Question

question ①
cyanide poisoning prevents the production of atp. if atp were not available to drive the sodium-potassium pump, which of the following is most likely?
answer
○ k⁺ concentrations inside the cell will increase causing hyperpolarization.
○ na⁺ will leak back into the cell and bring water with it causing cells to burst.
○ the voltage of the membrane will be unchanged.
○ na⁺ concentration inside the cell will decrease causing depolarization.
○ i dont know yet

Explanation:

The sodium - potassium pump ($\text{Na}^+-\text{K}^+$ pump) is an ATP - dependent pump. It actively transports 3 $\text{Na}^+$ out of the cell and 2 $\text{K}^+$ into the cell. If ATP is not available, the pump stops working. Normally, the cell has a higher $\text{Na}^+$ concentration outside and higher $\text{K}^+$ inside. Without the pump, the concentration gradients will start to dissipate. $\text{Na}^+$ will tend to move back into the cell (down its concentration gradient, through leak channels or other passive pathways). When $\text{Na}^+$ moves back into the cell, water will follow by osmosis (since $\text{Na}^+$ is a solute, and water moves to areas of higher solute concentration). This influx of water can cause the cell to swell and eventually burst.

• Option 1: Without ATP, the $\text{K}^+$ pump doesn't work, so $\text{K}^+$ won't be actively transported into the cell; in fact, $\text{K}^+$ might leak out, so this is incorrect.
• Option 3: The membrane voltage is maintained by the $\text{Na}^+-\text{K}^+$ pump and ion gradients. Without the pump, the ion gradients change, so the membrane voltage will change, making this option incorrect.
• Option 4: Without the pump, $\text{Na}^+$ will move into the cell (not decrease inside), so this is incorrect.

Answer:

B. $\text{Na}^+$ will leak back into the cell and bring water with it causing cells to burst.