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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
Which ABO blood group has both anti-A and anti-B antibodies in the plasma?
AGroup B
BGroup O
CGroup A
DGroup AB
4.Human physiologyHard
A patient with small-cell lung cancer has low plasma sodium and low plasma osmolality but inappropriately concentrated urine. What is the most likely explanation?
ASyndrome of inappropriate ADH secretion
BCranial diabetes insipidus
CPrimary hyperaldosteronism
DAddison's disease with high aldosterone
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.448 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 = 9.5 × 0.04 = 0.38 s
RR = 18 × 0.04 = 0.72 s
Bazett’s formula
QTc = 0.38/√0.72 = 0.38/0.8485 = 0.448 s
Answer
QTc = 0.448 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (448 ms)
2.ECG rhythm strip interpretation
Answer: P waves
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 = 22 small boxes, 1500 ÷ 22 ≈ 68 per minute.
2. Rhythm: regular, every R–R is 0.88 s.
3. P waves: inverted, just after each QRS.
4. PR interval: not measurable.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
P waves → accelerated junctional rhythm
Answer
P waves
3.Human physiology
Answer: B · Group O
Why this is the answer
Group O people have no A or B antigens and make both anti-A and anti-B. Group AB has both antigens and neither antibody.
Answer
B · Group O
4.Human physiology
Answer: A · Syndrome of inappropriate ADH secretion
Why this is the answer
In SIADH excess ADH makes the kidneys retain water, diluting the plasma while urine stays concentrated. Small-cell lung cancer can secrete ADH.