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Dynamics in a chemostat with an internal inhibitor and resistance mechanisms

We consider a competition model for three different strains wild, mutant and variant bacteria induced by an inhibitor created by one of the strains. We obtain two steady state regimes and we show that is possible the persistence of the three bacteria in the system.


Introduction
The Chemostat models are been used for modelling competition between species, Lenski and Hattingh [1], Hsu, Hubbel and Waltman [2], Hsu, Li and Waltman [3] and Hsu and Waltman [4] have studied competition between species in the presence of an inhibitor, the principal cause of the competition phenomenon.
In the environment, strains still sensitive to the inhibitor may persist [5]. Dominance by the resistant survivor strains, is directly linked to the "strength" of the intrinsic or acquired resistance to the inhibitor being used [6]. In [7], we considered two resistance mechanisms, phenotypic switching and mutation, assuming that cell replication does not occur in the phenotypic variant. Since it is characterized by neither growth nor significant reproduction. We generalized those results with the added assumption of the existence of a fraction of phenotypic variants capable of reproduction and we found the conditions to guarantee the persistence of the three bacteria in the system. These results are not according with Gause's law of competitive exclusion.
For these reason and considering that in nature it is known that micro-organisms produce inhibitors against their rivals [8], producing a selective medium in the same sense as the external inhibitor only "naturally", our problem setup is the following: in a constant environment, where resources are supplied at a constant rate, we introduce mutation into the dynamics of the bacterial population, an internal selective medium and as a consequence of this phenotypic switching, then we consider the competition of all bacterial strains for resources. Section 2 presents the construction of the model, Section 3 contains the model analysis. In Section 4 we present our numerical results and the conclusions are given in Section 5. In Appendix A we present some special cases; proofs are found in Appendix B.

General Model
Let S(t) denote the concentration of nutrient in the environment at time t, u(t) the concentration of wild-type bacteria sensitive to a biocide, v 0 (t) denotes a mutated strain concentration and the producer of the biocide whose concentration is denoted by p(t). Wild bacteria may mutate in a new mutated strain or undergo phenotypic switching to avoid the action of the biocide; to avoid the use of more variables, we assume the new mutated bacteria as the same v 0 (t) and v 1 (t) indicates the density of the phenotypic variant at time t. Our model takes the form where, we consider that nutrient is supplied at a constant rate S 0 D and is eliminated either by washout at a rate SD, or by consumption, through a Michaelis-Menten function with constants a i and m i , i = 1, 2. Strain u is sensitive to the biocide so its rate of consumption is decreased by the inhibitor. This effect is represented by the function φ(p), the degree of inhibition that the biocide exerts upon strain u, given by φ(p) = exp(−λp) [1]. The mutation rate is µ and the phenotypic switch occurs at a rate α(1 − φ(p)); mutation results in the increase of the resistant strain numbers v 0 and the phenotypic switching results in the population growth of the phenotypic variant v 1 . The parameter δ represents the portion of the nutrient consumption by v 0 that has been allocated to the production of the biocide, in consequence the growth rate corresponding has been debilitated. We generalize our previous model by assuming that the phenotypic variant reproduces and that a fraction γ regress to the wild stage. Finally, p indicates the concentration of biocide that is lost by washout or by absorption by resistant bacteria v 1 . Let As in [3], [4], [9], [10], letS = S/S (0 The parameters γ and δ are dimensionless. System (2), becomes (dropping the bars) Thus, the bacteria population is bounded. Moreover, at any omega limit point S + u + v 0 + v 1 + p ≤ 1, that is, the right hand side of (3) is bounded.
In Appendix A we analyze the special cases for γ, that is, the case when phenotypic strain is able to have total replication (γ = 0) and the case when there is not replication (γ = 1).
We claim that E 0 = (1, 0, 0, 0, 0) ∈ Ω, the bacteria-free equilibrium point is globally asymptotically stable as it will be proven shortly. Following [4], [10], using (3c) and (3d) we obtain the break-even parameters for v 0 and v 1 respectively: There is another equilibrium point given by By calculating the variational matrix of (3) and substituting into E 0 and E 1 we obtain Lemma 1.
-The equilibrium E 1 is locally asymptotically stable if it exists (0 < λ 2 < 1), and if where χ 1 and χ 2 are the positive real roots of The proof is a straight-forward computation (see Appendix B).
From Lemma 1 we see that competitive exclusion may take place. In order to prove the existence of a non trivial interior equilibrium point (one where all coordinates are positive), we consider equation (3e) and we assume that S, v 1 and v 0 are biologically feasible then we get that p is a root of and we can note that independently of the sign on the linear coefficient, we just have a positive solution (see Figure 1). In particular, for some fixed values for S, v 0 and v 1 we obtain a unique positive value (see Figure 2)

Global stability
Our model has two equilibrium points E 0 and E 1 , and depending on the biological feasibility of S, v 1 and v 0 , a third non-trivial one E c . When E c does not exist and λ 2 > 1 (that is, E 1 does not exist), then the unique equilibrium point is E 0 and in consequence, if λ 1 > 1 and Theorem 1. If 0 < λ 2 < 1, E c does not exist, the break-even parameter for γ = 0 that is m 1λ1 a 1 +λ 1 = 1 is such that either of the following two conditions hold where λ 2 is defined as in (5), and either of the following hold Conditions (i) and (ii) require the nutrient concentration of the phenotypic strain to be at equilibrium (λ 1 ). For the interpretation of (iii) and (iv) let f 1 = Sm 1 a 1 +S be the phenotypic variant consumption function and f 2 = Sm 2 a 2 +S be the one for the mutant strain. Then, (iii) implies f 2 > f 1 ; in case (iv) the behavior of f 1 and f 2 depends upon if m 1 > m 2 or m 1 < m 2 . Now, we will prove the theorem.
Proof. Consider the Lyapunov function Observe that Applying the same procedure to the second integral and given that c 1 , c 2 > 0 we have where So, defining c 2 = m 2 λ 2 a 2 we obtain A(S) < 0. Moreover, by replacing this value into D(S) we get D(S) = 0. Now, B(S, p) depends on c 1 and c 2 and thus we need to consider the cases c 1 = c 2 and c 1 = c 2 .
• For the case c 1 = c 2 , we obtain By assumption (iii) the expression within brackets in C(S) is positive for all S ∈ (0, 1). Therefore, C(S) and B(S, p) are negative, and thus we haveV ≤ 0 for all (S, u, v 0 , v 1 , p) ∈ ∆.
Numerical simulation of (3) shows the stability of the equilibrium point given by     Figure 5. Stability of u and v 1 . We note that the stability value for u is highly sensitive to changes in δ; for δ ∈ (0, 0.5) u tends to zero faster than for δ ∈ (0.5, 1). This behavior does not occur for v 1 that regardless on the value of δ tends to zero at the same rate.  Figure 6. Stability of v 0 . We can see that changes in δ are inversely proportional to the stability value of v 0 , this is because when δ tends to 1, the strain v 0 will increase due to mutation of u (see equation (3c)) which tends to zero.  Figure 7. Stability of p. Different from S, we observe that if δ ∈ (0, 0.5) then p has an equilibrium value different from zero, however, if δ ∈ (0.5, 1) then the equilibrium value for p tends to zero .  Figure 8. Numerical simulation of the non trivial equilibrium point. We note that equilibrium value of u is bigger than equilibrium values of v 0 and v 1 , it means that the cost on the inhibitor production is bigger than the inhibition on u growth.

Discusion and conclusion
Competition models in a chemostat in the presence of an inhibitor can serve as a model system for ecological processes. For example, we can think the resistant organism as a detoxifying agent in the environment. From the standpoint of competition, the question is whether the detoxifying organisms survive in the environment [10]. In contrast to the models studied in [2], [3], [4], [10], we consider three strains in competition under the effects of an internal inhibitor; one of the strains is a wild strain on which two different mechanisms of resistance, mutation and phenotypic switching are acting. We have proven that both resistant types may coexist together with the wild type: there is no competitive exclusion. This result is not general since it depends on parameter values (in fact, in our last simulation we observe that the cost on the inhibitor production is bigger than its inhibition on u growth but we can not proof this in general). With an appropriate selection of them, competitive exclusion can be obtained and, as before, the mutant strain is the one that survives in the environment.

Appendix A
A.1 Special case γ = 0. This special case of (3) considers that phenotypic strain is able to have replication as a whole and, as a consequence, we obtain the wash out of the wild strain.
When we consider in (3) the special case γ = 0 we get the model: About this model we get an important result: There is no non-trivial equilibrium point with all coordinates different to zero for the system (A.1).

Appendix B
Proof. of lemma 1 The Jacobian matrix of the system is, for a general point E = (S, u, v 0 , v 1 , p) in Ω, is given by: Where A 11 = 1+ a 1 m 1 (a 1 + S) 2 uφ(p)+v 1 + a 2 m 2 (a 2 + S) 2 v 0 A 21 = a 1 m 1 (a 1 + S) 2 uφ(p)+ γv 1 Replacing E 0 we obtain the characteristic polynomial in x: So, E 0 is locally asymptotically stable if and only if 1 < min {λ 1 , λ 2 } and m 1 a 1 + 1 < µ + 1 On the other hand, replacing E 1 into the Jacobian matrix, we obtain the characteristic polynomial: So, E 1 will be locally stable if the roots of x 2 + Ax + B have negative real part, that is A > 0 and B > 0. From  Now, we observe that B is a quadratic polynomial for χ whose determinant is given by then B has two positive real roots (χ 1 and χ 2 ) and in consequence will be positive if and only if 0 < χ < χ 1 or χ > χ 2 . (A.7) Therefore, from (A.6) and (A.7), E 1 will be locally stable if and only if 0 < χ < min χ 1 , µ + 2 + α(1 −φ)