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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 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
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
What is the main effect of aldosterone on the kidney?
AIncreased filtration in the glomerulus
BGlucose reabsorption in the proximal tubule
CSodium reabsorption and potassium secretion in the distal nephron
DWater reabsorption by inserting aquaporins only
4.Human physiologyHard
Why does breathing 100% oxygen correct hypoxaemia from a true right-to-left shunt only poorly?
AShunted blood is already fully saturated
BThe extra oxygen is all exhaled as CO₂
CThe shunted blood never comes into contact with alveolar gas
DPure oxygen damages haemoglobin
3Solutions
1.ECG rhythm strip interpretation
Answer: QT = 0.34 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
8.5 small boxes from the start of the QRS to the end of the T wave
Convert
QT = 8.5 × 0.04 s = 0.34 s
Answer
QT = 0.34 s
A rough rule: at normal rates the QT should be less than half the R–R interval (here R–R = 0.68 s).
2.ECG rhythm strip interpretation
Answer: rate
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 = 12 small boxes, 1500 ÷ 12 ≈ 125 per minute.
2. Rhythm: regular, every R–R is 0.48 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.12 s, constant, within 0.12–0.20 s.
5. QRS: 0.08 s, narrow (below 0.12 s).
First abnormal step
rate → sinus tachycardia
Answer
rate
3.Human physiology
Answer: C · Sodium reabsorption and potassium secretion in the distal nephron
Why this is the answer
Aldosterone acts on the distal tubule and collecting duct to increase Na⁺ reabsorption and K⁺ secretion. Water follows the sodium, raising blood volume.
Answer
C · Sodium reabsorption and potassium secretion in the distal nephron
4.Human physiology
Answer: C · The shunted blood never comes into contact with alveolar gas
Why this is the answer
Shunted blood bypasses ventilated alveoli and mixes deoxygenated blood into arterial blood. Extra inspired oxygen cannot reach it, unlike in V/Q mismatch or hypoventilation.
Answer
C · The shunted blood never comes into contact with alveolar gas