Modeling the Role of Feedback in the Adaptive Response of Bacterial Quorum Sensing

Abstract

Bacterial quorum sensing (QS) is a form of intercellular communication that relies on the production and detection of diffusive signaling molecules called autoinducers. Such a mechanism allows the bacteria to track their cell density in order to regulate group behavior, such as biofilm formation and bioluminescence. In a number of bacterial QS systems, including V. harveyi, multiple signaling pathways are integrated into a single phosphorylation–dephosphorylation cycle. In this paper, we propose a weight control mechanism, in which QS uses feedback loops to ‘decode’ the integrated signals by actively changing the sensitivity in different pathways. We first use a slow/fast analysis to reduce a single-cell model to a planar dynamical system involving the concentrations of phosphorylated signaling protein LuxU and a small non-coding RNA. In addition to identifying the weight control mechanism, we show that adding a feedback loop can lead to a bistable QS response in certain parameter regimes. We then combine the slow/fast analysis with a contraction mapping theorem in order to reduce a population model to an effective single-cell model, and show how the weight control mechanism allows bacteria to have a finer discrimination of their social and physical environment.

Appendix A

In this appendix, we show that a sufficient condition for the equilibrium of the single-cell model to be unique is given by Eq. (40). We already showed the uniqueness for the cases of $u_1>u_2$ and $u_1=u_2$, in Sect. 2.3. Next, we will work on the case of $u_1<u_2$. Since the equilibrium satisfies $f_1(\phi )=f_2(\phi )$, it follows after some algebra that

$$\begin{aligned} \frac{u_2-h(\phi )}{u_2+1}&=\Bigg [\frac{\phi C_2E_q}{(1+\eta -\eta \xi )\phi +\eta \xi }+ D_2\Bigg ]\frac{h(\phi )-u_1}{1+u_1}\nonumber \\&\equiv f_3(\phi ). \end{aligned}$$

(A.1)

Note that

$$\begin{aligned} \frac{\phi C_2E_q}{(1+\eta -\eta \xi )\phi +\eta \xi }+ D_2=\frac{\phi C_2E_q}{(1+\eta )\phi +\eta \xi (1-\phi )}+ D_2>0. \end{aligned}$$

(A.2)

Here, because the two sides of (A.1) have to be the same sign, we have the following cases,

$$\begin{aligned} u_1<h(\phi )<u_2,\quad u_1=h(\phi )=u_2\quad \text {or}\quad u_1>h(\phi )>u_2. \end{aligned}$$

(A.3)

Since we are studying the case $u_1<u_2$, it follows that $u_1<h(\phi )<u_2$, which implies $f_3(\phi )>0$.

By some algebra on Eq. (A.1), we get

$$\begin{aligned} u_2=f_4(\phi )\equiv \frac{h(\phi )+f_3(\phi )}{1-f_3(\phi )}. \end{aligned}$$

(A.4)

Recall from Eq. (33) that $h(\phi )\ge 0$. Since the AI concentration $u_2\ge 0$, we have $1-f_3(\phi )>0$. Uniqueness is guaranteed by the monotonicity of the function $f_4(\phi )$. By taking the derivative with respect to $\phi$ of $u_2=f_4(\phi )$, we find

$$\begin{aligned} f_4^\prime =\frac{\mathrm{d}f_4}{\mathrm{d}\phi }=\frac{h^\prime (1-f_3)+f_3^\prime (1+h)}{(1-f_3)^2}, \end{aligned}$$

(A.5)

where,

$$\begin{aligned} h^\prime&=\frac{\mathrm{d}h}{\mathrm{d}\phi }=-\frac{Q(\phi -1)^2+K\phi ^2+KQ}{[\phi ^2-(1+K)\phi ]^2}<0, \end{aligned}$$

(A.6a)

$$\begin{aligned} f_3^\prime&=\frac{\mathrm{d}f_3}{\mathrm{d}\phi }=\frac{C_2 E_q \eta \xi }{[(1+\eta -\eta \xi )\phi +\eta \xi ]^2}\frac{h(\phi )-u_1}{1+u_1}\nonumber \\&\quad +\Bigg (\frac{\phi C_2E_q}{(1+\eta -\eta \xi )\phi +\eta \xi }+ D_2\Bigg )\frac{h^\prime (\phi )}{1+u_1}. \end{aligned}$$

(A.6b)

Since $0<f_3<1$ and $1+h>0$, one sufficient condition for $f_4^\prime <0$ to be true is having $f_3^\prime <0$. Also,

$$\begin{aligned} f_3^\prime&<\frac{ C_2E_q \eta \xi }{[(1+\eta -\eta \xi )\phi +\eta \xi ]^2}\frac{h(\phi )}{1+u_1}+\frac{\phi C_2E_q}{(1+\eta -\eta \xi )\phi +\eta \xi }\frac{h^\prime (\phi )}{1+u_1},\nonumber \\&=\frac{ C_2E_q (1+\eta )\phi ^2h^\prime (\phi )}{[(1+\eta -\eta \xi )\phi +\eta \xi ]^2(1+u_1)} +\frac{ C_2E_q \eta \xi (h(\phi )-h^\prime (\phi )\phi ^2+\phi h^\prime (\phi ))}{[(1+\eta -\eta \xi )\phi +\eta \xi ]^2(1+u_1)},\nonumber \\&<\frac{C_2 E_q \eta \xi }{[(1+\eta -\eta \xi )\phi +\eta \xi ]^2(1+u_1)}[h(\phi )-h^\prime (\phi )\phi ^2+\phi h^\prime (\phi )]. \end{aligned}$$

(A.7)

Since $0\le \phi \le 1$, we require $h(\phi )-h^\prime (\phi )\phi ^2+\phi h^\prime (\phi )<0$. By direct substitution, we have

$$\begin{aligned} h(\phi )-h^\prime (\phi )\phi ^2+\phi h^\prime (\phi )=\frac{(\phi -1)[(1+Q+K)\phi -(1+K)]}{(\phi -1-K)^2}. \end{aligned}$$

(A.8)

Therefore, Eq. (40) is a sufficient condition for the response to be unique.