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?
AIncreased red cell production
BA sudden failure of the bone marrow
CA laboratory error in the first sample
DDilution of the blood by the IV fluid
2.Lab values & electrolytesHard
A patient with a stable INR on warfarin starts rifampin (rifampicin) for tuberculosis. Two weeks later the INR is 1.3. What is the most likely explanation?
ATuberculosis causes the INR to fall
BRifampin raises the INR, so it is wrong
CRifampin speeds up warfarin metabolism
DRifampin contains a lot of vitamin K
3.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 10 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 10 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.24, PaCO₂ 65 mmHg (8.7 kPa), HCO₃⁻ 27 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 7.24PaCO₂ 65 mmHg (8.7 kPa)HCO₃⁻ 27 mmol/L
1acute
2chronic
3acute on chronic
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). A 77 kg patient is to be loaded with digoxin tablets to reach a level of 1 mcg/L. The volume of distribution is 7 L/kg and the tablets have a bioavailability F = 0.7. Find the total oral loading dose LD in mcg.
77 kgVd = 7 L/kgC = 1 mcg/LF = 0.7
LD = mcg
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 61-year-old woman weighs 94 kg and has a serum creatinine of 2.2 mg/dL (194 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 ciprofloxacin dose is 500 mg for a creatinine clearance of 120 mL/min, and a fraction fe = 0.6 of the drug is excreted unchanged by the kidneys. Using the Dettli dose fraction Q = 1 − fe·(1 − CrCl/normal CrCl), find the adjusted dose D. Round to the nearest whole number.
A patient with dysphagia after a stroke has been cleared for a modified diet. How should they be positioned for meals?
ALying on their side with the knees bent
BSemi-reclined at about 20 degrees, with the head supported on pillows
CSitting fully upright, and staying upright for a time after eating
DLying flat on their back with the head turned to one side
8.Clinical nursingHard
Why is a patient kept under observation for several hours after successful treatment of anaphylaxis?
AAntihistamines cause dangerous drowsiness for 24 hours
BThe patient must stay until all allergy tests are complete
CSymptoms can return later in a biphasic reaction
DAdrenaline always causes a heart attack hours later
3Solutions
1.Lab values & electrolytes
Answer: D · 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
D · Dilution of the blood by the IV fluid
2.Lab values & electrolytes
Answer: C · Rifampin speeds up warfarin metabolism
Why this is the answer
Rifampin induces liver enzymes that break down warfarin, so warfarin's effect falls and the INR drops, increasing clot risk. The INR is monitored closely and the dose adjusted; the reverse happens when rifampin is stopped.
Answer
C · Rifampin speeds up warfarin metabolism
3.Arterial blood gas interpretation
Answer: PaCO₂ = 23 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 = 1.5·10 + 8 = 23 mmHg
range: 21–25 mmHg
Answer
PaCO₂ = 23 mmHg
4.Arterial blood gas interpretation
Answer: acute
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 + 1·2.5 = 26.5 mmol/L
chronic: 24 + 3.5·2.5 = 32.75 mmol/L
Compare
measured: 27 mmol/L → acute
Answer
acute
5.Clinical pharmacokinetics
Answer: LD = 770 mcg
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.
Volume of distribution
Vd = 7·77 = 539 L
Amount of active drug needed
Vd·C = 539·1 = 539 mcg
Loading dose
LD = 539/(1·0.7) = 770 mcg
Answer
LD = 770 mcg
In practice the dose is rounded to a convenient amount.
6.Clinical pharmacokinetics
Answer: D = 300 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.
Creatinine clearance (Cockcroft–Gault)
CrCl = (140 − 61)·94/(72·2.2)·0.85 = 39.85 mL/min
Dose fraction (Dettli)
Q = 1 − 0.6·(1 − 39.85/120) = 0.5992
Adjusted dose
D = 0.5992·500 = 299.6 ≈ 300 mg
Answer
D = 300 mg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: C · Sitting fully upright, and staying upright for a time after eating
Why this is the answer
Sitting upright during and after meals reduces the risk of aspiration. The patient should also be supervised and given the consistency recommended by the speech and language therapist.
Answer
C · Sitting fully upright, and staying upright for a time after eating
8.Clinical nursing
Answer: C · Symptoms can return later in a biphasic reaction
Why this is the answer
Some patients have a second wave of symptoms hours after the first, called a biphasic reaction. Observation, an adrenaline auto-injector where indicated and allergy referral are part of safe discharge.
Answer
C · Symptoms can return later in a biphasic reaction