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
110100t (h)CC₀t½ = 2.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, tobramycin has an elimination half-life of 2.5 h. Find the elimination rate constant k in h⁻¹.

t½ = 2.5 h
k = h⁻¹
2.Easy
Vd = 0.7 L/kgC = 25 mg/LLD = ?104 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 104 kg needs a loading dose of vancomycin to reach a level of 25 mg/L straight away. The volume of distribution is 0.7 L/kg. Find the loading dose LD in mg.

104 kgVd = 0.7 L/kgC = 25 mg/L
LD = mg
3.Medium
110100t (h)C (mg/L)1 h10 mg/L5 h3.3 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 10 mg/L at 1 h and 3.3 mg/L at 5 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 10 mg/L (1 h)C2 = 3.3 mg/L (5 h)
k = h⁻¹
4.Medium
10100t (h)C (mg/L)1 h35 mg/L11 h14.7 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 35 mg/L at 1 h and 14.7 mg/L at 11 h. Assume first-order elimination. Find the half-life t½.

C1 = 35 mg/L (1 h)C2 = 14.7 mg/L (11 h)
t½ = h
5.Hard
93 kg86 · manSCr 1.3 mg/dLfe​ = 0.7D = ? mcg

Practice exercise (made-up values, not clinical guidance). An 86-year-old man weighs 93 kg and has a serum creatinine of 1.3 mg/dL (115 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 digoxin dose is 250 mcg for a creatinine clearance of 100 mL/min, and a fraction fe = 0.7 of the drug is excreted unchanged by the kidneys. Using the Dettli dose fraction Q = 1 − fe·(1 − CrCl/normal CrCl), find the adjusted dose D. Round to the nearest whole number.

86 years93 kgmanSCr = 1.3 mg/dL250 mcgfe = 0.7CrCl normal 100 mL/min
D = mcg
6.Hard
VCss,avg​ = ?450 mg · q12hCL = 0.035 L/kg/h95 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 95 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.035 L/kg/h. Each dose is 450 mg of theophylline. Find the average steady-state level Css,avg.

CL = 0.035 L/kg/hτ = 12 hS = 1F = 195 kgD = 450 mg
Css,avg = mg/L

Answer key

  1. 1. k = 0.2773 h⁻¹
  2. 2. LD = 1820 mg
  3. 3. k = 0.2772 h⁻¹
  4. 4. t½ = 7.99 h
  5. 5. D = 169 mcg
  6. 6. Css,avg = 11.28 mg/L

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