How the nervous system works Flashcards
Explain how never impulses are conducted along a myelinated axon after an action potential has been initiated
- propagation of action potential
- section of membrane that has been depolarised has the opposite electrical charge to resting membrane
- some Na+ ions diffuse sideways attracted to the -ve charge of resting potential
- these are local currents
- this depolarises section of axon adjacent
- Na+ channels open in the section of resting membrane, Na+ diffuses in
- more Na+ ions move in so more Na+ channels are opened (positive feedback)
- inside axon becomes +40mv
- Na+ channels then close and K+ voltage dependent channels open, K+ moves out of axon down electrochemical and concentration gradient
- membrane repolarises
- membrane becomes hyperpolarised, K+ channels close and Na+/K+ pump restores resting potential
- during hyperpolarisation is the refractory period (more -ve than resting potential), no AP can be stimulated
- refractory period ensures impulses moves in one direction only (cannot spread backwards), impulses move away from parts of membrane in the refractory period as they cannot fire an AP
- wave of depolarisation, depolarisation-repolarisation process repeats itself along the axon
- myelin sheath speeds up conduction as it is an electrical insulator
- current jumps between nodes of Ranvier, saltatory conduction
How changes in the functioning of the synapse might bring about habituation
- after repeated stimulation
- less Ca2+ moves into the presynaptic membrane
- fewer vesicles containing neurotransmitter move to the presynaptic membrane and fuse with it
- less transmitter is released into the synaptic cleft by exocytosis
- less transmitter diffuses across synaptic cleft
- less transmitter binds to receptors on postsynaptic membrane
- fewer Na+ channels open, less Na+ enter axon
- less membrane depolarisation
- depolarisation doesn’t reach threshold
- no AP stimulated in the second neurone
Describe the effect of changing light intensity on the area of the pupil
- as light intensity increases, pupil area decreases
How does the eye bring about changing pupil diameter
- to decrease diameter (bright light), radial muscles relax, circular muscles contract
- to increase diameter (dim light), radial muscles contract, and circular muscles relax
Change in area of pupil is controlled by a reflex action, in the pupil reflex what is the receptor and effector
- receptors are photoreceptors rods and cones
- effect is the iris muscle (circular/radial muscle)
Why does the pupil reflex still occur in an unconscious person
- it is an involuntary response (autonomic)
Explain how light causes depolarisation of the bipolar cell
- when the rod cell absorbs light, retinal turns from cis to trans
- this causes rhodopsin to break up into opsin and retinal (bleaching)
- Na+ channels close
- inner segment continues to pump out Na+
- inside rod cell becomes hyperpolarised (more -ve)
- release of neurotransmitter glutamate stops (glutamate binds to the bipolar cell, stopping depolarisation), inhibition of bipolar cell lifted
- lack of glutamate results in depolarisation of bipolar cell, Na+ channels open
Describe how the nervous system controls the pupil reflex in a mammal in response to bright light
- light hits photoreceptors on the retina
- impulses pass to the brain, via sensory neurone
- this is an autonomic response (involuntary)
- impulses pass along parasympathetic nerve
- along motor neurone to effector
- circular muscles contract, radial muscles relax
- pupil constricts
Explain the advantage of myelination
- myeline acts as an electrical insulator
- schwann cells produce the myeline sheath
- depolarisation can only occur at nodes of Ranvier
- the jumping of current from node to node is saltatory conduction
- results in faster impulses
- better protection as reflex is a rapid response
Explain how neurotransmitters stimulate neurones
- CA2+ ions are required
- when presynaptic membrane depolarises, CA2+ channels open, CA2+ moves in
- vesicles containing transmitter move to presynaptic membrane
- vesicles fuse with membrane and release transmitter into synaptic cleft by exocytosis
- transmitter diffuses across synaptic cleft
- transmitter binds to receptors on postsynaptic membrane
- Na+ channels in postsynaptic membrane open, Na+ ions enter
- depolarisation of postsynaptic membrane
- if depolarisation reaches threshold, AP initiated
Describe the role of ATP in nerve impulse transmission
- ATP supplies energy for active transport (Na+/K+ pump)
- Na+ ions are pumped out of the axon against concentration gradient, restoring/maintaining membrane resting potential
Describe the events that begin depolarisation of the membrane of a neurone
- depolarisation of adjacent membrane
- change of potential difference across membrane
- Na+ gates open
- Na+ ions move into neurone
Explain the movement of ions during repolarisation
- voltage dependent K+ gates open
- Na+ gates close
- increased permeability to K+ ions
- K+ ions move down electrochemical/concentration gradient
Explain how the pd across the membrane is returned to resting potential during hyperpolarisation
- membrane remains permeable to K+ ions through K+ channels
- K+ ions move into axon because of charge difference- K+ ion is removing a positive charge from the outside
- equilibrium is established, concentration gradient is balanced by electrochemical gradient (diffusion gradient balanced by pd)
- voltage dependent K+ channels close
- Na+/K+ pump re-establishes resting potential
During depolarisation and repolarisation what ions is the membrane permeable to
- during depolarisation, membrane is permeable to Na+ ions
- during repolarisation, membrane is permeable to K+ ions
