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. How many flutter waves per QRS complex are there (the n of n:1 conduction)?

lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
flutter waves per QRS =
2.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mV

This lead II strip shows an AV block, so the atria and the ventricles beat at different rates. Measure the P–P interval and find the atrial rate in beats per minute (1500 method, whole number).

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
3.Human physiologyMedium

Which hormonal event triggers ovulation?

  1. AA rise in progesterone from the corpus luteum
  2. BA surge of prolactin
  3. CA fall in oestrogen
  4. DA surge of luteinising hormone (LH)
4.Human physiologyHard

Why does a person with a high spinal cord injury above T1 risk low blood pressure?

  1. ARenin release from the kidney stops completely
  2. BThe sympathetic outflow to the vessels and heart is cut off
  3. CThe vagus nerve is cut, so the heart rate rises
  4. DThe phrenic nerve can no longer control the tone of the arteries

3Solutions

1.ECG rhythm strip interpretation

Answer: flutter waves per QRS = 2

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 = 6 small boxes = 0.24 s → 1500 ÷ 6 = 250 per minute

QRS complexes

R–R = 12 small boxes = 0.48 s → 1500 ÷ 12 = 125 per minute
conduction ratio = 250 ÷ 125 = 2:1

Answer

flutter waves per QRS = 2
2.ECG rhythm strip interpretation

Answer: atrial rate = 65 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 = 23 small boxes = 0.92 s → 1500 ÷ 23 = 65.22 ≈ 65 per minute

Ventricular rate

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

Answer

atrial rate = 65 bpm

Answer

atrial rate = 65 bpm

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

3.Human physiology

Answer: D · A surge of luteinising hormone (LH)

Why this is the answer

Rising oestrogen from the dominant follicle triggers an LH surge from the anterior pituitary, which causes ovulation about a day later.

Answer

D · A surge of luteinising hormone (LH)
4.Human physiology

Answer: B · The sympathetic outflow to the vessels and heart is cut off

Why this is the answer

Sympathetic outflow leaves the cord at T1–L2. A lesion above it removes vasomotor tone and cardiac sympathetic drive while the vagus still works, causing hypotension and bradycardia (neurogenic shock).

Answer

B · The sympathetic outflow to the vessels and heart is cut off

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