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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
A monitored patient is found unresponsive and not breathing normally, and no carotid pulse can be felt. The monitor shows this organised rhythm. What is this situation called? 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
1asystole
2pulseless electrical activity (PEA)
3ventricular fibrillation
4sinus bradycardia
3.Human physiologyMedium
Low plasma albumin, as in severe liver disease, causes oedema mainly because:
ACapillary hydrostatic pressure falls
BLymphatic drainage increases
CPlasma oncotic pressure falls
DThe capillaries become impermeable
4.Human physiologyHard
Why can a patient with carbon monoxide poisoning have a normal PaO₂ but severe tissue hypoxia?
ACO lowers the barometric pressure in the alveoli
BCO occupies haemoglobin binding sites, so oxygen content falls
CCO stops oxygen from dissolving in plasma
DCO blocks diffusion across the alveolar wall
3Solutions
1.ECG rhythm strip interpretation
Answer: QTc = 0.451 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 = 11.5 × 0.04 = 0.46 s
RR = 26 × 0.04 = 1.04 s
Bazett’s formula
QTc = 0.46/√1.04 = 0.46/1.02 = 0.451 s
Answer
QTc = 0.451 s
Between the usual limits for men (about 0.45 s) and women (about 0.46 s): borderline, depending on the reference used. (451 ms)
2.ECG rhythm strip interpretation
Answer: pulseless electrical activity (PEA)
Pulseless electrical activity (PEA): PEA cannot be seen on a strip. The monitor shows organised electrical activity (a rhythm other than VF, VT or asystole) but the patient has no palpable pulse, so it is recognised by checking the patient. In adult resuscitation guidelines PEA and asystole are the non-shockable cardiac arrest rhythms: high-quality CPR, epinephrine as soon as possible and a search for reversible causes (the Hs and Ts, such as hypovolaemia, hypoxia, tension pneumothorax and cardiac tamponade). VF and pulseless VT are the shockable rhythms.
What the strip shows
1. Rate: R–R = 36 small boxes, 1500 ÷ 36 ≈ 42 per minute.
2. Rhythm: regular, every R–R is 1.44 s.
3. P waves: upright, one before every QRS, all alike.
4. PR interval: 0.18 s, constant, within 0.12–0.20 s.
5. QRS: 0.10 s, narrow (below 0.12 s).
sinus bradycardia
What the patient shows
no pulse despite organised electrical activity: pulseless electrical activity (PEA), a non-shockable cardiac arrest rhythm
Answer
pulseless electrical activity (PEA)
Answer
pulseless electrical activity (PEA)
PEA is recognised at the bedside, never from the strip alone. This is exam practice, not clinical guidance.
3.Human physiology
Answer: C · Plasma oncotic pressure falls
Why this is the answer
Albumin generates most of the plasma oncotic pressure that pulls fluid back into capillaries. When it falls, more fluid stays in the tissues.
Answer
C · Plasma oncotic pressure falls
4.Human physiology
Answer: B · CO occupies haemoglobin binding sites, so oxygen content falls
Why this is the answer
CO binds haemoglobin far more strongly than oxygen and shifts the curve left. Dissolved O₂ (PaO₂) is normal, but total oxygen content and release to tissues drop sharply.
Answer
B · CO occupies haemoglobin binding sites, so oxygen content falls