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/mVatrial flutter, 2:1 conductionatrial fibrillationsinus tachycardiasupraventricular tachycardia (SVT)

Interpret this lead II rhythm strip. Which rhythm is it? 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
  1. 1atrial flutter, 2:1 conduction
  2. 2atrial fibrillation
  3. 3sinus tachycardia
  4. 4supraventricular tachycardia (SVT)
2.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVsecond-degree AV block type IIthird-degree (complete) AV blocksecond-degree AV block type I (Wenckebach)first-degree AV block

This lead II strip shows a conduction problem between the atria and the ventricles. Which rhythm is it? 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
  1. 1second-degree AV block type II
  2. 2third-degree (complete) AV block
  3. 3second-degree AV block type I (Wenckebach)
  4. 4first-degree AV block
3.Human physiologyMedium

The Frank–Starling mechanism describes that:

  1. AHeart rate rises when arterial blood pressure falls suddenly
  2. BThe atria contract after the ventricles
  3. CStroke volume rises when ventricular filling increases
  4. DContractility falls when temperature rises
4.Human physiologyHard

According to the length–tension relationship, why does a skeletal muscle fibre produce less force when overstretched?

  1. AActin and myosin overlap less, so fewer cross-bridges form
  2. BThe muscle runs out of ATP when stretched
  3. CCalcium is no longer released from the sarcoplasmic reticulum when stretched
  4. DThe neuromuscular junction stops working

3Solutions

1.ECG rhythm strip interpretation

Answer: sinus tachycardia

Rhythm criteria (lead II): sinus rhythm has an upright P wave before every QRS, a constant PR of 0.12–0.20 s and a narrow QRS: normal sinus rhythm at 60–100/min, sinus bradycardia below 60, sinus tachycardia above 100. Atrial fibrillation: no P waves, a fibrillatory baseline and an irregularly irregular R–R. Atrial flutter: sawtooth flutter waves at about 250–350/min with a fixed conduction ratio such as 2:1 or 4:1. First-degree AV block: sinus rhythm with a constant PR above 0.20 s. Junctional rhythm: narrow QRS at 40–60/min with absent or inverted P waves (PR below 0.12 s, or P after the QRS). A PVC is an early, wide (0.12 s or more), bizarre QRS without a P wave, with the T wave in the opposite direction and usually a full compensatory pause; in bigeminy every second beat is a PVC.

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).

Interpretation

sinus tachycardia

Answer

sinus tachycardia
2.ECG rhythm strip interpretation

Answer: second-degree AV block type I (Wenckebach)

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.

Five steps

1. Rate: 5 QRS complexes in 6 s, × 10 = 50 per minute.
2. Rhythm: regularly irregular, the QRS complexes come in groups separated by a pause (group beating).
3. P waves: upright and regular (P–P 0.88 s, 68 per minute); one P in every 4 is not followed by a QRS.
4. PR interval: lengthens 0.20 → 0.32 → 0.38 s, then a P wave is not conducted.
5. QRS: 0.08 s, narrow (below 0.12 s).

Interpretation

second-degree AV block type I (Wenckebach)

Answer

second-degree AV block type I (Wenckebach)
3.Human physiology

Answer: C · Stroke volume rises when ventricular filling increases

Why this is the answer

Greater end-diastolic volume stretches the myocardium, which increases the force of contraction and the stroke volume, matching output to venous return.

Answer

C · Stroke volume rises when ventricular filling increases
4.Human physiology

Answer: A · Actin and myosin overlap less, so fewer cross-bridges form

Why this is the answer

Maximum force occurs at an optimal sarcomere length with best filament overlap. Overstretching reduces overlap; too short a length makes filaments collide.

Answer

A · Actin and myosin overlap less, so fewer cross-bridges form

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