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12-Lead ECG Interpretation Flashcards

7 cards from real CET practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. Which condition produces diffuse ST elevation in a saddle-shaped (concave-up) pattern across most leads without reciprocal depression?

    Answer: Pericarditis

    Acute pericarditis causes diffuse concave (saddle-shaped) ST elevation in nearly all leads, often with PR depression, without the focal or reciprocal changes of STEMI.

  2. A patient has a PR interval of 320 ms with all P waves followed by QRS complexes. This represents:

    Answer: First-degree AV block

    First-degree AV block is defined as a PR interval >200 ms (0.20 s) with every P wave still conducted to the ventricles.

  3. Which precordial lead placement is used to detect posterior STEMI when standard leads show only tall R waves and ST depression in V1–V2?

    Answer: V7, V8, V9 (posterior leads)

    Posterior leads V7–V9 placed at the back of the thorax reveal direct ST elevation in posterior STEMI, which appears as mirror-image ST depression in V1–V2.

  4. In Mobitz II (second-degree AV block), the ECG shows:

    Answer: Constant PR interval with occasional non-conducted P waves

    Mobitz II features a consistent PR interval with sudden, unexpected failure of a P wave to conduct, indicating block below the AV node.

  5. Peaked P waves exceeding 2.5 mm in lead II are called P pulmonale and suggest:

    Answer: Right atrial enlargement

    P pulmonale—tall, peaked P waves >2.5 mm in lead II—reflects right atrial enlargement, commonly from chronic pulmonary disease.

  6. A wide, deep S wave in lead I combined with a pathological Q wave and inverted T in lead III on a 12-lead ECG (S1Q3T3) is associated with acute:

    Answer: Right heart strain from pulmonary embolism

    The S1Q3T3 pattern reflects acute right heart strain from elevated pulmonary vascular resistance, classically seen in pulmonary embolism.

  7. On a standard 12-lead ECG recording at 25 mm/s, how many milliseconds does each small square (1 mm) on the horizontal axis represent?

    Answer: 40 ms

    At the standard speed of 25 mm/s, each 1 mm small square equals 40 ms (0.04 s), and each large square (5 mm) equals 200 ms.