Date post: | 19-Jan-2016 |
Category: |
Documents |
Upload: | eileen-jacobs |
View: | 219 times |
Download: | 0 times |
1
Biological Computing – DNA solution
Presented by Wooyoung Kim
4/8/09
CSc 8530 Parallel Algorithms, Spring 2009
Dr. Sushil K. Prasad
Outline
NP and NP-complete
Biological computation
Hamiltonian path problem (HPP)
Satisfaction problem
Generalized SAT
Discussion
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.
Biological computation – Adv. Speed of any computer is determined by:
1. How many parallel processes it has.
2. 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.
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.
Hamiltonian Path ProblemL.M. Adleman. "Molecular Computation of Solutions
To Combinatorial Problem," Science, vol. 266, 1994, pp 1021-1024.
Using DNA, solve Hamiltonian Path Problem efficiently.
Hamiltonian Path Problem
1
0
3
2 5
6
4
0 1 2 3 4 5 6
Algorithm for HPP
1. Generating random paths through the graph.
2. Keep only those paths that begin with vin and end with vout.
3. If the graph has n vertices, then keep only those paths that enter exactly
n vertices.
4. Keep only those paths that enter all of the vertices of the graph at least
once.
5. If any paths remain, say “Yes”; otherwise say “No”.
Implementing Step 11. 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.
TATCGGATCG GTATATCCGA GCTATTCGAG CTTAAAGCTA
GTATATCCGA GCTATTCGAG
Vertex 2 (O2) Vertex 3 (O3)
Edge 2->3
Implementing Step 22. 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
Implementing Step 33. 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.
Implementing Step 44. 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
Implementing Step 55. If any paths remain, say “Yes”; otherwise say “No”.
The product of step 4 was amplified by PCR and run on a gel.
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
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)
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.
mCCC ...21
kvvv ...21
)()( yxyxF
Satisfaction problem
The graph Gn encoding two-bit numbers
Graph formulation
• Suppose we have n variables in the formula, where ai represents 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 : unprimed1, primed0
Ex. A path a1x’ a2ya3 01 , that is, x is 0 and y is 1.
unprimed 1
primed 0
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;
)()( yxyxF
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).
Satisfaction problem
• Assign random pattern of DNA strings to each vertex. (ex. length 8)
• Then decide the pattern of DNA strings of each edge.
ATTCGGAA TTACGGGT GGATTCCA
TATCCCGA
GCTAAGCT
GGCTCGTT
CCCAATTA
CCTTATAG
CCTTCGAT TCGAAATG
GGCTAATG CCCACCGA
CCCAGGGT
GCAACCTA
TAATCCTA
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.
ATTCGGAA TTACGGGT GGATTCCA
TATCCCGA
GCTAAGCT
GGCTCGTT
CCCAATTA
TAAG AGGT
1. Let t0 be an initial test tube containing all the DNA strings of
vertices and edges.
2. 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
3. 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
4. Pour t0-1 and t0-2 together to form t1 test tubes.
5. Note that now the patterns of t1 is 01,10,11 and that is the solution of
the first clause.
Satisfaction problem
)( yx
)()( yxyxF
6. Repeat the same process for the second clause starting from t1.
7. Since the second clause is operate E(t1, 1, 0) to
extract it to the t1-1 test tube.
8. Put the remainder to t’1-1 and make t1-2 by operating E(t’1-1,
2,0).
9. Pour t1-1 and t1-2 into t2 test tube.
10.Check to see if there is any DNA in the last tube.
11.The satisfying assignments are exactly those in this final test
tube.
Satisfaction problem
)( yx
Satisfaction problem
Test tube OP Values
t0 initial 00, 01, 10, 11
t0-1 E(t0,1,1) 10, 11
t’0-1 Reminder of t0-1 00, 01
t0-2 E(t’0-1,2,1) 01
t1 Put t0-1 and t0-2 together 01, 10, 11
t1-1 E(t1, 1, 0) 01
t’1-1 Reminder of t1-1 10,11
t1-2 E(t’1-1,2,0) 10
t2 Put t1-1 and t1-2 together 01, 10
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.
Generalized SAT
Generalize this to consider problems that correspond to any
Boolean formula.
Formulas are defined by the recursive definition
1. Any variable x is a formula
2. If F is a formula, then so is F
3. If F and G are formulas, then so are and GF GF
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.
Generalized SAT – step1
1. 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.
x
For example, the above graph is 1 only if w=1 or x=y=z=1
Generalized SAT – step2
2. 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)
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.
Generalized SAT – claim 1
GFGF
GFGF
Existence of contact network for given formula: simple formula
Any formula can be placed into a normal form with DeMorgan’s laws.
Generalized SAT – claim 1
(A) The networks for
(B) The networks for FE
FE
Existence of contact network for given formula: general formula
G is a network for E, H is a network for F.
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 vu is an edge with the label x and that Pv is already
constructed. Then construct Pu by 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.
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
ReferenceR.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/