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Mark Chaskes and Paul Lazarescu Mentor: Tamar Ratner The Schulich Faculty of Chemistry

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. Objective.

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Mark Chaskes and Paul Lazarescu Mentor: Tamar Ratner The Schulich Faculty of Chemistry

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  1. Challenging Traditional Approaches to ComputationA 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. 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

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

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

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

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

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

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

  12. Molecular Design of the Input Encoding 2-0-0 in Ternary (18 in base ten) Terminator EagI Recognition Site BbvI Recognition Site Spacers 200 AATTCGGCCGTT..8 base..CTCCTCGCAGC..8 base..CTCGTTAGTCTTAGTCTTTGCTGAAATT TTAAGCCGGCAA..pairs ..GAGGAGCGTCG..pairs ..GAGCAATCAGAATCAGAAACGACTTTAA Plasmid BseRI Recognition Site

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

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

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

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

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

  18. AATTCGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGAAATT TTAAGCCGGCAATCAGAAGAGCAATCAGCTTTAA Divide-by-three Computation Final Restriction

  19. Detection Molecule TGCTGA...Reporter... AAACGACT....Gene 0....ACGA AATTCGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGAAATT TTAAGCCGGCAATCAGAAGAGCAATCAGCTTTAA Divide-by-three Computation Final Ligation

  20. Divide-by-three Computation Final Ligation AATTCGGCCGTTAGTCTTCTCGTTAGTCTTTGCTGA...Reporter...TGCTGAAATT TTAAGCCGGCAATCAGAAGAGCAATCAGAAACGACT....Gene 0....ACGACTTTAA 0 2 0 This transducer has printed 020, which is 6 in base ten (610). Check: 18/3 = 6? Yes.

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

  22. 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.

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

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

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

  26. Divide-by-two Computation Final Restriction AATTAGTCTTGGTATTAGTCTTTGCTGA...Reporter...TGCTGAAATT TTAATCAGAACCATAATCAG CT....Gene 0....ACGACTTTAA

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

  28. Divide-by-two Computation Final Ligation AATTAGTCTTGGTATTAGTCTTTGCTGA...Reporter...TGCTGA...Reporter...TGCTGAAATT TTAATCAGAACCATAATCAGAAACGACT....Gene 0....ACGACT....Gene 0....ACGACTTTAA 0 1 0 (310) This transducer has printed 010, which is 3 in base ten. Check: (18/3)/2 = 3? Yes.

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

  30. 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

  31. 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.

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