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 and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Fridericia’s formula QTc = QT/∛RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
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
During contraction of a skeletal muscle sarcomere, which band stays the same length?
- AThe whole sarcomere
- BA band
- CI band
- DH zone
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
- BPure oxygen damages haemoglobin
- CThe extra oxygen is all exhaled as CO₂
- DShunted blood is already fully saturated
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.487 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
QT = 12 × 0.04 = 0.48 s
RR = 24 × 0.04 = 0.96 s
Fridericia’s formula
QTc = 0.48/∛0.96 = 0.48/0.9865 = 0.487 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (487 ms)
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 = 20 small boxes = 0.80 s → 1500 ÷ 20 = 75 ≈ 75 per minute
Ventricular rate
5 QRS complexes in 6 s × 10 = 50 per minute
Answer
ventricular rate = 50 bpm
ventricular rate = 50 bpm
Second-degree type II: 3 P waves for every 2 QRS complexes, with a constant PR.
3.Human physiology
Answer: B · A band
Why this is the answer
In the sliding filament model, filaments slide past each other without shortening. The A band (myosin length) stays the same while the I band and H zone shorten.
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