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 = ?C₀/2

Practice exercise (made-up values, not clinical guidance). The elimination rate constant of vancomycin in a patient is 0.116 h⁻¹. Find the half-life t½.

k = 0.116 h⁻¹
t½ = h
2.Easy
110100t (h)CC₀t½ = 3 hC₀/2

Practice exercise (made-up values, not clinical guidance). In a patient, tobramycin has an elimination half-life of 3 h. Find the elimination rate constant k in h⁻¹.

t½ = 3 h
k = h⁻¹
3.Medium
Vd = 0.7 L/kgC = 15 mg/LLD = ? mg65 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 65 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.

65 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
4.Medium
Vd = 0.7 L/kgC = 20 mg/LLD = ? mg66 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 66 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 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.

66 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
5.Hard
VCss,avg​ = 15 mg/LD = ? mg · q12hCL = 5 L/hS = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A patient receives vancomycin IV every 12 h; the vancomycin clearance is 5 L/h. What dose D in mg gives an average steady-state level of 15 mg/L?

CL = 5 L/hτ = 12 hS = 1F = 1Css,avg = 15 mg/L
D = mg
6.Hard
Vd = 0.5 L/kg6 → 10 mg/LLD = ? mg51 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 51 kg patient has a theophylline level of 6 mg/L; the target is 10 mg/L. The drug is given intravenously as aminophylline (S = 0.8), and the volume of distribution is 0.5 L/kg. What extra loading dose LD raises the level to the target?

51 kgVd = 0.5 L/kgC = 6 → 10 mg/LS = 0.8
LD = mg

Answer key

  1. 1. t½ = 5.975 h
  2. 2. k = 0.231 h⁻¹
  3. 3. LD = 741.8 mg
  4. 4. LD = 1004 mg
  5. 5. D = 900 mg
  6. 6. LD = 127.5 mg

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