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)C (mg/L)10 mg/L6 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the tobramycin level is 10 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 = 10 mg/Lt½ = 2 ht = 6 h
C = mg/L
2.Easy
110100t (h)CC₀t½ = 8 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, theophylline has an elimination half-life of 8 h. Find the elimination rate constant k in h⁻¹.

t½ = 8 h
k = h⁻¹
3.Medium
110100t (h)C (mg/L)1 h10 mg/L4 h5.5 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 10 mg/L at 1 h and 5.5 mg/L at 4 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 10 mg/L (1 h)C2 = 5.5 mg/L (4 h)
k = h⁻¹
4.Medium
VCss​ = 10 mg/LR = ? mg/h · S = 0.8CL = 0.035 L/kg/h61 kg

Practice exercise (made-up values, not clinical guidance). A 61 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.035 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 10 mg/L?

61 kgCL = 0.035 L/kg/hCss = 10 mg/LS = 0.8
R = mg/h
5.Hard
t (h)C (mg/L)Css​t = ?80%

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion is started without a loading dose. The half-life in this patient is 6 h. How long t does it take to reach 80% of the steady-state level?

t½ = 6 h80% Css
t = h
6.Hard
0.1110t (h)C (mg/L)Cpeak​ = 6τ = ?Ctrough​ = 0.5

Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 3.5 h. The target is a peak of 6 mg/L and a trough of 0.5 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 3.5 hCpeak = 6 mg/LCtrough = 0.5 mg/L
τ = h

Answer key

  1. 1. C = 1.25 mg/L
  2. 2. k = 0.08664 h⁻¹
  3. 3. k = 0.1993 h⁻¹
  4. 4. R = 26.69 mg/h
  5. 5. t = 13.93 h
  6. 6. τ = 12.5 h

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