Final answer:
Myocardial depolarization travels to the ventricles via the atrioventricular bundle fibers. The AV node delays the impulse, ensuring proper cardiac cycle timing. The component with the slowest rate of firing is the atrioventricular node.
Step-by-step explanation:
Myocardial depolarization travels to the ventricles primarily via the atrioventricular bundle fibers, often referred to as the bundle of His. The process starts with the sinoatrial (SA) node generating an electrical impulse causing atrial contraction, followed by a delay at the atrioventricular (AV) node. This pause allows atria to empty into ventricles. Subsequently, the impulse moves through the atrioventricular bundle, reaches the left and right bundle branches, then finally spreads through the Purkinje fibers, resulting in ventricular contraction. Therefore, the correct answer to which component of the heart conduction system would have the slowest rate of firing is (a) atrioventricular node. The AV node's rate of firing is slower because it intentionally delays the impulse, ensuring that the ventricles contract only after the atria have emptied.