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
Patients taking metronidazole are usually advised to avoid:
ADairy products
BSunlight
CGrapefruit juice
DAlcohol
2.PharmacologyHard
A pregnant woman develops a deep vein thrombosis. Which anticoagulant is usually preferred?
AAspirin alone
BRivaroxaban, a direct oral factor Xa inhibitor
CWarfarin
DLow-molecular-weight heparin, such as enoxaparin
3.Dosage calculationsMedium
A patient weighs 18 kg. The prescribed dose is 2 mg/kg per day, given in 3 equal doses. How many milligrams D are given per dose?
18 kg2 mg/kg/day3× per day
D = mg
4.Dosage calculationsHard
The prescription is 200 micrograms. The ampoule contains 2 mg/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
200 mcg2 mg/mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 35 mg/L at 2 h and 10.2 mg/L at 18 h. Assume first-order elimination. Find the half-life t½.
C1 = 35 mg/L (2 h)C2 = 10.2 mg/L (18 h)
t½ = h
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 6 mg/L and a trough of 1 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 = 6 mg/LCtrough = 1 mg/L
τ = h
3Solutions
1.Pharmacology
Answer: D · Alcohol
Why this is the answer
Metronidazole can cause a disulfiram-like reaction with alcohol, with flushing, nausea and palpitations. Alcohol is avoided during treatment and for a short time after.
Answer
D · Alcohol
2.Pharmacology
Answer: D · 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
D · Low-molecular-weight heparin, such as enoxaparin
3.Dosage calculations
Answer: D = 12 mg
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.
Total per day
2·18 = 36 mg
Per dose
D = 36/3 = 12 mg
Answer
D = 12 mg
4.Dosage calculations
Answer: V = 0.1 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.
Same units
2 mg/mL = 2000 mcg/mL
Volume
V = 200/2000 = 0.1 mL
Answer
V = 0.1 mL
5.Clinical pharmacokinetics
Answer: t½ = 8.995 h
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.
Elimination rate constant
k = ln(35/10.2)/16 = 1.233/16 = 0.07706 h⁻¹
Half-life
t½ = 0.693/0.07706 = 8.995 h
Answer
t½ = 8.995 h
6.Clinical pharmacokinetics
Answer: τ = 10.3 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(6/1)/0.1733 = 1.792/0.1733 = 10.34 ≈ 10.3 h
Answer
τ = 10.3 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.