1.14.23: Enzyme-Substrate Interaction
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.