Biological Computing – DNA solution

1. Biological Computing – DNA solution Presented by Wooyoung Kim 4/8/09 CSc 8530 Parallel Algorithms, Spring 2009 Dr. Sushil K. Prasad

2. Outline • NP and NP-complete • Biological computation • Hamiltonian path problem (HPP) • Satisfaction problem • Generalized SAT • Discussion

3. NP and NP-complete • NP vs. NP-complete • NP problems: Non-deterministic Polynomial Time complexity. • NP-complete : all NP problems can be reduced to it, and if it has an efficient solution, then so do all NP problems. • No general efficient solution exists for any NP-complete problem.

4. Biological computation – Adv. • Speed of any computer is determined by: • How many parallel processes it has. • How many steps each can perform per unit time. • Biological computations could potentially have vastly more parallelism. • Ex: 3 g water contains approx. 1022 molecules. • The second factor favors conventional computers, since biological machine is limited to small fraction of a biological experiment. • However, the advantage in parallelism is so huge, the difference in the execution time is not a problem.

5. Biological computation – Disadv. • Even with parallelism, brute force approach is not always feasible, too inefficient. • The biological computer can solve any HPP of 70 or less edges. • Practically, there is not a great need, though.

6. Hamiltonian Path Problem L.M. Adleman. "Molecular Computation of Solutions To Combinatorial Problem," Science, vol. 266, 1994, pp 1021-1024. Using DNA, solve Hamiltonian Path Problem efficiently.

7. 4 3 1 0 6 2 5 Hamiltonian Path Problem 0  1  2  3  4  5  6

8. Algorithm for HPP • Generating random paths through the graph. • Keep only those paths that begin with vin and end with vout. • If the graph has n vertices, then keep only those paths that enter exactly n vertices. • Keep only those paths that enter all of the vertices of the graph at least once. • If any paths remain, say “Yes”; otherwise say “No”.

9. Implementing Step 1 • Generating random paths through the graph. • Ligation reaction (annealing) • Each vertex encoded by random 20bp sequences (Oi) • Approximately 3x1013 copies of the associated oligonucleotides (a short nucleic acid polymer) were added. Vertex 2 (O2) Vertex 3 (O3) TATCGGATCG GTATATCCGA GCTATTCGAG CTTAAAGCTA GTATATCCGAGCTATTCGAG Edge 2->3

10. Implementing Step 2 • Keep only those paths that begin with vin (O0)and end with vout(O6). • The product of step 1 were amplified by PCR (polymerase chain reaction) using O0(starting point) and O6(ending point) • Thus keep only those molecules encode paths which begin with vin and end with vout. O0 • O6

11. Implementing Step 3 • If the graph has n vertices, then keep only those paths that enter exactly n vertices. • The product of Step2 was run on an agarose gel. • The 140bp band (corresponding to double strand (ds) DNA encoding paths entering exactly seen vertices) was excised and soaked in ddH2O to extract DNA.

12. Implementing Step 4 • Keep only those paths that enter all of the vertices of the graph at least once. • The product of step 3 was affinity-purified with a biotin-avidin magnetic bead system, by • First generating single stranded (ss) DNA from the dsDNA of step3 • Then incubating the ssDNA with the O1 conjugated to magnetic beads. • Only those ssDNA containing O1 annealed to the bound O1, and were were retained. • It is repeated with O2 until O5

13. Implementing Step 5 • If any paths remain, say “Yes”; otherwise say “No”. • The product of step 4 was amplified by PCR and run on a gel.

14. Drawbacks • 7 days of lab work. • Step 4 (magnetic bead separation) is most labor-intensive work. • Possibility of errors • Pseudo-paths • Inexact reactions • Hairpin loops

15. Advantages • The number of different oligonucleotides required should grow linearly with the number of edges. • O(n) • The fastest supercomputer vs. DNA computer • 106 op/sec vs. 1014 op/sec • 109 op/J vs. 1019 op/J (in ligation step) • 1bit per 1012 nm3 vs. 1 bit per 1 nm3 (video tape vs. molecules)

16. Satisfaction problem • SAT consists of a Boolean formula of , , where each Cl is a clause of the form . Vi is a variable or its negation. Ex. • Problem : find values of the variables so that the formula is 1. • If we have n variables, then there are 2n choices to search.

17. Satisfaction problem • Graph formulation unprimed 1 primed 0 • Suppose we have n variables in the formula, where airepresents the variables. • This graph is constructed so that all paths from a1 to an+1 encode an n-bit binary number. • At each stage, a path has exactly two choices : unprimed1, primed0 • Ex. A patha1x’ a2ya3 01 , that is, x is 0 and y is 1. The graph Gn encoding two-bit numbers

18. Satisfaction problem Example • Number of variables : n=2 (x and y) • Number of clauses : m =2 • Construct a graph with (n+1) +2n nodes for each clause and connect them as the following;

19. Satisfaction problem Graph paths and SAT problem • If we have a path from a1 to an+1, that means each variable is represented by 0 or 1 and the formula satisfies. • If there is no path from start to end, then the formula does not have any solution (not satisfies). • Using the properties of DNA annealing (Watson-Crick complement binding), we can construct a graph representing the variables, and using test tubes, we can either obtain paths (satisfies) or no paths at all (not satisfies).

20. Satisfaction problem • Assign random pattern of DNA strings to each vertex. (ex. length 8) • Then decide the pattern of DNA strings of each edge. TATCCCGA GGCTCGTT GCAACCTA CCTTATAG GGCTAATG CCCACCGA ATTCGGAA TTACGGGT GGATTCCA CCCAGGGT TAATCCTA CCTTCGAT TCGAAATG CCCAATTA GCTAAGCT

21. Satisfaction problem • In an initial test tube t0, put many copies of the DNA strings corresponding the vertices and the edges. (many copies of each vertex and each edge) • Put a sequence of complement of the first half of a1 and complement of the last half of a3 : To show the start and end strings. TATCCCGA GGCTCGTT TAAG AGGT ATTCGGAA TTACGGGT GGATTCCA CCCAATTA GCTAAGCT

22. Satisfaction problem Let t0 be an initial test tube containing all the DNA strings of vertices and edges. Since the first clause is (that is, the first variable x is 1), operate E(t0,1,1) setting the first variable x to 1. Then extract only those corresponding patterns (10,11) and put it to t0-1 Put the remainder (pattern 00, 01), to t’0-1 and operate E(t’0-1,2,1) setting the second variable y to 1. Then extract only those corresponding patterns from t’0-1 and put them to t0-2 Pour t0-1 and t0-2 together to form t1 test tubes. Note that now the patterns of t1 is 01,10,11 and that is the solution of the first clause.

23. Satisfaction problem Repeat the same process for the second clause starting from t1. Since the second clause is operate E(t1, 1, 0) to extract it to the t1-1 test tube. Put the remainder to t’1-1 and make t1-2 by operating E(t’1-1, 2,0). Pour t1-1 and t1-2 into t2 test tube. Check to see if there is any DNA in the last tube. The satisfying assignments are exactly those in this final test tube.

24. Satisfaction problem

25. Satisfaction problem • For general formula with n variables and m clauses, we only need O(m) number of test tubes. (For each clause there are constant number of test tubes are additionally constructed) • The last tube are checked to see if there is any patterns (paths) left from the start vertex to the end vertex.

26. Generalized SAT • Generalize this to consider problems that correspond to any Boolean formula. • Formulas are defined by the recursive definition • Any variable x is a formula • If F is a formula, then so is F • If F and G are formulas, then so are and

27. Generalized SAT • Size of the formula S: the number of operations used to build the formula. • SAT problem: given a formula, find an assignment of Boolean values of variables so that the formula is true.  NP-complete. • Claim: A O(S) number of DNA experiments can solve this SAT problem.

28. Generalized SAT – step1 • Construct a contact network for a formula. • A contact network is a directed graph with source s and sink t • Each edge is x or • Given any assignment, an edge is connected if it is 1. For example, the above graph is 1 only if w=1 or x=y=z=1

29. Generalized SAT – step2 • Solve the SAT problem of a contact network by deciding: • Whether or not there is an assignment of values to the variables such that there is a directed connected path from s to t. • If two edges have the same label, they should be consistent. • How many of DNA experiments? – O(S)

30. Generalized SAT – claims • Note that the result follows from the two claims: • Given any formula of size S, there is a contact network of size linear in S , s.t. if the formula satisfies then the network satisfies. • Given any contact network of size S, the SAT problem for the network can be solved in O(S) DNA experiments.

31. Generalized SAT – claim 1 Existence of contact network for given formula: simple formula Any formula can be placed into a normal form with DeMorgan’s laws.

32. Generalized SAT – claim 1 Existence of contact network for given formula: general formula G is a network for E, H is a network for F. The networks for (B) The networks for

33. Generalized SAT – claim2 • Solve the SAT problem for any contact network using O(S) number of DNA experiments • Associate a test tube Pv with each node v in the contact network. • The test tube Pt associated with the sink t is the “answer” • Suppose that vu is an edge with the label x and that Pv is already constructed. Then construct Puby doing the extraction E(Pv, x,1) • If several edges leave a vertex v then use an amplify step to get multiple copies in test tube Pv • If several enter a vertex v, then pour the resulting test tubes together to form Pv.

34. Discussion • Can we actually build DNA computers? • All the methods described here assumes that all the operations are perfect without error. • However, the operations are not perfect. • In the future, the DNA-based computers are hoped to be a practical means of solving hard problems.

35. Reference R.J. Lipton. “DNA solution of hard computational problems,” Science, vol. 268, 1995, pp.542-545. L.M. Adleman. "Molecular Computation of Solutions To Combinatorial Problem," Science, vol. 266, 1994, pp 1021-1024. R.J. Lipton. “Speeding Up Computations via Molecular Biology,” unpublished manuscript, available at www.cs.princeton.edu/~rjl/