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 is the largest nerve in the human body?
AMedian nerve
BFemoral nerve
CSciatic nerve
DVagus nerve
2.Human anatomyHard
Which structure is found in the posterior mediastinum?
AHeart
BAscending aorta
CThymus
DOesophagus
3.Human physiologyMedium
Why can cardiac muscle not be tetanised like skeletal muscle?
AIt has no calcium in the sarcoplasmic reticulum
BIt is not striated
CIts long refractory period lasts almost as long as the contraction
DIt is controlled only by hormones
4.Human physiologyHard
Which change shifts the oxygen–haemoglobin dissociation curve to the right?
AHigher temperature
BLower CO₂
CHigher pH
DLower 2,3-BPG
5.MicrobiologyMedium
Which laboratory test separates Staphylococcus aureus from most other staphylococci?
AThe coagulase test
BThe catalase test
CThe acid-fast stain
DThe Gram stain
6.MicrobiologyHard
A positive blood culture shows Gram-positive cocci in clusters. Which organism is most likely?
AEnterococcus in short chains
BStaphylococcus
CStreptococcus
DNeisseria
7.PharmacologyMedium
Which adverse effect is typical of metoclopramide, especially in young people?
ALow blood glucose
BExtrapyramidal movement disorders such as acute dystonia
CDry cough
DSevere constipation
8.PharmacologyHard
Why is the TPMT enzyme often tested before azathioprine is started?
ALow TPMT activity causes toxic levels and marrow suppression
BTPMT testing shows whether the patient has tuberculosis
CTPMT is needed to absorb azathioprine from the gut
DHigh TPMT activity causes an allergic reaction to azathioprine
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.25, PaCO₂ 40 mmHg (5.3 kPa) and HCO₃⁻ 17 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.25PaCO₂ 40 mmHg (5.3 kPa)HCO₃⁻ 17 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 30% oxygen shows PaO₂ 82 mmHg (10.9 kPa) and PaCO₂ 60 mmHg (8.0 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Interpret this lead II rhythm strip. Which rhythm is it? 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 flutter, 4:1 conduction
2atrial flutter, 3:1 conduction
3atrial flutter, 2:1 conduction
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
Interpret this lead II strip with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1rate
2rhythm
3P waves
4PR interval
5QRS duration
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 420 patients received a new treatment and 170 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk reduction RRR as a percentage. Round to one decimal place.
treatment: 20/420control: 10/170
RRR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A treatment reduces the risk of an event by 34% in relative terms (RRR = 34%). In a patient group the risk without treatment (CER) is 4% over five years. Find the number needed to treat NNT for five years to prevent one event. Round up to a whole number of patients.
RRR = 34%CER = 4%
NNT =
3Solutions
1.Human anatomy
Answer: C · Sciatic nerve
Why this is the answer
The sciatic nerve (L4–S3) leaves the pelvis through the greater sciatic foramen and supplies the hamstrings and all the muscles below the knee. Skin on the inner side of the leg is supplied by the saphenous nerve instead.
Answer
C · Sciatic nerve
2.Human anatomy
Answer: D · Oesophagus
Why this is the answer
The posterior mediastinum contains the oesophagus, descending aorta, azygos veins and thoracic duct. The heart and ascending aorta are in the middle mediastinum; the thymus is in the superior and anterior mediastinum.
Answer
D · Oesophagus
3.Human physiology
Answer: C · Its long refractory period lasts almost as long as the contraction
Why this is the answer
The plateau of the cardiac action potential gives a long refractory period, so the muscle relaxes before it can be excited again. This allows the heart to refill.
Answer
C · Its long refractory period lasts almost as long as the contraction
4.Human physiology
Answer: A · Higher temperature
Why this is the answer
Higher temperature, CO₂, H⁺ (lower pH) and 2,3-BPG lower haemoglobin's affinity and shift the curve right, releasing more oxygen to active tissues.
Answer
A · Higher temperature
5.Microbiology
Answer: A · The coagulase test
Why this is the answer
S. aureus produces coagulase, which converts fibrinogen to fibrin. All staphylococci are Gram-positive and catalase-positive, so those tests do not separate them.
Answer
A · The coagulase test
6.Microbiology
Answer: B · Staphylococcus
Why this is the answer
Staphylococci grow in grape-like clusters; streptococci and enterococci in chains or pairs. Neisseria are Gram-negative diplococci.
Answer
B · Staphylococcus
7.Pharmacology
Answer: B · Extrapyramidal movement disorders such as acute dystonia
Why this is the answer
Metoclopramide is a dopamine D2 antagonist. Blocking dopamine in the basal ganglia can cause dystonia and other extrapyramidal effects, more often in young adults.
Answer
B · Extrapyramidal movement disorders such as acute dystonia
8.Pharmacology
Answer: A · Low TPMT activity causes toxic levels and marrow suppression
Why this is the answer
Thiopurine methyltransferase helps inactivate azathioprine and mercaptopurine. People with low or absent TPMT activity need much lower doses or another drug to avoid life-threatening myelosuppression.
Answer
A · Low TPMT activity causes toxic levels and marrow suppression
9.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 7.25: acidaemia (below 7.35)
PaCO₂ 40 mmHg: normal (35–45)
HCO₃⁻ 17 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 31.5–35.5 mmHg, so the measured 40 mmHg points to an additional respiratory acidosis (a mixed disorder). The answer above is the ROME exam classification.
uncompensated
Answer
uncompensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 56.9 mmHg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
Alveolar oxygen
PAO₂ = 0.3·713 − 60/0.8 = 213.9 − 75 = 138.9 mmHg
Gradient
A–a = 138.9 − 82 = 56.9 mmHg
Answer
A–a gradient = 56.9 mmHg
11.ECG rhythm strip interpretation
Answer: atrial flutter, 3:1 conduction
Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.
Five steps
1. Rate: R–R = 15 small boxes, 1500 ÷ 15 ≈ 100 per minute.
2. Rhythm: regular, every R–R is 0.60 s.
3. P waves: none; regular sawtooth flutter waves every 0.20 s (300 per minute), 3 for each QRS.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial flutter, 3:1 conduction
Answer
atrial flutter, 3:1 conduction
12.ECG rhythm strip interpretation
Answer: QRS duration
Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.
Five steps
1. Rate: R–R = 21 small boxes, 1500 ÷ 21 ≈ 71 per minute.
2. Rhythm: regular, every R–R is 0.84 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.12 s, wide (0.12 s or more).
First abnormal step
QRS duration → sinus rhythm with wide QRS
Answer
QRS duration
13.Evidence-based practice statistics
Answer: RRR = 19 %
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.
Event rates
EER = 20/420 = 0.04762
CER = 10/170 = 0.05882
Relative risk reduction
ARR = 0.05882 − 0.04762 = 0.0112
RRR = 0.0112/0.05882 = 0.1905 ≈ 19%
Answer
RRR = 19%
14.Evidence-based practice statistics
Answer: NNT = 74
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.
Absolute risk reduction
ARR = 0.04·0.34 = 0.0136
Number needed to treat
NNT = 1/0.0136 = 73.53 → 74
Answer
NNT = 74
The same relative effect gives a smaller NNT when the baseline risk is higher.