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
DLow hemoglobin, low platelets, low white cells, low reticulocytes
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 30 mmHg (4.0 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 30 mg/L at 1 h and 13.6 mg/L at 9 h. Assume first-order elimination. Find the half-life t½.
C1 = 30 mg/L (1 h)C2 = 13.6 mg/L (9 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 58-year-old man weighs 91 kg and has a serum creatinine of 1.2 mg/dL (106 µmol/L). Estimate the creatinine clearance with Cockcroft–Gault: CrCl (mL/min) = (140 − age) × weight (kg)/(72 × SCr in mg/dL), × 0.85 for women. The usual tobramycin regimen is 120 mg every 8 h for a creatinine clearance of 120 mL/min. Keeping the dose, the interval is lengthened in proportion to the fall in clearance. Find the new dosing interval τ. Round to one decimal place.
Which finding with a groin hernia needs urgent surgical review?
AThe lump gets bigger when the patient coughs
BThe lump has become painful and tender and cannot be pushed back
CThe lump disappears completely when the patient lies down flat
DThe lump has been present for several years without pain
8.Clinical nursingHard
A patient suddenly develops a painful red eye with blurred vision, haloes around lights and vomiting. What should be suspected?
ADry eye syndrome
BAcute angle-closure glaucoma
CSimple conjunctivitis from a viral infection
DA stye
3Solutions
1.Lab values & electrolytes
Answer: B · Hold the warfarin dose and inform the prescriber
Why this is the answer
An INR of 5.2 is far above the target and increases the bleeding risk. The nurse holds the dose, checks for signs of bleeding and informs the prescriber, who may prescribe a lower dose or vitamin K.Checked against:Prothrombin Time Test and INR (PT/INR) · MedlinePlus (NIH) (2024); Warfarin · MedlinePlus (NIH) (2026)
Answer
B · Hold the warfarin dose and inform the prescriber
2.Lab values & electrolytes
Answer: A · Low hemoglobin, high reticulocytes, high unconjugated bilirubin, high LDH
Why this is the answer
When red cells break down early, hemoglobin is turned into unconjugated bilirubin and LDH is released, while the marrow increases reticulocytes. Option B fits iron deficiency, C fits B12 deficiency and D fits marrow failure.Checked against:Hemolytic anemia · MedlinePlus (NIH) (2026); Reticulocyte Count · MedlinePlus (NIH) (2024); Bilirubin Blood Test · MedlinePlus (NIH) (2024)
Answer
A · Low hemoglobin, high reticulocytes, high unconjugated bilirubin, high LDH
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 22 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
|30 − 40| = 10 mmHg → 1 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 2·1 = 22 mmol/L
Answer
HCO₃⁻ = 22 mmol/L
4.Arterial blood gas interpretation
Answer: AGc = 11.75 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.
Anion gap
AG = 134 − (112 + 13) = 9
Correction for albumin
9 + 0.25·(40 − 29) = 11.75 mmol/L
Answer
AGc = 11.75 mmol/L
Still within about 8–12 after correction.
5.Clinical pharmacokinetics
Answer: t½ = 7.009 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(30/13.6)/8 = 0.7911/8 = 0.09889 h⁻¹
Half-life
t½ = 0.693/0.09889 = 7.009 h
Answer
t½ = 7.009 h
6.Clinical pharmacokinetics
Answer: τ = 11.1 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 58)·91/(72·1.2) = 86.37 mL/min
New interval
τ = 8·120/86.37 = 11.12 ≈ 11.1 h
Answer
τ = 11.1 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.
7.Clinical nursing
Answer: B · The lump has become painful and tender and cannot be pushed back
Why this is the answer
An irreducible, painful, tender hernia, especially with vomiting, suggests incarceration or strangulation. This is a surgical emergency because the trapped bowel can become ischaemic.
Answer
B · The lump has become painful and tender and cannot be pushed back
8.Clinical nursing
Answer: B · Acute angle-closure glaucoma
Why this is the answer
Acute angle-closure glaucoma causes a rapid rise in eye pressure, with pain, visual disturbance, haloes and vomiting. It is an emergency that can cause permanent sight loss without urgent treatment.