Quiescence, Activation, and Return to Quiescence

Why Cell-State Fractions Need Their Own Model

A quiescent cell can activate and enter the cycling population, while a cycling cell can return to quiescence. Cycling cells may also divide or die, and quiescent cells may die. These processes change both the number of cells and the fraction in each state. A constant fraction therefore does not, by itself, mean that the total population is constant: both state counts might be growing, or both might be declining, while their ratio remains stable.

The model below represents the two states with constant rates. It follows the cell-count accounting used in turnover models, but focuses on the long-run balance of quiescent and cycling cells. The key step is to derive an equation for the cycling fraction from the state-count equations rather than assuming that switching alone determines that fraction.

Definition: Let \(Q(t)\) be the number of quiescent cells and \(C(t)\) the number of cycling cells at time \(t\). Let \(a\) be the activation rate from quiescence to cycling, \(r\) the return rate from cycling to quiescence, \(\beta\) the division-event rate per cycling cell, and \(\mu_Q\) and \(\mu_C\) the death hazards in the two states. All rates are nonnegative and have units of inverse time. In this model, division replaces one cycling mother cell with two cycling daughters, so each event produces a net increase of one cycling cell.

State-Count Equations and Assumptions

Activation removes cells from \(Q\) at rate \(aQ\) and adds them to \(C\) at the same rate. Return does the reverse, at rate \(rC\). Death removes cells from the state in which it occurs. A division event in \(C\) has a net effect of adding one cycling cell, so its contribution to the cycling-cell count is \(\beta C\), not \(2\beta C\).

\frac{dQ}{dt}=-(a+\mu_Q)Q+rC, \qquad \frac{dC}{dt}=aQ+(\beta-\mu_C-r)C.

These equations assume a well-mixed population; constant rates that do not depend on cell counts or time; and per-cell event rates that apply equally to all cells within a state. Transitioning cells remain in the same modeled lineage. Division produces two cycling daughters, while death removes one cell. There is no external influx, and no additional density-dependent feedback or delay is represented. If, for example, division produces quiescent daughters, the division terms must be reassigned between the equations.

For positive activation and return rates, \(a>0\) and \(r>0\), both states can be reached from the other. The analysis below uses this two-way switching assumption when discussing a unique long-run composition. If either rate is zero, a state may become inaccessible or absorbing, and boundary compositions need separate consideration.

Worked Example: Writing the Cell-Count Equations

Suppose a hypothetical cell population has \(a=0.08\) per day, \(r=0.12\) per day, \(\beta=0.20\) per day, \(\mu_Q=0.05\) per day, and \(\mu_C=0.10\) per day. Write the model and evaluate the instantaneous changes when \(Q=600\) cells and \(C=400\) cells.

Using the stated assumptions, the quiescent-cell change is \(dQ/dt=-(0.08+0.05)Q+0.12C\). At the given counts, \[ \frac{dQ}{dt}=-(0.13)(600)+(0.12)(400)=-78+48=-30 \] cells per day.

For cycling cells, the net division contribution is \(\beta C\), and the full equation is \(dC/dt=0.08Q+(0.20-0.10-0.12)C\). Thus, \[ \frac{dC}{dt}=(0.08)(600)+(-0.02)(400)=48-8=40 \] cells per day. The total population is changing at \(d(Q+C)/dt=-30+40=10\) cells per day at this instant. Neither state count is at equilibrium, and the total’s instantaneous increase does not establish whether its long-run growth will be positive.

Deriving the Cycling-Fraction Equation

Define the total count \(N=Q+C\) and the cycling fraction \(x=C/N\), for \(N>0\). Then the quiescent fraction is \(1-x=Q/N\). The total count changes according to the net effects of death and division; activation and return only transfer cells between states:

\frac{dN}{dt}=-\mu_Q Q+(\beta-\mu_C)C.

To find how the composition changes, differentiate \(x=C/N\) using the quotient rule. Substituting the equations for \(dC/dt\) and \(dN/dt\), and writing \(Q/N=1-x\) and \(C/N=x\), gives

\begin{aligned} \frac{dx}{dt} &=\frac{1}{N}\frac{dC}{dt} -x\frac{1}{N}\frac{dN}{dt}\\ &=a(1-x)+(\beta-\mu_C-r)x -x\left[-\mu_Q(1-x)+(\beta-\mu_C)x\right]\\ &=a(1-x)-rx+(\beta-\mu_C+\mu_Q)x(1-x). \end{aligned}

The first two terms represent switching: activation raises the cycling fraction and return lowers it. The final term reflects the difference in net growth between the states. Write \(\delta=\beta-\mu_C+\mu_Q\) for this contrast. It compares cycling-cell net growth, \(\beta-\mu_C\), with quiescent-cell net growth, \(-\mu_Q\). A steady composition \(x^*\) satisfies

0=a(1-x^*)-rx^*+\delta x^*(1-x^*).

When \(a>0\) and \(r>0\), the right-hand side is positive at \(x=0\) and negative at \(x=1\). The model’s two-state linear system has a dominant real eigenvalue with an eigenvector whose entries are both positive. Its normalized entries give the long-run state fractions. Equivalently, \(x^*\) is the unique root of the fraction equation between zero and one; perturbations of the fraction toward either side move it back toward that root. The root describes composition, not necessarily a nonzero equilibrium of the cell counts.

Key result: Under constant rates, positive two-way switching, and the stated division-daughter assumption, the long-run cycling fraction is the root \(x^*\) in \(0<x^*<1\) of \(a(1-x^*)-rx^*+(\beta-\mu_C+\mu_Q)x^*(1-x^*)=0\). The quiescent fraction is \(1-x^*\). The total population can still grow or decline.

Worked Fraction Calculations

Worked Example: Switching With Equal Net Growth Across States

Consider hypothetical rates \(a=0.06\), \(r=0.14\), \(\beta=0.10\), \(\mu_C=0.20\), and \(\mu_Q=0.10\), all per day. Here cycling-cell net growth is \(\beta-\mu_C=-0.10\) per day, equal to quiescent-cell net growth \(-\mu_Q=-0.10\) per day. Therefore \(\delta=0\), and the fraction equation reduces to \(a(1-x)-rx=0\).

Solving gives \[ x^*=\frac{a}{a+r}=\frac{0.06}{0.06+0.14}=0.300. \] The long-run composition is 30.0% cycling and 70.0% quiescent. The total population’s per-capita growth rate at this composition is \[ -\mu_Q(1-x^*)+(\beta-\mu_C)x^* =(-0.10)(0.70)+(-0.10)(0.30)=-0.10 \] per day. Thus the composition can be stable while the population declines. Under this linear model, the total count declines exponentially at 0.10 per day once the composition is stable.

Worked Example: Cycling Cells Have a Net Growth Advantage

Suppose hypothetical rates are \(a=0.10\), \(r=0.20\), \(\beta=0.30\), \(\mu_C=0.10\), and \(\mu_Q=0.10\) per day. Then \(\delta=0.30-0.10+0.10=0.30\) per day. The steady-fraction equation becomes \[ 0=0.10(1-x)-0.20x+0.30x(1-x)=0.10-0.30x^2. \]

The root in the physical range is \[ x^*=\sqrt{\frac{0.10}{0.30}}=\sqrt{\frac{1}{3}}=0.5774. \] The long-run cycling fraction is about 57.7%, and the quiescent fraction is \(1-0.5774=0.4226\), or 42.3%. The per-capita growth rate of the total population at that composition is \[ -\mu_Q(1-x^*)+(\beta-\mu_C)x^* =(-0.10)(0.4226)+(0.20)(0.5774)=0.07321 \] per day. The population therefore grows in this example, even though some cells die in both states. The positive total growth reflects the division contribution; it does not alter the conclusion that the composition has a steady value.

When Is the Total Count Actually at Equilibrium?

A stationary total count requires its net growth to be zero. A stationary two-state population requires both state-count derivatives to be zero, which is more restrictive than requiring a stationary fraction. For a nonzero stationary state, both \(dQ/dt=0\) and \(dC/dt=0\) must hold. In a linear constant-rate model without density dependence, such a nonzero equilibrium generally requires a precise balance of rates; if it exists, its overall scale is not set by the equations.

Worked Example: A Balanced Nonzero Stationary Population

Use the hypothetical rates \(a=0.10\), \(r=0.20\), \(\mu_Q=0.10\), \(\beta=0.20\), and \(\mu_C=0.10\) per day. The cell-count equations are \[ \frac{dQ}{dt}=-0.20Q+0.20C,\qquad \frac{dC}{dt}=0.10Q-0.10C. \] Setting each derivative to zero gives \(C=Q\) from either equation. Hence \(x^*=C/(Q+C)=0.5\): half the cells are cycling and half are quiescent. For example, \(Q=C=500\) gives a stationary total of 1,000 cells, since both derivatives are zero.

This example has \(\delta=0.20-0.10+0.10=0.20\) per day. Substitution into the fraction equation confirms the composition: \[ 0.10(1-0.5)-0.20(0.5)+0.20(0.5)(1-0.5) =0.05-0.10+0.05=0. \] The equations permit any stationary scale with \(Q=C\); they do not select 500 cells in each state. A separate homeostatic or density-dependent mechanism would be needed to determine a particular population size.

Diagnostics, Alternatives, and Limitations

A practical check is to calculate both the fraction dynamics and total growth at the proposed composition. The fraction equation should equal zero at \(x^*\), and the total per-capita growth rate should equal \[ -\mu_Q(1-x^*)+(\beta-\mu_C)x^*. \] If this second quantity is not zero, the cell counts are not stationary, even though their fractions may be. Also check that all stated rates are nonnegative, \(0\leq x^*\leq1\), and the units match. A computed fraction outside this range signals an algebra or parameter-substitution error, or that the assumed model conditions do not hold.

When net growth is equal in the two states, \(\delta=0\), the fraction is the switching balance \(a/(a+r)\). This simpler expression is not generally valid when division and death create different net growth across states: the additional \(\delta x(1-x)\) term then shifts the composition. Conversely, solving only for the switching balance can be misleading even when activation and return rates are known precisely.

The state-count equations are useful when absolute counts and their rates of change matter. The fraction equation is more direct when the question is composition, because it removes population scale. Neither approach can make constant rates appropriate if the rates change substantially over time or depend on population size. In such settings, a fixed long-run fraction may not exist. The model also cannot represent multiple cycling substates, delays in activation, or division that places daughters in different states without adding states or changing the flux terms.

A common accounting error is to add \(2\beta C\) to the cycling equation because each division produces two daughters. The mother is already counted before division; replacing one mother with two daughters increases the count by one. Thus the net contribution is \((2-1)\beta C=\beta C\). Another error is to interpret a stable fraction as evidence of a stationary population. The total-growth calculation is a separate, necessary check.

Check Your Understanding

Use the model definitions above. Rates are per day unless otherwise stated.

If \(a=0.04\), \(r=0.16\), \(\beta=0.10\), and \(\mu_C=\mu_Q=0.10\), find the steady cycling fraction and say whether the total population grows, declines, or remains constant at that composition.
For \(a=0.10\), \(r=0.10\), \(\beta=0.20\), \(\mu_C=0.10\), and \(\mu_Q=0.10\), write the fraction equation and identify its root in the physical range.
Why does a nonzero steady fraction not prove that \(Q\) and \(C\) are individually constant?
In this model, what is the net count contribution of cycling-cell divisions, and why?

Solutions. (1) Here \(\delta=0.10-0.10+0.10=0.10\), so the equation is \(0=0.04-0.20x+0.10x(1-x)=0.04-0.10x-0.10x^2\). Rearranging gives \(x^2+x-0.4=0\), whose physical root is \((-1+\sqrt{2.6})/2=0.3062\). Total growth is \((-0.10)(1-0.3062)+0(0.3062)=-0.06938\) per day, so the population declines. (2) Here \(\delta=0.20\), giving \(0=0.10(1-x)-0.10x+0.20x(1-x)=0.10-0.20x^2\). Thus \(x^*=\sqrt{0.5}=0.7071\). (3) A steady fraction only means that the ratio \(C/(Q+C)\) is stable. Both counts can still change at the same proportional rate; stationarity of the counts additionally requires both state-count derivatives to be zero. (4) The net contribution is \(\beta C\): each event replaces one cycling mother with two cycling daughters, increasing the count by one, not two.