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 = 36 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, digoxin has an elimination half-life of 36 h. Find the elimination rate constant k in h⁻¹.

t½ = 36 h
k = h⁻¹
2.Easy
t/t½C/Css (%)100%597%

Practice exercise (made-up values, not clinical guidance). A patient starts regular doses of phenobarbital, which has a half-life of 100 h. Using the rule that steady state is reached after about five half-lives, after how many hours tss is steady state reached?

t½ = 100 h
tss = h
3.Medium
110t (h)C (mg/L)2 h8 mg/L10 h1.3 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 8 mg/L at 2 h and 1.3 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.

C1 = 8 mg/L (2 h)C2 = 1.3 mg/L (10 h)
t½ = h
4.Medium
110100t (h)C (mg/L)1 h30 mg/L17 h6.2 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 30 mg/L at 1 h and 6.2 mg/L at 17 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 30 mg/L (1 h)C2 = 6.2 mg/L (17 h)
k = h⁻¹
5.Hard
VCss,avg​ = 15 mg/LD = ? mg · q8hCL = 0.03 L/kg/h55 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 55 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.03 L/kg/h. What dose D in mg gives an average steady-state level of 15 mg/L?

CL = 0.03 L/kg/hτ = 8 hS = 1F = 155 kgCss,avg = 15 mg/L
D = mg
6.Hard
Vd = 0.25 L/kg1 → 6 mg/LD = ? mg52 kg

Practice exercise (made-up values, not clinical guidance). A 52 kg patient on regular tobramycin doses is at steady state. The target is a peak of 6 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.25 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?

52 kgVd = 0.25 L/kgCpeak = 6 mg/LCtrough = 1 mg/L
D = mg

Answer key

  1. 1. k = 0.01925 h⁻¹
  2. 2. tss = 500 h
  3. 3. t½ = 3.052 h
  4. 4. k = 0.09854 h⁻¹
  5. 5. D = 198 mg
  6. 6. D = 65 mg

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