Modelling compartmentalization towards elucidation and engineering of spatial organization in biochemical pathways

Compartmentalization is a fundamental ingredient, central to the functioning of biological systems at multiple levels. At the cellular level, compartmentalization is a key aspect of the functioning of biochemical pathways and an important element used in evolution. It is also being exploited in multiple contexts in synthetic biology. Accurate understanding of the role of compartments and designing compartmentalized systems needs reliable modelling/systems frameworks. We examine a series of building blocks of signalling and metabolic pathways with compartmental organization. We systematically analyze when compartmental ODE models can be used in these contexts, by comparing these models with detailed reaction-transport models, and establishing a correspondence between the two. We build on this to examine additional complexities associated with these pathways, and also examine sample problems in the engineering of these pathways. Our results indicate under which conditions compartmental models can and cannot be used, why this is the case, and what augmentations are needed to make them reliable and predictive. We also uncover other hidden consequences of employing compartmental models in these contexts. Or results contribute a number of insights relevant to the modelling, elucidation, and engineering of biochemical pathways with compartmentalization, at the core of systems and synthetic biology.


Models
The di↵erent biochemical pathways examined here are described below (for the two compartment scenario). Schematics for the various pathways are presented in the main text and Figure S1.

Single modification cycle
This is a closed system consisting of a single enzymatic modification cycle involving species X and X ⇤ . Both species di↵use across the whole domain, and may have di↵erent di↵usivites. E 1 , the enzyme catalyzing X ! X ⇤ is non-di↵using and uniformly distributed in compartment 1 (see Figure 1(a)). E 2 , the enzyme catalyzing X ⇤ ! X is non-di↵using and uniformly distributed in compartment 2. The enzyme substrate complexes, XE 1 and X ⇤ E 2 (referred to as C 1 and C 2 ) are also non-di↵using. The total amount of substrate within the whole domain, consisting of free substrate (X and X ⇤ ) and the complexes, is conserved.
PDE model: In compartment 1 (0  ✓  L 1 ), Between the compartments (L 1  ✓  L 1 + L), In compartment 2 (L 1 + L  ✓  L 1 + L + L 2 = l), With the conservation equations, where h·i represents the spatial average over the whole domain, and h·i i represents the spatial average over the compartment i. The corresponding compartmental model is as follows.
2 Compartmental ODE model: In compartment 1, In compartment 2, With corresponding conservation equations for the substrate and the enzymes. The form of the conservation equation for the di↵using substrate will be examined as part of our analysis. In the compartmental ODE equations, the variables [X i ] and [X ⇤ i ] represent the concentrations of species X and X ⇤ in compartment i.

Two-step cascade
This is a closed system consisting of two enzymatic modification cycles -the first involving species X and X ⇤ , and the second involving Y and Y ⇤ (See Fig. S1(a)). As before, the X cycle involves enzymes E 1 and E 2 . Both of these are non-di↵using and uniformly distributed in compartments 1 and 2 respectively. X and X ⇤ di↵use over the whole domain, with X ⇤ acting as the enzyme for Y ! Y ⇤ . Y and Y ⇤ are non-di↵using and confined to compartment 2. E 3 , the enzyme catalyzing Y ⇤ ! Y is also non-di↵using and uniformly distributed in compartment 2. All enzyme-substrate complexes are non-di↵using. The total amount of substrate over the whole domain, in each of the cycles, is conserved.

4
Compartmental ODE model: In compartment 1, With corresponding conservation equations for the substrates and the enzymes.

Two-site modification
This is a closed system involving interconversion between the species X and its modified forms, X ⇤ and X ⇤⇤ , with the three forms belonging to a conserved pool. As in the previous models, both enzymes and complexes are non-di↵using. Enzyme E 1 , uniformly distributed in compartment 1, catalyzes both the forward steps X ! X ⇤ and X ⇤ ! X ⇤⇤ , while E 2 , uniformly distributed in compartment 2, catalyzes both the reverse steps.
Between the compartments (L 1  ✓  L 1 + L), With the conservation equations, The corresponding compartmental model is as follows.

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Compartmental ODE model: In compartment 1, In compartment 2, With corresponding conservation equations for the substrates and the enzymes.

Simple open system
This is an open system involving a single species X that is produced in compartment 1 and degraded in compartment 2.

Metabolic pathway
This system consists of part of a metabolic pathway, involving two successive enzymatic conversions of metabolites. The two enzymes, as well as the initial metabolite R, are non-di↵usible, while the product of the first conversion X, is di↵usible across the whole domain. R is present in compartment 1, with the first enzyme P also confined to this compartment. The concentration of R is assumed to be maintained at a fixed level in compartment 1. The second enzyme E, that converts X to another metabolite Y , is distributed between the two compartments (uniformly in each compartment), with a possible di↵erence in its kinetics in compartments 1 and 2 . There are competing reactions, consuming X, in the second compartment. The enzymatic reactions are assumed to operate at the mass-action regime.
PDE model: In compartment 1 (0  ✓  L 1 ), @✓ 2 The total amount of the enzyme F is fixed, The corresponding compartmental model is as follows.
In compartment 2, with the total amount of the enzyme E, We also examine a variant of the above pathway, where the product metabolite X negatively regulates the enzyme E. This type of negative feedback by a metabolite on the upstream enzymes in a pathway may be realised through allosteric regulation. In this case, the production term for species X in compartment 1 is changed to k 1 [P ] [R] K+ [X] in the PDE and k 1 [P ][R] K+[X] L 1 in the compartmental model.

Analytical Investigations
In this section, we present a series of investigations to consolidate and further discuss points made in the main text. This includes solution of the relevant PDEs, correspondence with the relevant compartmental models (and obtaining relevant parameters therein, as well as other related discussions). In all cases, analytical solutions to the PDE are cross-validated against numerical solutions. Furthermore, parametrizing or modifying compartmental models to account for sequestration or changing transport parameters is done through a correspondence between compartmental ODE models and the PDE. This is subject to an additional level of cross-validation. Indeed, as discussed in the main text, these changes have demonstrated a marked improvement in matching the compartmental ODE model with the PDE in many di↵erent parameter regimes and scenarios. Numerical evidence in the text substantiates other analytical discussions.

Transport parameters in the thin compartment regime
For the kinds of compartmentalized reaction pathways examined in this study, compartmental ODE models naturally correspond to the limit of well mixed compartments connected by pure di↵usive transport. This limit is usually assumed when using compartmental ODE models. This limit may be realised in di↵erent ways -for the type of scenario examined here, this limit corresponds to having compartment sizes much smaller than the intervening space, i.e. the thin compartment regime.
Consider any 1-D, two compartment system (the following discussion is equally applicable to systems with more compartments) of the kind examined here (for example, see Figure 1). The net transport rate of a species X j out of compartment 1 may be expressed as the product of the di↵usivity D j and the concentration gradient of X j at the edge of the compartment. Since the intervening space involves only di↵usion, this transport rate, at the limit of well mixed compartments, is given by where X j1 and X j2 are the concentrations of X j in compartments 1 and 2, and L is the size of the space between compartments. Now, the net transport rate of the same species X out of compartment 1, in the corresponding compartmental ODE model is represented as From (1) and (2), we see that the transport parameters for X corresponding to the limit of well mixed compartments is given by In the following discussion, we use analytical solutions to the PDE (for the simple closed and open systems) to obtain expressions for the transport parameters, and show that the above values for the transport parameters are attained in the limit of thin compartments.

Accounting for sequestration: Modified conservation condition
Consider the simple closed system, the compartmentalized single modification cycle discussed in the text (see Figure 1(a)). In the thin compartment regime, the total amount of substrate in the compartments depends on kinetics, while the transport parameters may be approximated by D/LL 1 . In order for such a compartmental ODE model to completely capture the e↵ect of changing kinetics, it needs a way of accounting for the variation in total amount of substrate (in the compartments). Now, although the total amount in the compartments, i.e.X T otal , may depend on kinetics, the total amount in the whole domain is still conserved and remains invariant. That is, is fixed independent of the kinetics, where X Inter represents the total amount of X and X ⇤ in the intermediate space. Note that the enzyme-substrate complexes, being non-di↵usible, are not present in the intermediate space. Since the concentration profiles of X and X ⇤ are linear in the intermediate space (due to pure di↵usion in the intermediate space), X Inter can be computed exactly using the concentrations of X and X ⇤ at the edges of the compartments (i.e. at ✓ = L 1 and ✓ = L 1 + L).
In the thin compartment regime, in the absence of significant gradients within the compartments, these edge concentrations may be reasonably approximated by the compartmental averages. This means that, for the compartmental ODE, we can write a conservation condition for the whole domain: For the single modification cycle, there is only one enzyme-substrate complex in each compartment -C 1 in compartment 1 and C 2 in compartment 2. The above conservation condition involves only the compartmental variables, but still accounts for the sequestration of species in the intermediate space. Note that we have explicitly used steady state conditions in obtaining this conservation condition.

Accounting for sequestration: Using a third compartment for the intermediate space
The modified conservation condition described above approximates the total amount in the intermediate space well, for a system in the thin compartment regime. However, this approximation only describes the steady state. Away from steady state conditions, the total amount in the intermediate space may not be equal to the average of the compartmental concentrations multiplied by the size of the intermediate space. One way to overcome this limitation, is to use a third compartment to describe the intermediate space.
Consider a general pathway involving two compartments in 1-D. In the thin compartment regime, the basic (two) compartmental model is given by: where j = 1, 2, ..., n where the variables X ji represent the (average) concentration of species j in compartment i. The reaction kinetic terms are represented by the functions Each species j has a pair of transport parameters associated with ittr j 1!2 and tr j 1!2 . As seen above, in the thin compartment regime, these are well approximated by is the di↵usivity of species j. Thus, the transport rate of X j out of compartment 1 (into compartment 2) has the form (1). Now, instead of just using a modified conservation condition for each di↵using species that is conserved, we introduce a third compartment representing the intermediate space. The compartmental model is now given by where j = 1, 2, ..., n Here, X jm represents the (average) concentration of species j in the intermediate space denoted by m. Suppose that transport parameters are chosen in the following way: , and the transport rate of X j out of compartment 1 (into compartment 2) continues to have the form (1). Now, the conservation condition for a conserved, di↵using species may be correctly approximated by incorporating the corresponding amount in the third compartment, when computing the total amount of substrate. Note, that the above model fails completely if any reactions involving net production or removal of X j are present in the intermediate space, even if the corresponding kinetic terms are incorporated into (4). This is because the resulting system would no longer correspond to the limit of thin compartments. Now we discuss analytical solutions to the PDE models for the single modification cycle and the simple open system discussed in the main text.

Solutions to PDE models
Case 1: Single modification cycle(mass-action) 1-D PDE model: We assume that both phosphorylation and dephosphorylation occure through mass action kinetics. Note that this can be formally obtained from the model in the previous section by making the catalytic constants high. The kinase and phosphatase are uniformly distributed in the regions 0  ✓ < L 1 and L 1 + L  ✓ < l < L 1 + L + L 2 respectively (see Figure 1(a)). We assume that both the phosphorylation and dephosphorylation involve mass action kinetics. We first consider the case where X and X ⇤ are equally di↵usible.
In compartment 1 (0  ✓  L 1 ), Between the compartments (L 1  ✓  L 1 + L), @✓ 2 With the conservation condition, where h·i represents the spatial average over the whole domain. In the above, k 1 represents the product of the phosphorylation rate constant and the total kinase concentration, while k 2 represents the product of the dephosphorylation rate constant and the total phosphatase concentration Since X and X ⇤ are equally di↵usible, the PDE model implies that the quantity [X]+[X ⇤ ] must be constant throughout the domain at steady state. That is, [X] + [X ⇤ ] = X T , where X T = X T otal /l. This also means that the total amount of substrate sequestered in the middle is independent of the kinetic parameters in the compartments. Also note that the equations for X and X ⇤ are linear. Combined with the above condition, this means that [X] and [X ⇤ ] are linearly proportional to X T (and hence to X T otal ).

1-D PDE Solution
Implementing no-flux boundary conditions we find: Compartmental averages Since the di↵usivities of X and X ⇤ are equal, the compartmental averages must also satisfy the equations We now consider the corresponding compartmental ODE model for the case where the volumes of the two compartments are equal, i.e. L 1 = L 2 . The case where L 1 6 = L 2 is essentially no di↵erent. Since the volumes are equal, we have a single transport parameter associated with each speciestr X for X and tr X ⇤ for X ⇤ . The model is as follows.
In compartment 1, Transport parameter The values of the transport parameters that make the steady state of the compartmental model exactly match the PDE averages may then be computed as follows: As we show below, the two transport parameters must be in the same ratio as the di↵usivities of X and X ⇤ . Thus, for the case of equal di↵usivities, we have a single transport parameter tr. Using equations (5) and (6), we have Taking the limit L ! 1, keeping D L constant, we find that lim L!1 The compartmental model for the system is given by (7). The values of the transport parameters that produce an exact match with the PDE averages at steady state are computed as follows: where we have also used the fact that h e Xi and h e X ⇤ i satisfy equations of the form (6). From (9) and (10), we see that the transport parameters are in the same ratio as the di↵usivities.

Case 2: Simple open system
This is an open system involving a single species X that is produced in compartment 1 (rate constant k 0 ) and degraded/consumed in compartment 2 (rate constant k 1 ).

1-D PDE Solution
The solution is as follows: Compartmental averages We now consider the corresponding compartmental model for the case where the volumes of the two compartments are equal, i.e. L 1 = L 2 . The case where L 1 6 = L 2 is essentially no di↵erent. Since the volumes are equal, we have a single transport parameter associated the species X. The model is given by:

Transport parameter
The values of the transport parameters that make the steady state of the compartmental model exactly match the PDE averages may then be computed as follows: where hXi 1 and hXi 2 are given by (11). Substituting for hXi 1 and hXi 2 and taking the limit L ! 1, keeping D L constant, we find that lim L!1

Case 3: Simple open system (with degradation in both compartments)
This is an open system involving a single species X that is produced in compartment 1 (rate constant k 0 ) and degraded/consumed in both compartment 1 (rate constant k 1 ) and compartment 2 (rate constant k 2 ).

1-D PDE Model
In compartment 1 (0  ✓  L 1 ), @✓ 2 Between the compartments (L 1  ✓  L 1 + L), In compartment 2 (L 1 + L  ✓  L 1 + L + L 2 = l), The solution is as follows: Compartmental averages We now consider the corresponding compartmental model for the case where the volumes of the two compartments are equal, i.e. L 1 = L 2 . The case where L 1 6 = L 2 is essentially no di↵erent. Since the volumes are equal, we have a single transport parameter associated the species X. The model is given by: 20 Compartmental ODE model: In compartment 1, In compartment 2,

Transport parameter
The values of the transport parameters that make the steady state of the compartmental model exactly match the PDE averages may then be computed as follows:

Case 4: Compartmentalized two-site modification (Massaction)
We consider the system involving interconversion between the species X and its modified forms, X ⇤ and X ⇤⇤ , with the three forms belonging to a conserved pool. The phosphorylation reactions are assumed to be in the mass action regime, with kinases restricted to compartment 1 and phosphatases in compartment 2. All species are assumed to be equally di↵usible.

1-D PDE Model
In compartment 1 (0  ✓  L 1 ), Between the compartments (L 1  ✓  L 1 + L), @✓ 2 With the concentration profiles satisfying a conservation condition of the form, The constants c i (i = 1, ..., 8) that determine the solution, are obtained by using the boundary conditions -no-flux conditions at the outer boundaries of the domain, and continuity of the concentration profiles and their first derivatives, at the compartment boundaries. This yields eight linear equations in the variables c i , which are then solved by symbolic computation using Maple. The expressions thus obtained for the steady state concentration profiles of [X], [X ⇤ ], and [X ⇤⇤ ] are then validated by comparison with the steady state profiles obtained by numerical solution of the PDE system.

Thin compartment regime in a 2-D system
We have seen that a compartmental model, with fixed transport parameters of the form D/LL i , is a good approximation for a 1-D system in the thin compartment regime, at steady state. Here we show how a similar ODE model may be obtained for the same kind of system in a general 2-D case, with smooth boundaries for the compartments and the spatial domain. Consider the 2-D system shown in Fig. S1(b). Since the kinetics play no role in representing transport in the thin compartment regime, we can focus on a single species X to get the general picture. At steady state, let X(r) denote the concentration of species X outside the compartments. Within the compartments, and on the compartment boundaries ( 1 and 2 ), the concentrations are given by the compartmental averages -X 1 and X 2 .
Since the intermediate space involves only di↵usion of X, at steady state we have r 2 X(r) = 0 with the boundary conditions rX(r) = 0 on the outer boundary (noflux) X(r) = X 1 on the boundary 1 X(r) = X 2 on the boundary 2 The solution to this boundary value problem may be expressed as where the functions f 1 (r) and f 2 (r) are solutions to the above boundary problem for the basal cases X 1 = 1, X 2 = 0 and X 1 = 0, X 2 = 1 respectively. As long as the geometry of the system remains unchanged, these functions only need to be solved for once (perhaps numerically), and can be used in the following way to obtain a compartmental description in the thin compartment regime. Note that the gradient of X(r) is given by rX(r) = X 1 rf 1 (r) + X 2 rf 2 (r) ( 1 3 ) 23 At steady state, the net transport of X out of compartment 1 is given by averaging the flux over the boundary 1 : wheren is the unit outward normal to 1 . This can be rewritten as: In the compartmental model for this 2-D system, the net transport of X out of compartment 1 is represented by: where X i are the compartmental (average) concentrations and A i are the volumes (areas) of the compartments. Comparing (14) and (15), we see that the two transport parameters associated with the species X in the compartmental model are given by: rf 2 (r).nds ◆ 2.6 Finding a compartmental ODE description to match the PDE

1-D PDE at steady state
Consider a general pathway with first-order mass-action kinetics and possible zeroth order production, distributed between two equally sized compartments (L 1 = L 2 ), in a 1-D domain (see Figure 1 for the geometry). At steady state, the PDE model in vector form gives: Here, u is the vector of species concentrations at a point in the domain, and A 1 and A 2 are square matrices containing the rate constants corresponding to the mass-action reactions in compartments 1 and 2 respectively. The vectors b 1 and b 2 represent possible zeroth order production terms, and D is a diagonal matrix containing the di↵usivities of the di↵erent species. For example, for the single modification cycle discussed in the main text, with kinetics at the mass-action regime, No-flux boundaries are present at the ends of the domain. Averaging over compartments 1 and 2 at steady state, we have Here, hui 1 and hui 2 are vectors of compartmental averages for compartments 1 and 2. Since no reactions occur in the intermediate space, we also have ✓ @u @✓ This also means that Compartmental ODE For the system described by (16), we consider a compartmental ODE model where each species has a single transport parameter associated with it. The compartmental model has the form: Here,ū 1 andū 2 denote vectors of species concentrations corresponding to compartments 1 and 2 respectively. is a vector containing the transport parameters, and is a diagonal matrix having the components of the vector u 2 ū 1 as its diagonal.
Since we want the compartmental model to match the PDE compartmental averages at steady state, the steady state equations for the compartmental model must satisfy the equations: u is a diagonal matrix having the vector hui 2 hui 1 as its diagonal. Thus, the matrix u is non-singular, assuming that the system corresponds to a case where all di↵using species exhibit gradients between compartments. This fact, along with (20), allows us to compute the unique vector of transport parametersss , that produces an exact match between the compartmental model and the PDE averages at steady state.
Note that there are exceptional cases where the compartmental averages may be equal for certain species (i.e. certain components of the vector hui 2 hui 1 may be zero), even when there is a non-zero concentration gradient for this species between compartments. In such cases, the transport parameter associated with this species diverges (See main text).
Of course, in addition to using these transport parameters, the ability of the compartmental model to match the PDE at steady state also relies on correctly describing the conservation conditions for any conserved species, as seen previously.

Compartmental ODEs: Nonlinear kinetics
We now examine the case similar to (16), but where the kinetics can have arbitrary form, i.e. possibly including nonlinear terms such as second-order mass action steps or Michaelis-Menten kinetics. At steady state we have: where f 1 (u) and f 2 (u) are nonlinear, vector valued functions of u representing the sum of the kinetic terms in compartments 1 and 2. Averaging over the compartments, we find As before, we have Note however, that these are spatial averages of f 1 (u) and f 2 (u), and not functions of the average of u, i.e. f 1 (hui 1 ) and f 2 (hui 2 ). In general, these are not equal, and therefore, Now consider the compartmental model for the system (23), given by At steady state, the vectors of compartmental concentrations,ū 1 andū 2 must satisfy ( 2 6 ) From (24) and (26), we can conclude that the compartmental model cannot exactly match the PDE averages at steady state, when nonlinear kinetic terms are involved. The best that can be done in this case is to choose the vector of transport parameters such that the mismatch is minimized in some sense. In the present study, we choose to minimize the sum-squared error k(ū 1 hui 1 )k 2 + k(ū 2 hui 2 )k 2 . For the compartmental model (10), this gives us a nonlinear optimization problem with a quadratic objective function and nonlinear equality constraints: Any conservation conditions would form additional equality constraints in (27).

Transport Parameters and Di↵usivities
We now examine the possibility that, for the first-order mass-action system in (16), the transport parameters for the di↵erent species, that produce an exact steady state match, are in the same ratio as the di↵usivities of the corresponding species. For this to be true, at steady state we need the vectors of compartmental averages, hui 1 and hui 2 to satisfy an equation of the form: for some positive . Thus, from equations (17) and (28), we need Single modification cycle (X to X*). Since the kinetics in each compartment conserves the total amount of species, the species concentrations satisfy at all points in the domain. X T is constant within the domain but its value depends on both kinetic and spatial parameters (including compartment size, compartment separation, and di↵usivities). Thus, the compartmental averages satisfy, Now, the steady state mass balance for compartment 1 implies i.e. the net flux out of the compartment is zero.
Equations (31) and (32) imply that the di↵erences in compartmental averages are related to the concentration gradients in the intervening space, as follows Thus, the compartmental fluxes and compartmental averages satisfy an equation of the form (29). This means that a compartmental ODE model that exactly matches the PDE compartmental averages at steady state, must have transport parameters in the same ratio as the di↵usivites. For a given set of kinetic and spatial parameters, these transport parameters would be uniquely determined.
Two-site modification cycle (X to X* to X**). Since the kinetics in each compartment only results in an interconversion between species, adding the equations for X, X ⇤ , and X ⇤⇤ at steady state results in: at all points in the domain, where X T is constant across the domain As in the previous case, the compartmental averages for the di↵erent species satisfy Considering the steady state mass balance for compartment 1, we have However, unlike the case of the single modification cycle, equations (35) This means that a compartmental ODE, which matches the PDE compartmental averages at steady state, need not have transport parameters in the same ratio as the di↵usivities. However, the transport parameter associated with each species would still be uniquely determined. The transport parameters for the exact match, obtained numerically (solving the PDE and computing compartmental averages) for a sample case, are shown in Fig.  5(b).

Adding a third compartment
We examine the case of adding a third compartment to an existing two compartment open pathway (see schematic in Fig. 4(c)). The three compartments are of the same size, and are equally spaced. The species X is produced at a constant rate (zeroth order kinetics) in compartment 1, and di↵uses across the domain. X is consumed by reactions in compartments 1, 2 and 3 (represented by first order reactions). We obtain the analytical solution to this three compartment system at steady state (not shown), and use the steady state concentration profile of X to compute spatial averages within each compartment. We use these averages to calculate the two transport parameters tr X 12 and tr X 23 representing transport between compartments 1 and 2 and between 2 and 3 respectively, as shown in (22) for the general case. Each of these transport parameters depend on the di↵usivity, compartment sizes, compartment separation, and all the kinetic parameters. We fix the di↵usivity, compartment sizes, compartment separation, and the kinetic parameters in compartments 1 and 2 (see table below for parameter values), and examine the e↵ect of varying the rate constant of X consumption in compartment 3. The result is shown in Fig. 4(d). Note that the analytical solution to the three compartment system was cross-validated by comparison to the numerical solution of the PDE.

Bistable two-site modification
As discussed in the text, we consider a compartmentalized two-site modification pathway (see Fig. 5(a)), in a kinetic regime where it exhibits two stable steady states, over a range of values of E T otal 1 -the total amount of kinase in compartment 1. We consider this system in the thin compartment regime (L/L 1 = 18). The concentration profiles for the two steady states are obtained by numerical solution of the PDE model, with a fixed total amount of substrate X T otal = 10 and enzymes E T otal 1 = 0.6; E T otal 2 = 5. We note that the total amount of substrate (free substrate + complexes) in the compartments is di↵erent for the two steady states. Now, for a given steady state, there are two ways of calibrating the compartmental model with the correct total amount of substrate in the compartment. One way is to obtain this total amount from the concentration profiles of the given steady state. However, since this amount is di↵erent for the two steady states, the resulting compartmental description can only capture one of these accurately (i.e. the steady state used for the calibration). As shown in Fig. 5(f) this may not even capture the second steady state. The other way is to use the modified conservation condition discussed previously, which accounts for substrate in intermediate space as well. This is a good approximation because the system is in the thin compartment regime. By using this conservation condition, the compartmental model is able to accurately capture both steady states.