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

This lead II strip shows atrial flutter. Measure the flutter-wave interval and find the atrial 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
atrial rate = bpm
2.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mVthird-degree (complete) AV blockjunctional rhythmfirst-degree AV blocksecond-degree AV block type I (Wenckebach)

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. 1third-degree (complete) AV block
  2. 2junctional rhythm
  3. 3first-degree AV block
  4. 4second-degree AV block type I (Wenckebach)
3.Human physiologyMedium

According to Poiseuille's law, if the radius of an arteriole halves, its resistance to flow:

  1. ADoubles
  2. BIncreases 4-fold
  3. CIncreases about 16-fold
  4. DHalves
4.Human physiologyHard

Which spirometry finding is typical of chronic obstructive pulmonary disease (COPD)?

  1. AFEV1/FVC ratio below 0.7
  2. BIncreased peak expiratory flow
  3. CNormal FEV1 with low FVC
  4. DFEV1/FVC ratio above 0.9 with low FVC

3Solutions

1.ECG rhythm strip interpretation

Answer: atrial rate = 300 bpm

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.

Flutter waves

F–F = 5 small boxes = 0.20 s → 1500 ÷ 5 = 300 per minute

QRS complexes

R–R = 10 small boxes = 0.40 s → 1500 ÷ 10 = 150 per minute
conduction ratio = 300 ÷ 150 = 2:1

Answer

atrial rate = 300 bpm
2.ECG rhythm strip interpretation

Answer: third-degree (complete) AV block

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: R–R = 29 small boxes, 1500 ÷ 29 ≈ 52 per minute.
2. Rhythm: R–R regular (1.16 s) and P–P regular (0.68 s), but the two are independent.
3. P waves: upright and regular at 88 per minute, but not linked to the QRS complexes; some hide in a QRS or a T wave.
4. PR interval: varies from beat to beat; there is no fixed relation.
5. QRS: 0.08 s, narrow (below 0.12 s).

Interpretation

third-degree (complete) AV block

Answer

third-degree (complete) AV block

Type II and third-degree blocks are the ones that can suddenly progress, so textbooks rank them as more serious than type I.

3.Human physiology

Answer: C · Increases about 16-fold

Why this is the answer

Resistance is proportional to 1/r⁴, so halving the radius raises resistance by 2⁴ = 16. This is why small changes in arteriole diameter control blood flow so strongly.

Answer

C · Increases about 16-fold
4.Human physiology

Answer: A · FEV1/FVC ratio below 0.7

Why this is the answer

Narrowed airways slow expiration, so FEV1 falls more than FVC and the ratio drops below 0.7. In COPD this is not fully reversed by bronchodilators.

Answer

A · FEV1/FVC ratio below 0.7

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