Y S L N I F The translation selects 3 RNA basis and associates 1 amino acid. The translation rules are encoded in the universal code. The code contains stop symbol and some redundant RNA triplets. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 45 / 98

  48. Protein Structure Prediction Amino acids Proteins Amino acids Proteins are molecules made of a linear sequence of amino acids. Amino acids are combined through peptide bond . Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 46 / 98

  49. Protein Structure Prediction Amino acids Proteins Amino acids Proteins are molecules made of a linear sequence of amino acids. Amino acids are combined through peptide bond . The purple dots represent the side chains , that depend on the amino acid type Side chains have different shape, size, charge, polarity, etc. A side chain contains from 1 (Glycine) up to 18 (Tryptophan) atoms. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 46 / 98

  50. Protein Structure Prediction Amino acids Proteins Amino acids There are 2 degrees of freedom (black arrows) for each amino acid A protein with n amino acids has 2 n degrees of freedom (plus side chains)! Typical size range from 50 to 500 amino acids Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 47 / 98

  51. Protein Structure Prediction The PSP problem The structure prediction problem Given the primary structure of a protein (its amino acid sequence) For each amino acid, output its position in the space (tertiary structure of a protein) A L F W K L R R ... ? ⇓ ? Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 48 / 98

  52. Protein Structure Prediction The PSP problem The structure prediction problem Given the primary structure of a protein (its amino acid sequence) For each amino acid, output its position in the space (tertiary structure of a protein) A L F W K L R R ... ? ⇓ ? Secondary structures are rigid subparts (helices, sheets) that can be “easily” predicted Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 48 / 98

  53. Protein Structure Prediction The PSP problem Proteins Facts Folding is consistent ⇒ same protein folds in the same way [Anfinsen74] Folding is fast ⇒ 1ms – 1s Driven by non covalent forces: electrostatic interactions, volume constraints, Hydrogen Bonding, van der Waals, Salt/disulfide Bridges Backbone is rigid, interaction with water, ions and ligands There is a fixed distance (3.8Å) between the C α atoms of consecutive aminoacids. There are several statistics on (bend/torsional) angles. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 49 / 98

  54. Protein Structure Prediction The PSP problem The structure prediction problem ... and this is the hard part: In nature a protein has a unique/stable 3D conformation A cost function (that mimics physics laws) can be used to score each conformation Searching for the optimal score produces the best candidate is difficult (NP-complete even in extremely simplified modelings) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 50 / 98

  55. Protein Structure Prediction Modeling The protein structure prediction problem A first simplification (HP): Protein model: only one atom per amino acid, only 2 classes of amino acids (hydrophobic and polar) = ⇒ Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 51 / 98

  56. Protein Structure Prediction Modeling The protein structure prediction problem A first simplification (HP): Protein model: only one atom per amino acid, only 2 classes of amino acids (hydrophobic and polar) A second simplification: Spatial model: 2D square lattice to represent amino acid positions = ⇒ Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 51 / 98

  57. Protein Structure Prediction Modeling The protein structure prediction problem Model The input is a list S of amino acids S = s 1 , . . . , s n , where s i ∈ { h , p } Each s i is placed on a 2D grid with integer coordinates Any pair of two amino acids can’t occupy the same position If two amino acids are at distance 1, they are in contact Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 52 / 98

  58. Protein Structure Prediction Modeling The protein structure prediction problem Model → N 2 where A folding is a function ω : { 1 , . . . , n } − ∀ i next ( ω ( i ) , ω ( i + 1 )) and ∀ i , j ( i � = j → ω ( i ) � = ω ( j )) next ( � X 1 , Y 1 � , � X 2 , Y 2 � ) ⇐ ⇒ | X 1 − X 2 | + | Y 1 − Y 2 | = 1. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 53 / 98

  59. Protein Structure Prediction Modeling The protein structure prediction problem Model → N 2 where A folding is a function ω : { 1 , . . . , n } − ∀ i next ( ω ( i ) , ω ( i + 1 )) and ∀ i , j ( i � = j → ω ( i ) � = ω ( j )) next ( � X 1 , Y 1 � , � X 2 , Y 2 � ) ⇐ ⇒ | X 1 − X 2 | + | Y 1 − Y 2 | = 1. Find a folding that minimizes the (simplified) energy function: � Pot ( s i , s j ) · next ( ω ( i ) , ω ( j )) E ( S , ω ) = 1 ≤ i ≤ n − 2 i + 2 ≤ j ≤ n where Pot ( p , p ) = Pot ( h , p ) = Pot ( p , h ) = 0 and Pot ( h , h ) = − 1. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 53 / 98

  60. Protein Structure Prediction Modeling The protein structure prediction problem Complexity With N 2 and HP , establishing whether there is a folding with energy < k is NP-complete (Crescenzi, Goldman, Papadimitriou, Piccolboni, Yannakakis. On the Complexity of Protein Folding. Journal of Computational Biology 5(3): 423-466 (1998)) This formulation of the problem has a nice property: you can teach it to a children without speaking of proteins and so on: Write a folding using paper and pencil that maximizes the contacts between “H” aminoacids (black circles) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 54 / 98

  61. Protein Structure Prediction Modeling Example of PF HP N 2 Yellow: H, Grey: P. All foldings have energy -6 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 55 / 98

  62. Protein Structure Prediction Modeling HP on N 2 : FD encoding Primary = [ a 1 , . . . , a n ] = [ h / p , p / p , h / p , ... ] Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 56 / 98

  63. Protein Structure Prediction Modeling HP on N 2 : FD encoding Primary = [ a 1 , . . . , a n ] = [ h / p , p / p , h / p , ... ] Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] W.l.o.g., let X 1 = X 2 = Y 1 = n , Y 2 = n + 1. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 56 / 98

  64. Protein Structure Prediction Modeling HP on N 2 : FD encoding Primary = [ a 1 , . . . , a n ] = [ h / p , p / p , h / p , ... ] Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] W.l.o.g., let X 1 = X 2 = Y 1 = n , Y 2 = n + 1. Namely, we start with n + 1 ② ✻ n ② n − 1 n − 1 n n + 1 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 56 / 98

  65. Protein Structure Prediction Modeling HP on N 2 : FD encoding Primary = [ a 1 , . . . , a n ] = [ h / p , p / p , h / p , ... ] Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] W.l.o.g., let X 1 = X 2 = Y 1 = n , Y 2 = n + 1. Namely, we start with n + 1 ② ✻ n ② n − 1 n − 1 n n + 1 dom ( X 1 ) = · · · = dom ( X n ) = dom ( Y 1 ) = · · · = dom ( Y n ) = 1 .. 2 n Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 56 / 98

  66. Protein Structure Prediction Modeling HP on N 2 : FD encoding Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] contiguous: for i = 1 , . . . , n − 1: | X i − X i + 1 | + | Y i − Y i + 1 | = 1 no-overlap: for i = 1 , . . . , n − 1, for j = i + 1 , . . . , n : | X i − X i | + | Y i − Y j | ≥ 1 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 57 / 98

  67. Protein Structure Prediction Modeling HP on N 2 : FD encoding Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] contiguous: for i = 1 , . . . , n − 1: | X i − X i + 1 | + | Y i − Y i + 1 | = 1 no-overlap: for i = 1 , . . . , n − 1, for j = i + 1 , . . . , n : | X i − X i | + | Y i − Y j | ≥ 1 We want to express that ( X i , Y i ) � = ( X j , Y j ) . Can we use alldifferent? Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 57 / 98

  68. Protein Structure Prediction Modeling HP on N 2 : FD encoding Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] contiguous: for i = 1 , . . . , n − 1: | X i − X i + 1 | + | Y i − Y i + 1 | = 1 no-overlap: for i = 1 , . . . , n − 1, for j = i + 1 , . . . , n : | X i − X i | + | Y i − Y j | ≥ 1 We want to express that ( X i , Y i ) � = ( X j , Y j ) . Can we use alldifferent? Let [ P 1 , . . . , P n ] be a list and M a “big” integer (100 is ok for us). for i = 1 , . . . , n − 1: P i = X i + MY i . Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 57 / 98

  69. Protein Structure Prediction Modeling HP on N 2 : FD encoding Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] contiguous: for i = 1 , . . . , n − 1: | X i − X i + 1 | + | Y i − Y i + 1 | = 1 no-overlap: for i = 1 , . . . , n − 1, for j = i + 1 , . . . , n : | X i − X i | + | Y i − Y j | ≥ 1 We want to express that ( X i , Y i ) � = ( X j , Y j ) . Can we use alldifferent? Let [ P 1 , . . . , P n ] be a list and M a “big” integer (100 is ok for us). for i = 1 , . . . , n − 1: P i = X i + MY i . We can now post: alldifferent ([ P 1 , . . . , P n ]) . Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 57 / 98

  70. Protein Structure Prediction Modeling HP on N 2 : FD encoding Primary = [ a 1 , . . . , a n ] = [ h , p , p , h , p , p , h , ... ] Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 58 / 98

  71. Protein Structure Prediction Modeling HP on N 2 : FD encoding Primary = [ a 1 , . . . , a n ] = [ h , p , p , h , p , p , h , ... ] Tertiary x = [ X 1 , . . . , X n ] , Tertiary y = [ Y 1 , . . . , Y n ] energy: for i = 1 , . . . , n − 2, for j = i + 2 , . . . , n : c i , j ∈ { 0 , − 1 } c i , j = − 1 ↔ ( | X i − X i | + | Y i − Y j ) | = 1 ) ∧ ( a i = a j = h ) Energy = � n − 2 � n j = i + 2 c i , j i = 1 Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 58 / 98

  72. Protein Structure Prediction Modeling 3D Lattice models: Cube, FCC, Chess Knight Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 59 / 98

  73. Protein Structure Prediction Modeling The FCC lattice The Face Centered Cube lattice models the discrete space in which the protein can fold. It is proved to allow realistic conformations. The cube has size 2. Two points are connected (next) iff | x i − x j | 2 + | y i − y j | 2 + | z i − z j | 2 = 2, Each point has 12 neighbors (but 60 ◦ and 180 ◦ can be removed). Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 60 / 98

  74. Protein Structure Prediction Modeling The protein folding problem HP on FCC Backofen and Will fold HP-proteins up to length 200 on FCC using constraint programming Clever propagation, an idea of stratification and some geometrical results on the lattice. Drawbacks: It is only an abstraction. The solutions obtained are far from reality. For instance, helices and sheets are never obtained. Problems: ◦ Energy function too simple. ◦ Contact too strict. Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 61 / 98

  75. A 20 × 20 energy function Protein Structure Prediction The protein folding problem A more realistic Energy function A 20 × 20 potential matrix Pot storing the contribution for each pair of aminoacids is used. Values are either positive or negative. The notion of contact (easy) on lattice models is slightly extended: Pot ( a i , a j ) if distance ( a i , a j ) < k then Pot ( a i , a j ) else 2 distance Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 62 / 98

  76. A 20 × 20 energy function Protein Structure Prediction The protein folding problem A more realistic Energy function A 20 × 20 potential matrix Pot storing the contribution for each pair of aminoacids is used. Values are either positive or negative. The notion of contact (easy) on lattice models is slightly extended: Pot ( a i , a j ) if distance ( a i , a j ) < k then Pot ( a i , a j ) else 2 distance COLA (COnstraint solving on LAttices) can predict on FCC proteins of length 100–120 in reasonable time Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 62 / 98

  77. Protein Structure Prediction Global constraints Global constraints contiguous Let X 1 , . . . , X n be variables with domains D 1 , . . . , D n : contiguous ( X 1 , . . . , X n ) = ( D 1 × · · · × D n ) \ { ( a 1 , . . . , a n ) ∈ ( D 1 × · · · × D n ) : ∃ i . ( 1 ≤ i < n ∧ ( a i , a i + 1 ) / ∈ E ) } where E is the set of lattice edges. CON (consistency chcking) and GAC (generalized arc consistency filtering) are polynomial Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 63 / 98

  78. Protein Structure Prediction Global constraints Global constraints contiguous Let X 1 , . . . , X n be variables with domains D 1 , . . . , D n : contiguous ( X 1 , . . . , X n ) = ( D 1 × · · · × D n ) \ { ( a 1 , . . . , a n ) ∈ ( D 1 × · · · × D n ) : ∃ i . ( 1 ≤ i < n ∧ ( a i , a i + 1 ) / ∈ E ) } where E is the set of lattice edges. CON (consistency chcking) and GAC (generalized arc consistency filtering) are polynomial Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 63 / 98

  79. Protein Structure Prediction Global constraints Global constraints contiguous Let X 1 , . . . , X n be variables with domains D 1 , . . . , D n : contiguous ( X 1 , . . . , X n ) = ( D 1 × · · · × D n ) \ { ( a 1 , . . . , a n ) ∈ ( D 1 × · · · × D n ) : ∃ i . ( 1 ≤ i < n ∧ ( a i , a i + 1 ) / ∈ E ) } where E is the set of lattice edges. CON (consistency chcking) and GAC (generalized arc consistency filtering) are polynomial Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 63 / 98

  80. Protein Structure Prediction Global constraints Global constraints alldifferent Let X 1 , . . . , X n be variables with domains D 1 , . . . , D n : alldifferent ( X 1 , . . . , X n ) = ( D 1 × · · · × D n ) \ { ( a 1 , . . . , a n ) ∈ ( D 1 × · · · × D n ) : ∃ i , j . ( 1 ≤ i < j ≤ n ∧ a i = a j ) } CON and GAC are polynomial Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 64 / 98

  81. Protein Structure Prediction Global constraints Global constraints alldifferent Let X 1 , . . . , X n be variables with domains D 1 , . . . , D n : alldifferent ( X 1 , . . . , X n ) = ( D 1 × · · · × D n ) \ { ( a 1 , . . . , a n ) ∈ ( D 1 × · · · × D n ) : ∃ i , j . ( 1 ≤ i < j ≤ n ∧ a i = a j ) } CON and GAC are polynomial Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 64 / 98

  82. Protein Structure Prediction Global constraints Global constraints alldifferent Let X 1 , . . . , X n be variables with domains D 1 , . . . , D n : alldifferent ( X 1 , . . . , X n ) = ( D 1 × · · · × D n ) \ { ( a 1 , . . . , a n ) ∈ ( D 1 × · · · × D n ) : ∃ i , j . ( 1 ≤ i < j ≤ n ∧ a i = a j ) } CON and GAC are polynomial Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 64 / 98

  83. Protein Structure Prediction Global constraints Global constraints self avoiding walk Given n variables X 1 , . . . , X n , with domains D 1 , . . . , D n , the global constraint saw is the following: saw ( X 1 , . . . , X n ) = alldifferent ( X 1 , . . . , X n ) ∩ contiguous ( X 1 , . . . , X n ) CON (and GAC) are NP-complete (Dal Palù, Dovier, Pontelli. IJDMB 4(1), 2010) Other global constraints have been studied (all distant, chain, rigid block, density maps) Agostino Dovier (Univ. of Udine, DIMI) Constraints and Bioinformatics Cork, Sept. 4, 2015 65 / 98

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