Unknown Assay Scale and Biological Parameter Confounding

Why Assay Scale Can Be Confounded With Biology

A mechanistic model may describe a biological amount in concentration units, while an assay reports a signal in instrument units. If the conversion between those scales is unknown, a large biological amount measured with a low assay response can produce the same signal as a smaller amount measured with a high response. The central question is not simply whether the model fits the observations, but which combinations of biological and measurement parameters the observations can determine.

This tutorial focuses on a multiplicative assay calibration factor. It does not imply that every unknown assay parameter is confounded with every biological parameter. The ambiguity arises under specific conditions: the observations must depend on the assay factor and biological scale only through their product, and there must be no independent information that separates the two.

Definition: An unknown assay scale \(c\) and biological scale \(A_0\) are structurally confounded in the observations if distinct admissible pairs \((c,A_0)\) can produce exactly the same noiseless output functions because the model uses them only through a combination such as \(q=cA_0\). This means \(c\) and \(A_0\) are not separately identifiable from those observations. It does not, by itself, establish that \(q\) is identifiable.

Structural identifiability concerns uniqueness from ideal, noiseless observations under the stated model and design. A parameter is identifiable only if equality of the model-predicted observations implies equality of that parameter. A confounded pair violates this condition for the individual factors. Whether their product, or other parameters, is identifiable is a separate question about the full model and the observation design.

A Mechanistic Observation Model

Consider a biological amount \(x(t)\) that follows first-order loss with rate constant \(k\), beginning at unknown amount \(A_0\). Suppose the assay reports signal \(y(t)\) proportional to the amount, with unknown positive calibration factor \(c\):

\frac{dx}{dt}=-kx,\qquad x(0)=A_0,\qquad y(t)=c\,x(t),

Assume \(A_0>0\), \(c>0\), and \(k>0\). The model is deterministic and has no additive assay background, saturation, or other observation term. The exact solution of the differential equation is \(x(t)=A_0 e^{-kt}\), so

y(t)=cA_0e^{-kt}=q e^{-kt},\qquad q=cA_0.

Here \(A_0\) has units of biological amount, \(c\) has units of assay signal per biological amount, and \(q\) has units of assay signal. The time-dependent factor \(e^{-kt}\) describes the relative decay of the signal; the product \(q\) sets its signal scale.

Key result: In this model, if all observations depend on \(c\) and \(A_0\) only through \(q=cA_0\), then \(c\) and \(A_0\) cannot be separately identified from those observations. The product \(q\) is identifiable only if the observation times and the available information also identify the signal scale in the reduced model \(y(t)=q e^{-kt}\).

Why the Two Factors Cannot Be Separated

The ambiguity follows directly from the product form. For any \(\lambda>0\), define a new pair by multiplying \(c\) by \(\lambda\) and dividing \(A_0\) by \(\lambda\). The transformed pair remains positive and has the same product:

c'=\lambda c,\qquad A_0'=\frac{A_0}{\lambda}, \qquad c'A_0'=(\lambda c)\frac{A_0}{\lambda}=cA_0.

Consequently, at every observation time \(t\), the transformed parameters predict

y'(t)=c'A_0'e^{-kt}=cA_0e^{-kt}=y(t).

This is a complete nonuniqueness argument for the individual factors: for any admissible pair, there are infinitely many distinct positive pairs with the same output function. It does not rely on measurement noise or a particular finite set of sampling times. However, the argument does not prove that \(q\) is identifiable: unknown parameters such as \(k\) may create additional ambiguity in the reduced model.

The same reasoning applies when \(A_0\) is replaced by a biological gain \(g\), provided the gain enters the observations only through \(cg\). For example, if a known time course \(z(t)\) is multiplied by both factors, \(y(t)=cgz(t)\), the observations cannot separate \(c\) from \(g\). The condition matters: if \(g\) also changes the dynamics or appears elsewhere in the measured outputs, the full model may contain information that distinguishes it from \(c\).

Worked Example: One Observation Does Not Necessarily Identify the Product

Suppose a decay model has \(y(t)=q e^{-kt}\), with \(q=cA_0\), \(k>0\), and one noiseless observation \(y(1\ \mathrm{h})=1.000\) assay-signal unit. Determine whether this observation identifies \(q\).

At \(t=1\ \mathrm{h}\), the model gives \(1.000=q e^{-k(1\ \mathrm{h})}\). One admissible parameter pair is \(k=1\ \mathrm{h}^{-1}\) and \(q=e\) signal units, because

q e^{-k(1\ \mathrm{h})}=e\,e^{-1}=1.000.

A different admissible pair is \(k=2\ \mathrm{h}^{-1}\) and \(q=e^2\) signal units, because

q e^{-k(1\ \mathrm{h})}=e^2e^{-2}=1.000.

The observation therefore does not identify \(q\) when \(k\) is also unknown. Since neither \(q\) nor the factors that form it are identified by this single observation, it would be incorrect to conclude that the product is automatically identifiable merely because the model contains \(cA_0\).

When the Signal Scale Is Identifiable

For the noiseless monoexponential model, two distinct-time observations can identify \(q\) and \(k\), provided the signals are positive and the model assumptions hold. Let \(t_2>t_1\). Dividing the two model equations cancels \(q\):

\frac{y(t_2)}{y(t_1)} =\frac{q e^{-kt_2}}{q e^{-kt_1}} =e^{-k(t_2-t_1)}.

Taking logarithms gives a unique value of \(k\) for positive observed signals, after which either observation determines \(q\):

k=-\frac{\ln\!\left(y(t_2)/y(t_1)\right)}{t_2-t_1}, \qquad q=y(t_1)e^{kt_1}.

These equations establish identifiability of the reduced parameters \(k\) and \(q\) for this model and design. They do not identify \(c\) and \(A_0\) separately: every positive pair satisfying \(cA_0=q\) still gives the same signal trajectory.

Worked Example: Two Times Identify the Product and Decay Rate

For a hypothetical noiseless experiment, suppose the model \(y(t)=q e^{-kt}\) gives \(y(0\ \mathrm{h})=10.00\) signal units and \(y(2\ \mathrm{h})=10e^{-0.6}\approx5.488\) signal units. Assume the decay model is correct, there is no background signal, and the two times are distinct.

Because \(t_1=0\) and \(t_2=2\ \mathrm{h}\), the ratio is

\frac{y(2)}{y(0)} =\frac{10e^{-0.6}}{10} =e^{-0.6}.

Substituting into the rate formula gives

k=-\frac{\ln(e^{-0.6})}{2\ \mathrm{h}-0\ \mathrm{h}} =\frac{0.6}{2\ \mathrm{h}} =0.300\ \mathrm{h}^{-1}.

At \(t=0\), \(y(0)=q\), so \(q=10.00\) assay-signal units. Thus these observations identify the decay rate and product in the reduced model. They do not determine the factors: for example, \(c=2.00\) signal units per mg/L with \(A_0=5.00\) mg/L gives \(q=10.00\), as does \(c=4.00\) signal units per mg/L with \(A_0=2.50\) mg/L.

Information That Can Break the Confounding

To estimate both \(c\) and \(A_0\) separately, the model needs information that does not preserve their product-only dependence. An independent calibration experiment can determine \(c\), while an independently known initial amount can determine \(A_0\). Either source of information can convert an identified product into an estimate of the other factor. Extra measurements of the same type, all governed by \(y(t)=cA_0e^{-kt}\), may improve information about the decay shape or the product but cannot split \(c\) from \(A_0\).

The product-only condition must be checked against the complete observation model. A known additive background \(b\), for example, changes the equation to \(y(t)=b+cA_0e^{-kt}\). If \(b\) is known, subtracting it leaves the same product confounding. But an additional output that contains \(A_0\) without \(c\), or a known standard measured through the same assay, can supply separate scale information. Conversely, merely adding another output does not help if it is also multiplied by the same unknown assay factor and contains the same biological scale only as a product.

Worked Example: Independent Calibration Separates the Biological Amount

Consider the decay model with hypothetical noiseless measurements \(y(0\ \mathrm{h})=10.00\) signal units and \(y(1\ \mathrm{h})=6.7032\) signal units. Suppose a separate assay standard establishes \(c=2.00\) signal units per mg/L. Assume the same calibration factor applies to the samples and standard, the assay is linear in amount, and there is no additive background.

The observation at time zero gives \(y(0)=cA_0\), so the known calibration factor yields

A_0=\frac{y(0)}{c} =\frac{10.00\ \text{signal units}} {2.00\ \text{signal units per mg/L}} =5.00\ \text{mg/L}.

The ratio of the two sample signals eliminates both the calibration factor and \(A_0\), giving the decay rate:

k=-\frac{\ln(6.7032/10.00)}{1\ \mathrm{h}} \approx0.4000\ \mathrm{h}^{-1}.

Here the sample time course identifies \(k\) and the product \(cA_0\), while the independent calibration fixes \(c\) and therefore determines \(A_0\). Without that calibration, the sample observations alone would still permit infinitely many pairs with \(cA_0=10.00\) signal units.

Diagnostics, Alternatives, and Limitations

A useful diagnostic is to inspect how the predicted observations change when each scale parameter changes. For the decay model, the derivatives with respect to the logarithms of the positive parameters are identical:

\frac{\partial y(t)}{\partial \ln c} =y(t) =\frac{\partial y(t)}{\partial \ln A_0}.

Thus a proportional increase in either factor has the same effect on every predicted observation. This is a local indication of the same exact ambiguity proved by the transformation above. In practice, fitting may return one pair of values, or different starting values may lead to different pairs with essentially identical predictions. A good residual fit does not show that both factors have been estimated uniquely.

The main alternatives answer different questions. Collecting more time points can help identify the decay rate and signal scale if the time course is informative, but it does not remove a product-only ambiguity. An independent assay standard provides information about \(c\); an independently measured initial amount provides information about \(A_0\). A different output or observation model can also help, but only if it contributes a constraint that depends on the two factors differently.

These conclusions depend on the model structure. Assay saturation, nonlinear response, an unknown offset, or an assay factor that varies over time requires a different observation equation and can change identifiability. A known offset may be removed before assessing the multiplicative ambiguity; an unknown offset can introduce further ambiguity. Also, even when the reduced product \(q\) is structurally identifiable, finite noisy data may estimate it imprecisely. That is an uncertainty issue distinct from the exact nonuniqueness of \(c\) and \(A_0\).

Common pitfall: “The output contains \(cA_0\), so the data identify \(cA_0\).” The first part may show that \(c\) and \(A_0\) cannot be separated, but it does not prove that the product is identifiable. Unknown kinetic parameters and an insufficient observation design can leave the product undetermined too. First assess identifiability in the reduced model; then assess whether independent information separates the factors.

Check Your Understanding

Use the model assumptions stated in each question. Solutions are included below so that each identifiability claim can be checked explicitly.

For \(y(t)=cA_0e^{-kt}\), show why replacing \(c\) by \(\lambda c\) and \(A_0\) by \(A_0/\lambda\), for \(\lambda>0\), leaves the output unchanged.
One noiseless observation is \(y(1\ \mathrm{h})=1\) signal unit, and \(k\) is unknown. Give two distinct \((q,k)\) pairs that fit it, showing that \(q=cA_0\) need not be identifiable.
In the same decay model, two observations are \(y(0)=8.00\) signal units and \(y(2\ \mathrm{h})=8e^{-0.4}\) signal units. Find \(q\) and \(k\), and state whether \(c\) and \(A_0\) are separately identified.
If a second assay output is also \(y_2(t)=d\,cA_0e^{-kt}\), where \(d\) is known, does that output alone separate \(c\) and \(A_0\)? Explain.

Solutions. (1) The transformed product is \((\lambda c)(A_0/\lambda)=cA_0\), so \(y'(t)=cA_0e^{-kt}=y(t)\) for every \(t\). (2) At \(t=1\ \mathrm{h}\), both \((q,k)=(e\ \text{signal units},1\ \mathrm{h}^{-1})\) and \((q,k)=(e^2\ \text{signal units},2\ \mathrm{h}^{-1})\) give \(y=1\) signal unit; hence \(q\) is not determined by that observation. (3) Since \(y(0)=q\), \(q=8.00\) signal units. Also, \(k=-\ln(y(2)/y(0))/(2\ \mathrm{h})=-\ln(e^{-0.4})/(2\ \mathrm{h})=0.200\ \mathrm{h}^{-1}\). The product and rate are identified for this design, but any positive \(c,A_0\) pair with product \(8.00\) signal units fits, so the factors are not separately identified. (4) No. The second output is just the first multiplied by known \(d\); both still depend on \(c\) and \(A_0\) through the same product, so the transformation in part (1) leaves both outputs unchanged.