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 part of the brainstem contains the main cardiovascular and respiratory centres?
APons
BMidbrain
CCerebellum
DMedulla oblongata
2.Human anatomyHard
Which coronary artery most often supplies the sinoatrial (SA) node?
ALeft anterior descending artery
BCircumflex artery
CLeft main coronary artery
DRight coronary artery
3.Human physiologyMedium
What is the normal pH range of arterial blood?
A7.50 to 7.65
B7.00 to 7.20
C6.80 to 7.00
D7.35 to 7.45
4.Human physiologyHard
A patient with diabetic ketoacidosis breathes deeply and rapidly. What pattern is expected on the blood gas?
ALow pH, high bicarbonate and high PaCO₂
BLow pH, low bicarbonate and low PaCO₂
CHigh pH, low bicarbonate and low PaCO₂
DLow pH, normal bicarbonate and high PaCO₂
5.MicrobiologyMedium
MRSA is resistant to methicillin because it has:
AAn efflux pump for every antibiotic
BAn altered penicillin-binding protein (PBP2a)
CNo cell wall at all
DA polysaccharide capsule that blocks the entry of all beta-lactams
6.MicrobiologyHard
Joseph Lister reduced infection after surgery in the 1860s. What was his main innovation?
AVaccinating surgeons against staphylococci
BUsing steam autoclaves to sterilise all surgical instruments before each operation
CGiving sulfonamide antibiotics after every operation
DUsing carbolic acid (phenol) as an antiseptic on wounds and instruments
7.PharmacologyMedium
Why should doxycycline not be taken at the same time as milk or antacids?
AThe combination causes severe constipation
BAntacids turn it into a toxic compound
CCalcium and other metal ions bind it and reduce absorption
DMilk speeds up its breakdown in the liver by inducing enzymes
8.PharmacologyHard
Why can an ACE inhibitor cause acute kidney injury in a patient with bilateral renal artery stenosis?
AThe drug crystallises in the renal tubules
BACE inhibitors constrict the afferent arteriole
CACE inhibitors are directly toxic to the proximal tubule cells in the kidney
DGFR depends on angiotensin II constricting the efferent arteriole
9.Arterial blood gas interpretationMedium
A practice gas shows PaCO₂ 54 mmHg (7.2 kPa). Use the rule that in chronic respiratory acidosis HCO₃⁻ rises by 3.5 mmol/L for every 10 mmHg that PaCO₂ is above 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.
A practice gas shows a metabolic alkalosis with HCO₃⁻ 46 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 46 mmol/L
PaCO₂ = mmHg
11.ECG rhythm strip interpretationMedium
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
1sinus arrhythmia
2junctional rhythm
3atrial fibrillation
4atrial flutter, 2:1 conduction
12.ECG rhythm strip interpretationHard
This patient has a pacemaker. Interpret the 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
1third-degree (complete) AV block
2paced rhythm, failure to capture
3dual-chamber (AV) paced rhythm
4idioventricular rhythm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 760 patients received a new treatment and 800 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the absolute risk reduction ARR as a percentage. Round to one decimal place.
treatment: 173/760control: 304/800
ARR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 50% for a disease. A test with sensitivity 81% and specificity 90% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 50%Se = 81%Sp = 90%negative
p′ = %
3Solutions
1.Human anatomy
Answer: D · Medulla oblongata
Why this is the answer
The medulla oblongata contains the vital centres that regulate heart rate, blood pressure and breathing rhythm.
Answer
D · Medulla oblongata
2.Human anatomy
Answer: D · Right coronary artery
Why this is the answer
The SA nodal artery arises from the right coronary artery in roughly 60% of people and from the circumflex artery in the rest.
Answer
D · Right coronary artery
3.Human physiology
Answer: D · 7.35 to 7.45
Why this is the answer
Arterial pH is normally 7.35–7.45. Below 7.35 is acidaemia and above 7.45 alkalaemia.
Answer: B · An altered penicillin-binding protein (PBP2a)
Why this is the answer
The mecA gene codes for PBP2a, which has low affinity for beta-lactams, so the bacterium can keep building its cell wall despite methicillin.
Answer
B · An altered penicillin-binding protein (PBP2a)
6.Microbiology
Answer: D · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
Why this is the answer
Lister applied carbolic acid to wounds, dressings and instruments, which greatly reduced post-operative infection. This started antiseptic surgery.
Answer
D · Using carbolic acid (phenol) as an antiseptic on wounds and instruments
7.Pharmacology
Answer: C · Calcium and other metal ions bind it and reduce absorption
Why this is the answer
Tetracyclines form complexes with calcium, magnesium, aluminium and iron in the gut, which reduces absorption. Doses are separated from these products.
Answer
C · Calcium and other metal ions bind it and reduce absorption
8.Pharmacology
Answer: D · GFR depends on angiotensin II constricting the efferent arteriole
Why this is the answer
When renal perfusion is low, angiotensin II constricts the efferent arteriole to maintain glomerular pressure. Blocking it lets the pressure and GFR fall, so kidney function can drop sharply.
Answer
D · GFR depends on angiotensin II constricting the efferent arteriole
9.Arterial blood gas interpretation
Answer: HCO₃⁻ = 28.9 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
|54 − 40| = 14 mmHg → 1.4 × 10 mmHg
Expected HCO₃⁻
HCO₃⁻ = 24 + 3.5·1.4 = 28.9 mmol/L
Answer
HCO₃⁻ = 28.9 mmol/L
10.Arterial blood gas interpretation
Answer: PaCO₂ = 53.2 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
0.7·46 + 21 = 53.2 mmHg
Answer
PaCO₂ = 53.2 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: atrial fibrillation
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: 7 QRS complexes in 6 s, × 10 = 70 per minute.
2. Rhythm: irregularly irregular, R–R from 0.64 to 1.02 s with no pattern.
3. P waves: none; the baseline shows small irregular fibrillatory waves.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
atrial fibrillation
Answer
atrial fibrillation
12.ECG rhythm strip interpretation
Answer: dual-chamber (AV) paced rhythm
Paced rhythms: a pacemaker impulse shows as a narrow vertical pacing spike. After a ventricular spike the QRS is wide (0.12 s or more), because the impulse spreads from the lead through the muscle instead of the conduction system; with right-ventricular pacing it is often mostly negative in lead II. A spike before each P wave means atrial pacing, and a dual-chamber (AV) pacemaker shows an atrial spike and a P wave, then, after the programmed AV delay, a ventricular spike and a wide QRS. Capture means every spike is followed by a depolarisation. Failure to capture is a spike without a P wave or QRS after it; failure to sense (spikes that ignore the patient’s own beats) and failure to pace (no spike when one is due) are the other classic malfunctions.
Five steps
1. Rate: R–R = 24 small boxes, 1500 ÷ 24 ≈ 63 per minute.
2. Rhythm: regular, every R–R is 0.96 s.
3. P waves: each P wave follows its own pacing spike (atrial pacing).
4. AV interval: the second spike comes 0.16 s after the first one (the programmed AV delay).
5. QRS: 0.16 s, wide, right after the second spike (ventricular pacing).
Interpretation
dual-chamber (AV) paced rhythm
Answer
dual-chamber (AV) paced rhythm
The paced QRS is wide by nature; do not call it a bundle branch block or a ventricular rhythm.
13.Evidence-based practice statistics
Answer: ARR = 15.2 %
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 = 173/760 = 0.2276
CER = 304/800 = 0.38
Absolute risk reduction
ARR = 0.38 − 0.2276 = 0.1524 ≈ 15.2%
Answer
ARR = 15.2%
14.Evidence-based practice statistics
Answer: p′ = 17.4 %
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.