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

Practice exercise (made-up values, not clinical guidance). In a patient, vancomycin has an elimination half-life of 12 h. Find the elimination rate constant k in h⁻¹.

t½ = 12 h

k = h⁻¹

2.Easy

Practice exercise (made-up values, not clinical guidance). A patient who weighs 89 kg needs a loading dose of gentamicin to reach a level of 8 mg/L straight away. The volume of distribution is 0.25 L/kg. Find the loading dose LD in mg.

89 kgVd = 0.25 L/kgC = 8 mg/L

LD = mg

3.Medium

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 8 mg/L at 2 h and 3.2 mg/L at 6 h. Assume first-order elimination. Find the elimination rate constant k.

C₁ = 8 mg/L (2 h)C₂ = 3.2 mg/L (6 h)

k = h⁻¹

4.Medium

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 8 mg/L at 1 h and 2 mg/L at 5 h. Assume first-order elimination. Find the elimination rate constant k.

C₁ = 8 mg/L (1 h)C₂ = 2 mg/L (5 h)

k = h⁻¹

5.Hard

Practice exercise (made-up values, not clinical guidance). A 49-year-old woman weighs 73 kg and has a serum creatinine of 2.2 mg/dL (194 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 120 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.

49 years73 kgwomanSCr = 2.2 mg/dL250 mcgf_e = 0.7CrCl normal 120 mL/min

D = mcg

6.Hard

Practice exercise (made-up values, not clinical guidance). A theophylline infusion runs at 20 mg/h with no loading dose. The patient's clearance is 2 L/h and the half-life is 6 h. What is the level C after 10 h?

R = 20 mg/hCL = 2 L/ht½ = 6 ht = 10 h

C = mg/L

Answer key

1. k = 0.05776 h⁻¹
2. LD = 178 mg
3. k = 0.2291 h⁻¹
4. k = 0.3466 h⁻¹
5. D = 127 mcg
6. C = 6.85 mg/L

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