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 valve lies between the left atrium and the left ventricle?
AMitral (bicuspid) valve
BPulmonary valve
CAortic valve
DTricuspid valve
2.Human anatomyHard
What is the correct order of structures that urine passes through after leaving the collecting ducts?
AMajor calyx, minor calyx, renal pelvis, ureter
BMinor calyx, major calyx, renal pelvis, ureter
CMinor calyx, renal pelvis, major calyx, ureter
DRenal pelvis, minor calyx, major calyx, ureter
3.Human physiologyMedium
What is the main stimulus for breathing in a healthy person?
AA fall in blood pH caused by lactate only
BA fall in arterial O₂
CA rise in blood glucose
DA rise in arterial CO₂
4.Human physiologyHard
Why does breathing 100% oxygen correct hypoxaemia from a true right-to-left shunt only poorly?
AThe extra oxygen is all exhaled as CO₂
BThe shunted blood never comes into contact with alveolar gas
CPure oxygen damages haemoglobin
DShunted blood is already fully saturated
5.MicrobiologyMedium
Which bacterium is the most common cause of stomach ulcers?
AEscherichia coli
BStaphylococcus aureus
CClostridioides difficile
DHelicobacter pylori
6.MicrobiologyHard
How do high-risk HPV types contribute to cervical cancer?
AViral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
BThe virus kills cells so quickly that they are replaced by cancer cells
CThe virus produces an endotoxin that damages DNA
DThe virus destroys CD4 T cells in the cervix
7.PharmacologyMedium
A patient with Addison's disease on hydrocortisone develops a fever with flu. What is the usual advice?
AHalve the hydrocortisone to reduce side effects
BIncrease the dose as instructed (sick-day rules) and seek help if vomiting
CMake no change, as illness does not affect cortisol needs
DStop the hydrocortisone until the fever settles, to let the body recover
8.PharmacologyHard
A patient taking warfarin asks for something for a headache. Which drug is usually the safest choice?
AIbuprofen, at the lowest dose
BParacetamol (acetaminophen)
CNaproxen
DAspirin
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 6 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 6 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
A practice gas shows a metabolic alkalosis with HCO₃⁻ 45 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.
HCO₃⁻ 45 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
1atrial fibrillation
2sinus rhythm with PVCs
3ventricular bigeminy
4normal sinus rhythm
12.ECG rhythm strip interpretationHard
This lead II strip shows a conduction problem between the atria and the ventricles. 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
1second-degree AV block type II
2third-degree (complete) AV block
3second-degree AV block type I (Wenckebach)
4atrial fibrillation
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A test has a sensitivity of 78% and a specificity of 83%. Find the positive likelihood ratio LR+. Round to two decimal places.
Se = 78%Sp = 83%
LR+ =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). 10 000 people are screened with a test of sensitivity 72% and specificity 81%. The prevalence of the disease is 14%. How many false-positive results FP are expected?
N = 10 000p = 14%Se = 72%Sp = 81%
FP =
3Solutions
1.Human anatomy
Answer: A · Mitral (bicuspid) valve
Why this is the answer
The mitral (bicuspid) valve separates the left atrium and left ventricle; the tricuspid valve does the same on the right side.
Answer
A · Mitral (bicuspid) valve
2.Human anatomy
Answer: B · Minor calyx, major calyx, renal pelvis, ureter
Why this is the answer
Collecting ducts open at the renal papillae into minor calyces, which join to form major calyces, which empty into the renal pelvis and then the ureter.
Answer
B · Minor calyx, major calyx, renal pelvis, ureter
3.Human physiology
Answer: D · A rise in arterial CO₂
Why this is the answer
Central chemoreceptors respond to CO₂ (through the pH of the cerebrospinal fluid). Low O₂ adds strongly to the drive only when PaO₂ falls well below normal, for example at high altitude.
Answer
D · A rise in arterial CO₂
4.Human physiology
Answer: B · The shunted blood never comes into contact with alveolar gas
Why this is the answer
Shunted blood bypasses ventilated alveoli and mixes deoxygenated blood into arterial blood. Extra inspired oxygen cannot reach it, unlike in V/Q mismatch or hypoventilation.
Answer
B · The shunted blood never comes into contact with alveolar gas
5.Microbiology
Answer: D · Helicobacter pylori
Why this is the answer
H. pylori colonises the stomach lining and produces urease to neutralise acid locally. It causes most peptic ulcers and is treated with antibiotics plus a PPI.
Answer
D · Helicobacter pylori
6.Microbiology
Answer: A · Viral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
Why this is the answer
E6 promotes the breakdown of p53 and E7 binds Rb, removing controls on cell division and DNA repair. Infected cells can then accumulate mutations and become cancerous over years.
Answer
A · Viral E6 and E7 proteins inactivate the tumour suppressors p53 and Rb
7.Pharmacology
Answer: B · Increase the dose as instructed (sick-day rules) and seek help if vomiting
Why this is the answer
During illness, the body needs more cortisol, so patients usually double their glucocorticoid dose as instructed. If they cannot keep tablets down, they need injectable hydrocortisone and urgent care.
Answer
B · Increase the dose as instructed (sick-day rules) and seek help if vomiting
8.Pharmacology
Answer: B · Paracetamol (acetaminophen)
Why this is the answer
NSAIDs and aspirin impair platelet function and can cause GI bleeding, which adds to warfarin's bleeding risk. Paracetamol is usually preferred, though regular high doses can raise the INR.
Answer
B · Paracetamol (acetaminophen)
9.Arterial blood gas interpretation
Answer: PaCO₂ = 17 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 = 1.5·6 + 8 = 17 mmHg
range: 15–19 mmHg
Answer
PaCO₂ = 17 mmHg
10.Arterial blood gas interpretation
Answer: PaCO₂ = 52.5 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
0.7·45 + 21 = 52.5 mmHg
Answer
PaCO₂ = 52.5 mmHg
The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.
11.ECG rhythm strip interpretation
Answer: ventricular bigeminy
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: regularly irregular, a short and a long R–R alternate (0.46 s and 1.30 s).
3. P waves: upright before every narrow QRS; none before the early wide beats.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s in the sinus beats; the early beats are wide (0.16 s) and bizarre, with the T wave pointing the other way.
Interpretation
ventricular bigeminy
Answer
ventricular bigeminy
12.ECG rhythm strip interpretation
Answer: second-degree AV block type I (Wenckebach)
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.
Five steps
1. Rate: 4 QRS complexes in 6 s, × 10 = 40 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 0.92 s, 65 per minute); one P in every 3 is not followed by a QRS.
4. PR interval: lengthens 0.20 → 0.34 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
second-degree AV block type I (Wenckebach)
Answer
second-degree AV block type I (Wenckebach)
13.Evidence-based practice statistics
Answer: LR+ = 4.59
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.
Positive likelihood ratio
LR+ = 0.78/(1 − 0.83) = 4.588 ≈ 4.59
Answer
LR+ = 4.59
14.Evidence-based practice statistics
Answer: FP = 1634
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.
With and without the disease
10 000·0.14 = 1400
10 000 − 1400 = 8600
False positives
FP = 8600·0.19 = 1634
Answer
FP = 1634
Of the 2642 positives only 1008 have the disease (PPV 38.2%).