Optimal resource allocation enables mathematical exploration of microbial metabolic configurations

Abstract

Central to the functioning of any living cell, the metabolic network is a complex network of biochemical reactions. It may also be viewed as an elaborate production system, integrating a diversity of internal and external signals in order to efficiently produce the energy and the biochemical precursors to ensure all cellular functions. Even in simple organisms like bacteria, it shows a striking level of coordination, adapting to very different growth media. Constraint-based models constitute an efficient mathematical framework to compute optimal metabolic configurations, at the scale of a whole genome. Combining the constraint-based approach “Resource Balance Analysis” with combinatorial optimization techniques, we propose a general method to explore these configurations, based on the inference of logical rules governing the activation of metabolic fluxes in response to diverse extracellular media. Using the concept of partial Boolean functions, we notably introduce a novel tractable algorithm to infer monotone Boolean functions on a minimal support. Monotonicity seems particularly relevant in this context, since the orderliness exhibited by the metabolic network’s dynamical behavior is expected to give rise to relatively simple rules. First results are promising, as the application of the method on Bacillus subtilis central carbon metabolism allows to recover known regulations as well as to investigate lesser known parts of the global regulatory network.

Appendices

Appendix 1: Proof of Theorem 1

We start with the following technical lemma.

Lemma 1

Let $X\subseteq {\mathscr {B}}^n$. For all $x\in X$, either $x\in \min X$ or there exists $y\in \min X$ such that $y<x$.

Proof

Consider $x\in X\backslash \min X$. Negating (5), one obtains $x_1\in X$ such that $x_1 < x$. If $x_1\in \min X$ then define $y = x_1$ and the proof is over, otherwise the same process is iterated to obtain $x_2\in X$ such that $x_2<x_1$. X being finite, this iterative process eventually leads to $y = x_k< x_{k-1}<\dots< x_1 < x$, with $y\in \min X$. $\square$

Proof

(Theorem 1) The equivalence between assertions (i) and (ii) is classical (Crama and Hammer 2011, p. 35). The equivalence between (ii) and (iii) is easier to see using negations:

$$\begin{aligned} \lnot (ii)\Longleftrightarrow & {} \exists x,y\in {\mathscr {B}}^n,\; (x\le y) \wedge \lnot \left( f(x)\le f(y)\right) ,\\\Longleftrightarrow & {} \exists x,y\in {\mathscr {B}}^n,\; (x\le y) \wedge (f(x)=1) \wedge (f(y)=0),\\\Longleftrightarrow & {} \exists x\in T_f, \exists y\in F_f,\; x\le y ,\\\Longleftrightarrow & {} \lnot (iii). \end{aligned}$$

Finally, let us prove the implication (iii) $\Rightarrow$ (iv) (the converse being trivial). Suppose there exists $(x,y)\in T_f\times F_f$ such that $x\le y$. Lemma 1 ensures there exists $\xi \in \min T_f$ such that $\xi \le x$. A symmetrical reasoning gives a $\chi \in \max F_f$ such that $y\le \chi$. Therefore, we found $(\xi ,\chi )\in \min T_f \times \max F_f$ such that $\xi \le x\le y\le \chi$, completing the proof. $\square$

Appendix 2: Proofs of Theorems 2, 3 and Corollary 1

Proof

(Theorem 2) Suppose $T\cap F\ne \varnothing$, then an extension f cannot exist otherwise it would verify $f(x) = 0$ and $f(x) = 1$ for any $x\in T\cap F$.

Suppose now $T\cap F=\varnothing$ and define the function $f_{min}\in {\mathscr {F}}_n$ by $T_{f_{min}} = T$, then since $T\cap F=\varnothing$, $F_{f_{min}} = {\mathscr {B}}^n\backslash T \supseteq F$, therefore $f_{min}$ is an extension of (T, F). With a similar argument, the function $f_{max}$ defined by $F_{f_{max}} = F$ is also an extension of (T, F). To prove the last part of the theorem, it suffices to remark that the set of extensions is in fact the segment $[f_{min},f_{max}]$:

$$\begin{aligned} f\in {\mathscr {E}}(T,F)\Longleftrightarrow & {} T\subseteq T_f\text { and }F\subseteq F_f, \\\Longleftrightarrow & {} T_{f_{min}}\subseteq T_f\text { and }F_{f_{max}}\subseteq F_f, \\\Longleftrightarrow & {} f_{min}\le f\le f_{max}. \end{aligned}$$

In other words, choosing an extension of (T, F) consists in choosing arbitrary Boolean values on the set $U={\mathscr {B}}^n\backslash (T\cup F)$ of unknown examples. This means one has $\vert U\vert = 2^n - \vert T \vert - \vert F\vert = q$ degrees of freedom. From a computational point of view, note that the existence of an extension is simply solved by the computation of the intersection $T\cap F$. In the worst case it will involve $|T|\times |F|$ comparisons, which is polynomial in p the size of the PBF. $\square$

Proof

(Theorem 3) Suppose ${\mathscr {I}}\ne \varnothing$, then there are $x\in T$ and $y\in F$ such that $x\le y$. If a positive extension f existed, it would imply $f(x)=1>0=f(y)$ which is absurd.

Suppose now ${\mathscr {I}}=\varnothing$. Define $f^+_{min}\in {\mathscr {F}}_n$ by $T_{f^+_{min}} := \{x\in {\mathscr {B}}^n,\,\exists a\in T,\,a\le x\}$, then $f^+_{min}$ verifies three properties:

• (i) $f^+_{min}$ is an extension of (T, F),

• (ii) $f^+_{min}$ is positive, and

• (iii) any positive extension f of (T, F) verifies $f^+_{min}\le f$.

(i): By construction, $T\subseteq T_{f^+_{min}}$. Let $y\in F$ and suppose $y\in T_{f^+_{min}}$. Then there is $a\in T$ such that $a\le y$, implying (a, y) belongs to ${\mathscr {I}}$ which contradicts the assumption ${\mathscr {I}}=\varnothing$. So $F\subseteq F_{f^+_{min}}$ and f is an extension of (T, F).

(ii): Let $x,y\in {\mathscr {B}}^n$ such that $x\le y$, and suppose $x\in T_{f^+_{min}}$. Then there is $a\in T$ such that $a\le x$. Since $x\le y$, it implies that y necessarily belongs to $T_{f^+_{min}}$ as well, thus proving $f^+_{min}$ is positive.

(iii): Let $f\in {\mathscr {E}}^+(T,F)$ and $x\in T_{f^+_{min}}$. There is $a\in T$ such that $a\le x$; f being positive, $f(a) = 1\le f(x)$, thus proving that $T_{f^+_{min}}\subseteq T_f$, i.e. $f^+_{min}\le f$.

Now define $f^+_{max}\in {\mathscr {F}}_n$ by $F_{f^+_{max}} := \{x\in {\mathscr {B}}^n,\,\exists b\in F,\,x\le b\}$. With completely symmetrical arguments, it can be shown that $f^+_{max}$ is also a positive extension of (T, F), and that any positive extension f verifies $f\le f^+_{max}$.

To conclude the proof, consider any positive function f verifying $f^+_{min}\le f\le f^+_{max}$. Then, $T\subseteq T_{f^+_{min}}\subseteq T_f$ and $F\subseteq F_{f^+_{max}}\subseteq F_f$, ensuring $f\in {\mathscr {E}}^+(T,F)$. Therefore, the set of positive extension is in fact ${\mathscr {E}}^+(T,F) = [f^+_{min},f^+_{max}]\cap {\mathscr {F}}_n^+$, which is a sublattice of ${\mathscr {F}}_n^+$. As for Theorem 2, from a computational point of view deciding whether a positive extension exists is solved by the computation of the set ${\mathscr {I}}$, which amounts to $|T|\times |F|$ comparisons in the worst case. $\square$

Proof

(Corollary 1) Let f be a positive extension of (T, F), then $\min T \subseteq T\subseteq T_f$ and $\max F \subseteq F\subseteq F_f$, so f is a positive extension of $(\min T,\max F)$. Reciprocally, suppose f is a positive extension of $(\min T,\max F)$. Let $x\in T$, then there is $y\in \min T$ such that $y\le x$ and the positivity of f implies $f(x)\ge f(y) = 1$, so $T \subseteq T_f$. With a similar reasoning, $F \subseteq F_f$, and f is a positive extension of (T, F).

To prove the second part, recall the smallest positive extension $f_{min}^+$ of (T, F) is defined by $T_{f^+_{min}} = \{x\in {\mathscr {B}}^n,\,\exists a\in T,\,a\le x\}$. This definition immediately yields $T\subseteq T_{f^+_{min}}$. Assertion (i) is proved in two steps.

First, let $x\in \min T_{f^+_{min}}$, i.e. (1) $x\in T_{f^+_{min}}$ and (2) $(\forall y\in T_{f^+_{min}},\,y\le x\Rightarrow y=x)$. By definition, (1) implies there exists $a\in T$ such that $a\le x$. Since $T\subseteq T_{f^+_{min}}$, $a\in T_{f^+_{min}}$ so (2) leads to $a=x$ and $x\in T$. Now let $y\in T$ such that $y\le x$, $y\in T\subseteq T_{f^+_{min}}$ so (2) implies $y=x$. In conclusion, $\min T_{f^+_{min}} \subseteq \min T$.

Second, let $x\in \min T$, i.e. (1’) $x\in T$ and (2’) $(\forall y\in T,\,y\le x\Rightarrow y=x)$. Then clearly $x\in T_{f^+_{min}}$. Now let $y \in T_{f^+_{min}}$ such that $y\le x$. By definition there is $a\in T$ such that $a\le y$ and (2’) leads to $x=y=a$. Therefore, $\min T_{f^+_{min}} = \min T$. Equality (ii) is proved by symmetrical arguments. $\square$

Appendix 3: Proofs of Propositions 1, 2 and 3

Proof

(Proposition 1) Let $S\in \text{ Supp }(T,F)$. Since $T[S] \cap F[S]=\varnothing$, a direct application of Theorem 2 implies the existence of an extension f whose essential variables are in S. In other words, $S(f)\subseteq S$. By construction, S(f) is a support of (T, F) and since S is minimal, $S=S(f)$. Furthermore, there cannot be an extension g with $S(g)\subsetneq S$ as it would contradict the fact that S is minimal.

Conversely, if $S\in \min \{ S(f),f\in {\mathscr {E}}(T,F) \}$ then there is an extension f such that $S=S(f)$, implying S is a support of (T, F). Furthermore, it is minimal otherwise it would imply the existence of an extension with a support strictly included in S, which would lead to a contradiction. $\square$

Proof

(Proposition 2) The positivity of $\Sigma$ is straightforward: it is a direct application of Theorem 1 and of the fact that any superset of a support is itself a support (which is a natural consequence of Definition 5). The second part is simply a translation of Definition 6.

To prove Proposition 3 we will use the following technical lemma.

Lemma 2

For all $s\in {\mathscr {B}}^n$, the following equivalence is verified:

$$\begin{aligned} \Sigma (s)=0 \;\Longleftrightarrow \; \exists a\in {\mathscr {A}},\,s\le a. \end{aligned}$$

Proof

Let $s\in {\mathscr {B}}^n$ such that $\Sigma (s)=0$. By definition, S is not a support of (T, F), i.e. $T[S]\cap F[S]\ne \varnothing$. Let $z\in T[S]\cap F[S]$: there exist $x\in T$ and $y\in F$ such that $x[S] = y[S] = z$. Define $a = \overline{x\oplus y}$, a belongs to ${\mathscr {A}}$ and $x[S]=y[S]$ implies $a_i = 1$ for all $i\in S$, yielding $s\le a$.

Conversely, let $s\in {\mathscr {B}}^n$ and suppose there exists $a\in {\mathscr {A}}$ such that $s\le a$. By definition, $a = \overline{x\oplus y}$ for some $(x,y)\in T\times F$. Using the usual convention, let A denote the set $\{i,\,a_i=1\}\subseteq \{1,\dots ,n\}$. From $a = \overline{x\oplus y}$ we deduce that $x[A]=y[A]$, which implies $T[A]\cap F[A]\ne \varnothing$, i.e. $\Sigma (a)=0$. By positivity of $\Sigma$, $\Sigma (s)=0$. $\square$

Proof

(Proposition 3) Let $s\in \max {\mathscr {A}}$, i.e.: (i) $s\in {\mathscr {A}}$ and (ii) $(\forall s'\in {\mathscr {A}},\,s'\ge s\Rightarrow s'=s)$. Since $s\in {\mathscr {A}}$, s belongs to $F_\Sigma$. To show it is a maximal element of $F_\Sigma$, let $s'\in F_\Sigma$ such that $s'\ge s$. Using Lemma 2, $s'\le a$ for some $a\in {\mathscr {A}}$. So $s\le s'\le a$ and from (ii) we deduce that $s=a$, hence $s=s'$. Therefore, $\max {\mathscr {A}} \subseteq \max F_\Sigma$.

Now let $s\in \max F_\Sigma$, i.e.: (i’) $s\in F_\Sigma$ and (ii’) $(\forall s'\in F_\Sigma ,\,s'\ge s\Rightarrow s'=s)$. Since $s\in F_\Sigma$, Lemma 2 implies that $s\le a$ for some $a\in {\mathscr {A}}\subseteq F_\Sigma$. With (ii’) we deduce that $s = a$, so $s\in {\mathscr {A}}$. Now let $s'\in {\mathscr {A}}$ such that $s'\ge s$. Similarly, $s'\in F_\Sigma$ so (ii’) yields $s=s'$, which concludes the proof.

Let p designate the size of (T, F). The computation of ${\mathscr {A}}$ involves $O(p^2n)$ bitwise operations. For a set $X\subseteq {\mathscr {B}}^n$, the enumeration of its maximal elements is in $O(n\vert X\vert ^2)$ in the worst case, ensuring that the complexity of the computation of $\max F_\Sigma$ is indeed polynomial in the size of the instance. $\square$