Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
Check the result: levels fall by half every half-life, and steady state takes about five half-lives.
2Practice problems
1.PharmacologyMedium
During adult cardiac arrest, how often is IV adrenaline (epinephrine) usually repeated?
AEvery 15–20 minutes
BEvery 3–5 minutes
CEvery 30 seconds
DOnce only, at the start
2.PharmacologyHard
A pregnant woman develops a deep vein thrombosis. Which anticoagulant is usually preferred?
AWarfarin
BLow-molecular-weight heparin, such as enoxaparin
CRivaroxaban, a direct oral factor Xa inhibitor
DAspirin alone
3.Dosage calculationsMedium
A child weighing 29 kg is prescribed 4 mg/kg of a drug. The stock solution contains 20 mg/mL. Draw up the dose: what volume V do you give?
29 kg4 mg/kg20 mg/mL
V = mL
4.Dosage calculationsHard
A patient weighs 127.6 lb. The prescription is 2 mg/kg as a single dose, and the stock contains 40 mg in 1 mL. Using 1 kg = 2.2 lb, what volume V do you give?
127.6 lb2 mg/kg40 mg/1 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 93 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 15 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.
93 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 4 h. The target is a peak of 10 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 = 4 hCpeak = 10 mg/LCtrough = 0.5 mg/L
τ = h
3Solutions
1.Pharmacology
Answer: B · Every 3–5 minutes
Why this is the answer
In adult advanced life support, adrenaline 1 mg IV or IO is repeated every 3–5 minutes during the arrest. High-quality CPR and defibrillation for shockable rhythms remain the priorities.
Answer
B · Every 3–5 minutes
2.Pharmacology
Answer: B · Low-molecular-weight heparin, such as enoxaparin
Why this is the answer
Heparins are large molecules that do not cross the placenta. Warfarin is teratogenic, and direct oral anticoagulants are not recommended in pregnancy.
Answer
B · Low-molecular-weight heparin, such as enoxaparin
3.Dosage calculations
Answer: V = 5.8 mL
Drawing up a dose: volume = (prescribed dose/stock strength) × stock volume. Convert to the same units first, then check that the volume fits the syringe and lands on one of its graduations.
Dose from body weight
D = 4·29 = 116 mg
Volume
V = 116/20 = 5.8 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 5.8 mL
4.Dosage calculations
Answer: V = 2.9 mL
Dose calculation: the amount to give equals what is prescribed divided by what you have, times the quantity it comes in: amount = (desired dose/stock strength) × stock volume. Always keep the units the same (convert g to mg, mg to mcg first) and check that the answer is sensible.
Weight in kg
127.6/2.2 = 58 kg
Dose
2·58 = 116 mg
Volume
V = (116/40)·1 = 2.9 mL
Answer
V = 2.9 mL
5.Clinical pharmacokinetics
Answer: LD = 871.9 mg
Clearance and volume: at steady state the dosing rate equals the elimination rate, so the maintenance rate = CL × Css. A loading dose is LD = Vd × C/(S × F), with S the salt factor (fraction of active drug in the salt) and F the bioavailability (1 for IV). From two levels, k = ln(C₁/C₂)/Δt; for an IV bolus Vd = dose/C₀ and CL = k × Vd. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Volume of distribution
Vd = 0.5·93 = 46.5 L
Amount of active drug needed
Vd·C = 46.5·15 = 697.5 mg
Loading dose
LD = 697.5/(0.8·1) = 871.9 mg
Answer
LD = 871.9 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: τ = 17.3 h
Designing a regimen: with rapid IV doses the level falls from peak to trough during the interval, so τ = ln(Cpeak/Ctrough)/k and the dose that restores the peak is Vd × (Cpeak − Ctrough). The average steady-state level is Css,avg = S·F·dose/(CL·τ). An infusion rises as C = (R/CL)(1 − e−kt). For a drug cleared by the kidneys, the dose is scaled with the creatinine clearance. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.
Rate constant
k = 0.693/4 = 0.1733 h⁻¹
Dosing interval
τ = ln(10/0.5)/0.1733 = 2.996/0.1733 = 17.29 ≈ 17.3 h
Answer
τ = 17.3 h
In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.