write the steps transmission of nerve impulse
Answers
Explanation:
The following four steps describe the initiation of an impulse to the “resetting” of a neuron to prepare for a second stimulation:
Action potential. ...
Repolarization. ...
Hyperpolarization. ...
Refractory period
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The following four steps describe the initiation of an impulse to the “resetting” of a neuron to prepare for a second stimulation:
1. Action potential
2. Repolarization
3. Hyperpolarization
4. Refractory period
Explanation:
Action potential: Unlike a graded potential, an action potential is capable of traveling long distances. If a depolarizing graded potential is sufficiently large, Na + channels in the trigger zone open. In response, Na + on the outside of the membrane becomes depolarized (as in a graded potential). If the stimulus is strong enough—that is, if it is above a certain threshold level—additional Na + gates open, increasing the flow of Na + even more, causing an action potential, or complete depolarization (from –70 to about +30 millivolts). This in turn stimulates neighboring Na + gates, farther down the axon, to open. In this manner, the action potential travels down the length of the axon as opened Na + gates stimulate neighboring Na + gates to open. The action potential is an all‐or‐nothing event: When the stimulus fails to produce depolarization that exceeds the threshold value, no action potential results, but when threshold potential is exceeded, complete depolarization occurs.
Repolarization: In response to the inflow of Na +, K + channels open, this time allowing K + on the inside to rush out of the cell. The movement of K + out of the cell causes repolarization by restoring the original membrane polarization. Unlike the resting potential, however, in repolarization the K + are on the outside and the Na + are on the inside. Soon after the K + gates open, the Na + gates close.
Hyperpolarization: By the time the K + channels close, more K + have moved out of the cell than is actually necessary to establish the original polarized potential. Thus, the membrane becomes hyperpolarized (about –80 millivolts).
Refractory period: With the passage of the action potential, the cell membrane is in an unusual state of affairs. The membrane is polarized, but the Na + and K + are on the wrong sides of the membrane. During this refractory period, the axon will not respond to a new stimulus. To reestablish the original distribution of these ions, the Na + and K + are returned to their resting potential location by Na +/K + pumps in the cell membrane. Once these ions are completely returned to their resting potential location, the neuron is ready for another stimulus.