# Enzyme kinetics¶

This document explains the assumptions about enzyme kinetics that Maud uses.

## Modular rate law¶

This section outlines the assumptions made with the modular rate law, and includes a derivation of a 2 substrate (A, B), 2 product (P, Q) random mechanism with competitive inhibitor I. It also highlights the general structure of the modular rate law used in Maud. The modular rate law framework was taken from [1], and was adapted to suit our structure. The general rate structure for the modular rate law is given below

$v = E_t f \frac{T}{D + D^{reg}}$

where,

\begin{align}\begin{aligned}\begin{split}T = kcat_{1}\prod_{i, substrate} (\frac{X_i}{K_{m, i}})^{|n_i|} - kcat_{2}\prod_{i, product}(\frac{X_i}{K_{m, i}})^{|n_i|} \\\end{split}\\D = \prod_{i, substrate}(1 + \frac{X_i}{K_{m, i}})^{|n_i|} + \prod_{i, product}(1 + \frac{X_i}{K_{m, i}})^{|n_i|}.\end{aligned}\end{align}

Additionally, $$E_t$$ is the total enzyme concentration, $$f$$ is a regulatory function, and , $$D^{reg}$$ considers specific binding regulation [1]. $$n_i$$ is the stoichiometric value for each metabolite in the reaction, however, in [1] it is considered as a structure number accounting for cooperativity. In the Maud framework, cooperativity (allostery) is integrated using the generalised MWC model seperating the regulatory modulators from the catalytic rate.

### Assumptions¶

The assumptions used in the modular rate law are listed below:
• the metabolite binding occurs in a random order,

• binding does not occur simultanesouly,

• substrates and products cannot bind at the same time,

• metabolite binding rates are much higher than the interconversion of substrate to product (rapid equilibrium assumption),

• metabolite binding affinity is independent of order.

### Irreversible Modular Rate Law¶

When the Gibbs energy of the reaction becomes highly irreversible (e.g. $$\Delta G_r = -10 kJ/mol$$) the reverse rate becomes insignificant. This insignificant rate is not a problem to calculate, however, it can become detrimental to the sampling algorithm as it is essentially unresolvable. We can approximate the kinetics of these essentially irreversible mechanisms bu removing the reverse rate from the rate law. This is dangerous however, as it can lead to free energy cycles if used indiscriminately. For example, including this mechanism where a reaction is operating close to equilibrium and is part of a cycle could easily result in an infinite loop for energy production.

The one caveat is the product saturation of the enzyme is not natively captured. This can be fixed by including competitive inhibition from the product, which will result in the same mechanism.

#### Assumptions¶

Additional assumptions with the irreversible modular rate law formulation are as follows:
• product binding is insignificant and saturation of enzyme from products is neglected,

• the reaction has a very negative $$\Delta G_r$$, which will remove the reverse rate from the mechanism

### Example: 2 products and 2 substrate network¶

For a random Bi-Bi network with the above assumptions, the rate will be the following

$v = E_t \frac{kcat_1 a' b' - kcat_2 p' q'}{(1 + a')(1 + b') + (1 + p')(1 + q') -1}.$

where, for metabolite X the corresponding term is given by

$\begin{split}x' &= \frac{X}{K_m^{x}} \\ K_m^{x} &= \frac{X \bullet E_{X, unbound}}{[E_{X, bound}]}.\end{split}$

Because we assume binding to be independent of order of addition, there can be multiple relationships for a given $$K_m^{x}$$. For example, $$K_m^{A} = \frac{A \bullet E}{EA} = \frac{A \bullet EB}{EAB}$$

The rate is determined by the conversion from substrate to product and using elementary mass action kinetics is

$v = kcat_1 EAB - kcat_2 EPQ.$

Because of the rapid equilibrium assumption, the Michaelis-Menten constants are approximated by the dissociation constants. All enzyme state concentrations can be determined from the free enzyme concentration and metabolite concentrations. In this case

$\begin{split}EA &= a' E_0 \\ EB &= b' E_0 \\ EAB &= a' EB = b' EA = a' b' E_0 \\\\ EP &= p' E_0 \\ EQ &= q' E_0 \\ EPQ &= p' EQ = q' EP = p' q' E_0.\end{split}$

With the free enzyme concentration being a function of free enzyme ratio $$\theta$$ and total enzyme concentration

$\begin{split}E_0 &= E_t - \sum_{i, bound} E_i \\ &= E_t - E_0 (a' + b' + a' b' + p' + q' + p' q') \\ &= E_t \theta\end{split}$

where

$\theta = \frac{1}{1 + a' + b' + a' b' + p' + q' + p' q'}.$

After substituting the enzyme concentrations into the rate equation it becomes

$v = E_t \theta (kcat_1 a' b' - kcat_2 p' q').$

### Competitive inhibition¶

In the following case we will consider competitive inhibition where an inhibitor selectively binds to the free enzyme, preventing binding from either substrate or product.

As described in [1], competitive inhibition is accounted for in the denominator term of the rate equation. It’s easy to see how this occurs when you look at the free enzyme concentration

$EI = i' E_0.$

by using the previous

$\begin{split}E_0 &= E_t - \sum_{i,bound} E_i \\\end{split}$

and

$\theta = \frac{1}{1 + a' + b' + a' b' + p' + q' + p' q' + i'}.$

which can then be substituted into the original rate equation with the form

$v = E_t \frac{kcat_1 a' b' - kcat_2 p' q'}{(1 + a')(1 + b') + (1 + p')(1 + q') + i' -1}.$

### Allostery¶

Differing from the modular rate law defined in [1], allostery is considered using the generalised MWC form [see allostery link]. This requires the free enzyme amount - calculated above.

## Flux decomposition¶

In order to understand the enzyme mechanism, we can decompose the flux of an enzyme into several components:

• Enzyme - changes due to enzyme concentration.

• Allostery - changes due to the equilibrium between active and inactive states.

• Phosphorylation - changes due to actively phosphorylated enzymes.

• Saturation - how much of the enzyme is saturated by the substrates.

• Reversibility - what is the thermodynamic driving force of this reaction.

$\begin{split}v &= Enzyme * Allostery * Phosphorylation * Saturation * Reversibility \\ Eznyme &= [Enzyme] \\ Allostery &= \frac{1}{1 + L_0 (\theta \bullet \frac{\prod{(1 + \frac{[e_T]}{Km_T})}}{\prod{(1 + \frac{[e_R]}{Km_R})}})^{subunits}} \\ Phosphorylation &= \frac{1}{1 + (\frac{v_{kinase}}{v_{phosphatase}})^{subunits}} \\ Saturation &= \frac{\prod{\frac{[substrate]}{K_{substrate}}}}{\theta} \\ Reversibility &= (1 - exp(\frac{\Delta_{r}G}{RT})\end{split}$

## References¶

[1] Liebermeister, W., Uhlendorf, J. & Klipp, E. Modular rate laws for enzymatic reactions: thermodynamics, elasticities and implementation. Bioinformatics 26, 1528–1534 (2010).