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

A patient with insulin-treated diabetes is found unconscious with very low blood glucose. There is no IV access. Which drug can a trained helper give?

  1. AOral glucose gel poured into the mouth
  2. BGlucagon by injection or nasal spray
  3. CIntramuscular adrenaline
  4. DRapid-acting insulin under the skin
2.PharmacologyHard

A patient who takes clonidine stopped it suddenly two days ago. He is now agitated, has a headache and his blood pressure is very high. What is the most likely explanation?

  1. AAn allergic reaction to clonidine
  2. BClonidine toxicity due to slow excretion by the kidneys
  3. CNormal variation unrelated to the medication
  4. DRebound hypertension from abrupt clonidine withdrawal
3.Dosage calculationsMedium
18 kg2 mg/kg/day3× per dayD = ? mg

A patient weighs 18 kg. The prescribed dose is 2 mg/kg per day, given in 3 equal doses. How many milligrams D are given per dose?

18 kg2 mg/kg/day3× per day
D = mg
4.Dosage calculationsHard
100 mg/1 mLV = ? mL

A patient weighs 39.6 lb. The prescription is 10 mg/kg as a single dose, and the stock contains 100 mg in 1 mL. Using 1 kg = 2.2 lb, what volume V do you give?

39.6 lb10 mg/kg100 mg/1 mL1 kg = 2.2 lb
V = mL
5.Clinical pharmacokineticsMedium
Vd = 7 L/kgC = 1 mcg/LLD = ? mcg66 kg · S = 1 · F = 0.7

Practice exercise (made-up values, not clinical guidance). A 66 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.

66 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
t (h)C (mg/L)Css​5 hC = ?

Practice exercise (made-up values, not clinical guidance). A vancomycin infusion runs at 100 mg/h with no loading dose. The patient's clearance is 5 L/h and the half-life is 8 h. What is the level C after 5 h?

R = 100 mg/hCL = 5 L/ht½ = 8 ht = 5 h
C = mg/L

3Solutions

1.Pharmacology

Answer: B · Glucagon by injection or nasal spray

Why this is the answer

Glucagon breaks down liver glycogen and raises blood glucose. Nothing is given by mouth to an unconscious patient because of the risk of choking and aspiration.

Answer

B · Glucagon by injection or nasal spray
2.Pharmacology

Answer: D · Rebound hypertension from abrupt clonidine withdrawal

Why this is the answer

Clonidine is a central alpha-2 agonist. Stopping it suddenly lets sympathetic activity surge, causing rebound hypertension, so it is tapered gradually.

Answer

D · Rebound hypertension from abrupt clonidine withdrawal
3.Dosage calculations

Answer: D = 12 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

2·18 = 36 mg

Per dose

D = 36/3 = 12 mg

Answer

D = 12 mg
4.Dosage calculations

Answer: V = 1.8 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

39.6/2.2 = 18 kg

Dose

10·18 = 180 mg

Volume

V = (180/100)·1 = 1.8 mL

Answer

V = 1.8 mL
5.Clinical pharmacokinetics

Answer: LD = 660 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 = 7·66 = 462 L

Amount of active drug needed

Vd·C = 462·1 = 462 mcg

Loading dose

LD = 462/(1·0.7) = 660 mcg

Answer

LD = 660 mcg

In practice the dose is rounded to a convenient amount.

6.Clinical pharmacokinetics

Answer: C = 7.032 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 = 100/5 = 20 mg/L

Rate constant

k = 0.693/8 = 0.08664 h⁻¹

Level after the given time

C = 20·(1 − e−0.08664·5) = 7.032 mg/L

Answer

C = 7.032 mg/L

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