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.
ECG rhythm strip interpretation
Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)
Method
Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
Which cranial nerve supplies the trapezius and sternocleidomastoid muscles?
AHypoglossal nerve (XII)
BVagus nerve (X)
CFacial nerve (VII)
DAccessory nerve (XI)
2.Human anatomyHard
After a stroke, a patient speaks fluently but uses meaningless words and cannot understand speech. Where is the lesion most likely to be?
ACerebellar hemisphere
BPosterior superior temporal gyrus of the dominant hemisphere
CInferior frontal gyrus of the dominant hemisphere
DPrecentral gyrus of the non-dominant hemisphere
3.Human physiologyMedium
Which receptors respond mainly to a fall in arterial oxygen levels?
APeripheral chemoreceptors in the carotid and aortic bodies
BCentral chemoreceptors in the medulla
CBaroreceptors in the carotid sinus
DStretch receptors in the lungs
4.Human physiologyHard
Which adaptation develops over days to weeks at high altitude?
AA fall in haematocrit
BHigher erythropoietin levels and more red blood cells
CLower ventilation to save oxygen
DIncreased renal bicarbonate retention
5.MicrobiologyMedium
Which organism is a common cause of infection in burns and in people with cystic fibrosis, and can produce a green pigment?
AStreptococcus pyogenes
BHelicobacter pylori
CClostridium tetani
DPseudomonas aeruginosa
6.MicrobiologyHard
Staphylococcus aureus food poisoning starts only 1–6 hours after eating. Why so fast?
AThe bacteria multiply very fast inside the gut and invade the intestinal wall
BThe bacteria invade the bloodstream immediately
CThe bacteria made a heat-stable toxin in the food before it was eaten
DThe immune system reacts to the bacteria within minutes
7.PharmacologyMedium
How is adenosine given for supraventricular tachycardia?
AAs an intramuscular injection into the outer thigh muscle
BAs a slow infusion over one hour
CAs a rapid IV push followed by a fast saline flush
DAs a tablet under the tongue
8.PharmacologyHard
Codeine is converted to morphine by CYP2D6. What happens in a patient who is a CYP2D6 ultra-rapid metaboliser?
ACodeine gives almost no pain relief because little morphine is formed
BCodeine is converted to an inactive compound only
CMore morphine is formed, increasing the risk of opioid toxicity
DCodeine is excreted unchanged in the urine
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.37, PaCO₂ 64 mmHg (8.5 kPa) and HCO₃⁻ 36 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.
pH 7.37PaCO₂ 64 mmHg (8.5 kPa)HCO₃⁻ 36 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
A practice gas shows a respiratory acidosis: pH 7.27, PaCO₂ 67 mmHg (8.9 kPa), HCO₃⁻ 30 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 7.27PaCO₂ 67 mmHg (8.9 kPa)HCO₃⁻ 30 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Bazett’s formula QTc = QT/√RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
Identify the rhythm on this lead II strip. Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1atrial fibrillation
2supraventricular tachycardia (SVT)
3atrial flutter, 2:1 conduction
4ventricular tachycardia
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A new test was compared with the reference standard in a study (see the 2 × 2 table). Find the negative likelihood ratio LR−. Round to two decimal places.
a = 158b = 152c = 92d = 508
LR− =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 96% and a specificity of 92%. It is used in a population where the prevalence is 30%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 96%Sp = 92%p = 30%
PPV = %
3Solutions
1.Human anatomy
Answer: D · Accessory nerve (XI)
Why this is the answer
The accessory nerve (XI) supplies sternocleidomastoid and trapezius. Damage weakens head turning and shoulder shrugging.
Answer
D · Accessory nerve (XI)
2.Human anatomy
Answer: B · Posterior superior temporal gyrus of the dominant hemisphere
Why this is the answer
This is Wernicke's (receptive) aphasia, caused by damage to Wernicke's area in the posterior superior temporal gyrus. Broca's area in the inferior frontal gyrus causes non-fluent aphasia.
Answer
B · Posterior superior temporal gyrus of the dominant hemisphere
3.Human physiology
Answer: A · Peripheral chemoreceptors in the carotid and aortic bodies
Why this is the answer
The carotid and aortic bodies detect low PaO₂ (and also high CO₂ and H⁺). The central chemoreceptors respond to CO₂ but not directly to oxygen.
Answer
A · Peripheral chemoreceptors in the carotid and aortic bodies
4.Human physiology
Answer: B · Higher erythropoietin levels and more red blood cells
Why this is the answer
Hypoxia stimulates EPO and red cell production, raising oxygen-carrying capacity. Hyperventilation causes alkalosis, which the kidneys compensate by excreting bicarbonate.
Answer
B · Higher erythropoietin levels and more red blood cells
5.Microbiology
Answer: D · Pseudomonas aeruginosa
Why this is the answer
Pseudomonas aeruginosa is an opportunistic pathogen of burns, ventilated patients and cystic fibrosis lungs. It is naturally resistant to many antibiotics.
Answer
D · Pseudomonas aeruginosa
6.Microbiology
Answer: C · The bacteria made a heat-stable toxin in the food before it was eaten
Why this is the answer
S. aureus produces enterotoxins in food left at warm temperatures. The toxin is already there when the food is eaten and survives reheating, so symptoms start quickly.
Answer
C · The bacteria made a heat-stable toxin in the food before it was eaten
7.Pharmacology
Answer: C · As a rapid IV push followed by a fast saline flush
Why this is the answer
Adenosine must reach the heart quickly, so it is pushed rapidly into a large vein and flushed. Patients are warned about brief flushing, chest tightness and a short pause in the heartbeat.
Answer
C · As a rapid IV push followed by a fast saline flush
8.Pharmacology
Answer: C · More morphine is formed, increasing the risk of opioid toxicity
Why this is the answer
Ultra-rapid metabolisers convert more codeine to morphine, which can cause respiratory depression. Poor metabolisers form little morphine and get poor pain relief. This is a classic example of pharmacogenetics.
Answer
C · More morphine is formed, increasing the risk of opioid toxicity
9.Arterial blood gas interpretation
Answer: fully 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 7.37: within 7.35–7.45
PaCO₂ 64 mmHg: high (35–45)
HCO₃⁻ 36 mmol/L: high (22–26)
Compensation
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: acute on chronic
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 + 1·2.7 = 26.7 mmol/L
chronic: 24 + 3.5·2.7 = 33.45 mmol/L
Compare
measured: 30 mmol/L → acute on chronic
Answer
acute on chronic
11.ECG rhythm strip interpretation
Answer: QTc = 0.454 s
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 12 × 0.04 = 0.48 s
RR = 28 × 0.04 = 1.12 s
Bazett’s formula
QTc = 0.48/√1.12 = 0.48/1.058 = 0.454 s
Answer
QTc = 0.454 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (454 ms)
12.ECG rhythm strip interpretation
Answer: ventricular tachycardia
Ventricular rhythms: ventricular tachycardia is a run of three or more wide (0.12 s or more) complexes in a row at more than 100/min, usually 150–250/min, regular, without related P waves. Ventricular fibrillation is a chaotic, irregular waveform with no identifiable P waves, QRS complexes or T waves and no countable rate. Asystole is an almost flat line with no ventricular activity; a flat line is always confirmed by checking the patient, the leads and the gain and a second lead. In resuscitation guidelines VF and pulseless VT are the shockable rhythms and asystole is not.
Five steps
1. Rate: R–R = 10 small boxes, 1500 ÷ 10 ≈ 150 per minute.
2. Rhythm: regular, every R–R is 0.40 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.16 s, wide (0.12 s or more).
Interpretation
ventricular tachycardia
Answer
ventricular tachycardia
In resuscitation guidelines VF and pulseless VT are the shockable rhythms. This is identification practice only.
13.Evidence-based practice statistics
Answer: LR− = 0.48
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Sensitivity and specificity
Se = 158/250 = 0.632
Sp = 508/660 = 0.7697
Negative likelihood ratio
LR− = (1 − 0.632)/0.7697 = 0.4781 ≈ 0.48
Answer
LR− = 0.48
14.Evidence-based practice statistics
Answer: PPV = 83.7 %
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.
True and false positives (per person)
Se·p = 0.96·0.3 = 0.288
(1 − Sp)·(1 − p) = 0.08·0.7 = 0.056
Positive predictive value
PPV = 0.288/(0.288 + 0.056) = 0.8372 ≈ 83.7%
Answer
PPV = 83.7%
The lower the prevalence, the lower the PPV of the same test.