Question

One way of achieving an IPSP is to increase permeability to Cl^-. By what other mechanism can you achieve an IPSP (change permeability to what ion)? Explain why.

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Answer 1

An inhibitory postsynaptic potential (IPSP) is a kind of synaptic potential that makes a postsynaptic neuron less likely to generate an action potential. An IPSP can be produced by a localized increase in Cl^- transports

IPSP can also be produced by opening of K⁺ channels, with movement of K⁺out of the postsynaptic cell.

In addition, they can produce by closure of Na⁺ or Ca²⁺ channels.

Mechanism:

All cellular membranes are electrically charged due to the concentration of ions present in the extracellular and intracellular space. For neurons, sodium ions and chloride ions exists in large quantities in the extracellular space while the intracellular space contains many potassium ions and other organic anions. The cell membrane contains ions channels that regulate the passing of these various ions between the inside and outside of the cell. When the cell is at the resting potential (-70 mV), the intracellular space is more negative relative to the extracellular space and the voltage dependent Na⁺ and K⁺ channels are closed.

The opening of Na⁺ channels is triggered by the reduction of the membrane potential (depolarization) to the threshold of excitation. Once the ion channels open, the action potential begins as electrostatic pressure and the force of diffusion drive Na⁺ into the cell. The entry of positively charged ions into the cell actually reverses the electrical potential from a negative potential to a positive one (that is inside becomes positive). This depolarization is produced by the entry of Na⁺ which subsequently causes the K⁺ channels to open. The force of diffusion drives K⁺ out of the cell. In about 1 millisecond, the membrane potential reaches +40 mV because of the many Na⁺ ions entering the cell.

The entry of positively charged ions into the cell reduces the membrane potential to an extent that causes the inside to become positive. The depolarization caused by the entry of Na⁺ causes the K⁺ channels to open and the force of diffusion drives K⁺ out of the cell. In about 1 millisecond, the membrane potential reaches +40mV because of the Na⁺ entering the cell. At this point, the Na⁺ channels become blocked (refractory) and Na⁺ cannot enter the cell anymore. But the K⁺ channels remain open and are continually pumping K⁺ out of the cell. The outflow of K⁺ helps to bring the membrane potential back to its resting value. Once the resting potential is

Reached, K⁺ channels close.

Once the membrane potential reaches +40mV, the Na⁺ channels become blocked and Na⁺ cannot enter the cell anymore. But the K⁺ channels remain open and continually pump K⁺ out of the cell. The outflow of K⁺ helps to bring the membrane potential back to its resting value. Once the resting potential is reached, the K⁺ channels close. As you can see from picture below, the membrane actually overshoots the resting potential (hyperpolarization) and there is an excess of K⁺ outside of the membrane. The resting potential returns to -70mV as the excess K⁺ diffuses away (repolarization).