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 adverse effect of doxorubicin requires monitoring of heart function?
APericarditis caused by infection
BSevere hypertension during the infusion itself
CCumulative heart muscle damage leading to heart failure
DRapid rise in cholesterol
2.PharmacologyHard
Why is glibenclamide (glyburide) generally avoided in older adults?
AIt is long-acting and can cause prolonged, severe hypoglycaemia
BIt causes severe weight loss and muscle wasting in older people
CIt is not absorbed in people over 65
DIt causes high blood glucose in older adults
3.Dosage calculationsMedium
A patient weighs 45 kg. The prescribed dose is 15 mg/kg per day, given in 2 equal doses. How many milligrams D are given per dose?
45 kg15 mg/kg/day2× per day
D = mg
4.Dosage calculationsHard
The prescription is 0.35 mg. The ampoule contains 500 micrograms per mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
0.35 mg500 mcg/mL
V = 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 1 h and 6.3 mg/L at 17 h. Assume first-order elimination. Find the elimination rate constant k.
C1 = 25 mg/L (1 h)C2 = 6.3 mg/L (17 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 98 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.04 L/kg/h. Each dose is 300 mg of theophylline. Find the average steady-state level Css,avg.
CL = 0.04 L/kg/hτ = 8 hS = 1F = 198 kgD = 300 mg
Css,avg = mg/L
3Solutions
1.Pharmacology
Answer: C · Cumulative heart muscle damage leading to heart failure
Why this is the answer
Anthracyclines such as doxorubicin can damage heart muscle, and the risk increases with cumulative dose. Heart function is assessed before and during treatment; the drug can also turn urine red.
Answer
C · Cumulative heart muscle damage leading to heart failure
2.Pharmacology
Answer: A · It is long-acting and can cause prolonged, severe hypoglycaemia
Why this is the answer
Glibenclamide has a long action and active metabolites that accumulate when kidney function declines. Older adults are at high risk of prolonged hypoglycaemia, so shorter-acting agents or other drug classes are preferred.
Answer
A · It is long-acting and can cause prolonged, severe hypoglycaemia
3.Dosage calculations
Answer: D = 337.5 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
15·45 = 675 mg
Per dose
D = 675/2 = 337.5 mg
Answer
D = 337.5 mg
4.Dosage calculations
Answer: V = 0.7 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
0.35 mg = 350 mcg
Volume
V = 350/500 = 0.7 mL
Answer
V = 0.7 mL
5.Clinical pharmacokinetics
Answer: k = 0.08615 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(25/6.3)/16 = 1.378/16 = 0.08615 h⁻¹
Answer
k = 0.08615 h⁻¹
6.Clinical pharmacokinetics
Answer: Css,avg = 9.566 mg/L
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.