Persister Subpopulations and Relapsing Bacterial Burden

Why a Small Dormant Population Can Matter

Bacterial burden may fall substantially during treatment and later rise again. A model with only one bacterial state can describe the observed total, but it cannot represent a surviving subpopulation that contributes bacteria after treatment pressure changes. A useful QSP model distinguishes actively growing bacteria from persisters: bacteria in a reversible, relatively slow-growing or dormant physiological state that can survive conditions that remove many actively growing cells.

The distinction is about state, not necessarily a permanent genetic difference. A persister can reactivate and resume growth; a newly growing cell can also enter the persister state. Thus, a later increase in burden may be generated by bacteria already present in the host, rather than requiring a new exposure. The model can represent this possibility, but the trajectory alone does not prove that persistence caused a recurrence.

Definition: Let \(G(t)\) be the burden of actively growing bacteria and \(P(t)\) the burden of persister bacteria at time \(t\). Let \(k_{GP}\) be the rate per day at which growing bacteria enter the persister state, and \(k_{PG}\) the rate per day at which persisters reactivate and enter the growing state. Both are nonnegative transition rates. Let \(a_G(t)\) and \(a_P(t)\) be the net per-capita growth rates, in inverse days, within each state, excluding switching. A positive \(a\) means net growth; a negative \(a\) means net loss.

For a simple model, assume that switching and net growth or loss are first-order processes; rates are constant within each specified treatment phase; and the two states are well mixed, so each cell in a state has the same per-capita rates. Assume no new bacterial input after the initial infection, no density dependence, and no explicit spatial or host-cell compartments. The growing and persister burdens can be measured in the same units, such as viable CFU per milliliter.

A Two-State Model with Reversible Switching

The equations below separate changes within each state from transitions between states. The transition terms leaving one state appear as gains in the other state. An external input \(u(t)\), if included, represents newly introduced growing bacteria per unit time; setting \(u(t)=0\) represents no new inoculation during the modeled period.

\frac{dG}{dt}=[a_G(t)-k_{GP}]G+k_{PG}P+u(t), \qquad \frac{dP}{dt}=k_{GP}G+[a_P(t)-k_{PG}]P.

The terms \(-k_{GP}G\) and \(+k_{GP}G\) describe the same movement from growing to persister bacteria. Likewise, \(+k_{PG}P\) and \(-k_{PG}P\) describe reactivation. Adding the equations shows that switching alone does not create or remove bacteria from the total burden \(B(t)=G(t)+P(t)\):

\frac{dB}{dt}=a_G(t)G+a_P(t)P+u(t).

This cancellation is a useful check on model construction. If the switching terms do not cancel in the total-burden equation, a transition has likely been counted as a gain without its corresponding loss, or vice versa. Switching can still strongly affect the future total: it changes which net growth or loss rate applies to each bacterium.

Worked Example: How Many Persisters Reactivate Before They Die?

Consider a hypothetical treatment phase in which a persister can either reactivate at rate \(k_{PG}=0.080\) per day or die at rate \(0.020\) per day. For this calculation, treat those as competing constant hazards and ignore return to the persister state after reactivation. Find the probability that a persister reactivates within 14 days before dying.

The probability that neither event has occurred by time \(t\) is \(e^{-(0.080+0.020)t}\). The probability density for reactivation at time \(t\), before either event has occurred, is \(0.080e^{-0.100t}\). Therefore:

P(\text{reactivation by }T\text{ before death}) =\int_0^T 0.080e^{-0.100t}\,dt =\frac{0.080}{0.100}\left(1-e^{-0.100T}\right).

At \(T=14\) days, this is \(0.8(1-e^{-1.4})=0.8(1-0.2466)=0.6027\). Under these assumptions, about 60.3% of persisters reactivate within 14 days before dying. This is a probability for an individual persister in the stated competing-hazard calculation, not the fraction of total bacterial burden that must rebound.

Treatment, Reactivation, and Relapse-Like Increases

Treatment can change \(a_G(t)\) and \(a_P(t)\), as well as the switching rates. For instance, a drug may make \(a_G\) strongly negative while having a smaller effect on bacteria in the persister state. During treatment, surviving persisters may reactivate, adding bacteria to \(G\), even while the growing population is declining overall. After treatment pressure is reduced, \(a_G\) may become positive; reactivated bacteria can then expand. In this account, the recurrent burden is seeded by the persister population already present, without requiring a new initial infection.

A linear model with constant rates in a treatment phase can be written as a matrix system. Its long-term behavior is governed by the eigenvalues of the rate matrix. The eigenvalue with the largest real part determines the dominant exponential growth or decline mode, provided the corresponding mode is present in the initial state. If that eigenvalue is positive, the associated mode grows; if it is negative, it declines.

Worked Example: Estimating the Post-Treatment Rebound Rate

Suppose, hypothetically, that after treatment is stopped, growing bacteria have \(a_G=0.60\) per day and persisters have \(a_P=-0.020\) per day. Let \(k_{GP}=0.030\) per day and \(k_{PG}=0.080\) per day. Assume constant rates, no new inoculation, and that the measured burden is well represented by these two states. The rate matrix is:

A= \begin{pmatrix} 0.60-0.030 & 0.080\\ 0.030 & -0.020-0.080 \end{pmatrix} = \begin{pmatrix} 0.570 & 0.080\\ 0.030 & -0.100 \end{pmatrix}.

For a two-state matrix, the eigenvalues satisfy \(\lambda^2-\operatorname{tr}(A)\lambda+\det(A)=0\). Here, \(\operatorname{tr}(A)=0.470\) per day and \(\det(A)=(0.570)(-0.100)-(0.080)(0.030)=-0.0594\) per day squared. Thus:

\lambda^2-0.470\lambda-0.0594=0, \qquad \lambda_{\max} =\frac{0.470+\sqrt{(0.470)^2+4(0.0594)}}{2} =\frac{0.470+\sqrt{0.4585}}{2} \approx 0.5736\ \text{per day}.

The dominant mode therefore grows. Its implied doubling time is \(\ln(2)/0.5736=1.208\) days, if the dominant exponential mode describes the burden over the interval considered. For example, if \(G(0)=0\) and \(P(0)=100\) CFU per milliliter, the initial derivatives are \(dG/dt=0.080(100)=8.00\) CFU per milliliter per day and \(dP/dt=-0.100(100)=-10.0\) CFU per milliliter per day. Reactivation initially supplies growing bacteria even as the persister compartment falls. The calculated doubling time describes the dominant mode, not necessarily the immediate change from this starting composition.

What a Rising Burden Can Establish

A burden measurement is usually an observation of a total, not a direct measurement of \(G\) and \(P\) separately. Consequently, recurrence is consistent with persistence-driven reactivation but does not establish it by itself. An external inoculum, an unmodeled tissue compartment, changing drug exposure, or changing host-mediated killing can produce a similar pattern. The earlier material on bacterial growth under innate immune pressure and intracellular bacterial replication and host-cell lysis concerns additional mechanisms that may need to be represented when they are relevant to the data.

Worked Example: The Same Initial Rise from Reactivation or New Input

Use the post-treatment rates from the previous example and suppose, hypothetically, that at a particular time \(G=10\) and \(P=500\) CFU per milliliter. With no new input, the model predicts an instantaneous change in growing bacteria of:

\frac{dG}{dt}=(0.570)(10)+(0.080)(500)=5.70+40.0=45.7 \ \text{CFU per milliliter per day}.

Now consider a model with no persister contribution at that instant, but with \(G=10\) CFU per milliliter and an external input \(u=40.0\) CFU per milliliter per day. It predicts \(dG/dt=(0.570)(10)+40.0=45.7\) CFU per milliliter per day as well. Thus, a single observed slope cannot distinguish these sources. Evidence across time, measurements informative about the persister state, treatment and exposure history, or other independent biological evidence is needed to support one explanation over another.

Diagnostics, Alternatives, and Limitations

A fitted persistence model should be checked against both the biology and the observed trajectory. Useful checks include whether the model predicts nonnegative state burdens, whether rates have plausible units and signs, and whether the timing of treatment changes is aligned with the data. If a model predicts an increase in total burden while both \(a_G\) and \(a_P\) are negative and \(u=0\), inspect the equations: switching cannot by itself increase \(B\). A mismatch between predicted and observed rebounds may instead indicate changing rates, unmodeled compartments, measurement noise, an external source, or a model that is too simple.

State-specific burdens may be hard to observe, especially when \(P\) is a latent model compartment rather than a directly measured phenotype. Measurements near a detection limit can make apparent clearance uncertain; a later positive measurement does not on its own establish that burden was truly zero in between. Sparse sampling can also miss a rapid decline and rebound. These issues weaken attempts to infer switching rates or initial persister abundance from total burden alone.

Alternatives answer different questions. A one-state model can summarize net burden change but cannot identify reactivation from a hidden state. A model with an explicit input \(u(t)\) represents newly introduced bacteria, but fitting it in place of a persister source does not show that a new exposure occurred. An empirical biexponential curve can describe two apparent timescales without assigning either one to a biological subpopulation. The two-state model is useful when reversible persistence is a biologically motivated hypothesis and data can constrain its consequences; it is not automatically the best explanation for every recurrence.

Do not equate persistence with genetic resistance. Reversible physiological tolerance and stable inherited resistance are different mechanisms; this two-state model does not establish either from burden data.
Do not infer a new infection from a rebound alone. A rise can arise from reactivation, but alternative sources and changing conditions remain possible.
Do not interpret a dominant eigenvalue as the observed slope at every time. Transient behavior depends on the initial mixture of \(G\) and \(P\); treatment changes also change the rate matrix.
Do not treat fitted switching rates as uniquely identified by a total-burden curve. Different combinations of hidden-state size, transition rates, and net growth can produce similar observations.

Key takeaway: A persister compartment provides a mechanistic route from surviving bacteria to a later increase: persisters reactivate into the growing population, which can expand when conditions permit. The model makes that route explicit, but observations of total bacterial burden alone may not distinguish it from renewed input or other unmodeled causes.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If \(G=20\), \(P=100\), \(k_{PG}=0.05\) per day, and \(k_{GP}=0.02\) per day, what is the net contribution of switching to \(dB/dt\)?
With reactivation rate \(0.080\) per day and death rate \(0.020\) per day, what is the probability a persister reactivates before dying and within 7 days?
For the rate matrix in the rebound example, state the dominant eigenvalue to four significant figures and its interpretation.
Why can one instantaneous observed increase in \(G\) fail to distinguish reactivation from a new bacterial input?
Suppose both state-specific net rates are negative and there is no external input. Can switching alone cause total burden to increase? Explain.

Solutions. (1) The switching terms cancel: \(-k_{GP}G+k_{GP}G+k_{PG}P-k_{PG}P=0\), so their net contribution to \(dB/dt\) is zero. (2) The probability is \(0.8(1-e^{-0.100(7)})=0.8(1-e^{-0.7})\approx0.4027\), or 40.3%. (3) \(\lambda_{\max}\approx0.5736\) per day; it is the long-term exponential growth rate of the dominant mode under the stated constant rates. (4) The reactivation flux \(k_{PG}P\) and the external input \(u(t)\) both add to \(dG/dt\); different values of these sources can yield the same instantaneous slope. (5) No. With \(u=0\), \(dB/dt=a_GG+a_PP\), which is nonpositive when \(G,P\geq0\) and \(a_G,a_P<0\). Switching redistributes burden between states but does not increase the total.