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 is a well-known adverse effect of long-term oral corticosteroids?
AHypoglycaemia
BWeight loss
CLow blood pressure
DOsteoporosis
2.PharmacologyHard
A nurse gave a patient ten times the prescribed dose of hydromorphone. The patient is currently alert. What is the most appropriate sequence of actions?
ASay nothing because the patient is alert
BChange the chart so that it shows the correct dose, and continue to monitor
CMonitor closely, have naloxone ready, inform the prescriber and report
DGive another dose later to balance it out
3.Dosage calculationsMedium
A patient weighs 25 kg. The prescribed dose is 2 mg/kg per day, given in 3 equal doses. How many milligrams D are given per dose?
25 kg2 mg/kg/day3× per day
D = mg
4.Dosage calculationsHard
You must prepare 250 mL of a 2.5% (w/v) solution from a 10% (w/v) stock solution. What volume of stock V₁ do you need?
C₁ = 10%C₂ = 2.5%V₂ = 250 mL
V₁ = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 25 mg/L at 2 h and 10.9 mg/L at 14 h. Assume first-order elimination. Find the elimination rate constant k.
C1 = 25 mg/L (2 h)C2 = 10.9 mg/L (14 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 58 kg patient has a theophylline level of 4 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?
58 kgVd = 0.5 L/kgC = 4 → 10 mg/LS = 0.8
LD = mg
3Solutions
1.Pharmacology
Answer: D · Osteoporosis
Why this is the answer
Long-term corticosteroids reduce bone formation and raise fracture risk. They also cause weight gain, high blood glucose, hypertension and a higher infection risk.
Answer
D · Osteoporosis
2.Pharmacology
Answer: C · Monitor closely, have naloxone ready, inform the prescriber and report
Why this is the answer
The patient may become sedated and develop respiratory depression later, so close monitoring and a ready antidote are needed. The prescriber is informed, the patient is told honestly, and the error is reported to improve the system.
Answer
C · Monitor closely, have naloxone ready, inform the prescriber and report
3.Dosage calculations
Answer: D = 16.67 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
2·25 = 50 mg
Per dose
D = 50/3 = 16.67 mg
Answer
D = 16.67 mg
4.Dosage calculations
Answer: V₁ = 62.5 mL
Solution strengths: a percentage strength is grams per 100 mL, so 1% = 1 g/100 mL = 10 mg/mL. A ratio strength 1:1000 means 1 g in 1000 mL = 1 mg/mL; 1:10 000 is 0.1 mg/mL. For a dilution the amount of drug stays the same: C₁·V₁ = C₂·V₂.
Volume of stock
V₁ = 2.5·250/10 = 62.5 mL
Answer
V₁ = 62.5 mL
5.Clinical pharmacokinetics
Answer: k = 0.06918 h⁻¹
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(25/10.9)/12 = 0.8301/12 = 0.06918 h⁻¹
Answer
k = 0.06918 h⁻¹
6.Clinical pharmacokinetics
Answer: LD = 217.5 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 = 0.5·58 = 29 L
Rise needed
10 − 4 = 6 mg/L
Loading dose
LD = 29·6/(0.8·1) = 217.5 mg
Answer
LD = 217.5 mg
In practice the dose is rounded to a convenient amount.