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Team:Edinburgh/Model Comparison
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We found that the equations used for the deterministic modelling only gave sensible answers when the model parameters remained within certain limits. Outside those limits, results could be physically impossible; e.g. producing negative amounts of cellobiose therefore breaking the law of conservation of mass. The deterministic model always had reactants available, i.e cellulose able for every reaction, ensuing it would never reach zero. A 'stress test' was carried out simulating the model over one hundred thousand hours confirming this. Therefore, within the limits of differential equation based modelling, it will unlikely reach a mathematical steady state. An engineer defines steady state when 99% of the initial value of cellulose has been degraded, which was found to occur after 8000 hours. It can be seen from [https://2011.igem.org/Team:Edinburgh/Cellulases_(MATLAB_model) Figure 3], that the 'stress test' revealed neither a mathematical nor engineering steady state will be reached for cellobiose and glucose in the MATLAB model. | We found that the equations used for the deterministic modelling only gave sensible answers when the model parameters remained within certain limits. Outside those limits, results could be physically impossible; e.g. producing negative amounts of cellobiose therefore breaking the law of conservation of mass. The deterministic model always had reactants available, i.e cellulose able for every reaction, ensuing it would never reach zero. A 'stress test' was carried out simulating the model over one hundred thousand hours confirming this. Therefore, within the limits of differential equation based modelling, it will unlikely reach a mathematical steady state. An engineer defines steady state when 99% of the initial value of cellulose has been degraded, which was found to occur after 8000 hours. It can be seen from [https://2011.igem.org/Team:Edinburgh/Cellulases_(MATLAB_model) Figure 3], that the 'stress test' revealed neither a mathematical nor engineering steady state will be reached for cellobiose and glucose in the MATLAB model. | ||
| - | Whereas with the [ https://2011.igem.org/Team:Edinburgh/Cellulases_(C_model) C model] and the [https://2011.igem.org/Team:Edinburgh/Cellulases_(Kappa_model) Kappa model] it can be seen that cellulose, glucose and cellobiose all reach steady state within the | + | Whereas with the [https://2011.igem.org/Team:Edinburgh/Cellulases_(C_model) C model] and the [https://2011.igem.org/Team:Edinburgh/Cellulases_(Kappa_model) Kappa model] it can be seen that cellulose, glucose and cellobiose all reach steady state within the |
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Revision as of 13:20, 9 September 2011
Model Comparison
The approaches
- MATLAB - Within the reactor tasked with degrading cellulose into glucose in the biorefinery, temperature, enzyme concentration, substrate reactivity as well as xylose, cellobiose and glucose inhibition all govern the amount of glucose product. Deterministic modelling using a set of ordinary differential equations highlights the essential kinetic relationship among the enzymes, exo/endo-glucanase and β-glucosidase. By solving these governing equations using the numerical tool MATLAB the level of degradation is qualitatively predicted.
- C model
- Kappa - As an alternative, stochastic models were created using the Kappa language tool. These incorporate indeterminacy in the evolution of the state of the system. Rules are defined which describe how the model moves from one state to the next.
- Spacial kappa
Steady state
We found that the equations used for the deterministic modelling only gave sensible answers when the model parameters remained within certain limits. Outside those limits, results could be physically impossible; e.g. producing negative amounts of cellobiose therefore breaking the law of conservation of mass. The deterministic model always had reactants available, i.e cellulose able for every reaction, ensuing it would never reach zero. A 'stress test' was carried out simulating the model over one hundred thousand hours confirming this. Therefore, within the limits of differential equation based modelling, it will unlikely reach a mathematical steady state. An engineer defines steady state when 99% of the initial value of cellulose has been degraded, which was found to occur after 8000 hours. It can be seen from Figure 3, that the 'stress test' revealed neither a mathematical nor engineering steady state will be reached for cellobiose and glucose in the MATLAB model.
Whereas with the C model and the Kappa model it can be seen that cellulose, glucose and cellobiose all reach steady state within the
