1. Challenging Traditional Approaches to Computation
Challenging Traditional Approaches to Computation A Biomolecular Transducer Employing Ternary Language and Rendering a Biological Output
Mark Chaskes and Paul Lazarescu
Mentor: Tamar Ratner
The Schulich Faculty of Chemistry
Technion, Haifa, Israel, 32000

2. Objective
Design a theoretical biomolecular transducer to solve consecutive mathematical equations in ternary.
-First divide an input by three and then divide the yeild of that by two

3. What is biomolecular computing?
A biomolecular computer is a group of molecules that ‘read’ dsDNA and can ‘print’ an output.

4. What is a DNA based transducer?
A transducer is not a PC; it has unique capabilities that an ordinary computer does not.
Advantages include:
Direct interface with a biological system
Can release a biological output
Able to compute in parallel
Store large amounts of data in a small volume

5. Schematic Diagram S0 S1
Key
S0 and S1 are states
“a” and “b” are symbols
Begins in S0 and changes states depending on read symbols

6. Design on the Molecular Level
Symbols are dsDNA strands
Restriction enzymes cleave the sequence at recognition sites
States are determined by the location of cleavage within the symbol

7. Reading 2 from S0 prints 0 and goes to S2
Process
Divide by three transducer reading the input 2-0-0
Reading 2 from S0 prints 0 and goes to S2
State 0 2

8. Reading 0 from S2 prints 2 and goes to S0
Process
Divide by three transducer reading the input 2-0-0
Reading 0 from S2 prints 2 and goes to S0
State 2

9. Reading 0 from S0 prints 0 and encodes the output 0-2-0
Process
Divide by three transducer reading the input 2-0-0
Reading 0 from S0 prints 0 and encodes the output 0-2-0 2
State 0

10. Reading 0 from S0 prints 0 and goes to S0
Process
Divide by two transducer reading the input 0-2-0 S1
Reading 0 from S0 prints 0 and goes to S0
State 0 2

11. Reading 2 from S0 prints 1 and goes to S0
Process
Divide by two transducer reading the input 0-2-0 S1
Reading 2 from S0 prints 1 and goes to S0
State 0 2

12. Reading 0 from S0 prints 0 and goes to S0
Process
Divide by two transducer reading the input 0-2-0 S1
Reading 0 from S0 prints 0 and goes to S0 1
State 0

13. Molecular Design of the Input
Encoding in Ternary (18 in base ten)
Terminator
EagI Recognition Site
BbvI Recognition Site
Spacers
AATTCGGCCGTT..8 base..CTCCTCGCAGC..8 base..CTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGCAA..pairs ..GAGGAGCGTCG..pairs ..GAGCAATCAGAATCAGAAACGACTTTAA
Plasmid
BseRI Recognition Site

14. Divide-by-three Computation
First Restriction
AATTCGGCCGTT..8 base..CTCCTCGCAGC..8 base..CTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGCAA..pairs ..GAGGAGCGTCG..pairs ..GAGCAATCAGAATCAGAAACGACTTTAA
GGTATT...8 base...CTCCTCGCAGC...3 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CAGA
S1 to S0, read 0, print 1
CTCGTT...8 base...CTCCTCGCAGC...4 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...AGAA
S2 to S0, read 0, print 2
AGTCTT...8 base...CTCCTCGCAGC...2 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...TCAG
S0 to S0, read 0, print 0
AGTCTT...8 base...CTCCTCGCAGC...1 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...CCAT
S0 to S1, read 1, print 0
GGTATT...8 base...CTCCTCGCAGC...2 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CATA
S1 to S1, read 1, print 1
CTCGTT...8 base...CTCCTCGCAGC...3 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...ATAA
S2 to S1, read 1, print 2
GGTATT...8 base...CTCCTCGCAGC...1 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...AGCA
S1 to S2, read 2, print 1
AGTCTT...8 base...CTCCTCGCAGC
AATCAGAA...pairs ...GAGGAGCGTCGGAGC
S0 to S2, read 2, print 0
CTCGTT...8 base...CTCCTCGCAGC...2 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...GCAA
S2 to S2, read 2, print 2

15. Divide-by-three Computation
First Restriction
AATTCGGCCGTT CTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGC AATCAGAATCAGAAACGACTTTAA
GGTATT...8 base...CTCCTCGCAGC...3 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CAGA
S1 to S0, read 0, print 1
CTCGTT...8 base...CTCCTCGCAGC...4 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...AGAA
S2 to S0, read 0, print 2
AGTCTT...8 base...CTCCTCGCAGC...2 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...TCAG
S0 to S0, read 0, print 0
AGTCTT...8 base...CTCCTCGCAGC...1 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...CCAT
S0 to S1, read 1, print 0
GGTATT...8 base...CTCCTCGCAGC...2 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CATA
S1 to S1, read 1, print 1
CTCGTT...8 base...CTCCTCGCAGC...3 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...ATAA
S2 to S1, read 1, print 2
GGTATT...8 base...CTCCTCGCAGC...1 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...AGCA
S1 to S2, read 2, print 1
AGTCTT...8 base...CTCCTCGCAGC
AATCAGAA...pairs ...GAGGAGCGTCGGAGC
S0 to S2, read 2, print 0
CTCGTT...8 base...CTCCTCGCAGC...2 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...GCAA
S2 to S2, read 2, print 2

16. Divide-by-three Computation
First Ligation
AGTCTT...8 base...CTCCTCGCAGC
AATCAGAA...pairs ...GAGGAGCGTCGGAGC
Transition Molecule S0 to S2, reading 2, printing 0
AATTCGGCCGTT CTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGC AATCAGAATCAGAAACGACTTTAA
DNA Ligase
CTCGTT...8 base...CTCCTCGCAGC...4 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...AGAA
S2 to S0, read 0, print 2
AGTCTT...8 base...CTCCTCGCAGC...2 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...TCAG
S0 to S0, read 0, print 0
GGTATT...8 base...CTCCTCGCAGC...3 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CAGA
S1 to S0, read 0, print 1
AGTCTT...8 base...CTCCTCGCAGC...1 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...CCAT
S0 to S1, read 1, print 0
GGTATT...8 base...CTCCTCGCAGC...2 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CATA
S1 to S1, read 1, print 1
CTCGTT...8 base...CTCCTCGCAGC...3 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...ATAA
S2 to S1, read 1, print 2
GGTATT...8 base...CTCCTCGCAGC...1 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...AGCA
S1 to S2, read 2, print 1
S0 to S2, read 2, print 0
CTCGTT...8 base...CTCCTCGCAGC...2 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...GCAA
S2 to S2, read 2, print 2
AGTCTT...8 base...CTCCTCGCAGC
AATCAGAA...pairs ...GAGGAGCGTCGGAGC

17. Divide-by-three Computation
First Ligation
AATTCGGCCGTTAGTCTT...8 base...CTCCTCGCAGCCTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGCAATCAGAA...pairs ...GAGGAGCGTCGGAGCAATCAGAATCAGAAACGACTTTAA
AATTCGGCCGTTAGTCTT...8 base...CTCCTCGCAGCCTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGCAATCAGAA...pairs ...GAGGAGCGTCGGAGCAATCAGAATCAGAAACGACTTTAA
CTCGTT...8 base...CTCCTCGCAGC...4 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...AGAA
S2 to S0, read 0, print 2
AGTCTT...8 base...CTCCTCGCAGC...2 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...TCAG
S0 to S0, read 0, print 0
GGTATT...8 base...CTCCTCGCAGC...3 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CAGA
S1 to S0, read 0, print 1
AGTCTT...8 base...CTCCTCGCAGC...1 base
AATCAGAA...pairs ...GAGGAGCGTCG...pairs...CCAT
S0 to S1, read 1, print 0
GGTATT...8 base...CTCCTCGCAGC...2 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...CATA
S1 to S1, read 1, print 1
CTCGTT...8 base...CTCCTCGCAGC...3 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...ATAA
S2 to S1, read 1, print 2
GGTATT...8 base...CTCCTCGCAGC...1 base
AACCATAA...pairs ...GAGGAGCGTCG...pairs...AGCA
S1 to S2, read 2, print 1
S0 to S2, read 2, print 0
CTCGTT...8 base...CTCCTCGCAGC...2 base
AAGAGCAA...pairs ...GAGGAGCGTCG...pairs...GCAA
S2 to S2, read 2, print 2
AGTCTT...8 base...CTCCTCGCAGC
AATCAGAA...pairs ...GAGGAGCGTCGGAGC

18. Continue cycle of restriction, hybridization, and ligation until terminator is cleaved
AATTCGGCCGTTAGTCTT...8 base...CTCCTCGCAGCCTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGCAATCAGAA...pairs ...GAGGAGCGTCGGAGCAATCAGAATCAGAAACGACTTTAA
AATTCGGCCGTTAGTCTT...8 base...CTCCTCGCAGCCTCGTTAGTCTTAGTCTTTGCTGAAATT
TTAAGCCGGCAATCAGAA...pairs ...GAGGAGCGTCGGAGCAATCAGAATCAGAAACGACTTTAA

19. Divide-by-three Computation
Final Restriction
AATTCGGCCGTTAGTCTTCTCGTTAGTCTT TGCTGAAATT
TTAAGCCGGCAATCAGAAGAGCAATCAG CTTTAA

20. Divide-by-three Computation
Final Ligation
TGCTGA...Reporter...
AAACGACT....Gene 0....ACGA
Detection Molecule
AATTCGGCCGTTAGTCTTCTCGTTAGTCTT TGCTGAAATT
TTAAGCCGGCAATCAGAAGAGCAATCAG CTTTAA

21. Divide-by-three Computation
Final Ligation
AATTCGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT
TTAAGCCGGCAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA
This transducer has printed 020, which is 6 in base ten (610).
Check: 18/3 = 6? Yes.

22. Biological Function 0
AATTCGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT
TTAAGCCGGCAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA
Biological Function 0 could be releasing a drug, changing the bacteria phenotype, etc.

23. Divide-by-two Computation
Transition Stage
AATTCGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT
TTAAGCCGGCAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA
A third restriction enzyme that cleaves within its recognition site is necessary only when consecutive computation (using two separate transducers) occurs.

24. Divide-by-two Computation
Transition Stage
GGCCTTTCTCCTCGCAGCT
AAAGAGGAGCGTCGACCGG
Reinsertion Molecule
AATTC GGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT
TTAAGCCGG CAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA
Reinsertion of the recognition sites is also required for consecutive computation.

25. Divide-by-two Computation
Transition Stage
AATTCGGCCTTTCTCCTCGCAGCTGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT
TTAAGCCGGAAAGAGGAGCGTCGACCGGCAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA

26. Divide-by-two Computation
Entire cycle repeats again until terminator is cleaved once more
AATTCGGCCTTTCTCCTCGCAGCTGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT
TTAAGCCGGAAAGAGGAGCGTCGACCGGCAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA

27. Divide-by-two Computation
Final Restriction
AATTAGTCTTGGTATTAGTCTT TGCTGA...Reporter...TGCTGAAATT
TTAATCAGAACCATAATCAG CT....Gene 0....ACGACTTTAA

28. Divide-by-two Computation
Final Ligation
TGCTGA...Reporter...
AAACGACT....Gene 0....ACGA
Detection Molecule
AATTAGTCTTGGTATTAGTCTT TGCTGA...Reporter...TGCTGAAATT
TTAATCAGAACCATAATCAG CT....Gene 0....ACGACTTTAA

29. Divide-by-two Computation
Final Ligation
AATTAGTCTTGGTATTAGTCTTTGCTGA...Reporter...TGCTGA...Reporter...TGCTGAAATT
TTAATCAGAACCATAATCAGAAACGACT....Gene 0....ACGACT....Gene 0....ACGACTTTAA
(310)
This transducer has printed 010, which is 3 in base ten.
Check: (18/3)/2 = 3? Yes.

30. Biological Function 0
AATTAGTCTTGGTATTAGTCTTTGCTGA...Reporter...TGCTGA...Reporter...TGCTGAAATT
TTAATCAGAACCATAATCAGAAACGACT....Gene 0....ACGACT....Gene 0....ACGACTTTAA

31. Discussion & Conclusions
This project worked as expected.
18 ÷ 3= 6 ; 6 ÷ 2= 3
No molecule encoded the recognition site of an enzyme
Proof of concept worked however not done in practicality.
Transducers engineered functioned as coded

32. Acknowledgements
We would like to sincerely thank Mr. Russell N. Stern for his generosity and donation.
Thank you to the Louis Herman Israel Experience Fund for their contribution.
We would also like to thank our mentor Tamar Ratner, for her continued dedication and help.
Finally, we would like to thank Professor Ehud Keinan for allowing us to use his laboratory and his student.