Question

Nerve signals in the body occur when a small voltage, called an action potential, is applied across the membrane of a cell. When this action potential is applied across a region of the cell membrane called an ion channel, current in the form of moving potassium ions will be established across the cell. If, during an action potential, a single ion channel with a resistance of 1.8 GΩ, is opened for 1.1 ms, approximately 90000 singly-ionized potassium ions travel through the channel during this time. What was the voltage of the action potential

Answer 1

Answer:

A stimulus causes sodium channels to open, allowing sodium ions to enter the neuron.

The neuron becomes more positive.

Potassium  channels open, allowing potassium ions to exit the cell.

Sodium channels close and the cell gradually returns to resting levels.

Answer 2

Answer:  23.56 nV

Explanation:

Assuming that we apply Ohm's Law to this situation, we know that under this condition, the current resultant, is proportional to the voltage applied, and that the proportionality constant, is called the resistance.

Now, we define electric current, as the passage of a number of charges over time.

In this case, we can say that each singly-ionized potassium ion carries the charge equivalent to the one electron, which is q = 1.6. 10⁻¹⁹ coulombs.

So, as we have 90,000 ions, we will have a total charge as follows:

Q =90.10³. 1.6.10⁻¹⁹ coulombs

Also, we are told that this charge will be moving for 1.1 msec, so we can find the current I as follows:

I = Q/t ⇒ I = 13.09. 10⁻¹² A.

If we know that R= 1.8.10⁹ Ω, we can determine V, applying Ohm's Law, as follows:

V = 13.09. 10⁻¹² A .  1.8.10⁹ Ω = 23.56 nV