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

What makes an overdose of tricyclic antidepressants such as amitriptyline especially dangerous?

  1. ACardiac arrhythmias and seizures
  2. BSudden high blood glucose
  3. CKidney stones
  4. DSevere liver failure within hours
2.PharmacologyHard

A malnourished patient with low serum albumin shows nystagmus and unsteady gait. Her total phenytoin level is within range. What is the most likely explanation?

  1. AThe total level is high, but the laboratory is wrong
  2. BLow albumin speeds up phenytoin metabolism
  3. CLow albumin makes phenytoin bind more tightly, so the drug is less active
  4. DMore phenytoin is unbound when albumin is low, so free levels rise
3.Dosage calculationsMedium
1%100 mLm = ? mg

How many milligrams of drug m are in 100 mL of a 1% (w/v) solution?

1% w/v100 mL
m = mg
4.Dosage calculationsHard
25k units250 mL? mL/h1500 units/h

An infusion contains 25000 units of a drug in 250 mL. The prescription is 1500 units per hour. At what rate R do you set the pump, in mL/h?

25000 units/250 mL1500 units/h
R = mL/h
5.Clinical pharmacokineticsMedium
Vd = 0.7 L/kgC = 15 mg/LLD = ? mg51 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 51 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 15 mg/L. The volume of distribution is 0.7 L/kg and the salt factor is S = 0.92. Find the loading dose LD of phenytoin sodium in mg.

51 kgVd = 0.7 L/kgC = 15 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
Vd = 0.3 L/kg1 → 7 mg/LD = ? mg65 kg

Practice exercise (made-up values, not clinical guidance). A 65 kg patient on regular tobramycin doses is at steady state. The target is a peak of 7 mg/L and a trough of 1 mg/L, and the volume of distribution is 0.3 L/kg. Assuming rapid IV doses, what dose D raises the level from the trough to the peak?

65 kgVd = 0.3 L/kgCpeak = 7 mg/LCtrough = 1 mg/L
D = mg

3Solutions

1.Pharmacology

Answer: A · Cardiac arrhythmias and seizures

Why this is the answer

Tricyclics block cardiac sodium channels (widening the QRS complex) and have anticholinergic effects. In overdose they cause arrhythmias, hypotension, seizures and coma.

Answer

A · Cardiac arrhythmias and seizures
2.Pharmacology

Answer: D · More phenytoin is unbound when albumin is low, so free levels rise

Why this is the answer

Phenytoin is highly protein bound. When albumin is low, the free fraction rises, so toxicity can occur with a "normal" total level; a free phenytoin level or corrected level is more useful.

Answer

D · More phenytoin is unbound when albumin is low, so free levels rise
3.Dosage calculations

Answer: m = 1000 mg

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% = 1 g/100 mL = 10 mg/mL

Mass of drug

m = 10·100 = 1000 mg

Answer

m = 1000 mg
4.Dosage calculations

Answer: R = 15 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

25000/250 = 100 units/mL

Pump rate

R = 1500/100 = 15 mL/h

Answer

R = 15 mL/h
5.Clinical pharmacokinetics

Answer: LD = 582.1 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 = 0.7·51 = 35.7 L

Amount of active drug needed

Vd·C = 35.7·15 = 535.5 mg

Loading dose

LD = 535.5/(0.92·1) = 582.1 mg

Answer

LD = 582.1 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: D = 117 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 = 0.3·65 = 19.5 L

Dose

D = 19.5·(7 − 1) = 117 mg

Answer

D = 117 mg

In practice the dose is rounded to a convenient amount.

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