Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
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
Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
Check the result: levels fall by half every half-life, and steady state takes about five half-lives.
2Practice problems
1.Lab values & electrolytesMedium
A patient with joint aches has a positive antinuclear antibody (ANA) test. Which statement is correct?
AA positive ANA rules out all other autoimmune diseases
BA positive ANA proves the patient has rheumatoid arthritis
CA positive ANA means the patient has a viral infection
DA positive ANA supports lupus but does not diagnose it alone
2.Lab values & electrolytesHard
A patient has fatigue, weight gain and constipation. Results: TSH 24 mIU/L, free T4 low, total cholesterol 285 mg/dL (7.4 mmol/L), sodium 131 mEq/L. What is the best interpretation?
AHyperthyroidism, with suppressed thyroid hormone production
BPrimary hypothyroidism, which can raise lipids and lower sodium
CFamilial hypercholesterolemia, unrelated to the thyroid results
DSecondary hypothyroidism from failure of the pituitary gland
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 67 mmHg (8.9 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a respiratory acidosis: pH 7.25, PaCO₂ 73 mmHg (9.7 kPa), HCO₃⁻ 31 mmol/L. Use the rules ‘HCO₃⁻ rises 1 mmol/L per 10 mmHg (acute) or 3.5 mmol/L per 10 mmHg (chronic)’ to classify it. Click your answer.
pH 7.25PaCO₂ 73 mmHg (9.7 kPa)HCO₃⁻ 31 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 10 mg/L at 2 h and 1.6 mg/L at 10 h. Assume first-order elimination. Find the half-life t½.
C1 = 10 mg/L (2 h)C2 = 1.6 mg/L (10 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). An 81 kg patient takes oral aminophylline (S = 0.8, F = 1) every 8 h. The theophylline clearance is 0.035 L/kg/h. Each dose is 300 mg of aminophylline. Find the average steady-state level Css,avg.
Which is a recognised risk factor for postoperative nausea and vomiting?
ABeing male
BUse of opioids after surgery
CA short operation under local anaesthetic
DBeing a smoker
8.Clinical nursingHard
A patient's urine output has been very low for 12 hours and creatinine has risen sharply. Which drug on the chart is most likely to need review?
AParacetamol at usual doses
BAn emollient cream
CA vitamin supplement
DA non-steroidal anti-inflammatory drug such as ibuprofen
3Solutions
1.Lab values & electrolytes
Answer: D · A positive ANA supports lupus but does not diagnose it alone
Why this is the answer
Almost all patients with SLE have a positive ANA, so a negative test makes lupus unlikely. However, many healthy people and patients with other conditions also have a positive ANA, so diagnosis needs the clinical picture and more specific antibodies such as anti-dsDNA.Checked against:ANA (Antinuclear Antibody) Test · MedlinePlus (NIH) (2025)
Answer
D · A positive ANA supports lupus but does not diagnose it alone
2.Lab values & electrolytes
Answer: B · Primary hypothyroidism, which can raise lipids and lower sodium
B · Primary hypothyroidism, which can raise lipids and lower sodium
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 26.7 mmol/L
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Change in PaCO₂
|67 − 40| = 27 mmHg → 2.7 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·2.7 = 26.7 mmol/L
Answer
HCO₃⁻ = 26.7 mmol/L
4.Arterial blood gas interpretation
Answer: acute on chronic
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Expected HCO₃⁻
acute: 24 + 1·3.3 = 27.3 mmol/L
chronic: 24 + 3.5·3.3 = 35.55 mmol/L
Compare
measured: 31 mmol/L → acute on chronic
Answer
acute on chronic
5.Clinical pharmacokinetics
Answer: t½ = 3.026 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(10/1.6)/8 = 1.833/8 = 0.2291 h⁻¹
Half-life
t½ = 0.693/0.2291 = 3.026 h
Answer
t½ = 3.026 h
6.Clinical pharmacokinetics
Answer: Css,avg = 10.58 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.
Known risk factors include opioids, female sex, being a non-smoker, previous PONV or motion sickness and longer general anaesthetics. High-risk patients may be given preventive antiemetics.
Answer
B · Use of opioids after surgery
8.Clinical nursing
Answer: D · A non-steroidal anti-inflammatory drug such as ibuprofen
Why this is the answer
In acute kidney injury, nephrotoxic drugs (such as NSAIDs and aminoglycosides) and drugs that reduce renal perfusion (such as ACE inhibitors and diuretics) are reviewed and often stopped.
Answer
D · A non-steroidal anti-inflammatory drug such as ibuprofen