(A) Allosteric enzyme model

<figure><img src="image_151.png" /></figure>

At steady state:

$$ \underbrace{\frac{V_1 \cdot [S]}{K_1 + [S] + [S]^2/K_3}}_{\text{P production}} = \underbrace{\frac{V_2 \cdot (C - [S])}{K_2 + (C - [S])} + \alpha \cdot (S_0 - [S])}_{\text{P consumption}} $$

(B) Multi-site enzyme model

<figure><img src="image_152.png" /><figcaption>Figure A.2: Cartoon representations and brief analysis results of two enzymatic models capable of bistability. (A) Allosteric enzyme model. The first model considers an enzyme that can convert a substrate (S) into a product (P) and that is allosterically regulated by its own substrate. This regulation takes the form of inhibition and is implemented mathematically in the rate of the enzyme - black colored equation. This model results in a nonlinear curve for the relation between rate of production of P at steady state and the total concentration of substrate and product in the system, $S_{tot}$ (black curve on the top right panel). The intersections of this curve with the linear curve for the relation between rate of consumption of P at steady state and $S_{tot}$ (red curve on top right panel). We can see that the model is capable of resulting in three intersections, i.e. three steady states of the system. (B) Multi-site enzyme model. The second model considers instead of allostery, an enzyme that binds multiple substrates. This results in several enzyme-substrate complexes depending on the number of binding sites - 3 sites in the model shown. The resulting model can be solved for the steady state values of flux through each enzyme complex against $S_{tot}$ (shown in red and blue colors on the bottom right panel). The sum of these gives the rate of production of P at steady state (black curve on the bottom right panel). This model can also result in a non-linear production curve and three steady states. For further discussion of these models, see relevant citations.</figcaption></figure>

and proposed several possible mediators for such synchronization, including acetaldehyde, hydrogen sulfide, carbon dioxide, and media pH.

**Oscillations - intertwined negative and positive feedbacks** Several mathematical models of the reaction catalyzed by the enzyme phosphofructokinase (PFK) in the glycolysis pathway has shown that oscillations are possible to arise from the dynamics of this reaction alone. These models incorporate some of the observed allosteric regulation of PFK both by its substrates and products, resulting in intertwined negative and positive feedbacks.

It must be noted that some of these models, and others, use the same basic models that show bistable behavior (as discussed above) and extend them with in- and out-fluxes of involved metabolites, to display
oscillations. While these theoretical demonstrations of specific enzymatic schemes leading to oscillations have not been explored in detail experimentally, metabolic oscillations are readily observed both *in vivo* and *in vitro*, as discussed above. Models, involving some of these proposed synchronization molecules, were also developed and could reproduce experimental findings.

## A.2 Derivation of enzymatic rate laws

Enzymatic reactions can be modeled using a mechanistic model of enzyme binding and catalysis. The general approach is to develop a 'cartoon' model of the physical steps in a reaction. This cartoon model usually takes the form of a series of reactions, involving either binding / unbinding events or chemical conversions. Once a model is developed one can write down ordinary differential equations (ODEs) based on these reactions, and assuming each reaction to be governed by mass action kinetics (see Section 3.2.3). The ODEs can be simplified using certain assumptions, or sometimes just kept as is, before applying a quasi steady-state assumption (which states the enzyme-substrate complexes to be in steady-state). This assumption would allow us to solve the ODE for the enzyme-substrate complex($e_s$) at steady-state. We then enter these solutions into the ODE for the product, so to obtain a reduced system and a specific rate law for product formation. This approach forms the basis of obtaining simplified rate laws, that is, a reduced ODE for the rate of product formation, for enzymatic reactions.

### A.2.1 Derivation of the single substrate, irreversible rate law

This is the most generic model of an enzymatic reaction that has been developed/studied by Leonor Michaelis (1875 - 1947) and Maud Leonora Menten (1879 - 1960), and their contemporaries. It involves the following reaction scheme, where a substrate binds to an enzyme to form a enzyme-substrate complex, gets converted into a product, and then released from the enzyme:

$$S + E \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_4}{\overset{k_3}{\rightleftharpoons}} EP \underset{k_6}{\overset{k_5}{\rightleftharpoons}} P + E. \quad (A.1)$$

We can simplify this reaction system by assuming that (1) the transition between enzyme complexes ES and EP are instantaneous and are therefore considered as a single entity, e.g. $e_s$, and (2) that the release of product and enzyme is irreversible. The scheme now becomes:

$$S + E \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_3}{\overset{k_3}{\rightleftharpoons}} P + E. \quad (A.2)$$

We can now write a set of ODEs to describe the dynamics of this reaction system - using mass action kinetics. The ODEs are as follows:

$$\frac{ds}{dt} = -s \cdot e \cdot k_1 + e_s \cdot k_2$$

$$\frac{de}{dt} = -s \cdot e \cdot k_1 + e_s \cdot (k_2 + k_3)$$

$$\frac{dc}{dt} = s \cdot e \cdot k_1 - e_s \cdot (k_2 + k_3)$$

$$\frac{dp}{dt} = e_s \cdot k_3$$

where we used the small letter notation to represent the concentration of each species, e.g. "e" for the concentration of the enzyme, E, and "e_s" for the concentration of the enzyme-substrate complex, ES. At this stage, we can see that if we can formulate "e_s" as a function of "s", we can provide a simpler rate model that relates production of the product, P, to the level of the substrate, s. To achieve this we make several additional assumptions. First, we will assume that the total level of the enzyme is conserved, i.e. $e + e_s = C$, where C is a constant (referred to as $e_{tot}$ in the main text). This assumption effectively means that total enzyme levels are fixed in the timescale of reaction dynamics. This assumption already allows us to re-define the ODEs and reduce their number to three from four - since, we can now express e, as a
function of $e_s$. The new ODEs look like this:

$$ \frac{ds}{dt} = -s \cdot (C - e_s) \cdot k_1 + e_s \cdot k_2 $$

$$ \frac{de_s}{dt} = s \cdot (C - e_s) \cdot k_1 - e_s \cdot (k_2 + k_3) $$

$$ \frac{dp}{dt} = e_s \cdot k_3 $$

Second, we will assume that the binding/unbinding of substrate to the enzyme happens much faster than release of product from the enzyme-substrate complex. This assumption, together with the additional assumption that enzyme levels are much lower than substrate levels, allows us to consider the enzyme-substrate complex to remain constant throughout the reaction. In other words, we consider the enzyme-substrate complex to be in a 'quasi steady-state'. This allows us to solve the second ODE from above for steady-state:

$$ \frac{de_s}{dt} = 0 = s \cdot (C - e_s) \cdot k_1 - e_s \cdot (k_2 + k_3) $$

$$ e_s \cdot (k_2 + k_3) = s \cdot (C - e_s) \cdot k_1 $$

$$ e_s \cdot (k_2 + k_3) = sC \cdot k_1 - s \cdot e_s \cdot k_1 $$

$$ e_s \cdot (k_2 + k_3 + s \cdot k_1) = s \cdot C \cdot k_1 $$

$$ e_s = \frac{s \cdot C \cdot k_1}{(k_2 + k_3 + s \cdot k_1)} $$

We have now an expression for "$e_s$", which we can simply introduce to the ODE system. We have effectively reduced our ODE system from a three variable system into a two variable one:

$$ \frac{ds}{dt} = -s \cdot \left( C - \frac{s \cdot C \cdot k_1}{(k_2 + k_3 + s \cdot k_1)} \right) k_1 + \frac{s \cdot C \cdot k_1}{(k_2 + k_3 + s \cdot k_1)} \cdot k_2 $$

$$ \frac{dp}{dt} = \frac{s \cdot C \cdot k_1}{(k_2 + k_3 + s \cdot k_1)} \cdot k_3 $$

The second ODE describes the rate of change in product, P, as a function of substrate, S. It is a rate model for this enzymatic reaction, and holds under the assumptions we made in its derivation. It is known as the Michaelis-Menten kinetic rate model and is commonly expressed as:

$$ v = \frac{s \cdot e_{\text{tot}} \cdot k_{\text{cat}}}{K_{\text{M}} + s} $$

where $e_{tot}$ is equal to C and represents total enzyme concentration, $k_{cat}$ is equal to $k_3$ and is known as the maximal catalytic rate of an enzyme, and $K_M$ is equal to $(k_2 + k_3)/k_1$ and is known as the Michaelis-Menten coefficient of the enzyme. Plotting this rate against increasing substrate concentration would show that the rate is a 'saturating function' of s, i.e. the rate approaches a threshold point - given by $v^{max} = e_{tot} \cdot k_3$ as substrate increases. The enzymatic nature of the reaction introduces a limiting factor on the reaction rate! This saddle point is actually a underpinning point for some of the constraint-based methods discussed in this book.

### A.2.2 Derivation of a two substrate, irreversible rate law

See Problem 3.2

### A.2.3 Derivation of the single substrate, reversible rate law

We now return to the reaction scheme we considered in the above section:

$$ S + E \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_4}{\overset{k_3}{\rightleftharpoons}} EP \underset{k_6}{\overset{k_5}{\rightleftharpoons}} P + E. $$
The corresponding ODE system, written only for the key variables $e_s$, $e_p$, and $p$, is as follows:

$$ \frac{de_s}{dt} = e \cdot s \cdot k_1 + e_p \cdot k_4 - e_s \cdot (k_2 + k_3) $$

$$ \frac{de_p}{dt} = e \cdot p \cdot k_6 + e_s \cdot k_3 - e_p \cdot (k_4 + k_5) $$

$$ \frac{dp}{dt} = e_p \cdot k_5 - e \cdot p \cdot k_6 $$

As above, we will now introduce the assumptions of (1) total enzyme being conserved, and (2) the quasi steady-state, but this time for both of the enzyme-substrate and enzyme-product complexes. We will denote total enzyme concentration as $C$, as before, and use these two assumptions to express $e_s$ and $ep$ in terms of each other, and the other variables. Let us first proceed with $e_s$;

$$ \frac{de_s}{dt} = 0 = e \cdot s \cdot k_1 + ep \cdot k_4 - e_s \cdot (k_2 + k_3) $$

$$ e_s \cdot (k_2 + k_3) = (C - e_s - ep) \cdot s \cdot k_1 + ep \cdot k_4 $$

$$ e_s \cdot (k_2 + k_3 + s \cdot k_1) = (C - ep) \cdot s \cdot k_1 + ep \cdot k_4 $$

$$ e_s = \frac{C \cdot s \cdot k_1 + ep \cdot (k_4 - s \cdot k_1)}{(k_2 + k_3 + s \cdot k_1)} $$

We carry the same derivation for $ep$;

$$ \frac{dep}{dt} = 0 = e \cdot p \cdot k_6 + e_s \cdot k_3 - ep \cdot (k_4 + k_5) $$

$$ ep \cdot (k_4 + k_5) = (C - e_s - ep) \cdot p \cdot k_6 + e_s \cdot k_3 $$

$$ ep \cdot (k_4 + k_5 + p \cdot k_6) = (C - e_s) \cdot p \cdot k_6 + e_s \cdot k_3 $$

$$ ep = \frac{C \cdot p \cdot k_6 + e_s \cdot (k_3 - p \cdot k_6)}{(k_4 + k_5 + p \cdot k_6)} $$

We see that we have a symmetry in the expressions for $e_s$ and $ep$, in that the two expressions can be derived from each other by a replacement of variables $(k_1, k_4, k_2, s) \rightarrow (k_6, k_3, k_5, p)$. Keeping this symmetry in mind, we now attempt to eliminate one of the complexes from the equation for the other:

$$ ep \cdot (k_4 + k_5 + p \cdot k_6) = C \cdot p \cdot k_6 + e_s \cdot (k_3 - p \cdot k_6) $$

$$ ep \cdot (k_4 + k_5 + p \cdot k_6) = C \cdot p \cdot k_6 + \frac{C \cdot s \cdot k_1 + ep \cdot (k_4 - s \cdot k_1)}{(k_2 + k_3 + s \cdot k_1)} \cdot (k_3 - p \cdot k_6) $$

$$ ep \cdot (k_4 + k_5 + p \cdot k_6) = C \cdot p \cdot k_6 + \frac{C \cdot s \cdot k_1 k_3 - C \cdot s \cdot k_1 \cdot p \cdot k_6 + ep \cdot (k_4 - s \cdot k_1) \cdot (k_3 - p \cdot k_6)}{(k_2 + k_3 + s \cdot k_1)} $$

$$ ep \cdot (k_4 + k_5 + p \cdot k_6) \cdot (k_2 + k_3 + s \cdot k_1) = C \cdot p \cdot k_6 \cdot (k_2 + k_3 + s \cdot k_1) + C \cdot s \cdot k_1 k_3 - C \cdot s \cdot k_1 \cdot p \cdot k_6 + ep \cdot (k_4 - s \cdot k_1) \cdot (k_3 - p \cdot k_6) $$

$$ ep \cdot (k_4 + k_5 + p \cdot k_6) \cdot (k_2 + k_3 + s \cdot k_1) = C \cdot p \cdot k_6 k_2 + C \cdot p \cdot k_6 k_3 + C \cdot s \cdot k_1 k_3 + ep \cdot (k_4 - s \cdot k_1) \cdot (k_3 - p \cdot k_6) $$

$$ ep \cdot ((k_4 + k_5 + p \cdot k_6) \cdot (k_2 + k_3 + s \cdot k_1) - (k_4 - s \cdot k_1) \cdot (k_3 - p \cdot k_6)) = C \cdot p \cdot k_6 k_2 + C \cdot p \cdot k_6 k_3 + C \cdot s \cdot k_1 k_3 $$

$$ ep = \frac{C \cdot p \cdot k_6 \cdot (k_2 + k_3) + C \cdot s \cdot k_1 k_3}{(k_4 + k_5 + p \cdot k_6) \cdot (k_2 + k_3 + s \cdot k_1) - (k_4 - s \cdot k_1) \cdot (k_3 - p \cdot k_6)} $$

$$ ep = \frac{C \cdot p \cdot (k_6 k_2 + k_6 k_3) + C \cdot s \cdot k_1 k_3}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)} $$

Note that, in the above equation set, we have dropped the dot notation from multiplication of parameters for simplicity of expression. Based on the above argument of symmetry, or by following the same steps
for “e<sub>s</sub>”, we can show that we will have a similar expression with different parameters in the numerator:

$$e_s = \frac{C \cdot s \cdot (k_1 k_5 + k_1 k_4) + C \cdot p \cdot k_6 k_4}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)}$$

With these expressions for e<sub>s</sub> and ep at hand, we can now derive an expression for e:

$$e = C - e_s - ep$$

$$e = C - \frac{C \cdot s \cdot (k_1 k_5 + k_1 k_4) + C \cdot p \cdot k_6 k_4}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)} - \frac{C \cdot p \cdot (k_6 k_2 + k_6 k_3) + C \cdot s \cdot k_1 k_3}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)}$$

$$e = C - C \cdot \frac{s \cdot (k_1 k_3 + k_1 k_5 + k_1 k_4) + p \cdot (k_6 k_2 + k_6 k_3 + k_6 k_4)}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)}$$

$$e = C \cdot \frac{k_3 k_5 + k_2 k_5 + k_2 k_4}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)}$$

We are now ready to substitute all these expressions into the ODE for the product, so to obtain our rate law:

$$\frac{dp}{dt} = \frac{C \cdot p \cdot (k_6 k_2 + k_6 k_3) + C \cdot s \cdot k_1 k_3}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)} \cdot k_5 - C \cdot \frac{k_3 k_5 + k_2 k_5 + k_2 k_4}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)} \cdot p \cdot k_6$$

$$\frac{dp}{dt} = C \cdot \frac{s \cdot k_1 k_3 k_5 - p \cdot k_2 k_4 k_6}{(k_4 k_2 + k_5 k_2 + k_5 k_3 + s \cdot k_1 k_3 + s \cdot k_1 k_4 + s \cdot k_1 k_5 + p \cdot k_6 k_2 + p \cdot k_6 k_3 + p \cdot k_6 k_4)}$$

We can somewhat simplify this expression by defining the following composite rate constants:

$$K_S = \frac{k_3 k_5 + k_2 k_5 + k_2 k_4}{k_1 \cdot (k_3 + k_4 + k_5)}$$

$$K_P = \frac{k_3 k_5 + k_2 k_5 + k_2 k_4}{k_6 \cdot (k_2 + k_3 + k_4)}$$

$$k_{\text{cat}}^{+} = \frac{k_3 k_5}{k_3 + k_4 + k_5}$$

$$k_{\text{cat}}^{-} = \frac{k_2 k_4}{k_2 + k_3 + k_4}$$

and substituting them into the rate expression from above, to get:

$$\frac{dp}{dt} = v = C \cdot \frac{k_{\text{cat}}^{+}}{K_S} \cdot \frac{s - p \cdot \frac{k_{\text{cat}}^{-}}{K_P}}{1 + \frac{k_{\text{cat}}^{+}}{K_P} + \frac{s}{K_S}}$$

This reaction rate is referred to as the Haldane kinetic rate law, named after Jack Burden Sanderson Haldane (5 November 1892 – 1 December 1964). It can be re-expressed by recognizing the fact that the fraction entering as a multiplier for the product concentration is actually equivalent to the equilibrium constant of the reaction scheme drawn above, at the beginning of this section, when we assume the reac-
tion proceeding in the forward direction, i.e. towards product formation:

$$ \frac{k_{\text{cat}}^{-}}{K_{\text{P}}} = \frac{k_2 k_4 k_6}{k_1 k_3 k_5} = 1/K_{\text{eq}} $$

This allows us to re-express the Haldane rate law as:

$$ v = C \cdot k_{\text{cat}}^{+} \cdot \frac{s/K_{\text{S}}}{1 + \frac{p}{K_{\text{P}}} + \frac{s}{K_{\text{S}}}} \cdot \left(1 - \frac{p/s}{K_{\text{eq}}}\right) $$

This re-arranged expression is interesting because we can recognize that the last term is related to the thermodynamic Gibbs free energy of the reaction, allowing us to finally derive:

$$ v = C \cdot k_{\text{cat}}^{+} \cdot \frac{s/K_{\text{S}}}{1 + p/K_{\text{P}} + s/K_{\text{S}}} \cdot \left(1 - e^{\Delta_{\text{r}} G' / RT}\right) $$

where $\Delta_{\text{r}}G'$ is the Gibbs free energy of reaction for given substrate and product levels, considering forward direction, and R and T stand for the gas constant and temperature respectively. This rate law shows that forward reaction rate will be independent of thermodynamics, when the reaction free energy is highly negative (i.e. thermodynamically highly favored), but the reaction rate will decrease as Gibbs free energy gets close to zero.

A second, faster derivation of this rate law is found by noting that the ODEs for $\frac{de_s}{dt}$ and $\frac{de_p}{dt}$ are linear in $e, e_s$ and $e_p$, and can therefore be solved with linear matrix algebra. One may write:

$$ \begin{pmatrix} s k_1 & -(k_2 + k_3) & k_4 \\ p k_6 & k_3 & -(k_4 + k_5) \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} e \\ e_s \\ e_p \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ C \end{pmatrix}, \qquad (\text{A.3}) $$

where the first two rows of the matrix correspond to $\frac{de_s}{dt} = 0$ and $\frac{de_p}{dt} = 0$, and the last row represents conservation of total enzyme concentration. The equilibrium concentrations of $e, e_s$ and $e_p$ are then found by left-multiplying both sides of the equation by the inverse of this matrix. The obtained results are the same as given above.

### A.2.4 Derivation of two-substrate, reversible rate law with simultaneous binding

The two-substrate case is described by the following reaction scheme:

$$ S_1 + S_2 + E \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES_1S_2 \underset{k_4}{\overset{k_3}{\rightleftharpoons}} EP_1P_2 \underset{k_6}{\overset{k_5}{\rightleftharpoons}} P_1 + P_2 + E, $$

Where we assume that binding and unbinding of the substrates and products occurs simultaneously. Proceeding as above we let $e, es_1s_2, ep_1p_2, s_1, s_2, p_1$ and $p_2$ denote the concentrations of E, $ES_1S_2$, $EP_1P_2$, $S_1$, $S_2$, $P_1$ and $P_2$ respectively. The differential equations for $es_1s_2, ep_1p_2$ and $p_1 + p_2$ are:

$$ \frac{des_1s_2}{dt} = e \cdot s_1 \cdot s_2 \cdot k_1 + ep_1p_2 \cdot k_4 - es_1s_2 \cdot (k_2 + k_3) $$

$$ \frac{dep_1p_2}{dt} = e \cdot p_1 \cdot p_2 \cdot k_6 + es_1s_2 \cdot k_3 - ep_1p_2 \cdot (k_4 + k_5) $$

$$ \frac{d(p_1 + p_2)}{dt} = ep_1p_2 \cdot k_5 - e \cdot p_1 \cdot p_2 \cdot k_6. $$

Proceeding as in the single substrate case, we note that the ODEs for $\frac{des_1s_2}{dt}$ and $\frac{dep_1p_2}{dt}$ are linear in $e, es_1s_2$ and $ep_1p_2$, and that the total enzyme concentration $e + es_1s_2 + ep_1p_2$ is constant, denoted $C$.

$$ \begin{pmatrix} s_1s_2k_1 & -(k_2 + k_3) & k_4 \\ p_1p_2k_6 & k_3 & -(k_4 + k_5) \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} e \\ es_1s_2 \\ ep_1p_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ C \end{pmatrix}. \qquad (\text{A.4}) $$
We therefore see that the results for the two-substrate case are the same as for the single substrate case, with *s* replaced by *s*₁*s*₂ and *p* replaced by *p*₁*p*₂. This result is dependent on the assumption that binding/unbinding of substrates/products occurs simultaneously.

### A.3 A simple model illustrating product activation

This model demonstrates that allosteric regulation of an enzymatic reaction by its product can create a bistable system. In this simple example, we consider enzymatic production of a metabolite (labelled 'x') and its non-enzymatic consumption. It is assumed that the metabolite allosterically regulates the enzyme that produces it. The listing uses the Antimony format [552] which can be easily converted into SBML [278]. An online converter can be found at sys-bio.github.io/makesbml/

```sml
1 // The following model admits three steady-states at:
2 // x = 0.325, x = 1.671, and x = 0.873
3 // The first reaction step `-> x' uses a rate law that models
4 // positive feedback via the product x. The constant 0.2
5 // is to ensure that the lower steady-state is non-zero.
6 // The statement `ext Xo' indicates that the species Xo is fixed.
7
8 ext Xo
9 Xo -> x; (vo*x^n)/(1 + x^n) + 0.2
10 x ->; k1*x
11
12 k1 = 0.65
13 n = 4; vo = 1
14 x = 0
```

<caption>Listing A.1: Model illustrating bistability</caption>

```python
1 # Equivalent model as a differential equation in python:
2 def ode (x, t):
3     vo = 1
4     n = 4
5     k1 = 0.65
6     return [(vo*x**n)/(1 + x**n) + 0.2) - k1*x]
```

<caption>Listing A.2: Equivalent model as a differential equation in python</caption>
# Appendix B

## Appendix "Optimization of metabolic states"

### B.1 A model of central metabolism in *Escherichia coli*

<table><caption>Table B.1: Stoichiometry of biomass reaction - R70</caption><thead><tr><td>Metabolite name</td><td>Biomass stoichiometric coefficient</td></tr></thead><tbody><tr><td>AcCoA</td><td>-41</td></tr><tr><td>ADP</td><td>547</td></tr><tr><td>2-oxoglutarate</td><td>-14</td></tr><tr><td>ATP</td><td>-547</td></tr><tr><td>H<sub>2</sub>O</td><td>-547</td></tr><tr><td>P<sub>i</sub></td><td>547</td></tr><tr><td>CO<sub>2</sub></td><td>2</td></tr><tr><td>CoA</td><td>41</td></tr><tr><td>DHAP</td><td>-5</td></tr><tr><td>G6P</td><td>-4</td></tr><tr><td>NAD<sup>+</sup></td><td>178</td></tr><tr><td>NADH</td><td>-178</td></tr><tr><td>NH<sub>3</sub></td><td>-139</td></tr><tr><td>2-oxoglutarateAcetate</td><td>-24</td></tr><tr><td>PEP</td><td>-32</td></tr><tr><td>Pyruvate</td><td>-38</td></tr><tr><td>E4P</td><td>-5</td></tr><tr><td>R5P</td><td>-13</td></tr></tbody></table>

<table><caption>Table B.2: Glycolysis</caption><thead><tr><td>Reaction ID</td><td>EC number</td><td>Reaction name</td><td>Formula</td></tr></thead><tbody><tr><td>R1</td><td>2.7.1.69</td><td>pts</td><td>Glucose + PEP ⇌ G6P + Pyruvate</td></tr><tr><td>R2r</td><td>5.3.1.9</td><td>pgi</td><td>G6P ⇌ F6P</td></tr><tr><td>R3</td><td>2.7.1.11</td><td>pfk</td><td>F6P + ATP ⇌ FBP + ADP</td></tr><tr><td>R4</td><td>3.1.3.11</td><td>fbp</td><td>FBP + H<sub>2</sub>O ⇌ F6P + P<sub>i</sub></td></tr><tr><td>R5r</td><td>4.1.2.13</td><td>ald</td><td>FBP ⇌ DHAP + G3P</td></tr><tr><td>R6r</td><td>5.3.1.1</td><td>tim</td><td>G3P ⇌ DHAP</td></tr><tr><td>R7ra</td><td>1.2.1.12</td><td>gap</td><td>G3P + NAD<sup>+</sup> + P<sub>i</sub> ⇌ BPG + NADH</td></tr><tr><td>R7rb</td><td>2.7.2.3</td><td>pgk</td><td>BPG + ADP ⇌ 3PG + ATP</td></tr><tr><td>R7rc</td><td>5.4.2.11 / 5.4.2.12</td><td>pgm</td><td>3PG ⇌ 2PG</td></tr><tr><td>R8r</td><td>4.2.1.11</td><td>pgh</td><td>2PG ⇌ PEP</td></tr><tr><td>R9</td><td>2.7.1.40</td><td>pyk</td><td>PEP + ADP ⇌ Pyruvate + ATP</td></tr><tr><td>RR9</td><td>2.7.9.2</td><td>pps</td><td>Pyruvate + 2 ATP ⇌ PEP + 2 ADP + P<sub>i</sub></td></tr></tbody></table>
<table><caption>Table B.3: Pentose Phosphate Pathway</caption><thead><tr><th>Reaction ID</th><th>EC number</th><th>Reaction name</th><th>Formula</th></tr></thead><tbody><tr><td>R10a</td><td>1.1.1.49</td><td>zwf</td><td>G6P + NAD<sup>+</sup> ⇌ 6PGL + NADH</td></tr><tr><td>R10b</td><td>3.1.1.31</td><td>glh</td><td>6PGL ⇌ 6PGC</td></tr><tr><td>R10c</td><td>1.1.1.44</td><td>pgd</td><td>6PGC + NAD<sup>+</sup> ⇌ NADH + CO<sub>2</sub> + Ru5P</td></tr><tr><td>R11r</td><td>5.1.3.1</td><td>rpe</td><td>Ru5P ⇌ X5P</td></tr><tr><td>R12r</td><td>5.3.1.6</td><td>rpi</td><td>Ru5P ⇌ R5P</td></tr><tr><td>R13r</td><td>2.2.1.1</td><td>txt1</td><td>R5P + X5P ⇌ S7P + G3P</td></tr><tr><td>R14r</td><td>2.2.1.2</td><td>tal</td><td>G3P + S7P ⇌ E4P + F6P</td></tr><tr><td>R15r</td><td>2.2.1.1</td><td>txt2</td><td>E4P + X5P ⇌ G3P + F6P</td></tr><tr><td>R60</td><td>4.2.1.12</td><td>edd</td><td>6PGC ⇌ KDPG</td></tr><tr><td>R61r</td><td>4.1.2.14</td><td>eda</td><td>KDPG ⇌ G3P + Pyruvate</td></tr></tbody></table>

<table><caption>Table B.4: TCA Cycle</caption><thead><tr><th>Reaction ID</th><th>EC number</th><th>Reaction name</th><th>Formula</th></tr></thead><tbody><tr><td>R20</td><td>2.3.1.54</td><td>pfl</td><td>Pyruvate + CoA ⇌ AcCoA + Formate</td></tr><tr><td>R21</td><td>1.2.4.1 / 2.3.1.12</td><td>pdh</td><td>Pyruvate + NAD<sup>+</sup> + CoA ⇌ AcCoA + CO<sub>2</sub> + NADH</td></tr><tr><td>R22</td><td>2.3.3.1</td><td>csn</td><td>2-oxoglutarateacetate + AcCoA ⇌ Citrate + CoA</td></tr><tr><td>R23r</td><td>4.2.1.3</td><td>acn</td><td>Citrate ⇌ iso-Citrate</td></tr><tr><td>R24</td><td>1.1.1.41</td><td>icd</td><td>iso-Citrate + NAD<sup>+</sup> ⇌ 2-oxoglutarate + NADH + CO<sub>2</sub></td></tr><tr><td>R25</td><td>1.2.4.2</td><td>kgd</td><td>2-oxoglutarate + NAD<sup>+</sup> + CoA ⇌ NADH + Succinateinyl-CoA + CO<sub>2</sub></td></tr><tr><td>R26r</td><td>6.2.1.5</td><td>scs</td><td>Succinateinyl-CoA + ADP + P<sub>i</sub> ⇌ Succinateinate + ATP + CoA</td></tr><tr><td>R27</td><td>1.3.5.1</td><td>sdh</td><td>Succinateinate + ADP + O<sub>2</sub>[e] + P<sub>i</sub> ⇌ Fumarate + ATP</td></tr><tr><td>R27b</td><td>1.3.5.4</td><td>frd</td><td>Fumarate + NADH ⇌ Succinateinate + NAD<sup>+</sup></td></tr><tr><td>R28r</td><td>4.2.1.2</td><td>fum</td><td>Fumarate ⇌ Malate</td></tr><tr><td>R29r</td><td>1.1.1.37</td><td>mdh</td><td>Malate + NAD<sup>+</sup> ⇌ 2-oxoglutarateacetate + NADH</td></tr></tbody></table>

<table><caption>Table B.5: Anaplerotic Reactions</caption><thead><tr><th>Reaction ID</th><th>EC number</th><th>Reaction name</th><th>Formula</th></tr></thead><tbody><tr><td>R40</td><td>4.1.1.31</td><td>ppc</td><td>PEP + CO<sub>2</sub> ⇌ 2-oxoglutarateacetate + P<sub>i</sub></td></tr><tr><td>R41</td><td>1.1.1.38</td><td>me</td><td>Malate + NAD<sup>+</sup> ⇌ Pyruvate + NADH + CO<sub>2</sub></td></tr><tr><td>R42</td><td>4.1.1.49</td><td>ppck</td><td>2-oxoglutarateacetate + ATP ⇌ PEP + ADP + CO<sub>2</sub></td></tr></tbody></table>

<table><caption>Table B.6: Redox-associated reactions</caption><thead><tr><th>Reaction ID</th><th>EC number</th><th>Reaction name</th><th>Formula</th></tr></thead><tbody><tr><td>R53r</td><td>1.1.1.27</td><td>ldh</td><td>Pyruvate + NADH ⇌ Lactate + NAD<sup>+</sup></td></tr><tr><td>R54ra</td><td>1.2.1.10</td><td>ada</td><td>AcCoA + NADH ⇌ Acetaldehyde + NAD<sup>+</sup> + CoA</td></tr><tr><td>R54rb</td><td>1.1.1.1</td><td>adh</td><td>Acetaldehyde + NADH ⇌ ETOH + NAD<sup>+</sup></td></tr><tr><td>R55a</td><td>2.3.1.8</td><td>pta</td><td>AcCoA + P<sub>i</sub> ⇌ Acetyl-P + CoA</td></tr><tr><td>R55b</td><td>2.7.2.1</td><td>ack</td><td>Acetyl-P + ADP ⇌ Acetate + ATP</td></tr></tbody></table>

<table><caption>Table B.7: Oxidative phosphorylation</caption><thead><tr><th>Reaction ID</th><th>Reaction name</th><th>Formula</th></tr></thead><tbody><tr><td>R80</td><td>oxphos</td><td>NADH + 2 ADP + 0.5 O<sub>2</sub>[e] + 2 P<sub>i</sub> ⇌ NAD<sup>+</sup> + 2 ATP + 3 H<sub>2</sub>O</td></tr><tr><td>R82</td><td>atpmain</td><td>ATP + H<sub>2</sub>O ⇌ ADP + P<sub>i</sub> + ATP<sub>main</sub></td></tr></tbody></table>

<table><caption>Table B.8: Membrane Transport Reactions</caption><thead><tr><th>Reaction ID</th><th>Reaction name</th><th>Formula</th></tr></thead><tbody><tr><td>R90</td><td>ex<sub>etoh</sub></td><td>ETOH ⇌ ETOH[e]</td></tr><tr><td>R91</td><td>ex<sub>ace</sub></td><td>Acetate ⇌ Acetate[e]</td></tr><tr><td>R93</td><td>ex<sub>NH<sub>3</sub></sub></td><td>NH<sub>3</sub>[e] ⇌ NH<sub>3</sub></td></tr><tr><td>R94</td><td>ex<sub>lac</sub></td><td>Lactate ⇌ Lactate[e]</td></tr><tr><td>R95</td><td>ex<sub>suc</sub></td><td>Succinateinate ⇌ Succinateinate[e]</td></tr><tr><td>R96</td><td>ex<sub>for</sub></td><td>Formate ⇌ Formate[e]</td></tr><tr><td>R97r</td><td>ex<sub>CO<sub>2</sub></sub></td><td>CO<sub>2</sub> ⇌ CO<sub>2</sub>[e]</td></tr></tbody></table>
<table><caption>Table B.9: Kinetic parameters associated with reactions</caption><thead><tr><th>Reaction ID</th><th>k<sup>+</sup><sub>cat</sub> [1/s]</th><th>K<sub>eq</sub> [unitless]</th><th>Enzyme molecular weight [Da]</th></tr></thead><tbody><tr><td>R1</td><td>100</td><td>N/A</td><td>2.6·10<sup>5</sup></td></tr><tr><td>R10a</td><td>240</td><td>N/A</td><td>5.6·10<sup>4</sup></td></tr><tr><td>R10b</td><td>410</td><td>N/A</td><td>3.6·10<sup>4</sup></td></tr><tr><td>R10c</td><td>110</td><td>N/A</td><td>1.0·10<sup>5</sup></td></tr><tr><td>R11r</td><td>130</td><td>2.3</td><td>2.5·10<sup>4</sup></td></tr><tr><td>R12r</td><td>1400</td><td>2.3</td><td>1.9·10<sup>4</sup></td></tr><tr><td>R13r</td><td>46</td><td>3.7</td><td>7.3·10<sup>4</sup></td></tr><tr><td>R14r</td><td>17</td><td>0.9</td><td>3.5·10<sup>4</sup></td></tr><tr><td>R15r</td><td>75</td><td>38</td><td>7.3·10<sup>4</sup></td></tr><tr><td>R20</td><td>4800</td><td>N/A</td><td>8.5·10<sup>4</sup></td></tr><tr><td>R21</td><td>38</td><td>N/A</td><td>2.8·10<sup>5</sup></td></tr><tr><td>R22</td><td>360</td><td>N/A</td><td>9.6·10<sup>4</sup></td></tr><tr><td>R23r</td><td>33</td><td>0.074</td><td>9.6·10<sup>4</sup></td></tr><tr><td>R24</td><td>110</td><td>N/A</td><td>4.6·10<sup>4</sup></td></tr><tr><td>R25</td><td>150</td><td>N/A</td><td>1.2·10<sup>6</sup></td></tr><tr><td>R26r</td><td>89</td><td>0.52</td><td>7.1·10<sup>4</sup></td></tr><tr><td>R27</td><td>78</td><td>N/A</td><td>7.9·10<sup>5</sup></td></tr><tr><td>R27b</td><td>180</td><td>N/A</td><td>1.8·10<sup>5</sup></td></tr><tr><td>R28r</td><td>280</td><td>4.7</td><td>6.0·10<sup>4</sup></td></tr><tr><td>R29r</td><td>210</td><td>6.1·10<sup>-5</sup></td><td>3.2·10<sup>4</sup></td></tr><tr><td>R2r</td><td>320</td><td>0.51</td><td>6.2·10<sup>4</sup></td></tr><tr><td>R3</td><td>110</td><td>N/A</td><td>1.4·10<sup>5</sup></td></tr><tr><td>R4</td><td>25</td><td>N/A</td><td>3.7·10<sup>4</sup></td></tr><tr><td>R40</td><td>120</td><td>N/A</td><td>2.0·10<sup>5</sup></td></tr><tr><td>R41</td><td>76</td><td>N/A</td><td>6.3·10<sup>4</sup></td></tr><tr><td>R42</td><td>51</td><td>N/A</td><td>6.0·10<sup>4</sup></td></tr><tr><td>R53r</td><td>140</td><td>2.1·10<sup>4</sup></td><td>3.7·10<sup>4</sup></td></tr><tr><td>R54ra</td><td>0.35</td><td>2.3·10<sup>-3</sup></td><td>9.6·10<sup>4</sup></td></tr><tr><td>R54rb</td><td>320</td><td>2.8·10<sup>3</sup></td><td>9.6·10<sup>4</sup></td></tr><tr><td>R55a</td><td>91</td><td>N/A</td><td>7.7·10<sup>4</sup></td></tr><tr><td>R55b</td><td>59</td><td>N/A</td><td>4.3·10<sup>4</sup></td></tr><tr><td>R5r</td><td>8.0</td><td>3.0·10<sup>-4</sup></td><td>3.9·10<sup>4</sup></td></tr><tr><td>R60</td><td>250</td><td>N/A</td><td>6.5·10<sup>4</sup></td></tr><tr><td>R61r</td><td>80</td><td>9.6·10<sup>-3</sup></td><td>2.2·10<sup>4</sup></td></tr><tr><td>R6r</td><td>7800</td><td>11</td><td>5.4·10<sup>4</sup></td></tr><tr><td>R70</td><td>99</td><td>N/A</td><td>6.0·10<sup>4</sup></td></tr><tr><td>R7ra</td><td>230</td><td>0.088</td><td>3.6·10<sup>4</sup></td></tr><tr><td>R7rb</td><td>390</td><td>730</td><td>4.1·10<sup>4</sup></td></tr><tr><td>R7rc</td><td>53</td><td>0.16</td><td>2.9·10<sup>4</sup></td></tr><tr><td>R80</td><td>4.0·10<sup>6</sup></td><td>N/A</td><td>9.1·10<sup>5</sup></td></tr><tr><td>R82</td><td>180</td><td>N/A</td><td>6.0·10<sup>4</sup></td></tr><tr><td>R8r</td><td>210</td><td>3.5</td><td>4.6·10<sup>4</sup></td></tr><tr><td>R9</td><td>510</td><td>N/A</td><td>5.0·10<sup>4</sup></td></tr><tr><td>R90</td><td>100</td><td>N/A</td><td>N/A</td></tr><tr><td>R91</td><td>100</td><td>N/A</td><td>5.9·10<sup>4</sup></td></tr><tr><td>R93</td><td>100</td><td>N/A</td><td>4.5·10<sup>4</sup></td></tr><tr><td>R94</td><td>100</td><td>N/A</td><td>5.9·10<sup>4</sup></td></tr><tr><td>R95</td><td>100</td><td>N/A</td><td>4.5·10<sup>4</sup></td></tr><tr><td>R96</td><td>100</td><td>N/A</td><td>3.1·10<sup>4</sup></td></tr><tr><td>R97r</td><td>100</td><td>N/A</td><td>N/A</td></tr><tr><td>RR9</td><td>13</td><td>N/A</td><td>8.7·10<sup>4</sup></td></tr></tbody></table>
<table><thead><tr><th>Reaction ID</th><th>Metabolite name</th><th>K<sub>M</sub> [mM]</th></tr></thead><tbody><tr><td>R1</td><td>G6P</td><td>0.102</td></tr><tr><td>R1</td><td>Glucose</td><td>0.116</td></tr><tr><td>R1</td><td>PEP</td><td>0.0983</td></tr><tr><td>R1</td><td>Pyruvate</td><td>0.102</td></tr><tr><td>R10a</td><td>G6P</td><td>0.314</td></tr><tr><td>R10a</td><td>6PGL</td><td>0.129</td></tr><tr><td>R10a</td><td>NAD<sup>+</sup></td><td>0.863</td></tr><tr><td>R10a</td><td>NADH</td><td>0.129</td></tr><tr><td>R10b</td><td>6PGL</td><td>0.168</td></tr><tr><td>R10b</td><td>6PGC</td><td>0.0594</td></tr><tr><td>R10c</td><td>CO<sub>2</sub></td><td>0.0626</td></tr><tr><td>R10c</td><td>6PGC</td><td>0.101</td></tr><tr><td>R10c</td><td>Ru5P</td><td>0.0626</td></tr><tr><td>R10c</td><td>NAD<sup>+</sup></td><td>0.0591</td></tr><tr><td>R10c</td><td>NADH</td><td>0.0626</td></tr><tr><td>R11r</td><td>Ru5P</td><td>0.0878</td></tr><tr><td>R11r</td><td>X5P</td><td>0.114</td></tr><tr><td>R12r</td><td>R5P</td><td>1.25</td></tr><tr><td>R12r</td><td>Ru5P</td><td>0.558</td></tr><tr><td>R13r</td><td>G3P</td><td>1.23</td></tr><tr><td>R13r</td><td>R5P</td><td>0.972</td></tr><tr><td>R13r</td><td>S7P</td><td>2.11</td></tr><tr><td>R13r</td><td>X5P</td><td>0.157</td></tr><tr><td>R14r</td><td>E4P</td><td>0.175</td></tr><tr><td>R14r</td><td>F6P</td><td>0.888</td></tr><tr><td>R14r</td><td>G3P</td><td>0.578</td></tr><tr><td>R14r</td><td>S7P</td><td>0.206</td></tr><tr><td>R15r</td><td>E4P</td><td>0.0934</td></tr><tr><td>R15r</td><td>F6P</td><td>0.737</td></tr><tr><td>R15r</td><td>G3P</td><td>1.27</td></tr><tr><td>R15r</td><td>X5P</td><td>0.152</td></tr><tr><td>R20</td><td>AcCoA</td><td>0.0352</td></tr><tr><td>R20</td><td>CoA</td><td>0.0168</td></tr><tr><td>R20</td><td>Formate</td><td>6.35</td></tr><tr><td>R20</td><td>Pyruvate</td><td>2.18</td></tr><tr><td>R21</td><td>AcCoA</td><td>0.159</td></tr><tr><td>R21</td><td>CO<sub>2</sub></td><td>0.159</td></tr><tr><td>R21</td><td>CoA</td><td>0.0629</td></tr><tr><td>R21</td><td>Pyruvate</td><td>0.291</td></tr><tr><td>R21</td><td>NAD<sup>+</sup></td><td>0.0629</td></tr><tr><td>R21</td><td>NADH</td><td>0.159</td></tr><tr><td>R22</td><td>AcCoA</td><td>0.0867</td></tr><tr><td>R22</td><td>Citrate</td><td>0.0756</td></tr><tr><td>R22</td><td>CoA</td><td>0.0756</td></tr><tr><td>R22</td><td>2-oxoglutarate</td><td>0.0287</td></tr><tr><td>R23r</td><td>Citrate</td><td>3.49</td></tr><tr><td>R23r</td><td>iso-Citrate</td><td>2.42</td></tr><tr><td>R24</td><td>2-oxoglutarate</td><td>0.483</td></tr><tr><td>R24</td><td>CO<sub>2</sub></td><td>2.02</td></tr><tr><td>R24</td><td>iso-Citrate</td><td>0.0227</td></tr></tbody></table>

<table><thead><tr><th>Reaction ID</th><th>Metabolite name</th><th>K<sub>M</sub> [mM]</th></tr></thead><tbody><tr><td>R24</td><td>NAD<sup>+</sup></td><td>1.06</td></tr><tr><td>R24</td><td>NADH</td><td>0.0119</td></tr><tr><td>R25</td><td>2-oxoglutarate</td><td>0.0670</td></tr><tr><td>R25</td><td>CO<sub>2</sub></td><td>0.108</td></tr><tr><td>R25</td><td>CoA</td><td>0.0927</td></tr><tr><td>R25</td><td>Succinyl-CoA</td><td>0.108</td></tr><tr><td>R25</td><td>NAD<sup>+</sup></td><td>0.0927</td></tr><tr><td>R25</td><td>NADH</td><td>0.108</td></tr><tr><td>R26r</td><td>CoA</td><td>0.00731</td></tr><tr><td>R26r</td><td>Succinate</td><td>0.237</td></tr><tr><td>R26r</td><td>Succinyl-CoA</td><td>0.0105</td></tr><tr><td>R26r</td><td>ADP</td><td>0.0560</td></tr><tr><td>R26r</td><td>ATP</td><td>0.0812</td></tr><tr><td>R27</td><td>Fumarate</td><td>0.0812</td></tr><tr><td>R27</td><td>O<sub>2</sub>[e]</td><td>0.371</td></tr><tr><td>R27</td><td>Succinate</td><td>0.0756</td></tr><tr><td>R27</td><td>ADP</td><td>0.371</td></tr><tr><td>R27</td><td>ATP</td><td>0.0270</td></tr><tr><td>R27b</td><td>Fumarate</td><td>0.0201</td></tr><tr><td>R27b</td><td>Succinate</td><td>0.205</td></tr><tr><td>R27b</td><td>NAD<sup>+</sup></td><td>0.0431</td></tr><tr><td>R27b</td><td>NADH</td><td>0.232</td></tr><tr><td>R28r</td><td>Fumarate</td><td>0.314</td></tr><tr><td>R28r</td><td>Malate</td><td>0.615</td></tr><tr><td>R29r</td><td>Malate</td><td>3.19</td></tr><tr><td>R29r</td><td>2-oxoglutarate</td><td>0.0283</td></tr><tr><td>R29r</td><td>NAD<sup>+</sup></td><td>0.460</td></tr><tr><td>R29r</td><td>NADH</td><td>0.0321</td></tr><tr><td>R2r</td><td>F6P</td><td>0.162</td></tr><tr><td>R2r</td><td>G6P</td><td>0.273</td></tr><tr><td>R3</td><td>F6P</td><td>0.116</td></tr><tr><td>R3</td><td>FBP</td><td>0.113</td></tr><tr><td>R3</td><td>ADP</td><td>0.113</td></tr><tr><td>R3</td><td>ATP</td><td>0.141</td></tr><tr><td>R4</td><td>F6P</td><td>0.171</td></tr><tr><td>R4</td><td>FBP</td><td>0.0161</td></tr><tr><td>R40</td><td>CO<sub>2</sub></td><td>0.115</td></tr><tr><td>R40</td><td>2-oxoglutarate</td><td>0.0426</td></tr><tr><td>R40</td><td>PEP</td><td>0.364</td></tr><tr><td>R41</td><td>CO<sub>2</sub></td><td>0.0885</td></tr><tr><td>R41</td><td>Malate</td><td>0.361</td></tr><tr><td>R41</td><td>Pyruvate</td><td>0.0885</td></tr><tr><td>R41</td><td>NAD<sup>+</sup></td><td>0.0691</td></tr><tr><td>R41</td><td>NADH</td><td>0.0885</td></tr><tr><td>R42</td><td>CO<sub>2</sub></td><td>5.21</td></tr><tr><td>R42</td><td>2-oxoglutarate</td><td>0.571</td></tr><tr><td>R42</td><td>PEP</td><td>0.0643</td></tr><tr><td>R42</td><td>ADP</td><td>0.0484</td></tr><tr><td>R42</td><td>ATP</td><td>0.0750</td></tr><tr><td>R53r</td><td>LACTATE</td><td>0.517</td></tr></tbody></table>

Michaelis constants – part I
<table><thead><tr><th>Reaction ID</th><th>Metabolite name</th><th>K<sub>M</sub> [mM]</th></tr></thead><tbody><tr><td>R53r</td><td>Pyruvate</td><td>0.0193</td></tr><tr><td>R53r</td><td>NAD<sup>+</sup></td><td>0.517</td></tr><tr><td>R53r</td><td>NADH</td><td>0.0193</td></tr><tr><td>R54ra</td><td>AcCoA</td><td>0.0242</td></tr><tr><td>R54ra</td><td>Acetaldehyde</td><td>1.80</td></tr><tr><td>R54ra</td><td>CoA</td><td>0.00786</td></tr><tr><td>R54ra</td><td>NAD<sup>+</sup></td><td>0.0415</td></tr><tr><td>R54ra</td><td>NADH</td><td>0.113</td></tr><tr><td>R54rb</td><td>Acetaldehyde</td><td>0.0593</td></tr><tr><td>R54rb</td><td>ETOH</td><td>5.49</td></tr><tr><td>R54rb</td><td>NAD<sup>+</sup></td><td>0.169</td></tr><tr><td>R54rb</td><td>NADH</td><td>0.0593</td></tr><tr><td>R55a</td><td>AcCoA</td><td>0.0424</td></tr><tr><td>R55a</td><td>Acetyl-P</td><td>0.313</td></tr><tr><td>R55a</td><td>CoA</td><td>0.0860</td></tr><tr><td>R55b</td><td>Acetate</td><td>3.44</td></tr><tr><td>R55b</td><td>Acetyl-P</td><td>0.154</td></tr><tr><td>R55b</td><td>ADP</td><td>0.402</td></tr><tr><td>R55b</td><td>ATP</td><td>0.0714</td></tr><tr><td>R5r</td><td>DHAP</td><td>0.0782</td></tr><tr><td>R5r</td><td>FBP</td><td>0.204</td></tr><tr><td>R5r</td><td>G3P</td><td>0.0782</td></tr><tr><td>R60</td><td>6PGC</td><td>0.0434</td></tr><tr><td>R60</td><td>KDPG</td><td>0.150</td></tr><tr><td>R61r</td><td>G3P</td><td>0.00146</td></tr><tr><td>R61r</td><td>KDPG</td><td>0.561</td></tr><tr><td>R61r</td><td>Pyruvate</td><td>0.00146</td></tr><tr><td>R6r</td><td>DHAP</td><td>0.0750</td></tr><tr><td>R6r</td><td>G3P</td><td>0.745</td></tr><tr><td>R70</td><td>AcCoA</td><td>0.462</td></tr><tr><td>R70</td><td>2-oxoglutarate</td><td>0.352</td></tr><tr><td>R70</td><td>BIOMASS</td><td>0.0998</td></tr><tr><td>R70</td><td>CO<sub>2</sub></td><td>0.0996</td></tr><tr><td>R70</td><td>CoA</td><td>0.891</td></tr><tr><td>R70</td><td>E4P</td><td>0.0144</td></tr><tr><td>R70</td><td>G6P</td><td>4.31</td></tr><tr><td>R70</td><td>NH<sub>3</sub></td><td>0.0151</td></tr><tr><td>R70</td><td>2-oxoglutarate</td><td>0.00672</td></tr><tr><td>R70</td><td>PEP</td><td>0.169</td></tr><tr><td>R70</td><td>Pyruvate</td><td>0.319</td></tr><tr><td>R70</td><td>R5P</td><td>0.881</td></tr><tr><td>R70</td><td>ADP</td><td>0.0293</td></tr><tr><td>R70</td><td>ATP</td><td>0.342</td></tr><tr><td>R70</td><td>NAD<sup>+</sup></td><td>1.43</td></tr></tbody></table>

<table><thead><tr><th>Reaction ID</th><th>Metabolite name</th><th>K<sub>M</sub> [mM]</th></tr></thead><tbody><tr><td>R70</td><td>NADH</td><td>0.0913</td></tr><tr><td>R7ra</td><td>DPG</td><td>0.0576</td></tr><tr><td>R7ra</td><td>G3P</td><td>0.687</td></tr><tr><td>R7ra</td><td>NAD<sup>+</sup></td><td>0.0558</td></tr><tr><td>R7ra</td><td>NADH</td><td>0.0576</td></tr><tr><td>R7rb</td><td>DPG</td><td>0.0426</td></tr><tr><td>R7rb</td><td>3PG</td><td>0.235</td></tr><tr><td>R7rb</td><td>ADP</td><td>0.0426</td></tr><tr><td>R7rb</td><td>ATP</td><td>0.235</td></tr><tr><td>R7rc</td><td>3PG</td><td>0.132</td></tr><tr><td>R7rc</td><td>2PG</td><td>0.0755</td></tr><tr><td>R80</td><td>O<sub>2</sub>[e]</td><td>0.116</td></tr><tr><td>R80</td><td>ADP</td><td>0.136</td></tr><tr><td>R80</td><td>ATP</td><td>0.0737</td></tr><tr><td>R80</td><td>NAD<sup>+</sup></td><td>0.0859</td></tr><tr><td>R80</td><td>NADH</td><td>0.116</td></tr><tr><td>R82</td><td>ATP<sub>main</sub></td><td>0.130</td></tr><tr><td>R82</td><td>ADP</td><td>0.130</td></tr><tr><td>R82</td><td>ATP</td><td>0.0769</td></tr><tr><td>R8r</td><td>PEP</td><td>0.131</td></tr><tr><td>R8r</td><td>2PG</td><td>0.108</td></tr><tr><td>R9</td><td>PEP</td><td>0.291</td></tr><tr><td>R9</td><td>Pyruvate</td><td>0.0476</td></tr><tr><td>R9</td><td>ADP</td><td>0.218</td></tr><tr><td>R9</td><td>ATP</td><td>8.45</td></tr><tr><td>R90</td><td>ETOH</td><td>0.100</td></tr><tr><td>R90</td><td>ETOH[e]</td><td>0.100</td></tr><tr><td>R91</td><td>Acetate</td><td>0.100</td></tr><tr><td>R91</td><td>Acetate[e]</td><td>0.100</td></tr><tr><td>R93</td><td>NH<sub>3</sub></td><td>0.0999</td></tr><tr><td>R93</td><td>NH<sub>3</sub>[e]</td><td>0.100</td></tr><tr><td>R94</td><td>Lactate</td><td>0.100</td></tr><tr><td>R94</td><td>Lactate[e]</td><td>0.100</td></tr><tr><td>R95</td><td>Succinate</td><td>0.100</td></tr><tr><td>R95</td><td>Succinate[e]</td><td>0.100</td></tr><tr><td>R96</td><td>Formate</td><td>0.0999</td></tr><tr><td>R96</td><td>Formate[e]</td><td>0.100</td></tr><tr><td>R97r</td><td>CO<sub>2</sub></td><td>0.0999</td></tr><tr><td>R97r</td><td>CO<sub>2</sub>[e]</td><td>0.100</td></tr><tr><td>RR9</td><td>PEP</td><td>0.0934</td></tr><tr><td>RR9</td><td>Pyruvate</td><td>0.0864</td></tr><tr><td>RR9</td><td>ADP</td><td>0.0873</td></tr><tr><td>RR9</td><td>ATP</td><td>0.0350</td></tr></tbody></table>

Michaelis constants – part II
<figure><img src="image_153.png" /></figure>

Figure B.1: Metabolic strategies in the *E. coli* model, depending on external glucose and oxygen concentrations, In the EFM pahse diagram. Each region represents the winning EFM as explained in Figure 7.8. Here, the colors represent the flux in one specific reaction based on the winning EFM in that region. (A) The lactate secretion flux is strikingly equal to 0 in most regions. The only conditions where lactate is secreted is at low oxygen and medium/high glucose concentrations. (B) The biomass yield is, in general, high if and only if lactate is not secreted. This makes sense because the carbon coming from the glucose is often the limiting nutrient for growth, and there is a trade-off between using it for biomass versus fermentation products such as lactate. Interestingly, the region with high glucose and high oxygen levels (upper right quadrant) is occupied by an EFM that doesn't achieve the highest possible yield (i.e. *max-gr*). In low glucose and high oxygen, or in medium oxygen levels, the winning EFMs are the ones with relatively higher biomass yields.

## B.2 More results for the *E. coli* central metabolism model

This section contains additional results for the *E. coli* central metabolism model from Chapter 7, in particular, fluxes plotted in the EFM phase diagram and in flux space, as well as ideal and real enzyme costs for all EFMs.

Each point in the Monod landscape in Figure 7.8 (A) corresponds to a state of the model, and the calculations that lead to the growth rate and the "winning EFM" shown yield a full description of this state, including all fluxes, metabolite concentrations and enzyme levels. These data can be explored and visualized in many ways. For illustration, Figure B.2 shows a variant of Figure 7.8 (A) in the horizontal axes do not describe the external concentrations of glucose and oxygen, but their uptake rates, and the vertical axes shows the biomass production rate. Since uptake rates are a function of external concentrations, and the growth rate directly depends on the biomass production rate, we might have expected that this yields the same picture, just a bit stretched along each of the axes. However, the picture looks very different: instead of forming a continuous surface, the points now fall on disconnected rays, apparently with one ray for each colored region of the surface. In fact, when looking at the picture closely, we can see that each of the winning EFMs gives rise to exactly one ray. But this, after all, is logical. In the new plot, all axes refer to reaction rates, and for each EFM all rates come in fixed ratios, giving rise to a ray. So, if our solutions are EFMs, this picture cannot be continuous – in line with the fact that, in the original Monod landscape 7.8 (A), when moving from one region to the other one, one would notice a discrete jump of the reaction rates. But why is the new picture not a continuous surface, if uptake rates depends smoothly on external metabolite concentrations? In fact, they do not only depend on these concentrations, but also on resource allocation to the transporter. If this allocation shows a discrete jump (again, when moving from one region to another one), then also the rate shows a jump. The comparison between the two plots shows us what we gain by considering enzyme kinetics as compared to a pure stoichiometric model. With the biomass rate as a proxy for cell growth, each EFM defines fixed ratios between this growth rate and each of the metabolic fluxes, including the uptake rates. When continuously scaling an EFM, the glucose
<figure><img src="image_154.png" /></figure>

Figure B.2: Proportional scaling of fluxes within each EFM – In the diagram with glucose uptake, oxygen uptake, and biomass production rate on the axes, each colored line corresponds to one EFM, and shows the possible combinations of fluxes obtained from the model behind Figure 7.8 (which also shares the EFM colors with this figure). Importantly, here the *x* and *y* axes represent uptake rates and not substrate concentrations. Therefore, as expected, each EFM yields a straight line (because of the proportional scaling of different fluxes for each EFM). Since – according to our reasoning – optimal flux distributions must be EFM, only these combinations of fluxes are actually possible. When glucose and oxygen concentrations are varied smoothly in Figure 7.8, the corresponding movement in this plot would be along the lines and sometimes, jumps between different lines (when the system moves from one region to another one in Figure 7.8).

uptake, oxygen uptake, and biomass production will scale proportionally. So the rays in Figure B.2 reflect what we can know about possible metabolic fluxes based on network structure alone; but to get to the Monod landscape, as a function of concentrations, we had to use kinetic information and a extra principle of economical enzyme usage.

The phase diagram of “winning EFMs” can also be used to visualize other (optimized) quantities as functions of glucose and oxygen concentrations. Figure B.1 shows as example the (biomass-specific) lactate secretion (showing that also a number of other winning EFMs, apart from *ana-lac*, secrete lactate) and the biomass yield on glucose.

Finally, a statistics over all EFMs shows that the range of possible enzyme demands per biomass production
rate is quite large: as shown in Figure B.3, they vary over more than two orders of magnitude, making some
EFMs a hundred-fold more enzyme-expensive than others. The same plot also shows how enzyme costs
depend on the fact that enzymes do not operate at their full capacity (reaching their *k*<sub>cat</sub> value), but at best
at the enzyme efficiencies predicted by enzyme cost minimization. For our *E. coli* model and the aerobic
glucose conditions studied, if all enzymes could operate at their *k*<sub>cat</sub> values, this would decrease to overall
enzyme demand by a factor of at least 1.4, or maximally 4.7, depending on the EFM in question. But still,
in this case, for determining enzyme costs the choice of the right EFM (even assuming “ideal” enzymes)
is much more important than considering the actual, “non-ideal” way in which enzymes operate. But this
may not always hold: under low-oxygen conditions, the enzyme demands of some EFMs may increase
much more drastically.
<figure><img src="image_155.png" /><figcaption>Figure B.3: Ideal and real enzyme costs of elementary flux modes – For each EFM (shown as a cyan dot), the ideal enzyme cost per biomass production rate (i.e. assuming that all the enzymes are saturated) is compared to the actual cost (calculated using Enzyme Cost Minimization, assuming standard aerobic glucose conditions). The costs span a wide range from the most enzyme-efficient EFMs on the lower left to the least enzyme-efficient ones on the upper right. For different EFMs, the ratio of actual and ideal costs varies between 1.4 and 4.7. Here, the EFM with the minimal actual cost is among the top 5 in terms of ideal cost.</figcaption></figure>
# Appendix C

## Appendix “Strategies for cell cycle control”

### C.1 Equations for birth size

Here we derive the dynamic equations of the birth size $q_0^i$ across generations (indexed by i) in the discrete-time formalism. We define $\langle q_0 \rangle_\alpha$ as the average value of $q_0$, and the log size deviation $\delta q_0^i := q_0^i - \langle q_0 \rangle_\alpha$. The dynamics for the log-size deviation takes the form

$$ \delta q_0^{i+1} = g(\delta q_0^i, \alpha) + \zeta^i(\delta q_0^i, \alpha), \quad (C.1) $$

where $\zeta^i(\delta q_0^i, \alpha)$ is a random variable with zero mean. This equation has the same degree of generality of Eq. (14.8) and can express any arbitrary division control model (or equivalently any shapes of the hazard rate function). In order to make further mathematical (and biological) progress, we need to simplify the equation and make the comparison with data possible. There are several possible choices. In the following, for simplicity, we first neglect the fluctuation of the growth rate $\alpha$. Assume that the size at birth is the only variable influencing cell division ($g(\cdot)$ is a function of $\delta q_0^i$ only) will allow us to introduce a linear-response framework. We will then describe how to consider the heterogeneity of multiple growth parameters.

The main empirical observation that comes to our help is the fact that the coefficient of variation of $q_0^i$ is small (typically around 0.15) [432, 440, 450, 447, 449]. The small value of the coefficient of variation strongly suggests the possibility of Taylor-expanding the function $g(\delta q_0^i)$ around $\delta q_0^i = 0$ [450]. In this limit, the function $g(\delta q_0^i)$ is approximately linear and the random variable $\zeta^i(\delta q_0^i, \alpha)$ can be well approximated by a Gaussian random variable with zero mean and constant variance [449]. The resulting equation reads

$$ \delta q_0^{i+1} = (1 - \lambda)\delta q_0^i + \sigma\xi^i, \quad (C.2) $$

where $\xi^i$ is a Gaussian random variable with zero mean and unit variance. The two parameters $\lambda$ and $\sigma$ encode, respectively, the relevant information about the mechanism of size control and the level of stochasticity. The parameter $\sigma$ simply corresponds to $\zeta^i(0, \alpha)$. The parameter $\lambda$, which quantifies the strength of size control, has a direct relationship with the mechanism at its origin. It is defined as $\lambda = 1 - g'(0, \alpha)$. For instance, the sizer corresponds to $\lambda = 1$ and an adder to $\lambda = 1/2$. The case $\lambda = 0$ does not lead to a stationary process and corresponds to a timer. Consequently, this parameter can easily be inferred from the plots in Figure ??.

Eq. (C.2) can be solved analytically [449]. In particular one can show that the conditional probability of observing a log-size deviation $\delta q_0^i$ from the average at generation i given a deviation at generation 0, is a Gaussian with mean

$$ \langle \delta q_0^i \rangle_{\delta q_0^0} = (1 - \lambda)^i \delta q_0^0. \quad (C.3) $$
This result clearly shows how different mechanisms correspond to different strengths of cell-size homeostasis, leading to fluctuations persisting across a different number of generations. For a sizer, $\lambda = 1$, the expected deviation of the daughter cell is independent of the mother cell fluctuations. A timer, with $\lambda = 0$, does not lead to homeostasis, as the expected deviation of size at birth of a daughter cell is the same as the deviation of the mother. The adder, $\lambda = 1/2$, leads on average to a halving of the size at birth deviation at each generation, as approximately observed in experiments [440].

One can generalize the linear-response framework to consider fluctuations of different growth parameters [451]. In general, one can assume that the size at birth of the daughter cell depends on both size at birth of the mother and her individual growth rate fluctuations.

$$ \delta q_0^{i+1} = (1 - \lambda_{qq}) \delta q_0^i - \lambda_{q\alpha} \delta \alpha^i + \xi_q^i. \qquad (C.4) $$

Along the same lines, one can assume that the growth rate fluctuations obey a similar equation

$$ \delta \alpha^{i+1} = -\lambda_{\alpha q} \delta q_0^i - \lambda_{\alpha \alpha} \delta \alpha^i + \xi_{\alpha}^i. \qquad (C.5) $$

This kind of equation can be written in multiple forms, i.e. including multiple variables. For example, one can write an equation explicitly for the elongation rate between divisions $\delta G := \delta q_0^{i+1} - \delta q_0^i$ or for the division time. Since the linear-response equations assume that the fluctuations around the means of these variables are small, all these choices turn out to be mathematically equivalent. This is also the reason why the different plots in Figure ?? are equivalent. While a linear dependency of growth rate $\alpha$ and division time $\tau_d$ on (log-)size at birth $q_0$ would induce a non linear dependency of the elongation $G = \alpha \tau_d$ on the initial size, such non-linearities can be neglected in the limit of small fluctuations, leading always to linear dependencies [450, 451].

The values of the parameters $\lambda_{ab}$ can be easily inferred using the standard tools of linear regression. Notably, the best (maximum likelihood) estimates of these parameters can be directly obtained from the variable covariances [449, 451]. For instance, $\langle \delta q^{i+1} \delta q^i \rangle = \lambda_{qq} \sigma_q^2 + \lambda_{q\alpha} \langle \delta \alpha^i \delta q^i \rangle$. By writing the expressions for other correlations (e.g., $\langle \delta q^{i+1} \delta \alpha^i \rangle$ or $\langle \delta \alpha^{i+1} \delta q^{i+1} \rangle$) one can map the coefficient $\lambda_{ab}$ with the measured covariances.

## C.2 Growth laws

**Growth laws and trade-offs between protein sectors.** Prototypical predictions are the so-called "growth laws", general quantitative relationships linking proteome composition and rates of cellular processes. The reason why relationships of the kind $\lambda = \lambda(\phi_R, \phi_X, \dots)$ and $k_X(\phi_R, \phi_X, \dots)$ naturally emerge in the framework is due to cell growth and division rates being coupled to proteome allocation dynamics.

**Growth law for the ribosome sector.** For example, the first growth law, stating that the ribosome mass fraction increases linearly with the nutrient-imposed growth rate, that is $\lambda = \lambda(\phi_R) = K(\phi_R - \phi_R^{min})$, is obtained straightforwardly by noting that upon differentiation of Eq. (14.21) with respect to time and substitution of Eq. (14.19) and Eq. (14.20) one finds the dynamical relation $\lambda(t) = \frac{k_n P(t)}{M}$, which at equilibrium reads (neglecting degradation)

$$ \lambda^* = \frac{k_n P^*}{M} = \frac{ak_t R^* f_a}{M} = \frac{ak_t}{m_R} \frac{M_{prot}}{M} (\phi_R - \phi_R^{min}), \qquad (C.6) $$

since at equilibrium the amino-acid import flux $k_n P^*$ matches the biosynthesis flux $ak_t R^* f_a$ ($dA/dt = 0$ in absence of degradation). Note that we have used the definitions $\phi_i \equiv (m_i P_i)/M_{prot} = (m_i P_i)/(M - M_a)$ and $R f_a = R^{active} = R - R^{inactive}$ and we have identified $\phi_R^{inactive} = \phi_R^{min}$.

**Trade-Offs between Ribosomes and Division Protein Synthesis.** Following Refs. [487, 488], we re-write Eq. (C.6) as $k_n = \frac{ak_t}{m_R} m_P \frac{\phi_R - \phi_R^{min}}{\phi_P}$ and use the constraint $\phi_R^{max} = 1 - \phi_Q = \phi_R + \phi_P + \phi_X$ to obtain

$$ \phi_X = - \frac{K_n + K_t}{K_n} \phi_R + \frac{K_t \phi_R^{min} + K_n \phi_R^{max}}{K_n}, \qquad (C.7) $$
where $K_n \equiv k_n/m_P$ ($[K_n] = [T]^{-1}$) and $K_t \equiv ak_t/m_R$ ($[K_t] = [T]^{-1}$). Eq. (C.7) shows a negative correlation between the ribosome and division sectors under nutrient or translational perturbations, in agreement with recent published data [553]. Also, since the rates of growth and division protein synthesis are respectively proportional to the ribosome and the division sector, this negative correlation reflects a trade-offs between allocating ribosomal resources towards growth or division (see Fig.1F in Ref. [487]).

**Growth law for the division sector.** So, the larger the fraction of ribosomes making division proteins the smaller the fraction of ribosomes making ribosomes. In other words, there is a negative correlation between the growth rate and the division protein sector. Indeed, the ribosome sector is related to the growth rate via the first growth law $\phi_R = \frac{M\lambda}{M_{prot}K_t} + \phi_R^{min}$, but it is also related to $\phi_X$ via Eq. (C.7) $\phi_R = \frac{K_t\phi_R^{min}+K_n\phi_R^{max}}{K_n+K_t} - \frac{K_n}{K_n+K_t}\phi_X$. Equating the two terms yields

$$ \lambda = \frac{K_n K_t}{K_n + K_t} \frac{M_{prot}}{M} (\phi_R^{max} - \phi_R^{min} - \phi_X), \quad (C.8) $$

which is Eq. (9) in Ref. [487].

We now discuss how two known steady-growth size-related behaviors emerge in the unified framework from the interplay between cell growth and cell division.

**Adder mechanism.** As we discussed, *E. coli* cells regulate their size by adding a constant volume between consecutive cell divisions (adder mechanism). In a previous problem, we investigated with numerical simulations the range of validity of this property. In the following one, we instead show analytically that the adder property is naturally embedded in the unified framework.

It can be seen then that whenever $\lambda \gg d_X/m_X$ (e.g. fast growth conditions), $\Delta s_{1cycle} \approx \frac{\lambda}{k_X} X_{th} = \text{const}$ which is the adder property. Notably, in increasingly slower growth conditions, where degradation becomes with the growth rate, deviations from the adder are predicted, up to the point $\lambda \ll d_X/m_X$ where $s_d \approx X_{th}d_X/(k_X m_X) = \text{const}$.

**"Schaechter-Maaloe-Kjeldgaard" (SMK) growth law.** According to this law, the population-averaged cellular size scales with growth rate in an approximately exponential fashion [554]. Interestingly, deviations from the exponential trend have recently been reported, particularly at slow growth, leading to a different proposition [465]. Notably, deviations from this law are accounted in our framework. Indeed, in an exponentially expanding population the average cell size can be expressed as $\langle s \rangle = 2\log 2\langle s_0 \rangle$ [432], which, combined with Eq. (14.27) and $\langle s_d \rangle = 2\langle s_0 \rangle$ leads to

$$ \langle s \rangle = \frac{\lambda + \frac{d_X}{m_X}}{\tilde{k}_X \left( 2 - 2^{-\frac{d_X}{m_X \lambda}} \right)} \quad (C.9) $$

where, following Ref. [487], we have defined $\tilde{k}_X \equiv k_X/(2\log 2X_{th})$. Note that since $\lambda \propto \phi_R$ and $k_X \propto -\phi_R$ the average cell size increases with ribosome abundance, a trend observed in experiments. Notably, upon determining the model parameters and making full explicit the growth rate dependence, the authors in Ref. [487] with no further fitting showed that Eq. (C.9) recapitulates the experimental data [465, 487], a remarkable achievement of the unified framework.

**Non-steady relationships.** Finally, we contextualize within the unified framework some predictions of a model recently proposed to unify cell division and growth in non-steady growth conditions [478]. As we saw, although there is consensus on an inter-division adder at the phenomenological level, the mechanisms regulating cell division dynamics in the bacterium *E. coli* are still widely debated. In particular several mechanistic models based on different mechanisms for division control were proposed for the adder [472, 392, 463, 462]. In order to help selecting different scenarios, experiments beyond steady-state growth help comparing the specific causal relationships underlying different models with data. Following this philosophy, and aiming to shed more light on cell division dynamics, Panlilio et al. [478] ran multiple long-term *E. coli* microfluidics experiments jointly monitoring size-division dynamics and reporters of ribosomal and constitutive genes through nutritional up-shifts. The fluorescent reporters can be seen
as proxies for the dynamics of the *R* and *P* sectors during the shift. Remarkably, in their experiments they observed highly-complex multiple-timescale dynamics in different cell-division variables (particularly inter-division time, division rate, added volume and added-to-initial volume ratio) during the nutritional up-shift. Notably, in spite of this complex dynamics, they found the division control strategy to be unaffected by the shift. The transient observed division dynamics in their shift data falsifies several scenarios, such as the Harris-Theoriot septum-limited division and the classic scenario of replication-limited division. Instead, the authors found that a threshold accumulation model such as the one described by Eq. (14.19) could not be falsified,

$$ \frac{ds(t)}{dt} = \alpha(t)s(t) \qquad \frac{dN(t)}{dt} = r_X(t)s(t). \qquad (C.10) $$

This the usual scenario where a constitutive *X*-sector protein accumulates to a threshold value $N^*$ and at that point triggers cell division. The regulation of cell division from a constitutive sector is coherent with the observation that *ppGpp* is a cell size and cell division regulator [555]. These results are also in line with independent conclusions based on steady-state data [488, 463, 477] and isolate *FtsZ* as a likely candidate cell-division trigger, although the previous section has clarified how the complexity of the decision to divide is likely higher than described by the chromosome-agnostic cell-division models that are used in integrated frameworks. Future efforts will have to integrate this complexity in a description that also accounts for the interplay of different processes relevant for cell cycle progression with cell growth.
# Appendix D

## Solutions to problems

### D.1 Chapter “The inventory of a cell”

**Problem 2.1 (Intuition for biological numbers)**

Check the results at bionumbers.hms.harvard.edu, book.bionumbers.org or doi.org/10.1016/j.cell.2010.06.019.

**Problem 2.2 (Proteins per cell - estimate one)**

$$ \frac{\text{Proteins}}{\text{mL}} = 0.2 \frac{\text{g}}{\text{mL}} \cdot 6 \cdot 10^{23} \frac{1}{\text{mol}} \cdot \frac{1}{40000} \frac{\text{mol}}{\text{g}} = 3 \cdot 10^{18} \frac{1}{\text{mL}} $$

$$ \frac{\text{Proteins}}{\mu\text{m}^3} = 3 \cdot 10^{18} \frac{1}{\text{mL}} \cdot 10^{-12} \frac{\text{mL}}{\mu\text{m}^3} = 3 \cdot 10^6 \frac{1}{\mu\text{m}^3} $$

$$ \frac{\text{Proteins}}{\text{cell}} \approx \begin{cases} 3 \cdot 10^6 & \text{E. coli} \\ 2 \cdot 10^8 & \text{S. cerevisiae} \\ 9 \cdot 10^9 & \text{mammalian cells} \end{cases} $$

**Problem 2.3 (Proteins/ribosomes per cell - estimate two)**

$$ 10^9 \cdot 0.3/25 \approx 1.2 \cdot 10^7 \text{ proteins} $$

$$ 10^9 \cdot 0.3/3400 \approx 88000 \text{ ribosomes} $$

**Problem 2.4 (Buoyant cell density)**

$$ \text{Density} = 1 \cdot 0.7 + 1.3 \cdot 0.18 + 1.7 \cdot 0.08 + 1 \cdot 0.03 + 1.5 \cdot 0.01 = 1.115 $$

**Problem 2.5 (Concentrations enzymes and substrates)**

Concentration of one molecule per E. coli cell:

$$ \frac{1}{\mu\text{m}^3} \cdot 10^{15} \frac{\mu\text{m}^3}{\text{L}} \cdot \frac{1}{6 \cdot 10^{23}} \frac{\text{mol}}{1} = 1.7 \cdot 10^{-9} \frac{\text{mol}}{\text{L}} $$

Enzyme mass: 40 000 g mol⁻¹ (BNID 105861)

Glucose mass: 180 g mol⁻¹

$$ 40000/180 = 222 $$

Assuming that the mass of the free enzyme and the mass of the substrate should be equal for optimal flux, we need approximately 222 molecules of substrate (glucose) per one molecule of free enzyme. This corresponds to the following concentrations:
Enzyme: ~ 1.7 nmol L⁻¹

Substrate: ~ 370 nmol L⁻¹

### Problem 2.6 (Cell size in different dimensions)

<table><thead><tr><th></th><th>Cell 1</th><th>Cell 2</th><th>Ratio</th></tr></thead><tbody><tr><td>Diameter</td><td>1</td><td>2</td><td>2</td></tr><tr><td>Surface area</td><td>3.14</td><td>12.57</td><td>4</td></tr><tr><td>Volume</td><td>0.52</td><td>4.19</td><td>8</td></tr><tr><td>SA/V</td><td>6</td><td>3</td><td>0.5</td></tr></tbody></table>

As the cell diameter increases, the volume grows faster than the surface area, leading to a decrease in the surface area-to-volume ratio (SA/V). This decrease can impact nutrient uptake as there is less relative surface area available for membrane transporters. Therefore, larger cells with lower SA/V ratios may face challenges in obtaining enough nutrients to meet their metabolic demands.

### Problem 2.7 (Alien lifeforms)

We do not know the answer to this question. These are ideas from ChatGPT (openai.com, GPT-3.5, version 2023) – take them with a grain of salt:

When imagining alien lifeforms, it is difficult to make definitive statements about their nature since we have no direct knowledge of extraterrestrial life. However, we can explore some possibilities and consider the principles of biology and physics as we currently understand them.

**1. Composition of cells.** While cells are the fundamental building blocks of life on Earth, it is not certain that extraterrestrial life would be composed of cells. Cells are a product of evolution on our planet and are highly efficient structures for carrying out biological processes. However, alternative forms of life may exist with entirely different structures and organization. For example, hypothetical non-cellular life forms based on self-replicating molecules or other complex arrangements have been proposed.

**2. Different features of alien cells.** (a) **Membranes:** While cells on Earth have lipid bilayer membranes, it is conceivable that alien life forms might have different membrane compositions. For example, they could have membranes composed of different molecules or organized in alternative ways. (b) **Genetic Material:** On Earth, DNA and RNA are the primary carriers of genetic information. However, it is possible that alternative genetic molecules or information storage systems could exist in alien life. These could have different base pairs or use different coding mechanisms. (c) **Metabolism:** Alien life could have different metabolic processes compared to Earth organisms. They might utilize different energy sources, employ unique enzymatic reactions, or even rely on completely novel biochemical pathways. (d) **Size and Structure:** Cells on Earth exhibit a wide range of sizes, from microscopic bacteria to the largest known cells in organisms like ostrich eggs. It is conceivable that alien cells could differ significantly in size and overall structure, depending on the specific conditions and evolutionary paths of their respective environments.

**3. Features dictated by physics.** Certain fundamental principles of physics are likely to impose constraints on the functioning and structure of any kind of cell, including potential alien cells. These features include: (a) **Biochemistry:** Regardless of the specific molecular composition, alien cells would need a biochemistry that allows for the storage and utilization of energy, the replication and expression of genetic information, and the maintenance of internal equilibrium. (b) **Thermodynamics:** The laws of thermodynamics, such as energy conservation and entropy increase, are universal physical principles. Any living system, including alien cells, would need to adhere to these principles to maintain their internal processes. (c) **Water:** Water is a highly abundant molecule and a fundamental solvent for life on Earth. It provides a medium for biochemical reactions and allows for efficient transport of molecules within cells. It is possible that water or another suitable liquid would be essential for alien life, but alternative solvents cannot be ruled out entirely.

While these considerations provide a starting point for thinking about alien life, the possibilities are vast, and it is challenging to predict the specific characteristics of extraterrestrial organisms. Future discoveries and investigations in astrobiology will help refine our understanding of life beyond Earth.
**Problem 2.8 (Substrate demand to saturate an enzyme)** Take the irreversible Michaelis-Menten law, and plug in $v = 0.1 \times \nu^{\text{max}}$ and $v = 0.9 \times \nu^{\text{max}}$. By rearranging the terms to express $S$ in terms of $K_M$, the answers are $S_{0.1} = \frac{0.1}{0.9}K_M \approx 0.11K_M$ and $S_{0.9} = \frac{0.9}{0.1}K_M = 9K_M$. This is approx. 81-fold difference to go from $0.1 \times \nu^{\text{max}}$ to $0.9 \times \nu^{\text{max}}$!

## D.2 Chapter “Cell metabolism”

**Problem 3.1 (An irreversible reaction with simultaneous binding)**

$$ (a) \qquad E + S_1 + S_2 \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES_1S_2 \xrightarrow{k_3} E + P_1 + P_2 \qquad (D.1) $$

$$ (b) \qquad \frac{dp}{dt} = k_3 \frac{s_1 s_2 C}{s_1 s_2 + \frac{k_2+k_3}{k_1}}, \qquad (D.2) $$

where $p = [P_1 + P_2]$ and $C = [E] + [ES_1S_2]$.

**Problem 3.2 (A reversible reaction with simultaneous binding)**

$$ (a) \qquad E + S_1 + S_2 \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES_1S_2 \underset{k_4}{\overset{k_3}{\rightleftharpoons}} E + P_1 + P_2 \qquad (D.3) $$

$$ (b) \qquad \frac{dp}{dt} = k_3 \frac{C(s_1 s_2 - \frac{k_2 k_4}{k_1 k_3} p)}{s_1 s_2 + \frac{k_4}{k_1} p + \frac{k_2 + k_3}{k_1}}, \qquad (D.4) $$

where $p = [P_1 + P_2]$ and $C = [E] + [ES_1S_2]$.

**Problem 3.3 (An irreversible reaction with sequential binding)**

$$ (a) \qquad E + S_1 \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES_1 \\ ES_1 + S_2 \underset{k_4}{\overset{k_3}{\rightleftharpoons}} ES_1S_2 \xrightarrow{k_5} E + P_1 + P_2 \qquad (D.5) $$

$$ (b) \qquad \frac{dp}{dt} = k_5 \frac{s_1 s_2 C}{s_1 s_2 + s_1 \frac{k_4+k_5}{k_3} + s_2 \frac{k_5}{k_3} + \frac{k_2}{k_1 k_3} (k_1 + k_5)}, \qquad (D.6) $$

where $p = [P_1 + P_2]$ and $C = [E] + [ES_1] + [ES_1S_2]$.

**Problem 3.4 (An irreversible reaction with random-order binding)**

$$ (a) \qquad E + S_1 \underset{k_2}{\overset{k_1}{\rightleftharpoons}} ES_1 \\ ES_1 + S_2 \underset{k_4}{\overset{k_3}{\rightleftharpoons}} ES_1S_2 \\ E + S_2 \underset{k_6}{\overset{k_5}{\rightleftharpoons}} ES_2 \\ ES_2 + S_1 \underset{k_8}{\overset{k_7}{\rightleftharpoons}} ES_1S_2 \\ ES_1S_2 \xrightarrow{k_9} E + P_1 + P_2 \qquad (D.7) $$

$$ (b) \qquad \frac{dp}{dt} = k_9 \frac{C s_1 s_2 (k_1 k_3 (k_6 + k_7 s_1) + k_5 k_7 (k_2 + k_3 s_2))}{s_1 A(s_1) + s_2 B(s_2) + s_1 s_2 C(s_1, s_2) + D}, \qquad (D.8) $$
where $p = [P_1 + P_2]$, $C = [E] + [ES_1] + [ES_2] + [ES_1S_2]$, and

$$A(s_1) = k_1k_6(k_4 + k_8 + k_9) + k_7(k_0 + k_4)(k_2 + k_1s_1)$$

$$B(s_2) = k_2k_5(k_4 + k_8 + k_9) + k_3(k_0 + k_8)(k_6 + k_5s_2)$$

$$C(s_1, s_2) = k_1k_3(k_6 + k_8 + k_7s_1) + k_5k_7(k_2 + k_4 + k_3s_2) + k_3k_7k_9$$

$$D = k_2k_6(k_4 + k_8 + k_9)$$

## D.3 Chapter "Metabolic flux distributions"

### Problem 4.4 (Elementary Flux Modes (2))

EFMs containing forward fluxes only:

<figure><img src="image_156.png" /></figure>

EFMs containing forward and backward fluxes:

<figure><img src="image_157.png" /></figure>

## D.4 Chapter "The enzyme cost of metabolic fluxes"

### Problem 6.9 (Haldane kinetic rate law)

First, we add the constraint on the total enzyme concentration ($[E] + [ES] + [EP] = E_{tot}$) and rewrite the ODE system in matrix notation:

$$ \begin{pmatrix} 1 \\ [S]k_1 \\ [P]k_6 \\ -[S]k_1 - [P]k_6 \end{pmatrix} \begin{pmatrix} 1 \\ -(k_2 + k_3) \\ k_3 \\ k_2 \end{pmatrix} \begin{pmatrix} 1 \\ k_4 \\ -(k_4 + k_5) \\ k_5 \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = \begin{pmatrix} [E_0] \\ 0 \\ 0 \\ 0 \end{pmatrix}. \quad (D.9) $$

Note that the last row is linearly dependent on the two previous ones (it is minus their sum). Therefore, we can drop it from the system without losing information. Then, we will find explicit expressions for $[E]$, $[ES]$, and $[EP]$ by using Gaussian elimination – a process of eliminating off-diagonal values in the matrix until we reach the identity matrix, while at the same time applying the same operations to the vector on the right-hand side of the equality.

Step 1, elimination the off-diagonal elements on the first column (subtracting the first row times $[S]k_1$ from the 2nd row and the first row times $[P]k_6$ from the 3rd row)

$$ \begin{pmatrix} 1 & 1 & 1 \\ 0 & -(k_2 + k_3) - [S]k_1 & k_4 - [S]k_1 \\ 0 & k_3 - [P]k_6 & -(k_4 + k_5) - [P]k_6 \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{tot}] \begin{pmatrix} 1 \\ -[S]k_1 \\ -[P]k_6 \end{pmatrix}. $$
Step 2, dividing the second row by $-(k_2 + k_3 + [S]k_1)$ to have 1 on the diagonal:

$$ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & \frac{[S]k_1 - k_4}{k_2 + k_3 + [S]k_1} \\ 0 & k_3 - [P]k_6 & -(k_4 + k_5) - [P]k_6 \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{\text{tot}}] \begin{pmatrix} \frac{[S]k_1}{k_2 + k_3 + [S]k_1} \\ -[P]k_6 \end{pmatrix}. $$

Step 3, subtracting the second row from the 1st, and again from the 3rd (after multiplying by $k_3 - [P]k_6$):

$$ \begin{pmatrix} 1 & 0 & 1 - \frac{[S]k_1 - k_4}{k_2 + k_3 + [S]k_1} \\ 0 & 1 & \frac{[S]k_1 - k_4}{k_2 + k_3 + [S]k_1} \\ 0 & 0 & -(k_4 + k_5) - [P]k_6 - \frac{([S]k_1 - k_4)(k_3 - [P]k_6)}{k_2 + k_3 + [S]k_1} \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{\text{tot}}] \begin{pmatrix} 1 - \frac{[S]k_1}{k_2 + k_3 + [S]k_1} \\ \frac{[S]k_1}{k_2 + k_3 + [S]k_1} \\ -[P]k_6 - \frac{[S]k_1(k_3 - [P]k_6)}{k_2 + k_3 + [S]k_1} \end{pmatrix}. $$

which after simplifying becomes:

$$ \begin{pmatrix} 1 & 0 & \frac{k_2+k_3+k_4}{k_2+k_3+[S]k_1} \\ 0 & 1 & \frac{[S]k_1-k_4}{k_2+k_3+[S]k_1} \\ 0 & 0 & -\frac{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5}{k_2+k_3+[S]k_1} \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{\text{tot}}] \begin{pmatrix} \frac{k_2+k_3}{k_2+k_3+[S]k_1} \\ \frac{[S]k_1}{k_2+k_3+[S]k_1} \\ -\frac{[P]k_6k_2+[P]k_6k_3+[S]k_1k_3}{k_2+k_3+[S]k_1} \end{pmatrix}. $$

and we normalize the last row to have 1 on the diagonal:

$$ \begin{pmatrix} 1 & 0 & \frac{k_2+k_3+k_4}{k_2+k_3+[S]k_1} \\ 0 & 1 & \frac{[S]k_1-k_4}{k_2+k_3+[S]k_1} \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{\text{tot}}] \begin{pmatrix} \frac{k_2+k_3}{k_2+k_3+[S]k_1} \\ \frac{[S]k_1}{k_2+k_3+[S]k_1} \\ \frac{[P]k_6k_2+[P]k_6k_3+[S]k_1k_3}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5} \end{pmatrix}. $$

Step 4, we eliminate the off-diagonal values of the third column using the 3rd row:

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{\text{tot}}] \begin{pmatrix} \frac{k_2+k_3}{k_2+k_3+[S]k_1} & -\frac{k_2+k_3+k_4}{k_2+k_3+[S]k_1} & \frac{[P]k_6k_2+[P]k_6k_3+[S]k_1k_3}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5} \\ \frac{[S]k_1}{k_2+k_3+[S]k_1} & -\frac{[S]k_1-k_4}{k_2+k_3+[S]k_1} & \frac{[P]k_6k_2+[P]k_6k_3+[S]k_1k_3}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5} \\ \frac{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5} & \frac{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5} & \frac{[P]k_6k_2+[P]k_6k_3+[S]k_1k_3}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2k_4+k_2k_5+k_3k_5} \end{pmatrix} $$

Simplifying the expressions on the right-hand side is a lengthy process (which we do not show here) and in the end we get:

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} [E] \\ [ES] \\ [EP] \end{pmatrix} = [E_{\text{tot}}] \begin{pmatrix} \frac{k_2 k_4 + k_2 k_5 + k_3 k_5}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2 k_4+k_2 k_5+k_3 k_5} \\ \frac{[P]k_6 k_2 + [P]k_6 k_3 + [S]k_1 k_5}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2 k_4+k_2 k_5+k_3 k_5} \\ \frac{[P]k_6 k_2 + [P]k_6 k_3 + [S]k_1 k_3}{[S]k_1(k_3+k_4+k_5)+[P]k_6(k_2+k_3+k_4)+k_2 k_4+k_2 k_5+k_3 k_5} \end{pmatrix} $$

Therefore,

$$ [E] = [E_{\text{tot}}] \frac{k_2 k_4 + k_2 k_5 + k_3 k_5}{[S]k_1(k_3 + k_4 + k_5) + [P]k_6(k_2 + k_3 + k_4) + k_2 k_4 + k_2 k_5 + k_3 k_5} \quad (D.10) $$

$$ [ES] = [E_{\text{tot}}] \frac{[P]k_4 k_6 + [S]k_1 k_4 + [S]k_1 k_5}{[S]k_1(k_3 + k_4 + k_5) + [P]k_6(k_2 + k_3 + k_4) + k_2 k_4 + k_2 k_5 + k_3 k_5} \quad (D.11) $$

$$ [EP] = [E_{\text{tot}}] \frac{[P]k_2 k_6 + [P]k_3 k_6 + [S]k_1 k_3}{[S]k_1(k_3 + k_4 + k_5) + [P]k_6(k_2 + k_3 + k_4) + k_2 k_4 + k_2 k_5 + k_3 k_5} \quad (D.12) $$

## D.5 Chapter "Optimization of metabolic states"

### Problem 7.1 (Effect of oxygen concentration)

The oxygen concentration affects only the rate $v_3$, and an increase in oxygen increases this rate for the same enzyme concentration. Since EFM2 does not contain $v_3$, the enzyme cost of this EFM will not change.
EFM1 and EFM3 benefit from an increase of the oxygen concentration, because they will have to invest less enzyme in $v_3$ to obtain the same rate. Therefore, those EFMs can become more beneficial and the optimal EFM could shift to one of those EFMs.

### Problem 7.2 (Effect of external metabolites)

Increasing $[P_{ex}]$ will benefit EFM3 and decrease the benefit of EFM2 (because for EFM2 $[P_{ex}]$ will inhibit reaction $v_2$ and therefore more enzyme is needed for reaction $v_2$ and the enzyme cost of EFM2 will increase. Qualitatively, with increasing $[P_{ex}]$, EFM2 might become more expensive, and either EFM1 or EFM3 will become beneficial, or both at different $P_{ex}$ concentrations, depending on the kinetics of the reactions.

### Problem 7.3 (States of maximal growth rate)

$$ (a) \quad v_1 = e_1(k_1^+ s_1 - k_1^- X), v_2 = e_2(k_2^+ s_2 - k_2^- X) \text{ and } v_3 = e_3(k_3^+ X - k_3^- p) $$

$$ (b) \quad e_{\text{tot}} = \frac{v_1}{(2s_1 - x)} + \frac{v_2}{(30 - x)} + \frac{v_3}{x} $$

$$ (c) \quad \text{When } e_1 = 0, v_1 \text{ is also 0 and to achieve steady state } v_2 = v_3 \text{ and using the rate equations and filling in the parameters we get } e_{\text{tot}} = \frac{v_3}{(30-x)} + \frac{v_3}{x}. \text{ We now set } e_{\text{tot}} = 1 \text{ to obtain } v_3 = (1 - \frac{v_3}{30-x})x \text{ which we can rewrite to } v_3 = \frac{x}{1 + \frac{x}{30-x}}. \text{ We can take the derivative to } x \text{ and set it equal to 0 to find the optimum, which leads to } x = 15 \text{ and } v_3/e_{\text{tot}} = \frac{15}{2}. \text{ Note that in this specific case we did not need to set } e_{\text{tot}} = 1 \text{ and could have maximized } v_3/e_{\text{tot}} \text{ directly, but in general this does not always work. In the rest of the answers we assume we set } e_{\text{tot}} = 1 \text{ and therefore } v_3 = v_3/e_{\text{tot}}.} $$

$$ (d) \quad e_2 = 0 \text{ implies } v_2 = 0 \text{ and filling in } s_1 = 10 \text{ leads to } v_3 = (1 - \frac{v_3}{20-x})x, \text{ which can be rewritten to } v_3 = \frac{x}{1 + \frac{x}{20-x}}. \text{ This is optimal when } x = 10 \text{ and } v_3/e_{\text{tot}} = 5. $$

$$ (e) \quad e_1 = e_2 \text{ implies } \frac{v_1}{20-x} = \frac{v_2}{30-x}. \text{ From the steady state we know that } v_2 = v_3 - v_1. \text{ Filling this in and solving for } v_1 \text{ leads to } v_1 = v_3 \frac{x-20}{x-50}. \text{ From the total enzyme and by replacing } e_1 \text{ by } e_2 \text{ we get } e_3 = 1 - 2\frac{v_1}{20-x}. \text{ Putting this in the equation for } v_3 \text{ and using the previous equality to replace } v_1 \text{ leads to: } v_3 = (1 + \frac{v_3}{x-25})x. \text{ Solving this for } v_3 \text{ gives } v_3 = x - \frac{x^2}{25}, \text{ which is optimal for } x = 12.5 \text{ with } v_3/e_{\text{tot}} = \frac{25}{4}. $$

$$ (f) \quad \text{It was optimal to invest all enzyme in } e_2 \text{ and none in } e_1, \text{ because that lead to the highest specific flux } v_3 \text{ (namely } v_3 = 7.5). $$

$$ (g) \quad \text{Since } v_1 = 0, S_1 \text{ is not involved in any reaction and the solution is the same as above, } x = 15 \text{ and } v_3/e_{\text{tot}} = \frac{15}{2}. $$

$$ (h) \quad \text{Similar calculations as above but now with } s_1 = 50 \text{ lead to } v_3 = \frac{x}{1 + \frac{x}{100-x}}, \text{ which is optimal when } x = 50 \text{ and gives } v_3/e_{\text{tot}} = 25. $$

$$ (i) \quad \text{Similar calculations as above but now with } s_1 = 50 \text{ lead to } v_3 = x - \frac{x^2}{65}. v_3/e_{\text{tot}} \text{ is maximal at } x = 32.5 \text{ and takes the value } 16.25. $$

$$ (j) \quad \text{Now } s_1 \text{ increased the optimal strategy would be to invest all enzymes in } e_1, \text{ and have } e_2 = 0, \text{ because that leads to the highest specific flux of } v_3, \text{ namely } v_3/e_{\text{tot}} = 25. $$

$$ (k) \quad \text{The two EFMs in this pathway that produce } P \text{ are } v_1 = v_3 \text{ with } v_2 = 0 \text{ and } v_2 = v_3 \text{ with } v_1 = 0. \text{ In the problem we saw when we optimize the specific flux, we always obtained one of those EFMs as the best solution, from the options that we tested. Therefore, these results are in agreement with the proof outlined in this chapter.} $$

## D.6 Chapter "Universal features of autocatalytic systems"

### Problem 9.1 (Production function of a workstation with a single resource and single product)

(a) The equations follow directly from the definitions of the variables and an assumption similar to mass-action law kinetics.

(b) Since the total number of machines (busy machines and idling machines) is constant, we have

$$ u + v = u_0. \tag{D.13} $$

It follows from this that the steady state solution ($\dot{x} = \dot{v} = 0$) of the dynamical equations takes the
form

$$ 
\begin{aligned}
a \frac{\beta \bar{v}}{\alpha} + q \bar{x} &= r, \\
\alpha \bar{x}(u_0 - \bar{v}) &= \beta \bar{v},
\end{aligned}
\qquad (D.14) $$

which leads to

$$ \bar{v} = \frac{\alpha u_0 \bar{x}}{\beta + \alpha \bar{x}}. \qquad (D.15) $$

Therefore, the output rate dy/dt which equals $b\bar{v}$ saturates as a function of the amount of stock x. This saturation due to an excess of stock is similar to the one present in Michaelis-Menten enzyme kinetics.

(c) One finds from the previous questions that $\bar{v}$ satisfies a second order equation which can be written in the form

$$ \frac{\bar{v}}{u_0} = \frac{\rho + 1 + \gamma}{2} \pm \frac{1}{2} \sqrt{(\rho + 1 + \gamma)^2 - 4\rho}. \qquad (D.16) $$

As a result, to lowest order in $\gamma \ll 1$,

$$ \frac{\bar{v}}{u_0} = \min(1, \rho). \qquad (D.17) $$

Thus, the output approaches the Leontief production function in the limit $\gamma \to 0$. This function describes two separate regimes: when $\rho < 1$, only a fraction of machines $\rho$ are busy which limits the production. Instead when $\rho > 1$, all machines are busy, and the production is limited by the total number of machines $u_0$. Thus, the Leontief function can describe limitations due to fluxes (term due to $\rho$) but also due to the amount of stocks (term due to $u_0$).

(d) It follows that the above solution is below the Leontief function for finite $\gamma$ and approaches it as $\gamma \to 0$.

### Problem 9.2 (UPF model)

(a) The equations follow directly from the definitions of the variables and from the use of Leontief production function.

(b) In Regime I where $n_F \gg n_U$ and $n_f \gg n_P$, the equations simplify to:

$$ 
\begin{aligned}
\frac{dn_U}{dt} &= \frac{\alpha n_U}{\tau_a} - \frac{n_U}{\tau_L} \\
\frac{dn_P}{dt} &= \frac{(1 - \alpha)n_U}{\tau_a} \\
\frac{dn_F}{dt} &= \frac{n_P}{\tau_F} - \frac{n_U}{\tau_a}.
\end{aligned}
\qquad (D.18) $$

Therefore, the growth rate $\mu$ at which all species grow exponentially is $\mu = \alpha/\tau_a - 1/\tau_L$. Note that $\alpha$ is an increasing function of $\mu$.

Similarly, in regime II where $n_F \ll n_U$ and $n_f \gg n_P$, and the growth rate satisfies the second order equation

$$ \mu^2 + \frac{\mu}{\tau_a} - \frac{1 - \alpha}{\tau_a \tau_F} = 0, \qquad (D.19) $$

so that

$$ \mu = \frac{1}{2\tau_a} \left( 2\sqrt{1 + \frac{4(1-\alpha)\tau_a}{\tau_F}} - 1 \right), \qquad (D.20) $$

thus now, $\alpha$ is a decreasing function of $\mu$.

In regime III, where $n_F \gg n_U$ and $n_f \ll n_P$, the growth rate is the same as in regime I and in regime IV, where $n_F \ll n_U$ and $n_f \ll n_P$, and the growth rate is the same as in regime II.

(c) As a result, the plot of $\mu$ versus $\alpha$ contains a linear part and part of a parabola, and one can put the four regimes in a single table in which the columns are $n_f \ll n_P$ and $n_f \gg n_P$, while the rows are U starved or U non starved, as done in A. Roy et al., PNAS (2020).

### Problem 9.3 (MGF for two simple networks)
(a) Let $\mathbf{v} = (v_1, v_2)^T$ be the flux vector. After splitting the stoichiometric matrix into $\mathbf{N}^+$ and $\mathbf{N}^-$, we find for the first network:

$$ \alpha(\mathbf{v}) = \min(\frac{2v_2}{v_1}, \frac{v_1}{v_2}). \qquad (D.21) $$

Thus, either $\alpha(\mathbf{v}) = 2v_2/v_1$ if $2v_2/v_1 \le v_1/v_2$, then $v_2/v_1 \le 1/\sqrt{2}$ and $\alpha(\mathbf{v}) \le \sqrt{2}$, or $\alpha(\mathbf{v}) = v_1/v_2$ if $2v_2/v_1 \ge v_1/v_2$, then $v_1/v_2 < 1/\sqrt{2}$ and $\alpha(\mathbf{v}) \le 1/\sqrt{2}$. Thus in the end, $\alpha = \max_{\mathbf{v}} \alpha(\mathbf{v}) = \sqrt{2}$. The second network follows a similar calculation which leads to $\alpha = 1/\sqrt{2}$.

(b) For the first network, there is autocatalytic growth because $\alpha > 1$, while in the second network, there is degrowth (or shrinking) because $\alpha < 1$.

## D.7 Chapter “Cells in the face of uncertainty”

### Problem 13.1 (Kelly strategy with partial information)

(a) This is a direct application of Eq. (1.3) to the case $f(R, E) = f(E)$ if $R = E$ and $f(R, E) = 0$ otherwise after noticing that $q(S|E)p(E) = p(E|S)p(S) = p(E, S)$ (Bayes' formula).

(b) Without sensing, the strategy $u(E)$ that maximizes $\Lambda = \sum_E p(E) \ln(f(E)u(E))$ is proportional betting given by $u(E) = p(E)$. Formally, Eq. (1.15) has the form $\Lambda = \sum_S p(S)\Lambda(S)$, where $\Lambda(S)$ has the same form as without sensing, with $E$ replaced by $E|S$. The optimal strategy with sensing is therefore conditional proportional betting, $u(E|S) = p(E|S)$ or, using Bayes' formula, $u(E|S) = q(S|E)p(E)/p(S)$ with $p(S) = \sum_E q(S|E)p(E)$.

(c) In the absence of information, the optimal $\Lambda$ is $\Lambda_0 = \sum_E p(E) \ln(f(E)p(E))$ while it is $\Lambda = \sum_E q(S|E)p(E) \ln(f(E)p(E))$ in the presence of information. The difference is therefore

$$ \Lambda - \Lambda_0 = \sum_E q(S|E)p(E) \ln \left( \frac{p(E|S)}{p(E)} \right) $$

which is equivalent to Eq (1.16) since $p(E|S) = q(S|E)p(E)/q(S)$.

(d) The information channel is beneficial if $\Lambda - c > \Lambda_0$, i.e., $I(E, S) > c$.

(e) For a binary erasure channel, $S$ can take two values given $E$: $S = E$ with probability $q(S|E) = 1 - \epsilon$ and $S = *$ (erased information) with probability $q(*|E) = \epsilon$. The mutual information is therefore:

$$ I(E, S) = \sum_E \left( (1 - \epsilon)p(E) \ln \frac{(1 - \epsilon)}{(1 - \epsilon)p(E)} + \epsilon p(E) \ln \frac{\epsilon}{\epsilon} \right) = (1 - \epsilon) \sum_E p(E) \ln \frac{1}{p(E)} = (1 - \epsilon)H(E) $$

where $H(E) = -\sum_E p(E) \ln p(E)$ is the entropy of the environment. The dependence of the growth rate on $\epsilon$ is therefore controlled by $(1 - \epsilon)H(E) - c(\epsilon)$ and the optimal $\epsilon$ is the one that maximizes this quantity. This represents a trade-off since both $I(S, E) = (1 - \epsilon)H(E)$ and $c(\epsilon)$ decrease with increasing $\epsilon$.

### Problem 13.2 (Value of information beyond Kelly's model)

(a) Applying Eq. (1.3) we obtain

$$ \Lambda = \sum_S q(S|E = \text{low})(1 - p_a) \ln(2 - u(d|S)) + q(S|E = \text{high})p_a \ln u(d|S) $$

(b) We can apply again Eq. (1.15) from Problem 12.1 to obtain that the optimal $u(d|S)$ is given by the same formula as the optimal $u_d$ in the absence of information with $p_a = p(E = \text{high})$ replaced by $p_a = p(E = \text{high}|S)$, i.e. $u(d|S) = 2p(E|S)$ if $p(E|S) < 1/2$ and $1$ otherwise, where $p(E|S) = q(S|E = \text{high})p(E = \text{high})/p(S)$ with $p(S) = q(S|E = \text{high})p_a + q(S|E = \text{low})(1 - p_a)$.

(c) Let consider for instance the erasure channel introduced in problem 12.2 and assume $p_a < 1/2$. Then $u(d|E = \text{high}) = 1$, $u(d|E = \text{low}) = 0$ and $u(d|*) = 2p_a$, which leads to an optimal growth rate given by

$$ \Lambda = (1 - \epsilon)(1 - p_a) \ln 2 + \epsilon[\ln 2 - H(p)] $$

The case without information corresponds to $\epsilon = 1$, in which case $\Lambda_0 = \ln 2 - H(p)$. We therefore
have a gain in growth rate due to the information given by

$$ \Lambda - \Lambda_0 = (1 - \epsilon)[H(p_a) - p_a \ln 2] = I(S, E) - p_a \ln 2 $$

since $I(S, E) = (1 - \epsilon)H(p_a)$ for an erasure channel (see Problem 12.1). We verify that $\Lambda - \Lambda_0 < I(S, E)$ as soon as $p_a > 0$.

### Problem 13.3 (Stochastic sensing at the level of individual cells)

(a) If each individual has its own sensor, the information $S$ is formally part of the state of the individual and the strategy is of the form $u(R, S|E) = u(R|S)q(S|E)$ where $E$ is here treated as a common information. More generally, if the individual information $S_i$ were derived from a source $S_c$ that is common at the population level, we would have

$$ \Lambda = \sum_{E, S_c} q_c(S_c|E)p(E) \ln \left( \sum_{R, S_i} f(R, E)u(R|S_i)q_i(S_i|S_c) \right) $$

(b) Since $\sum_S q(S|E) = 1$, the concavity of the logarithm implies

$$ \Lambda = \sum_{E} p(E) \ln \left( \sum_{R,S} f(R, E) u(R|S) q(S|E) \right) \leq \sum_{E,S} q(S|E) p(E) \ln \left( \sum_{R} f(R, E) u(R|S) \right) $$

where the right-hand part is interpreted as the growth rate given an information $S$ that is common at the population level.

### Problem 13.4 (Pareto front for Kelly's model)

(a) Let the probability that the first horse wins (resp. loses) be $p$ (resp. $1-p$); the bet and the odd on the first (resp. second) horse are $b$ and $1/r$ (resp. $1-b$ and $1/(1-r)$). In this way, the odds are fair because the sum of the inverse of the odds is one. Using the definitions of the growth rate and volatility, we find:

$$ \langle W \rangle = p \ln \left( \frac{b}{r} \right) + (1-p) \ln \left( \frac{1-b}{1-r} \right), $$

and

$$ \sigma_W^2 = p(1-p) \ln^2 \left( \frac{b(1-r)}{(1-b)r} \right) = \left( \sigma \ln \left( \frac{b(1-r)}{(1-b)r} \right) \right)^2, $$

where we have introduced the parameter $\sigma = \sqrt{p(1-p)}$ to simplify the expression of the volatility. When $b = r$, it is obvious from these expressions that $\langle W \rangle = 0$, i.e. the strategy has no growth rate, and $\sigma_W = 0$, the strategy is also risk free.

(b) The maximization of the objective function

$$ J = \alpha \langle W \rangle - (1 - \alpha) \sigma_W, $$

interpolates between the maximization of the $\langle W \rangle$ when $\alpha = 1$ and the minimization of $\sigma_W$ when $\alpha = 0$. We find from the condition that $J$ be optimized with respect to $b$ the optimal strategy $b^{\pm}$:

$$ b^{\pm} = p \pm \gamma \sigma, $$

where $\gamma := \frac{1-\alpha}{\alpha}$ and the + (resp. -) sign corresponds to an overbetting (resp. underbetting) strategy with respect to Kelly's strategy where $b = p$. As shown in Fig. 12.3 of the main text, these two solutions form the two branches of the efficient border which meet at Kelly's point. When $p < r$, the lower blue solid line is the tradeoff branch associated with $b^+$, while the upper red solid line is the non-tradeoff branch, associated with $b^-$. The roles of $b^-$ and $b^+$ exchange when instead $p > r$.
(c) Let us first focus on the region where $\langle W \rangle \ge 0$ and let us assume e.g. $p > r$, in which case

$$b = p - \gamma\sigma.$$

Using the solutions for $\langle W \rangle$ and $\sigma_W^2$ from the step (1), we find

$$\frac{d\langle W \rangle}{db} = \frac{p-b}{b(1-b)} = \frac{\gamma\sigma}{b(1-b)},$$

and

$$\frac{1}{2} \frac{d(\sigma_W^2)}{db} = \frac{\sigma^2}{b(1-b)} \ln \left( \frac{b(1-r)}{(1-b)r} \right).$$

Hence by taking the ratio of the two equations,

$$\frac{1}{2} \frac{d(\sigma_W^2)}{d\langle W \rangle} = \frac{\sigma}{\gamma} \ln \left( \frac{b(1-r)}{(1-b)r} \right) = \frac{\sigma_W}{\gamma} > 0. \qquad (D.22)$$

Using the definition of $\gamma$, we deduce that the slope of the Pareto border is

$$\left. \frac{d\sigma_W}{d\langle W \rangle} \right|_{\gamma} = \frac{\sigma}{p-b},$$

where $b$ is equal to $b^-$ since we have assumed $r < p$. This equation shows that the slope becomes infinite at Kelly's point where $\gamma \to 0$ and $b^- \to p$; while it reaches a finite value near the null strategy, namely

$$\left. \frac{d\sigma_W}{d\langle W \rangle} \right|_{\gamma_c} = \frac{1}{\gamma_c} = \frac{\sigma}{p-r}.$$

This suggests that there is a phase transition between the null strategy and a mixed strategy at this critical value $\gamma_c$. To confirm this point, we need to check that the border is convex near the null strategy. To do so, we take the derivative of Eq.([?]) with respect to $\langle W \rangle$ as before:

$$\frac{1}{2} \frac{d^2(\sigma_W^2)}{d\langle W \rangle^2} = \frac{\frac{\sigma}{\gamma} \frac{d}{db} \ln\left(\frac{b}{1-b}\right) + \frac{\sigma_W}{\sigma\gamma^2}}{d\langle W \rangle / db} = \frac{1}{\gamma^2} + b(1-b) \frac{\sigma_W}{\sigma^2\gamma^3};$$

Finally, using the general formula

$$(\sqrt{f})'' = \frac{1}{2\sqrt{f}} f'' - \frac{1}{4} f^{-3/2} (f')^2,$$

we find the simple result

$$\frac{d^2\sigma_W}{d\langle W \rangle^2} = \frac{b(1-b)}{\sigma^2\gamma^3},$$

which is always positive, in particular near the null strategy where it takes the value

$$\left. \frac{d^2 \sigma_W}{d \langle W \rangle^2} \right|_{\gamma = \gamma_c} = \frac{r(1-r)}{\sigma^2 \gamma_c^3} > 0.$$

**Problem 13.5 (Evolution of an optimal strategy)**

The questions are already in the form of pseudocode.
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economic
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collective