Why Turnover Models Produce Delayed Responses
A drug can change a biomarker without changing the biomarker concentration instantaneously. For example, the drug may reduce the rate at which the biomarker is produced, while existing biomarker is removed gradually. Or it may change the rate of biomarker loss, so the amount present moves toward a new balance over time. These are indirect-response mechanisms: the drug acts on a rate of production or loss, and the measured response follows as the biomarker turns over.
In contrast to a direct-effect Emax model, which relates response directly to concentration at a given time, a turnover model specifies how the response changes through time. The central parameters are the baseline response, the rate constant for loss, and parameters describing how concentration modifies production or loss. The delay arises from the biomarker's turnover; the basic models here assume that the drug effect on a rate responds immediately to concentration.
Here \(R(t)\) has the units of the biomarker, \(k_{\mathrm{in}}\) has units of biomarker concentration per unit time, and \(k_{\mathrm{out}}\) has units of inverse time. First-order loss means that the amount lost per unit time is proportional to the amount currently present. At untreated steady state, production and loss are equal, not zero: \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\).
A turnover half-life can be defined as \(\ln(2)/k_{\mathrm{out}}\). It is the time for a deviation from the untreated steady state to halve under the untreated model. It is not necessarily the time to reach a new steady state, which is approached asymptotically.
Four Basic Indirect-Response Models
Let \(C(t)\) denote drug concentration. Effects on production are often represented by saturable inhibition or stimulation:
In these equations, \(I_{\max}\) is the maximum fractional inhibition of production, and \(S_{\max}\) is the maximum fractional stimulation. \(IC_{50}\) and \(SC_{50}\) are concentrations at which the corresponding concentration-dependent fraction reaches half its maximum. Concentration and these potency parameters must use the same units. A usual constraint is \(0\leq I_{\max}\leq1\), which keeps the production rate nonnegative.
Drug effects may instead modify first-order loss:
For loss inhibition, \(I_{\max}\) is the maximum fractional reduction in the loss-rate constant. For loss stimulation, the multiplier on \(k_{\mathrm{out}}\) increases above one. With nonnegative concentration and \(0\leq I_{\max}\leq1\), the effective loss rate remains nonnegative. The four models differ in which rate the drug modifies; they should not be treated as interchangeable merely because they can all yield an increase or decrease in \(R\).
Deriving the Response Trajectory
The key mathematical result follows from a simple balance equation. If concentration is constant and the drug changes production but leaves loss at \(k_{\mathrm{out}}\), write the drug-modified production rate as \(q\). Then \(\frac{dR}{dt}=q-k_{\mathrm{out}}R\). The steady-state response is \(R_{\mathrm{ss}}=q/k_{\mathrm{out}}\). Subtracting this steady state from both sides gives \(\frac{d}{dt}(R-R_{\mathrm{ss}})=-k_{\mathrm{out}}(R-R_{\mathrm{ss}})\). Solving this first-order equation yields:
Thus a production effect changes the destination, \(R_{\mathrm{ss}}\), while the rate of approach is governed by \(k_{\mathrm{out}}\). For an effect on loss, define the concentration-dependent multiplier \(g\) on \(k_{\mathrm{out}}\). At constant exposure, the same argument gives a new steady state \(k_{\mathrm{in}}/(k_{\mathrm{out}}g)\) and an approach rate \(k_{\mathrm{out}}g\).
For a constant concentration, define \(I(C)=I_{\max}C/(IC_{50}+C)\) and \(S(C)=S_{\max}C/(SC_{50}+C)\). If \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\) before dosing, the steady-state response relative to baseline is:
| Drug effect | Steady-state response | Approach rate |
|---|---|---|
| Inhibits production | \(R_0[1-I(C)]\) | \(k_{\mathrm{out}}\) |
| Stimulates production | \(R_0[1+S(C)]\) | \(k_{\mathrm{out}}\) |
| Inhibits loss | \(R_0/[1-I(C)]\) | \(k_{\mathrm{out}}[1-I(C)]\) |
| Stimulates loss | \(R_0/[1+S(C)]\) | \(k_{\mathrm{out}}[1+S(C)]\) |
These results assume a stable untreated baseline, constant concentration long enough for the steady-state calculation to apply, and the specified model structure. If concentration varies with time, the response generally depends on its history, not just the concentration at the observation time. For production modulation with production rate \(q(C(t))\), the exact solution is
The exponential weighting shows why earlier production changes still affect the current response, although their contribution diminishes over time. For loss modulation with time-varying multiplier \(g(C(t))\), the solution is
In practice, these time-varying models are evaluated numerically alongside the concentration-time profile. The formulas do not imply that the response instantly equilibrates to the steady-state value associated with the current concentration.
Worked Example: Inhibition of Biomarker Production
Suppose a hypothetical biomarker has baseline \(R_0=50\) pmol/L and \(k_{\mathrm{out}}=0.20\) day\(^{-1}\). Assume baseline steady state, so \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0=0.20\times50=10\) pmol/L/day. During constant exposure, \(C=4\) mg/L, \(IC_{50}=4\) mg/L, and \(I_{\max}=0.80\). Find the response after five days.
The inhibited fraction of production is \(I(C)=0.80\times4/(4+4)=0.40\). Production under drug is therefore \(10(1-0.40)=6\) pmol/L/day, and the new steady state is \(R_{\mathrm{ss}}=6/0.20=30\) pmol/L. Starting from baseline, the trajectory is
The response has fallen below baseline but remains above the new steady state because only five days have elapsed. The calculation assumes constant exposure, unchanged loss, and no other time-dependent change in biomarker production.
Worked Example: Inhibition of Biomarker Loss
Consider a hypothetical biomarker with \(R_0=80\) ng/mL and \(k_{\mathrm{out}}=0.10\) day\(^{-1}\), so \(k_{\mathrm{in}}=8\) ng/mL/day. Let \(C=9\) mg/L, \(IC_{50}=3\) mg/L, and \(I_{\max}=0.60\). Calculate the response after ten days of constant exposure.
First, \(I(C)=0.60\times9/(3+9)=0.45\). The loss-rate multiplier is \(1-0.45=0.55\), so the effective loss-rate constant is \(0.10\times0.55=0.055\) day\(^{-1}\). The steady state is \(R_{\mathrm{ss}}=8/0.055=145.5\) ng/mL. Starting from baseline:
The response increases because production continues while loss is reduced. The effective approach is slower than the untreated approach, as \(0.055\) day\(^{-1}\) is less than \(0.10\) day\(^{-1}\). The steady state is finite under the stated parameters because the loss-rate multiplier is positive.
Worked Example: Response After Drug Withdrawal
Suppose a hypothetical biomarker has \(R_0=60\) units/L and \(k_{\mathrm{out}}=0.25\) day\(^{-1}\), with constant baseline production \(k_{\mathrm{in}}=15\) units/L/day. A production-inhibiting drug produces \(I(C)=0.50\) for four days, then its concentration becomes negligible. Find the biomarker value at day eight.
During dosing, the new steady state is \(60(1-0.50)=30\) units/L. At the end of day four, \(R(4)=30+(60-30)e^{-0.25\times4}=30+30e^{-1}=41.04\) units/L. After withdrawal, production returns to baseline, so the response approaches 60 units/L with the untreated loss rate. Four more days later:
The response remains below baseline after withdrawal because the biomarker requires time to turn over. This calculation assumes that exposure falls to a negligible level at day four and that the drug effect follows concentration directly. The residual response does not, by itself, establish a delayed drug action at the site of effect.
Assumptions, Diagnostics, and Model Choice
Interpretation depends on the biological and mathematical assumptions. In the basic forms presented here, production and loss follow the specified rate laws; parameters are constant over the modeled period; concentration is an appropriate driver of the rate effect; and the baseline is stable or its time course is otherwise accounted for. A simple model also assumes that one rate modification is sufficient to describe the observed response. These are modeling assumptions to assess, not facts guaranteed by the equations.
Useful checks include asking whether untreated observations support a stable baseline, whether the predicted response direction matches the data, and whether the observed rise or fall is compatible with the proposed turnover rate. Inconsistent predictions across dose levels, systematic time-patterned residuals, or an apparent baseline drift can indicate that the model is inadequate or its assumptions do not hold. Sparse sampling can make the rate of approach difficult to distinguish from the magnitude of the drug effect; sampling both during exposure and after changes in exposure can be informative.
The main alternative at this stage is a direct-effect model, which relates concentration to response without representing production and loss explicitly. It may be adequate when the response closely tracks concentration. A turnover model is more appropriate when the biological mechanism plausibly changes a rate and the observed response changes gradually. Comparing the alternatives requires attention to the time course and biological plausibility, not only to whether each can fit the observed values.
Turnover models have limitations. The simple equations may misrepresent changing baseline physiology, multiple biomarker compartments, or effects that alter both production and loss. Their estimated parameters can be weakly identified when exposure changes little or observations cover only a short portion of the response trajectory. Further, a good fit does not prove the proposed biological mechanism: different mechanisms can produce similar measured trajectories. The parameters should be interpreted as features of the assumed model and data, with uncertainty considered in development decisions.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A biomarker is at a baseline of 40 units/L with \(k_{\mathrm{out}}=0.25\) day\(^{-1}\). What baseline production rate is implied by steady state?
- Under constant exposure, production is reduced to 70% of baseline, while loss is unchanged. What is the new steady-state response as a fraction of baseline?
- For the model in Question 2, the untreated baseline is 40 units/L and \(k_{\mathrm{out}}=0.25\) day\(^{-1}\). Starting at baseline, calculate the response after four days.
- A concentration reduces the loss-rate constant by 20%. If \(R_0=100\) units/L and production is unchanged, what is the new steady-state response?
- Why can a response remain changed after concentration has fallen to a negligible level?
Solutions. (1) At baseline steady state, \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0=0.25\times40=10\) units/L/day. (2) With production at 70% and unchanged loss, the steady-state response is \(0.70R_0\). (3) The new steady state is \(0.70\times40=28\) units/L. Thus \(R(4)=28+(40-28)e^{-0.25\times4}=28+12e^{-1}\approx32.41\) units/L. (4) The new loss-rate multiplier is \(0.80\), so \(R_{\mathrm{ss}}=k_{\mathrm{in}}/(0.80k_{\mathrm{out}})=R_0/0.80=125\) units/L. (5) The biomarker does not instantaneously return to its baseline balance. Once the drug effect on production or loss subsides, the remaining departure from baseline decays according to the turnover dynamics.