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An Indirect-Response Case for Antihypertensive Biomarkers

Work through two indirect-response models for blood pressure and learn how turnover dynamics, sampling times, and model diagnostics determine whether the data distinguish them.

Advanced 11 min read

What You'll Learn

  • Write inhibition-of-production and stimulation-of-loss models for a continuously measured blood-pressure biomarker.
  • Derive baseline and drug-exposed steady states from the turnover equations.
  • Show how the two mechanisms can produce identical steady-state concentration-response curves.
  • Calculate and compare their different time courses after a concentration change.
  • Assess mechanism distinguishability using sampling design, residuals, and model comparison.
  • Recognize when sparse or noisy data cannot support a mechanistic conclusion.

Why the Mechanism Matters

An antihypertensive drug may lower blood pressure by reducing the rate at which a physiological process produces the measured pressure signal, or by increasing the rate at which that signal is lost. If blood pressure is measured repeatedly while drug concentrations change, an indirect-response model represents both the biomarker’s natural turnover and the drug’s effect on that turnover. Comparing mechanisms is not simply a matter of finding which curve fits best: the sampling must contain information that separates their predicted time courses.

Let \(C(t)\) be the measured drug concentration at time \(t\), and let \(R(t)\) be the continuously measured blood-pressure biomarker, in mmHg. The examples below use hypothetical values. Concentration units are ng/mL and time units are hours. The two models use the same baseline turnover but place the drug effect on different parts of the turnover equation.

Definition: In an indirect-response model, a measured response changes over time because it is produced and removed. In the inhibition-of-production model, drug concentration reduces the production rate. In the stimulation-of-loss model, drug concentration increases the loss rate. The parameters \(k_{\mathrm{in}}\) and \(k_{\mathrm{out}}\) are the baseline production and loss rate constants; \(I_{\max}\) and \(S_{\max}\) are maximum fractional effects; and \(IC_{50}\) and \(SC_{50}\) are concentrations producing half of the corresponding maximum effects.

Specify the Competing Models

Assume that, without drug, production and loss are first-order with respect to the response: production is constant at \(k_{\mathrm{in}}\), while loss is \(k_{\mathrm{out}}R\). Under the inhibition-of-production hypothesis, a saturable concentration effect inhibits production:

$$ \frac{dR}{dt} = k_{\mathrm{in}}\left(1-\frac{I_{\max}C(t)}{IC_{50}+C(t)}\right) -k_{\mathrm{out}}R. $$

Under the stimulation-of-loss hypothesis, production remains at its baseline rate, while drug increases the loss rate:

$$ \frac{dR}{dt} = k_{\mathrm{in}} -k_{\mathrm{out}}\left(1+\frac{S_{\max}C(t)}{SC_{50}+C(t)}\right)R. $$

These equations assume nonnegative concentrations, positive baseline turnover rates, and a response whose change is adequately described by the specified production and loss processes. The fractions describe saturable drug effects. For the production model, \(0\leq I_{\max}\leq 1\) ensures that the drug does not make the modeled production rate negative. Both models assume that concentration affects the response through the stated instantaneous concentration-effect function; they do not add an unobserved delay between concentration and effect.

With no drug, the baseline equilibrium \(R_0\) satisfies \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\). Thus \(R_0=k_{\mathrm{in}}/k_{\mathrm{out}}\). At a fixed concentration \(C\), the production-inhibition model has equilibrium \(R_0\left(1-I_{\max}C/(IC_{50}+C)\right)\); the loss-stimulation model has equilibrium \(R_0/\left(1+S_{\max}C/(SC_{50}+C)\right)\). Those equilibrium expressions alone do not reveal which mechanism generated a measured response.

Worked Example: Establishing Baseline Turnover

Suppose a hypothetical participant has a stable pretreatment blood pressure of \(120\) mmHg, and the estimated baseline loss rate constant is \(k_{\mathrm{out}}=0.20\ \mathrm{h}^{-1}\). Under the no-drug equilibrium assumption, calculate the production rate.

At equilibrium, the derivative is zero, so \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\). Substituting the stated values gives

$$ k_{\mathrm{in}} = (0.20\ \mathrm{h}^{-1})(120\ \mathrm{mmHg}) = 24\ \mathrm{mmHg/h}. $$

The units are consistent: multiplying a rate constant per hour by a response in mmHg gives a production rate in mmHg per hour. This calculation also illustrates an assumption that must be checked: if pretreatment blood pressure is still trending, treating it as a stable baseline equilibrium may be inappropriate.

Deriving the Time Course at Constant Concentration

When concentration is held at a fixed value, either model can be written as \(dR/dt=a-bR\), where \(a\) is the drug-dependent production rate and \(b\) is the drug-dependent loss-rate constant. Its equilibrium is \(R_{\mathrm{ss}}=a/b\). The time-course solution is

$$ R(t)=R_{\mathrm{ss}}+\left(R(0)-R_{\mathrm{ss}}\right)e^{-bt}. $$

To justify the solution, first consider the equilibrium initial condition. If \(R(0)=R_{\mathrm{ss}}\), then \(R(t)=R_{\mathrm{ss}}\) for all \(t\), as direct substitution into the differential equation confirms. If \(R(0)\ne R_{\mathrm{ss}}\), write \(d(R-R_{\mathrm{ss}})/dt=-b(R-R_{\mathrm{ss}})\). Separating variables and integrating gives \(\log|R(t)-R_{\mathrm{ss}}|=-bt+\text{constant}\). Applying the initial condition yields the displayed solution. Thus the derivation covers both the equilibrium and nonequilibrium initial conditions.

The parameter \(b\) determines how quickly the response approaches equilibrium. In the production-inhibition model \(b=k_{\mathrm{out}}\), even during drug exposure. In the loss-stimulation model \(b=k_{\mathrm{out}}\left(1+S_{\max}C/(SC_{50}+C)\right)\), which is larger than baseline whenever the drug stimulates loss. Different \(b\) values can create different trajectories even when the models predict the same equilibrium.

Worked Example: Comparing Time Courses at the Same Concentration

Continue with \(R_0=120\) mmHg, \(k_{\mathrm{in}}=24\) mmHg/h, and \(k_{\mathrm{out}}=0.20\ \mathrm{h}^{-1}\). Suppose concentration is held at \(10\) ng/mL. For inhibition of production, set \(I_{\max}=0.20\) and \(IC_{50}=10\) ng/mL. For stimulation of loss, set \(S_{\max}=0.25\) and \(SC_{50}=12.5\) ng/mL. Calculate each equilibrium and the predicted blood pressure after 3 hours, starting from \(120\) mmHg.

For production inhibition, the fractional inhibition at \(10\) ng/mL is \(0.20(10)/(10+10)=0.10\). The production rate under drug is \(24(1-0.10)=21.6\) mmHg/h, so \(R_{\mathrm{ss}}=21.6/0.20=108\) mmHg. The loss rate constant remains \(0.20\ \mathrm{h}^{-1}\). Therefore,

$$ R(3)=108+(120-108)e^{-(0.20)(3)} =108+12e^{-0.60} \approx 114.6\ \mathrm{mmHg}. $$

For stimulation of loss, the fractional stimulation is \(0.25(10)/(12.5+10)=0.1111\). The effective loss rate constant is \(0.20(1+0.1111)=0.2222\ \mathrm{h}^{-1}\). Thus \(R_{\mathrm{ss}}=24/0.2222=108\) mmHg, and

$$ R(3)=108+12e^{-(0.2222)(3)} =108+12e^{-0.6667} \approx 114.2\ \mathrm{mmHg}. $$

Both mechanisms predict the same equilibrium, \(108\) mmHg, but their 3-hour predictions differ by only about \(0.425\) mmHg. That difference may be too small to distinguish from measurement error or natural within-person variation. A mathematical difference between predictions is not, by itself, evidence that a study can identify the mechanism.

Why Steady-State Data May Not Identify the Mechanism

In fact, the two models can produce exactly the same steady-state concentration-response curve. Write \(K_i=IC_{50}\), \(K_s=SC_{50}\), and \(r=1-I_{\max}\). The production-inhibition equilibrium relative to baseline is

$$ \frac{R_{\mathrm{ss}}}{R_0} = 1-\frac{I_{\max}C}{K_i+C} = \frac{K_i+rC}{K_i+C}. $$

For the loss-stimulation model, the corresponding ratio is \((K_s+C)/(K_s+(1+S_{\max})C)\). If \(0<I_{\max}<1\), choose \(S_{\max}=I_{\max}/r\) and \(K_s=K_i/r\). Then \(1+S_{\max}=1/r\), so the loss-model ratio becomes \((K_i/r+C)/(K_i/r+C/r)\). Multiplying numerator and denominator by \(r\) gives \((K_i+rC)/(K_i+C)\), exactly the production-model ratio. The two models are therefore indistinguishable from equilibrium response versus concentration alone under this parameter mapping. Dynamic observations, especially after concentration changes, are needed to exploit their different response rates.

Worked Example: Matching the Equilibrium Curves

Let the production model have \(I_{\max}=0.20\) and \(IC_{50}=10\) ng/mL. Then \(r=1-0.20=0.80\). Under the mapping above, the corresponding loss-model parameters are \(S_{\max}=0.20/0.80=0.25\) and \(SC_{50}=10/0.80=12.5\) ng/mL.

At \(C=10\) ng/mL, the production-model ratio is \(1-0.20(10)/(10+10)=0.9000\). The loss-model ratio is \(1/[1+0.25(10)/(12.5+10)]=1/1.1111=0.9000\). Each model therefore predicts \(0.9000(120)=108\) mmHg at equilibrium. Because the mapping makes the curves equal at every nonnegative concentration, adding more steady-state concentration levels does not, on its own, resolve the mechanism.

Assessing Distinguishability in a Study

A useful study needs accurate concentration measurements and response observations that cover the changing phase as well as, if feasible, the approach to equilibrium. Measurements clustered only before dosing and long after the response has stabilized mainly inform baseline and equilibrium. Samples taken while concentration or response is changing can reveal the different effective turnover rates. Replicate measurements and a suitable placebo or untreated reference may help characterize background variation, but they do not replace informative timing.

Model comparison should use the same response observations and a clearly specified measurement-error model. For example, if errors are independent, normally distributed, and have a common variance, the residual sum of squares is a basic measure of fit. When models have different numbers of estimated parameters, a complexity-adjusted criterion such as AIC can help compare them; lower AIC indicates a better trade-off between fit and parameter count within the candidate set. Neither criterion proves the biological mechanism, and a small difference may reflect inadequate information rather than a genuine preference.

Worked Example: Reading a Small Dynamic Comparison

Suppose the fixed parameters from the preceding time-course example are treated as given, and a hypothetical participant has measurements of \(114.4\) mmHg at 3 hours and \(110.3\) mmHg at 8 hours after concentration reaches and remains at \(10\) ng/mL. The baseline value is \(120\) mmHg. Compare squared residuals for the two sets of predictions.

At 8 hours, the production model predicts \(108+12e^{-1.60}\approx110.4\) mmHg. Its residuals, observed minus predicted, are approximately \(-0.186\) mmHg at 3 hours and \(-0.123\) mmHg at 8 hours. The sum of squared residuals is \((-0.186)^2+(-0.123)^2\approx0.0497\ \mathrm{mmHg}^2\).

The loss model predicts \(108+12e^{-(0.2222)(8)}\approx110.0\) mmHg at 8 hours. Its residuals are approximately \(0.239\) mmHg at 3 hours and \(0.272\) mmHg at 8 hours. Its sum of squared residuals is \((0.239)^2+(0.272)^2\approx0.131\ \mathrm{mmHg}^2\). The production model fits these two points more closely with the fixed parameters. However, these two residual sums do not establish a reliable mechanism: there are only two post-dose observations, no estimated measurement variance, and no allowance here for refitting parameters or individual variability.

Diagnostics, Limitations, and Common Pitfalls

Inspect residuals against time, predicted response, and concentration. Systematic runs of positive or negative residuals, or a pattern in which residuals change sign as concentration rises and falls, can indicate a misspecified turnover model, an inadequate concentration-effect function, or unmodeled variability. Residuals that grow with predicted blood pressure may call the constant-variance assumption into question. Also check whether fitted parameters are scientifically plausible and whether predictions remain sensible over the observed concentration range. A visually close curve is not reassuring if parameters are poorly estimated or markedly correlated.

The models are deliberately simple. They do not automatically account for measurement error in concentration, time-varying baseline blood pressure, adherence, covariates, or between-person differences. If concentration and response are measured at too few time points, or if measurement noise is large relative to the predicted difference, data may support an antihypertensive exposure-response relationship without supporting either specific mechanism. More flexible models may fit better but require additional information and can make the mechanism even harder to identify.

  • Pitfall: equating a good steady-state fit with mechanism identification. The wrong reasoning is that one mechanism must be correct because its equilibrium curve closely follows the observed concentration-response relationship. The correction is that the competing models can have exactly matching equilibrium curves; dynamic observations are needed to compare their different turnover rates.
  • Pitfall: treating a small fit advantage as proof. The wrong reasoning is that the model with the lower residual sum of squares has revealed the biological pathway. The correction is to consider measurement variability, parameter uncertainty, model complexity, residual patterns, and whether the study sampled the informative parts of the time course.

The practical comparison is between a constrained pair of mechanisms, not between two labels attached to the same curve. Inhibition of production changes the source term while leaving baseline loss unchanged; stimulation of loss changes the response-dependent removal term. Their distinct dynamic predictions are valuable only when the data are precise and timed to expose them.

Check Your Understanding

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

  1. In a no-drug equilibrium with \(R_0=130\) mmHg and \(k_{\mathrm{out}}=0.15\ \mathrm{h}^{-1}\), what is \(k_{\mathrm{in}}\)?
  2. Why can equilibrium measurements across many concentrations still fail to distinguish production inhibition from loss stimulation?
  3. At fixed concentration, what does the parameter \(b\) represent in the solution \(R(t)=R_{\mathrm{ss}}+(R(0)-R_{\mathrm{ss}})e^{-bt}\)?
  4. Name one sampling feature that can help distinguish these mechanisms and one diagnostic that can reveal a poor model fit.

Solutions. (1) \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0=(0.15)(130)=19.5\ \mathrm{mmHg/h}\). (2) A parameter mapping can make the models’ steady-state concentration-response curves identical, so equilibrium observations alone may contain no information about which turnover component the drug affects. (3) \(b\) is the effective loss-rate constant and determines the speed of approach to equilibrium; the time constant is \(1/b\). (4) Sampling during the response’s changing phase after concentration changes can help; systematic residual patterns against time or concentration can signal model misspecification.

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