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
Which type of diuretic acts on the thick ascending limb of the loop of Henle?
APotassium-sparing diuretics
BOsmotic diuretics
CLoop diuretics such as furosemide
DThiazide diuretics
2.PharmacologyHard
In myxoedema coma, why is hydrocortisone often given before or with thyroid hormone?
AHydrocortisone prevents the hypoglycaemia caused by levothyroxine
BThyroid hormone can trigger adrenal crisis if cortisol is also low
CHydrocortisone makes thyroid hormone work faster
DThyroid hormone is inactive unless hydrocortisone is given
3.Dosage calculationsMedium
A child weighing 10 kg is prescribed 4 mg/kg of a drug. The stock solution contains 10 mg/mL. Draw up the dose: what volume V do you give?
10 kg4 mg/kg10 mg/mL
V = mL
4.Dosage calculationsHard
The prescription is 230 micrograms. The ampoule contains 0.5 mg/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
230 mcg0.5 mg/mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 270 mg of tobramycin. The level back-extrapolated to time zero is 9 mg/L and the half-life is 2.5 h. Find the volume of distribution Vd.
D = 270 mgC₀ = 9 mg/Lt½ = 2.5 h
Vd = L
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 94 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?
94 kgVd = 0.5 L/kgC = 6 → 10 mg/LS = 0.8
LD = mg
3Solutions
1.Pharmacology
Answer: C · Loop diuretics such as furosemide
Why this is the answer
Loop diuretics block the Na⁺/K⁺/2Cl⁻ cotransporter in the thick ascending limb and are the most powerful diuretics.
Answer
C · Loop diuretics such as furosemide
2.Pharmacology
Answer: B · Thyroid hormone can trigger adrenal crisis if cortisol is also low
Why this is the answer
Restoring thyroid hormone increases metabolic rate and cortisol clearance. If adrenal reserve is also low, this can precipitate adrenal crisis, so steroid cover is given until adrenal insufficiency is excluded.
Answer
B · Thyroid hormone can trigger adrenal crisis if cortisol is also low
3.Dosage calculations
Answer: V = 4 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 = 4·10 = 40 mg
Volume
V = 40/10 = 4 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 4 mL
4.Dosage calculations
Answer: V = 0.46 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
0.5 mg/mL = 500 mcg/mL
Volume
V = 230/500 = 0.46 mL
Answer
V = 0.46 mL
5.Clinical pharmacokinetics
Answer: Vd = 30 L
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 = 270/9 = 30 L
Answer
Vd = 30 L
6.Clinical pharmacokinetics
Answer: LD = 235 mg
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.
Volume of distribution
Vd = 0.5·94 = 47 L
Rise needed
10 − 6 = 4 mg/L
Loading dose
LD = 47·4/(0.8·1) = 235 mg
Answer
LD = 235 mg
In practice the dose is rounded to a convenient amount.