Pharmacology: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

Which blood test is commonly used to monitor a continuous infusion of unfractionated heparin?

  1. AHaemoglobin A1c
  2. BINR (international normalised ratio, prothrombin time)
  3. CSerum potassium
  4. DActivated partial thromboplastin time (aPTT)
2.PharmacologyHard

Why must several weeks pass before starting an MAO inhibitor after stopping fluoxetine?

  1. AFluoxetine must first be converted into an MAO inhibitor
  2. BFluoxetine and its active metabolite have very long half-lives
  3. CFluoxetine permanently and irreversibly blocks MAO enzymes
  4. DMAO inhibitors are absorbed only after fluoxetine is gone
3.Dosage calculationsMedium
024681010 mL

A child weighing 16 kg is prescribed 5 mg/kg of a drug. The stock solution contains 10 mg/mL. Draw up the dose: what volume V do you give?

16 kg5 mg/kg10 mg/mL
V = mL
4.Dosage calculationsHard
10 mg/1 mLV = ? mL

A patient weighs 28.6 lb. The prescription is 15 mg/kg as a single dose, and the stock contains 10 mg in 1 mL. Using 1 kg = 2.2 lb, what volume V do you give?

28.6 lb15 mg/kg10 mg/1 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
10100t (h)C (mg/L)2 h30 mg/L14 h15 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 30 mg/L at 2 h and 15 mg/L at 14 h. Assume first-order elimination. Find the half-life t½.

C1 = 30 mg/L (2 h)C2 = 15 mg/L (14 h)
t½ = h
6.Clinical pharmacokineticsHard
VCss,avg​ = 10 mg/LD = ? mg · q8hCL = 0.03 L/kg/h63 kg · S = 1 · F = 1

Practice exercise (made-up values, not clinical guidance). A 63 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.03 L/kg/h. What dose D in mg gives an average steady-state level of 10 mg/L?

CL = 0.03 L/kg/hτ = 8 hS = 1F = 163 kgCss,avg = 10 mg/L
D = mg

3Solutions

1.Pharmacology

Answer: D · Activated partial thromboplastin time (aPTT)

Why this is the answer

Unfractionated heparin is often monitored with the aPTT, and the infusion rate is adjusted according to a protocol. Low-molecular-weight heparins usually do not need routine aPTT checks.

Answer

D · Activated partial thromboplastin time (aPTT)
2.Pharmacology

Answer: B · Fluoxetine and its active metabolite have very long half-lives

Why this is the answer

Norfluoxetine, the active metabolite of fluoxetine, stays in the body for weeks. A long washout is needed before an MAOI to avoid serotonin syndrome.

Answer

B · Fluoxetine and its active metabolite have very long half-lives
3.Dosage calculations

Answer: V = 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 = 5·16 = 80 mg

Volume

V = 80/10 = 8 mL

On the syringe

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

Answer

V = 8 mL
4.Dosage calculations

Answer: V = 19.5 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.

Weight in kg

28.6/2.2 = 13 kg

Dose

15·13 = 195 mg

Volume

V = (195/10)·1 = 19.5 mL

Answer

V = 19.5 mL
5.Clinical pharmacokinetics

Answer: t½ = 12 h

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.

Elimination rate constant

k = ln(30/15)/12 = 0.6931/12 = 0.05776 h⁻¹

Half-life

t½ = 0.693/0.05776 = 12 h

Answer

t½ = 12 h
6.Clinical pharmacokinetics

Answer: D = 151.2 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.

Clearance

CL = 0.03·63 = 1.89 L/h

Dose

D = Css,avg·CL·τ/(S·F) = 10·1.89·8/(1·1) = 151.2 mg

Answer

D = 151.2 mg

ExamProblems · examproblems.com/courses/pharmacology