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
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
2.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
3.Human physiologyMedium
Cardiac output equals heart rate multiplied by:
- APeripheral resistance
- BStroke volume
- CEjection fraction
- DBlood pressure
4.Human physiologyHard
Why does breathing 100% oxygen correct hypoxaemia from a true right-to-left shunt only poorly?
- AThe shunted blood never comes into contact with alveolar gas
- BShunted blood is already fully saturated
- CPure oxygen damages haemoglobin
- DThe extra oxygen is all exhaled as CO₂
3Solutions
1.ECG rhythm strip interpretation
Answer: QT = 0.48 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
12 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 12 × 0.04 s = 0.48 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 1.00 s).
2.ECG rhythm strip interpretation
Answer: atrial rate = 100 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 = 15 small boxes = 0.60 s → 1500 ÷ 15 = 100 ≈ 100 per minute
Ventricular rate
R–R = 52 small boxes = 2.08 s → 1500 ÷ 52 = 28.85 ≈ 29 per minute
Third-degree block with a wide-complex ventricular escape (usually 20–40 per minute).
3.Human physiology
Answer: B · Stroke volume
Why this is the answer
CO = HR × SV. At rest about 70 beats/min × 70 mL ≈ 5 L/min.
4.Human physiology
Answer: A · 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.
A · The shunted blood never comes into contact with alveolar gas