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

During adult cardiac arrest, how often is IV adrenaline (epinephrine) usually repeated?

  1. AEvery 15–20 minutes
  2. BEvery 3–5 minutes
  3. CEvery 30 seconds
  4. DOnce only, at the start
2.PharmacologyHard

A pregnant woman develops a deep vein thrombosis. Which anticoagulant is usually preferred?

  1. AWarfarin
  2. BLow-molecular-weight heparin, such as enoxaparin
  3. CRivaroxaban, a direct oral factor Xa inhibitor
  4. DAspirin alone
3.Dosage calculationsMedium
024681010 mL

A child weighing 29 kg is prescribed 4 mg/kg of a drug. The stock solution contains 20 mg/mL. Draw up the dose: what volume V do you give?

29 kg4 mg/kg20 mg/mL
V = mL
4.Dosage calculationsHard
40 mg/1 mLV = ? mL

A patient weighs 127.6 lb. The prescription is 2 mg/kg as a single dose, and the stock contains 40 mg in 1 mL. Using 1 kg = 2.2 lb, what volume V do you give?

127.6 lb2 mg/kg40 mg/1 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
Vd = 0.5 L/kgC = 15 mg/LLD = ? mg93 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 93 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.

93 kgVd = 0.5 L/kgC = 15 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
0.1110100t (h)C (mg/L)Cpeak​ = 10τ = ?Ctrough​ = 0.5

Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 4 h. The target is a peak of 10 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 = 4 hCpeak = 10 mg/LCtrough = 0.5 mg/L
τ = h

3Solutions

1.Pharmacology

Answer: B · Every 3–5 minutes

Why this is the answer

In adult advanced life support, adrenaline 1 mg IV or IO is repeated every 3–5 minutes during the arrest. High-quality CPR and defibrillation for shockable rhythms remain the priorities.

Answer

B · Every 3–5 minutes
2.Pharmacology

Answer: B · Low-molecular-weight heparin, such as enoxaparin

Why this is the answer

Heparins are large molecules that do not cross the placenta. Warfarin is teratogenic, and direct oral anticoagulants are not recommended in pregnancy.

Answer

B · Low-molecular-weight heparin, such as enoxaparin
3.Dosage calculations

Answer: V = 5.8 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 = 4·29 = 116 mg

Volume

V = 116/20 = 5.8 mL

On the syringe

On this 10 mL syringe every small line is 0.2 mL.

Answer

V = 5.8 mL
4.Dosage calculations

Answer: V = 2.9 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

127.6/2.2 = 58 kg

Dose

2·58 = 116 mg

Volume

V = (116/40)·1 = 2.9 mL

Answer

V = 2.9 mL
5.Clinical pharmacokinetics

Answer: LD = 871.9 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 = 0.5·93 = 46.5 L

Amount of active drug needed

Vd·C = 46.5·15 = 697.5 mg

Loading dose

LD = 697.5/(0.8·1) = 871.9 mg

Answer

LD = 871.9 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: τ = 17.3 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 = 0.693/4 = 0.1733 h⁻¹

Dosing interval

τ = ln(10/0.5)/0.1733 = 2.996/0.1733 = 17.29 ≈ 17.3 h

Answer

τ = 17.3 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.

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