Branching Models of Cell Lineages

Representing a Cell Lineage as a Branching Process

A lineage model separates three questions that can otherwise be conflated: what offspring each cell produces, how the expected number of each cell type changes across generations, and whether the lineage ultimately disappears. A multitype branching process gives a precise framework for these questions. Its generations count rounds of reproduction, not elapsed calendar time; division times and cell ages are not modeled here.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let the type set be \(S=\{S,P,D\}\), and let \((Z_n)_{n\ge0}\) be the population process in discrete generations \(n\ge0\). Here \(S\), \(P\), and \(D\) denote stem, progenitor, and differentiated types. Write \(Z_n=(Z_{n,S},Z_{n,P},Z_{n,D})\) as a row vector. The filtration \(\mathcal{F}_n\) records the population and reproduction information through generation \(n\). The branching assumption is that, conditional on \(\mathcal{F}_n\), individuals reproduce independently, and an individual's offspring law depends only on its type.

The mean progeny matrix is not a transition matrix. Its entries are expected offspring counts, so rows need not sum to \(1\). A row can sum to less than \(1\) or more than \(1\); that total is the expected number of offspring of the corresponding parent type. The probabilities in each offspring law, in contrast, must sum to \(1\).

For this model, define lineage extinction to mean that some generation has no modeled descendants: \(Z_n=(0,0,0)\) for some \(n\). A differentiated cell may persist physically after it stops dividing, but a sterile \(D\) type does not produce descendants in this branching model. Thus extinction here concerns the modeled reproducing lineage, not the survival of every terminal cell in tissue.

Two Calculations That Organize the Model

Proof. Given \(\mathcal{F}_n\), each of the \(Z_{n,i}\) type-\(i\) parents has expected \(m_{ij}\) type-\(j\) offspring. Adding their conditional expectations gives \(\mathbb{E}[Z_{n+1,j}\mid\mathcal{F}_n]=\sum_i Z_{n,i}m_{ij}\), which is the \(j\)th coordinate of \(Z_nM\). Taking expectations and using the defining property of conditional expectation gives \(\mathbb{E}[Z_{n+1}]=\mathbb{E}[Z_n]M\). Iterating this identity yields \(\mathbb{E}[Z_n]=\mathbb{E}[Z_0]M^n\). \(\square\)

The extinction probabilities require the full offspring laws, not just their means. Let \(q_i\) be the probability of eventual extinction from one initial individual of type \(i\). The next result explains how the offspring generating functions determine these probabilities.

Proof. Starting from one parent of type \(i\), extinction occurs exactly when the descendant lineages of all its offspring become extinct. Conditional on its offspring counts \(\xi^{(i)}\), the branching independence assumption makes that probability \(\prod_j q_j^{\xi^{(i)}_j}\). Taking its expectation gives \(q_i=G_i(q)\).

To prove the least-solution claim, let \(q^{(n)}_i\) be the probability that the population is zero by generation \(n\), starting from one type-\(i\) individual. Set \(q^{(0)}=(0,0,0)\), since an initially present individual means extinction has not occurred at generation zero. Conditioning on the first offspring gives \(q^{(n+1)}=G(q^{(n)})\). These probabilities increase with \(n\), and their limit is the probability of eventual extinction, so \(q^{(n)}\) increases coordinatewise to \(q\). The functions \(G_i\) are continuous on \([0,1]^3\): their defining expectations are limits of finite offspring sums, or equivalently continuity follows by bounded convergence since the products lie in \([0,1]\). Therefore \(q=G(q)\). If \(x\in[0,1]^3\) is any other solution, then \(q^{(0)}\le x\). Each \(G_i\) is nondecreasing in every coordinate on \([0,1]^3\), so induction gives \(q^{(n)}\le x\) for every \(n\). Taking limits gives \(q\le x\), as required. \(\square\)

The mean matrix describes expected growth through powers of \(M\). Its spectral radius, the largest absolute value of its eigenvalues, is useful for assessing long-run mean growth; Perron–Frobenius theory gives a more specific growth interpretation under its hypotheses. But a mean is not an extinction probability: determining extinction requires the offspring generating functions and the least-fixed-point calculation above.

Interpretation and a Common Matrix Error

A cell-lineage model is useful for comparing division programs, identifying which type contributes most to expected expansion, and separating terminal differentiation from self-renewal. The mean matrix summarizes expected production type by type; the generating functions retain the variation in offspring counts that determines extinction. In particular, a spectral radius above \(1\) signals potential growth in the mean, but does not say that every realization grows or survives.

Common mistake: transposing the population update. With the matrix in the mixed-population example, a common incorrect calculation treats the population as a column and computes \(M(1,0,0)^\mathsf{T}=(1,0,0)^\mathsf{T}\), thereby missing progenitors. The convention here uses row population vectors: \((1,0,0)M=(1,\tfrac12,0)\). The result is consistent with the offspring law: one stem cell produces one progenitor with probability \(\tfrac12\), so the expected number of progenitors after one generation is \(\tfrac12\), not zero.

Before interpreting a cell model, check that every offspring law is a valid probability distribution, that each mean-matrix row uses the parent type as its row index, and that the event called extinction matches the biological question. If terminal cells are to be counted as persisting cells rather than as the end of a reproducing lineage, the state description and extinction event must be adjusted accordingly.