Difference between revisions of "Team:Wageningen UR/Notebook/MetabolicModelling"
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| + | <p> These experiments were performed by Ronald de Jongh.</p> | ||
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Latest revision as of 21:22, 19 October 2016
These experiments were performed by Ronald de Jongh.
July
Week 1
Learning to code with cobrapy and understanding what flux balance analysis (FBA) exactly is. Found out how to modify the exchange reactions of the entire medium.
Week 2
Working on improving our understanding of FBA and cobrapy. Also going through the literature to see what Escherichia coli might do in the sugar water. Gitlab was set up.
Week 3
Continuing to improve our understanding of FBA and cobrapy. Realization that the objective function can be changed to something else. The main question to answer is:“Will the chassis survive in the sugar water?”.
Week 4
Thinking on the effect of sugar water on the bacteria. The model says the E. coli can survive on sucrose as a carbon source just fine.
May
Week 5
Realization that the trace metals will not be a problem as they will be present in high enough amounts (for the chassis bacterium) as impurities in regular tap water.
Week 6
In order to answer the question of “Will the chassis survive in the sugar water?”. We thought of using the rate of ATP loss in a high osmotic pressure system, and somehow using the maintenance reaction as a source. This idea turned into using the ATP maintenance reaction as the objective function to maximize, as the E. coli will go into a kind of starvation mode.
Week 7
According to this bionumbers page the average water content of an E. coli cell is about 6.7x10^-13 grams. This could be useful in determining how fast the cell might drain.
Week 8
Creation of the table to be used for the sugar water modelling as a new medium, though using H2Otex as the exchange reaction. Example calculation of some output: 1 mol of H2O = 18g
Minimally survivable efflux (in sugar water) = 482 (mmol*gDW^-1*h^-1)
482 (mmol*gDW^-1*h^-1) * 2.8×10^-13 (gDW) = 1.3496 * 10^-10 (mmol*h^-1)
1000mmol h2o = 18g h2o -> 3.722*10^-11 (mmol h2o per cell) = 6.7*10^-13 (g h2o per cell)
3.722*10^-11 (mmol) / 1.3496*10^-10 (mmol * h^-1) = 0.27578 (hour)
0.27578 (hour) * 60 = 16.54 minutes.
