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 healthy woman at 28 weeks of pregnancy has a mildly lower hemoglobin than before pregnancy, with a normal MCV and ferritin. What is the most likely explanation?
AThe fetus is destroying the mother's red cells in the circulation
BPlasma volume rises more than red cell mass, diluting the blood
CIron is being lost through the kidneys during the pregnancy
DPregnancy hormones stop the bone marrow from making red cells
2.Lab values & electrolytesHard
A patient on warfarin with a stable INR of 2.4 starts a course of an antibiotic such as ciprofloxacin. A week later the INR is 4.6. What is the most likely explanation?
AThe antibiotic has cured the patient’s need for anticoagulation
BThe patient has eaten more vitamin K–rich food
CA drug interaction that increases the effect of warfarin
DA laboratory error is the only possible explanation
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 6 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 6 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 35% oxygen shows PaO₂ 66 mmHg (8.8 kPa) and PaCO₂ 35 mmHg (4.7 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 25 mg/L at 1 h and 12.5 mg/L at 9 h. Assume first-order elimination. Find the half-life t½.
C1 = 25 mg/L (1 h)C2 = 12.5 mg/L (9 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 59 kg patient takes oral theophylline (F = 1) every 8 h. The clearance is 0.04 L/kg/h. What dose D in mg gives an average steady-state level of 10 mg/L?
What is the main benefit of pulmonary rehabilitation for patients with COPD?
AReversal of the lung damage caused by smoking
BBetter exercise capacity and quality of life
CNo further need for inhalers
DA cure for the disease within months
8.Clinical nursingHard
A patient faints during cannulation, with a slow pulse and pallor, and recovers quickly when laid flat. There is no rash, wheeze or swelling. What is most likely?
AAnaphylaxis
BVasovagal syncope
CA stroke
DA hypoglycaemic coma
3Solutions
1.Lab values & electrolytes
Answer: B · Plasma volume rises more than red cell mass, diluting the blood
Why this is the answer
In pregnancy, plasma volume increases by more than red cell mass, so the hemoglobin falls slightly (physiologic or dilutional anemia). Iron deficiency is still common in pregnancy and is checked with the MCV and ferritin.Checked against:Anemia · StatPearls (NCBI Bookshelf) (2026)
Answer
B · Plasma volume rises more than red cell mass, diluting the blood
2.Lab values & electrolytes
Answer: C · A drug interaction that increases the effect of warfarin
Why this is the answer
Several antibiotics increase warfarin’s effect by reducing its breakdown or by changing gut bacteria that make vitamin K. The INR is monitored more often when such drugs are started. More vitamin K in the diet would lower the INR.Checked against:Taking warfarin · MedlinePlus (NIH) (2025)
Answer
C · A drug interaction that increases the effect of warfarin
3.Arterial blood gas interpretation
Answer: PaCO₂ = 17 mmHg
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.
Winter’s formula
PaCO₂ = 1.5·6 + 8 = 17 mmHg
range: 15–19 mmHg
Answer
PaCO₂ = 17 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 139.8 mmHg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
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(25/12.5)/8 = 0.6931/8 = 0.08664 h⁻¹
Half-life
t½ = 0.693/0.08664 = 8 h
Answer
t½ = 8 h
6.Clinical pharmacokinetics
Answer: D = 188.8 mg
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.
Clearance
CL = 0.04·59 = 2.36 L/h
Dose
D = Css,avg·CL·τ/(S·F) = 10·2.36·8/(1·1) = 188.8 mg
Answer
D = 188.8 mg
7.Clinical nursing
Answer: B · Better exercise capacity and quality of life
Why this is the answer
Pulmonary rehabilitation improves breathlessness, exercise tolerance and well-being, and reduces hospital admissions. It does not reverse the underlying lung damage.
Answer
B · Better exercise capacity and quality of life
8.Clinical nursing
Answer: B · Vasovagal syncope
Why this is the answer
Vasovagal syncope is a reflex drop in heart rate and blood pressure triggered by pain, fear or needles. Anaphylaxis would usually include skin, airway or breathing features and a fast pulse.