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 patient takes high-dose furosemide. Besides potassium, which electrolyte is most commonly lowered?
AMagnesium
BPhosphate
CUric acid
DSodium bicarbonate
2.Lab values & electrolytesHard
A patient with a serum albumin of 1.9 g/dL (19 g/L) has signs of phenytoin toxicity, although the total phenytoin level is in the usual range. What is the most likely explanation?
ALow albumin makes the liver clear phenytoin much faster
BLow albumin raises the free (active) fraction of phenytoin
CLow albumin makes the total level falsely high in all cases
DLow albumin stops phenytoin from crossing into the brain
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.30, PaCO₂ 21 mmHg (2.8 kPa) and HCO₃⁻ 10 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.30PaCO₂ 21 mmHg (2.8 kPa)HCO₃⁻ 10 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
Practice values: Na⁺ 142, Cl⁻ 112, HCO₃⁻ 10 mmol/L and albumin 35 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Practice exercise (made-up values, not clinical guidance). An 85 kg patient receives a theophylline infusion. The clearance is 0.05 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 12 mg/L?
85 kgCL = 0.05 L/kg/hCss = 12 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A 42-year-old man weighs 87 kg and has a serum creatinine of 1.7 mg/dL (150 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 cefazolin dose is 1000 mg every 8 h for a creatinine clearance of 120 mL/min. Keeping the interval, the dose is reduced in proportion to the creatinine clearance. Find the adjusted dose D. Round to the nearest whole number.
Which advice is correct for a patient taking levothyroxine?
ATake it only on days when feeling tired
BTake it with a large meal to avoid nausea and stomach upset
CStop taking it once the symptoms improve
DTake it on an empty stomach, usually before breakfast
8.Clinical nursingHard
Which set of findings best distinguishes cardiogenic shock from hypovolaemic shock?
ARaised jugular venous pressure and lung crackles
BTachycardia, hypotension and a weak, thready peripheral pulse
CCool, pale skin
DReduced urine output
3Solutions
1.Lab values & electrolytes
Answer: A · Magnesium
Why this is the answer
Loop diuretics block the Na-K-2Cl transporter in the loop of Henle, where much magnesium and calcium are also reabsorbed. They therefore commonly cause hypokalemia and hypomagnesemia, and they can raise uric acid and bicarbonate.
Answer
A · Magnesium
2.Lab values & electrolytes
Answer: B · Low albumin raises the free (active) fraction of phenytoin
Why this is the answer
Phenytoin is highly protein bound, and only the free drug is active. When albumin is low, more phenytoin is free, so toxicity can occur with a "normal" total level. A free phenytoin level or a corrected level helps.
Answer
B · Low albumin raises the free (active) fraction of phenytoin
3.Arterial blood gas interpretation
Answer: partially compensated
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 7.30: acidaemia (below 7.35)
PaCO₂ 21 mmHg: low (35–45)
HCO₃⁻ 10 mmol/L: low (22–26)
Compensation
partially compensated
Answer
partially compensated
4.Arterial blood gas interpretation
Answer: AGc = 21.25 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 = 142 − (112 + 10) = 20
Correction for albumin
20 + 0.25·(40 − 35) = 21.25 mmol/L
Answer
AGc = 21.25 mmol/L
Above about 12 after correction: a high anion gap.
5.Clinical pharmacokinetics
Answer: R = 51 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.
Clearance
CL = 0.05·85 = 4.25 L/h
Infusion rate
R = 4.25·12 = 51 mg/h
Answer
R = 51 mg/h
6.Clinical pharmacokinetics
Answer: D = 580 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 − 42)·87/(72·1.7) = 69.66 mL/min
Adjusted dose
D = 1000·69.66/120 = 580.5 ≈ 580 mg
Answer
D = 580 mg
In practice the dose is rounded to a convenient amount and checked against the product information.
7.Clinical nursing
Answer: D · Take it on an empty stomach, usually before breakfast
Why this is the answer
Levothyroxine is absorbed best on an empty stomach and is usually taken at the same time each morning. It is normally lifelong, with doses adjusted by thyroid blood tests.
Answer
D · Take it on an empty stomach, usually before breakfast
8.Clinical nursing
Answer: A · Raised jugular venous pressure and lung crackles
Why this is the answer
Both types cause tachycardia, hypotension, cool skin and low urine output. In cardiogenic shock the failing heart causes venous congestion and pulmonary oedema, so large fluid boluses may worsen it.
Answer
A · Raised jugular venous pressure and lung crackles