Clinical Pharmacokinetics: practice problems and exam questions

Half-lives, clearance and volume of distribution, loading and maintenance doses, peak and trough levels, and dose adjustment for kidney function, as practice exercises for pharmacy, medicine and nursing students. All values are made up for practice and are not clinical guidance.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Clinical pharmacokineticshalf-life, levels and loading doses · clearance, Vd, S and F · regimen design and renal adjustmentcalculations
  2. Pharmacy calculationspercentage strength and ratio strength · reconstitution and concentrations · alligationcalculations
  3. Pediatric & critical-care dosingpediatric weight-based doses (mg/kg/day divided) · heparin and insulin infusions (units/h, mL/h) · titrated drips (mcg/kg/min to mL/h)calculations
  4. Pharmacologykey terms and facts · applying the concepts · cases and analysismultiple choice

Exam tips for Clinical Pharmacokinetics

  • Convert the half-life to k = 0.693/t½ before anything else, and use ln, not log₁₀.
  • Divide by the salt factor S and the bioavailability F; never multiply.
  • Check the answer with half-lives: the level halves every t½ and reaches steady state after about five.

Worked examples

Clinical pharmacokineticsEasy
110100t (h)C (mg/L)10 mg/L9 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the gentamicin level is 10 mg/L. The half-life is 3 h, and the drug is eliminated by first-order kinetics. What is the level C 9 h later?

C₀ = 10 mg/Lt½ = 3 ht = 9 h

Find C (mg/L)

Show hints
  1. Number of half-lives: n = t/t½.
  2. C = C₀·(1/2)ⁿ, or C = C₀·e^(−kt) with k = 0.693/t½.
  3. n = 9/3 = 3.
Show the answer and the step-by-step solution

First-order elimination: a constant fraction of the drug is removed per hour, so C = C₀·e−kt with k = ln 2/t½ = 0.693/t½. After n half-lives C = C₀·(1/2)ⁿ. With regular doses or an infusion, the level reaches 1 − (1/2)ⁿ of steady state after n half-lives: about 97% after 5. A loading dose fills the volume of distribution: LD = Vd × C. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.

Number of half-lives

n = 9/3 = 3

Level

C = 10·(1/2)3 = 1.25 mg/L

Answer

C = 1.25 mg/L
Pharmacy calculationsMedium
024681010 mL

A solution has a ratio strength of 1:10 000. You need 0.64 mg. Draw up the volume in the 10 mL syringe: what volume V is that?

1:10 0000.64 mg

Find V (mL)

Show hints
  1. A ratio strength 1:n means 1 g in n mL.
  2. 1 g in 10 000 mL is 0.1 mg/mL.
  3. On this 10 mL syringe every small line is 0.2 mL.
Show the answer and the step-by-step solution

Strength to volume: turn the strength into mg per mL first. A w/v percentage is grams per 100 mL, so x% = 10x mg/mL. A ratio strength 1:n is 1 g in n mL. After reconstitution the concentration is the powder's dose over the final volume (diluent + powder volume). Then volume = dose/concentration.

Strength in mg/mL

1000 mg/10 000 mL = 0.1 mg/mL

Volume

V = 0.64/0.1 = 6.4 mL

Answer

V = 6.4 mL

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

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