Tugbainanc (Talk | contribs) |
Tugbainanc (Talk | contribs) |
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<p> | <p> | ||
The cell growth rate of bacteria depends on the number of cells and their specific growth rate. Thus, it can be expressed as: | The cell growth rate of bacteria depends on the number of cells and their specific growth rate. Thus, it can be expressed as: | ||
− | |||
</p> | </p> | ||
+ | |||
+ | <h6>dX/dt = µ * X</h6> | ||
+ | |||
+ | <p> | ||
+ | when we arranged the equation and integrated the both sides we get: | ||
+ | </p> | ||
+ | |||
+ | <h6>X = e ^ µ*t</h6> | ||
+ | |||
+ | <p> | ||
+ | We decided to use the Monod kinetics to make substrate and cell concentration-dependent equation. It is shown below: | ||
+ | </p> | ||
+ | |||
+ | <h6>dX/dt = (µmax * X * S) / Ks + S</h6> | ||
+ | |||
+ | <p> | ||
+ | “Ks” represents the saturation concentration where “µmax” represents the maximum specific growth rate. In the Monod equation, we can observe the lag, exponential and stationary phase. However, because our system also includes inhibitor by-products, the Monod equation turned into a new form where the cell growth also depends on the inhibitor concentration: | ||
+ | </p> | ||
+ | |||
+ | <h6>dX/dt = (1 - (C / Cm)) * [(µmax * X * S) / (Ks * ( 1 - (C / Cm))]</h6> | ||
+ | |||
+ | <table> | ||
+ | <tr> | ||
+ | <td>X = Number of Cell,</td> | ||
+ | </tr> | ||
+ | <tr> | ||
+ | <td>C = inhibitor concentration</td> | ||
+ | </tr> | ||
+ | <tr> | ||
+ | <td>Cm = Maximum inhibitor concentration cell can live</td> | ||
+ | </tr> | ||
+ | <tr> | ||
+ | <td>µmax = Maximum specific growth rate</td> | ||
+ | </tr> | ||
+ | <tr> | ||
+ | <td>Ks = Saturation concentration</td> | ||
+ | </tr> | ||
+ | <tr> | ||
+ | <td> S = Sugar concentration.</td> | ||
+ | </tr> | ||
+ | </table> | ||
+ | |||
+ | <h3>Enzymatic Reaction Kinetics:</h3> | ||
+ | |||
+ | <p> | ||
+ | Enzymatic reactions were one of the main components of our project where we aimed to catalyze the furfural reduction. The enzymatic reactions depend on the presence of the substrate and the enzyme which are multiplied by forward and reverse reaction rate constants. We can show the reaction by using mass kinetics. | ||
+ | </p> | ||
+ | |||
+ | <ul> | ||
+ | <ol>E + S → ES → P + E</ol> | ||
+ | <ol>d[S]/ dt = -k1 * [E] * [S] + k-1* [ES]</ol> | ||
+ | <ol>d[E]/dt = -k1[E]*[S] + (k-1 + k2) * [ES] - (k-2 * [E]*[P])</ol> | ||
+ | <ol>d[P] /dt = k2 * [ES] - k-2 * [E] * [P]</ol> | ||
+ | </ul> | ||
+ | |||
+ | <p> | ||
+ | However, like in the Monod model, the formula can be developed by adding different constants like “Km” and “Vmax” to form Michaelis and Menten equations which represent the enzyme reaction rate. | ||
+ | </p> | ||
+ | |||
+ | <h6>V = (Vmax * [S]) / Km + [S])</h6> | ||
Revision as of 20:39, 10 October 2018