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 virus has been completely cured and removed
BThe treatment can now be stopped without any risk
CThe CD4 count must now be dangerously low too
DThe treatment is suppressing the virus very well
2.Lab values & electrolytesHard
A patient taking apixaban asks why they do not need regular INR tests like their neighbor on warfarin. What is the best explanation?
AApixaban does not actually thin the blood
BThe INR is checked secretly together with the other routine blood tests
CIts effect is predictable, so routine clotting tests are not needed
DApixaban is monitored with an aPTT every day instead
3.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.09, PaCO₂ 41 mmHg (5.5 kPa) and HCO₃⁻ 12 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.09PaCO₂ 41 mmHg (5.5 kPa)HCO₃⁻ 12 mmol/L
1uncompensated
2partially compensated
3fully compensated
4.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 39 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 39 mmol/L
PaCO₂ = mmHg
5.Clinical pharmacokineticsMedium
Practice exercise (made-up values, not clinical guidance). After an IV dose of vancomycin, the level is 35 mg/L at 1 h and 8.8 mg/L at 17 h. Assume first-order elimination. Find the half-life t½.
C1 = 35 mg/L (1 h)C2 = 8.8 mg/L (17 h)
t½ = h
6.Clinical pharmacokineticsHard
Practice exercise (made-up values, not clinical guidance). A vancomycin infusion runs at 100 mg/h with no loading dose. The patient's clearance is 5 L/h and the half-life is 6 h. What is the level C after 4 h?
R = 100 mg/hCL = 5 L/ht½ = 6 ht = 4 h
C = mg/L
7.Clinical nursingMedium
A patient's NEWS2 score has risen to 7. What is the appropriate response?
AContinue observations every 12 hours
BEmergency assessment by a clinical team with critical care skills
CGive a sedative and recheck later
DRecord the score, increase observations and wait for the morning ward round
8.Clinical nursingHard
A patient has a full-thickness burn all the way round the forearm. Which complication must be watched for closely?
ALoss of circulation to the hand as swelling builds under tight burned skin
BExcessive bleeding from the burn surface needing an urgent blood transfusion
CDelayed hair growth on the arm
DHypoglycaemia caused by the body using glucose to heal the burn
3Solutions
1.Lab values & electrolytes
Answer: D · 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
D · The treatment is suppressing the virus very well
2.Lab values & electrolytes
Answer: C · Its effect is predictable, so routine clotting tests are not needed
Why this is the answer
Direct oral anticoagulants such as apixaban have predictable effects at fixed doses, so routine INR testing is not needed. Kidney and liver function, adherence and signs of bleeding are monitored, and the INR does not reliably measure their effect.
Answer
C · Its effect is predictable, so routine clotting tests are not needed
3.Arterial blood gas interpretation
Answer: uncompensated
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 7.09: acidaemia (below 7.35)
PaCO₂ 41 mmHg: normal (35–45)
HCO₃⁻ 12 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 24–28 mmHg, so the measured 41 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
4.Arterial blood gas interpretation
Answer: PaCO₂ = 48.3 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.
Expected PaCO₂
0.7·39 + 21 = 48.3 mmHg
Answer
PaCO₂ = 48.3 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
5.Clinical pharmacokinetics
Answer: t½ = 8.033 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(35/8.8)/16 = 1.381/16 = 0.08629 h⁻¹
Half-life
t½ = 0.693/0.08629 = 8.033 h
Answer
t½ = 8.033 h
6.Clinical pharmacokinetics
Answer: C = 7.401 mg/L
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.
Steady-state level
Css = 100/5 = 20 mg/L
Rate constant
k = 0.693/6 = 0.1155 h⁻¹
Level after the given time
C = 20·(1 − e−0.1155·4) = 7.401 mg/L
Answer
C = 7.401 mg/L
7.Clinical nursing
Answer: B · Emergency assessment by a clinical team with critical care skills
Why this is the answer
In NEWS2, a score of 7 or more triggers an emergency response, usually continuous monitoring and urgent review by a team competent in managing critically ill patients.
Answer
B · Emergency assessment by a clinical team with critical care skills
8.Clinical nursing
Answer: A · Loss of circulation to the hand as swelling builds under tight burned skin
Why this is the answer
Circumferential full-thickness burns form a tight eschar that acts like a tourniquet as oedema develops. Neurovascular checks are needed, and escharotomy may be required.
Answer
A · Loss of circulation to the hand as swelling builds under tight burned skin