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

The Hamiltonian Path Problem

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?

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): 1021-1024.

Flipping DNA with Hin/hixC

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

Using Hin/hixC to Solve the HPP hixC Sites

Using Hin/hixC to Solve the HPP

Using Hin/hixC to Solve the HPP

Using Hin/hixC to Solve the HPP Solved Hamiltonian Path

What Genes Can Be Split? GFP before hixC insertion

What Genes Can Be Split? GFP displaying hixC insertion point

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.

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

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

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

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

False Positives Extra Edge

False Positives PCR Fragment Length PCR Fragment Length

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

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 -

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

Where Are We Now? Will switches back with Amber.

First Bacterial Computer Starting Arrangement

First Bacterial Computer Starting Arrangement Solved Arrangement

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

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

Extra Slides

Traveling Salesperson Problem

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

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

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

Path 5 nodes / 8 edges HP -1 12 25 54 43 2 3 4 1 T 5

Path 6 nodes / 10 edges HP-1 12 25 56 64 43 2 3 4 1 T 5 6

Path 7 nodes / 12 edges HP-1 12 25 56 67 74 43 2 3 4 1 T 5 6 7

More Gene-Splitter Output

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

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

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

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

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

Splitting Cre Recombinase

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

Gene Splitter Software