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 Bazett’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, so the atria and the ventricles beat at different rates. The QRS complexes are regular: find the ventricular 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
ventricular rate = bpm
3.Human physiologyMedium
During contraction of a skeletal muscle sarcomere, which band stays the same length?
- AH zone
- BA band
- CI band
- DThe whole sarcomere
4.Human physiologyHard
In Otto Loewi's classic experiment, fluid from a vagus-stimulated frog heart slowed a second heart. What did this show?
- ANerve impulses spread through the fluid directly as electricity
- BThe heart beats without any nerve supply
- CNerves pass signals to their targets by releasing a chemical
- DThe vagus nerve speeds up the heart
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.451 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 = 11.5 × 0.04 = 0.46 s
RR = 26 × 0.04 = 1.04 s
Bazett’s formula
QTc = 0.46/√1.04 = 0.46/1.02 = 0.451 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (451 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 = 21 small boxes = 0.84 s → 1500 ÷ 21 = 71.43 ≈ 71 per minute
Ventricular rate
R–R = 30 small boxes = 1.20 s → 1500 ÷ 30 = 50 ≈ 50 per minute
Answer
ventricular rate = 50 bpm
ventricular rate = 50 bpm
Third-degree block with a narrow-complex junctional escape (usually 40–60 per minute).
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: C · Nerves pass signals to their targets by releasing a chemical
Why this is the answer
The fluid carried "Vagusstoff", later identified as acetylcholine. This was the first clear evidence for chemical neurotransmission.
C · Nerves pass signals to their targets by releasing a chemical