4.3: Quantitative Pharmacokinetics (Vd, Clearance, Half-life, Steady-state)

4.3: Quantitative Pharmacokinetics (Vd, Clearance, Half-life, Steady-state)

Quantitative pharmacokinetics involves the mathematical application of pharmacokinetic models to drug dosing, helping clinicians design safe and effective therapeutic regimens.

Volume of Distribution (Vd) and Loading Dose Calculations

The apparent Volume of Distribution ($V_d$) is defined as:

$V_d = \frac{\text{Amount of drug in the body (Dose)}}{C}$

Where $C$ is the plasma concentration of the drug. For an extravascular dose with bioavailability $F$:

$V_d = \frac{F \cdot \text{Dose}}{C_0}$

Loading Dose (LD)

A loading dose is a large initial dose administered to rapidly achieve therapeutic plasma concentrations, bypassing the delay associated with reaching steady state (which takes 4–5 half-lives). Loading doses are used for drugs with long half-lives when an immediate therapeutic effect is critical (e.g., phenytoin in status epilepticus, digoxin in atrial fibrillation, amiodarone).

$\text{LD} = \frac{V_d \cdot C_{\text{target}}}{F}$

Where:

• $V_d$ is the volume of distribution (expressed in L or L/kg).
• $C_{\text{target}}$ is the desired therapeutic plasma concentration.
• $F$ is the bioavailability of the drug formulation. For intravenous administration, $F = 1$ (or $F = S$, the salt factor of the drug form).

Worked Clinical Example (Phenytoin):

A patient weighing 80 kg presents with status epilepticus. The target therapeutic concentration of phenytoin is 18 mg/L. The volume of distribution ($V_d$) of phenytoin is 0.75 L/kg. Phenytoin sodium injection has a salt factor ($S$) of 0.92. (For IV phenytoin sodium, $F = S = 0.92$).

1. Calculate total $V_d$: $V_d = 80 \text{ kg} \cdot 0.75 \text{ L/kg} = 60 \text{ L}$
2. Calculate the intravenous loading dose: $\text{LD} = \frac{60 \text{ L} \cdot 18 \text{ mg/L}}{0.92} = \frac{1080 \text{ mg}}{0.92} \approx 1174 \text{ mg}$ Clinical Context: In practice, this would be rounded to 1100–1200 mg and administered intravenously at a rate not exceeding 50 mg/minute (25 mg/minute in elderly patients) in normal saline, with continuous ECG and blood pressure monitoring to avoid severe cardiotoxicity (bradyarrhythmias, hypotension) caused by the propylene glycol solvent.

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Clearance (Cl)

Clearance ($Cl$) is the volume of plasma completely cleared of a drug per unit time (expressed in L/h or mL/min). Unlike elimination rate, which varies with concentration in first-order kinetics, clearance remains constant.

$Cl = \frac{\text{Rate of elimination}}{C}$

Clearance and AUC

Clearance can be calculated from the total systemic exposure, represented by the Area Under the Curve (AUC):

$Cl = \frac{F \cdot \text{Dose}}{\text{AUC}}$

Organ Clearance

Total systemic clearance ($Cl_{\text{total}}$) is the sum of clearances by individual eliminating organs:

$Cl_{\text{total}} = Cl_{\text{renal}} + Cl_{\text{hepatic}} + Cl_{\text{other}}$

• Hepating Clearance ($Cl_H$): $Cl_H = Q_H \cdot E_H$ Where $Q_H$ is hepatic blood flow (approx. 1.5 L/min or 90 L/h) and $E_H$ is the hepatic extraction ratio.
  • High Extraction Drugs ($E_H > 0.7$): (e.g., morphine, propranolol, verapamil). Clearance is blood-flow limited. Reductions in liver blood flow (e.g., in congestive heart failure, beta-blocker therapy) directly decrease clearance.
  • Low Extraction Drugs ($E_H < 0.3$): (e.g., phenytoin, warfarin). Clearance is capacity-limited and depends on intrinsic liver enzyme activity and the drug's unbound fraction ($f_u$).
• Renal Clearance ($Cl_R$): Renal clearance is calculated using: $Cl_R = \frac{U_x \cdot V}{P_x}$ Where $U_x$ is urine concentration, $V$ is urine flow rate, and $P_x$ is plasma concentration. In Australian pharmacy practice, renal function is estimated using the Cockcroft-Gault equation to calculate creatinine clearance (CrCl): $\text{CrCl (mL/min)} = \frac{(140 - \text{Age}) \cdot \text{Weight (kg)} \cdot \text{Constant}}{\text{Serum Creatinine (micromol/L)}}$ Where the Constant is 1.23 for males and 1.04 for females. Weight is typically actual body weight unless the patient is obese, in which case adjusted body weight is used.

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Half-life (t1/2) and Elimination Rate Constant (ke)

For drugs eliminated by first-order kinetics, a constant fraction of the drug is eliminated per unit time. The elimination rate constant ($k_e$) represents the fractional rate of drug removal:

$k_e = \frac{Cl}{V_d}$

The elimination half-life ($t_{1/2}$) is the time required for the plasma concentration to decrease by $50%$:

$t_{1/2} = \frac{\ln(2)}{k_e} \approx \frac{0.693}{k_e}$

By substituting $k_e$:

$t_{1/2} = \frac{0.693 \cdot V_d}{Cl}$

Important Clinical Concept: Half-life is a dependent variable, determined by the primary independent variables $V_d$ and $Cl$. If $V_d$ increases (e.g., in sepsis or ascites, where fluid shift expands the distribution volume of hydrophilic drugs like gentamicin) or if $Cl$ decreases (e.g., in acute kidney injury), the half-life will prolong.

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Steady-state Concentration (Css) and Maintenance Dose Calculations

Steady state ($C_{ss}$) is reached when the rate of drug administration equals the rate of drug elimination:

$\text{Rate of Drug Input} = \text{Rate of Drug Output}$

Time to Reach Steady State

The time to reach steady state is determined solely by the drug's half-life ($t_{1/2}$). It is independent of the dose size or dosing frequency.

• 1 half-life: $50%$ of steady state
• 2 half-lives: $75%$ of steady state
• 3 half-lives: $87.5%$ of steady state
• 4 half-lives: $93.75%$ of steady state
• 5 half-lives: $96.9%$ of steady state (clinically accepted as steady state)

Clinical Rule: A drug is considered fully washed out of the body after approximately 4–5 half-lives.

Average Steady-state Concentration ($Css,avg$)

For multiple dosing regimens:

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

Where $\tau$ is the dosing interval.

Maintenance Dose (MD)

The maintenance dose is designed to maintain steady-state therapeutic concentrations:

$\text{MD} = \frac{Cl \cdot C_{\text{target}} \cdot \tau}{F}$

Worked Clinical Example (Digoxin):

A 70-year-old female patient with a creatinine clearance of 40 mL/min (estimated digoxin clearance $Cl = 2.5 \text{ L/h}$) requires oral digoxin tablets ($F = 0.70$) for heart failure. The target steady-state concentration is 0.8 micrograms/L. The dosing interval is 24 hours ($\tau = 24 \text{ h}$).

1. Calculate the daily maintenance dose: $\text{MD} = \frac{2.5 \text{ L/h} \cdot 0.8 \text{ micrograms/L} \cdot 24 \text{ h}}{0.70}$ $\text{MD} = \frac{48 \text{ micrograms}}{0.70} \approx 68.6 \text{ micrograms}$ Clinical Context: Standard digoxin tablet sizes in Australia are 100 micrograms (0.1 mg) and 250 micrograms (0.25 mg). To approximate this calculated dose, the clinician may prescribe 100 micrograms daily, or 50 micrograms daily (using half a 100-microgram tablet) depending on age and clinical response, with subsequent therapeutic drug monitoring (TDM).

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Accumulation Factor (R) and Area Under the Curve (AUC)

Accumulation Factor (R)

When drug doses are repeated at intervals ($\tau$) shorter than 4–5 half-lives, the drug accumulates in the body. The accumulation factor ($R$) describes the ratio of the steady-state concentration to that observed after the first dose:

$R = \frac{1}{1 - e^{-k_e \cdot \tau}}$

• If the dosing interval equals the half-life ($\tau = t_{1/2}$), then $e^{-k_e \cdot \tau} = e^{-\ln(2)} = 0.5$: $R = \frac{1}{1 - 0.5} = 2$ This means that peak and trough levels at steady state will be exactly double the peak and trough levels observed after the first dose.
• Relationship between Loading Dose and Maintenance Dose: $\text{LD} = \text{MD} \cdot R = \frac{\text{MD}}{1 - e^{-k_e \cdot \tau}}$ If $\tau = t_{1/2}$, then $\text{LD} = 2 \cdot \text{MD}$.

Area Under the Curve (AUC)

AUC represents the total drug exposure over time and is measured in units of concentration $\cdot$ time (e.g., mg$\cdot$h/L).

• It is calculated using the trapezoidal rule from plasma concentration-time profiles.
• AUC is directly proportional to bioavailability and dose, and inversely proportional to clearance: $\text{AUC} = \frac{F \cdot \text{Dose}}{Cl}$