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
Benzathine benzylpenicillin (penicillin G benzathine) is ordered. Which route is correct?
AIntravenous infusion over 30 minutes
BDeep intramuscular only
CIntravenous push
DSubcutaneous in the abdomen
2.PharmacologyHard
A patient taking hydralazine for several months develops joint pain, fever and a rash. Anti-histone antibodies are positive. What is the most likely diagnosis?
ADrug-induced lupus
BRheumatoid arthritis
CAn acute viral infection
DGout
3.Dosage calculationsMedium
A patient weighs 70 kg. The prescribed dose is 25 mg/kg per day, given in 2 equal doses. How many milligrams D are given per dose?
70 kg25 mg/kg/day2× per day
D = mg
4.Dosage calculationsHard
You must prepare 250 mL of a 2.5% (w/v) solution from a 20% (w/v) stock solution. What volume of stock V₁ do you need?
C₁ = 20%C₂ = 2.5%V₂ = 250 mL
V₁ = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). An 88 kg patient is to be loaded with digoxin tablets to reach a level of 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.
88 kgVd = 7 L/kgC = 0.8 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). An 89 kg patient has a digoxin level of 0.5 mcg/L; the target is 0.8 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?
89 kgVd = 7 L/kgC = 0.5 → 0.8 mcg/LF = 0.7
LD = mcg
3Solutions
1.Pharmacology
Answer: B · Deep intramuscular only
Why this is the answer
Benzathine penicillin is a depot suspension given deep IM. Accidental intravenous injection can cause cardiorespiratory arrest and death.
Answer
B · Deep intramuscular only
2.Pharmacology
Answer: A · Drug-induced lupus
Why this is the answer
Hydralazine and procainamide are classic causes of drug-induced lupus, especially in slow acetylators. Symptoms usually resolve after the drug is stopped.
Answer
A · Drug-induced lupus
3.Dosage calculations
Answer: D = 875 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
25·70 = 1750 mg
Per dose
D = 1750/2 = 875 mg
Answer
D = 875 mg
4.Dosage calculations
Answer: V₁ = 31.25 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 = 2.5·250/20 = 31.25 mL
Answer
V₁ = 31.25 mL
5.Clinical pharmacokinetics
Answer: LD = 704 mcg
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 = 7·88 = 616 L
Amount of active drug needed
Vd·C = 616·0.8 = 492.8 mcg
Loading dose
LD = 492.8/(1·0.7) = 704 mcg
Answer
LD = 704 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: LD = 267 mcg
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 = 7·89 = 623 L
Rise needed
0.8 − 0.5 = 0.3 mcg/L
Loading dose
LD = 623·0.3/(1·0.7) = 267 mcg
Answer
LD = 267 mcg
In practice the dose is rounded to a convenient amount.