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
Why can cardiac muscle not be tetanised like skeletal muscle?
- AIts long refractory period lasts almost as long as the contraction
- BIt is not striated
- CIt has no calcium in the sarcoplasmic reticulum
- DIt is controlled only by hormones
4.Human physiologyHard
In Graves' disease, what thyroid function results are expected and why?
- ALow T4 and high TSH, because antibodies destroy the gland
- BHigh T4 and low TSH, because antibodies stimulate the TSH receptor
- CLow T4 and low TSH, because the hypothalamus fails
- DHigh T4 and high TSH, because the anterior pituitary gland is overactive
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 = 0.96 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): 3 P waves for every 2 QRS complexes.
3.Human physiology
Answer: A · Its long refractory period lasts almost as long as the contraction
Why this is the answer
The plateau of the cardiac action potential gives a long refractory period, so the muscle relaxes before it can be excited again. This allows the heart to refill.
A · Its long refractory period lasts almost as long as the contraction
4.Human physiology
Answer: B · High T4 and low TSH, because antibodies stimulate the TSH receptor
Why this is the answer
TSH-receptor antibodies stimulate the thyroid independently of TSH. The resulting high T4 and T3 suppress pituitary TSH by negative feedback.
B · High T4 and low TSH, because antibodies stimulate the TSH receptor