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 vessels carry oxygenated blood from the lungs to the heart?
APulmonary veins
BVenae cavae
CBronchial arteries
DPulmonary arteries
2.Human anatomyHard
A patient has a drooping eyelid, a dilated pupil and an eye that rests "down and out". Which nerve is most likely damaged?
AOptic nerve (II)
BTrochlear nerve (IV)
COculomotor nerve (III)
DAbducens nerve (VI)
3.Human physiologyMedium
Vital capacity is the sum of which lung volumes?
ATidal volume and residual volume
BInspiratory reserve volume and residual volume
CResidual volume and expiratory reserve volume
DTidal volume, inspiratory reserve volume and expiratory reserve volume
4.Human physiologyHard
Which adaptation develops over days to weeks at high altitude?
ALower ventilation to save oxygen
BIncreased renal bicarbonate retention
CA fall in haematocrit
DHigher erythropoietin levels and more red blood cells
5.MicrobiologyMedium
Which is a function of the complement system?
AProducing antibodies against newly encountered antigens
BPresenting antigens to T cells
CMaking new red blood cells
DForming pores in the membranes of some bacteria
6.MicrobiologyHard
Which hepatitis viruses are mainly spread through contaminated food and water?
AHepatitis B and C
BHepatitis A and E
CHepatitis B and D
DHepatitis C and E
7.PharmacologyMedium
Why do patients taking clozapine need regular blood counts?
AIt often causes iron-deficiency anaemia
BIt raises the platelet count dangerously
CIt changes the blood group
DIt can cause agranulocytosis
8.PharmacologyHard
A patient taking dabigatran has a life-threatening bleed. Which specific reversal agent is used?
AAndexanet alfa
BProtamine sulfate
CIdarucizumab
DVitamin K
9.Arterial blood gas interpretationMedium
A practice arterial blood gas shows pH 7.38, PaCO₂ 33 mmHg (4.4 kPa) and HCO₃⁻ 19 mmol/L. The primary disorder is metabolic acidosis. How far is it compensated? Click your answer.
pH 7.38PaCO₂ 33 mmHg (4.4 kPa)HCO₃⁻ 19 mmol/L
1uncompensated
2partially compensated
3fully compensated
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 28% oxygen shows PaO₂ 59 mmHg (7.9 kPa) and PaCO₂ 34 mmHg (4.5 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
This lead II strip shows sinus rhythm with one pause. Measure the pause, the R–R interval across it, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
pause = s
12.ECG rhythm strip interpretationHard
This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. Measure the P–P interval and find the atrial rate in beats per minute (1500 method, whole number).
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
atrial rate = bpm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 580 patients received a new treatment and 660 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the number needed to treat NNT to prevent one event. Round up to a whole number of patients.
treatment: 32/580control: 92/660
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 74% and a specificity of 98%. It is used in a population where the prevalence is 10%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 74%Sp = 98%p = 10%
PPV = %
3Solutions
1.Human anatomy
Answer: A · Pulmonary veins
Why this is the answer
Four pulmonary veins drain oxygenated blood into the left atrium. They are the main exception to the rule that veins carry deoxygenated blood.
Answer
A · Pulmonary veins
2.Human anatomy
Answer: C · Oculomotor nerve (III)
Why this is the answer
The oculomotor nerve supplies most eye muscles, levator palpebrae and the pupil constrictor. Lesion leaves lateral rectus and superior oblique unopposed, so the eye goes down and out.
Answer
C · Oculomotor nerve (III)
3.Human physiology
Answer: D · Tidal volume, inspiratory reserve volume and expiratory reserve volume
Why this is the answer
Vital capacity = IRV + TV + ERV. It excludes the residual volume, which always stays in the lungs.
Answer
D · Tidal volume, inspiratory reserve volume and expiratory reserve volume
4.Human physiology
Answer: D · 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
D · Higher erythropoietin levels and more red blood cells
5.Microbiology
Answer: D · Forming pores in the membranes of some bacteria
Why this is the answer
Complement proteins coat microbes for phagocytosis (opsonisation), attract immune cells and form the membrane attack complex that punches holes in some bacteria.
Answer
D · Forming pores in the membranes of some bacteria
6.Microbiology
Answer: B · Hepatitis A and E
Why this is the answer
Hepatitis A and E spread by the faecal–oral route. B, C and D spread through blood and body fluids and can cause chronic infection.
Answer
B · Hepatitis A and E
7.Pharmacology
Answer: D · It can cause agranulocytosis
Why this is the answer
Clozapine can cause severe neutropenia (agranulocytosis). Regular white cell and neutrophil counts allow treatment to be stopped before life-threatening infection occurs.
Answer
D · It can cause agranulocytosis
8.Pharmacology
Answer: C · Idarucizumab
Why this is the answer
Idarucizumab is an antibody fragment that binds dabigatran. Andexanet alfa reverses factor Xa inhibitors, protamine reverses heparin, and vitamin K reverses warfarin.
Answer
C · Idarucizumab
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 7.38: within 7.35–7.45
PaCO₂ 33 mmHg: low (35–45)
HCO₃⁻ 19 mmol/L: low (22–26)
Compensation
Clinically, also check the expected compensation: Winter’s formula expects PaCO₂ 34.5–38.5 mmHg, so the measured 33 mmHg points to an additional respiratory alkalosis (a mixed disorder). The answer above is the ROME exam classification.
fully compensated
Answer
fully compensated
10.Arterial blood gas interpretation
Answer: A–a gradient = 98.14 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.
Narrow-complex tachycardia and pauses: supraventricular tachycardia (SVT) is a regular tachycardia at about 150–250/min with narrow QRS complexes; the P waves are usually not visible because they are hidden in the QRS complex or the T wave. Sinus tachycardia in an adult at rest is usually below about 150/min and has a visible P wave before every QRS; atrial flutter with 2:1 conduction often runs at about 150/min and shows sawtooth waves. In sinus arrest the sinus node stops firing for a moment: a pause with no P wave and no QRS whose length is not a whole number of P–P cycles (a pause of exactly two or three cycles points to sinus exit block instead).
Measure
45 small boxes from R to R across the pause
Convert
pause = 45 × 0.04 s = 1.8 s
Compare with P–P
1.80 ÷ 0.72 ≈ 2.5 P–P cycles, not a whole number
Answer
pause = 1.8 s
No P wave appears in the pause, and the pause is not a whole number of P–P cycles: sinus arrest. A pause of exactly two or three cycles would point to sinus exit block.
12.ECG rhythm strip interpretation
Answer: atrial rate = 79 bpm
AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.
Atrial rate
P–P = 19 small boxes = 0.76 s → 1500 ÷ 19 = 78.95 ≈ 79 per minute
Ventricular rate
R–R = 38 small boxes = 1.52 s → 1500 ÷ 38 = 39.47 ≈ 39 per minute
Answer
atrial rate = 79 bpm
Answer
atrial rate = 79 bpm
2:1 AV block: every second P wave is not conducted, so the atrial rate is twice the ventricular rate.
13.Evidence-based practice statistics
Answer: NNT = 12
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 = 32/580 = 0.05517
CER = 92/660 = 0.1394
Number needed to treat
ARR = 0.1394 − 0.05517 = 0.08422
NNT = 1/0.08422 = 11.87 → 12
Answer
NNT = 12
Treating 12 patients prevents about one event.
14.Evidence-based practice statistics
Answer: PPV = 80.4 %
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 = 0.74·0.1 = 0.074
(1 − Sp)·(1 − p) = 0.02·0.9 = 0.018
Positive predictive value
PPV = 0.074/(0.074 + 0.018) = 0.8043 ≈ 80.4%
Answer
PPV = 80.4%
The lower the prevalence, the lower the PPV of the same test.