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Living Hardware to Solve the Hamiltonian Path Problem Faculty: Drs. Malcolm Campbell, Laurie Heyer, Karmella Haynes Students: Oyinade Adefuye, Will DeLoache, Jim Dickson, Andrew Martens, Amber Shoecraft, and Mike Waters Faculty: Drs. Todd Eckdahl, Jeff Poet Students: Jordan Baumgardner, Tom Crowley, Lane H. Heard, Nickolaus Morton, Michelle Ritter, Jessica Treece, Matthew Unzicker, Amanda Valencia
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The Hamiltonian Path Problem 1 3 5 4 2
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1 3 5 4 2
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Advantages of Bacterial Computation SoftwareHardwareComputation
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Advantages of Bacterial Computation SoftwareHardwareComputation
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http://www.turbosquid.com http://www.dnamnd.med.usyd.edu.au/ Advantages of Bacterial Computation SoftwareHardwareComputation
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Computational Complexity Non-Polynomial (NP) No Efficient Algorithms
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Computational Complexity Cell Division Non-Polynomial (NP) No Efficient Algorithms
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Flipping DNA with Hin/hixC
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Using Hin/hixC to Solve the HPP 1 3 5 4 2
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hixC Sites 1 3 5 4 2 Using Hin/hixC to Solve the HPP
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1 3 5 4 2
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1 3 5 4 2
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Solved Hamiltonian Path 1 3 5 4 2
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How to Split a Gene RBS Promoter Reporter Detectable Phenotype
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How to Split a Gene RBS Promoter Reporter hixC RBS Promoter Repo- rter Detectable Phenotype Detectable Phenotype ?
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Gene Splitter Software InputOutput 1. Gene Sequence (cut and paste) 2. Where do you want your hixC site? 3. Pick an extra base to avoid a frameshift. 1.Generates 4 Primers (optimized for Tm). 2. Biobrick ends are added to primers. 3. Frameshift is eliminated. http://gcat.davidson.edu/iGEM07/genesplitter.html
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Gene-Splitter Output Note: Oligos are optimized for Tm.
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Can We Detect A Solution?
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Starting Arrangement Probability of HPP Solution Number of Flips 4 Nodes & 3 Edges
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Elements in the shaded region can be arranged in any order. True Positives Number of True Positives = (Edges-(Nodes-1))! * 2 (Edges-Nodes+1) 1 3 5 4 2
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k = actual number of occurrences λ = expected number of occurrences λ = m plasmids * # solved permutations of edges ÷ # permutations of edges Cumulative Poisson Distribution: How Many Plasmids Do We Need? P(# of solutions ≥ k) = k = 151020 m = 10,000,000.0697000 50,000,000.3032.0000400 100,000,000.5145.000900 200,000,000.7643.0161.0000030 500,000,000.973.2961.00410 1,000,000,000.9992.8466.1932.00007 Probability of at least k solutions on m plasmids for a 14-edge graph
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False Positives Extra Edge 1 3 5 4 2
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False Positives PCR Fragment Length 1 3 5 4 2
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Detection of True Positives Total # of Positives # of True Positives ÷ Total # of Positives # of Nodes / # of Edges
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Building a Bacterial Computer
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Splitting Reporter Genes Green Fluorescent ProteinRed Fluorescent Protein
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Splitting Reporter Genes Green Fluorescent Protein Red Fluorescent Protein
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3-Node Graphs Graph AGraph B
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HPP Constructs Graph A Constructs: Graph B Construct: HPP 0 HPP-A 0 HPP-A 1 HPP-A 2 HPP-B 1 Positive Control Construct: Graph B Graph A
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Green Fluorescence Double Fluorescence HPP 0 T7 RNAP
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HPP-A 0 T7 RNAP Green Fluorescence Yellow Fluorescence Double Fluorescence HPP 0 T7 RNAP
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Fluorometer Measurements GFP Excitation Spectra for 4 HPP Constructs (at an Emission Wavelength of 515nm) 450 nm chosen as excitation wavelength to measure GFP
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Fluorometer Measurements RFP Excitation Spectra for 4 HPP Constructs (at an Emission Wavelength of 608nm) 560 nm chosen as excitation wavelength to measure RFP
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Normalized Fluorometer Measurements ConstructFluorescent Color on UV Box GreenRed pLac-RBS-RFPRed7263 pLac-RBS-RFP-RBS-GFPRed144370 pLac-GFP1-hixC-GFP2Green1360 pLac-RBS-RFP1-hixC-RFP2None0147 pLac-RBS-GFP1-hixC-RFP2None112 pLac-RBS-RFP1-hixC-GFP2None132 HPP-B 0 Green7218 HPP-A 0 Yellow340255 HPP-A 1 Red1143 HPP-A 2 None113 HPP-B 1 Hybrid green153
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Flipping Detected by Phenotype HPP-A 0 HPP-A 1 HPP-A 2
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Flipping Detected by Phenotype Hin-Mediated Flipping HPP-A 0 HPP-A 1 HPP-A 2
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Flipping Detected by PCR HPP-A 0 HPP-A 1 HPP-A 2 UnflippedFlipped
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Pending Experiments Test clonal colonies that contain flipped HPP and have the solution sequenced. Perform a false-positive screen for HPP-B 1 Split 2 antibiotic resistance genes using a reading frame shift just after the RBS Solve larger graphs Solve the Traveling Salesperson Problem
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Living Hardware to Solve the Hamiltonian Path Problem Support: Davidson College Missouri Western State University The Duke Endowment HHMI NSF Genome Consortium for Active Teaching James G. Martin Genomics Program Thanks to: Karen Acker, Davidson College ‘07
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Extra Slides
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Traveling Salesperson Problem
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Processivity 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 Problem: Solution : T7 bacteriophage RNA polymerase Highly processive single subunit viral polymerase which maintains processivity in vivo and in vitro
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Path at 3 nodes / 3 edges HP- 1 12 23 1 2 3 T T
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Path at 4 nodes / 6 edges HP-1 12 24 43 2 3 4 1 T T
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Path 5 nodes / 8 edges HP -1 12 25 54 43 2 3 4 1 T T 5
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Path 6 nodes / 10 edges HP-1 12 25 56 64 43 2 3 4 1 T T 56
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Path 7 nodes / 12 edges HP-1 12 25 56 67 74 43 2 34 1 T T 5 6 7
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Promoter Tester RBS:Kan:RBS:Tet:RBS:RFP Tested promoter-promoter tester-pSBIA7 on varying concentration plates KanamycinTetKan-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
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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
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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
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What Genes Can Be Split? GFP before hixC insertionGFP displaying hixC insertion point
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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 Splitting Kanamycin Nucleotidyltransferase
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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
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Splitting Cre Recombinase
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More Gene-Splitter Output
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Gene Splitter Software
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