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

Long-term use of metformin can cause a deficiency of which vitamin?

  1. AVitamin D
  2. BVitamin B12
  3. CVitamin K
  4. DVitamin A
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. ALarge swings between peak and trough levels
  2. BGradual accumulation to toxic levels
  3. CSteady state is never reached
  4. DVery stable levels with almost no fluctuation
3.Dosage calculationsMedium
024681010 mL

A child weighing 23 kg is prescribed 20 mg/kg of a drug. The stock solution contains 50 mg/mL. Draw up the dose: what volume V do you give?

23 kg20 mg/kg50 mg/mL
V = mL
4.Dosage calculationsHard
125 mg/5 mLV = ? mL

A patient weighs 59.4 lb. The prescription is 15 mg/kg as a single dose, and the stock contains 125 mg in 5 mL. Using 1 kg = 2.2 lb, what volume V do you give?

59.4 lb15 mg/kg125 mg/5 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
110t (h)C (mg/L)C₀ = 7 mg/Lt½ = 3.5 hC₀/2

Practice exercise (made-up values, not clinical guidance). A patient receives an IV bolus of 200 mg of gentamicin. The level back-extrapolated to time zero is 7 mg/L and the half-life is 3.5 h. Find the clearance CL in L/h.

D = 200 mgC₀ = 7 mg/Lt½ = 3.5 h
CL = L/h
6.Clinical pharmacokineticsHard
VCss,avg​ = ?300 mg · q12hCL = 0.035 L/kg/h54 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 54 kg patient takes oral aminophylline (S = 0.8, F = 1) every 12 h. The theophylline clearance is 0.035 L/kg/h. Each dose is 300 mg of aminophylline. Find the average steady-state level Css,avg.

CL = 0.035 L/kg/hτ = 12 hS = 0.8F = 154 kgD = 300 mg
Css,avg = mg/L

3Solutions

1.Pharmacology

Answer: B · Vitamin B12

Why this is the answer

Metformin reduces vitamin B12 absorption in the ileum. Deficiency can cause anaemia and neuropathy, so levels may be checked in long-term users.

Answer

B · Vitamin B12
2.Pharmacology

Answer: A · 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

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

Answer: V = 9.2 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 = 20·23 = 460 mg

Volume

V = 460/50 = 9.2 mL

On the syringe

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

Answer

V = 9.2 mL
4.Dosage calculations

Answer: V = 16.2 mL

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.

Weight in kg

59.4/2.2 = 27 kg

Dose

15·27 = 405 mg

Volume

V = (405/125)·5 = 16.2 mL

Answer

V = 16.2 mL
5.Clinical pharmacokinetics

Answer: CL = 5.658 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 = 200/7 = 28.57 L

Rate constant

k = 0.693/3.5 = 0.198 h⁻¹

Clearance

CL = 0.198·28.57 = 5.658 L/h

Answer

CL = 5.658 L/h
6.Clinical pharmacokinetics

Answer: Css,avg = 10.58 mg/L

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.

Clearance

CL = 0.035·54 = 1.89 L/h

Average steady-state level

Css,avg = S·F·D/(CL·τ) = 0.8·1·300/(1.89·12) = 10.58 mg/L

Answer

Css,avg = 10.58 mg/L

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