Pharmacology: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 many micrograms are in 0.5 mg?

  1. A50 micrograms
  2. B500 micrograms
  3. C5000 micrograms
  4. D5 micrograms
2.PharmacologyHard

A patient on simvastatin is prescribed gemfibrozil. What is the main concern?

  1. ALoss of all lipid-lowering effect
  2. BSevere hypoglycaemia
  3. CA much higher risk of myopathy and rhabdomyolysis
  4. DA sharp rise in LDL cholesterol and triglycerides
3.Dosage calculationsMedium
18 kg20 mg/kg/day4× per dayD = ? mg

A patient weighs 18 kg. The prescribed dose is 20 mg/kg per day, given in 4 equal doses. How many milligrams D are given per dose?

18 kg20 mg/kg/day4× per day
D = mg
4.Dosage calculationsHard
66 kg155 cmBSA = √(h·w/3600)60 mg/m²D = ? mg

A patient is 155 cm tall and weighs 66 kg. The prescription is 60 mg/m². Using the Mosteller formula BSA = √(height (cm) × weight (kg)/3600), find the dose D in mg.

155 cm66 kg60 mg/m²
D = mg
5.Clinical pharmacokineticsMedium
Vd = 7 L/kgC = 1 mcg/LLD = ? mcg90 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 90 kg patient is to be loaded with digoxin tablets to reach a level of 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.

90 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
0.1110t (h)C (mg/L)Cpeak​ = 6τ = ?Ctrough​ = 1

Practice exercise (made-up values, not clinical guidance). A patient's tobramycin half-life is 3 h. The target is a peak of 6 mg/L and a trough of 1 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 hCpeak = 6 mg/LCtrough = 1 mg/L
τ = h

3Solutions

1.Pharmacology

Answer: B · 500 micrograms

Why this is the answer

0.5 mg × 1000 = 500 micrograms. Writing "micrograms" in full (or "mcg" in US practice), not "µg", helps avoid confusion with "mg".

Answer

B · 500 micrograms
2.Pharmacology

Answer: C · A much higher risk of myopathy and rhabdomyolysis

Why this is the answer

Gemfibrozil reduces statin clearance and adds its own muscle toxicity. The combination greatly increases the risk of rhabdomyolysis and is generally avoided.

Answer

C · A much higher risk of myopathy and rhabdomyolysis
3.Dosage calculations

Answer: D = 90 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

20·18 = 360 mg

Per dose

D = 360/4 = 90 mg

Answer

D = 90 mg
4.Dosage calculations

Answer: D = 101.1 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 = √(155·66/3600) = √2.842 = 1.686 m²

Dose

D = 60·1.686 = 101.1 mg

Answer

D = 101.1 mg
5.Clinical pharmacokinetics

Answer: LD = 900 mcg

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 = 7·90 = 630 L

Amount of active drug needed

Vd·C = 630·1 = 630 mcg

Loading dose

LD = 630/(1·0.7) = 900 mcg

Answer

LD = 900 mcg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: τ = 7.8 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 = 0.231 h⁻¹

Dosing interval

τ = ln(6/1)/0.231 = 1.792/0.231 = 7.755 ≈ 7.8 h

Answer

τ = 7.8 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.

ExamProblems · examproblems.com/courses/pharmacology