Clinical pharmacokinetics worksheet with answers

Free printable Clinical pharmacokinetics worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
t/t½C/Css (%)100%597%

Practice exercise (made-up values, not clinical guidance). A patient starts regular doses of theophylline, which has a half-life of 9 h. Using the rule that steady state is reached after about five half-lives, after how many hours tss is steady state reached?

t½ = 9 h
tss = h
2.Easy
110100t (h)CC₀t½ = 2 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, lidocaine has an elimination half-life of 2 h. Find the elimination rate constant k in h⁻¹.

t½ = 2 h
k = h⁻¹
3.Medium
110100t (h)C (mg/L)1 h35 mg/L13 h12.4 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 35 mg/L at 1 h and 12.4 mg/L at 13 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 35 mg/L (1 h)C2 = 12.4 mg/L (13 h)
k = h⁻¹
4.Medium
VCss​ = 15 mg/LR = ? mg/hCL = 0.04 L/kg/h83 kg

Practice exercise (made-up values, not clinical guidance). An 83 kg patient receives a theophylline infusion. The clearance is 0.04 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 15 mg/L?

83 kgCL = 0.04 L/kg/hCss = 15 mg/L
R = mg/h
5.Hard
0.1110t (h)C (mg/L)Cpeak​ = 6τ = ?Ctrough​ = 0.5

Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 2.5 h. The target is a peak of 6 mg/L and a trough of 0.5 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 2.5 hCpeak = 6 mg/LCtrough = 0.5 mg/L
τ = h
6.Hard
Vd = 0.7 L/kg4.5 → 15 mg/LLD = ? mg74 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 74 kg patient has a phenytoin level of 4.5 mg/L; the target is 15 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?

74 kgVd = 0.7 L/kgC = 4.5 → 15 mg/LS = 0.92
LD = mg

Answer key

  1. 1. tss = 45 h
  2. 2. k = 0.3466 h⁻¹
  3. 3. k = 0.08647 h⁻¹
  4. 4. R = 49.8 mg/h
  5. 5. τ = 9 h
  6. 6. LD = 591.2 mg

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