From Drug Concentration to a Changing Response
An indirect response model describes a measured response that changes because a drug alters the process producing that response, the process removing it, or both. The response might be a biomarker concentration, a physiological measurement, or another quantity that changes over time. The model represents these changes with an ordinary differential equation (ODE): the response’s instantaneous rate of change is the rate entering its state minus the rate leaving it.
This distinction matters because drug action on a production or loss process is not the same as an immediate additive change in the measured response. A drug can alter the rate at which response is generated or removed, while the response itself evolves over time. The equations below make those rates explicit.
The subscript 0 in \(k_{\mathrm{in},0}\) identifies the production rate in the absence of drug, not the response value at time zero. Likewise, \(k_{\mathrm{out}}\) is a rate constant, not a response loss rate by itself: the loss rate depends on the current response through the product \(k_{\mathrm{out}}R\). This model assumes first-order loss unless its loss term is explicitly replaced by another function.
The Four Canonical State Equations
An indirect response ODE combines a production term and a loss term. Drug effects are expressed as dimensionless functions of the driving concentration. Let \(I_p(C)\) and \(I_l(C)\) denote inhibition functions for production and loss, and let \(S_p(C)\) and \(S_l(C)\) denote stimulation functions for production and loss. The subscripts distinguish the process affected; the functions need not have the same parameters or form.
These are four alternatives, not four terms that must all be included in one equation. Each equation specifies which rate drug modifies. The brackets multiplying \(k_{\mathrm{in},0}\) scale production; the brackets multiplying \(k_{\mathrm{out}}R\) scale loss. With no drug effect, each multiplier is one, and all four equations reduce to \(dR/dt=k_{\mathrm{in},0}-k_{\mathrm{out}}R\).
A common concentration-effect specification is a saturable function. For inhibition, one standard form is \(I(C)=I_{\max}C/(IC_{50}+C)\), with \(C\geq 0\), \(0\leq I_{\max}\leq 1\), and \(IC_{50}>0\). For stimulation, one standard form is \(S(C)=S_{\max}C/(SC_{50}+C)\), with \(S_{\max}\geq 0\) and \(SC_{50}>0\). The concentration and half-maximal concentration must use the same units. At \(C=0\), both functions are zero. As concentration increases, inhibition approaches \(I_{\max}\), while stimulation approaches \(S_{\max}\).
Why the Equations Have This Form
The state equation follows from a rate balance. Over a sufficiently short time interval, the response changes by the amount produced minus the amount lost. Dividing that change by the interval and taking the limit gives the derivative \(dR/dt\). Under the stated first-order loss assumption, the zero-drug loss rate is proportional to the current response, so it is \(k_{\mathrm{out}}R\). A drug multiplier on production or loss modifies that rate without changing the units of the response.
For example, if the drug inhibits production by a fraction \(I_p(C)\), the production rate is \(k_{\mathrm{in},0}-k_{\mathrm{in},0}I_p(C)=k_{\mathrm{in},0}[1-I_p(C)]\). Subtracting the unchanged loss rate yields the inhibited-production equation. If instead the drug inhibits loss, the loss rate becomes \(k_{\mathrm{out}}R-k_{\mathrm{out}}R I_l(C)=k_{\mathrm{out}}[1-I_l(C)]R\), while production remains \(k_{\mathrm{in},0}\). The stimulation equations follow by adding the corresponding fraction of the zero-drug rate rather than subtracting it.
The units provide a basic check. Both \(k_{\mathrm{in},0}\) and \(k_{\mathrm{out}}R\) have units of response per time, and each drug-effect multiplier is dimensionless. Thus the right-hand side has the same units as \(dR/dt\). A proposed equation that subtracts a dimensionless inhibition function directly from a response rate, or adds a concentration to a response rate, is not dimensionally consistent.
Worked Calculations
Worked Example: Calculating an Inhibited Production Rate
Suppose a hypothetical response is measured in units, with \(k_{\mathrm{in},0}=12\) units/h and \(k_{\mathrm{out}}=0.10\) h\(^{-1}\). At the current time, \(R=80\) units and the driving concentration is \(C=2\) mg/L. Assume \(I_p(C)=I_{\max}C/(IC_{50}+C)\), with \(I_{\max}=0.80\) and \(IC_{50}=3\) mg/L. Find the instantaneous rate of change under inhibited production.
First calculate inhibition: \(I_p(2)=0.80(2)/(3+2)=0.32\). Therefore, the production rate is \(12[1-0.32]=8.16\) units/h. The loss rate is \(0.10(80)=8.00\) units/h. Substitution into the state equation gives
At this instant, the response is increasing at 0.160 units/h. This is a local rate, not a prediction that the response will continue to increase at that rate; \(R\) and potentially \(C\) can change over time.
Worked Example: Comparing the Four Rate Balances
Consider hypothetical values \(k_{\mathrm{in},0}=10\) response units/h, \(k_{\mathrm{out}}=0.20\) h\(^{-1}\), \(R=40\) response units, and \(C=4\) mg/L. Use \(I(C)=0.75C/(2+C)\) for inhibition and \(S(C)=1.20C/(3+C)\) for stimulation. Calculate the instantaneous rate under each of the four equations.
At \(C=4\) mg/L, \(I(4)=0.75(4)/(2+4)=0.500\), and \(S(4)=1.20(4)/(3+4)=0.6857\). The unmodified loss rate is \(k_{\mathrm{out}}R=(0.20)(40)=8.00\) response units/h.
For inhibited production, the production rate is \(10(1-0.500)=5.00\) units/h, so \(dR/dt=5.00-8.00=-3.00\) units/h. For inhibited loss, the loss rate is \(0.20(1-0.500)(40)=4.00\) units/h, so \(dR/dt=10-4.00=6.00\) units/h.
For stimulated production, the production rate is \(10(1+0.6857)=16.86\) units/h, so \(dR/dt=16.86-8.00=8.86\) units/h. For stimulated loss, the loss rate is \(0.20(1+0.6857)(40)=13.49\) units/h, so \(dR/dt=10-13.49=-3.49\) units/h. The positive or negative sign describes whether the response is rising or falling at this instant, not whether the drug effect itself is beneficial or harmful.
Worked Example: Checking Zero-Drug Behavior
Suppose \(k_{\mathrm{in},0}=6\) units/h, \(k_{\mathrm{out}}=0.15\) h\(^{-1}\), and the current response is \(R=30\) units. The driving concentration is zero, and the inhibition and stimulation functions are concentration-dependent with value zero at \(C=0\). What does each canonical equation predict?
At zero concentration, every drug-effect multiplier is one. Thus, under each equation,
All four equations agree because none has a drug effect at \(C=0\). The response is increasing at 1.5 units/h for this specified state. This check verifies the zero-drug reduction and the arithmetic; it does not establish that the stated response is an equilibrium value.
Assumptions, Diagnostics, and Common Pitfalls
The equations are useful only when their assumptions are plausible for the intended application. In their basic form, they assume a single response state, a production rate that is constant in the absence of drug, and a loss rate proportional to the current response in the absence of drug. They also assume that the concentration-effect functions adequately describe the drug’s modulation of the selected rate. Concentration may vary over time, and then the effect functions are evaluated at the contemporaneous driving concentration.
Before fitting or interpreting an equation, check that the response scale and units are defined, that the initial response is specified, and that concentrations are expressed in units compatible with the concentration-effect parameters. Inspect whether the response can be meaningfully represented as a single state and whether first-order loss is credible over the observed range. During model evaluation, compare predictions with observed time courses and inspect residuals over time and concentration. Persistent systematic deviations, implausible parameter estimates, or predicted responses outside physically possible bounds may indicate a poor structural model, unsuitable effect function, or inadequate data. A good fit alone does not establish that the chosen production or loss mechanism is physiologically correct.
A frequent mistake is to treat inhibition of loss as a direct negative response term, such as subtracting \(I_l(C)\) from \(R\). That confuses a dimensionless rate modifier with the response state and gives inconsistent units. The correct inhibited-loss term is \(k_{\mathrm{out}}[1-I_l(C)]R\), which has units of response per time. Similarly, stimulation of production multiplies the production rate; it does not add \(S_p(C)\) directly to \(R\).
Another pitfall is interpreting \(k_{\mathrm{in},0}\) as a concentration or interpreting \(k_{\mathrm{out}}\) as the total amount lost per time. The former is a response production rate; the latter is a rate constant whose product with \(R\) gives the loss rate. Keeping the state, rates, and dimensionless effect functions distinct prevents many equation and interpretation errors.
How This Model Relates to Alternatives
A direct-response model relates the drug driver to the measured response without representing production and loss as separate processes. An effect-compartment model represents delayed equilibration between a measured concentration and a concentration driving effect; the earlier material on effect-compartment models addresses that delay structure. An indirect response model instead represents how an effect on a production or loss rate changes a response state over time. These structures may be combined when justified, but they represent different sources of delay and should not be treated as interchangeable explanations.
The canonical equations are intentionally simple. A real response may have nonlinear or saturable loss, multiple response states, feedback, time-varying production, or more than one drug action. Those features require a different or expanded state model. The four equations specify a useful starting structure, not proof that production or loss is the true biological mechanism.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In the basic equations, what are the units of \(k_{\mathrm{in},0}\), \(k_{\mathrm{out}}\), and \(k_{\mathrm{out}}R\)?
- For inhibited loss, let \(k_{\mathrm{in},0}=8\) mg/h, \(k_{\mathrm{out}}=0.25\) h\(^{-1}\), \(R=20\) mg, and \(I_l(C)=0.40\). Calculate \(dR/dt\), including units.
- Why is \(S_p(C)\) not added directly to \(R\) in the stimulated-production equation?
- If each drug-effect function equals zero at \(C=0\), what do all four equations reduce to?
Solutions. (1) \(k_{\mathrm{in},0}\) has units of response per time, \(k_{\mathrm{out}}\) has units of inverse time, and \(k_{\mathrm{out}}R\) has units of response per time. (2) The inhibited loss rate is \(0.25(1-0.40)(20)=3.00\) mg/h. Therefore, \(dR/dt=8-3.00=5.00\) mg/h. (3) \(S_p(C)\) is dimensionless and modifies the production rate, so the production term is \(k_{\mathrm{in},0}[1+S_p(C)]\), with units of response per time. Adding \(S_p(C)\) directly to \(R\) would mix incompatible units. (4) Every multiplier becomes one, leaving \(dR/dt=k_{\mathrm{in},0}-k_{\mathrm{out}}R\).