Team:Evry/ChemicalTools
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+ | <p> | ||
+ | There are 6 enzyms involved in the natural process of the enterobactin production: | ||
+ | <ul> | ||
+ | <li>EntA : </li> | ||
+ | <li>EntB : </li> | ||
+ | <li>EntC : </li> | ||
+ | <li>EntD : </li> | ||
+ | <li>EntE : </li> | ||
+ | <li>EntF : </li> | ||
+ | </ul> | ||
+ | The production of each one of those enzymes can thus become a rate-limiting-step, when it comes to mass enterobactin production.</p> | ||
+ | <img src="https://static.igem.org/mediawiki/2013/1/1d/Entero_pathway.jpg"/> | ||
+ | For now, we consider each one of these steps as a simple chemical reaction: | ||
+ | |||
+ | S → P : E | ||
+ | |||
+ | We are using the enzymatic kinetic model of Michaelis-Menten, which divides each reaction into two consecutives steps: | ||
+ | |||
+ | E + S ↔ ES → E + P : k1, k-1, k2 | ||
+ | |||
+ | The speed of the reaction is calculated as below: | ||
+ | |||
+ | v = dP/dt = k2[ES] | ||
+ | |||
+ | The steady state approximation gives us: | ||
+ | |||
+ | d[ES]/dt = k1[E] [S] – k-1[ES] – k2[ES] = 0 | ||
+ | |||
+ | And thus, | ||
+ | |||
+ | v = k2k1/(k-1+k2)* [E][S] = Kcat/Km*[E][S] | ||
+ | |||
+ | where Km = (k-1+k2)/k1 and Kcat = k2 are classic kinetic parameters. | ||
+ | |||
+ | Conclusion: | ||
+ | |||
+ | For simple enzymatic reactions (one reagent, one product and one enzyme) with the steady state approximation, we can directly apply the formula above-written. | ||
</div> | </div> |
Revision as of 07:34, 2 October 2013
Chemical tools
There are 6 enzyms involved in the natural process of the enterobactin production:
- EntA :
- EntB :
- EntC :
- EntD :
- EntE :
- EntF :