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%4f = ?

Practice exercise (made-up values, not clinical guidance). A patient receives lidocaine as a constant IV infusion without a loading dose. The half-life is 2 h. What percentage f of the steady-state level has been reached after 8 h?

t½ = 2 ht = 8 h
f = %
2.Easy
Vd = 0.5 L/kgC = 10 mg/LLD = ?58 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 58 kg needs a loading dose of theophylline to reach a level of 10 mg/L straight away. The volume of distribution is 0.5 L/kg. Find the loading dose LD in mg.

58 kgVd = 0.5 L/kgC = 10 mg/L
LD = mg
3.Medium
Vd = 0.5 L/kgC = 10 mg/LLD = ? mg82 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). An 82 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 10 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.

82 kgVd = 0.5 L/kgC = 10 mg/LS = 0.8
LD = mg
4.Medium
110100t (h)C (mg/L)1 h25 mg/L17 h9.9 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 25 mg/L at 1 h and 9.9 mg/L at 17 h. Assume first-order elimination. Find the half-life t½.

C1 = 25 mg/L (1 h)C2 = 9.9 mg/L (17 h)
t½ = h
5.Hard
t (h)C (mg/L)Css​27 hC = ?

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion runs at 75 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 27 h?

R = 75 mg/hCL = 3 L/ht½ = 10 ht = 27 h
C = mg/L
6.Hard
VCss,avg​ = 15 mg/LD = ? mg · q12hCL = 0.05 L/kg/h66 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 66 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.05 L/kg/h. What dose D in mg gives an average steady-state level of 15 mg/L?

CL = 0.05 L/kg/hτ = 12 hS = 1F = 166 kgCss,avg = 15 mg/L
D = mg

Answer key

  1. 1. f = 93.75 %
  2. 2. LD = 290 mg
  3. 3. LD = 512.5 mg
  4. 4. t½ = 11.97 h
  5. 5. C = 21.15 mg/L
  6. 6. D = 594 mg

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