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
A patient given haloperidol develops a sudden painful twisting of the neck and the eyes roll upward. Which drug is usually given?
ANaloxone
BAn anticholinergic such as benzatropine (benztropine)
CA second dose of haloperidol
DDantrolene by mouth to relax the twisted neck muscles
2.PharmacologyHard
A malnourished patient with low serum albumin shows nystagmus and unsteady gait. Her total phenytoin level is within range. What is the most likely explanation?
ALow albumin makes phenytoin bind more tightly, so the drug is less active
BLow albumin speeds up phenytoin metabolism
CThe total level is high, but the laboratory is wrong
DMore phenytoin is unbound when albumin is low, so free levels rise
3.Dosage calculationsMedium
A vial contains 500 mg of powder. The label says: add 1.8 mL of water for injection to give a total volume of 2 mL. The prescription is 150 mg. What volume V of the reconstituted solution do you give?
500 mg+ 1.8 mL → 2 mL150 mg
V = mL
4.Dosage calculationsHard
A child weighing 14 kg is prescribed 10 mg/kg per dose of an oral suspension that contains 40 mg in 1 mL. What volume V do you give?
14 kg10 mg/kg40 mg/1 mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 110 mg of gentamicin. The level back-extrapolated to time zero is 7 mg/L and the half-life is 3.5 h. Find the volume of distribution Vd.
D = 110 mgC₀ = 7 mg/Lt½ = 3.5 h
Vd = L
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 2 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 = 2 hCpeak = 6 mg/LCtrough = 1 mg/L
τ = h
3Solutions
1.Pharmacology
Answer: B · An anticholinergic such as benzatropine (benztropine)
Why this is the answer
Acute dystonia is caused by dopamine blockade, which leaves acetylcholine activity relatively high. An IM or IV anticholinergic usually relieves it within minutes.
Answer
B · An anticholinergic such as benzatropine (benztropine)
2.Pharmacology
Answer: D · More phenytoin is unbound when albumin is low, so free levels rise
Why this is the answer
Phenytoin is highly protein bound. When albumin is low, the free fraction rises, so toxicity can occur with a "normal" total level; a free phenytoin level or corrected level is more useful.
Answer
D · More phenytoin is unbound when albumin is low, so free levels rise
3.Dosage calculations
Answer: V = 0.6 mL
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.
Concentration after mixing
500 mg/2 mL = 250 mg/mL
Volume
V = 150/250 = 0.6 mL
Answer
V = 0.6 mL
4.Dosage calculations
Answer: V = 3.5 mL
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.
Dose from weight
10·14 = 140 mg
Volume
V = (140/40)·1 = 3.5 mL
Answer
V = 3.5 mL
5.Clinical pharmacokinetics
Answer: Vd = 15.71 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 = 110/7 = 15.71 L
Answer
Vd = 15.71 L
6.Clinical pharmacokinetics
Answer: τ = 5.2 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 = 0.693/2 = 0.3466 h⁻¹
Dosing interval
τ = ln(6/1)/0.3466 = 1.792/0.3466 = 5.17 ≈ 5.2 h
Answer
τ = 5.2 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.