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 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
- BH zone
- CI band
- DA band
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 becomes very high, acting as dead space
- BIt becomes low, like in pneumonia
- CIt stays normal because ventilation falls too
- DIt becomes zero, acting as a shunt
3Solutions
1.ECG rhythm strip interpretation
Answer: QT = 0.38 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
9.5 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 9.5 × 0.04 s = 0.38 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: 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
ventricular rate = 50 bpm
Second-degree type I (Wenckebach): 4 P waves for every 3 QRS complexes.
3.Human physiology
Answer: D · 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 · 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.
A · It becomes very high, acting as dead space