Pharmacology: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

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1Formulas and methods

Dosage calculations

Dose from strength
volume = dose / strength
Weight-based dose
dose = mg/kg × body weight
Tablets
number = dose / tablet strength
Units
1 g = 1000 mg, 1 mg = 1000 mcg

Method

  1. Convert the ordered dose and the stock strength to the same unit.
  2. Use desired over have times quantity: (D/H) × Q.
  3. For weight-based doses, multiply by the weight in kg first, then divide the daily dose by the number of doses.
  4. Estimate first and check that the answer is reasonable.

Clinical pharmacokinetics

Rate constant and half-life
k = ln 2/t½ = 0.693/t½
First-order decay
C = C₀·e−kt = C₀·(1/2)t/t½
Two levels
k = ln(C₁/C₂)/Δt
Volume and clearance
Vd = dose/C₀, CL = k·Vd
Loading dose
LD = Vd·C/(S·F)
Maintenance rate
R = CL·Css/(S·F)
Average steady state
Css,avg = S·F·D/(CL·τ)
Steady state
1 − (1/2)ⁿ after n half-lives; ≈ 97% after 5
Infusion
C = (R/CL)(1 − e−kt)
Interval and dose from peak and trough
τ = ln(Cpeak/Ctrough)/k, D = Vd·(Cpeak − Ctrough)
Cockcroft–Gault
CrCl = (140 − age) × kg/(72 × SCr mg/dL), × 0.85 for women
Renal adjustment
dose × CrCl/normal; Dettli Q = 1 − fe·(1 − CrCl/normal)

Method

  1. Convert the half-life to a rate constant: k = 0.693/t½.
  2. Turn per-kg values into totals: Vd (L) = L/kg × weight, CL (L/h) = L/kg/h × weight.
  3. Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
  4. For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
  5. Check the result: levels fall by half every half-life, and steady state takes about five half-lives.

2Practice problems

1.PharmacologyMedium

When should a patient take a once-daily proton pump inhibitor such as omeprazole?

  1. AOnly when heartburn is severe
  2. BWith a glass of milk at bedtime, when acid output is highest
  3. CAbout 30–60 minutes before the first meal of the day
  4. DStraight after a large evening meal
2.PharmacologyHard

The volume of distribution of a drug is very large (for example 500 L in an adult). What does this suggest?

  1. AThe drug stays almost entirely in the plasma compartment
  2. BThe drug is extensively taken up into tissues
  3. CThe drug cannot be given intravenously
  4. DThe drug is quickly excreted by the kidneys
3.Dosage calculationsMedium
22 kg25 mg/kg/day4× per dayD = ? mg

A patient weighs 22 kg. The prescribed dose is 25 mg/kg per day, given in 4 equal doses. How many milligrams D are given per dose?

22 kg25 mg/kg/day4× per day
D = mg
4.Dosage calculationsHard
100 mg/5 mLV = ? mL

A child weighing 28 kg is prescribed 12.5 mg/kg per dose of an oral suspension that contains 100 mg in 5 mL. What volume V do you give?

28 kg12.5 mg/kg100 mg/5 mL
V = mL
5.Clinical pharmacokineticsMedium
Vd = 0.7 L/kgC = 20 mg/LLD = ? mg96 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 96 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.

96 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Vd = 0.7 L/kg9 → 15 mg/LLD = ? mg61 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 61 kg patient has a phenytoin level of 9 mg/L; the target is 15 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?

61 kgVd = 0.7 L/kgC = 9 → 15 mg/LS = 0.92
LD = mg

3Solutions

1.Pharmacology

Answer: C · About 30–60 minutes before the first meal of the day

Why this is the answer

Proton pump inhibitors block pumps that are active, so they work best when taken before breakfast. Food then activates the pumps when the drug level is high.

Answer

C · About 30–60 minutes before the first meal of the day
2.Pharmacology

Answer: B · The drug is extensively taken up into tissues

Why this is the answer

A Vd far above total body water means the plasma concentration is low because most of the drug sits in tissues (for example fat or muscle), as with digoxin or amiodarone.

Answer

B · The drug is extensively taken up into tissues
3.Dosage calculations

Answer: D = 137.5 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

25·22 = 550 mg

Per dose

D = 550/4 = 137.5 mg

Answer

D = 137.5 mg
4.Dosage calculations

Answer: V = 17.5 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

12.5·28 = 350 mg

Volume

V = (350/100)·5 = 17.5 mL

Answer

V = 17.5 mL
5.Clinical pharmacokinetics

Answer: LD = 1461 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 = 0.7·96 = 67.2 L

Amount of active drug needed

Vd·C = 67.2·20 = 1344 mg

Loading dose

LD = 1344/(0.92·1) = 1461 mg

Answer

LD = 1461 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: LD = 278.5 mg

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 = 0.7·61 = 42.7 L

Rise needed

15 − 9 = 6 mg/L

Loading dose

LD = 42.7·6/(0.92·1) = 278.5 mg

Answer

LD = 278.5 mg

In practice the dose is rounded to a convenient amount.

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