From Division Outcomes to Population Change
A stem-cell division does more than add daughters: the mother cell that divided is no longer present as a mother. The lineage outcome therefore depends on the balance between the cells produced and the stem cell consumed in the event. In particular, two stem cells produced does not mean that the stem-cell population increases by two; it increases by one relative to the pre-division population.
We will track two populations: stem cells, denoted \(S\), and progenitors, denoted \(P\). A stem cell can divide to produce two stem cells, one stem cell and one progenitor, or two progenitors. These outcomes are often called symmetric self-renewal, asymmetric division, and symmetric differentiation, respectively. The purpose of the balance equations is to make their distinct effects on lineage size explicit.
For the basic balances, assume that division events occur at rate \(aS\), the three outcome probabilities apply to each event, and the parameters are constant over the time interval considered. Treat the populations as well-mixed and the balances as deterministic population or expected-count balances. Each event replaces exactly one stem-cell mother with the specified daughters. Progenitors do not divide, die, or leave the tracked population in this basic model; any such process would need its own balance term.
Accounting for Each Daughter Outcome
The safest way to build the equations is to calculate the net change from one event, not simply count the daughters. A two-stem outcome replaces one stem cell with two, so \(S\) changes by \(+1\). A stem-plus-progenitor outcome replaces one stem cell with one stem cell, so \(S\) changes by \(0\). A two-progenitor outcome removes the mother without replacing it with a stem cell, so \(S\) changes by \(-1\). The corresponding changes in \(P\) are 0, \(+1\), and \(+2\).
| Division outcome | Probability | Net change in \(S\) | Net change in \(P\) | Net change in \(S+P\) |
|---|---|---|---|---|
| Two stem cells | \(p_{SS}\) | \(+1\) | 0 | \(+1\) |
| One stem cell, one progenitor | \(p_{SP}\) | 0 | \(+1\) | \(+1\) |
| Two progenitors | \(p_{PP}\) | \(-1\) | \(+2\) | \(+1\) |
The expected net change in stem-cell number per division is \(p_{SS}-p_{PP}\). Multiplying by the event rate \(aS\) gives the stem-cell balance. The expected number of progenitors produced per division is \(p_{SP}+2p_{PP}\), which gives the progenitor production rate.
These equations describe net changes in the tracked populations. In particular, \(dP/dt\) here is the production rate before accounting for any progenitor exit, loss, or further division. If progenitors leave the tracked population, their outflow must be included separately. That added term does not change the daughter accounting at the stem-cell division.
Worked Example: Stem-Cell Maintenance With Ongoing Progenitor Production
Suppose a hypothetical population has \(S=80\) stem cells, each dividing at \(a=0.5\) per day. Let \(p_{SS}=0.20\), \(p_{SP}=0.60\), and \(p_{PP}=0.20\). Calculate the instantaneous stem-cell change and progenitor production rate.
The division-event rate is \(aS=(0.5\ \text{day}^{-1})(80\ \text{cells})=40\) events per day. The probabilities sum to 1.00, as required. The expected stem-cell change per event is \(0.20-0.20=0\), so
The expected progenitor production per event is \(0.60+2(0.20)=1.00\). Therefore,
The stem-cell count is maintained instantaneously in this model, while progenitors are produced. This does not mean that every division is asymmetric: the symmetric outcomes balance in their average effect on \(S\).
Interpreting Stem-Cell Balance and Lineage Size
Because \(aS\) is nonnegative, the sign of the stem-cell change is determined by \(p_{SS}-p_{PP}\), provided divisions are occurring. If \(p_{SS}>p_{PP}\), the stem-cell population increases; if \(p_{SS}<p_{PP}\), it decreases; and if the two probabilities are equal, its expected net change is zero. The asymmetric outcome contributes no net change to \(S\), but it does contribute to progenitor production.
The three probabilities also reveal an important distinction between maintaining the stem-cell pool and growing the lineage. In every listed outcome, one mother becomes two daughters, so the total number of tracked cells \(S+P\) increases by one per division. Adding the two balances confirms this result:
This identity assumes that only \(S\) and \(P\) are being counted and that no cells leave or die. It says that the total lineage grows at the rate of stem-cell division events, even when the division outcomes have very different effects on the stem-cell pool.
For constant \(a\) and probabilities, define \(g=a(p_{SS}-p_{PP})\) and \(q=p_{SP}+2p_{PP}\). Then \(dS/dt=gS\), while \(dP/dt=aqS\). The standard solution of the first equation is \(S(t)=S_0e^{gt}\). Integrating the progenitor production rate gives the trajectory below; when \(g=0\), progenitor production is constant because \(S\) is constant.
Worked Example: Comparing Stem-Cell and Lineage Growth
Consider a hypothetical population with \(S_0=50\) stem cells and \(P_0=20\) progenitors at time zero. Let \(a=0.30\) per day, \(p_{SS}=0.50\), \(p_{SP}=0.30\), and \(p_{PP}=0.20\). Find the predicted counts after 2 days under the basic model.
First, \(g=0.30(0.50-0.20)=0.0900\) per day and \(q=0.30+2(0.20)=0.700\). Thus \(S(t)=50e^{0.09t}\), giving
The progenitor production rate at time \(t\) is \(aqS(t)\). Integrating, or using the solution above,
The total is \(59.86+43.01=102.87\) cells, compared with \(70\) initially. The increase is \(32.87\) cells. It also follows from integrating \(d(S+P)/dt=aS\): \(0.30(50)(e^{0.18}-1)/0.0900=32.87\). The match checks the daughter-cell accounting.
Derivation, Diagnostics, and Model Choice
The population equations follow directly from event rates and expected net changes. In a short interval \(\Delta t\), the expected number of stem-cell divisions is approximately \(aS\Delta t\). Each event has expected stem-cell change \(p_{SS}(+1)+p_{SP}(0)+p_{PP}(-1)=p_{SS}-p_{PP}\). Therefore the expected change in \(S\) over the interval is approximately \(aS\Delta t(p_{SS}-p_{PP})\). Dividing by \(\Delta t\) and taking the limit as \(\Delta t\) approaches zero gives \(dS/dt=aS(p_{SS}-p_{PP})\). The same calculation for progenitors uses expected change \(p_{SS}(0)+p_{SP}(1)+p_{PP}(2)=p_{SP}+2p_{PP}\), yielding the second balance.
A useful diagnostic is to calculate the net change for each event before forming a rate. The probabilities must be nonnegative and sum to one; otherwise they do not describe the three mutually exclusive outcomes specified by the model. The division rate must have units of inverse time, so \(aS\) has units of cells per time. Each balance must therefore have units of cells per time. Finally, adding the equations should give \(aS\) for the total-cell rate under the stated no-loss assumptions. A different result signals an accounting error or an unlisted flow.
The deterministic balances are useful when the goal is to predict average population changes from division rates and outcome frequencies. They do not, by themselves, describe the distribution of outcomes across individual cells or the probability that a small clone disappears. Two systems can have the same mean stem-cell balance but different event-level variability. For example, equal frequencies of the two symmetric outcomes can maintain the mean stem-cell count, while every individual division still changes that count by either \(+1\) or \(-1\). When variability, extinction, or small clone sizes are central, an individual-based or stochastic branching description may be more informative. A simpler net-growth model can summarize the average change in \(S\), but it hides how division outcomes create progenitors.
Worked Example: Same Mean Stem-Cell Balance, Different Outcomes
Compare two hypothetical division schemes, both with \(a=0.40\) per day and \(S=100\): Scheme A has \((p_{SS},p_{SP},p_{PP})=(0.30,0.40,0.30)\), while Scheme B has \((0,1,0)\). Both sets of probabilities sum to 1.
For Scheme A, the net stem-cell change per event is \(0.30-0.30=0\), and the expected progenitor production is \(0.40+2(0.30)=1.00\). For Scheme B, the corresponding values are \(0-0=0\) and \(1+2(0)=1.00\). The event rate in either case is \((0.40)(100)=40\) per day. Thus both schemes predict \(dS/dt=0\) and \(dP/dt=40\) progenitors per day.
Their mean balances agree, but the division outcomes differ. Scheme B always replaces one stem cell with one stem cell and one progenitor. Scheme A includes divisions that increase or decrease the stem-cell count. The mean equations alone cannot distinguish those event-level patterns.
Common Mistake
A frequent error is to count the daughters as additions without subtracting the mother. For example, the reasoning “two stem daughters means \(+2\) stem cells” gives the wrong change in \(S\): the event starts with one stem cell and ends with two, so its net change is \(2-1=+1\). Likewise, a stem-plus-progenitor division has net stem-cell change \(1-1=0\), not \(+1\). Always compare the population immediately before and after one event, then multiply the expected net change by the event rate.
A second interpretive error is to infer that a stable stem-cell count implies a stable lineage size. When \(p_{SS}=p_{PP}\), the stem-cell balance is zero, but divisions still add one cell to \(S+P\) in this no-loss model. Stem-cell maintenance and total lineage growth are related but are not the same quantity.
Check Your Understanding
For each question, use net changes per event before converting to population rates. Solutions are included below.
- For one stem-cell division that produces two progenitors, what are the net changes in \(S\), \(P\), and \(S+P\)?
- If \(a=0.25\) per day, \(S=40\), and the outcome probabilities are \(p_{SS}=0.10\), \(p_{SP}=0.60\), and \(p_{PP}=0.30\), calculate \(dS/dt\) and \(dP/dt\).
- What condition on the division probabilities maintains the expected stem-cell count when divisions occur?
- Why does \(d(S+P)/dt=aS\) hold in the basic model, regardless of the three outcome probabilities?
- Name one situation in which the deterministic balances might not be sufficient for the scientific question.
Solutions. (1) The mother stem cell is replaced by two progenitors, so the changes are \(-1\) stem cell, \(+2\) progenitors, and \(+1\) total cell. (2) The division rate is \(0.25(40)=10\) events per day. Thus \(dS/dt=10(0.10-0.30)=-2.00\) stem cells per day, and \(dP/dt=10(0.60+2(0.30))=12.0\) progenitors per day. (3) The condition is \(p_{SS}=p_{PP}\). (4) Every division replaces one mother with two daughters, adding one tracked cell; the probabilities sum to one, so the average total increase is one per event. (5) If the goal is to quantify clone-extinction probabilities or cell-to-cell variability in a small population, a stochastic branching model may be needed.