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

How long does it take, approximately, for a drug given at regular intervals to reach a steady state?

  1. A4 to 5 half-lives
  2. B10 half-lives
  3. CIt depends only on the dose size
  4. D1 half-life
2.PharmacologyHard

A patient taking azathioprine is prescribed allopurinol for gout. What is the main danger?

  1. ASevere bone marrow suppression, because azathioprine breakdown is reduced
  2. BGout becomes worse because allopurinol is inactivated by azathioprine
  3. CSevere hypertension from the combination
  4. DLiver enzyme induction, making azathioprine ineffective
3.Dosage calculationsMedium
45 kg5 mg/kg/day4× per dayD = ? mg

A patient weighs 45 kg. The prescribed dose is 5 mg/kg per day, given in 4 equal doses. How many milligrams D are given per dose?

45 kg5 mg/kg/day4× per day
D = mg
4.Dosage calculationsHard
20k units250 mL? mL/h1000 units/h

An infusion contains 20000 units of a drug in 250 mL. The prescription is 1000 units per hour. At what rate R do you set the pump, in mL/h?

20000 units/250 mL1000 units/h
R = mL/h
5.Clinical pharmacokineticsMedium
Vd = 0.5 L/kgC = 12 mg/LLD = ? mg53 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 53 kg patient is to be loaded intravenously with aminophylline to reach a theophylline level of 12 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.

53 kgVd = 0.5 L/kgC = 12 mg/LS = 0.8
LD = mg
6.Clinical pharmacokineticsHard
Vd = 0.7 L/kg9 → 15 mg/LLD = ? mg67 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 67 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?

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

3Solutions

1.Pharmacology

Answer: A · 4 to 5 half-lives

Why this is the answer

With repeated dosing, the level approaches steady state after about 4–5 half-lives, whatever the dose. A loading dose can be used to reach it faster.

Answer

A · 4 to 5 half-lives
2.Pharmacology

Answer: A · Severe bone marrow suppression, because azathioprine breakdown is reduced

Why this is the answer

Azathioprine is converted to mercaptopurine, which is partly inactivated by xanthine oxidase. Allopurinol blocks this enzyme, so levels rise and dangerous myelosuppression can occur unless the dose is greatly reduced or the combination avoided.

Answer

A · Severe bone marrow suppression, because azathioprine breakdown is reduced
3.Dosage calculations

Answer: D = 56.25 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

5·45 = 225 mg

Per dose

D = 225/4 = 56.25 mg

Answer

D = 56.25 mg
4.Dosage calculations

Answer: R = 12.5 mL/h

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.

Concentration

20000/250 = 80 units/mL

Pump rate

R = 1000/80 = 12.5 mL/h

Answer

R = 12.5 mL/h
5.Clinical pharmacokinetics

Answer: LD = 397.5 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 = 0.5·53 = 26.5 L

Amount of active drug needed

Vd·C = 26.5·12 = 318 mg

Loading dose

LD = 318/(0.8·1) = 397.5 mg

Answer

LD = 397.5 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: LD = 305.9 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 = 0.7·67 = 46.9 L

Rise needed

15 − 9 = 6 mg/L

Loading dose

LD = 46.9·6/(0.92·1) = 305.9 mg

Answer

LD = 305.9 mg

In practice the dose is rounded to a convenient amount.

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