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 blood sample drawn just after a patient ate a fatty meal is reported as "lipemic." What does this mean for the results?
AHigh fat in the sample may interfere with some test results
BThe sample was contaminated by an IV fluid with dextrose
CThe results are more accurate than in a fasting sample
DThe sample shows the patient has diabetic ketoacidosis
2.Lab values & electrolytesHard
A healthy woman at 10 weeks of pregnancy has a slightly low TSH with a normal free T4 and no symptoms. What is the most likely explanation?
AThe result shows severe primary hypothyroidism
BShe has developed Graves disease in pregnancy
CShe has pituitary failure from the pregnancy
DhCG stimulates the thyroid in early pregnancy
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.21, PaCO₂ 41 mmHg (5.5 kPa) and HCO₃⁻ 16 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.21PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 16 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 32 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 32 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 4 L/h. What infusion rate R in mg/h gives a steady-state level of 20 mg/L?
CL = 4 L/hCss = 20 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 95 kg patient has a phenytoin level of 8 mg/L; the target is 20 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?
95 kgVd = 0.7 L/kgC = 8 → 20 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium
What is the recommended first aid for a thermal burn?
ASpread butter or oil over the burn
BApply ice directly to the burn until the pain stops completely
CCool it under cool running water for about 20 minutes
DBurst any blisters to release the fluid
8.Clinical nursingHard
A patient with lung cancer develops swelling of the face and arms, distended neck veins and breathlessness that is worse lying down. What should be suspected?
AAn allergic reaction to food
BSuperior vena cava obstruction
CHypothyroidism
DRight-sided heart failure from valve disease
3Solutions
1.Lab values & electrolytes
Answer: A · High fat in the sample may interfere with some test results
Why this is the answer
Lipemia makes serum cloudy and can interfere with several laboratory methods, giving inaccurate results. The test may need to be repeated, often after a fast, according to laboratory advice.
Answer
A · High fat in the sample may interfere with some test results
2.Lab values & electrolytes
Answer: D · hCG stimulates the thyroid in early pregnancy
Why this is the answer
hCG is structurally similar to TSH and stimulates the thyroid, peaking at the end of the first trimester, so TSH often falls slightly. With a normal free T4 and no symptoms, this is usually physiologic and is interpreted with pregnancy-specific ranges.Checked against:TSH (Thyroid-stimulating hormone) Test · MedlinePlus (NIH) (2024)
Answer
D · hCG stimulates the thyroid in early pregnancy
3.Arterial blood gas interpretation
Answer: uncompensated
Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.
ROME, value by value
pH 7.21: acidaemia (below 7.35)
PaCO₂ 41 mmHg: normal (35–45)
HCO₃⁻ 16 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 30–34 mmHg, so the measured 41 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: PaCO₂ = 43.4 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.
Expected PaCO₂
0.7·32 + 21 = 43.4 mmHg
Answer
PaCO₂ = 43.4 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: R = 80 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.
Infusion rate
R = 4·20 = 80 mg/h
Answer
R = 80 mg/h
6.Clinical pharmacokinetics
Answer: LD = 867.4 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.
Volume of distribution
Vd = 0.7·95 = 66.5 L
Rise needed
20 − 8 = 12 mg/L
Loading dose
LD = 66.5·12/(0.92·1) = 867.4 mg
Answer
LD = 867.4 mg
In practice the dose is rounded to a convenient amount.
7.Clinical nursing
Answer: C · Cool it under cool running water for about 20 minutes
Why this is the answer
Cooling with running water for around 20 minutes limits tissue damage and pain. Ice can cause further injury, and the patient should be kept warm while the burn is cooled.
Answer
C · Cool it under cool running water for about 20 minutes
8.Clinical nursing
Answer: B · Superior vena cava obstruction
Why this is the answer
Superior vena cava obstruction prevents venous drainage from the head and arms. It is an oncological emergency; patients are positioned upright and need urgent specialist treatment.