- 390 Views
- Updated On :

DNA CODES GENERATION USING AN IMPROVED METRIC. Vyacheslav V. Rykov. Outline. DNA Hybridization/Cross Hybridization DNA Codes Nearest Neighbor Thermodynamics complete computations bounds Overview of Applications and Purposes DNA Bitstring Library Biomolecular Computing.

Related searches for Tutorial 2

Download Presentation
## PowerPoint Slideshow about 'Tutorial 2' - Audrey

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

C A A C C A A A A A A- T T A C C T C A A A C C- T T C A A T C C A C A A- T C A C T C T C T C A A - C A T C T C A C C A T C

Outline

- DNA Hybridization/Cross Hybridization
- DNA Codes
- Nearest Neighbor Thermodynamics
- complete computations
- bounds

- Overview of Applications and Purposes
- DNA Bitstring Library
- Biomolecular Computing

- DNA strands are modeled by directed 3’--> 5’ sequences of letters from the alphabet {A, C, G, T}
- (A, T) and (C, G) are complementarypairs.
- Two oppositely directed DNA sequences are capable of coalescing into a duplex.
- Because an A (C) in one strand can (usually) only bind to a T (G) in the oppositely directed strand, the greatest energy of duplex formation is obtained when the two sequences are reverse-complements (complements)

Orientation of single DNA strands is important for hybridization.

A DNA Code hybridization.

Coding Strands

for Ligation

Probing Complement Strands

for Reading

TACGCGACTTTC GAAAGTCGCGTA

ATCAAACGATGC GCATCGTTTGAT

TGTGTGCTCGTC GACGAGCACACA

ATTTTTGCGTTA, TAACGCAAAAAT

CACTAAATACAA TTGTATTTAGTG

GAAAAAGAAGAA, TTCTTCTTTTTC

5’ 3’ 5’ 3’

Watson Crick

(WC) Duplexes

5’TACGCGACTTTC3’

ATCAAACGATGC

Must Have

5’GAAAGTCGCGTA3’

GCATCGTTTGAT

TACGCGACTTTC

Cross Hybridized

(CH) Duplexes

ATTTTTGCGTTA

Must Avoid

GCATCGTTTGAT

GAAAAAGAAGAA

x= hybridization. 5’ggCaCaTcatAct3’

5’ggCaCaTcatAct3’

5’ AggTTaaCcatct3’

y=5’agatgGttAAccT3’

5’ggCaCaTcatAct3’

3’ TccAAttGgtaga5’

5’agatgGttAAccT3’ =y

DNA codes hybridization. serve as universal components for biomolecular computing. DNA codes are closed under reverse-complementation. The strands in a DNA code have such binding specificity that a code strand will only hybridize with its reverse-complement and will not cross hybridize with any other code strand in the DNA code

Such collections of strands are crucial to the success biomolecular computing and biomolecular nanotechnology.

Basic idea is to have correct, parallel and autonomous addressing

Characterization of synthetic DNA bar codes in Saccharomyces cervisiae gene-deletion strains”

(Eason et al., PNAS).

DNA codes for self-assembly of any components that can be attached

to DNA. Their size presents the potential for increased complexity and location control in nanostructures produced by assembly that is driven by DNA duplex formation. Fundamentalphysical limits and increasing costs of fabrication facilities will force alternatives to conventional microelectronics manufacturing to be developed.

In self-assembly, weak, local interactions among molecular components spontaneously organize those components into aggregates with properties that range from simple to complex

DNA memory:The capacity and storage density of such memories is potentially very large. Information could

be mined through massively parallel template-matching reactions. In addition, information could be processed based upon context, and information matched associatively based upon content.

DNA Computing cervisiae gene-deletion strains”

Interest into DNA computing was sparked in 1994 by Len Adleman.

Adleman showed how we can use DNA molecules to solve a mathematical problem. (Hamiltonian path problem).

DNA computing relies on the fact that DNA strands can be represented as sequences of bases (4-ary sequences) and the property of hybridization.

In Hybridization, errors can occur. Thus, error-correcting codes are required for efficient synthesis of DNA strands to be used in computing.

DNA Computing Strand Engineering cervisiae gene-deletion strains”

No codeword-codewode CH (cc-CH)

No codeword-probe CH (cp-CH)

No probe-probe CH (pp-CH)

A A A A A A A A C C=T1

G G T T T T T T T T =BEAD PROBE (T1)

T T T C C A A A A A =F1

T T T T T G G A A A = BEAD PROBE (F1)

T T T C T T A A C C=T2

G G T T A A G A A A= BEAD PROBE (T2)

A C T A A C A A A A=F2

T T T T G T T A G T= BEAD PROBE (F2)

C A T A A A A C A C=T3

G T G T T T T A T G= BEAD PROBE (T3)

A T C T T T T C A A=F3

T T G A A A A G A T= BEAD PROBE (F3)

C A A T C C A T T A=T4

T A A T G G A T T G= BEAD PROBE (T4)

C C T T C T A A A T=F4

A T T T A G A A G G= BEAD PROBE (F4)

A C T C C T A A T A=T5

T A T T A G G A G T= BEAD PROBE (T5)

T C T C T C T A C T=F1

A G T A G A G A G A= BEAD PROBE (F5)

Only Allowed Hybridizations

T T T T T T G G T T G G=Probe(T1)

G G T G G T T T T T T T=Probe(F1)

T G G A A G G A A A A A=Probe(T2)

G G T T T G A G G T A A =Probe(F2)

G G A G T T G T G A A A=Probe(T3)

C C A A C C A A A A A A = T1

A A A A A A A C C A C C=F1

T T T T T C C T T C C A =T2

T T A C C T C A A A C C =F2

T T T C A C A A C T C C=T3

No cp-CH

T T G T G G A T T G A A=Probe(F3)

T T G A G A G A G T G A=Probe(T4)

A G A G G A G A A A G A=Probe(F4)

G A T G G T G A G A T G=Probe(T5)

G T G T G T A G T G T T=Probe(F5)

T T C A A T C C A C A A =F3

T C A C T C T C T C A A =T4

T C T T T C T C C T C T=F4

C A T C T C A C C A T C =T5

A A C A C T A C A C A C =F5

No cc-CH

No pp-CH

DNA Computing Strand Engineering cervisiae gene-deletion strains”

No codeword cp-CH

T T C A A T C C A C A A =F3

T T G T G G A T T G A A=Probe(F3)

T C A C T C T C T C A A =T4

T T G A G A G A G T G A=Probe(T4)

T C T T T C T C C T C T=F4

A G A G G A G A A A G A=Probe(F4)

C A T C T C A C C A T C =T5

G A T G G T G A G A T G=Probe(T5)

A A C A C T A C A C A C =F5

G T G T G T A G T G T T=Probe(F5)

C C A A C C A A A A A A = T1

T T T T T T G G T T G G=Probe(T1)

A A A A A A A C C A C C=F1

G G T G G T T T T T T T=Probe(F1)

T T T T T C C T T C C A =T2

T G G A A G G A A A A A=Probe(T2)

T T A C C T C A A A C C =F2

G G T T T G A G G T A A =Probe(F2)

T T T C A C A A C T C C=T3

G G A G T T G T G A A A=Probe(T3)

PROBE(F2)

G G T T T G A G G T A A

C A A C C A A A A A A- T T A C C T C A A A C C- T T C A A T C C A C A A- T C A C T C T C T C A A - C A T C T C A C C A T C

Yes WC bonding

Yes, bitstring is F2

Good read

T1-F2-F3-T4-T5

1 0 0 1 1

PROBE(T2)

G G A G T T G T G A A

C A A C C A A A A A A- T T A C C T C A A A C C- T T C A A T C C A C A A- T C A C T C T C T C A A - C A T C T C A C C A T C

Darn! CH bonding

No, bitstring is not T2 Bad read

T1-F2-F3-T4-T5

DNA Computing Strand Engineering cervisiae gene-deletion strains”

No codeword pp-CH, cc-CH

PROBE(F2)

pp-CH

interferes with reading

G G T T T G A G G T A A

T T G A G A G A GT G

PROBE(T4)

PROBE(F2)

G G T T T G A G G T A A

bonding site

competition

T T G A G A G A GT G

C A A C C A A A A A A- T T A C C T C A A A C C- T T C A A T C C A C A A- T C A C T C T C T C A A - C A T C T C A C C A T C

cc-CH

interferes with separation and leads to unwanted library strand interaction

T1-F2-F3-T4-T5

F1-F2-T3-T4-f5

T T T C C A A A A-AT T A C C T C A A A C C- T T T C A C A A C T C C-T C A C T C T C T C A A - A A C A C T A C A C A C

Watson-Crick Nearest Neighbor Computation cervisiae gene-deletion strains”

1.44

2.24

WC

Duplex

5’g g c a c a3’

3’c c g t g t 5’

5’g g c a c a3’

5’g g c a c a3’

NNFE=8.42

5’g g c a c a3’

5’g g c a c a3’

1.84

1.45

1.45

Cross Hybridized Nearest Neighbor Upper Bound Computation cervisiae gene-deletion strains”

1.45

1.28

5’ggCaCaTcatAct3’

3’ TccAAttGgtaga5’

5’g gC aCaTcatAct3’

3’ Tc cA AttGgtaga5’

5’g gC aC a T c a t A ct3’

5’ A g g T T a a C c a t ct3’

.27

1.84

0.88

NNFE~<5.45

5’ggCaCaTcatAct3’

5’ AggTTaaCcatct3’

NNFE~<5.72

Intermolecular Interactions cervisiae gene-deletion strains”

Duplexes

loop

symmetric

loop

asymmetric

2.90 (5.3)

2.20 (5.3)

Intramolecular Interactions cervisiae gene-deletion strains”

CAAGACTTTTTGGTAGTAAA

***TTTCCC*********GGAA***GGGAAA***********TTCC***

NNFE~< cervisiae gene-deletion strains” 5.66

NNFE~<5.66+ .59 + .32=6.57

5’ cervisiae gene-deletion strains” ggC aC aT c a tA ct3’

3’ T cc AA t t G g t aga5’

5’ggC a C a T c a t A ct3’

5’ A ggTT a aC c a t ct3’

5’ggCaCaTcatAct3’

3’ TccAAttGgtaga5’

Virtual Stacked Pairs

Virtual Duplex

5’ggCaCaTcatAct3’

5’ AggTTaaCcatc3’

5’ cervisiae gene-deletion strains” GGCACATCATACT3’

5’AGTATGATGTGCC 3’

5’AGGTTAACCATCT3’

5’AGATGGTTAACCT3’

5’GGCACATCATACT3’

Neareast Neighbor Appr. Free Energy

of duplex formation (WC)

5’AGTATGATGTGCC 3’

…= 18.8

2.24

1.84

1.45

5’GGCACATCATACT3’

5’AGGTTAACCATCT3’

5’ggCaCaTcatAct3’

3’ TccAAttGgtaga5’

5’ggC aC aT c a tA ct3’

3’ T cc AA t t G g t aga5’

1.28

1.45

5’ggCaCaTcatAct3’

5’ AggTTaaCcatct3’

5’ggC a C a T c a t A ct3’

5’ A ggTT a aC c a t ct3’

NNFE

CH =6.45

0.88

1.84

Basic Notations cervisiae gene-deletion strains”

Let denote a set consisting of all vectors (codewords) of length n built over

i.e.

Let such that:

1)

2)

3)

Let be such that:

is referred to as a Code of length n, size M, and minimum distance d.

A sphere in centered at cervisiae gene-deletion strains” x having radius d:

Volume of the sphere around x, of radius d:

Spaces

A space is HOMOGENEOUS when the volume of a sphere does not depend on where it is centered i.e.

A space is NON - HOMOGENEOUS when the volume of a sphere does depend on where it is centered.

Similarity cervisiae gene-deletion strains”

Sequence

is a subsequence of

if and only if there exists a strictly increasing sequence of indices:

Such that:

is defined to be the set of longest common subsequences of

and

is defined to be the length of the longest common subsequenceof

and

Example of LCS cervisiae gene-deletion strains”

Just what it says:

x = <A C G T C G A G C>

y = <G A C G C T G A G>

LCS(x, y) = {<A C G C G A G>}

|LCS(x, y)| = 7

Insertion-Deletion Metric cervisiae gene-deletion strains”

Original Insertion-Deletion metric (Levenshtein 1966):

This metric results from the number of deletions and insertions that need to be made to obtain ‘ y ’from ‘ x ’.

For vectors that have the same length:

the number of deletions that will be made is:

likewise, the number of insertions that will be made is:

Better Metric ? cervisiae gene-deletion strains”

- LCS is simple and easy to compute.
- LCS essentially is a count of the number of base pairings between two sequences, and thus does approximate bonding energy.
- Clue: if two base pairs bond, but neither their neighbors to the right or left bond, it really doesn’t contribute much.
- We might call such inconsequential bonds “lone” bonds.

“Lone Bonds” cervisiae gene-deletion strains”

B B B B B B B B B B B B B B B B

B B B B B B B B B B B B B B B B

The red bonds are “lone bonds” that don’t contribute to the binding energy.

Block LCS cervisiae gene-deletion strains”

The longest common subsequence SUCH THAT:

If xi is matched to yj, THEN EITHER

xi-1 is matched to yj-1, OR

xi+1 is matched to yj+1

Longest Common Stacked Pair Subsequence cervisiae gene-deletion strains”

A common subsequence is called a common stacked pair subsequence of length between x and y if two elements , are consecutive inx and consecutive inyor if they are non -consecutive in xand ornon-consecutiveiny, then and are consecutive in xandy.

Let , denote the length of the longest sequence occurring as a common stacked pair subsequence subsequence zbetween sequences x and y. The number , is called a similarityof blocks between xand y.The metric is defined to be

Bounds in Coding Theory cervisiae gene-deletion strains”

We will be working in a NON-HOMOGENEOUS space making the obtainment of exact formulas for sphere volumes and code sizes VERY HARD.

[6] L. M. G. M. Tolhuizen (1997): The Generalized Varshamov-Gilbert Bound is Implied

by Turan’s Theorem, IEEE Transactions on Information Theory, 43:05.

Varshamov-Gilbert Lower Bound on Code Size in with any metric:

Turan's Theorem cervisiae gene-deletion strains”

Let G be a simple graph on vertices and e edges. G contains an M-clique if:

CLIQUES:

The edge set of G is constructed as follows; an edge (x, y) exists in G if and only if

d(x, y) > d. The first question is; how many edges does G have? This can be found by taking spheres of radius d − 1 around each

vector and counting how many vectors are outside the particular sphere. Since

edges will be double counted, we must divide by 2:

Stacked Pair Metric Bounds exists in G if and only if

The upper bound for the average sphere volume in this metric will be:

The Varshamov-Gilbert bound becomes:

Insertion-deletion stacked pair exists in G if and only if

thermodynamic metric

Thermodynamic weight of virtual stacked pairs.

- Can use statistical estimation of sphere volume.

A C G C G T T A exists in G if and only if

C T G A T A C A

Get LCS of this and add 1 for the

A’s that have to match

Case 1:sequences end with the same symbolA C G C G T T A exists in G if and only if

C T G A T A C C

Take the best LCS of these two

Case 2:sequences end with different symbolsSolve Problem Recursively exists in G if and only if

- If x(i) and y(j) end with the same symbol, say A, then: LCS(x(i), y(j)) = LCS(x(i – 1), y(j – 1) + A
- If xi and yj do NOT end with the same symbol, then: LCS(x(i), y(j)) = max[LCS(x(i – 1), y(j)), LCS(x(i), y(j – 1))]

Inefficient: we keep evaluating the same LCS(i, j) over and over.

Instead, use dynamic programming.

Fill in a table of LCS(i, j) values by i and j.

You only have to figure each LCS(i, j) once.

O(n2).

Dynamic Programming

Stacked pair metric over.

Algorithm for Stacked Pair Metric

The longest common subsequence SUCH THAT there are no lone bonds.

If xi is matched to yj, THEN EITHER

xi-1 is matched to yj-1, OR

xi+1 is matched to yj+1

Cannot “break” a block LCS over.

Big regular LCS:

A C T G C T

G A C G C T

Break to get two smaller regular LCS’s:

A C T G C T

G A C G C T

Cannot “break” a block LCS over.

Big block LCS:

G G T A G G

C C T A C C

CANNOT break to get two smaller block LCS’s:

G G T A G G

C C T A C C

Adding a single symbol to a string can have effects arbitrarily far back

A C T C C C C T

G G G G G A C T G

A C T C C C C T G

G G G G G A C T G

These three bonds make the LCSP.

Add just one symbol, G, and the red bond must be moved to make the new LCSP.

Two algorithms arbitrarily far back

Tail equality arbitrarily far back of two sequences

Tail equality 3:

A G C T C

A T C T C

Tail equality 0:

A G C T G

A T C T A

End count of a matching

End count 2:

A G C T C

A T C T C

End count 0:

A G C T G

A T C T A

- The end count of a matching between x and y cannot exceed the tail equality of x and y.
- Let LCSP(k)(i, j) be the length of the longest LCSP(i, j) achievable with a matching of end count k.
- where e is the tail equality of x and y.

Example: Figure LCSP the tail equality of x and y.(3):

A C T G C T A T

A C T G C T A T

best of these two + 3

Substituting: the tail equality of x and y.

- O(n) worst case for one cell.
- O(n3) for algorithm.

- In practice, only 56% more time.
- “Efficient” algorithm takes O(n) memory.
- “Simple” algorithm takes O(n2) memory.

Code Generation the tail equality of x and y.

- Start with empty code.
- Repeatedly generate random codewords and add them if they meet the distance requirement.

When to stop?

- After “n” trials?
- When “n” trials in a row have failed?
- When fewer than “i” of the last “n” trials have succeeded?
- When the size of the code is near a maximum predicted by theory?

Empirical Relation Between Codeword Length and Code Size the tail equality of x and y.

A DNA Computing Paradigm the tail equality of x and y.

The identification of maximal frequent sets in data fields are the computational bottleneck in association rule discovery. This is an important problem and the independent sets and maximal cliques problems fit this paradigm.

DNA Code and DNA Bitstring Library the tail equality of x and y.

A A A A A A A A C C=T1

G G T T T T T T T T =BEAD PROBE (T1)

T T T C C A A A A A =F1

T T T T T G G A A A = BEAD PROBE (F1)

T T T C T T A A C C=T2

G G T T A A G A A A= BEAD PROBE (T2)

A C T A A C A A A A=F2

T T T T G T T A G T= BEAD PROBE (F2)

C A T A A A A C A C=T3

G T G T T T T A T G= BEAD PROBE (T3)

A T C T T T T C A A=F3

T T G A A A A G A T= BEAD PROBE (F3)

C A A T C C A T T A=T4

T A A T G G A T T G= BEAD PROBE (T4)

C C T T C T A A A T=F4

A T T T A G A A G G= BEAD PROBE (F4)

A C T C C T A A T A=T5

T A T T A G G A G T= BEAD PROBE (T5)

T C T C T C T A C T=F5

A G T A G A G A G A= BEAD PROBE (F5)

1. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-T4-T5

2. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-T4-F5

3. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-F4-T5

4. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-F4-F5

5. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-T4-T5

6. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-T4-F5

7. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-F4-T5

8. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-F4-F5

9. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-T5

10. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-F5

11. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-F4-T5

12. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-F4-F5

13. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-T4-T5

14. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-T4-F5

15. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-F4-T5

16. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-F4-F5

17. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-T4-T5

18. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-T4-F5

19. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-F4-T5

20. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-F4-F5

21. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-T4-T5

22. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-T4-F5

23. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-T5

24. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-F5

25. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-T4-T5

26. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-T4-F5

27. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-F4-T5

28. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-F4-F5

29. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-T5

30. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-F5

31. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-F4-T5

32. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-F4-F5

DNA LIBRARY=

DNA BITSTRINGS

DNA CODE

Example: Independent Sets and Cliques the tail equality of x and y.

3

Edges in G are

{1,2}, {2,3}, {3,4},

{4,5},{1,4},{2,5}

Edges in G’ are

{1,3}, {1,5}, {2,4},

{3,5},

4

2

3

3

2

2

4

G =

G’=

4

5

1

1

5

1

5

An independent set is a collection of vertices that contains no edge.

A clique is a subgraph were every pair of vertices has an edge between them.

For a graph G, its complement G’ is the set of edges not in G

A maximal independent set in G is a maximal clique in G’, e.g., {1,3,5}.

3

1

5

DNA Computing for Independent Sets and Cliques the tail equality of x and y.

3

3

2

2

4

G =

G’=

4

5

1

1

5

1. the tail equality of x and y.A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-T4-T5

2. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-T4-F5

3. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-F4-T5

4. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-F4-F5

5. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-T4-T5

6. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-T4-F5

7. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-F4-T5

8. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-F4-F5

9. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-T5

10. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-F5

11. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-F4-T5

12. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-F4-F5

13. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-T4-T5

14. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-T4-F5

15. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-F4-T5

16. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-F4-F5

17. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-T4-T5

18. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-T4-F5

19. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-F4-T5

20. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-F4-F5

21. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-T4-T5

22. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-T4-F5

23. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-T5

24. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-F5

25. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-T4-T5

26. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-T4-F5

27. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-F4-T5

28. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-F4-F5

29. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-T5

30. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-F5

31. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-F4-T5

32. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-F4-F5

DNA Library

2^( # Coding Strands / 2)

# Coding Strands / 2 Bits

T T T T T G G A A A

24. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-F5

T T T T G T T A G T

10.A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-F5

X1=F or X2=F

T T T T G T T A G T

T T T T T G G A A A

29. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-T5

All subsets

not containing

{1,2}

T T T T T G G A A A=Probe(F1)

T T T T G T T A G T=Probe(F2)

Edge {1,2} STM

9. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-T5

10. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-T4-F5

11. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-F4-T5

12. A A A A A A A A C C-A C T A A C A A A A-C A T A A A A C A C-F4-F5

13. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-T4-T5

14. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-T4-F5

15. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-F4-T5

16. A A A A A A A A C C-A C T A A C A A A A-A T C T T T T C A A-F4-F5

17. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-T4-T5

18. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-T4-F5

19. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-F4-T5

20. T T T C C A A A A A -T T T C T T A A C C-C A T A A A A C A C-F4-F5

21. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-T4-T5

22. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-T4-F5

23. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-T5

24. T T T C C A A A A A -T T T C T T A A C C -A T C T T T T C A A-F4-F5

25. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-T4-T5

26. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-T4-F5

27. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-F4-T5

28. T T T C C A A A A A -A C T A A C A A A A-C A T A A A A C A C-F4-F5

29. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-T5

30. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-T4-F5

31. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-F4-T5

32. T T T C C A A A A A -A C T A A C A A A A-A T C T T T T C A A-F4-F5

X1=T and X2=T

1. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-T4-T5

2. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-T4-F5

3. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-F4-T5

4. A A A A A A A A C C -T T T C T T A A C C-C A T A A A A C A C-F4-F5

5. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-T4-T5

6. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-T4-F5

7. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-F4-T5

8. A A A A A A A A C C -T T T C T T A A C C -A T C T T T T C A A-F4-F5

DNA Library the tail equality of x and y.

{1,2}

{2,4}

{1,3}

{2,5}

{1,4}

{3,4}

{1,5}

{3,5}

{2,3}

{4,5}

Black ON, Red OFF =Independent Sets in G

Black OFF, Red ON =Cliques in G

Universal DNA Computer for any the tail equality of x and y.

Graph on n Vertices

DNA Library

Every Graph G on n vertices has

G union G’= all possible pairs on

n vertices. This enables the construction

of a universal device.

{1,2}

{1,3}

Each possible edge is an STM. Then depending

on the problem, the flow is directed by the edges

present (or absent) in the given graph

.

{n-2,n}

{n-1,n}

Edges in G ON, Edges in G’ OFF =Independent Sets in G

when flow completed

Edges in G OFF, Edges in G’ ON =Cliques in G

when flow completed

Download Presentation

Connecting to Server..