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

Which factor increases the risk of lithium toxicity?

  1. ADrinking extra water
  2. BTaking the dose with food
  3. CA high-salt diet
  4. DDehydration
2.PharmacologyHard

During an alteplase infusion for acute ischaemic stroke, the patient develops a sudden severe headache and her level of consciousness falls. What is the nurse's first action?

  1. AGive a bolus of heparin to prevent a new clot
  2. BSlow the infusion to half the rate and recheck in 30 minutes
  3. CStop the alteplase infusion and notify the provider urgently
  4. DGive paracetamol for the headache and continue the infusion
3.Dosage calculationsMedium
0.5%20 mLm = ? mg

How many milligrams of drug m are in 20 mL of a 0.5% (w/v) solution?

0.5% w/v20 mL
m = mg
4.Dosage calculationsHard
00.10.20.30.40.50.60.70.80.911 mL

The prescription is 630 units of a drug that comes as 1000 units/mL. Draw up the dose in the 1 mL syringe: what volume V do you give?

630 units1000 units/mL
V = mL
5.Clinical pharmacokineticsMedium
VCss​ = 10 mg/LR = ? mg/h · S = 0.8CL = 0.05 L/kg/h54 kg

Practice exercise (made-up values, not clinical guidance). A 54 kg patient is to receive an aminophylline infusion (salt factor S = 0.8: 80% of aminophylline is theophylline). The theophylline clearance is 0.05 L/kg/h. What aminophylline infusion rate R in mg/h gives a steady-state theophylline level of 10 mg/L?

54 kgCL = 0.05 L/kg/hCss = 10 mg/LS = 0.8
R = mg/h
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​22 hC = ?

Practice exercise (made-up values, not clinical guidance). A theophylline infusion runs at 20 mg/h with no loading dose. The patient's clearance is 2 L/h and the half-life is 8 h. What is the level C after 22 h?

R = 20 mg/hCL = 2 L/ht½ = 8 ht = 22 h
C = mg/L

3Solutions

1.Pharmacology

Answer: D · Dehydration

Why this is the answer

When the body is short of water or sodium, the kidney reabsorbs more lithium, so its level rises. Thiazides, ACE inhibitors and NSAIDs can have the same effect.

Answer

D · Dehydration
2.Pharmacology

Answer: C · Stop the alteplase infusion and notify the provider urgently

Why this is the answer

A sudden headache and neurological decline during thrombolysis suggest intracranial haemorrhage. The infusion is stopped at once and the provider is called for urgent assessment and imaging.

Answer

C · Stop the alteplase infusion and notify the provider urgently
3.Dosage calculations

Answer: m = 100 mg

Solution strengths: a percentage strength is grams per 100 mL, so 1% = 1 g/100 mL = 10 mg/mL. A ratio strength 1:1000 means 1 g in 1000 mL = 1 mg/mL; 1:10 000 is 0.1 mg/mL. For a dilution the amount of drug stays the same: C₁·V₁ = C₂·V₂.

Concentration

0.5% = 0.5 g/100 mL = 5 mg/mL

Mass of drug

m = 5·20 = 100 mg

Answer

m = 100 mg
4.Dosage calculations

Answer: V = 0.63 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 = 630/1000 = 0.63 mL

On the syringe

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

Answer

V = 0.63 mL
5.Clinical pharmacokinetics

Answer: R = 33.75 mg/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.

Clearance

CL = 0.05·54 = 2.7 L/h

Infusion rate

R = CL·Css/S = 2.7·10/0.8 = 33.75 mg/h

Answer

R = 33.75 mg/h
6.Clinical pharmacokinetics

Answer: C = 8.513 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.

Steady-state level

Css = 20/2 = 10 mg/L

Rate constant

k = 0.693/8 = 0.08664 h⁻¹

Level after the given time

C = 10·(1 − e−0.08664·22) = 8.513 mg/L

Answer

C = 8.513 mg/L

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