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

Newborns given chloramphenicol can develop "grey baby syndrome". What is the main reason?

  1. ANewborns produce antibodies against the drug
  2. BThe newborn kidney excretes the drug too quickly, so higher doses are given
  3. CChloramphenicol is converted to vitamin K in neonates
  4. DImmature glucuronidation in the liver lets the drug accumulate
3.Dosage calculationsMedium
60 kg5 mg/kgD = ? mg

A patient weighs 60 kg and is prescribed 5 mg/kg as a single dose. Find the dose D in mg.

60 kg5 mg/kg
D = mg
4.Dosage calculationsHard
10%stockV₁ = ? mL

You must prepare 1000 mL of a 1% (w/v) solution from a 10% (w/v) stock solution. What volume of stock V₁ do you need?

C₁ = 10%C₂ = 1%V₂ = 1000 mL
V₁ = mL
5.Clinical pharmacokineticsMedium
Vd = 7 L/kgC = 1 mcg/LLD = ? mcg95 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 95 kg patient is to be loaded with digoxin tablets to reach a level of 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.

95 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​t = ?75%

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion is started without a loading dose. The half-life in this patient is 6 h. How long t does it take to reach 75% of the steady-state level?

t½ = 6 h75% Css
t = h

3Solutions

1.Pharmacology

Answer: C · 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

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

Answer: D · Immature glucuronidation in the liver lets the drug accumulate

Why this is the answer

Neonates have low glucuronyl transferase activity, so chloramphenicol is not conjugated and builds up. High levels cause circulatory collapse with grey skin. It shows how organ immaturity changes pharmacokinetics.

Answer

D · Immature glucuronidation in the liver lets the drug accumulate
3.Dosage calculations

Answer: D = 300 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.

Dose

5·60 = 300 mg

Answer

D = 300 mg
4.Dosage calculations

Answer: V₁ = 100 mL

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.

Volume of stock

V₁ = 1·1000/10 = 100 mL

Answer

V₁ = 100 mL
5.Clinical pharmacokinetics

Answer: LD = 950 mcg

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 = 7·95 = 665 L

Amount of active drug needed

Vd·C = 665·1 = 665 mcg

Loading dose

LD = 665/(1·0.7) = 950 mcg

Answer

LD = 950 mcg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: t = 12 h

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.

Rate constant

k = 0.693/6 = 0.1155 h⁻¹

Time to the given fraction of steady state

t = −ln(1 − 0.75)/0.1155 = 1.386/0.1155 = 12 h

Answer

t = 12 h

That is 2 half-lives.

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