Identifiability of Formation and Conversion Rates in Linked Pools

Why a downstream response may not reveal every rate

A measured biomarker can show a clear delayed response while leaving the underlying biology uncertain. In a linked-pool system, material first enters an unobserved precursor pool, then converts into a measured product. The observed product response reflects both steps, as well as losses from both pools. A good fit to the measured curve does not by itself show that each rate has been separately determined.

The central question here is identifiability: which model parameters, or combinations of them, can be uniquely learned from the available output? This matters when interpreting estimates from sparse biomarker sampling, a setting considered in the earlier material on turnover-rate estimation with sparse biomarker sampling. The issue is not simply whether an optimizer returns a number; it is whether the observations distinguish that number from alternatives.

A two-pool model and its observable response

Definition: Let $P(t)$ be the amount in an unobserved precursor pool and $B(t)$ the amount in the measured downstream pool. For a known input $u(t)$, consider $$ \frac{dP}{dt}=k_f u(t)-(k_c+k_p)P,\qquad \frac{dB}{dt}=k_cP-k_bB. $$ Here $k_f$ is the precursor formation-rate scale, $k_c$ is the first-order conversion rate, $k_p$ is the competing precursor-loss rate, and $k_b$ is the downstream loss rate. The rate constants $k_c,k_p,k_b$ have units of inverse time; if $u(t)$ is dimensionless, $k_f$ has units of amount per time.

Assume the rates are constant and positive, the model structure is correct, and the input $u(t)$ is known. Define $a=k_c+k_p$ and $q=k_fk_c$. When $u(t)$ is dimensionless, $q$ has units of amount per time squared. For a known input and zero initial pool amounts, the Laplace transform of the downstream response is

\frac{B(s)}{U(s)} = \frac{k_fk_c}{(s+k_c+k_p)(s+k_b)} = \frac{q}{(s+a)(s+k_b)}.

Thus the ideal downstream response contains a numerator $q$ and two decay rates, $a$ and $k_b$. It does not contain $k_f$ and $k_c$ as separate numerator terms. The denominator identifies its two rates as an unordered pair: exchanging $a$ and $k_b$ leaves the transfer function unchanged. If initial pool amounts are nonzero, their contributions add transient terms; the same two decay rates govern those terms, but unknown initial conditions can add further ambiguity.

Key result: Under informative known input and ideal continuous, noise-free observation of $B(t)$, the transfer function identifies $q=k_fk_c$ and the unordered pair $\{k_c+k_p,\ k_b\}$. In general it does not uniquely identify $k_f$, $k_c$, $k_p$, and $k_b$ separately.

Structural identifiability: a derivation

Structural identifiability asks whether distinct parameter values can produce exactly the same output under ideal observation. Starting from the transfer function, expand its denominator:

$$ (s+a)(s+k_b)=s^2+(a+k_b)s+ak_b. $$

An exact response therefore determines the numerator $q$, the sum $a+k_b$, and the product $ak_b$. The roots of $r^2-(a+k_b)r+ak_b=0$ are $a$ and $k_b$; the output alone does not label which root belongs to which process. Even if external information assigns the roots, knowing $a$ only gives $k_c+k_p$. For any $k_c$ satisfying $0<k_c<a$, one can set

k_p=a-k_c,\qquad k_f=\frac{q}{k_c},

and preserve both the numerator and denominator. This is a continuum of positive parameter sets with exactly the same downstream response. The conclusion relies on an input that excites the system sufficiently to reveal its dynamics; a constant input observed only at equilibrium generally supplies less information than a transient response.

A useful distinction is that structural non-identifiability persists even with perfect data, whereas practical non-identifiability occurs when the parameters are theoretically distinguishable but the available data are too noisy, sparse, or weakly informative to distinguish them reliably. More samples can help with practical limitations, but cannot remove an exact structural ambiguity without additional information or a changed model or experiment.

Worked Example: Two Rate Sets With the Same Downstream Curve

Suppose a hypothetical system receives a unit step input, $u(t)=1$ for $t\geq 0$, and starts with $P(0)=B(0)=0$. Let $k_f=4.74$ amount/time, $k_c=0.80$ per time, $k_p=0.20$ per time, and $k_b=0.80$ per time. Then $a=1.00$ per time and $q=3.792$ amount/time².

For this step input, integration of the transfer function gives, when $a\ne k_b$, $$ B(t)=\frac{q}{ak_b}\left[1-\frac{k_b e^{-at}-a e^{-k_bt}}{k_b-a}\right]. $$ Taking the repeated-pole limit when $a=k_b$ gives $$ B(t)=\frac{q}{a^2}\left[1-(1+at)e^{-at}\right]. $$ At $t=1$ time unit, substitute the values: $$ B(1)=\frac{3.792}{1.00(0.80)} \left[1-\frac{0.80e^{-1.00}-1.00e^{-0.80}}{0.80-1.00}\right]. $$

The exponential terms are $0.80e^{-1}=0.294304$ and $1.00e^{-0.80}=0.449329$. Therefore, $$ B(1)=4.74\left[1-\frac{0.294304-0.449329}{-0.20}\right] =4.74(0.224875)\approx1.0659\ \text{amount units}, $$ which rounds to 1.066 amount units.

Now take a second parameter set: $k_c=0.40$, $k_p=0.60$, $k_f=9.48$, and $k_b=0.80$, with the same units as above. It has $a=1.00$ and $q=9.48(0.40)=3.792$, exactly as before. Consequently, its entire downstream response—not just the value at one time—is identical. The downstream data cannot decide between these formation and conversion rates.

What extra information can resolve the ambiguity?

The ambiguity can be reduced by information that constrains a rate or observes another part of the pathway. For example, if $k_p$ is known independently and the two fitted poles can be assigned to $a$ and $k_b$, then $k_c=a-k_p$, after which $k_f=q/k_c$. An independent measurement of $P$ can provide a different constraint. These are not properties of the downstream curve alone: they rely on added information.

Worked Example: Using a Known Precursor-Loss Rate

Suppose a hypothetical, well-excited downstream response identifies $q=0.120$ amount/time² and decay rates $0.50$ and $0.20$ per time. Independent information establishes that $k_b=0.20$ per time and $k_p=0.20$ per time. The remaining pole must then be $a=0.50$ per time.

The precursor balance gives $$ k_c=a-k_p=0.50-0.20=0.30\ \text{per time}. $$ Since $q=k_fk_c$, $$ k_f=\frac{q}{k_c} =\frac{0.120\ \text{amount/time}^2}{0.30\ \text{per time}} =0.400\ \text{amount/time}. $$

With these constraints, the rates are determined as $k_c=0.30$, $k_f=0.400$, $k_p=0.20$, and $k_b=0.20$. Without the independent knowledge of $k_p$ and the assignment of the downstream pole, the same output would not justify these separate estimates.

A precursor measurement can also break the formation-conversion trade-off. At steady state under a known constant input $u$, the model gives $P_{\mathrm{ss}}=k_fu/a$ and $B_{\mathrm{ss}}=k_cP_{\mathrm{ss}}/k_b$. These relations are useful only when their inputs and rates are sufficiently constrained; equilibrium measurements alone should not be treated as if they were a fully informative time course.

Worked Example: An Additional Precursor Measurement

Consider a hypothetical model with known constant input $u=1$, and downstream response information giving $q=0.120$ amount/time², $a=0.50$ per time, and $k_b=0.20$ per time. Suppose a separate assay measures the steady-state precursor amount as $P_{\mathrm{ss}}=0.80$ amount.

From $P_{\mathrm{ss}}=k_fu/a$, the formation scale is $$ k_f=\frac{aP_{\mathrm{ss}}}{u} =\frac{(0.50)(0.80)}{1} =0.400\ \text{amount/time}. $$ Then the product $q=k_fk_c$ gives $$ k_c=\frac{q}{k_f} =\frac{0.120}{0.400} =0.300\ \text{per time}, \qquad k_p=a-k_c=0.50-0.30=0.20\ \text{per time}. $$

As a consistency check, the predicted downstream steady state is $B_{\mathrm{ss}}=k_cP_{\mathrm{ss}}/k_b=(0.300)(0.80)/0.20=1.20$ amount units. The transfer-function steady-state gain gives the same result: $q/(ak_b)=0.120/[(0.50)(0.20)]=1.20$ amount units.

Practical identifiability and diagnostics

Even when external constraints make the parameters structurally identifiable, the observed response may not contain enough information to estimate them precisely. Closely spaced decay rates produce similar time courses, and sparse sampling can miss the part of the response that distinguishes the modes. A short observation window may capture the initial rise but not the slower return toward baseline; a long interval between samples may miss the early curvature.

Worked Example: When Two Decay Modes Are Hard to Separate

Suppose a hypothetical fitted model has two decay rates $a=0.50$ and $k_b=0.48$ per hour. At one hour, the corresponding exponential terms are $e^{-0.50}=0.6065$ and $e^{-0.48}=0.6188$. Their difference is only $0.0123$.

If each mode has an amplitude of about 1 amount unit, the change attributable to their differing exponential values at that time is about $0.0123$ amount units. If assay residuals have a standard deviation of $0.050$ amount units, this difference is smaller than the measurement noise. A fit may then return noticeably different rates across repeated data sets or starting values, even though one particular fit looks smooth.

A practical check is to examine whether plausible parameter changes leave predictions almost unchanged, especially over the actual sampling times. Also inspect parameter uncertainty and whether estimates depend strongly on starting values or follow-up duration. Such signs indicate weak information; they do not make the parameters structurally identifiable if the underlying model has an exact ambiguity.

Modeling choices, limitations, and a common pitfall

Fitting the primitive rates directly can be convenient when external constraints are available, but it may hide the combinations the output actually supports. A complementary approach is to fit the observable transfer-function quantities $q$, $a+k_b$, and $ak_b$, then ask which biological rates can be recovered using independent knowledge. This reparameterization makes the ambiguity visible, but it does not create new information or uniquely label the two poles.

Other routes to greater identifiability include measuring the precursor, obtaining an independent estimate of a loss rate, or using a designed input and sampling schedule that better reveals the dynamics. Each route has limitations: an added measurement may be imprecise, an assumed known rate may be uncertain, and richer sampling cannot resolve a structural ambiguity on its own. The analysis also depends on the assumed two-pool structure, constant rates, and correct input description; a uniquely estimated parameter under a wrong model is not necessarily a meaningful biological rate.

Common mistake: “The optimizer returned separate estimates of $k_f$ and $k_c$, so the data identified both rates.”
Correction: A numerical solution can be one member of a continuum of equivalent solutions. Check which parameter combinations determine the output, and identify what independent information is required before interpreting individual rates.
Another warning: Two fitted decay rates should not automatically be assigned to precursor and product processes by their numerical size. The output identifies the poles without necessarily identifying their biological labels.

Key takeaway: A downstream biomarker can identify a dynamic response precisely while leaving its component rates unresolved. Report the estimable combinations, state the assumptions and external constraints used to separate rates, and avoid interpreting a convenient fitted parameterization as uniquely established by the biomarker alone.

Check Your Understanding

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

For the stated linked-pool model, what numerator and decay rates appear in the downstream transfer function?
If the fitted values are $q=0.18$ amount/time² and $a=0.60$ per time, name one of the infinitely many positive $(k_f,k_c,k_p)$ combinations consistent with them.
Why does an exact downstream response generally fail to label the two fitted decay poles?
Does collecting many more downstream samples necessarily separate $k_f$ from $k_c$? Explain.
Give one type of additional information that can help separate formation and conversion rates.

Solutions. (1) The numerator is $q=k_fk_c$; the poles are $a=k_c+k_p$ and $k_b$. (2) Choose $k_c=0.30$ per time, so $k_p=0.60-0.30=0.30$ per time and $k_f=0.18/0.30=0.60$ amount/time. These values give the required product and sum. (3) The denominator is $(s+a)(s+k_b)$, which is unchanged when the two rates are exchanged; assigning biological labels requires other information. (4) No. More precise downstream observations can improve practical precision, but the output still contains $k_fk_c$, not the two rates separately. (5) Examples include a precursor-pool measurement, an independently known precursor-loss rate, or other external evidence that constrains a rate or pole assignment.