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 trauma patient receives 2 L of IV crystalloid. The hemoglobin falls from 13.0 to 11.4 g/dL (130 to 114 g/L), and the vital signs are now stable. Which explanation is most likely?
AA laboratory error in the first sample
BIncreased red cell production
CDilution of the blood by the IV fluid
DA sudden failure of the bone marrow
2.Lab values & electrolytesHard
A patient who had pituitary surgery has fatigue and cold intolerance. The free T4 is low and the TSH is low. What does this pattern suggest?
AHyperthyroidism from overactivity of the thyroid gland
BPrimary hypothyroidism from failure of the thyroid gland
CSecondary (central) hypothyroidism from pituitary failure
DNormal thyroid function with a laboratory error in T4
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 16 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 16 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 21% oxygen shows PaO₂ 68 mmHg (9.1 kPa) and PaCO₂ 45 mmHg (6.0 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Practice exercise (made-up values, not clinical guidance). Vancomycin is given as a continuous infusion to a patient whose vancomycin clearance is 85 mL/min. What infusion rate R in mg/h gives a steady-state level of 20 mg/L?
CL = 85 mL/minCss = 20 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 47-year-old man weighs 51 kg and has a serum creatinine of 2.9 mg/dL (256 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 100 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.
A patient with type 1 diabetes has blood glucose of 25 mmol/L, deep rapid breathing and ketones in the urine. What is the likely diagnosis?
ADiabetic ketoacidosis
BHypoglycaemia
CHyperosmolar hyperglycaemic state without ketones
DNormal glucose fluctuation
8.Clinical nursingHard
Which route and site are recommended for the first dose of adrenaline (epinephrine) in anaphylaxis on a ward?
AOral tablet dissolved under the tongue
BSubcutaneous injection into the upper arm
CIntramuscular injection into the anterolateral thigh
DIntravenous bolus given by the nurse without monitoring
3Solutions
1.Lab values & electrolytes
Answer: C · Dilution of the blood by the IV fluid
Why this is the answer
IV fluids dilute the blood, so the hemoglobin falls even without further bleeding. Ongoing bleeding must still be considered, so vital signs, wounds and repeat hemoglobin values are monitored.
Answer
C · Dilution of the blood by the IV fluid
2.Lab values & electrolytes
Answer: C · Secondary (central) hypothyroidism from pituitary failure
Why this is the answer
In primary hypothyroidism, a low T4 causes a high TSH. A low T4 with a low or inappropriately normal TSH means the pituitary (or hypothalamus) is not producing enough TSH, so the TSH cannot be used alone to monitor treatment.Checked against:Thyroxine (T4) Test · MedlinePlus (NIH) (2024); Hypothyroidism · MedlinePlus (NIH) (2024)
Answer
C · Secondary (central) hypothyroidism from pituitary failure
3.Arterial blood gas interpretation
Answer: PaCO₂ = 32 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 = 1.5·16 + 8 = 32 mmHg
range: 30–34 mmHg
Answer
PaCO₂ = 32 mmHg
4.Arterial blood gas interpretation
Answer: A–a gradient = 25.48 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.
Clearance in L/h
85·60/1000 = 5.1 L/h
Infusion rate
R = 5.1·20 = 102 mg/h
Answer
R = 102 mg/h
6.Clinical pharmacokinetics
Answer: τ = 35.2 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 − 47)·51/(72·2.9) = 22.72 mL/min
New interval
τ = 8·100/22.72 = 35.22 ≈ 35.2 h
Answer
τ = 35.2 h
In practice the interval is rounded to a convenient value such as 12, 24, 36 or 48 h.
7.Clinical nursing
Answer: A · Diabetic ketoacidosis
Why this is the answer
High glucose, ketones and acidosis with Kussmaul breathing indicate DKA. It is treated with fluids, insulin and close potassium monitoring.
Answer
A · Diabetic ketoacidosis
8.Clinical nursing
Answer: C · Intramuscular injection into the anterolateral thigh
Why this is the answer
IM adrenaline into the outer thigh is absorbed quickly and is the first-line route. IV adrenaline is reserved for experienced specialists with full monitoring because of the risk of arrhythmias.
Answer
C · Intramuscular injection into the anterolateral thigh