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
110100t (h)CC₀t½ = ?C₀/2

Practice exercise (made-up values, not clinical guidance). The elimination rate constant of tobramycin in a patient is 0.347 h⁻¹. Find the half-life t½.

k = 0.347 h⁻¹
t½ = h
2.Easy
0.1110t (h)C (mg/L)8 mg/L10.5 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the gentamicin level is 8 mg/L. The half-life is 3.5 h, and the drug is eliminated by first-order kinetics. What is the level C 10.5 h later?

C₀ = 8 mg/Lt½ = 3.5 ht = 10.5 h
C = mg/L
3.Medium
Vd = 0.5 L/kgC = 15 mg/LLD = ? mg94 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 94 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 15 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.

94 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
4.Medium
Vd = 0.5 L/kgC = 15 mg/LLD = ? mg75 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 75 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 15 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.

75 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
5.Hard
Vd = 0.5 L/kg3.6 → 12 mg/LLD = ? mg63 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 63 kg patient has a theophylline level of 3.6 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?

63 kgVd = 0.5 L/kgC = 3.6 → 12 mg/LS = 0.8
LD = mg
6.Hard
t (h)C (mg/L)Css​t = ?95%

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion is started without a loading dose. The half-life in this patient is 6 h. How long t does it take to reach 95% of the steady-state level?

t½ = 6 h95% Css
t = h

Answer key

  1. 1. t½ = 1.998 h
  2. 2. C = 1 mg/L
  3. 3. LD = 881.3 mg
  4. 4. LD = 703.1 mg
  5. 5. LD = 330.8 mg
  6. 6. t = 25.93 h

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