1. Living Hardware to Solve the Hamiltonian Path Problem
Professors: Dr. Malcolm Campbell and Dr. Laurie Heyer
Students: Oyinade Adefuye, Will DeLoache, Jim Dickson, Andrew Martens, Amber Shoecraft, and Mike Waters

2. The Hamiltonian Path Problem

3. Computational Complexity
Millennium Problem
P=NP?
Brute Force required
This graph has 100 nodes that each connect to 4 other nodes (degree 4).
Does a Hamiltonian Path exist in this graph?

4. Why Should We Use Bacteria?
In vitro vs. in vivo
Adleman would elimate some paths, run a gel, elimnate some paths, run a gel
Mention exponential growth of E. coli cells.
VS.
Adleman LM (1994). Science 266 (11):

5. Flipping DNA with Hin/hixC

6. Using Hin/hixC to Solve the HPP
5 DNA elements
5 = TT

7. Using Hin/hixC to Solve the HPP
hixC Sites

8. Using Hin/hixC to Solve the HPP

9. Using Hin/hixC to Solve the HPP

10. Using Hin/hixC to Solve the HPP
Solved Hamiltonian Path

11. What Genes Can Be Split?
GFP before hixC insertion

12. What Genes Can Be Split?
GFP displaying hixC insertion point

13. Gene Splitter Software
Input
Output
1. Gene Sequence
2. Where do you want your hixC site?
3. Pick an extra base to avoid a frameshift
Generates 4 Primers (optimized for melting temperature).
2. Biobrick ends are added to primers.
3. Frameshift is eliminated.

14. Gene-Splitter Output
Note: Oligos are optimized for melting temperatures.

15. Red Fluorescent Protein
Use GFP to Split RFP
Green Fluorescent Protein
Red Fluorescent Protein

16. Can We Detect A Solution?
Will poses question. Amber responds to it.

17. True Positives
Elements in the shaded region can be arranged in any order.
(Edges-Nodes+1)
Number of True Positives = (Edges-Nodes+1)! * 2

18. False Positives
Extra Edge

19. False Positives
PCR Fragment Length
PCR Fragment Length

20. Detection of True Positives
Total # of Positives
# of Nodes / # of Edges
# of True Positives ÷
Total # of Positives
# of Nodes / # of Edges

21. How Many Plasmids Do We Need?
1 mL of culture = 10 cells 9
k = actual number of occurrences
λ = expected number of occurrences
The Poisson distribution gives the probability of k occurrences when the expected number of occurrences is λ as (e^-λ)*(λ^k)/(k!) by taking 1 minus the sum of the Poisson probabilities from 0 to k-1 we get the probability of at least k occurrences.
λ = m plasmids * # solved permutations of edges ÷ # permutations of edges
Cumulative Poisson Distribution:
_____ e -λ λ X! x . ∑
k-1
X=0
P(# of solutions ≥ k) = 1 -

22. Probability of HPP Solution
Starting Arrangement
4 Nodes & 3 Edges
Probability of HPP Solution
Markov chain model
Unbiased
Easy vs.. Hard
Convergence
Probability (solved / f flips)
# of paths of length f to any solution =
(edges+1 choose 2) ^ f
Lim Probability (solved / f flips)
f --> ∞
# permutations of edges that are solutions =
# permutations of edges
Number of Flips

23. Where Are We Now?
Will switches back with Amber.

24. First Bacterial Computer
Starting Arrangement

25. First Bacterial Computer
Starting Arrangement
Solved Arrangement

26. Future Directions Split additional genes: Make more complex graphs:
Solve other problems such as the Traveling Salesperson Problem:

27. Living Hardware to Solve the Hamiltonian Path Problem
Collaborators at MWSU:
Dr. Todd Eckdahl, Dr. Jeff Poet, Jordan Baumgardner,Tom Crowley, Lane H. Heard, Nickolaus Morton, Michelle Ritter, Jessica Treece, Matthew Unzicker, Amanda Valencia
Additional Thanks:
Karen Acker, Davidson College ‘07
Support:
Davidson College
The Duke Endowment
HHMI
NSF
Genome Consortium for Active Teaching
James G. Martin Genomics Program

28. Extra Slides

29. Traveling Salesperson Problem

30. Processivity Problem:
The nature of our construct requires a stable transcription mechanism that can read through multiple genes in vivo
Initiation Complex vs. Elongation Complex
Formal manipulation of gene expression (through promoter sequence and availability of accessory proteins) is out of the picture
Solution : T7 bacteriophage RNA polymerase
Highly processive single subunit viral polymerase which maintains processivity in vivo and in vitro

31. Path at 3 nodes / 3 edges HP- 1 12 23

32. Path at 4 nodes / 6 edges HP-1 12 24 43

33. Path 5 nodes / 8 edges HP 2 3 4 1 T 5

34. Path 6 nodes / 10 edges HP 2 3 4 1 T 5 6

35. Path 7 nodes / 12 edges HP 2 3 4 1 T 5 6 7

36. More Gene-Splitter Output

37. Promoter Tester RBS:Kan:RBS:Tet:RBS:RFP
Tested promoter-promoter tester-pSBIA7 on varying concentration plates
Kanamycin
Tet
Kan-Tet 50
50 / 50 75
75 / 75
100
100 / 100
125
125 / 125
Used Promoter Tester-pSB1A7 and Promoter Tester-pSB1A2 without promoters as control
Further evidence that pSB1A7 isn’t completely insulated

38. Promoters Tested
Selected “strong” promoters that were also repressible from biobrick registry
ompC porin (BBa_R0082)
“Lambda phage”(BBa_R0051)
pLac (BBa_R0010)
Hybrid pLac (BBa_R0011)
None of the promoters “glowed red”
Rus (BBa_J3902) and CMV (BBa_J52034) not the parts that are listed in the registry

39. Splitting Kanamycin Nucleotidyltransferase
Determined hixC site insertion at AA 125 in each monomer subunit
-AA 190 is involved in catalysis
-AA 195 and 208 are involved in Mg2+ binding
-Mutant Enzymes 190, 205, 210 all showed changes in mg+2 binding from the WT
-Substitution of AA 210 (conserved) reduced enzyme activity
-AA 166 serves to catalyze reactions involving ATP
-AA 44 is involved in ATP binding
-AA 60 is involved in orientation of AA 44 and ATP binding
-We did not consider any Amino Acids near the N or C terminus
-Substitution of AA 190 caused 650-fold decrease in enzyme activity
-We did not consider any residues near ß-sheets or ∂-helices close to the active site because hydrogen bonding plays an active role in substrate stabilization and the polarity of our hix site could disrupt the secondary structure and therefore the hydrogen bonding ability of KNTase)
Did not split

40. Plasmid Insulation
“Insulated” plasmid was designed to block read-through transcription
Read-through = transcription without a promoter
Tested a “promoter-tester” construct
RBS:Kan:RBS:Tet:RBS:RFP
Plated on different concentrations of Kan, Tet, and Kan-Tet plates
Growth in pSB1A7 was stunted
No plate exhibited cell growth in uninsulated plasmid and cell death in the insulated plasmid

41. Tetracycline Resistance Protein
Did not split
Transmembrane protein
Structure hasn’t been crystallized
determined by computer modeling
Vital residues for resistance are in transmemebrane domains (efflux function)
HixC inserted a periplasmic domains AA 37/38 and AA 299/300
Cytoplasmic domains allow for interaction with N and C terminus

42. Splitting Cre Recombinase

43. What Genes Can Be Split? GFP before hixC insertion
GFP displaying hixC insertion point

44. Gene Splitter Software