Worked Case: Reversible Inhibition During Repeated Dosing

Why Repeated Dosing Requires a Time Course

A reversible competitive inhibitor does not necessarily inhibit a hepatic enzyme by the same amount throughout a dosing interval. If inhibitor concentrations rise after each dose and then decline, enzyme activity changes as well. A calculation based only on one concentration can therefore misrepresent the interval-wide effect, even when the inhibitor concentration profile is predictable.

This worked case considers a hypothetical inhibitor administered every 12 hours. The goal is to calculate its repeated-dose hepatic concentration profile, convert that profile to enzyme activity, and compare the time-varying calculation with a constant-exposure approximation. The calculation is deliberately explicit about the concentration at the enzyme site and the assumptions needed to connect it to inhibition.

Definition: For rapid, reversible competitive inhibition at an enzyme, the inhibitor increases the apparent substrate \(K_m\) by the factor \(1+C_{I,u}/K_{i,u}\), where \(C_{I,u}\) is unbound inhibitor concentration at the enzyme site and \(K_{i,u}\) is the unbound inhibition constant. When substrate concentration is small relative to \(K_m\), the remaining activity relative to no inhibitor is approximately \(f_I(t)=1/[1+C_{I,u}(t)/K_{i,u}]\). Thus \(f_I=1\) means no inhibition, while a smaller value means less activity.

This activity multiplier is not itself the whole hepatic clearance. It applies to the enzyme-mediated pathway under the stated substrate condition. Other clearance pathways, hepatic blood flow, binding, and substrate concentrations may also affect the change in systemic drug exposure. The case below first isolates the enzyme-mediated component.

Set Up the Repeated-Dose Concentration Profile

Assume each inhibitor dose instantaneously adds \(C_{\text{dose}}=8.00\ \text{micromoles per litre}\) to the unbound hepatic concentration. Between doses, inhibitor concentration declines by first-order elimination with half-life \(t_{1/2}=6.00\ \text{hours}\). Doses are given every \(\tau=12.0\ \text{hours}\). These are hypothetical values, and the instantaneous increment is a simplifying representation of delivery to the hepatic site.

Let \(k=\ln(2)/t_{1/2}\) be the elimination rate constant. Between doses, concentration is multiplied by \(e^{-k\tau}\) over one dosing interval. For this case, \(e^{-k\tau}=e^{-\ln(2)(12/6)}=1/4\). At repeated-dose steady state, the concentration just after a dose is the dose increment plus the fraction remaining from the previous peak. The concentration just before the next dose is the current peak multiplied by the interval decay factor.

Worked Example: Peak and Trough at Repeated-Dose Steady State

Calculate the steady-state post-dose peak and pre-dose trough inhibitor concentrations.

At steady state, the peak \(C_{\max,ss}\) satisfies \(C_{\max,ss}=C_{\text{dose}}+rC_{\max,ss}\), where \(r=e^{-k\tau}=0.25\). Rearranging gives \(C_{\max,ss}=C_{\text{dose}}/(1-r)\). Therefore,

C_{\max,ss}=\frac{8.00}{1-0.25} =\frac{8.00}{0.75} =10.67\ \text{micromoles per litre}.

The trough is \(C_{\min,ss}=rC_{\max,ss}=0.25(10.67)=2.667\ \text{micromoles per litre}\). As a check, adding the next dose to the trough returns the peak: \(2.667+8.00=10.67\ \text{micromoles per litre}\), subject to rounding.

The result assumes the system has reached periodic steady state. After the first dose, the peak would be only \(8.00\ \text{micromoles per litre}\); subsequent doses accumulate because some inhibitor remains at each dosing time.

Within a steady-state interval, let \(t=0\) denote the instant just after a dose. The concentration is \(C_{I,u}(t)=C_{\max,ss}e^{-kt}\), for \(0\leq t<\tau\). This profile describes the assumed unbound concentration at the hepatic enzyme site, not automatically the measured plasma concentration. Substituting a plasma concentration is appropriate only if the model provides a justified relationship between plasma and the site concentration.

Calculate Time-Varying Competitive Inhibition

For the case, take \(K_{i,u}=4.00\ \text{micromoles per litre}\). Under the low-substrate approximation in the definition, enzyme activity at a given time is \(f_I(t)=1/[1+C_{I,u}(t)/K_{i,u}]\). The inhibitor concentration and \(K_{i,u}\) have the same units, so their ratio is dimensionless. At the steady-state peak, \(f_I=1/[1+(10.67/4.00)]=0.2727\); at the trough, \(f_I=1/[1+(2.667/4.00)]=0.6000\). Thus the enzyme’s remaining activity ranges from about 27.3% just after dosing to 60.0% just before the next dose in this simplified model.

Worked Example: Inhibition at a Specific Time After a Dose

Calculate inhibitor concentration and the remaining enzyme activity 6.00 hours after the steady-state peak.

Using the unrounded expression for \(k\), \(k=\ln(2)/6.00\ \text{hours}^{-1}\). After 6.00 hours, one half-life has elapsed:

C_{I,u}(6)=10.6667e^{-[\ln(2)/6.00](6.00)} =10.6667e^{-\ln(2)} =5.333\ \text{micromoles per litre}.

The corresponding activity multiplier is

f_I(6)=\frac{1}{1+5.333/4.00} =\frac{1}{2.333} =0.4286.

At this time, the enzyme retains about 42.9% of its uninhibited activity under the low-substrate assumption. This value lies between the calculated peak and trough activity, as expected from the declining inhibitor concentration.

Average Activity Over the Dosing Interval

A useful comparison is the mean of the instantaneous activity multiplier across the interval. It is not generally the same as activity evaluated at the mean inhibitor concentration, because the relationship between concentration and activity is nonlinear. For the exponential concentration profile, write \(A=C_{\max,ss}/K_{i,u}\). Then \(f_I(t)=1/(1+Ae^{-kt})\). Integrating this function from zero to \(\tau\) gives an exact interval average:

Result: Under the assumed exponential inhibitor decline and competitive activity multiplier, the interval-average activity is

\bar f_I =\frac{1}{\tau}\int_0^\tau \frac{1}{1+Ae^{-kt}}\,dt =\frac{\ln\!\left(\frac{e^{k\tau}+A}{1+A}\right)}{k\tau}, \qquad A=\frac{C_{\max,ss}}{K_{i,u}}.

For justification, multiply the integrand’s numerator and denominator by \(e^{kt}\) to obtain \(e^{kt}/(e^{kt}+A)\). This is the derivative of \(\ln(e^{kt}+A)\) divided by \(k\), since

\frac{d}{dt}\ln(e^{kt}+A) =\frac{ke^{kt}}{e^{kt}+A}.

Evaluating the integral at \(t=\tau\) and \(t=0\), then dividing by \(\tau\), yields the stated expression. The result averages activity over time; it does not by itself calculate a complete drug concentration-time profile.

Worked Example: Exact Mean Activity Versus Mean-Concentration Approximation

Calculate the steady-state mean inhibitor concentration and compare the exact interval-average activity with the activity evaluated at that mean concentration.

The mean concentration follows by integrating the exponential profile. Retain \(k=\ln(2)/6.00\) without rounding in the calculation:

\bar C_{I,u} =\frac{1}{\tau}\int_0^\tau C_{\max,ss}e^{-kt}\,dt =\frac{C_{\max,ss}(1-e^{-k\tau})}{k\tau} =\frac{C_{\text{dose}}}{k\tau} =\frac{8.00}{\ln(4)} =5.771\ \text{micromoles per litre}.

The equality \(C_{\max,ss}(1-e^{-k\tau})=C_{\text{dose}}\) follows from the steady-state peak formula. The constant-mean-concentration approximation therefore predicts

f_I(\bar C_{I,u}) =\frac{1}{1+5.770780/4.00} =0.4094.

For the time-varying calculation, \(A=10.6667/4.00=2.6667\) and \(e^{k\tau}=4\). Hence

\bar f_I =\frac{\ln[(4+2.6667)/(1+2.6667)]}{\ln(4)} =\frac{\ln(20/11)}{\ln(4)} =0.4312.

The mean-concentration approximation gives 0.4094, below the exact mean activity of 0.4312. It therefore implies more average inhibition in this example. The mean concentration itself is correct under the assumed profile; the difference arises from applying a nonlinear activity function to that single mean rather than averaging the activity over the changing concentrations.

Translate Activity Into an Enzyme Clearance Contribution

Suppose the uninhibited intrinsic clearance attributable to this enzyme is \(80.0\ \text{litres per hour}\), and other intrinsic-clearance pathways contribute \(20.0\ \text{litres per hour}\) and are unaffected by the inhibitor. If substrate concentration is sufficiently low relative to \(K_m\) throughout the interval, the inhibited enzyme contribution is its baseline value multiplied by \(f_I(t)\). The mean intrinsic clearance across the interval is then the unaffected component plus the enzyme contribution multiplied by mean activity.

Worked Example: Mean Intrinsic Clearance Under the Two Approximations

Calculate mean intrinsic clearance using the time-varying activity and using the constant mean-concentration approximation.

With the time-varying calculation, the mean enzyme-specific contribution is \(80.0(0.4312)=34.50\ \text{litres per hour}\). Adding the unaffected pathway gives

\overline{CL_{\mathrm{int}}} =20.0+80.0(0.4312) =54.50\ \text{litres per hour}.

With the constant mean-concentration approximation, the enzyme-specific contribution is \(80.0(0.4094)=32.75\ \text{litres per hour}\), giving

CL_{\mathrm{int,mean\ concentration}} =20.0+80.0(0.4094) =52.75\ \text{litres per hour}.

Thus the constant mean-concentration approximation is lower by \(54.50-52.75=1.75\ \text{litres per hour}\), or approximately \(1.75/54.50=3.21\%\) of the interval-average intrinsic clearance from the time-varying calculation. This comparison concerns the defined intrinsic-clearance quantity. It is not a direct prediction of a 3.21% change in plasma exposure: translating time-varying hepatic clearance into exposure requires the relevant PBPK model and its concentration-time dynamics.

Assumptions, Diagnostics, and Alternatives

The calculation rests on several linked assumptions. The inhibitor concentration at the enzyme site receives the same fixed increment after each dose and declines by first-order elimination between doses; the system is at repeated-dose steady state; and inhibitor binding is rapid, reversible, and competitive. The \(K_{i,u}\) value must describe inhibition under conditions consistent with the unbound concentration used. The activity multiplier also uses the low-substrate approximation. If substrate concentration is not small relative to \(K_m\), the competitive rate ratio is instead

\frac{v_I}{v_0} =\frac{K_m+[S]}{K_m(1+C_{I,u}/K_{i,u})+[S]},

where \([S]\) is substrate concentration, and the simpler \(1/(1+C_{I,u}/K_{i,u})\) form need not apply. In a full PBPK simulation, \([S]\), inhibitor concentration, and enzyme-mediated removal may all change with time.

Useful checks include confirming that the predicted trough is below the peak, that the steady-state trough plus the dose increment reproduces the next peak, and that \(0<f_I(t)\leq1\). The computed activity should increase as inhibitor concentration falls. Large inconsistencies between predicted and observed probe time courses can indicate that the assumed hepatic site concentration, elimination rate, dose input, or inhibition mechanism is inadequate. A plasma time course alone may not identify which assumption is wrong.

The constant mean-concentration approximation is convenient when a single summary is needed, but it discards the time course and generally does not equal the time average of the nonlinear activity function. Using the peak or trough as if it applied throughout the interval is another coarse alternative and can overstate or understate average inhibition. A time-resolved PBPK model is more appropriate when dosing, inhibitor disposition, substrate concentrations, or hepatic extraction change materially over the interval. Even that approach is limited by the quality of its site-concentration and enzyme-kinetic inputs.

Do not confuse total and unbound concentration. The concentration used in the inhibition equation must match the definition of \(K_i\); mixing total concentration with an unbound \(K_i\) can distort inhibition.
Do not average before applying a nonlinear function without checking. In this case, \(f_I(\bar C_{I,u})\) is not \(\bar f_I\), as the worked calculation demonstrates.
Do not treat the enzyme multiplier as whole-body clearance. It modifies one enzyme-mediated pathway under stated conditions; other pathways and liver blood flow remain relevant.
Do not assume steady state without evidence. Before accumulation reaches its periodic pattern, peak and trough concentrations differ from the steady-state values used here.

Check Your Understanding

Use the assumptions and values in the worked case unless a question states otherwise. Solutions are included so each calculation can be checked.

If the interval decay factor were 0.50 instead of 0.25, what would the steady-state peak be for an \(8.00\ \text{micromoles per litre}\) dose increment?
At the original steady-state trough of \(2.667\ \text{micromoles per litre}\), calculate the activity multiplier for \(K_{i,u}=4.00\ \text{micromoles per litre}\).
Why does activity evaluated at the mean inhibitor concentration differ from mean activity in the worked case?
Name two assumptions that would need reconsideration if the inhibitor were time-dependent rather than rapidly reversible.

Solutions. (1) \(C_{\max,ss}=8.00/(1-0.50)=16.0\ \text{micromoles per litre}\). (2) \(f_I=1/[1+(2.667/4.00)]=0.6000\). (3) The activity multiplier is a nonlinear function of concentration, so applying it to the mean concentration does not generally equal averaging its value across the concentration profile; here the two results are 0.4094 and 0.4312. (4) The rapid reversible-equilibrium assumption and the absence of time-dependent changes in enzyme activity would no longer be appropriate; inhibition and recovery would need to be represented over time.