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
How does glyceryl trinitrate (GTN) relieve angina?
AIt dissolves the clot in the coronary artery
BIt releases nitric oxide, mainly dilating veins and reducing cardiac workload
CIt blocks beta-1 receptors and slows the heart
DIt increases the force of contraction so the heart pumps blood more efficiently
2.PharmacologyHard
A drug with a short half-life is given at a dosing interval much longer than its half-life. What is the result at steady state?
AVery stable levels with almost no fluctuation
BLarge swings between peak and trough levels
CSteady state is never reached
DGradual accumulation to toxic levels
3.Dosage calculationsMedium
A patient weighs 30 kg. The prescribed dose is 30 mg/kg per day, given in 2 equal doses. How many milligrams D are given per dose?
30 kg30 mg/kg/day2× per day
D = mg
4.Dosage calculationsHard
The prescription is 1440 micrograms. The ampoule contains 2 mg/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
1440 mcg2 mg/mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 90 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.
90 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 3.5 h. The target is a peak of 7 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 = 3.5 hCpeak = 7 mg/LCtrough = 0.5 mg/L
τ = h
3Solutions
1.Pharmacology
Answer: B · It releases nitric oxide, mainly dilating veins and reducing cardiac workload
Why this is the answer
GTN is converted to nitric oxide, which relaxes vascular smooth muscle through cGMP. Venodilation lowers preload, so the heart works less and needs less oxygen.
Answer
B · It releases nitric oxide, mainly dilating veins and reducing cardiac workload
2.Pharmacology
Answer: B · Large swings between peak and trough levels
Why this is the answer
When the interval is long compared with the half-life, most of each dose is eliminated before the next one. Levels rise high after each dose and fall low before the next, so there may be toxic peaks or ineffective troughs.
Answer
B · Large swings between peak and trough levels
3.Dosage calculations
Answer: D = 450 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
30·30 = 900 mg
Per dose
D = 900/2 = 450 mg
Answer
D = 450 mg
4.Dosage calculations
Answer: V = 0.72 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 = 1440/2000 = 0.72 mL
Answer
V = 0.72 mL
5.Clinical pharmacokinetics
Answer: LD = 1370 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 = 0.7·90 = 63 L
Amount of active drug needed
Vd·C = 63·20 = 1260 mg
Loading dose
LD = 1260/(0.92·1) = 1370 mg
Answer
LD = 1370 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: τ = 13.3 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 = 0.693/3.5 = 0.198 h⁻¹
Dosing interval
τ = ln(7/0.5)/0.198 = 2.639/0.198 = 13.33 ≈ 13.3 h
Answer
τ = 13.3 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.