6.5 RNA Secondary Structure

Transcript

1 6.5 RNA Secondary Structure

2 Struttura di una biomolecola Biomolecola: DNA, RNA Struttura primaria: descrizione esatta della sua composizione atomica e dei legami presenti fra gli atomi Struttura secondaria: capacità di assumere una struttura spaziale regolare e ripetitiva

3 DNA Struttura secondaria: doppia elica (Watson e Crick). Ogni catena è composta da nucleotidi: A, C, G, T A (adenina), C (citosina), G (guanina), T (timina) Le catene sono connesse da basi complementari: A-T, C-G

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5 Per una stessa stringa di RNA possono esistere più strutture secondarie

6 Two views of RNA secondary structure

7 RNA Secondary Structure: Examples Examples. C G G U G G G C G G U C G C G C U A U A U A G U A U A U A base pair A U G U G G C C A U A U G G G G C A U A G U U G G C C A U 4 ok sharp turn: no! crossing: no! 7

8 RNA Secondary Structure Secondary structure. A set of pairs S = { (b i, b j ) } that satisfy: [Watson-Crick.] S is a matching and each pair in S is a Watson-Crick complement: A-U, U-A, C-G, or G-C. [No sharp turns.] The ends of each pair are separated by at least 4 intervening bases. If (b i, b j ) S, then i < j - 4. [Non-crossing.] If (b i, b j ) and (b k, b l ) are two pairs in S, then we cannot have i < k < j < l. Free energy. Usual hypothesis is that an RNA molecule will form the secondary structure with the optimum total free energy. approximate by number of base pairs Goal. Given an RNA molecule B = b 1 b 2 b n, find a secondary structure S that maximizes the number of base pairs. 8

9 RNA Secondary Structure: Subproblems First attempt. OPT(j) = maximum number of base pairs in a secondary structure of the substring b 1 b 2 b j. Case 1: b j is not involved in a pair : OPT(j)=OPT(j-1) Case 2: b j matches b t for some 1 t<j-4 1 t j Difficulty. Results in two sub-problems: Finding secondary structure in: b 1 b 2 b t-1. Finding secondary structure in: b t+1 b t+2 b j-1. OPT(t-1) need different sub-problems 9

10 Dynamic Programming Over Intervals Notation. OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+1 b j. Case 1. If i j - 4. OPT(i, j) = 0 by no-sharp turns condition. Case 2. Base b j is not involved in a pair. OPT(i, j) = OPT(i, j-1) Case 3. Base b j pairs with b t for some i t < j - 4. non-crossing constraint (no match over t) decouples resulting sub-problems OPT(i, j) = 1 + max t { OPT(i, t-1) + OPT(t+1, j-1) } take max over t such that i t < j-4 and b t and b j are Watson-Crick complements 10 i t-1 t t+1 j-1 j

11 Dynamic Programming Over Intervals Notation. OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+1 b j. Case 1. If i j - 4. OPT(i, j) = 0 by no-sharp turns condition. Case 2. Base b j is not involved in a pair. OPT(i, j) = OPT(i, j-1) Case 3. Base b j pairs with b t for some i t < j - 4. non-crossing constraint (no match over t) decouples resulting sub-problems OPT(i, j) = 1 + max t { OPT(i, t-1) + OPT(t+1, j-1) } i t-1 t t+1 j-1 j Proprietà della sottostruttura ottima: Caso 3: una soluzione ottima O per b i b i+1 b j è formata dalla coppia b t - b j più un insieme L di coppie scelte da b i b t-1 e un insieme R di coppie scelte da b t+1 b j. Inoltre L è una struttura secondaria (Watson-Crick, no-sharp, non-crossing) ed è ottima (massima cardinalità) per b i b t-1, altrimenti potrei sostituirne una migliore di L in O, ottenendo una soluzione per b i b i+1 b j migliore di O: assurdo! Analogo per R. 11

12 Relazione di ricorrenza OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+1 with i, j = 1, 2,, n b j Caso generale: i< j 4 ovvero i j-5, ovvero j i+5 i = 1 allora j 6 i = 2 allora j 7 i = n-5 allora j n i = n-4 allora j n+1 i = n allora j n+5 n = 9 allora i 9-5=4, j i j

13 Bottom Up Dynamic Programming Over Intervals Q. What order to solve the sub-problems? A. Do shortest intervals first (from length 5 up to n-1). RNA(b 1,,b n ) { for k = 5, 6,, n-1 for i = 1, 2,, n-k j = i + k Compute M[i, j] } return M[1, n] using recurrence i n=9, 4=n j j = i+5, i = 1,, n-5 j = i+6, i = 1,, n-6 j = i+7, i = 1,., n-7 j = i+k, i = 1,, n-k j = n, i = 1 Running time. O(n 3 ). k = 5, 6,, n-1 13

14 Esempio

15 b 4 =G does not match b 9 =U b 2 =C does not match b 7 =A

16 0 1,8) ( 1) (3, 1 max 1 (3,8) max (3,9)., t OPT t OPT OPT OPT compl b U b t t 0 1,6) ( 1) (1, 1 max 1 (1,6) max (1,7)., t OPT t OPT OPT OPT compl A b b t t 1 1,7) ( 1) (2, 1 max 0 (2,7) max (2,8)., t OPT t OPT OPT OPT compl G b b t t

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21 Dynamic Programming Summary Recipe. Characterize structure of problem. Recursively define value of optimal solution. Compute value of optimal solution. Construct optimal solution from computed information. Dynamic programming techniques. Binary choice: weighted interval scheduling. Multi-way choice: segmented least squares. Adding a new variable: knapsack. Dynamic programming over intervals: RNA secondary structure. Top-down vs. bottom-up: different people have different intuitions. 21

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23 Esercizi: varianti al problema dello zaino

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25 Dall elenco

26 Dall elenco

27 Esercizio 1 pag. 312 KT punto a) [continua ]

28 Esercizio 1 pag. 312 KT punti b) e c)