A patient in regular rhythm presents with absent P waves on ECG Leads II, III and AVF reveal a Saw-Tooth pattern. Which of the following is the most likely diagnosis
**Core Concept**
The question tests the ability to identify a specific ECG pattern associated with a particular cardiac condition. The absent P waves on ECG Lead II, III, and AVF reveal a saw-tooth pattern, indicating a disruption in the normal atrial electrical activity.
**Why the Correct Answer is Right**
The saw-tooth pattern on ECG is characteristic of atrial flutter, a type of supraventricular tachycardia. Atrial flutter is caused by a re-entrant circuit in the right atrium, typically around the tricuspid valve. This circuit creates a rapid, regular atrial activation pattern, leading to the characteristic saw-tooth appearance on ECG. The absence of P waves in Lead II, III, and AVF is due to the simultaneous activation of the right atrium, which masks the normal P wave.
**Why Each Wrong Option is Incorrect**
* **Option A:** Atrial fibrillation is characterized by a disorganized, irregular atrial activation pattern, leading to a "fibrillatory" or "irregularly irregular" ECG pattern. It does not typically present with a saw-tooth pattern.
* **Option B:** Sinus tachycardia is a normal heart rate response to various stimuli, and is characterized by a normal P wave and regular rhythm on ECG.
* **Option C:** Atrial septal defect (ASD) is a congenital heart defect that can lead to right atrial enlargement and abnormal ECG findings, but is not typically associated with a saw-tooth pattern.
**Clinical Pearl / High-Yield Fact**
A saw-tooth pattern on ECG is highly suggestive of atrial flutter, and should prompt the clinician to consider this diagnosis in patients presenting with a regular, rapid rhythm and absent P waves.
**Correct Answer:** C. Atrial flutter.