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

Aciclovir is mainly used to treat infections caused by:

  1. ABacteria that produce beta-lactamase
  2. BHIV
  3. CInfluenza A virus
  4. DHerpes viruses such as herpes simplex and varicella-zoster
2.PharmacologyHard

Two days after starting chemotherapy for a leukaemia with a high white cell count, a patient has reduced urine output. Which lab pattern suggests tumour lysis syndrome?

  1. AHigh potassium, high phosphate, high uric acid and low calcium
  2. BLow potassium, low phosphate, low uric acid and high calcium
  3. CNormal electrolytes with a high haemoglobin
  4. DHigh sodium, high glucose and low potassium
3.Dosage calculationsMedium
024681010 mL

A child weighing 34 kg is prescribed 5 mg/kg of a drug. The stock solution contains 25 mg/mL. Draw up the dose: what volume V do you give?

34 kg5 mg/kg25 mg/mL
V = mL
4.Dosage calculationsHard
20k units1000 mL? mL/h1000 units/h

An infusion contains 20000 units of a drug in 1000 mL. The prescription is 1000 units per hour. At what rate R do you set the pump, in mL/h?

20000 units/1000 mL1000 units/h
R = mL/h
5.Clinical pharmacokineticsMedium
110100t (h)C (mg/L)C₀ = 12 mg/Lt½ = 2.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 270 mg of gentamicin. The level back-extrapolated to time zero is 12 mg/L and the half-life is 2.5 h. Find the clearance CL in L/h.

D = 270 mgC₀ = 12 mg/Lt½ = 2.5 h
CL = L/h
6.Clinical pharmacokineticsHard
Vd = 7 L/kg0.6 → 1 mcg/LLD = ? mcg69 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 69 kg patient has a digoxin level of 0.6 mcg/L; the target is 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have F = 0.7. What extra loading dose LD raises the level to the target?

69 kgVd = 7 L/kgC = 0.6 → 1 mcg/LF = 0.7
LD = mcg

3Solutions

1.Pharmacology

Answer: D · Herpes viruses such as herpes simplex and varicella-zoster

Why this is the answer

Aciclovir is phosphorylated by viral thymidine kinase in infected cells and then blocks viral DNA polymerase. It is used for cold sores, genital herpes, chickenpox and shingles.

Answer

D · Herpes viruses such as herpes simplex and varicella-zoster
2.Pharmacology

Answer: A · High potassium, high phosphate, high uric acid and low calcium

Why this is the answer

Rapid tumour cell death releases potassium, phosphate and nucleic acids (which become uric acid). Phosphate binds calcium, lowering it; prevention includes IV fluids and drugs that lower uric acid.

Answer

A · High potassium, high phosphate, high uric acid and low calcium
3.Dosage calculations

Answer: V = 6.8 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.

Dose from body weight

D = 5·34 = 170 mg

Volume

V = 170/25 = 6.8 mL

On the syringe

On this 10 mL syringe every small line is 0.2 mL.

Answer

V = 6.8 mL
4.Dosage calculations

Answer: R = 50 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

20000/1000 = 20 units/mL

Pump rate

R = 1000/20 = 50 mL/h

Answer

R = 50 mL/h
5.Clinical pharmacokinetics

Answer: CL = 6.238 L/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.

Volume of distribution

Vd = 270/12 = 22.5 L

Rate constant

k = 0.693/2.5 = 0.2773 h⁻¹

Clearance

CL = 0.2773·22.5 = 6.238 L/h

Answer

CL = 6.238 L/h
6.Clinical pharmacokinetics

Answer: LD = 276 mcg

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 = 7·69 = 483 L

Rise needed

1 − 0.6 = 0.4 mcg/L

Loading dose

LD = 483·0.4/(1·0.7) = 276 mcg

Answer

LD = 276 mcg

In practice the dose is rounded to a convenient amount.

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