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
Why should loperamide usually be avoided in bloody diarrhoea with fever?
ASlowing the gut may keep bacteria and toxins in the bowel longer
BLoperamide causes severe hypertension when there is an infection
CLoperamide makes the bleeding worse by blocking platelets
DLoperamide is an antibiotic that causes resistance
2.PharmacologyHard
Why is sildenafil contraindicated in a patient who uses nitrates for angina?
ANitrates block the metabolism of sildenafil in the liver
BBoth raise cGMP in vascular smooth muscle and can cause severe hypotension
CSildenafil blocks the conversion of nitrates into nitric oxide in the vessels
DThe combination causes dangerous hyperkalaemia
3.Dosage calculationsMedium
A child weighing 8 kg is prescribed 20 mg/kg of a drug. The stock solution contains 25 mg/mL. Draw up the dose: what volume V do you give?
8 kg20 mg/kg25 mg/mL
V = mL
4.Dosage calculationsHard
A patient is 104 cm tall and weighs 32 kg. The prescription is 175 mg/m². Using the Mosteller formula BSA = √(height (cm) × weight (kg)/3600), find the dose D in mg.
104 cm32 kg175 mg/m²
D = mg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 74 kg patient receives a theophylline infusion. The clearance is 0.05 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 12 mg/L?
74 kgCL = 0.05 L/kg/hCss = 12 mg/L
R = mg/h
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 7 mg/L and a trough of 0.5 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 = 7 mg/LCtrough = 0.5 mg/L
τ = h
3Solutions
1.Pharmacology
Answer: A · Slowing the gut may keep bacteria and toxins in the bowel longer
Why this is the answer
In dysentery or suspected C. difficile infection, antimotility drugs may prolong illness and increase complications such as toxic megacolon. The cause is assessed and fluids are replaced first.
Answer
A · Slowing the gut may keep bacteria and toxins in the bowel longer
2.Pharmacology
Answer: B · Both raise cGMP in vascular smooth muscle and can cause severe hypotension
Why this is the answer
Nitrates increase cGMP production through nitric oxide and sildenafil blocks its breakdown by PDE5. Together they cause excessive vasodilation and a dangerous fall in blood pressure.
Answer
B · Both raise cGMP in vascular smooth muscle and can cause severe hypotension
3.Dosage calculations
Answer: V = 6.4 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.
Dose from body weight
D = 20·8 = 160 mg
Volume
V = 160/25 = 6.4 mL
On the syringe
On this 10 mL syringe every small line is 0.2 mL.
Answer
V = 6.4 mL
4.Dosage calculations
Answer: D = 168.3 mg
Body-surface-area dosing: many cytotoxic and some paediatric doses are prescribed per square metre of body surface area: dose (mg) = dose per m² × BSA (m²). The Mosteller formula estimates BSA = √(height (cm) × weight (kg)/3600).
Body surface area
BSA = √(104·32/3600) = √0.9244 = 0.9615 m²
Dose
D = 175·0.9615 = 168.3 mg
Answer
D = 168.3 mg
5.Clinical pharmacokinetics
Answer: R = 44.4 mg/h
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.
Clearance
CL = 0.05·74 = 3.7 L/h
Infusion rate
R = 3.7·12 = 44.4 mg/h
Answer
R = 44.4 mg/h
6.Clinical pharmacokinetics
Answer: τ = 7.6 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(7/0.5)/0.3466 = 2.639/0.3466 = 7.615 ≈ 7.6 h
Answer
τ = 7.6 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.