Turnover Models With Indirect-Response Observations

Why Turnover Models Require an Identifiability Check

In an indirect-response model, a drug changes the production or removal of a measured biological quantity, and the quantity then changes over time according to its turnover. The response therefore carries information about both the drug effect and the underlying production and degradation rates. But these parameters are not automatically distinguishable: a baseline measurement alone, for example, gives a ratio of production to degradation rather than either rate separately.

This tutorial uses inhibition of production as a representative drug-modulated turnover model. The same reasoning applies to stimulation of production after replacing the inhibition function with the appropriate stimulation function. The goal is to work out what ideal response observations can identify, given known drug concentrations and a specified model.

Model and assumptions: Let $R(t)$ be the observed response, in response units, and let $C(t)$ be a known drug concentration, in mg/L. The model is $$ \frac{dR}{dt}=k_{\mathrm{in}}\left(1-\frac{I_{\max}C(t)}{IC_{50}+C(t)}\right)-k_{\mathrm{out}}R(t). $$ Here $k_{\mathrm{in}}>0$ is the zero-drug production rate, in response units per unit time; $k_{\mathrm{out}}>0$ is the first-order degradation rate, in inverse time; $0\leq I_{\max}\leq 1$ is the maximum fractional inhibition; and $IC_{50}>0$ is the concentration giving half of that maximum inhibition. Assume the model form is correct, $C(t)$ is known, the response is measured on an absolute scale without error for the structural-identifiability argument, and no unmodelled input or feedback affects $R$.

With no drug, a steady response satisfies $dR/dt=0$. Its baseline value $R_0$ therefore obeys

R_0=\frac{k_{\mathrm{in}}}{k_{\mathrm{out}}}.

This relation is the first identifiability warning: knowing $R_0$ alone does not determine $k_{\mathrm{in}}$ and $k_{\mathrm{out}}$ separately. A time course, or additional information, is needed to distinguish them. Structural identifiability asks whether ideal observations under the stated model can uniquely determine a parameter (apart from exceptional parameter values); it is not a claim that finite, noisy data will estimate it precisely.

Deriving the Ideal Output

Define the fractional inhibition at concentration $C$ as $E(C)=I_{\max}C/(IC_{50}+C)$. The production rate is then $k_{\mathrm{in}}[1-E(C)]$. For any known concentration profile, multiply the differential equation by the integrating factor $e^{k_{\mathrm{out}}t}$, integrate from $0$ to $t$, and divide by $e^{k_{\mathrm{out}}t}$. This gives the ideal response output:

R(t)=R(0)e^{-k_{\mathrm{out}}t} +k_{\mathrm{in}}\int_0^t e^{-k_{\mathrm{out}}(t-s)} \left(1-\frac{I_{\max}C(s)}{IC_{50}+C(s)}\right)\,ds.

The expression shows why the response is indirect: the current observation depends on the history of production, with older input contributions attenuated by degradation. If concentration is constant at $C$ over an interval, the production rate on that interval is constant. Solving the resulting first-order equation yields

R(t)=R_{\mathrm{ss}}(C)+\left[R(0)-R_{\mathrm{ss}}(C)\right]e^{-k_{\mathrm{out}}t}, \qquad R_{\mathrm{ss}}(C)=\frac{k_{\mathrm{in}}}{k_{\mathrm{out}}} \left(1-\frac{I_{\max}C}{IC_{50}+C}\right).

The response approaches the drug-specific steady state $R_{\mathrm{ss}}(C)$ at a rate set by $k_{\mathrm{out}}$. When a concentration step begins from an untreated steady state, $R(0)=R_0$, so the baseline and the post-step trajectory jointly contain more information than either steady state alone.

Worked Example: Solving a Constant-Concentration Response

Suppose a hypothetical system has $k_{\mathrm{in}}=12$ response units/day, $k_{\mathrm{out}}=0.40$ day$^{-1}$, $I_{\max}=0.80$, and $IC_{50}=2.0$ mg/L. Before dosing, the system is at its untreated steady state. At time zero, concentration steps to $4.0$ mg/L and remains there. Find the baseline and the response at 2 days.

The baseline is $$ R_0=\frac{12\ \text{response units/day}}{0.40\ \text{day}^{-1}} =30\ \text{response units}. $$ At $C=4.0$ mg/L, fractional inhibition is $$ E(4)=\frac{0.80(4.0)}{2.0+4.0}=0.5333. $$ Thus the remaining production rate is $12(1-0.5333)=5.600$ response units/day, and the drug-specific steady state is $5.600/0.40=14.00$ response units. The step-response solution is $$ R(t)=14.00+(30.00-14.00)e^{-0.40t} =14.00+16.00e^{-0.40t}. $$ At 2 days, $$ R(2)=14.00+16.00e^{-0.80} =14.00+16.00(0.4493) =21.19\ \text{response units}. $$ The declining response is consistent with inhibition of production: the new equilibrium is below baseline, and the response moves toward it with a time constant $1/k_{\mathrm{out}}=2.50$ days.

When the Parameters Separate

Under ideal continuous observation, absolute response values, a known concentration profile, and a known untreated steady-state baseline, the baseline identifies the ratio $k_{\mathrm{in}}/k_{\mathrm{out}}$. A resolved transient after a concentration change identifies the exponential rate $k_{\mathrm{out}}$. Once $k_{\mathrm{out}}$ is known, the baseline gives $k_{\mathrm{in}}=R_0k_{\mathrm{out}}$. This separation requires a visible transient: if the only information is a single steady-state value, the exponential rate is absent.

Drug-effect parameters require informative exposure variation too. At a known constant concentration $C>0$, if $k_{\mathrm{in}}$ and $k_{\mathrm{out}}$ have been determined, the response plateau determines the fractional inhibition $E(C)$. It does not generally determine $I_{\max}$ and $IC_{50}$ separately. Indeed, every pair satisfying $$ I_{\max}=E(C)\frac{IC_{50}+C}{C} $$ produces the same inhibition at that one concentration. Two distinct informative concentrations can, generically, separate the pair. Let $E_j=E(C_j)$ and $q_j=E_j/C_j$. Then $$ q_j=\frac{I_{\max}}{IC_{50}+C_j}, \qquad IC_{50}=\frac{q_2C_2-q_1C_1}{q_1-q_2}, \qquad I_{\max}=q_1(IC_{50}+C_1), $$ provided the concentrations and observed effects make the denominator nonzero. These formulas describe generic identification; values at zero concentration, identical concentrations, or other degenerate cases do not provide the same information.

Worked Example: Separating Maximum Inhibition and IC50

In a hypothetical experiment, other model parameters are already determined. At two constant concentrations, plateau responses imply fractional inhibitions $E_1=0.400$ at $C_1=2.0$ mg/L and $E_2=0.600$ at $C_2=6.0$ mg/L. Determine $IC_{50}$ and $I_{\max}$.

Calculate the concentration-scaled effects: $$ q_1=\frac{E_1}{C_1}=\frac{0.400}{2.0}=0.200\ (\text{mg/L})^{-1}, \qquad q_2=\frac{E_2}{C_2}=\frac{0.600}{6.0}=0.100\ (\text{mg/L})^{-1}. $$ Then $$ IC_{50}=\frac{(0.100)(6.0)-(0.200)(2.0)}{0.200-0.100} =\frac{0.600-0.400}{0.100}=2.0\ \text{mg/L}. $$ Finally, $$ I_{\max}=q_1(IC_{50}+C_1) =(0.200)(2.0+2.0)=0.800. $$ The recovered values are $IC_{50}=2.0$ mg/L and $I_{\max}=0.800$, consistent with both observed fractional inhibitions. In an actual turnover analysis, those inhibitions must be inferred from the response plateaus using the model and the already identified baseline and degradation parameters.

Worked Example: Why One Concentration Is Not Enough

Suppose a hypothetical constant-concentration experiment at $C=4.0$ mg/L identifies $E(C)=0.5333$. Show that this observation does not uniquely determine the drug-effect parameters.

One parameter pair is $I_{\max}=0.800,\ IC_{50}=2.0$ mg/L, since $$ \frac{0.800(4.0)}{2.0+4.0}=0.5333. $$ A different pair, $I_{\max}=0.6667,\ IC_{50}=1.0$ mg/L, gives $$ \frac{0.6667(4.0)}{1.0+4.0}=0.5334, $$ with the small difference due only to rounding; using the exact value $I_{\max}=2/3$ gives $E=8/15=0.5333$. Both pairs therefore predict the same response trajectory at this fixed concentration when the turnover parameters and initial response are held fixed. A more precise observation of that same trajectory cannot distinguish the pairs; exposure at additional informative concentrations is needed.

Worked Example: Normalization Hides Production Scale

Consider the parameters from the first example, but suppose the only recorded output is the normalized response $Z(t)=R(t)/R_0$. Show what information is lost.

Because $R_0=k_{\mathrm{in}}/k_{\mathrm{out}}$, differentiating $Z=R/R_0$ gives $$ \frac{dZ}{dt}=k_{\mathrm{out}}\left(1-E(C)-Z\right). $$ The production rate $k_{\mathrm{in}}$ no longer appears. At $C=4.0$ mg/L, the normalized trajectory is $$ Z(t)=1-E(4)+[1-(1-E(4))]e^{-0.40t} =0.4667+0.5333e^{-0.40t}. $$ At 2 days, $Z(2)=0.7063$. Multiplying both $k_{\mathrm{in}}$ and the absolute response scale by 2 would leave this normalized trajectory unchanged, while changing $k_{\mathrm{in}}$ from 12 to 24 response units/day. Thus normalized observations can retain turnover-rate and drug-effect information under suitable inputs, but cannot recover the absolute production scale without an absolute response measurement or other scale information.

Diagnostics, Alternatives, and Limits

The ideal derivation also suggests how to assess whether the experiment can support the intended conclusions. First check whether the input actually excites the model. An untreated baseline alone identifies only $k_{\mathrm{in}}/k_{\mathrm{out}}$; a drug response observed only after it has reached a plateau lacks the transient needed to estimate $k_{\mathrm{out}}$ from its time course. Conversely, dense early observations after a concentration step help characterize the transient, while observations late enough to approach the new plateau help characterize the drug effect.

Exposure range matters for $I_{\max}$ and $IC_{50}$. If all concentrations are much smaller than $IC_{50}$, then $$ E(C)\approx \frac{I_{\max}}{IC_{50}}C. $$ The data then mainly inform the ratio $I_{\max}/IC_{50}$, not the two parameters separately. If all concentrations are much larger than $IC_{50}$, then $E(C)\approx I_{\max}$: maximum inhibition may be informed, but the concentration scale $IC_{50}$ is weakly informed. These are often practical-identifiability problems even when exact structural identifiability holds at finite, distinct concentrations.

A useful alternative is to estimate parameters from the full time course under one joint model fit, rather than first estimating plateaus and then fitting a concentration-effect curve. Joint fitting uses transient and plateau information together and avoids treating estimated plateaus as error-free. It does not create information absent from the input: a single constant concentration still cannot generically separate $I_{\max}$ from $IC_{50}$. Multiple concentration levels, or a sufficiently informative varying concentration profile, provide the needed excitation. If an independently justified value for one drug-effect parameter is available, one concentration may then suffice to determine the other.

The conclusions depend on the model assumptions. A delayed response, nonlinear degradation, time-varying baseline production, uncertain concentration input, or an unknown assay scale can change the output equation and the identifiability conclusions. The observed response must also be the state $R$, or have a known relationship to it. A good fit alone does not establish that the parameters are structurally identifiable; examine whether distinct parameter sets could generate the same ideal output under the actual input design.

Common pitfall: “The response changed after dosing, so the data identify $I_{\max}$ and $IC_{50}$.” A change at one fixed concentration may identify only $E(C)=I_{\max}C/(IC_{50}+C)$. To claim that the two drug-effect parameters are separately identifiable, show that the input and observations distinguish their effects at multiple informative concentrations, and that the turnover parameters and response scale are also resolved.

Key takeaway: An absolute untreated baseline identifies $k_{\mathrm{in}}/k_{\mathrm{out}}$; a resolved turnover transient can identify $k_{\mathrm{out}}$, allowing $k_{\mathrm{in}}$ to be recovered. Separating $I_{\max}$ from $IC_{50}$ additionally requires informative concentration variation. Structural identifiability is a property of the model and experimental input, not a guarantee of precise estimates from noisy data.

Check Your Understanding

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

If the only observation is an untreated steady-state response of 18 response units, what relationship between $k_{\mathrm{in}}$ and $k_{\mathrm{out}}$ is determined?
Why can a resolved post-dose transient help distinguish production from degradation?
At one nonzero concentration, what drug-effect quantity is generally identified from the plateau if turnover parameters are known?
What information about $I_{\max}$ and $IC_{50}$ is weak when every observed concentration is far below $IC_{50}$?

Solutions. (1) The steady-state equation gives $k_{\mathrm{in}}/k_{\mathrm{out}}=18$ response units; it does not determine either rate alone. (2) The transient contains the factor $e^{-k_{\mathrm{out}}t}$, so its rate of approach identifies $k_{\mathrm{out}}$ under the model; combining that with the baseline ratio gives $k_{\mathrm{in}}$. (3) It generally identifies the fractional inhibition $E(C)=I_{\max}C/(IC_{50}+C)$, not the two parameters separately. (4) Since $E(C)\approx (I_{\max}/IC_{50})C$, the data mainly inform the ratio $I_{\max}/IC_{50}$, leaving the separate values weakly determined.