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 10 h. What percentage f of the steady-state level has been reached after 25 h?

t½ = 10 ht = 25 h
f = %
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 9 h. What percentage f of the steady-state level has been reached after 22.5 h?

t½ = 9 ht = 22.5 h
f = %
3.Medium
110t (h)C (mg/L)C₀ = 8 mg/Lt½ = 2 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 140 mg of tobramycin. The level back-extrapolated to time zero is 8 mg/L and the half-life is 2 h. Find the clearance CL in L/h.

D = 140 mgC₀ = 8 mg/Lt½ = 2 h
CL = L/h
4.Medium
Vd = 0.7 L/kgC = 20 mg/LLD = ? mg50 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 50 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.

50 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
5.Hard
VCss,avg​ = 12 mg/LD = ? mg · q12hCL = 0.045 L/kg/h78 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 78 kg patient takes oral aminophylline (S = 0.8, F = 1) every 12 h. The theophylline clearance is 0.045 L/kg/h. What dose D in mg gives an average steady-state level of 12 mg/L?

CL = 0.045 L/kg/hτ = 12 hS = 0.8F = 178 kgCss,avg = 12 mg/L
D = 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 10 h. How long t does it take to reach 95% of the steady-state level?

t½ = 10 h95% Css
t = h

Answer key

  1. 1. f = 82.32 %
  2. 2. f = 82.32 %
  3. 3. CL = 6.065 L/h
  4. 4. LD = 760.9 mg
  5. 5. D = 631.8 mg
  6. 6. t = 43.22 h

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