Module 12 of 14 · 16 min
Pharmacokinetics
Aminoglycoside and vancomycin monitoring, loading doses, IV-to-oral switches and phenytoin adjustments all rest on a few first-order equations — and the exam rewards knowing which one applies.
🧠 "Half-lives to steady state: 1-2-3-4-5 → 50-75-88-94-97%."
📌 Memorize these
- $t_{1/2}$
- $0.693\div k$
- Steady state
- ≈ 90% at 3.3 t½ · ≈ 97% at 5 t½
- Logs
- use ln, not log₁₀
- Phenytoin
- non-linear (Michaelis–Menten)small dose rise → big level rise
Most drugs follow first-order elimination: a constant fraction leaves per hour. Two levels on the same decline give $k$; use the natural log (ln), not log₁₀.
Steady state: after $n$ half-lives the fraction reached is $1-0.5^{n}$ (≈ 90% at 3.3, ≈ 97% at 5). With repeated doses every $\tau$, the accumulation factor is $R=\dfrac{1}{1-e^{-k\tau}}$.
Loading dose $=\dfrac{C_{target}\times V_d}{S\times F}$; maintenance dose $=\dfrac{C_{ss}\times CL\times\tau}{S\times F}$, where $S$ = salt fraction and $F$ = bioavailability $=\dfrac{AUC_{oral}/\text{Dose}_{oral}}{AUC_{IV}/\text{Dose}_{IV}}$.
Linear drugs: level is proportional to dose. Phenytoin is non-linear (saturable, Michaelis–Menten): $\text{Dose}=\dfrac{V_{max}\times C_{ss}}{K_m+C_{ss}}$, so small dose increases cause disproportionate rises.
🧭 The method
- 1Identify the question type: k/t½ from levels, time to a level, steady state, loading/maintenance dose, F, or non-linear.
- 2Write the matching equation and list the given values with units (h, L, mg/L).
- 3Use ln for any log step; keep k in h⁻¹ and time in h.
- 4Apply salt factor S and bioavailability F where given.
- 5Sense-check: t½ should be plausible for the drug; levels should fall, not rise, between doses.
Before each reveal, predict the next step — then check.
After a gentamicin dose, the level 1 h after the start of the infusion is 8 mg/L and 9 h after the start is 2 mg/L. Find k and t½, and how long after the first sample the level reaches 1 mg/L (useful for choosing the dosing interval).
- 1
Elimination rate constant from two levels 8 h apart (natural log).
$$k=\frac{\ln(8/2)}{9-1\ \text{h}}=\frac{1.386}{8\ \text{h}}=0.173\ \text{h}^{-1}$$ Half-life.
$$t_{1/2}=\frac{0.693}{0.173\ \text{h}^{-1}}=4.0\ \text{h}$$Time to fall from 8 to 1 mg/L: rearrange the same equation for Δt.
$$\Delta t=\frac{\ln(8/1)}{k}=\frac{2.079}{0.173}=12\ \text{h}$$Check with half-lives: 8 → 4 → 2 → 1 is three half-lives = 12 h. ✔
$$3\times4\ \text{h}=12\ \text{h}$$
Same method — the last steps are yours. Work on paper, then reveal.
Two vancomycin levels on the same decline are 30 mg/L and 12 mg/L, drawn 6 h apart. What is the half-life?
- 1
Ratio of the levels and its natural log.
$$\ln\frac{30}{12}=\ln2.5=0.916$$ - 2
Divide by the time between the levels.
$$k=\frac{0.916}{6\ \text{h}}=0.153\ \text{h}^{-1}$$ - 3
Your turn
Now you: half-life.
Work it out on paper, then reveal.
$$t_{1/2}=\frac{0.693}{0.153}=4.54\ \text{h}$$
Type your number. Your first try decides when this module comes back for review.
A drug has a half-life of 8 h and is given by constant infusion. How many hours until it reaches 90% of steady state?
Solve $1-0.5^n=0.9$ for the number of half-lives $n$.
= 26.6 h
Load a 60 kg patient to a theophylline level of 10 mg/L using IV aminophylline (salt factor S = 0.8; Vd = 0.5 L/kg; F = 1). What aminophylline dose (mg)?
Vd in litres first; divide by S because aminophylline is only 80% theophylline.
= 375 mg
A drug has CL = 3 L/h and F = 0.8 orally. What oral dose every 12 h gives an average steady-state level of 10 mg/L?
Dose $=C_{ss}\times CL\times\tau/F$.
= 450 mg
An IV dose of 100 mg gives AUC 40 mg·h/L; an oral dose of 200 mg gives AUC 56 mg·h/L. What is the oral bioavailability (%)?
Normalize each AUC to its dose before comparing.
= 70 %
For phenytoin, a patient's Vmax = 500 mg/day and Km = 4 mg/L. What daily dose gives Css = 15 mg/L? (Ignore the salt factor.)
Non-linear: use Michaelis–Menten, not a straight proportion.
= 394.7 mg/day
A drug with t½ = 6 h is given every 12 h. What is the accumulation factor R (steady-state level ÷ level after the first dose)?
$k=0.693/t_{1/2}$; then $R=1/(1-e^{-k\tau})$.
= 1.33
⚠️ Tricky
A patient takes phenytoin 300 mg/day and has a steady-state level of 8 mg/L. Assume Km = 4 mg/L. Using Michaelis–Menten kinetics, what daily dose should give a level of 15 mg/L? (Ignore the salt factor.)
First solve for this patient's Vmax from the current dose and level, then use it for the new target.
= 355.3 mg/day
⚠️ The trap: A straight proportion (300 × 15/8) gives 562.5 mg/day; with saturable metabolism that dose would produce a toxic level.
⚠️ Tricky
After a gentamicin dose, levels are 6.4 mg/L at 30 min and 1.6 mg/L at 270 min after the end of the infusion. What is the elimination rate constant k in h⁻¹?
The time between the levels is 240 minutes — express it in hours because the answer is asked per hour.
= 0.347 h⁻¹
⚠️ The trap: Dividing by 240 (minutes) gives 0.0058, which is k per minute, not per hour.
⚠️ Tricky
An 80 kg patient already has a theophylline level of 6 mg/L. You want to raise it to 14 mg/L with an IV aminophylline bolus (S = 0.8, Vd = 0.5 L/kg). What aminophylline dose (mg)?
Load only the difference between target and current level.
= 400 mg
⚠️ The trap: Ignoring the level already present (loading to 14 from zero) gives 700 mg and overshoots to about 20 mg/L.
⚠️ Tricky
A tobramycin level drawn 1 hour after the end of a 30-minute infusion is 8.0 mg/L. The patient's k is 0.25 h⁻¹. What was the true peak concentration at the end of the infusion (mg/L)?
The level was drawn late; extrapolate it back 1 hour with $C_0=C\,e^{k\,\Delta t}$.
= 10.27 mg/L
⚠️ The trap: Reporting the measured 8.0 mg/L as the peak ignores the hour of elimination since the infusion ended.
⚠️ Tricky
A patient is stable on IV digoxin 250 mcg daily. The oral tablets have F = 0.7. What oral daily dose (mcg) gives the same amount of drug in the body?
Less of an oral dose reaches the circulation, so the oral dose must be larger: divide by F.
= 357.1 mcg
⚠️ The trap: Multiplying by F instead of dividing gives 175 mcg, about half the exposure needed.
🕵️ Spot the mistake
Levels fall from 16 mg/L to 4 mg/L over 6 h. What is the half-life?
A student wrote
⚠️ Traps examiners set
- Using log₁₀ instead of ln when calculating k.
- Scaling phenytoin levels linearly with dose (it is saturable).
- Forgetting the salt factor (aminophylline 0.8, phenytoin sodium 0.92) or F when switching routes.
- Comparing AUCs without dividing each by its dose.
- Counting time from the dose instead of between the two levels.
💬 Explain it in your own words
Why does the time to reach steady state depend only on the half-life and not on the dose or the dosing interval?