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 glyceryl trinitrate (GTN) relieve angina?

  1. AIt dissolves the clot in the coronary artery
  2. BIt releases nitric oxide, mainly dilating veins and reducing cardiac workload
  3. CIt blocks beta-1 receptors and slows the heart
  4. DIt increases the force of contraction so the heart pumps blood more efficiently
2.PharmacologyHard

A drug with a short half-life is given at a dosing interval much longer than its half-life. What is the result at steady state?

  1. AVery stable levels with almost no fluctuation
  2. BLarge swings between peak and trough levels
  3. CSteady state is never reached
  4. DGradual accumulation to toxic levels
3.Dosage calculationsMedium
30 kg30 mg/kg/day2× per dayD = ? mg

A patient weighs 30 kg. The prescribed dose is 30 mg/kg per day, given in 2 equal doses. How many milligrams D are given per dose?

30 kg30 mg/kg/day2× per day
D = mg
4.Dosage calculationsHard
00.10.20.30.40.50.60.70.80.911 mL

The prescription is 1440 micrograms. The ampoule contains 2 mg/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?

1440 mcg2 mg/mL
V = mL
5.Clinical pharmacokineticsMedium
Vd = 0.7 L/kgC = 20 mg/LLD = ? mg90 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). A 90 kg patient is to be loaded intravenously with phenytoin sodium to reach a phenytoin level of 20 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.

90 kgVd = 0.7 L/kgC = 20 mg/LS = 0.92
LD = mg
6.Clinical pharmacokineticsHard
0.1110t (h)C (mg/L)Cpeak​ = 7τ = ?Ctrough​ = 0.5

Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 3.5 h. The target is a peak of 7 mg/L and a trough of 0.5 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.

t½ = 3.5 hCpeak = 7 mg/LCtrough = 0.5 mg/L
τ = h

3Solutions

1.Pharmacology

Answer: B · It releases nitric oxide, mainly dilating veins and reducing cardiac workload

Why this is the answer

GTN is converted to nitric oxide, which relaxes vascular smooth muscle through cGMP. Venodilation lowers preload, so the heart works less and needs less oxygen.

Answer

B · It releases nitric oxide, mainly dilating veins and reducing cardiac workload
2.Pharmacology

Answer: B · Large swings between peak and trough levels

Why this is the answer

When the interval is long compared with the half-life, most of each dose is eliminated before the next one. Levels rise high after each dose and fall low before the next, so there may be toxic peaks or ineffective troughs.

Answer

B · Large swings between peak and trough levels
3.Dosage calculations

Answer: D = 450 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.

Total per day

30·30 = 900 mg

Per dose

D = 900/2 = 450 mg

Answer

D = 450 mg
4.Dosage calculations

Answer: V = 0.72 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.

Same units

2 mg/mL = 2000 mcg/mL

Volume

V = 1440/2000 = 0.72 mL

Answer

V = 0.72 mL
5.Clinical pharmacokinetics

Answer: LD = 1370 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 = 0.7·90 = 63 L

Amount of active drug needed

Vd·C = 63·20 = 1260 mg

Loading dose

LD = 1260/(0.92·1) = 1370 mg

Answer

LD = 1370 mg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: τ = 13.3 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 = 0.693/3.5 = 0.198 h⁻¹

Dosing interval

τ = ln(7/0.5)/0.198 = 2.639/0.198 = 13.33 ≈ 13.3 h

Answer

τ = 13.3 h

In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.

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