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 = 0.5 L/kgC = 10 mg/LLD = ?102 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 102 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.

102 kgVd = 0.5 L/kgC = 10 mg/L
LD = mg
2.Easy
t/t½C/Css (%)100%2f = ?

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 8 h. What percentage f of the steady-state level has been reached after 16 h?

t½ = 8 ht = 16 h
f = %
3.Medium
VCss​ = 25 mg/LR = ? mg/hCL = 80 mL/min

Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 80 mL/min. What infusion rate R in mg/h gives a steady-state level of 25 mg/L?

CL = 80 mL/minCss = 25 mg/L
R = mg/h
4.Medium
110100t (h)C (mg/L)1 h30 mg/L17 h8.7 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 30 mg/L at 1 h and 8.7 mg/L at 17 h. Assume first-order elimination. Find the half-life t½.

C1 = 30 mg/L (1 h)C2 = 8.7 mg/L (17 h)
t½ = h
5.Hard
95 kg68 · womanSCr 2.4 mg/dLCrCl → τ = ? h

Practice exercise (made-up values, not clinical guidance). A 68-year-old woman weighs 95 kg and has a serum creatinine of 2.4 mg/dL (212 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual gentamicin regimen is 80 mg every 8 h for a creatinine clearance of 100 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.

68 years95 kgwomanSCr = 2.4 mg/dL80 mg q8hCrCl normal 100 mL/min
τ = h
6.Hard
Vd = 7 L/kg0.4 → 1 mcg/LLD = ? mcg93 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 93 kg patient has a digoxin level of 0.4 mcg/L; the target is 1 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?

93 kgVd = 7 L/kgC = 0.4 → 1 mcg/LF = 0.7
LD = mcg

Answer key

  1. 1. LD = 510 mg
  2. 2. f = 75 %
  3. 3. R = 120 mg/h
  4. 4. t½ = 8.959 h
  5. 5. τ = 23.8 h
  6. 6. LD = 558 mcg

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