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 with HIV on antiretroviral therapy has an undetectable viral load. What does this mean?
AThe treatment is suppressing the virus very well
BThe CD4 count must now be dangerously low too
CThe virus has been completely cured and removed
DThe treatment can now be stopped without any risk
2.Lab values & electrolytesHard
A man of Mediterranean descent develops dark urine, jaundice and a falling hemoglobin a few days after eating fava beans. Which condition is most likely?
AG6PD deficiency with acute hemolysis
BIron-deficiency anemia from poor diet
CSickle cell crisis with vaso-occlusion
DVitamin B12 deficiency anemia
3.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 74 mmHg (9.9 kPa). Use the rule that in acute respiratory acidosis HCO₃⁻ rises by 1 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic acidosis: pH 7.31, PaCO₂ 29 mmHg (3.9 kPa), HCO₃⁻ 14 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.
pH 7.31PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 14 mmol/L
1appropriate compensation
2added respiratory acidosis
3added respiratory alkalosis
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). An 89 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 15 mg/L?
89 kgCL = 0.05 L/kg/hCss = 15 mg/L
R = mg/h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A patient's gentamicin half-life is 2.5 h. The target is a peak of 8 mg/L and a trough of 1 mg/L. Assuming rapid IV doses and first-order elimination, what dosing interval τ lets the level fall from the peak to the trough? Round to one decimal place.
t½ = 2.5 hCpeak = 8 mg/LCtrough = 1 mg/L
τ = h
7.Clinical nursingMedium
How does ulcerative colitis differ from Crohn's disease?
AUlcerative colitis typically causes patchy "skip" lesions
BUlcerative colitis always affects the full thickness of the bowel wall
CUlcerative colitis affects only the colon, continuously from the rectum
DUlcerative colitis can affect any part of the gut from mouth to anus
8.Clinical nursingHard
A patient has new facial weakness. Which feature suggests a stroke rather than Bell's palsy?
AThe whole of one side of the face, including the forehead, is weak
BThe patient has pain behind the ear
CThe patient has had a recent cold
DThe forehead can still be wrinkled on the affected side
3Solutions
1.Lab values & electrolytes
Answer: A · The treatment is suppressing the virus very well
Why this is the answer
An undetectable viral load shows that antiretroviral therapy is suppressing viral replication. The virus remains in the body, so treatment must continue. People with a sustained undetectable viral load do not transmit HIV sexually.
Answer
A · The treatment is suppressing the virus very well
2.Lab values & electrolytes
Answer: A · G6PD deficiency with acute hemolysis
Why this is the answer
G6PD deficiency makes red cells vulnerable to oxidative damage. Fava beans, some drugs (such as primaquine and some sulfonamides) and infections can trigger acute hemolysis, with dark urine, jaundice and anemia. Triggers are avoided.
Answer
A · G6PD deficiency with acute hemolysis
3.Arterial blood gas interpretation
Answer: HCO₃⁻ = 27.4 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
|74 − 40| = 34 mmHg → 3.4 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 1·3.4 = 27.4 mmol/L
Answer
HCO₃⁻ = 27.4 mmol/L
4.Arterial blood gas interpretation
Answer: appropriate compensation
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
1.5·14 + 8 = 29 → 27–31 mmHg
Compare
measured: 29 mmHg → appropriate compensation
Answer
appropriate compensation
5.Clinical pharmacokinetics
Answer: R = 66.75 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 = 0.05·89 = 4.45 L/h
Infusion rate
R = 4.45·15 = 66.75 mg/h
Answer
R = 66.75 mg/h
6.Clinical pharmacokinetics
Answer: τ = 7.5 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.
Rate constant
k = 0.693/2.5 = 0.2773 h⁻¹
Dosing interval
τ = ln(8/1)/0.2773 = 2.079/0.2773 = 7.5 h
Answer
τ = 7.5 h
In practice the interval is rounded to a convenient value such as 8, 12, 24 or 36 h, and the levels are checked again.
7.Clinical nursing
Answer: C · Ulcerative colitis affects only the colon, continuously from the rectum
Why this is the answer
Ulcerative colitis involves the mucosa of the colon, spreading continuously from the rectum. Crohn's disease can affect any part of the GI tract, is transmural and has skip lesions.
Answer
C · Ulcerative colitis affects only the colon, continuously from the rectum
8.Clinical nursing
Answer: D · The forehead can still be wrinkled on the affected side
Why this is the answer
In a stroke (upper motor neurone lesion), the forehead is usually spared because it has bilateral cortical supply. Bell's palsy (lower motor neurone) weakens the whole side, including the forehead.
Answer
D · The forehead can still be wrinkled on the affected side