Question

The Na/K/ATPase or ""pump"" is inhibited by ouabain. Which of the following consequences may be expected in response to this inhibition?

A. Depletion of intracellular sodium
B. Increased cell volume
C. Accumulation of potassium within cells
D. Hyperpolarization

Answer 1

Inhibition of the Na/K/ATPase by ouabain is expected to cause an accumulation of sodium inside the cell, a depletion of potassium within the cell, an increased cell volume due to water influx, and depolarization of the cell membrane.

Step-by-step explanation:

The inhibition of the Na/K/ATPase or "pump" by ouabain can lead to several consequences. Normally, the pump moves three sodium ions out of the cell and two potassium ions inside, which contributes to establishing an electrochemical gradient and membrane potential. If ouabain inhibits the pump, the expected consequences include:

• An accumulation of sodium (Na+) inside the cell, as its expulsion is hindered.
• A depletion of potassium (K+) within the cell, because its import is also blocked.
• Due to the accumulation of sodium ions inside the cell, water will follow inside, potentially leading to an increased cell volume.
• The cell will become less negative inside compared to the outside, a condition which is opposite to hyperpolarization. Thus, hyperpolarization is unlikely to occur; rather, the cell will experience depolarization or become less polarized.