
The one way or unidirectional transmission of nerve cells is due to
A. Synapses
B. Myelin sheath
C. Membrane polarity
D. Interneurons
Answer
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Hint: The released neurotransmitters bind to their specific receptors, present on the postsynaptic membrane. After generating a change in the permeability of the postsynaptic membrane, the neurotransmitter is immediately lost from the synaptic cleft.
Complete answer: At the point when an impulse (action potential) shows up at the axon terminal, it stimulates the development of the synaptic vesicles towards the plasma membrane where they fuse with the plasma membrane layer and deliver neurotransmitter in the synaptic cleft. This binding of the neurotransmitter to the postsynaptic membrane opens ion channels and allows the entry of ions which generates a new potential in the postsynaptic neuron. After generating a change in the permeability of the postsynaptic membrane, the neurotransmitter is immediately lost from the synaptic cleft. The synapse cannot transmit an impulse in the reverse direction as dendrites cannot secrete neurotransmitters. Thus, synapse acts as a one-way valve, allowing the conduction of impulse from axon to dendrite only.
The transmission of an impulse across a neuron is faster than across the synapse because the release of neurotransmitter from a presynaptic neuron requires time. On the basis of the number of axons and dendrites neurons are of three types:
a) Unipolar neurons: These have a single process arising from cyton. E.g. early embryos of invertebrates and vertebrates.
b) Bipolar neurons: These neurons have only two processes, an axon at one end and a dendrite at another end. E.g. retina of an eye, olfactory epithelium and cochlear and vestibular ganglia.
c) Multipolar neuron: These neurons have several dendrites and one axon. E.g. cerebral cortex.
So, the correct answer is option A.
Note: During the time of a nerve impulse, a nerve undergoes three phases viz., depolarization phase, repolarization phase and hyperpolarization phase. The resting membrane potential of a neuron is -70mV, whereas depolarization and repolarization phases have +30 mV and -90 mV.
Complete answer: At the point when an impulse (action potential) shows up at the axon terminal, it stimulates the development of the synaptic vesicles towards the plasma membrane where they fuse with the plasma membrane layer and deliver neurotransmitter in the synaptic cleft. This binding of the neurotransmitter to the postsynaptic membrane opens ion channels and allows the entry of ions which generates a new potential in the postsynaptic neuron. After generating a change in the permeability of the postsynaptic membrane, the neurotransmitter is immediately lost from the synaptic cleft. The synapse cannot transmit an impulse in the reverse direction as dendrites cannot secrete neurotransmitters. Thus, synapse acts as a one-way valve, allowing the conduction of impulse from axon to dendrite only.
The transmission of an impulse across a neuron is faster than across the synapse because the release of neurotransmitter from a presynaptic neuron requires time. On the basis of the number of axons and dendrites neurons are of three types:
a) Unipolar neurons: These have a single process arising from cyton. E.g. early embryos of invertebrates and vertebrates.
b) Bipolar neurons: These neurons have only two processes, an axon at one end and a dendrite at another end. E.g. retina of an eye, olfactory epithelium and cochlear and vestibular ganglia.
c) Multipolar neuron: These neurons have several dendrites and one axon. E.g. cerebral cortex.
So, the correct answer is option A.
Note: During the time of a nerve impulse, a nerve undergoes three phases viz., depolarization phase, repolarization phase and hyperpolarization phase. The resting membrane potential of a neuron is -70mV, whereas depolarization and repolarization phases have +30 mV and -90 mV.
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