Effect-Site Dynamics After an Intravenous Bolus

Why the Effect-Site Peak Occurs After a Bolus

After an intravenous bolus, plasma concentration begins declining as soon as the dose has entered the circulation. An effect-site compartment does not instantaneously match plasma: its concentration approaches plasma concentration at a finite rate. Consequently, effect-site concentration can still be rising while plasma concentration is falling. The peak of the effect-site concentration is the point at which those two tendencies balance.

This tutorial derives that peak for a bolus with first-order plasma elimination. The target is the concentration $C_e(t)$ and its time of maximum, not a predicted maximum pharmacological response. Translating effect-site concentration into response requires a concentration-effect model and is a separate step.

Model and assumptions: At time zero, an intravenous bolus produces plasma concentration $C_{p,0}=D/V$, where $D$ is dose and $V$ is the apparent volume of distribution in this one-compartment model. For $t\geq 0$, plasma concentration follows $C_p(t)=C_{p,0}e^{-k_{el}t}$, with $k_{el}>0$. Effect-site concentration starts at $C_e(0)=0$ and follows $dC_e/dt=k_{e0}(C_p-C_e)$, with $k_{e0}>0$. The effect site does not alter plasma concentration, and there is no other input or loss process in the effect-site equation.

Here $k_{el}$ is the plasma elimination rate constant and $k_{e0}$ is the effect-site equilibration rate constant. Both have units of inverse time. Concentrations such as $C_p$ and $C_e$ have the same units, for example mg/L. The effect-site equation says that $C_e$ rises when $C_p>C_e$, falls when $C_p<C_e$, and has zero instantaneous rate of change when the two concentrations are equal.

Deriving the Bolus Solution

Substitute the bolus plasma profile into the effect-site equation. This is a first-order linear differential equation; multiplying by the integrating factor $e^{k_{e0}t}$ and integrating from zero to $t$ gives the solution.

\begin{aligned} \frac{dC_e}{dt}+k_{e0}C_e &=k_{e0}C_{p,0}e^{-k_{el}t},\\ \frac{d}{dt}\left(e^{k_{e0}t}C_e(t)\right) &=k_{e0}C_{p,0}e^{(k_{e0}-k_{el})t},\\ C_e(t) &=\frac{k_{e0}C_{p,0}}{k_{e0}-k_{el}} \left(e^{-k_{el}t}-e^{-k_{e0}t}\right), \qquad k_{e0}\ne k_{el}. \end{aligned}

The initial condition is satisfied because the two exponentials are both one at $t=0$, so their difference is zero. The expression is positive for $t>0$, whether $k_{e0}$ is greater or less than $k_{el}$: in either ordering, the signs of the denominator and exponential difference match. At long times both exponentials tend to zero, so $C_e(t)$ also tends to zero.

Definition: The effect-site peak time $t_{\max}$ is the time $t\geq 0$ at which $C_e(t)$ is greatest. For unequal positive rates, the peak time and concentration are $$ t_{\max}=\frac{\ln(k_{e0}/k_{el})}{k_{e0}-k_{el}}, \qquad C_{e,\max}=C_{p,0}e^{-k_{el}t_{\max}}. $$ The logarithm has a dimensionless argument, and $t_{\max}$ has units of time.

Why This Is the Maximum

For unequal rates, differentiate the closed-form concentration. A stationary point occurs when the positive and negative terms in the derivative balance. Taking logarithms yields the stated peak time.

\begin{aligned} \frac{dC_e}{dt} &=\frac{k_{e0}C_{p,0}}{k_{e0}-k_{el}} \left(-k_{el}e^{-k_{el}t}+k_{e0}e^{-k_{e0}t}\right),\\ \frac{dC_e}{dt}=0 &\Longleftrightarrow k_{e0}e^{-k_{e0}t}=k_{el}e^{-k_{el}t},\\ (k_{e0}-k_{el})t &=\ln(k_{e0}/k_{el}). \end{aligned}

The peak time is positive: if $k_{e0}>k_{el}$, both the numerator and denominator of its expression are positive; if $k_{e0}<k_{el}$, both are negative. The stationary point is a maximum rather than a minimum. At that point, the derivative condition gives $k_{e0}e^{-k_{e0}t}=k_{el}e^{-k_{el}t}$, so the second derivative simplifies as follows.

\left.\frac{d^2C_e}{dt^2}\right|_{t=t_{\max}} = \frac{k_{e0}C_{p,0}}{k_{e0}-k_{el}}\, k_{el}(k_{el}-k_{e0})e^{-k_{el}t_{\max}} = -k_{e0}k_{el}C_{p,0}e^{-k_{el}t_{\max}}<0.

There is only one stationary point: the equation for it reduces to a linear equation in $t$, with one solution. Together with the zero initial concentration, positive concentrations after the bolus, and the return to zero at long times, this confirms that the stationary point is the global peak.

At the peak, the effect-site derivative is zero, so the original differential equation also gives $C_e(t_{\max})=C_p(t_{\max})$. Substituting the plasma profile gives $C_{e,\max}=C_{p,0}e^{-k_{el}t_{\max}}$. This equality is specific to the peak instant; it does not mean that effect-site and plasma concentrations are equal throughout the profile.

Worked Calculations

Worked Example: Peak With Faster Equilibration Than Elimination

Suppose a hypothetical bolus produces $C_{p,0}=10.0$ mg/L, with $k_{el}=0.200$ h$^{-1}$ and $k_{e0}=0.500$ h$^{-1}$. Calculate the time and value of the effect-site peak.

The rates are positive and unequal, and the stated model assumes the bolus is instantaneous, the plasma profile is a single exponential, and the effect site starts at zero. Therefore, the unequal-rate formulas apply. First calculate the time:

t_{\max} =\frac{\ln(0.500/0.200)}{0.500-0.200} =\frac{\ln(2.5)}{0.300} =\frac{0.9162907}{0.300} =3.054302\ldots\ \text{h}.

Using the exact peak-time expression in the peak-concentration calculation,

\begin{aligned} C_{e,\max} &=10.0\exp\left[-(0.200) \frac{\ln(0.500/0.200)}{0.500-0.200}\right]\ \text{mg/L}\\ &=10.0e^{-0.6108605}\ \text{mg/L}\\ &\approx 5.429\ \text{mg/L}. \end{aligned}

The predicted peak occurs about 3.054 h after the bolus and is 5.429 mg/L. The value is below the initial plasma concentration of 10.0 mg/L, as expected for this model: the effect site begins at zero and approaches a declining plasma concentration rather than receiving the bolus directly.

Worked Example: Peak With Slower Equilibration Than Elimination

For a second hypothetical bolus, let $C_{p,0}=8.00$ mg/L, $k_{el}=0.600$ h$^{-1}$, and $k_{e0}=0.150$ h$^{-1}$. Find the peak time and peak concentration.

The assumptions remain a zero initial effect-site concentration, positive unequal rates, and a single first-order plasma decline. Here the equilibration rate is smaller than the elimination rate. The peak-time expression is still positive because both its numerator and denominator are negative:

t_{\max} =\frac{\ln(0.150/0.600)}{0.150-0.600} =\frac{\ln(0.25)}{-0.450} =\frac{-1.386294}{-0.450} =3.080654\ldots\ \text{h}.

Then the peak is

\begin{aligned} C_{e,\max} &=8.00e^{-(0.600)(3.080654\ldots)}\ \text{mg/L}\\ &=8.00e^{-1.848392\ldots}\ \text{mg/L}\\ &\approx 1.260\ \text{mg/L}. \end{aligned}

The effect-site concentration peaks about 3.081 h after the bolus at approximately 1.260 mg/L. A slower equilibration rate does not mean that the peak formula changes sign or ceases to apply. It does mean that the effect-site concentration responds more slowly while plasma is declining.

Worked Example: Equal Elimination and Equilibration Rates

Suppose a hypothetical bolus gives $C_{p,0}=12.0$ mg/L and both $k_{el}$ and $k_{e0}$ equal $0.300$ h$^{-1}$. The unequal-rate expression has a zero denominator, so it must not be used by direct substitution.

Use the effect-site equation with $C_p(t)=12.0e^{-0.300t}$ mg/L. Multiplying by the integrating factor $e^{0.300t}$ and applying $C_e(0)=0$ gives

C_e(t)=C_{p,0}k_{e0}t e^{-k_{e0}t} =12.0(0.300)t e^{-0.300t}\ \text{mg/L}.

Differentiating this expression gives a derivative proportional to $1-k_{e0}t$, so its maximum occurs when $t=1/k_{e0}$. Thus,

t_{\max}=\frac{1}{0.300}=3.333\ \text{h}, \qquad C_{e,\max}=\frac{12.0}{e}=4.415\ \text{mg/L}.

This is the continuous equal-rate limit of the unequal-rate solution. The formula with a denominator $k_{e0}-k_{el}$ appears singular at equality, but the concentration itself remains finite.

Calculation Checks and Common Pitfalls

A useful diagnostic is to verify the shape implied by the assumptions. The predicted effect-site concentration is zero at $t=0$, rises initially because $C_p(0)>C_e(0)$, reaches one peak, and declines toward zero. At the peak, $C_e=C_p$; before it, the effect-site concentration is rising, and after it, the effect-site concentration is falling. A computed negative peak time or negative concentration signals an algebraic, unit, or input error under the stated positive-rate model.

A common mistake is to use $\ln(k_{el}/k_{e0})/(k_{e0}-k_{el})$ for the peak time. That expression has the wrong logarithm and does not solve the derivative equation. The balance condition is $k_{e0}e^{-k_{e0}t}=k_{el}e^{-k_{el}t}$; taking its logarithm gives $t_{\max}=\ln(k_{e0}/k_{el})/(k_{e0}-k_{el})$. Another pitfall is to insert rounded rates, or a rounded peak time, into intermediate calculations and then report more precision than those inputs support. Keep full precision through the calculation and round the final result to a sensible number of significant figures.

When the two rates are very close, evaluating the unequal-rate concentration formula can be numerically unstable: both the denominator and the difference of exponentials are small. The model is not singular, but direct computation can lose precision through subtraction of nearly equal values. Use the equal-rate expression when the rates are exactly equal, and for near equality use a stable limiting or numerical implementation rather than interpreting a computational artifact as unusual PK/PD behavior.

The closed-form solution is preferable when plasma concentration follows the assumed single exponential, since it provides an exact peak time and value without time-grid error. The effect-site differential equation can also be solved numerically, which is useful when the plasma input is more complicated; numerical peak estimates then depend on solver accuracy and time resolution. The convolution representation of the effect-site state is another equivalent description of the linear model, but it does not remove the need to check the input profile and initial condition. None of these approaches compensates for a misspecified plasma profile or an effect site that does not follow first-order equilibration.

Scope and Interpretation

These formulas describe concentration at a modeled effect site, not receptor occupancy, clinical response, or a measured concentration in a separate anatomical compartment. Their use assumes an instantaneous bolus, first-order elimination with a single $k_{el}$, constant $k_{e0}$, and no initial effect-site drug. A multicompartment plasma profile, a nonzero initial effect-site concentration, ongoing input, or time-varying rate constants changes the calculation. The next analytical step in a PK/PD model may connect $C_e$ to response, but a concentration peak alone does not establish when or how large a response peak will be.

Before interpreting a calculated peak, check that the dose and volume produce the stated $C_{p,0}$, both rate constants use the same time unit, and the plasma curve actually supports a single exponential decline over the time range of interest. A mismatch in any of these conditions can yield a mathematically correct answer to the wrong model.

Check Your Understanding

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

For a bolus with $k_{el}=0.400$ h$^{-1}$ and $k_{e0}=0.100$ h$^{-1}$, is the unequal-rate peak time positive? Explain using the signs in its formula.
If $C_{p,0}=6.00$ mg/L, $k_{el}=0.200$ h$^{-1}$, and $k_{e0}=0.200$ h$^{-1}$, calculate $t_{\max}$ and $C_{e,\max}$.
At the effect-site concentration peak, what relationship must hold between $C_p$ and $C_e$, and why?
A calculation gives $t_{\max}=-2.0$ h for positive unequal rate constants. Name one equation-level check that can reveal the error.

Solutions. (1) Yes. The numerator $\ln(0.100/0.400)=\ln(0.25)$ and denominator $0.100-0.400$ are both negative, so their ratio is positive. (2) For equal rates, $t_{\max}=1/k_{e0}=1/0.200=5.00$ h and $C_{e,\max}=C_{p,0}/e=6.00/e=2.207$ mg/L. (3) At a peak, $dC_e/dt=0$; the model equation $dC_e/dt=k_{e0}(C_p-C_e)$, with $k_{e0}>0$, therefore requires $C_p=C_e$. (4) Differentiate the closed-form concentration and set its derivative to zero. The resulting condition must be $k_{e0}e^{-k_{e0}t}=k_{el}e^{-k_{el}t}$, which gives $t_{\max}=\ln(k_{e0}/k_{el})/(k_{e0}-k_{el})$. For positive unequal rates the numerator and denominator have the same sign, so a negative result indicates a sign, ratio, or substitution error.