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 two regions of the adult spine normally show a lordosis (curve concave towards the back)?
ACervical and lumbar
BThoracic and lumbar
CThoracic and sacral
DCervical and thoracic
2.Human anatomyHard
Why does a displaced fracture of the femoral neck carry a high risk of avascular necrosis of the femoral head?
AIt damages the sciatic nerve that controls blood flow
BIt tears the retinacular branches of the medial circumflex femoral artery
CIt cuts the main trunk of the femoral artery
DIt blocks the femoral vein and causes congestion
3.Human physiologyMedium
Which receptors respond mainly to a fall in arterial oxygen levels?
ABaroreceptors in the carotid sinus
BPeripheral chemoreceptors in the carotid and aortic bodies
CCentral chemoreceptors in the medulla
DStretch receptors in the lungs
4.Human physiologyHard
What is the main reason the ventricles do not contract at the same moment as the atria?
AThe Purkinje fibres conduct slowly
BConduction is delayed at the atrioventricular node
CThe ventricles have a longer refractory period only
DThe SA node fires twice per beat
5.MicrobiologyMedium
Which white blood cells are the main phagocytes in an acute bacterial infection and the main component of pus?
ANeutrophils
BEosinophils
CB lymphocytes
DBasophils
6.MicrobiologyHard
Which mechanism of antibiotic resistance does a bacterium use when it pumps the drug out of the cell?
AEnzymatic inactivation
BTarget modification
CEfflux pumps
DReduced porin uptake only
7.PharmacologyMedium
A child weighing 20 kg is prescribed a drug at 15 mg/kg per dose. What is the dose?
A150 mg
B300 mg
C35 mg
D3000 mg
8.PharmacologyHard
A patient taking metoprolol is also started on verapamil. Which adverse effect is most likely?
ASevere bradycardia and heart block
BHyperkalaemia and muscle weakness
CHypoglycaemia and weight gain
DRebound tachycardia and hypertension
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 10 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 10 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
Practice values: Na⁺ 143, Cl⁻ 106, HCO₃⁻ 19 mmol/L and albumin 35 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
Measure the QT interval on this lead II strip, in seconds.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QT = 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, 280 patients received a new treatment and 350 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: 13/280control: 42/350
NNT =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 82% and specificity 78%. The prevalence of the disease is 24%. How many people with the disease are missed by the test (false negatives, FN)?
N = 10 000p = 24%Se = 82%Sp = 78%
FN =
3Solutions
1.Human anatomy
Answer: A · Cervical and lumbar
Why this is the answer
The cervical and lumbar regions are lordotic (secondary curves). The thoracic and sacral regions are kyphotic (primary curves).
Answer
A · Cervical and lumbar
2.Human anatomy
Answer: B · It tears the retinacular branches of the medial circumflex femoral artery
Why this is the answer
The femoral head is supplied mainly by retinacular branches of the medial circumflex femoral artery running along the neck. A displaced neck fracture tears them and the head can die.
Answer
B · It tears the retinacular branches of the medial circumflex femoral artery
3.Human physiology
Answer: B · 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
B · Peripheral chemoreceptors in the carotid and aortic bodies
4.Human physiology
Answer: B · Conduction is delayed at the atrioventricular node
Why this is the answer
The AV node conducts slowly (about 0.1 s delay), so atrial contraction completes ventricular filling before ventricular depolarisation begins.
Answer
B · Conduction is delayed at the atrioventricular node
5.Microbiology
Answer: A · Neutrophils
Why this is the answer
Neutrophils arrive quickly at sites of infection and engulf bacteria. Dead neutrophils, bacteria and tissue debris form pus.
Answer
A · Neutrophils
6.Microbiology
Answer: C · Efflux pumps
Why this is the answer
Efflux pumps actively export antibiotics such as tetracyclines and fluoroquinolones, keeping the inside concentration too low to work.
Answer
C · Efflux pumps
7.Pharmacology
Answer: B · 300 mg
Why this is the answer
20 kg × 15 mg/kg = 300 mg. The result is also checked against the safe dose range and the adult maximum.
Answer
B · 300 mg
8.Pharmacology
Answer: A · Severe bradycardia and heart block
Why this is the answer
Beta blockers and non-dihydropyridine calcium channel blockers (verapamil, diltiazem) both slow heart rate and AV conduction. Together they can cause severe bradycardia, heart block and heart failure.
Answer
A · Severe bradycardia and heart block
9.Arterial blood gas interpretation
Answer: PaCO₂ = 23 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.
Winter’s formula
PaCO₂ = 1.5·10 + 8 = 23 mmHg
range: 21–25 mmHg
Answer
PaCO₂ = 23 mmHg
10.Arterial blood gas interpretation
Answer: AGc = 19.25 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.
Anion gap
AG = 143 − (106 + 19) = 18
Correction for albumin
18 + 0.25·(40 − 35) = 19.25 mmol/L
Answer
AGc = 19.25 mmol/L
Above about 12 after correction: a high anion gap.
11.ECG rhythm strip interpretation
Answer: QT = 0.42 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
10.5 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 10.5 × 0.04 s = 0.42 s
Answer
QT = 0.42 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.80 s).
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 = 42 small boxes = 1.68 s → 1500 ÷ 42 = 35.71 ≈ 36 per minute
Answer
atrial rate = 79 bpm
Answer
atrial rate = 79 bpm
Third-degree block with a wide-complex ventricular escape (usually 20–40 per minute).
13.Evidence-based practice statistics
Answer: NNT = 14
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 = 13/280 = 0.04643
CER = 42/350 = 0.12
Number needed to treat
ARR = 0.12 − 0.04643 = 0.07357
NNT = 1/0.07357 = 13.59 → 14
Answer
NNT = 14
Treating 14 patients prevents about one event.
14.Evidence-based practice statistics
Answer: FN = 432
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.