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1.14.23: Enzyme-Substrate Interaction

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    We consider the formation in aqueous solution of an enzyme –substrate complex \(\mathrm{ES}\) by an enzyme \(\mathrm{E}\) and substrate \(\mathrm{S}\). The system is prepared using \(\mathrm{n}^{0} (\mathrm{E})\) moles of enzyme and \(\mathrm{n}^{0} (\mathrm{S}\)) moles of substrate; equation (a)

      \(\mathrm{E}(\mathrm{aq}) +\) \(\mathrm{S}(\mathrm{aq}) \Leftrightarrow\) \(\mathrm{ES}(\mathrm{aq})\)
    At \(t = 0\) \(\mathrm{n}^{0}(\mathrm{E})\) \(\mathrm{n}^{0}(\mathrm{S})\) \(0 \mathrm{~mol}\)
    At \(t = \infty\) \(n^{0}(E)-\xi\) \(\mathrm{n}^{0}(\mathrm{~S})-\xi\) \(\xi \mathrm{~mol}\)

    The upper limit of the extent of interaction \(\xi\) is controlled by whichever is the smallest amount, either \(\mathrm{n}^{0}(\mathrm{E})\) or \(\mathrm{n}^{0}(\mathrm{S})\). The latter two variables determine the total amount of \(\mathrm{ES}(\mathrm{aq})\) which can be formed in the limit of tight binding.

    The approach described above can be extended to more complicated schemes involving multip-step reactions. In the following we consider the case where enzyme \(\mathrm{E}\) converts substrate \(\mathrm{A}\) into product \(\mathrm{D}\). The system is prepared using \(\mathrm{n}^{0}(\mathrm{E})\) moles of enzyme and \(\mathrm{n}^{0}(\mathrm{A})\) moles of substrate \(\mathrm{A}\) such that there are two intermediates \(\mathrm{EB}(\mathrm{aq})\) and \(\mathrm{EC}(\mathrm{aq})\), product \(\mathrm{D}\) being liberated from the bound state in the final step. The equilibrium state can be represented by the following scheme.

    \[\begin{aligned}
    &\mathrm{E}(\mathrm{aq}) \quad+\mathrm{A}(\mathrm{aq}) \Leftrightarrow \mathrm{EB}(\mathrm{aq}) \Leftrightarrow \mathrm{EC}(\mathrm{aq}) \Leftrightarrow \mathrm{E}(\mathrm{aq})+\mathrm{D}(\mathrm{aq})\\
    &\mathrm{n}^{0}(\mathrm{E})-\xi_{1}+\xi_{3} \mathrm{n}^{0}(\mathrm{~A})-\xi_{1} \quad \xi_{1}-\xi_{2} \quad \xi_{2}-\xi_{3} \quad \mathrm{n}^{0}(\mathrm{E})-\xi_{1}+\xi_{3} \xi_{3}
    \end{aligned} \nonumber \]

    The key point is that at equilibrium the amounts of enzyme \(\mathrm{E}(\mathrm{aq})\) identified at both ends of the reaction must be the same. Further a mass balance shows that the total amount of enzyme present equals \(\mathrm{n}^{0}(\mathrm{E}) \mathrm{~mol}\). Other features are interesting; \(\xi_{3}\) must be zero if \(\xi_{2}\) is zero. The method can be applied to more complicated reaction schemes including those where the path from reactant to product involves parallel reactions.

    This page titled 1.14.23: Enzyme-Substrate Interaction was last modified on Mon, 20 May 2024 07:02:58 GMT and is shared under a Public Domain license and was authored, remixed, and/or curated by Michael J Blandamer and Joao Carlos R Reis via source content that was edited to the style and standards of the LibreTexts platform.