Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-20T07:13:54Z https://chemistry.stackexchange.com/feeds/question/85916 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/85916 1 Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state? PascalIv https://chemistry.stackexchange.com/users/54973 2017-11-16T12:17:48Z 2019-08-04T13:53:10Z <p>Suppose an enzyme $\ce{E}$ can catalyze two reactions:</p> <p>\begin{align} \ce{S1 + E &amp;&lt;=&gt; S1E -&gt; P1 + E} \tag{R1} \\ \ce{S2 + E &amp;&lt;=&gt; S2E -&gt; P2 + E} \tag{R2} \end{align}</p> <p>I want to derive a rate law. Can I assume that</p> <p>\begin{align} \frac{d[\ce{S1E}]}{dt} &amp;= 0 \tag{1}\\ \frac{d[\ce{S2E}]}{dt} &amp;= 0 \tag{2} \end{align}</p> <p>like in the derivation of the <a href="https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Derivation" rel="nofollow noreferrer">Michaelis-Menten rate law</a>?</p> https://chemistry.stackexchange.com/questions/85916/-/118855#118855 2 Answer by blu potatos for Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state? blu potatos https://chemistry.stackexchange.com/users/30951 2019-08-04T13:46:48Z 2019-08-04T13:53:10Z <p><a href="https://i.stack.imgur.com/uJgA3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uJgA3.png" alt="enter image description here"></a> </p> <p>In the <strong>steady-state reaction</strong>, the intermediate concentration [ES] is assumed to remain at a small constant value. So <strong>in this case only if k2 >> k1</strong> and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P.</p> <p><br></p> <p><span class="math-container">\begin{align} \frac{d[\ce{S1E}]}{dt} &amp;= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}] - \ce{k2}[\ce{S1E}] = 0\\ \end{align}</span></p> <p>Therfore:</p> <p><span class="math-container">\begin{align} [\ce{S1E}] &amp;= \frac{\ce{k1}}{k_{-1}+\ce{k2}}[\ce{S1}][\ce{E}] = K_a[\ce{S1}][\ce{E}] \\ \end{align}</span></p> <p>Similarly for the second reaction:</p> <p><span class="math-container">\begin{align} [\ce{S2E}] &amp;= \frac{\ce{k3}}{k_{-3}+\ce{k4}}[\ce{S2}][\ce{E}] = K_b[\ce{S2}][\ce{E}] \\ \end{align}</span></p> <p>The enzyme E is involved in both reactions and its total concentration(bound and unbound) is constant. This total concentration of enzyme <span class="math-container">$[E]_{0}$</span> is equivalent to the concentration of the free enzyme before adding the substrates. The concentration of the free enzyme at a certain time t is [E]:</p> <p><span class="math-container">\begin{align} [E]_{0} &amp;= [E] + [\ce{S1E}] + [\ce{S2E}] \\ \end{align}</span></p> <p>if you substitute <span class="math-container">$[\ce{S1E}]$</span> and <span class="math-container">$[\ce{S2E}]$</span> with the previous expressions then:</p> <p><span class="math-container">\begin{align} [E] &amp;= \frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]} \\ \end{align}</span></p> <p>Eventually: <span class="math-container">\begin{align} \frac{d[P_{1}]}{dt} = k_2[\ce{S1E}] = \ce{k2}(K_a[\ce{S1}][\ce{E}]) = K_p[\ce{S1}][\ce{E}] = K_p[\ce{S1}]\frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]} \\ \end{align}</span></p> <p>and similarly for the rate of production of <span class="math-container">$P_2$</span></p>