Question

Na+ is released from its binding site on the Na+/K+ pump as a result of the

A) interaction of the Na+ and K+ binding sites.

B) exposure of the Na+ binding site to the inside of the cell.

C) enzyme on the inside of the cell which cleaves the Na+ from its binding site.

D) conformational change decreasing the affinity of the Na+ binding site.

E) binding of K+ to its binding site

Answer 1

Sodium ions are released from the Na+/K+ pump when ATP induces a conformational change, decreasing the binding site's affinity for Na+. option D is the correct answer.

Step-by-step explanation:

Na+ is released from its binding site on the Na+/K+ pump as a result of the D) conformational change decreasing the affinity of the Na+ binding site. When ATP binds to the intracellular domain of the Na+/K+ pump, a conformational change is triggered. This change decreases the affinity of the binding site for Na+, causing the Na+ ions to be released into the extracellular space.

Meanwhile, K+ ions are taken up by the pump on the inner surface of the cell when the pump reveals its binding sites for K+ as a result of another conformational change.

The entire process is part of the active transport mechanism that maintains the essential sodium and potassium ion gradient across the cell membrane, crucial for numerous cellular processes including the maintenance of the resting membrane potential.