A linear biochemical pathway is a chain of enzyme-catalyzed reaction steps where the product of one reaction becomes the substrate for the next reaction in a linear sequence. The molecules progress through the pathway in a straight line from the starting substrate to the final product. Each step in the pathway is usually facilitated by a different specific enzyme that catalyzes the chemical transformation.

Biological cells consume nutrients to sustain life. These nutrients are broken down to smaller molecules which are then reassembled into more complex structures required for life. The breakdown and reassembly of nutrients is called metabolism. An individual cell will contain thousands of different kinds of small molecules, such as sugars, lipids, and amino acids. The interconversion of these molecules is carried out by catalysts called enzymes. E. coli contains 2,338 metabolic enzymes.^[1] These enzymes form a complex web of reactions forming pathways by which nutrients are converted. These pathways come in various forms. Enzymes can form simple linear sequences of reactions forming a linear biochemical pathway. Pathways can also show other patterns such as branches or cycles. Some famous cyclic pathways include Kreb's cycle and Calvin's cycle. On the other hand, linear biochemical pathways include DNA replication, which connects the starting substrate and the end product in a straightforward sequence.

The figure below shows a four step pathway, with intermediates, ${\displaystyle S_{1},S_{2},}$ and ${\displaystyle S_{3}}$. To sustain a steady-state, the boundary species ${\displaystyle X_{o}}$ and ${\displaystyle X_{1}}$ are fixed. Each step is catalyzed by an enzyme, ${\displaystyle e_{i}}$.

Linear pathways follow a step-by-step sequence, where each enzymatic reaction results in the transformation of a substrate into an intermediate product, further processed by subsequent enzymes until the final product is synthesized.

A linear chain of four enzyme-catalyzed steps.

A linear pathway can be studied in various ways. Multiple computer simulations can be run to try to understand the pathway's behavior. Another way to understand the properties of a linear pathway is to take a more analytical approach. Analytical solutions can be derived for the steady-state if we assume simple mass-action kinetics.^[2][3][4] Analytical solutions for the steady-state when assuming Michaelis-Menten kinetics can be obtained^[5][6] but are quite often avoided. Instead, such models are linearized. The three approaches that are usually used are therefore:

• Computer simulation
• Analytical solutions using a linear mathematical model
• Linearization of a non-linear model

Computer simulation

It is possible to build a computer simulation of a linear biochemical pathway. This can be done by building a simple model that describes each intermediate in terms of a differential equation. The differential equations can be written by invoking mass conservation. For example, for the linear pathway:

${\displaystyle X_{o}{\stackrel {v_{1}}{\longrightarrow }}S_{1}{\stackrel {v_{2}}{\longrightarrow }}S_{2}{\stackrel {v_{3}}{\longrightarrow }}S_{3}{\stackrel {v_{4}}{\longrightarrow }}X_{1}}$

where ${\displaystyle X_{o}}$ and ${\displaystyle X_{1}}$ are fixed boundary species, the non-fixed intermediate ${\displaystyle S_{1}}$ can be described using the differential equation:

${\displaystyle {\frac {dS_{1}}{dt}}=v_{1}-v_{2}}$

The rate of change of the non-fixed intermediates ${\displaystyle S_{2}}$ and ${\displaystyle S_{3}}$ can be written in the same way:

${\displaystyle {\frac {dS_{2}}{dt}}=v_{2}-v_{3}}$

${\displaystyle {\frac {dS_{3}}{dt}}=v_{3}-v_{4}}$

To run a simulation the rates, ${\displaystyle v_{i}}$ need to be defined. If we assume mass-action kinetics for the reaction rates, then the differential equation can be written as:

${\displaystyle {\begin{array}{lcl}{\dfrac {dS_{1}}{dt}}&=&k_{1}X_{o}-k_{2}S_{1}\\[4pt]{\dfrac {dS_{2}}{dt}}&=&k_{2}S_{1}-k_{3}S_{2}\\[4pt]{\dfrac {dS_{3}}{dt}}&=&k_{3}S_{2}-k_{4}S_{3}\end{array}}}$

If values are assigned to the rate constants, ${\displaystyle k_{i}}$, and the fixed species ${\displaystyle X_{o}}$ and ${\displaystyle X_{1}}$the differential equations can be solved.

Plot shows a simulation of three intermediates from a four step pathway. The boundary species are fixed enabling the pathway to reach a steady state. Values k1 = 0.1; k2 = 0.15; k3 = 0.34; k4 = 0.1, Xo = 10, X1 = 0. S1, S2 and S3 are set to zero at time zero.

Analytical solutions

Computer simulations can only yield so much insight because one would be required to run simulations on a wide range of parameter values. which would become unwieldy. A more powerful way to understand the properties of a model is to solve the differential equations analytically.

Analytical solutions are possible if simple mass-action kinetics on each reaction step are assumed:

${\displaystyle v_{i}=k_{i}s_{i-1}-k_{-i}s_{i}}$

where ${\displaystyle k_{i}}$ and ${\displaystyle k_{-1}}$ are the forward and reverse rate-constants respectively. ${\displaystyle s_{i-1}}$ is the substrate and ${\displaystyle s_{i}}$ the product. If we recall that the equilibrium constant for this simple reaction is:

${\displaystyle K_{eq}=q_{i}={\frac {k_{i}}{k_{-i}}}={\frac {s_{i}}{s_{i-1}}}}$

we can modify the mass-action kinetic equation to be:

${\displaystyle v_{i}=k_{i}\left(s_{i-1}-{\frac {s_{i}}{q_{i}}}\right)}$

Given the reaction rates, the differential equations describing the rates of change of the species can be described. For example, the rate of change of ${\displaystyle s_{1}}$ will equal:

${\displaystyle {\frac {ds_{1}}{dt}}=k_{1}\left(x_{0}-{\frac {s_{1}}{q_{1}}}\right)-k_{2}\left(s_{1}-{\frac {s_{2}}{q_{2}}}\right)}$

By setting the differential equations to zero, the steady-state concentration for the species can be derived, from which the pathway flux equation can also be determined. For the three-step pathway, the steady-state concentrations of ${\displaystyle s_{1}}$ and ${\displaystyle s_{2}}$ are given by:

${\displaystyle {\begin{aligned}&s_{1}={\frac {q_{1}}{q_{3}}}{\frac {k_{2}k_{3}x_{1}+k_{1}k_{2}q_{3}x_{o}+k_{1}k_{3}q_{2}q_{3}x_{o}}{k_{1}k_{2}+k_{1}k_{3}q_{2}+k_{2}k_{3}q_{1}q_{2}}}\\[6pt]&s_{2}={\frac {q_{2}}{q_{3}}}{\frac {k_{1}k_{3}x_{1}+k_{2}k_{3}q_{1}x_{1}+k_{1}k_{2}q_{1}q_{3}x_{o}}{k_{1}k_{2}+k_{1}k_{3}q_{2}+k_{2}k_{3}q_{1}q_{2}}}\end{aligned}}}$

Inserting either ${\displaystyle s_{1}}$ or ${\displaystyle s_{2}}$ into one of the rate laws will give the steady-state pathway flux, ${\displaystyle J}$:

${\displaystyle J={\frac {x_{o}q_{1}q_{2}q_{3}-x_{1}}{{\frac {1}{k_{1}}}q_{1}q_{2}q_{3}+{\frac {1}{k_{2}}}q_{2}q_{3}+{\frac {1}{k_{3}}}q_{3}}}}$

A pattern can be seen in this equation such that, in general, for a linear pathway of ${\displaystyle n}$ steps, the steady-state pathway flux is given by:

${\displaystyle J={\frac {x_{o}\prod _{i=1}^{n}q_{i}-x_{1}}{\sum _{i=1}^{n}{\frac {1}{k_{i}}}\left(\prod _{j=i}^{n}q_{j}\right)}}}$

Note that the pathway flux is a function of all the kinetic and thermodynamic parameters. This means there is no single parameter that determines the flux completely. If ${\displaystyle k_{i}}$ is equated to enzyme activity, then every enzyme in the pathway has some influence over the flux.

Linearized model: deriving control coefficients

Given the flux expression, it is possible to derive the flux control coefficients by differentiation and scaling of the flux expression. This can be done for the general case of ${\displaystyle n}$ steps:

${\displaystyle C_{i}^{J}={\frac {{\frac {1}{k_{i}}}\prod _{j=i}^{n}q_{j}}{\sum _{j=1}^{n}{\frac {1}{k_{j}}}\prod _{k=j}^{n}q_{k}}}}$

This result yields two corollaries:

• The sum of the flux control coefficients is one. This confirms the summation theorem.
• The value of an individual flux control coefficient in a linear reaction chain is greater than 0 or less than one: ${\displaystyle 0\leq C_{i}^{J}\leq 1}$

For the three-step linear chain, the flux control coefficients are given by:

${\displaystyle C_{1}^{J}={\frac {1}{k_{1}}}{\frac {q_{1}q_{2}q_{3}}{d}};\quad C_{2}^{J}={\frac {1}{k_{2}}}{\frac {q_{2}q_{3}}{d}};\quad C_{3}^{J}={\frac {1}{k_{3}}}{\frac {q_{3}}{d}}}$

where ${\displaystyle d}$ is given by:

${\displaystyle d={\frac {1}{k_{1}}}q_{1}q_{2}q_{3}+{\frac {1}{k_{2}}}q_{2}q_{3}+{\frac {1}{k_{3}}}q_{3}}$

Given these results, there are some immediate observations:

• If all three steps have large equilibrium constants, that is ${\displaystyle q_{i}\gg 1}$, then ${\displaystyle C_{1}^{J}}$ tends to one and the remaining coefficients tend to zero.
• If the equilibrium constants are smaller, control tends to get distributed across all three steps.

The reason why control gets more distributed is that with more moderate equilibrium constants, perturbations can more easily travel upstream as well as downstream. For example, a perturbation at the last step, ${\displaystyle k_{3}}$, is better able to influence the reaction rates upstream, which results in an alteration in the steady-state flux.

An important result can be obtained if we set all ${\displaystyle k_{i}}$ equal to each other. Under these conditions, the flux control coefficient is proportional to the numerator. That is:

${\displaystyle {\begin{aligned}C_{1}^{J}&\propto q_{1}q_{2}q_{3}\\C_{2}^{J}&\propto q_{2}q_{3}\\C_{3}^{J}&\propto q_{3}\\\end{aligned}}}$

If we assume that the equilibrium constants are all greater than 1.0, then since earlier steps have more ${\displaystyle q_{i}}$ terms, it must mean that earlier steps will, in general, have high larger flux control coefficients. In a linear chain of reaction steps, flux control will tend to be biased towards the front of the pathway. From a metabolic engineering or drug-targeting perspective, preference should be given to targeting the earlier steps in a pathway since they have the greatest effect on pathway flux. Note that this rule only applies to pathways without negative feedback loops.^[7]

References