ECG Interpretation: cram pack

Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.

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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

  1. 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.
  2. Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
  3. 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.
  4. 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.
  5. 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
II25 mm/s · 10 mm/mV

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
II25 mm/s · 10 mm/mV

This lead II strip shows an AV block with an irregular ventricular rhythm. Use the 6-second method to find the ventricular rate in beats per minute.

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
ventricular rate = bpm
3.Human physiologyMedium

Cardiac output equals heart rate multiplied by:

  1. APeripheral resistance
  2. BStroke volume
  3. CEjection fraction
  4. DBlood pressure
4.Human physiologyHard

In pernicious anaemia, autoimmune destruction of gastric parietal cells leads to anaemia because:

  1. APepsin is lost, so haemoglobin cannot be made
  2. BAcid is lost, so iron is never absorbed in any form
  3. CGastrin is lost, so the bone marrow is no longer stimulated
  4. DIntrinsic factor is lost, so vitamin B12 is not absorbed

3Solutions

1.ECG rhythm strip interpretation

Answer: QTc = 0.388 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 = 20 × 0.04 = 0.80 s

Fridericia’s formula

QTc = 0.36/∛0.80 = 0.36/0.9283 = 0.388 s

Answer

QTc = 0.388 s

Within the usual limits (about 0.45 s for men, 0.46 s for women). (388 ms)

2.ECG rhythm strip interpretation

Answer: ventricular rate = 40 bpm

AV blocks: march out the P waves and check which ones are followed by a QRS. First-degree: every P conducted, PR constant and above 0.20 s. Second-degree type I (Mobitz I, Wenckebach): the PR lengthens from beat to beat until one P wave is not conducted, then the cycle starts again (group beating). Second-degree type II (Mobitz II): the PR of the conducted beats stays constant, and P waves are suddenly not conducted; the QRS is often wide. Third-degree (complete) block: P waves and QRS complexes are each regular but independent, the PR varies at random and the atrial rate is faster than the escape rate (junctional 40–60/min with a narrow QRS, ventricular 20–40/min with a wide QRS). A 2:1 block cannot be called type I or type II from one strip.

Atrial rate

P–P = 25 small boxes = 1.00 s → 1500 ÷ 25 = 60 ≈ 60 per minute

Ventricular rate

4 QRS complexes in 6 s × 10 = 40 per minute

Answer

ventricular rate = 40 bpm

Answer

ventricular rate = 40 bpm

Second-degree type I (Wenckebach): 3 P waves for every 2 QRS complexes.

3.Human physiology

Answer: B · Stroke volume

Why this is the answer

CO = HR × SV. At rest about 70 beats/min × 70 mL ≈ 5 L/min.

Answer

B · Stroke volume
4.Human physiology

Answer: D · Intrinsic factor is lost, so vitamin B12 is not absorbed

Why this is the answer

Parietal cells secrete intrinsic factor, needed for B12 absorption in the terminal ileum. B12 deficiency impairs DNA synthesis and causes megaloblastic anaemia.

Answer

D · Intrinsic factor is lost, so vitamin B12 is not absorbed

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