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 with a fixed conduction ratio. 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
2.ECG rhythm strip interpretationHard
II25 mm/s · 10 mm/mV

This lead II strip shows second-degree AV block type I (Wenckebach). Measure the PR interval of the last conducted beat before a non-conducted P wave (the longest PR), in seconds.

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

Which is a classic example of positive feedback in the body?

  1. AADH release when plasma osmolality rises
  2. BSweating when body temperature rises
  3. CInsulin release after a meal
  4. DOxytocin release during labour
4.Human physiologyHard

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

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

3Solutions

1.ECG rhythm strip interpretation

Answer: ventricular rate = 150 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

ventricular rate = 150 bpm
2.ECG rhythm strip interpretation

Answer: longest PR = 0.4 s

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.

March out the P waves

P–P = 0.88 s, regular; one P wave in every 4 is not conducted

PR in each group

PR₁ = 0.18 s, PR₂ = 0.32 s, PR₃ = 0.40 s

Answer

longest PR = 0.4 s

The biggest step in PR comes between the first and second beats of a group; later steps are smaller, which is why the R–R intervals shorten before the pause.

3.Human physiology

Answer: D · Oxytocin release during labour

Why this is the answer

Stretch of the cervix releases oxytocin, which strengthens contractions and causes more stretch. The loop ends with delivery. The other examples are negative feedback.

Answer

D · Oxytocin release during labour
4.Human physiology

Answer: D · 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

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

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