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1Formulas and methods
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.
2Practice problems
1.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
1normal sinus rhythm
2sinus rhythm with sinus arrest
3second-degree AV block type II
4sinus bradycardia
2.ECG rhythm strip interpretationHard
This lead II strip shows an AV block with an irregular ventricular rhythm. Use the 6-second method to 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
3.Human physiologyMedium
What is the main stimulus for breathing in a healthy person?
AA rise in arterial CO₂
BA rise in blood glucose
CA fall in arterial O₂
DA fall in blood pH caused by lactate only
4.Human physiologyHard
Why can a patient with carbon monoxide poisoning have a normal PaO₂ but severe tissue hypoxia?
ACO lowers the barometric pressure in the alveoli
BCO occupies haemoglobin binding sites, so oxygen content falls
CCO stops oxygen from dissolving in plasma
DCO blocks diffusion across the alveolar wall
3Solutions
1.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: 6 QRS complexes in 6 s, × 10 = 60 per minute.
2. Rhythm: regular (R–R 0.88 s) except for one pause of 2.00 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
2.ECG rhythm strip interpretation
Answer: ventricular rate = 50 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 = 22 small boxes = 0.88 s → 1500 ÷ 22 = 68.18 ≈ 68 per minute
Ventricular rate
5 QRS complexes in 6 s × 10 = 50 per minute
Answer
ventricular rate = 50 bpm
Answer
ventricular rate = 50 bpm
Second-degree type I (Wenckebach): 4 P waves for every 3 QRS complexes.
3.Human physiology
Answer: A · 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
A · A rise in arterial CO₂
4.Human physiology
Answer: B · CO occupies haemoglobin binding sites, so oxygen content falls
Why this is the answer
CO binds haemoglobin far more strongly than oxygen and shifts the curve left. Dissolved O₂ (PaO₂) is normal, but total oxygen content and release to tissues drop sharply.
Answer
B · CO occupies haemoglobin binding sites, so oxygen content falls