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2 years ago

Which is the correct path of an electrical excitation from the pacemaker to a cardiomyocyte in the left ventricle (LV)?

Medicine

1 answer:

lys-0071 [83]2 years ago

5 0

Answer:

The question is lacking the options, below is the completely stated question and options:

This is the correct path of an electrical excitation from the pacemaker to a cardiocyte in the left ventricle (LV).

A. sinoatrial (SA) node → atrioventricular (AV) node → atrioventricular (AV) bundle → Purkinje fibers → cardiocyte in LV

B. atrioventricular (AV) node → sinoatrial (SA) node → atrioventricular (AV) bundle → Purkinje fibers → cardiocyte in LV

C. sinoatrial (SA) node → atrioventricular (AV) bundle → atrioventricular (AV) node → Purkinje fibers → cardiocyte in LV

D. atrioventricular (AV) node → Purkinje fibers → atrioventricular (AV) bundle → sinoatrial (SA) node → cardiocyte in LV

E. sinoatrial (SA) node → atrioventricular (AV) node → Purkinje fibers → atrioventricular (AV) bundle → cardiocyte in LV

Answer:

The correct answer is:

sinoatrial (SA) node → atrioventricular (AV) node → atrioventricular (AV) bundle → Purkinje fibers → cardiocyte in LV (A.)

Explanation:

The electrical impulse conduction system of the heart refers to the pathways through which electrical signals pass, to cause excitation of specific paths of the heart, leading to depolarization, which in turn leads to contraction and relaxation of the heart. The correct excitation pathway is as follows:

1. Normal excitation originates in the sinoatrial (SA) node, from which depolarization spreads throughout the atria, causing impulses (excitation) to spread from the SA node to the atria. It is believed that this depolarization spreads to the atrial cells through adjacent myocardial cells and myofibrils.

2. Next, the depolarization from the atria reaches the atrioventricular (AV) node, causing excitation of the AV node. The AV node is located on the floor of the right ventricle, close to the interventricular septum. The excitation leaves this site by two pathways; the fast and slow pathways which vary based on the time of transmission of impulses to the next phase.

3. Next, from the AV node, the excitation is passed to the two branches of the bundle of His, which form a network of fibers known as purkinje fibers, from where excitation passes to the apical region of the left and right ventricles causing depolarization of the ventricular myocardium.

Attached to this answer is a picture to show what the pathway looks like.

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