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Reaction Efficiency in an Open Michaelis-Menten System

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Introduction

The closed Michaelis-Menten model can easily be modified to allow for a more biologically realistic open system, by introducing a substrate input rate b_{1} and a product outflow flux v_{3}, such that:

{\xrightarrow  {b_{1}}}S+E{\xleftarrow  {v_{{-1}}}}{\xrightarrow  {v_{1}}}C{\xrightarrow  {v_{2}}}E+P{\xrightarrow  {v_{3}}}


This simple Michaelis-Menten system is being studied because its behavior in response to certain inputs offers a great deal of insight into reactant behavior in much more complex pathways, including the complement system of the human immune system, which I will study later as a biological application.


My focus will be on the open Michaelis-Menten system, in which traditional analysis of the closed Michaelis-Menten system will prove important. In this case, the stoichiometric matrix for the open reaction (as depicted in the introduction) is given by:


{\textbf  {S}}={\begin{pmatrix}1&-1&1&0&0\\0&-1&1&1&0\\0&1&-1&-1&0\\0&0&0&1&-1\end{pmatrix}}, where {\textbf  {v}}={\begin{pmatrix}b_{1}\\k_{1}ES\\k_{{-1}}C\\k_{2}C\\k_{3}P\end{pmatrix}} and {\textbf  {x}}={\begin{pmatrix}S&E&C&P\end{pmatrix}}.


Using the Law of Mass Action, this corresponds to the following set of differential equations:


{\begin{cases}{\frac  {dS}{dt}}=b_{1}-k_{1}SE+k_{{-1}}C\\{\frac  {dE}{dt}}=(k_{{-1}}+k_{2})C-k_{1}SE\\{\frac  {dC}{dt}}=k_{1}SE-(k_{{-1}}+k_{2})C\\{\frac  {dP}{dt}}=k_{2}C-k_{3}P\end{cases}}



Maximum Allowable Reaction Rate

One particular question we would like to answer is the maximum allowable substrate input rate such that the substrate does not grow out of control over time, but instead approaches a steady state. Implicit in this requirement is that the complex and product both approach a steady-state as well.

By taking the (right) null space of the stoichiometric matrix, we can find spanning vectors for the steady state flux vector {\textbf  {v}}_{{ss}}={\begin{pmatrix}b_{1}\\k_{1}E_{{ss}}S_{{ss}}\\k_{{-1}}C_{{ss}}\\k_{2}C_{{ss}}\\k_{3}P_{{ss}}\end{pmatrix}}. The rank of the matrix is 3, so the dimension of null(S) must be 2. The two biologically significant spanning vectors are {\begin{pmatrix}1\\1\\0\\1\\1\end{pmatrix}} and {\begin{pmatrix}0\\1\\1\\0\\0\end{pmatrix}}, which correspond to a pathway through the system and the reversible reaction, respectively. In essence, these vectors tell us that the steady-state flux balance for the system is as follows:


b_{1}=v_{1}-v_{{-1}}=v_{2}=v_{3}


This makes sense: the system will have reached equilibrium if the fluxes are balanced as such. Thus, it follows that the maximum substrate input rate b_{1}below but not equal to which the substrate will approach a steady-state and not grow out of control is:


b_{{1,max}}=v_{2}=k_{2}E_{0}
.


The flux v_{2} is the most logical to use, as we can make use of the conservation relationship E+C=E_{0} to note that b_{{1,max}} cannot approach a steady state if the system has all complex and no enzyme in the conservation relationship at any time; that is, if C=E_{0}. In this case, the enzyme will not be able to convert substrate to product quickly enough, so the substrate input rate must be set at a value lower than k_{2}E_{0} for the system to approach a steady-state.


Steady-State Concentrations

Algebraic expressions for the steady-state concentrations may be useful. By setting all four differential equations in the open system to 0 to simulate the steady-state, we can algebraically solve for all four steady-state concentrations.


{\begin{cases}C_{{ss}}={\frac  {b_{1}}{k_{2}}}\\P_{{ss}}={\frac  {b_{1}}{k_{3}}}\\E_{{ss}}=E_{0}-C_{{ss}}=E_{0}-{\frac  {b_{1}}{k_{2}}}\\S_{{ss}}=({\frac  {k_{2}}{k_{1}}})({\frac  {{\frac  {k_{{-1}}}{k_{2}}}+1}{{\frac  {E_{0}k_{2}}{b_{1}}}-1}})\end{cases}}