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)C (mg/L)30 mg/L11 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the vancomycin level is 30 mg/L. The half-life is 6 h, and the drug is eliminated by first-order kinetics. What is the level C 11 h later?

C₀ = 30 mg/Lt½ = 6 ht = 11 h
C = mg/L
2.Easy
0.1110100t (h)C (mg/L)9 mg/L9 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the gentamicin level is 9 mg/L. The half-life is 3 h, and the drug is eliminated by first-order kinetics. What is the level C 9 h later?

C₀ = 9 mg/Lt½ = 3 ht = 9 h
C = mg/L
3.Medium
VCss​ = 10 mg/LR = ? mg/hCL = 0.03 L/kg/h70 kg

Practice exercise (made-up values, not clinical guidance). A 70 kg patient receives a theophylline infusion. The clearance is 0.03 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 10 mg/L?

70 kgCL = 0.03 L/kg/hCss = 10 mg/L
R = mg/h
4.Medium
110100t (h)C (mg/L)1 h9 mg/L7 h2.3 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 9 mg/L at 1 h and 2.3 mg/L at 7 h. Assume first-order elimination. Find the half-life t½.

C1 = 9 mg/L (1 h)C2 = 2.3 mg/L (7 h)
t½ = h
5.Hard
t (h)C (mg/L)Css​t = ?90%

Practice exercise (made-up values, not clinical guidance). A theophylline infusion is started without a loading dose. The half-life in this patient is 9 h. How long t does it take to reach 90% of the steady-state level?

t½ = 9 h90% Css
t = h
6.Hard
Vd = 0.5 L/kg3.6 → 12 mg/LLD = ? mg95 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 95 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?

95 kgVd = 0.5 L/kgC = 3.6 → 12 mg/LS = 0.8
LD = mg

Answer key

  1. 1. C = 8.418 mg/L
  2. 2. C = 1.125 mg/L
  3. 3. R = 21 mg/h
  4. 4. t½ = 3.048 h
  5. 5. t = 29.9 h
  6. 6. LD = 498.8 mg

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