If an enzyme follows this mechanism what rate law is...

Name: if an enzyme follows this mechanism what rate law is expected for the reaction use ee for ee ss for ss eses for eses and kk for the rate constant 95531
Uploaded: 2021-10-08T01:44:20-08:00
Duration: 1 min 30 s
Channel: Lizabeth Tumminello
Description: if an enzyme follows this mechanism what rate law is expected for the reaction use ee for ee ss for ss eses for eses and kk for the rate constant 95531

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Enzymes are often described as following the two-step mechanism: $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{ES} \text { (fast) } \\ \mathrm{ES} &-\rightarrow \mathrm{E}+\mathrm{P} \text { (slow) } \end{aligned} $$ Where $\mathrm{E}=$ enzyme, $\mathrm{S}=$ substrate, and $\mathrm{P}=$ product. If an enzyme follows this mechanism, what rate law is expected for the reaction?

Consider the following mechanism for the enzymecatalyzed reaction: $$ \begin{array}{r} \mathrm{E}+\mathrm{S} \underset{k-1}{\stackrel{k_{1}}{\sum_{-1}}} \mathrm{ES} \\ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \end{array} $$ Derive an expression for the rate law of the reaction in terms of the concentrations of $\mathrm{E}$ and $\mathrm{S}$. (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of forward reaction is equal to the rate of the reverse reaction.

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Consider the following mechanism for the enzymecatalyzed reaction: $$\mathrm{E}+\mathrm{S} \frac{k_{1}}{k_{-1}} \mathrm{ES} \quad \text { (fast equilibrium) }$$ $$\mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad \text { (slow) }$$ Derive an expression for the rate law of the reaction in terms of the concentrations of E and S. (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of forward reaction is equal to the rate of the reverse reaction.)

Transcript

00:01 If we are looking at a generic reaction, which has two steps, and the second step is the slow step, that means that everything from the slow step and above is included in the rate law expression.

00:46 Remember that the general format is rate equals k times concentrations of reactants to the powers of their orders, and we do cancel out our intermediates first...

Question

If an enzyme follows this mechanism, what rate law is expected for the reaction? Use EE for [E][E], SS for [S][S], EsEs for [ES][ES] and kk for the rate constant.

Question

If an enzyme follows this mechanism, what rate law is expected for the reaction? Use EE for [E][E], SS for [S][S], EsEs for [ES][ES] and kk for the rate constant.

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