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 nerve supplies the diaphragm?
AAccessory nerve
BPhrenic nerve
CIntercostal nerves
DVagus nerve
2.Human anatomyHard
Which ligament of the knee prevents the tibia from sliding forward relative to the femur?
AAnterior cruciate ligament
BPatellar ligament
CMedial collateral ligament
DPosterior cruciate ligament
3.Human physiologyMedium
Why do myelinated axons conduct action potentials faster than unmyelinated ones?
AMyelin contains more sodium channels
BMyelin makes the axon thinner
CMyelin releases neurotransmitter along the axon
DThe impulse jumps between nodes of Ranvier
4.Human physiologyHard
A pulmonary embolus blocks blood flow to a well-ventilated region of lung. How does the ventilation–perfusion (V/Q) ratio of that region change?
AIt becomes zero, acting as a shunt
BIt stays normal because ventilation falls too
CIt becomes very high, acting as dead space
DIt becomes low, like in pneumonia
5.MicrobiologyMedium
Which is a function of the complement system?
AMaking new red blood cells
BProducing antibodies against newly encountered antigens
CPresenting antigens to T cells
DForming pores in the membranes of some bacteria
6.MicrobiologyHard
After eating home-canned vegetables, a patient develops double vision, difficulty swallowing and a descending flaccid paralysis. What is the most likely cause?
AA viral infection of the motor neurons
BEndotoxin from Gram-negative bacteria
CBotulinum toxin, which blocks acetylcholine release
DTetanus toxin, which blocks inhibitory neurotransmitters
7.PharmacologyMedium
After instilling eye drops, why should the patient press gently on the inner corner of the eye for a minute or two?
ATo spread the drop evenly over the eyelid
BTo reduce drainage into the nose and absorption into blood
CTo make the pupil dilate faster
DTo stop the eye from producing tears for about an hour
8.PharmacologyHard
Why is intravenous insulin given with glucose in severe hyperkalaemia?
AInsulin moves potassium from the blood into cells
BGlucose binds potassium in the blood
CInsulin protects the heart muscle directly from potassium
DInsulin increases potassium excretion by the kidney
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 22 mmHg (2.9 kPa). Use the rule that in chronic respiratory alkalosis HCO₃⁻ falls by 5 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a respiratory acidosis: pH 7.23, PaCO₂ 66 mmHg (8.8 kPa), HCO₃⁻ 27 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 7.23PaCO₂ 66 mmHg (8.8 kPa)HCO₃⁻ 27 mmol/L
1acute
2chronic
3acute on chronic
11.ECG rhythm strip interpretationMedium
This lead II strip shows atrial flutter with a fixed conduction ratio. Find the ventricular rate in beats per minute.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
ventricular rate = bpm
12.ECG rhythm strip interpretationHard
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? 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
1asystole
2sinus tachycardia
3pulseless electrical activity (PEA)
4ventricular fibrillation
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 310 patients with a disease (cases) and 180 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 238b = 103c = 72d = 77
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A screening test has a sensitivity of 71% and a specificity of 82%. It is used in a population where the prevalence is 5%. Find the positive predictive value PPV as a percentage. Round to one decimal place.
Se = 71%Sp = 82%p = 5%
PPV = %
3Solutions
1.Human anatomy
Answer: B · Phrenic nerve
Why this is the answer
The phrenic nerve (roots C3–C5) is the motor supply to the diaphragm. High cervical cord injury above C3 therefore stops spontaneous breathing.
Answer
B · Phrenic nerve
2.Human anatomy
Answer: A · Anterior cruciate ligament
Why this is the answer
The anterior cruciate ligament limits anterior translation of the tibia; a positive anterior drawer or Lachman test suggests an ACL tear.
Answer
A · Anterior cruciate ligament
3.Human physiology
Answer: D · The impulse jumps between nodes of Ranvier
Why this is the answer
Myelin insulates the axon, so action potentials are only generated at the nodes of Ranvier. This saltatory conduction is much faster.
Answer
D · The impulse jumps between nodes of Ranvier
4.Human physiology
Answer: C · It becomes very high, acting as dead space
Why this is the answer
Ventilation without perfusion gives a very high V/Q ratio (dead space). A shunt, such as in pneumonia, is the opposite: perfusion without ventilation.
Answer
C · It becomes very high, acting as dead space
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: C · Botulinum toxin, which blocks acetylcholine release
Why this is the answer
Clostridium botulinum can grow in poorly canned, low-acid foods and produce botulinum toxin. The toxin blocks acetylcholine release at nerve endings, causing descending flaccid paralysis.
Answer
C · Botulinum toxin, which blocks acetylcholine release
7.Pharmacology
Answer: B · To reduce drainage into the nose and absorption into blood
Why this is the answer
Nasolacrimal occlusion keeps the drop in the eye longer and reduces systemic absorption through the nasal mucosa. This matters for drugs such as timolol, which can cause bradycardia or bronchospasm.
Answer
B · To reduce drainage into the nose and absorption into blood
8.Pharmacology
Answer: A · Insulin moves potassium from the blood into cells
Why this is the answer
Insulin activates the Na⁺/K⁺-ATPase and drives potassium into cells, lowering plasma levels within minutes. Glucose is given to prevent hypoglycaemia. The effect is temporary; potassium must still be removed from the body.
Answer
A · Insulin moves potassium from the blood into cells
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 15 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
|22 − 40| = 18 mmHg → 1.8 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 − 5·1.8 = 15 mmol/L
Answer
HCO₃⁻ = 15 mmol/L
10.Arterial blood gas interpretation
Answer: acute
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 + 1·2.6 = 26.6 mmol/L
chronic: 24 + 3.5·2.6 = 33.1 mmol/L
Compare
measured: 27 mmol/L → acute
Answer
acute
11.ECG rhythm strip interpretation
Answer: ventricular rate = 150 bpm
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.
Flutter waves
F–F = 5 small boxes = 0.20 s → 1500 ÷ 5 = 300 per minute
QRS complexes
R–R = 10 small boxes = 0.40 s → 1500 ÷ 10 = 150 per minute
conduction ratio = 300 ÷ 150 = 2:1
Answer
ventricular rate = 150 bpm
12.ECG rhythm strip interpretation
Answer: pulseless electrical activity (PEA)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 12 small boxes, 1500 ÷ 12 ≈ 125 per minute.
2. Rhythm: regular, every R–R is 0.48 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.06 s, narrow (below 0.12 s).
sinus tachycardia
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
pulseless electrical activity (PEA)
Answer
pulseless electrical activity (PEA)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
13.Evidence-based practice statistics
Answer: OR = 2.47
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.
Odds of exposure
cases: 238/72 = 3.306
controls: 103/77 = 1.338
Odds ratio
OR = (238·77)/(103·72) = 2.471 ≈ 2.47
Answer
OR = 2.47
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: PPV = 17.2 %
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 = 0.71·0.05 = 0.0355
(1 − Sp)·(1 − p) = 0.18·0.95 = 0.171
Positive predictive value
PPV = 0.0355/(0.0355 + 0.171) = 0.1719 ≈ 17.2%
Answer
PPV = 17.2%
The lower the prevalence, the lower the PPV of the same test.