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
110t (h)C (mg/L)7 mg/L3 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the gentamicin level is 7 mg/L. The half-life is 3 h, and the drug is eliminated by first-order kinetics. What is the level C 3 h later?

C₀ = 7 mg/Lt½ = 3 ht = 3 h
C = mg/L
2.Easy
0.1110t (h)C (mg/L)6 mg/L6 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the gentamicin level is 6 mg/L. The half-life is 2 h, and the drug is eliminated by first-order kinetics. What is the level C 6 h later?

C₀ = 6 mg/Lt½ = 2 ht = 6 h
C = mg/L
3.Medium
Vd = 0.7 L/kgC = 15 mg/LLD = ? mg51 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 51 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 15 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.

51 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
4.Medium
110t (h)C (mg/L)C₀ = 8 mg/Lt½ = 2.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 120 mg of tobramycin. The level back-extrapolated to time zero is 8 mg/L and the half-life is 2.5 h. Find the clearance CL in L/h.

D = 120 mgC₀ = 8 mg/Lt½ = 2.5 h
CL = L/h
5.Hard
80 kg68 · manSCr 1.6 mg/dLCrCl → D = ? mg

Practice exercise (made-up values, not clinical guidance). A 68-year-old man weighs 80 kg and has a serum creatinine of 1.6 mg/dL (141 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 cefazolin dose is 1000 mg every 8 h for a creatinine clearance of 100 mL/min. Keeping the interval, the dose is reduced in proportion to the creatinine clearance. Find the adjusted dose D. Round to the nearest whole number.

68 years80 kgmanSCr = 1.6 mg/dL1000 mg q8hCrCl normal 100 mL/min
D = mg
6.Hard
VCss,avg​ = 15 mg/LD = ? mg · q8hCL = 0.035 L/kg/h99 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 99 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.035 L/kg/h. What dose D in mg gives an average steady-state level of 15 mg/L?

CL = 0.035 L/kg/hτ = 8 hS = 0.8F = 199 kgCss,avg = 15 mg/L
D = mg

Answer key

  1. 1. C = 3.5 mg/L
  2. 2. C = 0.75 mg/L
  3. 3. LD = 582.1 mg
  4. 4. CL = 4.159 L/h
  5. 5. D = 500 mg
  6. 6. D = 519.8 mg

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