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
Vd = 0.3 L/kgC = 6 mg/LLD = ?76 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 76 kg needs a loading dose of tobramycin to reach a level of 6 mg/L straight away. The volume of distribution is 0.3 L/kg. Find the loading dose LD in mg.

76 kgVd = 0.3 L/kgC = 6 mg/L
LD = mg
2.Easy
110100t (h)CC₀t½ = 48 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, digoxin has an elimination half-life of 48 h. Find the elimination rate constant k in h⁻¹.

t½ = 48 h
k = h⁻¹
3.Medium
VCss​ = 15 mg/LR = ? mg/hCL = 0.035 L/kg/h76 kg

Practice exercise (made-up values, not clinical guidance). A 76 kg patient receives a theophylline infusion. The clearance is 0.035 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 15 mg/L?

76 kgCL = 0.035 L/kg/hCss = 15 mg/L
R = mg/h
4.Medium
110100t (h)C (mg/L)1 h35 mg/L17 h8.8 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 8.8 mg/L at 17 h. Assume first-order elimination. Find the half-life t½.

C1 = 35 mg/L (1 h)C2 = 8.8 mg/L (17 h)
t½ = h
5.Hard
Vd = 0.5 L/kg7.2 → 12 mg/LLD = ? mg80 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). An 80 kg patient has a theophylline level of 7.2 mg/L; the target is 12 mg/L. The drug is given intravenously as aminophylline (S = 0.8), and the volume of distribution is 0.5 L/kg. What extra loading dose LD raises the level to the target?

80 kgVd = 0.5 L/kgC = 7.2 → 12 mg/LS = 0.8
LD = mg
6.Hard
VCss,avg​ = ?300 mg · q8hCL = 0.03 L/kg/h77 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 77 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.03 L/kg/h. Each dose is 300 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.03 L/kg/hτ = 8 hS = 1F = 177 kgD = 300 mg
Css,avg = mg/L

Answer key

  1. 1. LD = 136.8 mg
  2. 2. k = 0.01444 h⁻¹
  3. 3. R = 39.9 mg/h
  4. 4. t½ = 8.033 h
  5. 5. LD = 240 mg
  6. 6. Css,avg = 16.23 mg/L

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