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

A patient is starting olanzapine. Which monitoring is most important over the following months?

  1. APeak expiratory flow
  2. BWeight, blood glucose and blood lipids
  3. CHearing tests and vision tests every month
  4. DINR and platelet count
2.PharmacologyHard

A woman has received a high-dose oxytocin infusion with large volumes of hypotonic fluid for many hours. She now has a headache, confusion and a low serum sodium. What is the most likely cause?

  1. AWater intoxication from the antidiuretic effect of oxytocin
  2. BAn allergic reaction to oxytocin causing brain swelling
  3. CExcessive salt in the infusion fluid
  4. DNormal tiredness after a long labour
3.Dosage calculationsMedium
1%w/vV = ? mL

A dose of 10 mg is needed from a 1% (w/v) solution. What volume V contains this dose?

10 mg1% w/v
V = mL
4.Dosage calculationsHard
60 kg163 cmBSA = √(h·w/3600)25 mg/m²D = ? mg

A patient is 163 cm tall and weighs 60 kg. The prescription is 25 mg/m². Using the Mosteller formula BSA = √(height (cm) × weight (kg)/3600), find the dose D in mg.

163 cm60 kg25 mg/m²
D = mg
5.Clinical pharmacokineticsMedium
10100t (h)C (mg/L)1 h35 mg/L13 h13.9 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 35 mg/L at 1 h and 13.9 mg/L at 13 h. Assume first-order elimination. Find the half-life t½.

C1 = 35 mg/L (1 h)C2 = 13.9 mg/L (13 h)
t½ = h
6.Clinical pharmacokineticsHard
Vd = 0.7 L/kg7.5 → 15 mg/LLD = ? mg85 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). An 85 kg patient has a phenytoin level of 7.5 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?

85 kgVd = 0.7 L/kgC = 7.5 → 15 mg/LS = 0.92
LD = mg

3Solutions

1.Pharmacology

Answer: B · Weight, blood glucose and blood lipids

Why this is the answer

Olanzapine and several other atypical antipsychotics cause weight gain, insulin resistance and dyslipidaemia. Weight, waist size, glucose and lipids are checked regularly.

Answer

B · Weight, blood glucose and blood lipids
2.Pharmacology

Answer: A · Water intoxication from the antidiuretic effect of oxytocin

Why this is the answer

Oxytocin has antidiuretic activity at high doses. With large volumes of fluid this can cause water retention and dilutional hyponatraemia, so fluid balance and sodium are monitored.

Answer

A · Water intoxication from the antidiuretic effect of oxytocin
3.Dosage calculations

Answer: V = 1 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% = 10 mg/mL

Volume

V = 10/10 = 1 mL

Answer

V = 1 mL
4.Dosage calculations

Answer: D = 41.21 mg

Body-surface-area dosing: many cytotoxic and some paediatric doses are prescribed per square metre of body surface area: dose (mg) = dose per m² × BSA (m²). The Mosteller formula estimates BSA = √(height (cm) × weight (kg)/3600).

Body surface area

BSA = √(163·60/3600) = √2.717 = 1.648 m²

Dose

D = 25·1.648 = 41.21 mg

Answer

D = 41.21 mg
5.Clinical pharmacokinetics

Answer: t½ = 9.007 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(35/13.9)/12 = 0.9235/12 = 0.07695 h⁻¹

Half-life

t½ = 0.693/0.07695 = 9.007 h

Answer

t½ = 9.007 h
6.Clinical pharmacokinetics

Answer: LD = 485.1 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 = 0.7·85 = 59.5 L

Rise needed

15 − 7.5 = 7.5 mg/L

Loading dose

LD = 59.5·7.5/(0.92·1) = 485.1 mg

Answer

LD = 485.1 mg

In practice the dose is rounded to a convenient amount.

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