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Intermediate [ edit ]
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{\displaystyle {\begin{aligned}E+S{\underset {k_{-1}}{\overset {k_{1}}{\begin{smallmatrix}\displaystyle \longrightarrow \\\displaystyle \longleftarrow \end{smallmatrix}}}}ES{\overset {k_{2}}{\longrightarrow }}EI{\overset {k_{3}}{\longrightarrow }}E+P\end{aligned}}}
So we can write down the following kinetic equations where we use the quasi-steady-state assumptions for both intermediate enzyme complexes:
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{\displaystyle {\begin{aligned}{\frac {d{[}P{]}}{dt}}&=k_{3}{[}EI{]}\\{\frac {d{[}EI{]}}{dt}}&=k_{2}{[}ES{]}-k_{3}{[}EI{]}\;{\overset {!}{=}}0\;\\{\frac {d{[}ES{]}}{dt}}&=k_{1}{[}E{]}{[}S{]}-{[}ES{]}(k_{-1}+k_{2})\;{\overset {!}{=}}0\;\\\end{aligned}}}
Again we can write down the total enzyme concentration and assume it to be constant over the time scale in question
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{\displaystyle {\begin{aligned}{[}E{]}_{0}&={[}E{]}+{[}ES{]}+{[}EI{]}\;{\overset {!}{=}}\;{\text{const}}\end{aligned}}}
Now again we express [ES] in terms of the other variables
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{\displaystyle {\begin{aligned}0&=k_{1}{[}S{]}({[}E{]}_{0}-{[}ES{]}-{[}EI{]})-{[}ES{]}(k_{-1}+k_{2})\\k_{1}{[}S{]}({[}E{]}_{0}-{[}EI{]})&=k_{1}{[}S{]}{[}ES{]}+{[}ES{]}(k_{-1}+k_{2})\\{[}S{]}({[}E{]}_{0}-{[}EI{]})&={[}S{]}{[}ES{]}+{[}ES{]}\underbrace {\frac {(k_{-1}+k_{2})}{k_{-1}}} _{K_{M}}\\{[}S{]}({[}E{]}_{0}-{[}EI{]})&=(K_{M}+{[}S{]}){[}ES{]}\\{[}ES{]}&={\frac {{[}S{]}({[}E{]}_{0}-{[}EI{]})}{K_{M}+{[}S{]}}}\end{aligned}}}
Now we plug this into the equation for
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{\displaystyle d{[}EI{]}/dt}
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{\displaystyle {\begin{aligned}0&=k_{2}{[}ES{]}-k_{3}{[}EI{]}\\0&=k_{2}{\frac {{[}S{]}({[}E{]}_{0}-{[}EI{]})}{K_{M}+{[}S{]}}}-k_{3}{[}EI{]}\\0&=k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{K_{M}+{[}S{]}}}-k_{2}{\frac {{[}S{]}{[}EI{]}}{K_{M}+{[}S{]}}}-k_{3}{[}EI{]}\\k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{K_{M}+{[}S{]}}}&=(k_{2}{\frac {{[}S{]}}{K_{M}+{[}S{]}}}+k_{3}){[}EI{]}\\{[}EI{]}&=k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{(K_{M}+{[}S{]})(k_{2}{\frac {{[}S{]}}{K_{M}+{[}S{]}}}+k_{3})}}\\{[}EI{]}&=k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{k_{2}{[}S{]}+k_{3}(K_{M}+{[}S{]})}}\end{aligned}}}
Finally we can find again an expression for the rate of substrate conversion to product
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{\displaystyle {\begin{aligned}{\frac {d{[}P{]}}{dt}}&=v_{0}=k_{3}{[}EI{]}=k_{3}k_{2}{\frac {{[}E{]}_{0}{[}S{]}}{k_{3}K_{M}+{[}S{]}(k_{2}+k_{3})}}\\{\frac {d{[}P{]}}{dt}}&=\underbrace {\dfrac {k_{3}k_{2}}{k_{2}+k_{3}}} _{k_{cat}}\cdot {\frac {{[}E{]}_{0}{[}S{]}}{\underbrace {{\frac {k_{3}}{k_{2}+k_{3}}}K_{M}} _{K_{M}^{\prime }}+{[}S{]}}}\\v_{0}&=k_{cat}{\frac {{[}S{]}{[}E{]}_{0}}{K_{M}^{\prime }+{[}S{]}}}\end{aligned}}}
In the last equation we have written a Michaelis-Menten like formula with modified constants, thus in this case we have for the measured constants the following functions
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{\displaystyle {\begin{aligned}K_{M}^{\prime }\ &{\stackrel {\mathrm {def} }{=}}\ {\frac {k_{3}}{k_{2}+k_{3}}}K_{M}={\frac {k_{3}}{k_{2}+k_{3}}}\cdot {\frac {k_{2}+k_{-1}}{k_{1}}}\\k_{cat}\ &{\stackrel {\mathrm {def} }{=}}\ {\dfrac {k_{3}k_{2}}{k_{2}+k_{3}}}\end{aligned}}}
We see that for the limiting case
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EI to E + P is much faster than the previous step, we get again the original equation. Mathematically we have then
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{\displaystyle K_{M}^{\prime }\approx K_{M}}
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{\displaystyle k_{cat}\approx k_{2}}
![{\displaystyle {\begin{aligned}{\frac {d{[}P{]}}{dt}}&=k_{3}{[}EI{]}\\{\frac {d{[}EI{]}}{dt}}&=k_{2}{[}ES{]}-k_{3}{[}EI{]}\;{\overset {!}{=}}0\;\\{\frac {d{[}ES{]}}{dt}}&=k_{1}{[}E{]}{[}S{]}-{[}ES{]}(k_{-1}+k_{2})\;{\overset {!}{=}}0\;\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09e833a3b60b2455bc2bac9bccfce4c2babe2faf)
![{\displaystyle {\begin{aligned}{[}E{]}_{0}&={[}E{]}+{[}ES{]}+{[}EI{]}\;{\overset {!}{=}}\;{\text{const}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68c381b6f49877023b744cb68b0822106d53a80a)
![{\displaystyle {\begin{aligned}0&=k_{1}{[}S{]}({[}E{]}_{0}-{[}ES{]}-{[}EI{]})-{[}ES{]}(k_{-1}+k_{2})\\k_{1}{[}S{]}({[}E{]}_{0}-{[}EI{]})&=k_{1}{[}S{]}{[}ES{]}+{[}ES{]}(k_{-1}+k_{2})\\{[}S{]}({[}E{]}_{0}-{[}EI{]})&={[}S{]}{[}ES{]}+{[}ES{]}\underbrace {\frac {(k_{-1}+k_{2})}{k_{-1}}} _{K_{M}}\\{[}S{]}({[}E{]}_{0}-{[}EI{]})&=(K_{M}+{[}S{]}){[}ES{]}\\{[}ES{]}&={\frac {{[}S{]}({[}E{]}_{0}-{[}EI{]})}{K_{M}+{[}S{]}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3918a38b97d3acff9dcbc9cc94e4587adc40ff0)
![{\displaystyle d{[}EI{]}/dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e693ec716ca8210bab89a0a2f31deeb1e15f9b36)
![{\displaystyle {\begin{aligned}0&=k_{2}{[}ES{]}-k_{3}{[}EI{]}\\0&=k_{2}{\frac {{[}S{]}({[}E{]}_{0}-{[}EI{]})}{K_{M}+{[}S{]}}}-k_{3}{[}EI{]}\\0&=k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{K_{M}+{[}S{]}}}-k_{2}{\frac {{[}S{]}{[}EI{]}}{K_{M}+{[}S{]}}}-k_{3}{[}EI{]}\\k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{K_{M}+{[}S{]}}}&=(k_{2}{\frac {{[}S{]}}{K_{M}+{[}S{]}}}+k_{3}){[}EI{]}\\{[}EI{]}&=k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{(K_{M}+{[}S{]})(k_{2}{\frac {{[}S{]}}{K_{M}+{[}S{]}}}+k_{3})}}\\{[}EI{]}&=k_{2}{\frac {{[}S{]}{[}E{]}_{0}}{k_{2}{[}S{]}+k_{3}(K_{M}+{[}S{]})}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7492bac61c9f3d4c87d6662f1fb6074e11936e71)
![{\displaystyle {\begin{aligned}{\frac {d{[}P{]}}{dt}}&=v_{0}=k_{3}{[}EI{]}=k_{3}k_{2}{\frac {{[}E{]}_{0}{[}S{]}}{k_{3}K_{M}+{[}S{]}(k_{2}+k_{3})}}\\{\frac {d{[}P{]}}{dt}}&=\underbrace {\dfrac {k_{3}k_{2}}{k_{2}+k_{3}}} _{k_{cat}}\cdot {\frac {{[}E{]}_{0}{[}S{]}}{\underbrace {{\frac {k_{3}}{k_{2}+k_{3}}}K_{M}} _{K_{M}^{\prime }}+{[}S{]}}}\\v_{0}&=k_{cat}{\frac {{[}S{]}{[}E{]}_{0}}{K_{M}^{\prime }+{[}S{]}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea6664bc7579b7ba40045a1fe4a056eca6ef5aa9)
