User:Rhodydog/sandbox

Linear biochemical pathway
A linear biochemical pathway is a chain of enzyme-catalyzed reaction steps. The figure below shows a four step pathway, with intermediates, $$S_1, S_2, $$ and $$S_3$$. In order to sustain a steady-state, the boundary species $$X_o$$ and $$X_1$$ are fixed. Each step is catalyzed by an enzyme, $$e_i$$.

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 pathway can be studied in various ways. One way is to build a computer simulation and run many simulations 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 is we assume simple mass-action kinetics. Although analytical solutions for the steady-state when assuming Michaelis-Menten kinetics can be obtained (ref, sauro) but they are unwieldy and of little practical use. Instead, such models are linearized and the linearized equations become amenable to study.

Analytical Solutions
Analytical solutions for the can be obtained if we assume simple mass-action kinetics on each reaction step:


 * $$ v_i = k_i s_{i-1} - k_{-i} s_{i} $$

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


 * $$K_{eq} = q_i = \frac{k_i}{k_{-i}} = \frac{s_i}{s_{i-1}} $$

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


 * $$ v_i = k_i \left( s_{i-1} - \frac{s_i}{q_i} \right) $$

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 $$ s_1$$ will equal:


 * $$ \frac{ds_1}{dt} = k_1 \left( x_0 - \frac{s_1}{q_1} \right) - k_2 \left( s_1 - \frac{s_2}{q_2} \right) $$

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 $$s_1$$ and $$s_2$$ are given by:

$$ \begin{aligned} &s_1=\frac{q_1}{q_3} \frac{k_2 k_3 x_1+k_1 k_2 q_3 x_o+k_1 k_3 q_2 q_3 x_o}{k_1 k_2+k_1 k_3 q_2+k_2 k_3 q_1 q_2} \\[6pt] &s_2=\frac{q_2}{q_3} \frac{k_1 k_3 x_1+k_2 k_3 q_1 x_1+k_1 k_2 q_1 q_3 x_o}{k_1 k_2+k_1 k_3 q_2+k_2 k_3 q_1 q_2} \end{aligned} $$

Inserting either $$ s_1$$ or $$s_2$$ into one of the rate laws will give the steady-state pathway flux, $$ J$$:


 * $$J=\frac{x_o q_1 q_2 q_3-x_1}{\frac{1}{k_1} q_1 q_2 q_3+\frac{1}{k_2} q_2 q_3+\frac{1}{k_3} q_3} $$

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

$$J=\frac{x_o \prod_{i=1}^n q_i-x_1}{\sum_{i=1}^n \frac{1}{k_i}\left(\prod_{j=i}^n q_j\right)}$$

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 $$k_i$$ is equated to enzyme activity, then every enzyme in the pathway has some influence over the flux.

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 $$n$$ steps:

$$C_i^J=\frac{\frac{1}{k_i} \prod_{j=i}^n q_j}{\sum_{j=1}^n \frac{1}{k_j} \prod_{k=j}^n q_k}$$

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: $$0 \leq C^J_i \leq 1$$

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

$$ C_1^J=\frac{1}{k_1} \frac{q_1 q_2 q_3}{d} ; \quad C_2^J=\frac{1}{k_2} \frac{q_2 q_3}{d} ; \quad C_3^J=\frac{1}{k_3} \frac{q_3}{d} $$

where $$d$$ is given by:

$$ d=\frac{1}{k_1} q_1 q_2 q_3+\frac{1}{k_2} q_2 q_3+\frac{1}{k_3} q_3 $$

Given these results, there are some immediate observations:


 * If all three steps have large equilibrium constants, that is $$q_i \gg 1$$, then $$ C^J_{1}$$ 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, $$k_3$$, 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 $$k_i$$ equal to each other. Under these conditions, the flux control coefficient is proportional to the numerator. That is:

$$ \begin{aligned} C^J_1 &\propto q_1 q_2 q_ 3\\ C^J_2 &\propto q_2 q_ 3\\ C^J_3 &\propto q_ 3\\ \end{aligned} $$

If we assume that the equilibrium constants are all greater than 1.0, then since earlier steps have more $$q_i$$ 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.