Clonal Selection Under Differential Growth

How Composition Changes Without Mutation

A cell population can change its clonal composition even when no new clone is created. If clones have different net growth rates, faster-growing clones contribute an increasing share of the total population, while slower-growing clones may contribute a decreasing share. This change in relative abundance is clonal selection under differential growth: it describes a compositional consequence of growth, not a change in the identity of a clone.

The distinction between counts and fractions matters. A clone’s fraction can decline while its absolute cell count still rises, provided the rest of the population grows faster. Conversely, a growing fraction does not by itself tell us the clone’s absolute count. The calculations below keep both quantities visible.

Model and Assumptions

Consider a fixed set of \(K\) clones, indexed by \(i=1,\ldots,K\). Let \(N_i(t)\) be the number of cells in clone \(i\) at time \(t\), and let \(r_i\) be that clone’s constant net growth rate. A net rate combines cell production and loss; it may be positive, zero, or negative. The model assumes each clone follows an independent exponential count equation:

\frac{dN_i}{dt}=r_iN_i,\qquad i=1,\ldots,K.

The assumptions are that the set of clone identities is fixed; there is no mutation, immigration, or transition between clones; each \(r_i\) is constant over the time interval; and the growth of one clone does not directly alter another clone’s rate. Thus this model does not include shared capacity or density-dependent growth. It is a useful baseline for isolating the consequence of differences in constant net rates.

Definition: When the total population \(N(t)=\sum_{j=1}^{K}N_j(t)\) is positive, the fraction of the population in clone \(i\) is \(f_i(t)=N_i(t)/N(t)\). The population’s fraction-weighted mean net growth rate is \(\bar r(t)=\sum_{j=1}^{K}f_j(t)r_j\). The fractions are nonnegative and sum to one.

The units of \(r_i\) are inverse time. For example, if time is measured in days, \(r_i=0.03\ \text{day}^{-1}\). Counts are cell numbers, while fractions have no units.

Count and Fraction Equations

For a clone with \(N_i(0)>0\), separation and integration give \(\ln N_i(t)-\ln N_i(0)=r_it\), so \(N_i(t)=N_i(0)e^{r_it}\). This logarithmic derivation assumes a positive initial count. If \(N_i(0)=0\), the count equation instead has the solution \(N_i(t)=0\); the exponential formula also gives zero in that case. Hence the count solution applies for nonnegative initial counts:

N_i(t)=N_i(0)e^{r_it}.

If the initial total count is positive, the total remains positive under this model, so clone fractions are defined. Summing the count equations gives \(dN/dt=\sum_j r_jN_j=N\bar r\). Differentiate \(f_i=N_i/N\) using the quotient rule:

\frac{df_i}{dt} =\frac{(dN_i/dt)N-N_i(dN/dt)}{N^2} =\frac{r_iN_iN-N_iN\bar r}{N^2} =f_i(r_i-\bar r).

This equation says that the fraction’s instantaneous direction of change depends on how the clone’s rate compares with the current population mean rate. The comparison is with \(\bar r(t)\), not necessarily with a fixed benchmark or with every other clone individually.

Key result: A clone with \(f_i(t)>0\) increases in fraction when \(r_i>\bar r(t)\), decreases when \(r_i<\bar r(t)\), and has zero instantaneous fraction change when \(r_i=\bar r(t)\). A clone with \(f_i(t)=0\) remains at zero fraction under this model, regardless of its rate.

The count solution also gives the fraction directly:

f_i(t)= \frac{N_i(0)e^{r_it}} {\sum_{j=1}^{K}N_j(0)e^{r_jt}}.

A useful two-clone consequence is the odds relation. If both clones begin with positive counts, their fraction ratio is \(f_1(t)/f_2(t)=[N_1(0)/N_2(0)]e^{(r_1-r_2)t}\). The relative growth rate \(r_1-r_2\), rather than either absolute rate alone, controls how their odds change.

Worked Example: Two Clones With Different Net Rates

Suppose a hypothetical population begins with 80 cells of clone 1 and 20 cells of clone 2. Their constant net growth rates are \(r_1=0.03\ \text{day}^{-1}\) and \(r_2=0.01\ \text{day}^{-1}\). Under the stated model, both initial counts are positive, there are no clone transitions, and the rates remain constant. Find the composition after 20 days.

The count solutions are \(N_1(20)=80e^{0.03(20)}=80e^{0.6}\approx80(1.82212)=145.770\) cells and \(N_2(20)=20e^{0.01(20)}=20e^{0.2}\approx20(1.22140)=24.4281\) cells. Thus \(N(20)\approx145.770+24.4281=170.198\) cells. The clone fractions are

f_1(20)=\frac{145.770}{170.198}\approx0.8565, \qquad f_2(20)=\frac{24.4281}{170.198}\approx0.1435.

Clone 1’s share rises from \(80/100=0.800\) to about 0.8565. Clone 2’s share falls, even though its count increases from 20 to about 24.4281 cells. The odds check gives \((80/20)e^{(0.03-0.01)(20)}=4e^{0.4}\approx5.9673\), which agrees with \(145.770/24.4281\). This is a change in composition through differential growth; no mutation is needed in the model.

More Than Two Clones

With several clones, the mean rate changes as the composition changes. Consequently, a clone with a rate greater than the mean at one time need not remain above the mean forever. One way to describe this is to differentiate the mean rate, treating each \(r_i\) as constant:

\frac{d\bar r}{dt} =\sum_i r_i\frac{df_i}{dt} =\sum_i f_i r_i(r_i-\bar r) =\sum_i f_i(r_i-\bar r)^2 \geq 0.

In the last equality, expanding the squared term uses \(\sum_i f_i=1\) and \(\sum_i f_ir_i=\bar r\). Thus the mean rate is nondecreasing, and it increases whenever the currently represented clones have differing rates. With three or more clones, an intermediate-rate clone can initially gain fraction and later lose it as the mean rate moves past its rate. This is why the fraction equation is more informative than a blanket claim that any particular clone must always rise or fall.

Worked Example: Three-Clone Composition

Consider a hypothetical population with initial counts \(N_1(0)=50\), \(N_2(0)=30\), and \(N_3(0)=20\) cells. Let the constant rates be \(0.04\), \(0.02\), and \(-0.01\ \text{day}^{-1}\), respectively. The assumptions of no transitions and constant rates are imposed for this calculation. Find the fractions after 10 days and the initial direction of change for clone 1.

The count predictions are \(N_1(10)=50e^{0.4}\approx74.5912\), \(N_2(10)=30e^{0.2}\approx36.6421\), and \(N_3(10)=20e^{-0.1}\approx18.0967\) cells. Their sum is approximately \(129.330\) cells. Therefore,

f_1(10)=\frac{74.5912}{129.330}\approx0.5768,\quad f_2(10)=\frac{36.6421}{129.330}\approx0.2833,\quad f_3(10)=\frac{18.0967}{129.330}\approx0.1399.

Initially, the fractions are \(0.50\), \(0.30\), and \(0.20\), so \(\bar r(0)=0.50(0.04)+0.30(0.02)+0.20(-0.01)=0.024\ \text{day}^{-1}\). It follows that \(df_1/dt\vert_{t=0}=0.50(0.04-0.024)=0.008\ \text{day}^{-1}\). Clone 1 initially gains fraction. This is an instantaneous rate of fraction change, not a claim that its fraction rises by exactly 0.008 over every day.

Worked Example: A Clone Absent at Baseline

Suppose a hypothetical population starts with 100 cells of clone A and zero cells of clone B. Their assigned net rates are \(0.01\) and \(0.05\ \text{day}^{-1}\), respectively. Assume the fixed-clone model: no mutation, immigration, or transitions occur. After 30 days, \(N_A(30)=100e^{0.3}\approx134.986\) cells, while \(N_B(30)=0e^{1.5}=0\) cells. Therefore \(f_B(30)=0\) and \(f_A(30)=1\).

Although clone B has the larger rate, it cannot increase from a zero count in this model. Its zero fraction is invariant. The model can describe selection among represented clones, but it cannot generate a clone that was absent initially. A measured appearance of a new clone would require processes outside these assumptions or a different account of what was present but undetected.

Diagnostics, Comparisons, and Limits

For a proposed application, compare predicted and observed counts as well as fractions. The model requires each clone’s log count to be linear in time when its count is positive: \(\ln N_i(t)=\ln N_i(0)+r_it\). Systematic curvature, changing estimated slopes, or inconsistent rate differences across intervals can signal that constant net rates are inadequate. Also check that predicted fractions remain between zero and one and sum to one; violations indicate an arithmetic or implementation error.

The direct count approach and the fraction equation answer related but distinct questions. Solving \(dN_i/dt=r_iN_i\) predicts absolute counts and then yields fractions by normalization. The equation \(df_i/dt=f_i(r_i-\bar r)\) focuses on composition and explains its local direction of change. The odds equation is especially compact for two represented clones. None of these forms repairs a misspecified assumption: a changing rate, a clone transition, or an unmodeled influx can produce trajectories that the constant-rate fixed-clone model does not represent.

A common mistake is to conclude that a declining fraction means a declining cell count. The fraction is \(N_i/N\), so it can fall even while \(N_i\) rises, as clone 2 did in the first example. Another mistake is to infer mutation solely because a clone’s fraction increased. Under this model, differential growth alone changes fractions; observed composition by itself does not establish which mechanism produced the change. Conversely, an exactly absent clone stays absent here, so the model cannot explain its emergence.

Key takeaway: Under constant, clone-specific net growth with no transitions, counts change exponentially and fractions follow \(df_i/dt=f_i(r_i-\bar r)\). Differential growth can alter composition without mutation, but the model cannot create an initially absent clone or account for processes it excludes.

Check Your Understanding

Answer each question, then compare with the solutions below.

A clone has fraction \(0.25\), rate \(0.03\ \text{day}^{-1}\), and the population mean rate is \(0.02\ \text{day}^{-1}\). What is its instantaneous fraction-change rate?
Two represented clones have rate difference \(r_1-r_2=0.01\ \text{day}^{-1}\). By what factor do their odds change over 15 days?
Can an initially absent clone increase in fraction under the fixed-clone model, even if its assigned rate is the largest?
A clone’s count rises from 40 to 50 cells while its fraction falls from 0.40 to 0.30. Is that contradictory? Explain.

Solutions. 1. \(df_i/dt=f_i(r_i-\bar r)=0.25(0.03-0.02)=0.0025\ \text{day}^{-1}\), so its fraction is increasing instantaneously. 2. The odds multiplier is \(e^{(r_1-r_2)t}=e^{0.01(15)}=e^{0.15}\approx1.162\). 3. No. Its count remains zero by \(N_i(t)=N_i(0)e^{r_it}\), so its fraction remains zero while the total is positive. 4. No. Fraction is count divided by total; a sufficiently faster increase in the total can make the clone’s share fall even as its own count rises.