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
After instilling eye drops, why should the patient press gently on the inner corner of the eye for a minute or two?
ATo reduce drainage into the nose and absorption into blood
BTo stop the eye from producing tears for about an hour
CTo spread the drop evenly over the eyelid
DTo make the pupil dilate faster
2.PharmacologyHard
A patient receiving a phenylephrine infusion has a rise in blood pressure and a fall in heart rate. What is the best explanation?
AAlpha-1 vasoconstriction raises blood pressure and causes reflex bradycardia
BPhenylephrine directly blocks beta-1 receptors in the heart's conduction system
CThe fall in heart rate shows phenylephrine toxicity of the AV node
DPhenylephrine stimulates muscarinic receptors
3.Dosage calculationsMedium
A prefilled syringe of epinephrine (adrenaline) is labelled 1:10 000 (0.1 mg/mL). The prescription is 1 mg IV in cardiac arrest. What volume V do you give?
1:10 0001 mg
V = mL
4.Dosage calculationsHard
A patient weighs 55 lb. The prescription is 2 mg/kg as a single dose, and the stock contains 200 mg in 5 mL. Using 1 kg = 2.2 lb, what volume V do you give? Round to one decimal place.
55 lb2 mg/kg200 mg/5 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 55 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 15 mg/L. The volume of distribution is 0.5 L/kg and the salt factor is S = 0.8. Find the loading dose LD of aminophylline in mg.
55 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 98 kg patient takes oral theophylline (F = 1) every 12 h. The clearance is 0.05 L/kg/h. Each dose is 200 mg of theophylline. Find the average steady-state level Css,avg.
CL = 0.05 L/kg/hτ = 12 hS = 1F = 198 kgD = 200 mg
Css,avg = mg/L
3Solutions
1.Pharmacology
Answer: A · To reduce drainage into the nose and absorption into blood
Why this is the answer
Nasolacrimal occlusion keeps the drop in the eye longer and reduces systemic absorption through the nasal mucosa. This matters for drugs such as timolol, which can cause bradycardia or bronchospasm.
Answer
A · To reduce drainage into the nose and absorption into blood
2.Pharmacology
Answer: A · Alpha-1 vasoconstriction raises blood pressure and causes reflex bradycardia
Why this is the answer
Phenylephrine is a pure alpha-1 agonist with little direct cardiac action. The higher blood pressure activates baroreceptors, which increase vagal tone and slow the heart.
Answer
A · Alpha-1 vasoconstriction raises blood pressure and causes reflex bradycardia
3.Dosage calculations
Answer: V = 10 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.
Concentration
1:10 000 = 1 g/10000 mL = 0.1 mg/mL
Volume
V = 1/0.1 = 10 mL
Answer
V = 10 mL
US labels now state mg/mL (1 mg/mL = 1:1000; 0.1 mg/mL = 1:10 000). Never give 1 mg/mL epinephrine IV push.
4.Dosage calculations
Answer: V = 1.3 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.
Weight in kg
55/2.2 = 25 kg
Dose
2·25 = 50 mg
Volume
V = (50/200)·5 = 1.25 ≈ 1.3 mL
Answer
V = 1.3 mL
5.Clinical pharmacokinetics
Answer: LD = 515.6 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 = 0.5·55 = 27.5 L
Amount of active drug needed
Vd·C = 27.5·15 = 412.5 mg
Loading dose
LD = 412.5/(0.8·1) = 515.6 mg
Answer
LD = 515.6 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: Css,avg = 3.401 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.