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 = 8 mg/LLD = ?81 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 81 kg needs a loading dose of tobramycin to reach a level of 8 mg/L straight away. The volume of distribution is 0.3 L/kg. Find the loading dose LD in mg.

81 kgVd = 0.3 L/kgC = 8 mg/L
LD = mg
2.Easy
t/t½C/Css (%)100%2.5f = ?

Practice exercise (made-up values, not clinical guidance). A patient receives aminophylline as a constant IV infusion without a loading dose. The half-life is 6 h. What percentage f of the steady-state level has been reached after 15 h?

t½ = 6 ht = 15 h
f = %
3.Medium
110t (h)C (mg/L)1 h8 mg/L5 h2 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 8 mg/L at 1 h and 2 mg/L at 5 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 8 mg/L (1 h)C2 = 2 mg/L (5 h)
k = h⁻¹
4.Medium
110t (h)C (mg/L)1 h8 mg/L5 h3.2 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 8 mg/L at 1 h and 3.2 mg/L at 5 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 8 mg/L (1 h)C2 = 3.2 mg/L (5 h)
k = h⁻¹
5.Hard
Vd = 7 L/kg0.4 → 0.8 mcg/LLD = ? mcg62 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 62 kg patient has a digoxin level of 0.4 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?

62 kgVd = 7 L/kgC = 0.4 → 0.8 mcg/LF = 0.7
LD = mcg
6.Hard
t (h)C (mg/L)Css​19 hC = ?

Practice exercise (made-up values, not clinical guidance). A theophylline infusion runs at 52.5 mg/h with no loading dose. The patient's clearance is 3.5 L/h and the half-life is 9 h. What is the level C after 19 h?

R = 52.5 mg/hCL = 3.5 L/ht½ = 9 ht = 19 h
C = mg/L

Answer key

  1. 1. LD = 194.4 mg
  2. 2. f = 82.32 %
  3. 3. k = 0.3466 h⁻¹
  4. 4. k = 0.2291 h⁻¹
  5. 5. LD = 248 mcg
  6. 6. C = 11.53 mg/L

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