2.5 Pharmacokinetic Models

Pharmacokinetic models describe how drug concentrations change over time in the body. From simple one-compartment IV bolus to nonlinear Michaelis-Menten kinetics, these mathematical frameworks enable dose optimization, therapeutic drug monitoring, and prediction of drug accumulation.

Historical Development

Torsten Teorell (1937) published the first comprehensive pharmacokinetic model, often considered the foundation of modern PK. Dost (1953) formalized the one-compartment model. Wagner and Nelson (1964) developed methods for oral absorption analysis. Sheiner and Beal (1980) introduced population PK with the NONMEM software.

Derivation 1: One-Compartment IV Bolus Model

The simplest PK model: drug is injected directly into the bloodstream, distributes instantaneously, and is eliminated by first-order kinetics.

Step 1: Mass Balance

The body is treated as a single well-stirred compartment of volume $ V_d $. The rate of change of drug amount $ X = C \cdot V_d $:

$ \frac{dX}{dt} = -k_{el} \cdot X $

where $ k_{el} $ is the first-order elimination rate constant.

Step 2: Solve the ODE

Separating variables and integrating:

$ \int_{X_0}^{X} \frac{dX'}{X'} = -k_{el} \int_0^t dt' $

$ \ln X - \ln X_0 = -k_{el} t $

$ X(t) = X_0 \cdot e^{-k_{el} t} $

Dividing both sides by $ V_d $ (since $ C = X/V_d $ and $ C_0 = X_0/V_d = \text{Dose}/V_d $):

One-Compartment IV Bolus

$ C(t) = C_0 \cdot e^{-k_{el} t} = \frac{Dose}{V_d} \cdot e^{-k_{el} t} $

Step 3: Derive Key Relationships

The elimination rate constant relates to clearance and volume:

$ k_{el} = \frac{CL}{V_d} $

The half-life is derived by setting $ C(t_{1/2}) = C_0/2 $:

$ \frac{1}{2} = e^{-k_{el} t_{1/2}} \quad \Rightarrow \quad t_{1/2} = \frac{\ln 2}{k_{el}} = \frac{0.693}{k_{el}} $

The area under the curve (AUC) from 0 to infinity:

$ AUC = \int_0^{\infty} C_0 e^{-k_{el}t} dt = \frac{C_0}{k_{el}} = \frac{Dose}{CL} $

Derivation 2: One-Compartment Oral Absorption

For oral administration, absorption follows first-order kinetics from the GI tract into the systemic compartment.

Step 1: Two Simultaneous Processes

The rate of change in systemic drug amount reflects absorption in minus elimination out:

$ \frac{dX}{dt} = k_a \cdot X_{GI} - k_{el} \cdot X $

where $ k_a $ is the absorption rate constant and $ X_{GI} = F \cdot \text{Dose} \cdot e^{-k_a t} $ is the remaining drug at the absorption site.

Step 2: Solve by Laplace Transform or Direct Integration

The solution (Bateman equation) is:

$ C(t) = \frac{F \cdot Dose \cdot k_a}{V_d(k_a - k_{el})} \left( e^{-k_{el} t} - e^{-k_a t} \right) $

where F is the bioavailability fraction.

Step 3: Derive t_max and C_max

Setting dC/dt = 0 to find the time of peak concentration:

$ t_{max} = \frac{\ln(k_a/k_{el})}{k_a - k_{el}} $

The peak concentration is obtained by substituting t_max back:

$ C_{max} = \frac{F \cdot Dose}{V_d} \cdot e^{-k_{el} \cdot t_{max}} $

Derivation 3: Two-Compartment IV Bolus Model

Many drugs show a biphasic decline: rapid distribution into tissues followed by slower elimination. This requires two compartments.

Step 1: Coupled ODEs

Central compartment (1) and peripheral compartment (2) with transfer rate constants k_12, k_21 and elimination k_10:

$ \frac{dX_1}{dt} = -(k_{12} + k_{10})X_1 + k_{21}X_2 $

$ \frac{dX_2}{dt} = k_{12}X_1 - k_{21}X_2 $

Step 2: Solve as Eigenvalue Problem

The system is linear, so the solution is a sum of two exponentials. The eigenvalues alpha and beta of the coefficient matrix are:

$ \alpha, \beta = \frac{1}{2}\left[(k_{12}+k_{21}+k_{10}) \pm \sqrt{(k_{12}+k_{21}+k_{10})^2 - 4k_{21}k_{10}}\right] $

where alpha > beta (alpha is the faster distribution phase).

Biexponential Equation

$ C(t) = A \cdot e^{-\alpha t} + B \cdot e^{-\beta t} $

A and B are macro-constants determined by initial conditions. The alpha phase (distribution) dominates early; the beta phase (elimination) dominates at later times. On a semi-log plot, this appears as a curve that "bends" to reveal two distinct slopes.

Alpha Phase (Distribution)

Rapid decline. Drug distributing from plasma to tissue. Half-life = 0.693/alpha (typically minutes to hours).

Beta Phase (Elimination)

Slower decline. True elimination from the body. Half-life = 0.693/beta (hours to days). This is the terminal half-life.

Derivation 4: Multiple Dosing & Accumulation

When a drug is given repeatedly at a fixed interval tau, it accumulates until steady state is reached.

Step 1: Superposition of Doses

After the n-th dose (IV bolus), the concentration is the sum of residuals from all previous doses:

$ C_n(t) = C_0 \sum_{k=0}^{n-1} e^{-k_{el}(t + k\tau)} = C_0 e^{-k_{el}t} \cdot \frac{1 - e^{-n k_{el}\tau}}{1 - e^{-k_{el}\tau}} $

(geometric series formula applied)

Step 2: Accumulation Factor

As n approaches infinity, the accumulation factor R represents the ratio of steady-state to first-dose levels:

$ R = \lim_{n \to \infty} \frac{1 - e^{-n k_{el}\tau}}{1 - e^{-k_{el}\tau}} = \frac{1}{1 - e^{-k_{el}\tau}} $

Step 3: Steady-State Concentrations

At steady state, the peak and trough concentrations are:

$ C_{ss,max} = \frac{C_0}{1 - e^{-k_{el}\tau}} $

$ C_{ss,min} = C_{ss,max} \cdot e^{-k_{el}\tau} $

$ C_{ss,avg} = \frac{F \cdot Dose}{CL \cdot \tau} $

Steady state is reached after approximately 4-5 half-lives (at which point 93.75-96.87% of the ultimate accumulation has occurred).

Loading dose: To achieve target concentration immediately: $ LD = C_{\text{target}} \cdot V_d / F $. This bypasses the 4–5 half-life approach to steady state.

Derivation 5: Nonlinear (Michaelis-Menten) Pharmacokinetics

Some drugs saturate their metabolic enzymes at therapeutic concentrations. Elimination becomes capacity-limited rather than first-order.

Step 1: Michaelis-Menten Elimination

When enzyme capacity is saturated, the rate of elimination follows Michaelis-Menten kinetics:

$ \frac{dC}{dt} = -\frac{V_{max} \cdot C}{K_m + C} $

V_max is the maximum elimination rate and K_m is the concentration at half-maximal elimination rate.

Step 2: Limiting Cases

When C is much less than K_m (low concentrations), the equation reduces to first-order:

$ \frac{dC}{dt} \approx -\frac{V_{max}}{K_m} \cdot C = -k_{el} \cdot C \quad (\text{first-order}) $

When C is much greater than K_m (high concentrations), elimination becomes zero-order:

$ \frac{dC}{dt} \approx -V_{max} \quad (\text{zero-order, constant rate}) $

Step 3: Clinical Consequences

At steady state with nonlinear kinetics:

$ C_{ss} = \frac{K_m \cdot R_0}{V_{max} - R_0} $

where R_0 = F * Dose/tau is the dosing rate. As R_0 approaches V_max, C_ss rises disproportionately. A small dose increase near saturation produces a large concentration increase.

Phenytoin

K_m approximately 4 mg/L, V_max approximately 7 mg/kg/day. Therapeutic range 10-20 mg/L (well above K_m). Small dose changes cause large, unpredictable concentration changes.

Ethanol

Eliminated at approximately 7-10 g/h regardless of concentration (zero-order). V_max of alcohol dehydrogenase is exceeded at blood alcohol concentrations above approximately 0.02 g/dL.

Compartment Model Diagrams

Python Simulation: PK Models

PK Models — IV Bolus, Oral, Two-Compartment, Multiple Dosing & Nonlinear PK

Python

script.py98 lines

import numpy as np

t = np.linspace(0, 24, 300)

# ── Panel 1: One-Compartment IV Bolus ──
Dose = 500  # mg
Vd = 50     # L
kel = 0.2   # 1/h
C0 = Dose / Vd
C_iv = C0 * np.exp(-kel * t)
half_life = 0.693 / kel

print("CHART1_TITLE: One-Compartment IV Bolus (t1/2 = {:.1f} h)".format(half_life))
print("CHART1_XLABEL: Time (h)")
print("CHART1_YLABEL: Concentration (mg/L)")
for i in range(0, len(t), 6):
    print(f"t={t[i]:.2f} conc={C_iv[i]:.3f}")

# ── Panel 2: One-Compartment Oral ──
F = 0.8
ka = 1.5  # 1/h
C_oral = (F * Dose * ka) / (Vd * (ka - kel)) * (np.exp(-kel * t) - np.exp(-ka * t))
tmax = np.log(ka / kel) / (ka - kel)
Cmax = (F * Dose / Vd) * np.exp(-kel * tmax)

print("\nCHART2_TITLE: Oral Absorption (tmax = {:.1f} h, Cmax = {:.1f} mg/L)".format(tmax, Cmax))
print("CHART2_XLABEL: Time (h)")
print("CHART2_YLABEL: Concentration (mg/L)")
for i in range(0, len(t), 6):
    print(f"t={t[i]:.2f} IV={C_iv[i]:.3f} Oral={max(0, C_oral[i]):.3f}")

# ── Panel 3: Two-Compartment Model ──
k12 = 0.8
k21 = 0.3
k10 = 0.15
s = k12 + k21 + k10
alpha = 0.5 * (s + np.sqrt(s**2 - 4 * k21 * k10))
beta = 0.5 * (s - np.sqrt(s**2 - 4 * k21 * k10))
A_coeff = C0 * (alpha - k21) / (alpha - beta)
B_coeff = C0 * (k21 - beta) / (alpha - beta)
C_2comp = A_coeff * np.exp(-alpha * t) + B_coeff * np.exp(-beta * t)

print("\nCHART3_TITLE: Two-Compartment Model (alpha={:.2f}, beta={:.2f})".format(alpha, beta))
print("CHART3_XLABEL: Time (h)")
print("CHART3_YLABEL: Concentration (mg/L)")
for i in range(0, len(t), 6):
    print(f"t={t[i]:.2f} C_total={C_2comp[i]:.3f} dist_phase={A_coeff * np.exp(-alpha * t[i]):.3f} elim_phase={B_coeff * np.exp(-beta * t[i]):.3f}")

# ── Panel 4: Multiple Dosing (approach to steady state) ──
tau = 6.0  # dosing interval (h)
n_doses = 8
t_multi = np.linspace(0, n_doses * tau, 500)
C_multi = np.zeros_like(t_multi)
for dose_n in range(n_doses):
    t_since_dose = t_multi - dose_n * tau
    mask = t_since_dose >= 0
    C_multi[mask] += C0 * np.exp(-kel * t_since_dose[mask])

R_accum = 1 / (1 - np.exp(-kel * tau))
print("\nCHART4_TITLE: Multiple Dosing (tau={:.0f}h, R={:.2f})".format(tau, R_accum))
print("CHART4_XLABEL: Time (h)")
print("CHART4_YLABEL: Concentration (mg/L)")
for i in range(0, len(t_multi), 10):
    print(f"t={t_multi[i]:.2f} conc={C_multi[i]:.3f}")

# ── Panel 5: Nonlinear PK (Michaelis-Menten) ──
Vmax = 500  # mg/day
Km = 4.0    # mg/L
doses_daily = [200, 300, 350, 400, 450]
print("\nCHART5_TITLE: Nonlinear PK: Css vs Dose Rate (Phenytoin)")
print("CHART5_XLABEL: Daily Dose (mg)")
print("CHART5_YLABEL: Css (mg/L)")
for d in np.linspace(100, 480, 50):
    R0 = d
    if R0 < Vmax:
        Css = Km * R0 / (Vmax - R0)
        print(f"dose={d:.1f} Css={Css:.2f}")

# ── Panel 6: Linear vs Nonlinear Elimination ──
print("\nCHART6_TITLE: Linear vs Nonlinear Elimination Over Time")
print("CHART6_XLABEL: Time (h)")
print("CHART6_YLABEL: Concentration (mg/L)")
t_nl = np.linspace(0, 30, 300)
C_lin = 20.0 * np.exp(-0.15 * t_nl)

# Numerical integration for nonlinear
C_nl = np.zeros(len(t_nl))
C_nl[0] = 20.0
dt_step = t_nl[1] - t_nl[0]
Vmax_h = 500 / 24  # convert to per hour
for j in range(1, len(t_nl)):
    dCdt = -Vmax_h * C_nl[j-1] / (Km + C_nl[j-1])
    C_nl[j] = max(0, C_nl[j-1] + dCdt * dt_step)

for i in range(0, len(t_nl), 6):
    print(f"t={t_nl[i]:.2f} linear={C_lin[i]:.3f} nonlinear={C_nl[i]:.3f}")

print("\nSimulation complete: 6 panels generated")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Clinical Applications

Aminoglycoside Dosing

One-compartment model guides once-daily aminoglycoside dosing. Target: high C_max/MIC ratio (concentration-dependent killing) with C_trough below 2 mg/L (minimize nephrotoxicity).

Vancomycin Two-Compartment PK

Vancomycin follows two-compartment kinetics. The distribution phase (alpha) lasts 1-2 hours. Trough levels must be drawn after distribution is complete (before next dose) to avoid falsely elevated readings.

Phenytoin Dose Adjustment

Due to Michaelis-Menten kinetics, phenytoin dose adjustments must be small (25-50 mg increments) near therapeutic levels. A 10% dose increase can produce a 40% concentration increase.

Loading Doses in Emergency

Digoxin ($ t_{1/2} \approx 40 $ hours) would take 8 days to reach steady state without a loading dose. IV loading (0.5 mg then 0.25 mg q6h) achieves therapeutic levels within hours.

Key Takeaways

1.
One-compartment IV bolus: $ C(t) = \frac{\text{Dose}}{V_d} \cdot e^{-k_{el} t} $, with $ t_{1/2} = \frac{0.693}{k_{el}} $ and $ \text{AUC} = \frac{\text{Dose}}{CL} $.
2.
Oral absorption adds a rising phase: $ C(t) = \frac{F \cdot \text{Dose} \cdot k_a}{V_d(k_a - k_{el})} \left( e^{-k_{el}t} - e^{-k_a t} \right) $.
3.
Two-compartment model $ C(t) = A e^{-\alpha t} + B e^{-\beta t} $ describes distribution and elimination phases.
4.
Multiple dosing leads to accumulation factor $ R = \frac{1}{1 - e^{-k_{el}\tau}} $; steady state at 4–5 half-lives.
5.
Nonlinear (Michaelis-Menten) kinetics cause disproportionate concentration increases near enzyme saturation (phenytoin, ethanol).

Practice Problems

Problem 1: One-Compartment Half-Life from Plasma DataAfter a 500 mg IV bolus, plasma concentrations are 25 mg/L at $t = 0$ and 6.25 mg/L at $t = 6$ h. Determine $V_d$, $k_{el}$, $t_{1/2}$, and clearance.

Solution:

1. Volume of distribution from the initial concentration:

$$V_d = \frac{\text{Dose}}{C_0} = \frac{500\;\text{mg}}{25\;\text{mg/L}} = 20\;\text{L}$$

2. Using $C(t) = C_0 e^{-k_{el}t}$, solve for $k_{el}$:

$$6.25 = 25\,e^{-k_{el}(6)} \implies e^{-6k_{el}} = 0.25 \implies k_{el} = \frac{\ln 4}{6} = \frac{1.386}{6}$$

3. The elimination rate constant is:

$$k_{el} = 0.231\;\text{h}^{-1}$$

4. The half-life is:

$$\boxed{t_{1/2} = \frac{0.693}{k_{el}} = \frac{0.693}{0.231} = 3.0\;\text{h}}$$

5. Clearance is:

$$CL = k_{el} \times V_d = 0.231 \times 20 = 4.62\;\text{L/h}$$

Verification: after 6 h = 2 half-lives, concentration should drop by factor of $2^2 = 4$: $25/4 = 6.25$ mg/L. Confirmed.

Problem 2: Steady-State ConcentrationA drug with $t_{1/2} = 8$ h and $V_d = 40$ L is given as 200 mg IV every 8 hours. Calculate the steady-state peak and trough concentrations, and the accumulation factor.

Solution:

1. Compute $k_{el}$ and the fraction remaining per dosing interval:

$$k_{el} = \frac{0.693}{8} = 0.0866\;\text{h}^{-1}, \quad e^{-k_{el}\tau} = e^{-0.693} = 0.5$$

2. The accumulation factor is:

$$R = \frac{1}{1 - e^{-k_{el}\tau}} = \frac{1}{1 - 0.5} = 2.0$$

3. The increment per dose is $\Delta C = \text{Dose}/V_d = 200/40 = 5.0$ mg/L.

4. Steady-state peak (immediately after a dose):

$$\boxed{C_{ss,\text{peak}} = R \times \Delta C = 2.0 \times 5.0 = 10.0\;\text{mg/L}}$$

5. Steady-state trough (just before the next dose):

$$\boxed{C_{ss,\text{trough}} = C_{ss,\text{peak}} \times e^{-k_{el}\tau} = 10.0 \times 0.5 = 5.0\;\text{mg/L}}$$

With $\tau = t_{1/2}$, the peak-to-trough ratio is exactly 2:1 and the accumulation factor is 2. Steady state is reached after 4–5 half-lives ($\approx 32\text{--}40$ hours).

Problem 3: Loading Dose CalculationA drug has $V_d = 70$ L, $CL = 3.5$ L/h, and the target steady-state concentration is $C_{ss} = 15$ mg/L (IV infusion). Calculate the loading dose and the maintenance infusion rate.

Solution:

1. The loading dose is designed to immediately achieve the target concentration:

$$\boxed{D_L = C_{ss} \times V_d = 15\;\text{mg/L} \times 70\;\text{L} = 1050\;\text{mg}}$$

2. The maintenance infusion rate to sustain $C_{ss}$:

$$\boxed{R_0 = C_{ss} \times CL = 15\;\text{mg/L} \times 3.5\;\text{L/h} = 52.5\;\text{mg/h}}$$

3. Verify the half-life:

$$t_{1/2} = \frac{0.693 \times V_d}{CL} = \frac{0.693 \times 70}{3.5} = 13.86\;\text{h}$$

4. Without a loading dose, reaching 90% of $C_{ss}$ would take $3.32 \times t_{1/2} \approx 46$ hours.

5. The loading dose eliminates this delay. Note that $D_L$ depends on $V_d$ (not clearance), while maintenance rate depends on $CL$ (not $V_d$). This is clinically important: obese patients may need larger loading doses (larger $V_d$) but similar maintenance rates if CL is unchanged.

Problem 4: Oral Bioavailability from AUCA 100 mg IV dose gives AUC$_{\text{IV}} = 500$ mg$\cdot$h/L. A 250 mg oral dose gives AUC$_{\text{oral}} = 400$ mg$\cdot$h/L. Calculate the absolute bioavailability and estimate the fraction absorbed if the hepatic extraction ratio is 0.3.

Solution:

1. Absolute bioavailability compares dose-normalized AUCs:

$$F = \frac{AUC_{\text{oral}} / D_{\text{oral}}}{AUC_{\text{IV}} / D_{\text{IV}}} = \frac{400/250}{500/100} = \frac{1.6}{5.0}$$

2. Therefore:

$$\boxed{F = 0.32 = 32\%}$$

3. Bioavailability is the product of fraction absorbed ($f_a$) and fraction escaping first-pass metabolism:

$$F = f_a \times (1 - E_H)$$

4. With hepatic extraction ratio $E_H = 0.3$:

$$f_a = \frac{F}{1 - E_H} = \frac{0.32}{0.70} \approx 0.457$$

5. Only 45.7% of the oral dose is absorbed from the GI tract, and of that, 30% is removed by first-pass hepatic metabolism:

$$\boxed{f_a \approx 0.46, \quad F = 0.46 \times 0.70 = 0.32}$$

The low bioavailability is dominated by poor absorption (not first-pass metabolism), suggesting formulation improvements or a different route may be beneficial.

Problem 5: Two-Compartment Model ParametersAfter a 1000 mg IV bolus, the plasma concentration-time profile fits $C(t) = 30\,e^{-2t} + 10\,e^{-0.2t}$ (mg/L, hours). Determine the micro-constants $A$, $B$, $\alpha$, $\beta$, the terminal half-life, and $V_d$ at steady state.

Solution:

1. Directly from the biexponential equation:

$$A = 30\;\text{mg/L},\; \alpha = 2\;\text{h}^{-1},\; B = 10\;\text{mg/L},\; \beta = 0.2\;\text{h}^{-1}$$

2. The terminal (elimination) half-life is governed by $\beta$:

$$\boxed{t_{1/2,\beta} = \frac{0.693}{\beta} = \frac{0.693}{0.2} = 3.47\;\text{h}}$$

3. The distribution half-life is:

$$t_{1/2,\alpha} = \frac{0.693}{\alpha} = \frac{0.693}{2} = 0.347\;\text{h} \approx 20.8\;\text{min}$$

4. The volume of distribution at steady state ($V_{ss}$) uses the moment method. First compute AUC and AUMC:

$$AUC = \frac{A}{\alpha} + \frac{B}{\beta} = \frac{30}{2} + \frac{10}{0.2} = 15 + 50 = 65\;\text{mg·h/L}$$

$$AUMC = \frac{A}{\alpha^2} + \frac{B}{\beta^2} = \frac{30}{4} + \frac{10}{0.04} = 7.5 + 250 = 257.5\;\text{mg·h}^2\text{/L}$$

5. Mean residence time and $V_{ss}$:

$$MRT = \frac{AUMC}{AUC} = \frac{257.5}{65} = 3.96\;\text{h}$$

$$\boxed{V_{ss} = CL \times MRT = \frac{D}{AUC} \times MRT = \frac{1000}{65} \times 3.96 = 60.9\;\text{L}}$$

The rapid distribution phase ($\alpha \gg \beta$) indicates extensive tissue uptake, and $V_{ss} = 60.9$ L confirms distribution beyond the plasma compartment.

← 2.4 Excretion 2.6 Drug Interactions →

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