May 21, 2018, Monday

Looking for a Steady-State Flux Distribution

Introduction

As seen in our diffusion approximation, a biologically realistic steady-state flux distribution does not exist for our C5a production model. We are interested studying cases where a steady-state flux distribution arises when diffusion is considered in addition to reaction stoichiometry, so we now turn to the canonical Michaelis-Menten model and add diffusion using our "flux" method.

Classical Michaelis-Menten Model With Diffusion

We will begin by looking at the classical model considered earlier, in which it is assumed that substrate flows into some domain containing enzyme to form the substrate-enzyme complex. The product formed leaves the domain, while the enzyme remains internal to the spatial domain. We will continue to consider diffusion along a line as in our C5a production model and reaction occurring at the surface of the enzyme (bound by substrate). We approximate the reaction space as a single point and simplify the system so that it effectively involves the substrate flowing into this point from the left, binding with enzyme at our single point, and then undergoing a conformational change to produce a product, which flows out of the domain to the right. We label the reaction space as point 0, the points to the left of the reaction space where the substrate diffuses in as ..., -3, -2, -1, and the points to the right of the reaction space where the product leaves as +1, +2, +3, ... However, we will begin by approximating this system at three discrete points: -1, some distance $h$to the left of the reaction space, 0, at the reaction space, and +1, some distance $h$ to the right of the reaction space. Then our system of differential equations is:

${\frac {\partial [S,0]}{\partial t}}=k_{{-1}}[C]-k_{1}[S,0][E]+{\frac {D_{{S}}}{h^{{2}}}}([S,-1]-[S,0])$
${\frac {\partial [S,-1]}{\partial t}}={\frac {D_{{S}}}{h^{{2}}}}([S,0]-[S,-1])$
${\frac {\partial [E]}{\partial t}}=(k_{{-1}}+k_{2})C-k_{1}[S,0][E]$
${\frac {\partial [C]}{\partial t}}=k_{1}[S,0][E]-(k_{{-1}}+k_{2})[C]$
${\frac {\partial [P,0]}{\partial t}}=k_{2}[C]+{\frac {D_{{P}}}{h^{{2}}}}([P,+1]-[P,0])$
${\frac {\partial [P,+1]}{\partial t}}={\frac {D_{{P}}}{h^{{2}}}}([P,0]-[P,+1])$

Where $D_{{S}}$ is the diffusion rate of the substrate into the reaction and $D_{{P}}$ is the diffusion rate of the product out of the reaction. Note that we assume some nonzero initial concentration $S_{{0}}$ at point -1 that flows into the reaction space, which has zero initial concentration of the substrate. We also assume that the enzyme and complex remain internal to the reaction space.

We now construct the stoichiometric-diffusion matrix. As seen in the C5a production case, note that the diffusion is considered as a gradient, such that the diffusion flux terms seen in S, 0 and S, -1 are equal but opposite. Thus, we have created an $6\times 5$ matrix $S:D$ such that

${\frac {\partial \phi }{\partial t}}=(S:D){\textbf {v}}$,

where

$S:D={\begin{pmatrix}-1&1&1&0&0\\1&-1&-1&0&0\\-1&1&0&1&0\\0&0&0&-1&0\\0&0&1&0&-1\\0&0&0&0&1\end{pmatrix}}$

$\phi =(E,C,S_{{0}},S_{{-1}},P_{{0}},P_{{+1}})$

${\textbf {v}}=(k_{{1}}S_{{0}}E,k_{{-1}}C,k_{{2}}C,{\frac {D_{{S}}}{h^{{2}}}}([S,-1]-[S,0],{\frac {D_{{P}}}{h^{{2}}}}([P,0]-[P,+1]))$

We now conduct our SVD analysis to identify the steady-state flux distribution. As seen in our C5a production example, our results for a finite number of points at well-chosen locations can be generalized to the real infinite case.

The rank $r=4$, so the dimension of the right null space is 1. By SVD, a basis for this space is given by:

$V_{{5}}=(1,1,0,0,0)^{{T}}$

This corresponds to a steady-state arising when $k_{{1}}S_{{0}}E=k_{{-1}}C$, or when the substrate-enzyme binding reaction is equal and opposite to the complex decomposition with all other reaction fluxes zero. Of course, this is not a biologically realistic scenario. A similar result had been found for the closed case previously.

Considering the Reaction Space Only

However, note that this SVD method takes into account steady-states for the complex at point -1 and for the product at point +1, which we are not necessarily concerned with. We will see what happens when these two differential equations are removed from our stoichiometric-diffusion matrix:

$(S:D)_{{new}}={\begin{pmatrix}-1&1&1&0&0\\1&-1&-1&0&0\\-1&1&0&1&0\\0&0&1&0&-1\\\end{pmatrix}}$

$\phi =(E,C,S_{{0}},P_{{0}})$

${\textbf {v}}=(k_{{1}}S_{{0}}E,k_{{-1}}C,k_{{2}}C,{\frac {D_{{S}}}{h^{{2}}}}([S,-1]-[S,0]),{\frac {D_{{P}}}{h^{{2}}}}([P,0]-[P,+1]))$

We now have a $4\times 5$ matrix with rank $r=3$, so the dimension of the right null space is 2, given by the following orthonormal basis:

$V_{{4}}=(-0.7477,-0.6344,-0.1133,-0.1133,-0.1133)^{{T}}$
$V_{{5}}=(0.1114,-0.4110,0.5224,0.5224,0.5224)^{{T}}$

Any linear combination of these vectors will give a steady-state flux distribution; our goal is now to find biological relevance from these results. We begin by trying $-V_{{4}}+V_{{5}}$, as this will give us all positive entries.

$-V_{{4}}+V_{{5}}=(0.8591,0.2234,0.6357,0.6357,0.6357)^{{T}}$

And a mathematical check shows that this flux distribution indeed gives rise to a steady-state flux distribution. The biological significance of this particular reaction stoichiometry is uncertain, but would indeed be biologically realistic. It is worth noting, however, that as part of this steady-state we have substrate flowing into the reaction at the same rate that product flows out of it, as expected.