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 blood test is commonly used to monitor a continuous infusion of unfractionated heparin?

  1. AActivated partial thromboplastin time (aPTT)
  2. BHaemoglobin A1c
  3. CINR (international normalised ratio, prothrombin time)
  4. DSerum potassium
2.PharmacologyHard

A nurse gives 2 mg from a 4 mg ampoule of morphine. What should happen to the remaining 2 mg?

  1. AIt is given to the next patient who needs morphine
  2. BIt is poured down the sink without telling anyone
  3. CIt is discarded in front of a witness, and both sign the record
  4. DIt is kept labelled in the patient's drawer for the next dose
3.Dosage calculationsMedium
5%50 mLm = ? mg

How many milligrams of drug m are in 50 mL of a 5% (w/v) solution?

5% w/v50 mL
m = mg
4.Dosage calculationsHard
125 mg/5 mLV = ? mL

A patient weighs 55 lb. The prescription is 1 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?

55 lb1 mg/kg125 mg/5 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
VCss​ = 12 mg/LR = ? mg/hCL = 0.03 L/kg/h75 kg

Practice exercise (made-up values, not clinical guidance). A 75 kg patient receives a theophylline infusion. The clearance is 0.03 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 12 mg/L?

75 kgCL = 0.03 L/kg/hCss = 12 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​9 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 9 h?

R = 20 mg/hCL = 2 L/ht½ = 8 ht = 9 h
C = mg/L

3Solutions

1.Pharmacology

Answer: A · Activated partial thromboplastin time (aPTT)

Why this is the answer

Unfractionated heparin is often monitored with the aPTT, and the infusion rate is adjusted according to a protocol. Low-molecular-weight heparins usually do not need routine aPTT checks.

Answer

A · Activated partial thromboplastin time (aPTT)
2.Pharmacology

Answer: C · It is discarded in front of a witness, and both sign the record

Why this is the answer

Unused controlled drugs are wasted in front of a witness according to policy, and both staff document the waste. This prevents diversion and keeps the stock count accurate.

Answer

C · It is discarded in front of a witness, and both sign the record
3.Dosage calculations

Answer: m = 2500 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

5% = 5 g/100 mL = 50 mg/mL

Mass of drug

m = 50·50 = 2500 mg

Answer

m = 2500 mg
4.Dosage calculations

Answer: V = 1 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

55/2.2 = 25 kg

Dose

1·25 = 25 mg

Volume

V = (25/125)·5 = 1 mL

Answer

V = 1 mL
5.Clinical pharmacokinetics

Answer: R = 27 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 = 0.03·75 = 2.25 L/h

Infusion rate

R = 2.25·12 = 27 mg/h

Answer

R = 27 mg/h
6.Clinical pharmacokinetics

Answer: C = 5.415 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−0.08664·9) = 5.415 mg/L

Answer

C = 5.415 mg/L

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