# General Enzyme Kinetics

The applicability of Michaelis Menten Kinetics is restricted to chemical reactions with the rate equation $$E+S \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} ES \overset{k_2}{\rightarrow}E+P$$ However, the procedure for deriving Michaelis Menten Kinetics by Separation of Timescales is still the same for general enzyme kinetics. To elaborate on this, the steps of the method are again summarised and demonstrated at a specific example.

## How to Derive the Steady State Enzyme Kinetics for General Rate Equations

Enzymes do not only catalyse reactions, they are also regulatory units, that control the metabolic flux according to environmental conditions. In general the kinetics of an enzymatic reaction is influenced by effectors that change the rate at which substrate is converted and/or product formed. Effectors are compounds (e.g. proteins or small molecules) that interfere with the performance of the enzymatic reaction.

1. Draw a wiring diagram for the considered rate equation. Think about which steps of the reaction are reversible. A complete wiring diagram contains all substrates, products and $n$ free enzymes $E$ and enzyme complexes $ES$.
2. Write down the rate law considering mass action kinetics for each enzyme. Think about what the "input" (substrate) and what the "output" (product) of the reaction is. Keep in mind you want the rate of product formation $dP / dt$ as a function of the substrate(s) (the rate constants and enzyme concentration). $\frac{dP}{dt}= f(S_1,S_2, \dots)$ write down this equation for the change in product explicitly for your problem.
3. Separate timescales and assume steady state for all $n$ enzyme(complex) rates except the one that is formed with the product $P$.
4. Step 3 results in $n-1$ linear equations. Together with the conservation of total amount of enzyme $E_{total}=E+ES+ \dots$ this gives a solvable system of $n$ equations and $n$ unknown. Arrange the resulting system of equations neatly in matrix form. Solve the system by Matrix Elimination or with Mathematica.
5. Plug in the respective result in the equation for the change in product form point 2. to obtain the final result.

Example

Below you see a reaction map from the KEGG database. We are interested in the reactions involving Arginine $A$, Ornithine $O$ and Urea $U$ (reactions of interest marked i blue). These reactions are catalysed by the enzyme Arginase, which belongs to the Hydrolases (Hydrolases catalyse reactions $\mathrm{A{-}B + H_2O \leftrightharpoons A{-}H + B{-}OH}$). Like the name suggests water is involved in the reaction. Another popular example for a Hydrolase is Beta-galactosidase which catalyses the hydrolysis of Lactose.

• To draw a wiring diagram it is helpful to first write down the rate equation

\begin{align} \quad &EA \underset{k_{-3}}{\overset{k_3}{\rightleftarrows}} E+A \\ & \hspace{0.1cm} \downarrow \scriptsize{k_2} \\ E+O \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} E&O + U \end{align}

Notice that $E$ appears twice, we can write it just once if we make a circular wiring diagram
• The change in the amount of $E$ is given by $E$ consuming flux characterised by $k_1$, the $E$ forming flux $k_{-1}$, the $E$ consuming flux characterised by $k_3$ and the $E$ forming flux $k_{-3}$. Similar considerations can be made on the enzyme complexes $EA$ and $EO$, whereby the flux characterised by $k_2$ is unidirectional, only $EA$ is consumed and $EO+U$ formed, the reverse reaction does not occur.

\begin{align} \frac{dE}{dt}&=-v_1 + v_{-1}+v_3 - v_{-3} \\ \frac{dEA}{dt}&= v_1 - v_{-1} - v_{2} \\ \frac{dEO}{dt}&=v_2 - v_{3}+ v_{-3} \\ \frac{dO}{dt}&=v_3 \\ \frac{dU}{dt}&=v_2 \\ \frac{dA}{dt}&=v_{-1} \\ \end{align}

The inputs are Arginine ($A$) and Ornithine $O$, the output is Urea ($U$), so we are interested in an equation equation

\begin{align} \frac{dU}{dt}&=f(A,O) \\ \end{align}

where we want to determine the function $f$. Using the mass action law, we write the balance equations as

\begin{align} \frac{dE}{dt}&=-k_1 E \cdot A + k_{-1} EA + k_3 EO - k_{-3} E \cdot O\\ \frac{dEA}{dt}&= k_1 E \cdot A - k_{-1} EA - k_{2} EA\\ \frac{dEO}{dt}&=k_2 EA - k_{3} EO + k_{-3} E \cdot O \\ \frac{dO}{dt}&=k_{3} EO \\ \frac{dU}{dt}&=k_2 EA \\ \frac{dA}{dt}&=k_{-1} EA \\ \end{align}

• We separate timescales by assuming steady state for all reactions involving enzymes except the one that is produced with the product Urea $U$. The flux $v_2$ forms both $EO$ and the product $U$ from $EA$. So we don't assume steady state form the change in $EO$. There are $n=3$ enzyme/enzymecomplexes in involved, so have $n-1=2$ equations

\begin{align} 0&=\frac{dE}{dt}=-k_1 E \cdot A + k_{-1} EA + k_3 EO - k_{-3} E \cdot O\\ 0&=\frac{dEA}{dt}= k_1 E \cdot A - k_{-1} EA - k_{2} EA\\ \end{align}

• As we are interested in a function of $A$ and $O$ only, our goal is to eliminate all expressions including $E$ ($E,EA,EO$) from the equations. We can do this by solving for $E,EA,EO$ in terms of the other variables. If we assume that the rate constants are known we still only have two equations and three unknowns ($E,EA,EO$), so we need an additional equation. Therefore we assume the conservation of the total amount of enzyme $E_T$ during the timescale considered.

\begin{align} E_{T}=E+EA+EO \end{align}

To solve the system of equations, we first write it in matrix form

\begin{align} \begin{bmatrix} -k_1 A - k_{-3} O & k_{-1} & k_3\\ k_1 A & - k_{-1} - k_2 & 0\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} E \\ EA \\ EO \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ E_T \end{bmatrix} \end{align}

and perform Matrix Elimination. Alternatively we can just use Mathematica to find the solution.
• By plugging in $EA$ in the equation for the change in $U$ we find the explicit form of $f(A,O)$

\begin{align} \frac{dU}{dt}&=k_2 EA = -\frac{k_1 k_2 k_3 E_T \, A}{k_1 \left(k_{-1}-k_3\right) A + \left(k_{-1}+k_2\right) \left(-k_{-3} O -k_1 A-k_3\right)} \\ &=-\frac{k_1 k_2 k_3 E_T \, A}{- \left(k_{-1}+k_2\right)k_3 - \left(k_{-1}+k_2\right) k_{-3} O + \left[ \left(k_{-1}-k_3 \right) + \left(-k_{-1}-k_2\right) \right]k_1 A} \\ &=\frac{k_1 k_2 k_3 E_T \, A}{\left(k_{-1}+k_2 \right) k_3 + \left( k_{-1}+k_2 \right) k_{-3} O + \left( k_3 + k_2 \right) k_1 A} \\ \end{align} which can be further rearranged to the form \begin{align} \frac{dU}{dt}&=\frac{k_1 k_2 k_3 / \left[ \left( k_3 + k_2 \right) k_1 \right] E_T \, A}{\left(k_{-1}+k_2 \right) k_3 / \left[ \left( k_3 + k_2 \right) k_1 \right] + \left( k_{-1}+k_2 \right) k_{-3} / \left[ \left( k_3 + k_2 \right) k_1 \right] O + A} \\ \end{align} Comparing with $v=\frac{v_{max} \cdot S}{S + K_m}$, where $A$ plays the role of $S$, we find \begin{align} k_{cat}^f&=\frac{v_{max}^f}{E_T}= \frac{k_2 k_3}{\left( k_3 + k_2 \right)} \\ K_{m,A}&= \frac{\left(k_{-1}+k_2 \right) k_3}{ \left( k_3 + k_2 \right) k_1 }\\ K_{i,O}&= \frac{k_3}{k_{-3}} \end{align} and the factor in front of $O$ is the ratio $K_{m,A} / K_{i,O}$. Notice, the larger $O$ is the smaller is $\frac{dU}{dt}$, thus $O$ is an Inhibitor (generally denoted $I$). For $O=0$ the equation reduces to the Michaelis Menten Equation \begin{align} \frac{dU}{dt}&=\frac{k_{cat}^f E_T \, A}{ K_{m,A} + K_{m,A} / K_{i,O} O + A} \\ \end{align} In the case of $O=0$ the equation reduces to the Michaelis Menten Equation. If we compress all the rate constants we can write the steady state equation in the concise form \begin{align} \frac{dU}{dt}&=\frac{k_{cat}^f E_T \, A}{ K_{m,A} (1+ O / K_{i,O}) + A} \\ \end{align} This is a special example of Competitive Inhibition with the Substrate $A$ and the Inhibitor $O$ \begin{align} \frac{dU}{dt}&=\frac{k_{cat}^f E_T \,S}{ K_{m,A} (1+ I / K_{i,O}) + S} \\ \end{align}

• Philip W. Kuchel $\&$ Peter J. Mulquiney - Modelling Metabolism with Mathematica: Analysis of Human Erythrocyte[2]