4. (20 points) Derive the rate law of product for the following enzymatic reaction. E + S \iff E \cdot S \iff E \cdot P \to P + E Assuming: (a) Pseudo-steady state hypothesis (PSSH) (b) Quasi-equilibrium assumption (QEA)
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The reaction equation for the enzymatic reaction is: E + SE.SE.PP + E Show more…
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Consider the following mechanism for the enzyme-catalyzed reaction: E+S ⇄ ES (fast equilibrium) ES E+P (slow) Derive an expression for the rate law of this reaction in terms of the concentrations of E and S. To solve for [ES], make use of the fact that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction.
David C.
Consider the following mechanism for the enzyme-catalyzed reaction: 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 the forward reactionlis equal to the rate of the reverse reaction.).
Consider the following mechanism for the enzymecatalyzed reaction: (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 $\mathrm{E}$ 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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