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 = 0.25 L/kgC = 8 mg/LLD = ?81 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 81 kg needs a loading dose of gentamicin to reach a level of 8 mg/L straight away. The volume of distribution is 0.25 L/kg. Find the loading dose LD in mg.

81 kgVd = 0.25 L/kgC = 8 mg/L
LD = mg
2.Easy
110100t (h)CC₀t½ = 1.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, lidocaine has an elimination half-life of 1.5 h. Find the elimination rate constant k in h⁻¹.

t½ = 1.5 h
k = h⁻¹
3.Medium
Vd = 0.5 L/kgC = 15 mg/LLD = ? mg51 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 51 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.

51 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
4.Medium
Vd = 7 L/kgC = 0.8 mcg/LLD = ? mcg82 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). An 82 kg patient is to be loaded with digoxin tablets to reach a level of 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.

82 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
5.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 8 h. How long t does it take to reach 95% of the steady-state level?

t½ = 8 h95% Css
t = h
6.Hard
t (h)C (mg/L)Css​5 hC = ?

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion runs at 60 mg/h with no loading dose. The patient's clearance is 3 L/h and the half-life is 10 h. What is the level C after 5 h?

R = 60 mg/hCL = 3 L/ht½ = 10 ht = 5 h
C = mg/L

Answer key

  1. 1. LD = 162 mg
  2. 2. k = 0.4621 h⁻¹
  3. 3. LD = 478.1 mg
  4. 4. LD = 656 mcg
  5. 5. t = 34.58 h
  6. 6. C = 5.858 mg/L

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