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

Why should long-term oral corticosteroids not be stopped suddenly?

  1. AThe immune system would immediately attack the liver and adrenal glands
  2. BThe drug forms toxic metabolites when stopped
  3. CThe adrenal glands are suppressed and may not make enough cortisol
  4. DStopping causes a sudden rise in blood glucose
2.PharmacologyHard

Which pair of adverse effects is characteristic of long-term amiodarone treatment?

  1. AGum overgrowth and excess hair growth
  2. BOtotoxicity and nephrotoxicity
  3. CThyroid dysfunction and pulmonary fibrosis
  4. DDry cough and angioedema
3.Dosage calculationsMedium
12 kg20 mg/kg/day3× per dayD = ? mg

A patient weighs 12 kg. The prescribed dose is 20 mg/kg per day, given in 3 equal doses. How many milligrams D are given per dose?

12 kg20 mg/kg/day3× per day
D = mg
4.Dosage calculationsHard
00.10.20.30.40.50.60.70.80.911 mL

The prescription is 220 micrograms. The ampoule contains 1 mg/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?

220 mcg1 mg/mL
V = mL
5.Clinical pharmacokineticsMedium
110100t (h)C (mg/L)C₀ = 9 mg/Lt½ = 2 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 210 mg of gentamicin. The level back-extrapolated to time zero is 9 mg/L and the half-life is 2 h. Find the volume of distribution Vd.

D = 210 mgC₀ = 9 mg/Lt½ = 2 h
Vd = L
6.Clinical pharmacokineticsHard
0.1110t (h)C (mg/L)Cpeak​ = 7τ = ?Ctrough​ = 1

Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 3.5 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 = 3.5 hCpeak = 7 mg/LCtrough = 1 mg/L
τ = h

3Solutions

1.Pharmacology

Answer: C · The adrenal glands are suppressed and may not make enough cortisol

Why this is the answer

Steroid therapy suppresses the hypothalamic–pituitary–adrenal axis. Sudden withdrawal can cause an adrenal crisis, so doses are reduced gradually.

Answer

C · The adrenal glands are suppressed and may not make enough cortisol
2.Pharmacology

Answer: C · Thyroid dysfunction and pulmonary fibrosis

Why this is the answer

Amiodarone is iodine-rich and accumulates in tissues, causing hypo- or hyperthyroidism, pulmonary fibrosis, liver injury, photosensitivity and corneal deposits. The other pairs fit aminoglycosides, phenytoin and ACE inhibitors.

Answer

C · Thyroid dysfunction and pulmonary fibrosis
3.Dosage calculations

Answer: D = 80 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

20·12 = 240 mg

Per dose

D = 240/3 = 80 mg

Answer

D = 80 mg
4.Dosage calculations

Answer: V = 0.22 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.

Same units

1 mg/mL = 1000 mcg/mL

Volume

V = 220/1000 = 0.22 mL

Answer

V = 0.22 mL
5.Clinical pharmacokinetics

Answer: Vd = 23.33 L

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 = 210/9 = 23.33 L

Answer

Vd = 23.33 L
6.Clinical pharmacokinetics

Answer: τ = 9.8 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 = 0.693/3.5 = 0.198 h⁻¹

Dosing interval

τ = ln(7/1)/0.198 = 1.946/0.198 = 9.826 ≈ 9.8 h

Answer

τ = 9.8 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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