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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 with the five steps in order: rate, rhythm, P waves, PR interval, QRS duration. At which step do you find the first abnormal finding? Click it.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1rate
2rhythm
3P waves
4PR interval
5QRS duration
3.Human physiologyMedium
During contraction of a skeletal muscle sarcomere, which band stays the same length?
AI band
BThe whole sarcomere
CA band
DH zone
4.Human physiologyHard
During exercise, blood flow to active skeletal muscle rises greatly. What is the main cause?
AThe parasympathetic system dilates all arteries
BSympathetic stimulation constricts the muscle arterioles via alpha receptors
CThe heart pumps blood only to the muscles
DLocal metabolites such as CO₂, H⁺, K⁺ and adenosine dilate the arterioles
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.437 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 = 9 × 0.04 = 0.36 s
RR = 17 × 0.04 = 0.68 s
Bazett’s formula
QTc = 0.36/√0.68 = 0.36/0.8246 = 0.437 s
Answer
QTc = 0.437 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (437 ms)
2.ECG rhythm strip interpretation
Answer: PR interval
Five steps for every strip: 1. rate (300 or 1500 method, or the 6-second method if irregular); 2. rhythm: are the R–R intervals equal; 3. P waves: present, upright in lead II, one before every QRS and all alike; 4. PR interval: 0.12–0.20 s and constant; 5. QRS duration: below 0.12 s. Only after the five steps name the rhythm. Normal sinus rhythm passes all five: 60–100/min, regular, upright P before every QRS, PR 0.12–0.20 s, QRS below 0.12 s.
Five steps
1. Rate: R–R = 21 small boxes, 1500 ÷ 21 ≈ 71 per minute.
2. Rhythm: regular, every R–R is 0.84 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.34 s, constant, longer than 0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
PR interval → first-degree AV block
Answer
PR interval
3.Human physiology
Answer: C · 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.
Answer
C · A band
4.Human physiology
Answer: D · Local metabolites such as CO₂, H⁺, K⁺ and adenosine dilate the arterioles
Why this is the answer
Active hyperaemia: metabolites produced by working muscle relax arteriolar smooth muscle locally, overriding sympathetic vasoconstriction in that tissue.
Answer
D · Local metabolites such as CO₂, H⁺, K⁺ and adenosine dilate the arterioles