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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 Fridericia’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
How is the ejection fraction calculated?
AStroke volume ÷ end-diastolic volume
BEnd-systolic volume ÷ stroke volume
CCardiac output ÷ heart rate
DEnd-diastolic volume ÷ end-systolic volume
4.Human physiologyHard
After a head injury, a patient passes large volumes of very dilute urine and plasma sodium rises. What is the most likely cause?
AExcess ADH release (SIADH)
BUncontrolled diabetes mellitus
CLack of ADH release (cranial diabetes insipidus)
DExcess aldosterone
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.403 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 = 8.5 × 0.04 = 0.34 s
RR = 15 × 0.04 = 0.60 s
Fridericia’s formula
QTc = 0.34/∛0.60 = 0.34/0.8434 = 0.403 s
Answer
QTc = 0.403 s
Within the usual limits (about 0.45 s for men, 0.46 s for women). (403 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 = 19 small boxes, 1500 ÷ 19 ≈ 79 per minute.
2. Rhythm: regular, every R–R is 0.76 s.
3. P waves: none visible.
4. PR interval: not measurable.
5. QRS: 0.06 s, narrow (below 0.12 s).
First abnormal step
P waves → accelerated junctional rhythm
Answer
P waves
3.Human physiology
Answer: A · Stroke volume ÷ end-diastolic volume
Why this is the answer
Ejection fraction is the proportion of blood in the full ventricle that is ejected per beat. A normal value is about 55–70%.
Answer
A · Stroke volume ÷ end-diastolic volume
4.Human physiology
Answer: C · Lack of ADH release (cranial diabetes insipidus)
Why this is the answer
Without ADH the collecting ducts stay impermeable to water, so dilute urine is produced and the patient loses free water, raising plasma sodium.
Answer
C · Lack of ADH release (cranial diabetes insipidus)