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Human Physiology - Neurons and Synapses

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Neurons are specialized cells that transmit electrical impulses. The basic structure includes dendrites (receive stimuli), the cell body (soma), and the axon (transmits the impulse).

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Myelination by Schwann cells provides electrical insulation. This enables saltatory conduction, where the action potential 'jumps' between the Nodes of Ranvier, significantly increasing the speed of transmission from approximately 1 m/s1\text{ m/s} to 100 m/s100\text{ m/s}.

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The resting potential (approximately −70 mV-70\text{ mV}) is maintained by the Na+/K+Na^{+}/K^{+}-ATPase pump, which actively transports 3Na+3\text{Na}^{+} ions out of the cell for every 2K+2\text{K}^{+} ions pumped in, creating an electrochemical gradient.

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An action potential consists of depolarization (opening of voltage-gated Na+Na^{+} channels, causing Na+Na^{+} influx and a potential rise to +30 mV+30\text{ mV}) and repolarization (opening of voltage-gated K+K^{+} channels, causing K+K^{+} efflux and a potential drop).

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The 'All-or-nothing' principle states that an action potential is only propagated if the stimulus is strong enough to reach the threshold potential of approximately −55 mV-55\text{ mV}.

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Synaptic transmission involves the release of neurotransmitters from the pre-synaptic neuron into the synaptic cleft. Ca2+Ca^{2+} ions enter the pre-synaptic knob following an action potential, triggering exocytosis of neurotransmitter vesicles.

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Acetylcholine is a common neurotransmitter. It is broken down in the synaptic cleft by the enzyme acetylcholinesterase to prevent continuous stimulation of the post-synaptic neuron.

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Neonicotinoids are synthetic compounds that bind to acetylcholine receptors in insect central nervous systems. Because acetylcholinesterase cannot break them down, they cause overstimulation, paralysis, and death.

📐Formulae

Vresting≈−70 mVV_{resting} \approx -70\text{ mV}

Vthreshold≈−55 mVV_{threshold} \approx -55\text{ mV}

Ratio of Naout+:Kin+=3:2\text{Ratio of } Na^{+}_{out} : K^{+}_{in} = 3 : 2

Velocity(v)=ΔdΔt\text{Velocity} (v) = \frac{\Delta d}{\Delta t}

💡Examples

Problem 1:

Explain why the resting potential of a neuron is negative (−70 mV-70\text{ mV}) relative to the outside.

Solution:

The interior is more negative due to the Na+/K+Na^{+}/K^{+} pump and membrane permeability.

Explanation:

The pump moves 3Na+3\text{Na}^{+} out for every 2K+2\text{K}^{+} in. Additionally, the membrane is more permeable to K+K^{+} than Na+Na^{+}, allowing K+K^{+} to leak out more easily. Large negatively charged proteins (A−A^{-}) inside the cytoplasm further contribute to the negative charge.

Problem 2:

Calculate the time taken for a nerve impulse to travel down a myelinated axon of length 1.2 m1.2\text{ m} if the conduction velocity is 100 m/s100\text{ m/s}.

Solution:

t=0.012 st = 0.012\text{ s} or 12 ms12\text{ ms}

Explanation:

Using the formula v=dtv = \frac{d}{t}, we rearrange to solve for tt: t=dv=1.2 m100 m/s=0.012 st = \frac{d}{v} = \frac{1.2\text{ m}}{100\text{ m/s}} = 0.012\text{ s}.

Problem 3:

What happens to the membrane potential during the refractory period?

Solution:

Hyperpolarization occurs, where the potential drops below −70 mV-70\text{ mV}.

Explanation:

During repolarization, K+K^{+} channels remain open slightly longer than necessary to reach resting potential, causing the potential to reach approximately −80 mV-80\text{ mV}. This ensures the impulse travels in only one direction.