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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
Interpret this lead II strip. Which interpretation can you defend from this single strip? Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
12:1 AV block (type undetermined)
2second-degree AV block type II
3second-degree AV block type I (Wenckebach)
4sinus bradycardia
3.Human physiologyMedium
Which receptors respond mainly to a fall in arterial oxygen levels?
ACentral chemoreceptors in the medulla
BPeripheral chemoreceptors in the carotid and aortic bodies
CStretch receptors in the lungs
DBaroreceptors in the carotid sinus
4.Human physiologyHard
In long-standing aortic stenosis, how does the left ventricle adapt?
AAtrophy of the muscle due to less work
BConcentric hypertrophy with a thicker wall
CThinning of the wall with dilation
DNo change because the valve is not part of the ventricle
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.392 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 = 10 × 0.04 = 0.40 s
RR = 26 × 0.04 = 1.04 s
Bazett’s formula
QTc = 0.40/√1.04 = 0.40/1.02 = 0.392 s
Answer
QTc = 0.392 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (392 ms)
2.ECG rhythm strip interpretation
Answer: 2:1 AV block (type undetermined)
2:1 AV block: with two P waves for every QRS, no two conducted P waves follow each other, so you cannot see whether the PR would lengthen (type I) or stay constant (type II) before a non-conducted P wave. From one strip it is called a 2:1 AV block; a longer strip or a change in the conduction ratio decides the type. A narrow QRS makes type I more likely and a wide QRS type II, but neither proves it.
Five steps
1. Rate: R–R = 38 small boxes, 1500 ÷ 38 ≈ 39 per minute.
2. Rhythm: regular, every R–R is 1.52 s.
3. P waves: upright and regular (P–P 0.76 s, 79 per minute), two for every QRS.
4. PR interval: constant at 0.18 s in the conducted beats.
5. QRS: 0.14 s, wide (0.12 s or more).
Why the type stays open
Every second P wave is not conducted, so no two conducted beats follow each other and a lengthening PR could never show.
2:1 AV block (type undetermined)
Answer
2:1 AV block (type undetermined)
A narrow QRS makes type I more likely and a wide QRS type II, but only a longer strip or a change in the conduction ratio decides.
3.Human physiology
Answer: B · Peripheral chemoreceptors in the carotid and aortic bodies
Why this is the answer
The carotid and aortic bodies detect low PaO₂ (and also high CO₂ and H⁺). The central chemoreceptors respond to CO₂ but not directly to oxygen.
Answer
B · Peripheral chemoreceptors in the carotid and aortic bodies
4.Human physiology
Answer: B · Concentric hypertrophy with a thicker wall
Why this is the answer
Pressure overload makes the ventricular wall thicken (concentric hypertrophy) to keep wall stress normal. Volume overload, as in regurgitation, causes dilation instead.