How mutation accumulation depends on the structure of the cell lineage tree

All the cells of a multicellular organism are the product of cell divisions that trace out a single binary tree, the so-called cell lineage tree. Because cell divisions are accompanied by replication errors, the shape of the cell lineage tree is a key determinant of how somatic evolution, which can potentially lead to cancer, proceeds. Carcinogenesis requires the accumulation of a certain number of driver mutations. By mapping the accumulation of mutations into a graph theoretical problem, we present an exact numerical method to calculate the probability of collecting a given number of mutations and show that for low mutation rates it can be approximated with a simple analytical formula, which depends only on the distribution of the lineage lengths, and is dominated by the longest lineages. Our results are crucial in understanding how natural selection can shape the cell lineage trees of multicellular organisms and curtail somatic evolution.

Figure 1

Illustration of a cell lineage tree with mutation accumulation. (a) A cell lineage tree $\mathcal{T}$ with $L$ leaf nodes (white circles) and $L-1$ internal nodes (gray circles). The lineage length (or divisional load) $D_i$ corresponding to leaf node $i$ is the number of edges (cell divisions, denoted by arrows) leading from the root (bottom-most) node to the leaf node. Mutations are indicated by red stars. (b) Cell lineage tree ${\mathcal{T}}^{\prime }$ obtained by making leaf node $L$ in tree $\mathcal{T}$ divide (indicated by blue dashed arrows and circles).

Figure 2

Examples for the probability of accumulating at least $m$ mutations in a series of cell lineage trees. The top panel illustrates how a series of lineage trees that interpolate between the most skewed binary tree (a linear chain of all the internal nodes, $n_{\text{b}}=0$) and the most balanced one (the perfect binary tree, $n_{\text{c}}=0$) is generated: Identical perfect binary subtrees (indicated with thin arrows) of depths (lineage lengths) $n_{\text{b}}$ are joined (with thick gray arrows) to an $n_{\text{c}}$ long linear chain (indicated with thick black arrows). The number of leaf nodes can be expressed as $L=(n_{\text{c}}+2)2^{n_{\text{b}}}$. The four plots in the bottom panel show the exact probabilities $P_{\mathcal{T}}(\mu ,m)$ (blue lines, obtained form the merging process) that the lineage trees (for $L=2^{16}$ and for four different values of $n_{\text{b}}=0$, 5, 10, and 15) have a lineage with at least $m$ mutations (ranging between $m=1$ and 10 from left to right) as a function of the mutation rate $\mu$, as well as their approximation $P_{\mathcal{T}}^{\text{appr}}(\mu ,m)$ (red dashed lines) using Eq. (16). The approximation is very accurate as long as $P_{\mathcal{T}}^{\text{appr}}(\mu ,m)\ll 1$.

Figure 3

Examples for two basic population dynamics models of $N=3$ cells in discrete time. The left panel shows an ensemble of $N$ parallel linear chains. At each time step a randomly selected cell divides, and one of its daughter cells becomes lost. The right panel shows a Moran process, in which at each time step two cells are chosen randomly, one for division and one for loss (with $1/N$ chance the same cell is chosen for both events, which means that after division one of its daughters becomes lost). Lost cells are explicitly marked with a cross.