Cell cycle dynamics: clustering is universal in negative feedback systems

Abstract

We study a model of cell cycle ensemble dynamics with cell–cell feedback in which cells in one fixed phase of the cycle $S$ (Signaling) produce chemical agents that affect the growth and development rate of cells that are in another phase $R$ (Responsive). For this type of system there are special periodic solutions that we call $k$-cyclic or clustered. Biologically, a $k$-cyclic solution represents $k$ cohorts of synchronized cells spaced nearly evenly around the cell cycle. We show, under very general nonlinear feedback, that for a fixed $k$ the stability of the $k$-cyclic solutions can be characterized completely in parameter space, a 2 dimensional triangle $T$. We show that $T$ is naturally partitioned into $k^2$ sub-triangles on each of which the $k$-cyclic solutions all have the same stability type. For negative feedback we observe that while the synchronous solution ($k=1$) is unstable, regions of stability of $k \ge 2$ clustered solutions seem to occupy all of $T$. We also observe bi-stability or multi-stability for many parameter values in negative feedback systems. Thus in systems with negative feedback we should expect to observe cyclic solutions for some $k$. This is in contrast to the case of positive feedback, where we observe that the only asymptotically stable periodic orbit is the synchronous solution.

Appendix A: verification for \(k=3\)

Referring to the labeling in Fig. 6, we present calculations verifying rigorously the conclusions of the numerically generated plot in Fig. 9 for the case $k=3$.

Regions 2, 6, 7:

In Fig. 9 for the case $k=3$ we see that regions 2 and 7 have neutral stability of the 3 cohort cyclic solution and in region 6 it is stable. These three regions are covered by Theorem 2.4.

Region 1:

We wish to confirm that for all parameter values in this region, the $3$ cohort cyclic solution is neutrally stable.

For parameter values in region 1 both $x_1$ and $x_2$ are initially in $S$. The order of events is that first $x_2$ reaches $s$, then it reaches $r$ and finally it reaches $1$. The speed of $x_2$ will be $1$ until it reaches $r$. While in $R$ the cohort $x_2$ will experience feedback due to $x_1, x_1 \in S$, and so it’s speed will be $1+\beta _2$. The time required for $x_2$ to reach $1$ is thus:

$$\begin{aligned} t^* = r-x_2 + \frac{1}{1+\beta _2} (1-r). \end{aligned}$$

The positions of $x_1$ and $x_1$ at this time will be:

$$\begin{aligned} x_1(t^*) = t^* \quad x_1(t^*) = x_1 + t^*. \end{aligned}$$

We observe that this map is affine in the variables $x_1$, $x_2$. Thus the derivative of the map $F$ in a neighborhood of the initial condition of the cyclic solution is:

$$\begin{aligned} DF = \left( \begin{array}{c@{\quad }c} 0 &{} -1 \\ 1 &{} -1 \end{array} \right) . \end{aligned}$$

(7.1)

The eigenvalues of this matrix are $-\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$ which implies that the solution is linearly neutrally stable. Since the map is affine in a neighborhood of the initial condition, the cyclic solution is neutral.

Region 3:

In this region we wish to confirm that the 3 cyclic solution is stable.

For these parameter values first $x_2$ reaches $r$, then $x_1$ reaches $s$ and finally $x_2$ reaches $1$. Thus $x_2$ will experience feedback of $\beta _2$ for the times $r-x_2 < t < s - x_1$. For $t > s-x_1$ it will be subject to feedback $\beta _1$. The trajectory of $x_2$ is thus:

$$\begin{aligned} x_2(t) = {\left\{ \begin{array}{ll} x_2 + t &{}\quad \text {for }\quad 0 \le t \le r - x_2, \\ r + (1+\beta _2)(t - r+x_2) &{}\quad \text {for }\quad r-x_2 \le t \le s - x_1, \\ r + (1+\beta _2)(s - x_1 - r+x_2) &{}\quad \text {for }\quad s-x_1 \le t \le t^* .\\ \quad + (1 + \beta _1)(t - s + x_1) \end{array}\right. } \end{aligned}$$

It follows that $t^*$ satisfies:

$$\begin{aligned} r + (1+\beta _2)(s - x_1 - r+x_2) + (1 + \beta _1)(t ^*- s + x_1) = 1 \end{aligned}$$

and so

$$\begin{aligned} t^* = \frac{\beta _2 - \beta _1}{1+ \beta _1} x_1 - \frac{1 + \beta _2}{1 + \beta _1} x_2. \end{aligned}$$

The local affine map is given by:

$$\begin{aligned} x_1(t^*) = t^*, \qquad x_1(t^*) = x_1 + t^*, \end{aligned}$$

which gives us:

$$\begin{aligned} DF = \left( \begin{array}{c@{\quad }c} \frac{\beta _2 - \beta _1}{1+ \beta _1} &{} - \frac{1 + \beta _2}{1 + \beta _1} \\ 1+ \frac{\beta _2 - \beta _1}{1+ \beta _1} &{} - \frac{1 + \beta _2}{1 + \beta _1} \end{array} \right) = \left( \begin{array}{c@{\quad }c} \frac{\beta _2 - \beta _1}{1+ \beta _1} &{} - \frac{1 + \beta _2}{1 + \beta _1} \\ \frac{1+ \beta _2}{1+ \beta _1} &{} - \frac{1 + \beta _2}{1 + \beta _1} \end{array} \right) . \end{aligned}$$

(7.2)

The characteristic polynomial for this matrix simplifies to:

$$\begin{aligned} \lambda ^2 + \lambda + \frac{1 + \beta _2 }{1+\beta _1} = 0. \end{aligned}$$

One can check that the roots of this polynomial are less than 1 in modulus if and only if $\beta _2 < \beta _1$. This is the case for monotone negative feedback and so the cyclic solution is stable.

Region 4:

We confirm that the 3-cyclic solution is neutrally stable in this region. We first compute the time it takes $x_2$ to reach 1; $x_2$ travels with speed $(1+\beta _1)$ until time $t=s-x_1$, then travels at speed $(1+\beta _1)$, allowing us to set up the equation:

$$\begin{aligned} 1 = x_2 + (1+\beta _2)(s-x_1) + (1+\beta _2)(t^* - s + x_1). \end{aligned}$$

Solving, we find $t^* = \displaystyle \frac{1+(\beta _1 - \beta _2)s}{1+\beta _1} - \displaystyle \frac{1}{1+\beta _1}x_2 + \displaystyle \frac{\beta _2 - \beta _1}{1+\beta _1}x_1$.

cohort $x_1$ travels at speed 1; $x_1$ travels at speed 1 until $t=r-x_1$, then at speed $(1+\beta _1)$:

$$\begin{aligned} x_1(t^*) = t^* \quad x_1(t^*) = r+(1+\beta _1)(t^* - r + x_1) \end{aligned}$$

Gathering for convenience the constant terms as a single $C$, $x_1(t^*) = (1+\beta _1)x_1 + (\beta _2 - \beta _1)x_1 - x_2 + C= (1+\beta _2)x_1 - x_2 + C$; thus

$$\begin{aligned} DF = \left( \begin{array}{c@{\quad }c} \frac{\beta _2 - \beta _1}{1+\beta _1} &{} -\frac{1}{1+\beta _1} \\ 1+\beta _2 &{} -1 \end{array} \right) . \end{aligned}$$

The characteristic polynomial of DF is given by $\lambda ^2 + \lambda (\frac{\beta _1 - \beta _2}{1+\beta _1} + 1) + 1$. To simplify the following expressions, we denote $w = \frac{\beta _2 - \beta _1}{1+\beta _1}$. The roots of the polynomial are then $\lambda _{\pm }=\frac{-w-1 \pm \sqrt{(w+1)^2 - 4}}{2}$. Observing that $w<1$ iff $-1<\beta _2$, which is true, we see that the roots are complex. We compute the norm of $\lambda _{+}$ as $\sqrt{\frac{1}{4}w^2 + \frac{1}{4}(4-w^2)} = 1$

Region 5:

We wish to confirm that the 3 cyclic solution is neutrally stable in this region.

For this region $x_1$ reaches $s$ before $x_1$ or $x_2$ reach $r$. This implies that no cell actually experiences any feedback. The map $F$ in a neighborhood of the cyclic solution is thus given by the trivial affine expression:

$$\begin{aligned} x_1(t^*) = t^* = 1-x_2, \quad x_1(t^*) = x_1 +1 - x_2. \end{aligned}$$

The derivative $DF$ is then the same as (6.1) and the solution is thus neutrally stable.

Region 8:

For parameter values in this region $x_2$ travels with speed $1+\beta _1$ until time $t=s$, then at speed 1. The time it takes to reach 1 can thus be calculated:

$$\begin{aligned} 1 = x_2 + (1+\beta _1)s+t^* - s \rightarrow 1 - x_2 - \beta _1 s = t^*. \end{aligned}$$

Clearly $x_1(t^*) = t^*$; bearing in mind that $x_1$ enters $R$ before $x_1$ leaves $S$, we calculate:

$$\begin{aligned} x_1(t^*) = r+ (1+\beta _1)(s-r+x_1)+t^* - s. \end{aligned}$$

This gives rise to the matrix

$$\begin{aligned} DF = \left( \begin{array}{c@{\quad }c} 0 &{} -1\\ 1+\beta _1 &{} -1 \end{array} \right) , \end{aligned}$$

yielding a characteristic polynomial of $\lambda ^2 + \lambda + (1+\beta _ 1)$ and roots of $\lambda _{\pm } = \frac{-1 \pm \sqrt{-3 - 4\beta _1}}{2}$. If the roots are real, the maximum root (in absolute value), $\lambda _{-}$, is clearly contained in the interval $(-1,-\frac{1}{2})$; if the roots are complex, their norm can be explicitly calculated as $\sqrt{1+\beta _1}$.

Region 9:

cohort $x_2$ travels at rate $(1+\beta _1)$ until time $t=s$, whereafter it travels with speed 1. We can then calculate the time it takes $x_2$ to reach 1:

$$\begin{aligned} 1 = x_2 + (1+\beta _1)s+t^* -s, \end{aligned}$$

yielding $t^* = 1 - x_2 - \beta _1 s$.

cohort $x_1$ experiences feedback exactly when $x_2$ does, and thus travels the same distance. The local affine map is given by:

$$\begin{aligned} x_1(t^*) = t^* \quad x_1(t^*) = x_1+1 -x_2. \end{aligned}$$

The resultant map DF is exactly that in (A.1), so the solution is neutrally stable.