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
Besides lithium levels, which blood tests are monitored in long-term lithium treatment?
ALiver enzymes, serum amylase and lipase
BKidney function and thyroid function
CINR and fibrinogen
DTroponin and CK-MB
2.PharmacologyHard
A patient who uses insulin is taught sick-day rules. Which statement shows a need for further teaching?
A"I will keep drinking fluids, even if I cannot eat much."
B"I will seek help if I keep vomiting or have ketones."
C"If I am ill and not eating, I will stop my insulin completely."
D"I will check my blood glucose and ketones more often while I am ill."
3.Dosage calculationsMedium
A patient weighs 70 kg. The prescribed dose is 15 mg/kg per day, given in 3 equal doses. How many milligrams D are given per dose?
70 kg15 mg/kg/day3× per day
D = mg
4.Dosage calculationsHard
The prescription is 4900 units of a drug that comes as 10000 units/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?
4900 units10000 units/mL
V = mL
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 60 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.
60 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 2 h. The target is a peak of 7 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 = 7 mg/LCtrough = 1 mg/L
τ = h
3Solutions
1.Pharmacology
Answer: B · Kidney function and thyroid function
Why this is the answer
Lithium can impair kidney function and cause hypothyroidism. Kidney function, thyroid function and calcium are checked regularly, along with lithium levels.
Answer
B · Kidney function and thyroid function
2.Pharmacology
Answer: C · "If I am ill and not eating, I will stop my insulin completely."
Why this is the answer
Stress hormones during illness often raise glucose, so basal insulin is usually continued and doses may need adjusting. Stopping insulin in type 1 diabetes can lead to ketoacidosis.
Answer
C · "If I am ill and not eating, I will stop my insulin completely."
3.Dosage calculations
Answer: D = 350 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
15·70 = 1050 mg
Per dose
D = 1050/3 = 350 mg
Answer
D = 350 mg
4.Dosage calculations
Answer: V = 0.49 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.
Volume
V = 4900/10000 = 0.49 mL
On the syringe
On this 1 mL syringe every small line is 0.01 mL.
Answer
V = 0.49 mL
5.Clinical pharmacokinetics
Answer: LD = 913 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 = 0.7·60 = 42 L
Amount of active drug needed
Vd·C = 42·20 = 840 mg
Loading dose
LD = 840/(0.92·1) = 913 mg
Answer
LD = 913 mg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: τ = 5.6 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(7/1)/0.3466 = 1.946/0.3466 = 5.615 ≈ 5.6 h
Answer
τ = 5.6 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.