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%2.5f = ?

Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin 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 = %
2.Easy
110100t (h)C (mg/L)25 mg/L23 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the vancomycin level is 25 mg/L. The half-life is 7 h, and the drug is eliminated by first-order kinetics. What is the level C 23 h later?

C₀ = 25 mg/Lt½ = 7 ht = 23 h
C = mg/L
3.Medium
Vd = 7 L/kgC = 0.8 mcg/LLD = ? mcg53 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 53 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.

53 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
4.Medium
110t (h)C (mg/L)1 h6 mg/L5 h1.5 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 6 mg/L at 1 h and 1.5 mg/L at 5 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 6 mg/L (1 h)C2 = 1.5 mg/L (5 h)
k = h⁻¹
5.Hard
Vd = 0.5 L/kg4.5 → 15 mg/LLD = ? mg56 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 56 kg patient has a theophylline level of 4.5 mg/L; the target is 15 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?

56 kgVd = 0.5 L/kgC = 4.5 → 15 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 8 h. How long t does it take to reach 95% of the steady-state level?

t½ = 8 h95% Css
t = h

Answer key

  1. 1. f = 82.32 %
  2. 2. C = 2.564 mg/L
  3. 3. LD = 424 mcg
  4. 4. k = 0.3466 h⁻¹
  5. 5. LD = 367.5 mg
  6. 6. t = 34.58 h

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