COMPUTATIONAL PROCESSES IN LIVING CELLS

Transcription

1 COMPUTATIONAL PROCESSES IN LIVING CELLS Lecture 7: Formal Systems for Gene Assembly in Ciliates: the String Pointer Reduction System March 31, 2010

2 MDS-descriptors MDS descriptors: strings over the following alphabet: Γ κ = { (b, e), (e, b) } { (i, j), (j, i) 2 i < j κ } { (b, i), (i, b), (i, e), (e, i) 2 i κ }. The realistic MDS descriptor associated to actin I gene in O.nova (3, 4)(4, 5)(6, 7)(5, 6)(7, 8)(9, e)(3, 2)(b, 2)(8, 9) A realistic MDS descriptor: (2, 3)(5, 4)(b, 2)(5, e)(3, 4) A non-realistic MDS descriptor: (b, 2)(e, b)(2, 3) 1

3 LD, HI, DLAD on realistic MDS descriptors For each p Π κ, the ld-rule for p is defined as follows: ld p (δ 1 (q, p)(p, r)δ 2 ) = δ 1 (q, r)δ 2, ld p ((p, m 1 )(m 2, p)) = (m 2, m 1 ), where q, r Π ex,κ, m 1, m 2 are markers and δ 1, δ 2 (Γ κ ). The first case is a simple application of ld: the two occurrences of p are separated by one IES only The second case is a boundary application of ld: all the other MDSs are between the two occurrences of p. We can always assume that the boundary ld is the last step of the assembly: (p, m 1 )(m 2, p), with m 1, m 2 markers. 2

4 LD, HI, DLAD on realistic MDS descriptors For each p Π κ, the hi-rule for p is defined as follows: hi p (δ 1 (p, q)δ 2 (p, r)δ 3 ) = δ 1 δ 2 (q, r)δ 3, hi p (δ 1 (q, p)δ 2 (r, p)δ 3 ) = δ 1 (q, r)δ 2 δ 3, where q, r Π ex,κ and δ i (Γ κ ) for each i = 1, 2, 3. Note that here the pointer p must be positive in the MDS descriptor to which hi p is applicable. 3

5 LD, HI, DLAD on realistic MDS descriptors dlad p,q (δ 1 (p, r 1 )δ 2 (q, r 2 )δ 3 (r 3, p)δ 4 (r 4, q)δ 5 ) = δ 1 δ 4 (r 4, r 2 )δ 3 (r 3, r 1 )δ 2 δ 5, dlad p,q (δ 1 (p, r 1 )δ 2 (r 2, q)δ 3 (r 3, p)δ 4 (q, r 4 )δ 5 ) = δ 1 δ 4 δ 3 (r 3, r 1 )δ 2 (r 2, r 4 )δ 5, dlad p,q (δ 1 (r 1, p)δ 2 (q, r 2 )δ 3 (p, r 3 )δ 4 (r 4, q)δ 5 ) = δ 1 (r 1, r 3 )δ 4 (r 4, r 2 )δ 3 δ 2 δ 5, dlad p,q (δ 1 (r 1, p)δ 2 (r 2, q)δ 3 (p, r 3 )δ 4 (q, r 4 )δ 5 ) = δ 1 (r 1, r 3 )δ 4 δ 3 δ 2 (r 2, r 4 )δ 5, dlad p,q (δ 1 (p, r 1 )δ 2 (q, p)δ 4 (r 4, q)δ 5 ) = δ 1 δ 4 (r 4, r 1 )δ 2 δ 5, dlad p,q (δ 1 (p, q)δ 3 (r 3, p)δ 4 (q, r 4 )δ 5 ) = δ 1 δ 4 δ 3 (r 3, r 4 )δ 5, dlad p,q (δ 1 (r 1, p)δ 2 (q, r 2 )δ 3 (p, q)δ 5 ) = δ 1 (r 1, r 2 )δ 3 δ 2 δ 5, where r i Π ex,κ and δ i (Γ κ ) for each i. The pointers p and q overlap and they both must be negative. 4

6 Next level of abstraction: strings MDS descriptors: faithful representation of the genomic structure in ciliates (only the IESs are dropped from the representation) Difficult to work on this level because of the involved definitions (too many details preserved) Idea: Make further simplifications in the representation 5

7 Simplifying the representation Gene assembly in ciliates A process of MDS ordering and assembling: reduce the number of MDSs with each step of the assembly A process of reducing pointers: a number of pointers in the beginning; 2 or 4 pointers are eliminated with each step of the process Question: Could it be possible to represent the genomic structure using pointers only, without essential loss of information? Remarkable simplification: at this level, the gene would only be represented by a linear sequence of symbols 6

8 Plan Simplify the structure by keeping only the sequence of pointers New framework: signed double occurrence strings Define the operations: The String Pointer Reduction System Prove formally that no essential information is lost, as far as the gene assembly is concerned 7

9 Some definitions Let δ = (x 1, x 2 )(x 3, x 4 )... (x 2n 1, x 2n ) be a realistic MDS descriptor. If r is a pointer and x i, x j {r, r}, then the set {x i, x i+1,..., x j } is the r-interval of r If both r and r occur in δ, then r is positive in δ If r has two occurrences in δ, then r is negative in δ Left occurrence: p in (p, q) Right occurrence: q in (p, q) 8

10 Some definitions Let δ = (p, q) Γ κ. If p, q κ, then p is an incoming pointer and q is an outgoing pointer in δ. If p, q κ, then p is an outgoing pointer and q is an incoming pointer in δ. Example: (2, 3)(5, 4)(b, 2)(5, e)(3, 4) 2 is incoming and 3 is outgoing; 5 is outgoing and 4 is incoming 9

11 On MDS descriptors Let δ Γ κ be a realistic MDS descriptor. Then: For any marker m Ψ, δ has exactly one occurrence from {m, m}. Each pointer p Π κ of δ has exactly two occurrences in δ. If p is negative, then one occurrence is left and one is right. If p is positive, then both occurrences are left or both are right. 10

12 If δ = δ 1 (x, y)δ 2, where (x, y) = (b, e) or (x, y) = (e, b) and δ 1, δ 2 Γ κ, then δ 1 = δ 2 = Λ. Similarly, if δ = δ 1 (p, m 1 )(m 2, p)δ 2, with m 1, m 2 markers, then δ 1 = δ 2 = Λ.

13 Second level of abstraction: strings Idea: model the MDS structure of the gene by its sequence of pointers only: signed double occurrence strings Compared with the MDS descriptors, we get rid of the parenthesis and the markers Note: the loss of information appears to be huge; e.g., there are different MDS descriptors having the same sequence of pointers Example: (b, 2)(3, 4)(e, 4)(3, 2) (2, 3)(4, e)(4, 3)(2, b)

14 Second level of abstraction: strings Idea: model the MDS structure of the gene by its sequence of pointers only: signed double occurrence strings Compared with the MDS descriptors, we get rid of the parenthesis and the markers Note: the loss of information appears to be huge; e.g., there are different MDS descriptors having the same sequence of pointers Nevertheless, as far as the gene assembly process is concerned, no essential information is lost; on the other hand, we simplify the model by removing the unnecessary details 12

15 The framework: legal and realistic strings A string v Σ over an (unsigned) alphabet Σ is a double occurrence string, if every letter a dom(v) occurs exactly twice in v. A signing of a nonempty double occurrence string is called a legal string. Notation: Σ = (Σ Σ) is the set of signed strings over Σ Let a Σ Σ and let u Σ be a legal string. If u contains both a and a, then a is positive in u; otherwise, a is negative in u. 13

16 Example u = u is legal. Pointers 2 and 5 are positive in u, while 3 and 4 are negative in u. w = w is not legal:??? 14

17 Example u = u is legal. Pointers 2 and 5 are positive in u, while 3 and 4 are negative in u. w = w is not legal: ξ(w) = is not a double occurrence string. (Recall that ξ removes the bars from the strings.) 15

18 Intervals and overlaps u = a 1 a 2... a n a legal string over Σ, where a i Σ Σ for each i. For each letter a occurring in u, there are 1 i < j n such that ξ(a i ) = a = ξ(a j ). Then is the a-interval of u. u (a) = a i a i+1... a j a, b Σ are overlapping in u if the a-interval and the b-interval of u overlap: if u (a) = a i1... a j1 and u (b) = a i2... a j2, then either i 1 < i 2 < j 1 < j 2 or i 2 < i 1 < j 2 < j 1. 16

19 Intervals and overlaps Denote: O u (a) = {b Σ b overlaps with a in u} {a}. Similarly, denote by O u + (a) (Ou (a), resp.) the set of all positive letters (negative letters, resp.) in O u (a). Note: it is convenient to let a O u (a). If a is positive in u, then a O u + (a), and if a is negative in u, then a Ou (a). 17

20 From MDS descriptors to legal strings Idea: drop the parenthesis and the markers, retaining only the sequence of pointers. The result is a signed double occurrence string - legal string More formally, rem κ: Γ κ κ is the morphism that removes the parentheses and the markers from each MDS descriptor, defined as rem κ (x, y) = for all (x, y) Γ κ. Λ if x, y Ψ, x if x κ κ, y Ψ, y if x Ψ, y κ κ, xy if x, y κ κ 18

21 Example The realistic MDS descriptor associated to actin I gene in O.nova (3, 4)(4, 5)(6, 7)(5, 6)(7, 8)(9, e)(3, 2)(b, 2)(8, 9) Its corresponding legal string is 19

22 Example The realistic MDS descriptor associated to actin I gene in O.nova (3, 4)(4, 5)(6, 7)(5, 6)(7, 8)(9, e)(3, 2)(b, 2)(8, 9) Its corresponding legal string is

23 Realistic strings A legal string u is realistic if there exists a realistic MDS descriptor δ such that u = rem κ (δ) for some κ Previous example: is realistic Question: Are there strings that are not realistic? Yes! Example: u = is legal, but it is not realistic Proof: 21

24 Realistic strings A legal string u is realistic if there exists a realistic MDS descriptor δ such that u = rem κ (δ) for some κ Previous example: is realistic Question: Are there strings that are not realistic? Yes! Example: u = is legal, but it is not realistic Proof: An easy analysis shows that it is impossible to construct a corresponding MDS descriptor 22

25 Analyzing the loss of information Some structural information about the gene does get lost in the model simplification: there are different MDS descriptors with the same associated realistic string. Characterization of those MDS descriptors: consider M 1 M 2... M κ Let and (k 1, k 2 ) = (κ, κ 1), if κ is odd, (κ 1, κ), if κ is even. α o = M 1 M 3... M k1 and α e = M 2 M 4... M k2. 23

26 The loss of information: ambiguous MDS descriptors Let α 1 and α 2 be two different micronuclear sequences of size κ. Then α 1 and α 2 have the same associated legal string if and only if one of the following four conditions holds: α 1 = α o α e and α 2 = α e α o, (i) α 1 = α o α e and α 2 = α e α o, (ii) α 1 = α o α e and α 2 = α e α o, (iii) α 1 = α o α e and α 2 = α e α o. (iv) 24

27 The loss of information: ambiguous MDS descriptors Let α 1 and α 2 be two different micronuclear sequences of size κ. Then α 1 and α 2 have the same associated legal string if and only if one of the following four conditions holds: α 1 = α o α e and α 2 = α e α o, (i) κ κ α 1 = α o α e and α 2 = α e α o, (ii) α 1 = α o α e and α 2 = α e α o, (iii) α 1 = α o α e and α 2 = α e α o. (iv) 25

28 The loss of information: ambiguous MDS descriptors Let α 1 and α 2 be two different micronuclear sequences of size κ. Then α 1 and α 2 have the same associated legal string if and only if one of the following four conditions holds: α 1 = α o α e and α 2 = α e α o, (i) α 1 = α o α e and α 2 = α e α o, (ii) κ κ α 1 = α o α e and α 2 = α e α o, (iii) α 1 = α o α e and α 2 = α e α o. (iv) 26

29 The loss of information: ambiguous MDS descriptors Let α 1 and α 2 be two different micronuclear sequences of size κ. Then α 1 and α 2 have the same associated legal string if and only if one of the following four conditions holds: α 1 = α o α e and α 2 = α e α o, (i) α 1 = α o α e and α 2 = α e α o, (ii) α 1 = α o α e and α 2 = α e α o, (iii) κ κ α 1 = α o α e and α 2 = α e α o. (iv) 27

30 The loss of information: ambiguous MDS descriptors Let α 1 and α 2 be two different micronuclear sequences of size κ. Then α 1 and α 2 have the same associated legal string if and only if one of the following four conditions holds: α 1 = α o α e and α 2 = α e α o, (i) α 1 = α o α e and α 2 = α e α o, (ii) α 1 = α o α e and α 2 = α e α o, (iii) α 1 = α o α e and α 2 = α e α o. (iv) κ κ

31 The only ambiguous realistic strings are the following: κ κ (i) κ κ (ii) κ κ (iii) κ κ (iv)

32 String Pointer Reduction System Second level of abstraction: legal words and string pointer reduction systems Translate the MDS descriptor rules to legal strings Result: the string pointer reduction system modelling the transformation of pointers during gene assembly. This system consists of three operations, the equivalent for legal strings of ld, hi, dlad Prove the equivalence with the MDS descriptor rewriting system 29

33 String Pointer Reduction System The equivalents for strings of the three molecular operations ld, hi, dlad are snr, spr, sdr We use sometimes also the notation ld, hi, dlad to denote the operations on strings, if there is no risk of confusion Each of the string operations below maps legal strings to legal strings. 30

34 String Pointer Reduction System The string negative rule snr p for a pointer p Π is applicable to a legal string of the form u = u 1 ppu 2, where u 1, u 2. The result of this operation is snr p (u 1 p p u 2 ) = u 1 u 2. The string positive rule spr p for a pointer p Π is applicable to a legal string of the form u = u 1 pu 2 pu 3, where u 1, u 2, u 3. The result of this operation is spr p (u 1 p u 2 p u 3 ) = u 1 u 2 u 3. 31

35 The string double rule sdr p,q for pointers p, q Π with p q is applicable to a legal string of the form u = u 1 pu 2 qu 3 pu 4 qu 5, where u i for each i. The result of this operation is sdr p,q (u 1 p u 2 q u 3 p u 4 q u 5 ) = u 1 u 4 u 3 u 2 u 5.

36 String Pointer Reduction System Only three types of rules MDS descriptors: 11 cases of rules (2 for ld, 2 for hi, and 7 for dlad) Therefore, the string pointer reduction system is considerably simpler than the MDS pointer reduction system 32

37 String Pointer Reduction System A composition ϕ = ϕ n... ϕ 1 of operations from {snr p, spr p, sdr p,q p, q Π} is applicable to a legal string u, if ϕ 1 is applicable to u, and ϕ i is applicable to ϕ i 1... ϕ 1 (u) for all 1 < i n. ϕ is a string reduction of u, if ϕ is applicable to u. ϕ is successful for u, if ϕ(u) = Λ, the empty string. Looking for a gene assembly strategy means now to look for a successful reduction of the corresponding realistic string. 33

38 Example The realistic string corresponding to the actin I gene in O.nova: A successful reduction:

39 Example: assembling gene actin I in S.Nova Step 1: LD 38

40 Example: assembling the actini gene in S.nova Step 2: DLAD 39

41 Example: assembling the actini gene in S.nova Step 3: LD 40

42 Example: assembling the actini gene in S.nova Step 4: DLAD 41

43 Example: assembling the actini gene in S.nova Step 5: HI 42

44 Example: assembling the actini gene in S.nova Step 6: HI 43

45 Example The realistic string corresponding to the actin I gene in O.nova: A successful reduction: snr sdr 5, snr sdr 8, spr spr 3 Λ 41

46 Correctness of the reduction from MDS descriptors to legal strings Question: is the string pointer reduction system equivalent to the MDS pointer reduction system? In other words: For a realistic MDS descriptor δ, can we translate a successful assembly of δ into a successful reduction of its corresponding realistic string? Also the other way around: for a realistic string u, can we translate a successful reduction of u into a successful assembly of its corresponding realistic MDS descriptors? 42

47 BONUS TOPIC Recovering the gene structure from the pointer sequence Problem: We get the sequence of pointers from the lab, but we do not know which is the first pointer, which is the second, etc. Deduce this information. Note: In general the problem is in fact more complicated because one does not know which of the reported sequences are inversions of ciliate pointers. For simplicity, assume here that we do have the information on the orientation of the pointers. 43

48 Recovering the gene structure from the pointer sequence The problem in formal terms A legal string u over Σ is said to be realizable if u is isomorphic to a realistic legal string, i.e., if there exists an injective morphism τ : Σ such that τ(σ) and τ(u) is realistic. Decide if a legal string is realizable and if it is, construct a corresponding MDS descriptor 44

49 Example u = is legal. u is not realistic: can never be a substring of a realistic legal string. u is realizable: define τ(2) = 2, τ(3) = 5, τ(4) = 4, and τ(5) = 3: is realistic. τ(u) =

50 Solution: graphs and alternating hamiltonian paths We associate to a legal string w = a 1 a 2... a 2κ a graph A w, with labels both on vertices and edges Vertices: V = {1, 2,..., 2κ} with labels l(i) = a i. Edges: E = E 0 E 1, where with the labels E 0 = { (i, j) a i = a j, 1 i < j 2κ }, E 1 = { (i, j) j i = 1, 1 i < 2κ }, h(e) = 0 if e E 0, 1 if e E 1. 46

51 Example Let w = be a legal string. Its associated graph is drawn below: a solid line represents the color 1 and a dashed line the color 0. The labels of the vertices are written beneath them

52 Alternating hamiltonian path A path e 1 e 2...e k in the graph A w is alternating if h(e 2i+1 )=0and h(e 2(i+1) )=1for each i. An alternating path is called alternating hamiltonian if it visits every vertex of the graph exactly once. 38

53 Example The previous example visited. The graph presented in the previous example has an alternating hamiltonian path, drawn in the second picture below

54 Relabelling a realizable string to get a realistic one Based on the alternating Hamiltonian path, one can easily define a relabelling transforming the string into a realistic one, see bellow an example (the previous example continued)

55 Relabelling a realizable string to get a realistic one For a legal string and a pointer x occurring, one can define a unique alternating path starting at x in the following way: Start with the edge x 1 0 x2, where x 1 = x and l(x 2 )=l(x 1 ). Iterate then the next step: Suppose the alternating path x 1 0 x xi 1 0 xi has been constructed so that the color of the edge {x i 1,x i } is 0. If x i is labelled with a signed letter in w, then the next edges are {x i,x i 1} (of color 1) and {x i 1,x i+1 } (of color 0), where l(x i+1 )=l(x i ). 41

56 Relabelling a realizable string to get a realistic one Result: Let v be a legal string. Then v is realizable if and only if there exists a vertex x of A v such that the induced alternating path alt(v) is an alternating hamiltonian path of A v. 42