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

100 mL of an antibiotic is to be given over 30 minutes using a giving set of 20 drops/mL. What is the drip rate?

  1. AAbout 67 drops per minute
  2. BAbout 20 drops per minute
  3. CAbout 100 drops per minute
  4. DAbout 33 drops per minute
2.PharmacologyHard

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

  1. AMAO inhibitors are absorbed only after fluoxetine is gone
  2. BFluoxetine and its active metabolite have very long half-lives
  3. CFluoxetine permanently and irreversibly blocks MAO enzymes
  4. DFluoxetine must first be converted into an MAO inhibitor
3.Dosage calculationsMedium
22 kg15 mg/kgD = ? mg

A patient weighs 22 kg and is prescribed 15 mg/kg as a single dose. Find the dose D in mg.

22 kg15 mg/kg
D = mg
4.Dosage calculationsHard
25k units500 mL? mL/h1000 units/h

An infusion contains 25000 units of a drug in 500 mL. The prescription is 1000 units per hour. At what rate R do you set the pump, in mL/h?

25000 units/500 mL1000 units/h
R = mL/h
5.Clinical pharmacokineticsMedium
10100t (h)C (mg/L)1 h40 mg/L11 h12.6 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 40 mg/L at 1 h and 12.6 mg/L at 11 h. Assume first-order elimination. Find the half-life t½.

C1 = 40 mg/L (1 h)C2 = 12.6 mg/L (11 h)
t½ = h
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​t = ?90%

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion is started without a loading dose. The half-life in this patient is 6 h. How long t does it take to reach 90% of the steady-state level?

t½ = 6 h90% Css
t = h

3Solutions

1.Pharmacology

Answer: A · About 67 drops per minute

Why this is the answer

100 mL × 20 drops/mL ÷ 30 min = 66.7, so about 67 drops per minute. Gravity infusions are counted for a full minute and checked regularly.

Answer

A · About 67 drops per minute
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: D = 330 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.

Dose

15·22 = 330 mg

Answer

D = 330 mg
4.Dosage calculations

Answer: R = 20 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

25000/500 = 50 units/mL

Pump rate

R = 1000/50 = 20 mL/h

Answer

R = 20 mL/h
5.Clinical pharmacokinetics

Answer: t½ = 6 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(40/12.6)/10 = 1.155/10 = 0.1155 h⁻¹

Half-life

t½ = 0.693/0.1155 = 6 h

Answer

t½ = 6 h
6.Clinical pharmacokinetics

Answer: t = 19.93 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 = 0.693/6 = 0.1155 h⁻¹

Time to the given fraction of steady state

t = −ln(1 − 0.9)/0.1155 = 2.303/0.1155 = 19.93 h

Answer

t = 19.93 h

That is 3.322 half-lives.

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