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

After instilling eye drops, why should the patient press gently on the inner corner of the eye for a minute or two?

  1. ATo stop the eye from producing tears for about an hour
  2. BTo spread the drop evenly over the eyelid
  3. CTo reduce drainage into the nose and absorption into blood
  4. DTo make the pupil dilate faster
2.PharmacologyHard

A patient taking azathioprine is prescribed allopurinol for gout. What is the main danger?

  1. AGout becomes worse because allopurinol is inactivated by azathioprine
  2. BSevere bone marrow suppression, because azathioprine breakdown is reduced
  3. CSevere hypertension from the combination
  4. DLiver enzyme induction, making azathioprine ineffective
3.Dosage calculationsMedium
12 kg20 mg/kg/day3× per dayD = ? mg

A patient weighs 12 kg. The prescribed dose is 20 mg/kg per day, given in 3 equal doses. How many milligrams D are given per dose?

12 kg20 mg/kg/day3× per day
D = mg
4.Dosage calculationsHard
00.10.20.30.40.50.60.70.80.911 mL

The prescription is 4500 units of a drug that comes as 5000 units/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?

4500 units5000 units/mL
V = mL
5.Clinical pharmacokineticsMedium
110t (h)C (mg/L)2 h6 mg/L6 h1.5 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 6 mg/L at 2 h and 1.5 mg/L at 6 h. Assume first-order elimination. Find the half-life t½.

C1 = 6 mg/L (2 h)C2 = 1.5 mg/L (6 h)
t½ = h
6.Clinical pharmacokineticsHard
VCss,avg​ = 15 mg/LD = ? mg · q8hCL = 0.04 L/kg/h69 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). A 69 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.04 L/kg/h. What dose D in mg gives an average steady-state level of 15 mg/L?

CL = 0.04 L/kg/hτ = 8 hS = 0.8F = 169 kgCss,avg = 15 mg/L
D = mg

3Solutions

1.Pharmacology

Answer: C · To reduce drainage into the nose and absorption into blood

Why this is the answer

Nasolacrimal occlusion keeps the drop in the eye longer and reduces systemic absorption through the nasal mucosa. This matters for drugs such as timolol, which can cause bradycardia or bronchospasm.

Answer

C · To reduce drainage into the nose and absorption into blood
2.Pharmacology

Answer: B · Severe bone marrow suppression, because azathioprine breakdown is reduced

Why this is the answer

Azathioprine is converted to mercaptopurine, which is partly inactivated by xanthine oxidase. Allopurinol blocks this enzyme, so levels rise and dangerous myelosuppression can occur unless the dose is greatly reduced or the combination avoided.

Answer

B · Severe bone marrow suppression, because azathioprine breakdown is reduced
3.Dosage calculations

Answer: D = 80 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

20·12 = 240 mg

Per dose

D = 240/3 = 80 mg

Answer

D = 80 mg
4.Dosage calculations

Answer: V = 0.9 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.

Volume

V = 4500/5000 = 0.9 mL

On the syringe

On this 1 mL syringe every small line is 0.01 mL.

Answer

V = 0.9 mL
5.Clinical pharmacokinetics

Answer: t½ = 2 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.

Elimination rate constant

k = ln(6/1.5)/4 = 1.386/4 = 0.3466 h⁻¹

Half-life

t½ = 0.693/0.3466 = 2 h

Answer

t½ = 2 h
6.Clinical pharmacokinetics

Answer: D = 414 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.

Clearance

CL = 0.04·69 = 2.76 L/h

Dose

D = Css,avg·CL·τ/(S·F) = 15·2.76·8/(0.8·1) = 414 mg

Answer

D = 414 mg

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