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 does clopidogrel reduce platelet aggregation?

  1. AIt blocks the glycoprotein IIb/IIIa receptor directly
  2. BIt inhibits platelet COX-1
  3. CIt blocks the platelet P2Y12 receptor for ADP
  4. DIt activates antithrombin
2.PharmacologyHard

A patient with type 2 diabetes starts a high dose of prednisolone for a flare of COPD. Which change should the nurse expect?

  1. AHigher blood glucose, so more frequent glucose checks are needed
  2. BNo change in glucose, as steroids act only in the lungs
  3. CLower potassium only, with no effect on glucose
  4. DLower blood glucose, so the diabetes medicines should be stopped
3.Dosage calculationsMedium
1:1000ampouleV = ? mL

An ampoule of epinephrine (adrenaline) is labelled 1:1000 (1 mg/mL). The prescription is 0.5 mg IM for anaphylaxis. What volume V do you draw up?

1:10000.5 mg
V = mL
4.Dosage calculationsHard
25%stockwater = ? mL

You must prepare 1000 mL of a 5% (w/v) solution from a 25% (w/v) stock solution. How much water W must be added to the stock?

C₁ = 25%C₂ = 5%V₂ = 1000 mL
W = mL
5.Clinical pharmacokineticsMedium
110t (h)C (mg/L)1 h8 mg/L9 h1.6 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of gentamicin, the level is 8 mg/L at 1 h and 1.6 mg/L at 9 h. Assume first-order elimination. Find the half-life t½.

C1 = 8 mg/L (1 h)C2 = 1.6 mg/L (9 h)
t½ = h
6.Clinical pharmacokineticsHard
Vd = 0.7 L/kg12 → 20 mg/LLD = ? mg88 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). An 88 kg patient has a phenytoin level of 12 mg/L; the target is 20 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?

88 kgVd = 0.7 L/kgC = 12 → 20 mg/LS = 0.92
LD = mg

3Solutions

1.Pharmacology

Answer: C · It blocks the platelet P2Y12 receptor for ADP

Why this is the answer

Clopidogrel is activated in the liver and irreversibly blocks the P2Y12 ADP receptor on platelets. It is often combined with aspirin after a heart attack or stent.

Answer

C · It blocks the platelet P2Y12 receptor for ADP
2.Pharmacology

Answer: A · Higher blood glucose, so more frequent glucose checks are needed

Why this is the answer

Glucocorticoids raise blood glucose by increasing liver glucose production and insulin resistance. Glucose is monitored more often, and diabetes treatment may need to be increased during the course.

Answer

A · Higher blood glucose, so more frequent glucose checks are needed
3.Dosage calculations

Answer: V = 0.5 mL

Solution strengths: a percentage strength is grams per 100 mL, so 1% = 1 g/100 mL = 10 mg/mL. A ratio strength 1:1000 means 1 g in 1000 mL = 1 mg/mL; 1:10 000 is 0.1 mg/mL. For a dilution the amount of drug stays the same: C₁·V₁ = C₂·V₂.

Concentration

1:1000 = 1 g/1000 mL = 1 mg/mL

Volume

V = 0.5/1 = 0.5 mL

Answer

V = 0.5 mL

US labels now state mg/mL (1 mg/mL = 1:1000; 0.1 mg/mL = 1:10 000). Never give 1 mg/mL epinephrine IV push.

4.Dosage calculations

Answer: W = 800 mL

Solution strengths: a percentage strength is grams per 100 mL, so 1% = 1 g/100 mL = 10 mg/mL. A ratio strength 1:1000 means 1 g in 1000 mL = 1 mg/mL; 1:10 000 is 0.1 mg/mL. For a dilution the amount of drug stays the same: C₁·V₁ = C₂·V₂.

Volume of stock

V₁ = 5·1000/25 = 200 mL

Water to add

W = 1000 − 200 = 800 mL

Answer

W = 800 mL
5.Clinical pharmacokinetics

Answer: t½ = 3.445 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(8/1.6)/8 = 1.609/8 = 0.2012 h⁻¹

Half-life

t½ = 0.693/0.2012 = 3.445 h

Answer

t½ = 3.445 h
6.Clinical pharmacokinetics

Answer: LD = 535.7 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 = 0.7·88 = 61.6 L

Rise needed

20 − 12 = 8 mg/L

Loading dose

LD = 61.6·8/(0.92·1) = 535.7 mg

Answer

LD = 535.7 mg

In practice the dose is rounded to a convenient amount.

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