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 superficial vein in the front of the elbow is most often used for venepuncture?
AGreat saphenous vein
BAxillary vein
CBrachial vein
DMedian cubital vein
2.Human anatomyHard
An indirect inguinal hernia enters the inguinal canal at which point?
AThrough the superficial ring only, without entering the canal
BBelow the inguinal ligament, through the femoral canal
CThrough the posterior wall of the canal, medial to the inferior epigastric vessels
DThrough the deep inguinal ring, lateral to the inferior epigastric vessels
3.Human physiologyMedium
Which receptors respond mainly to a fall in arterial oxygen levels?
ACentral chemoreceptors in the medulla
BStretch receptors in the lungs
CBaroreceptors in the carotid sinus
DPeripheral chemoreceptors in the carotid and aortic bodies
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 stays normal because ventilation falls too
BIt becomes low, like in pneumonia
CIt becomes zero, acting as a shunt
DIt becomes very high, acting as dead space
5.MicrobiologyMedium
Which bacterium is the most common cause of stomach ulcers?
AStaphylococcus aureus
BClostridioides difficile
CEscherichia coli
DHelicobacter pylori
6.MicrobiologyHard
A deep, dirty wound becomes swollen and very painful, and gas can be felt under the skin. Which organism is most likely?
ABordetella pertussis
BNeisseria gonorrhoeae
CStreptococcus pneumoniae
DClostridium perfringens
7.PharmacologyMedium
A patient is starting lisinopril. Which rare but serious adverse effect should the nurse teach the patient to report immediately?
AA harmless rash on the palms of the hands
BSwelling of the lips, tongue or face
CDarker skin after exposure to sunlight
DAn increase in appetite in the evening
8.PharmacologyHard
Which finding most favours serotonin syndrome over neuroleptic malignant syndrome?
AA very high creatine kinase
BSlow onset over days with lead-pipe rigidity
CRecent start of an antipsychotic drug
DRapid onset with hyperreflexia and clonus
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⁺ 139, Cl⁻ 109, HCO₃⁻ 19 mmol/L and albumin 37 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).
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 rhythm with sinus arrest
2normal sinus rhythm
3second-degree AV block type II
4sinus arrhythmia
12.ECG rhythm strip interpretationHard
Interpret this lead II strip with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1rate
2rhythm
3P waves
4PR interval
5QRS duration
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). In a randomised trial, 880 patients received a new treatment and 540 received the control treatment. The table shows how many in each group had the outcome (an unwanted event such as a stroke). Find the relative risk reduction RRR as a percentage. Round to one decimal place.
treatment: 25/880control: 38/540
RRR = %
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 15% for a disease. A test with sensitivity 65% and specificity 93% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 15%Se = 65%Sp = 93%negative
p′ = %
3Solutions
1.Human anatomy
Answer: D · Median cubital vein
Why this is the answer
The median cubital vein crosses the cubital fossa between the cephalic and basilic veins. It is superficial, large and fairly fixed, so it is easy to puncture.
Answer
D · Median cubital vein
2.Human anatomy
Answer: D · Through the deep inguinal ring, lateral to the inferior epigastric vessels
Why this is the answer
An indirect hernia passes through the deep ring, lateral to the inferior epigastric vessels, and runs along the canal. A direct hernia pushes through the posterior wall medial to them.
Answer
D · Through the deep inguinal ring, lateral to the inferior epigastric vessels
3.Human physiology
Answer: D · 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
D · Peripheral chemoreceptors in the carotid and aortic bodies
4.Human physiology
Answer: D · 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
D · It becomes very high, acting as dead space
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: D · Clostridium perfringens
Why this is the answer
Clostridium perfringens causes gas gangrene in poorly oxygenated wounds. Its toxins destroy muscle and its metabolism produces gas, felt as crepitus. Urgent surgery and antibiotics are needed.
Answer
D · Clostridium perfringens
7.Pharmacology
Answer: B · Swelling of the lips, tongue or face
Why this is the answer
ACE inhibitors can cause angioedema, which can obstruct the airway and needs emergency care. It is linked to raised bradykinin levels and can occur even after long use.
Answer
B · Swelling of the lips, tongue or face
8.Pharmacology
Answer: D · Rapid onset with hyperreflexia and clonus
Why this is the answer
Serotonin syndrome develops within hours of a serotonergic drug and causes clonus, hyperreflexia, agitation, dilated pupils and diarrhoea. NMS develops over days after a dopamine antagonist, with lead-pipe rigidity and slow reflexes.
Answer
D · Rapid onset with hyperreflexia and clonus
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 = 11.75 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 = 139 − (109 + 19) = 11
Correction for albumin
11 + 0.25·(40 − 37) = 11.75 mmol/L
Answer
AGc = 11.75 mmol/L
Still within about 8–12 after correction.
11.ECG rhythm strip interpretation
Answer: sinus rhythm with sinus arrest
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. 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).
Five steps
1. Rate: 7 QRS complexes in 6 s, × 10 = 70 per minute.
2. Rhythm: regular (R–R 0.72 s) except for one pause of 1.72 s, which is not a whole number of P–P cycles.
3. P waves: upright, one before every QRS; no P wave at all during the pause.
4. PR interval: 0.14 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
Interpretation
sinus rhythm with sinus arrest
Answer
sinus rhythm with sinus arrest
12.ECG rhythm strip interpretation
Answer: P waves
Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.
Five steps
1. Rate: R–R = 23 small boxes, 1500 ÷ 23 ≈ 65 per minute.
2. Rhythm: regular, every R–R is 0.92 s.
3. P waves: inverted, just after each QRS.
4. PR interval: not measurable.
5. QRS: 0.10 s, narrow (below 0.12 s).
First abnormal step
P waves → accelerated junctional rhythm
Answer
P waves
13.Evidence-based practice statistics
Answer: RRR = 59.6 %
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 = 25/880 = 0.02841
CER = 38/540 = 0.07037
Relative risk reduction
ARR = 0.07037 − 0.02841 = 0.04196
RRR = 0.04196/0.07037 = 0.5963 ≈ 59.6%
Answer
RRR = 59.6%
14.Evidence-based practice statistics
Answer: p′ = 6.2 %
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.