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explain the ion current in phototransduction
- In the dark, Na channels in the outer segment are open
- There is a net flux of Na into the outer segment and this is removed by exchange for K in the region of the cell body
- Electrical neutrality is maintained by an efflux of K through “leakage” channels
- There is, therefore, a dark current which keeps the cell in a relatively depolarised state.
- The transition from 11-cis to all-trans retinal in the light causes a conformational change in the opsin, which, through a series of biochemical events leads to the closure of the Na channels in the outer segment.
-K continues to leave the cell through the leakage channels, driving the membrane potential towards the equilibrium potential for K (said to be hyperpolarising).
Under the relative depolarisation of the dark-adapted state, there is continuous release of neurotransmitter from the synapses on to the bipolar cells, and this is reduced on hyperpolarisation in the light.
- There is a net flux of Na into the outer segment and this is removed by exchange for K in the region of the cell body
- Electrical neutrality is maintained by an efflux of K through “leakage” channels
- There is, therefore, a dark current which keeps the cell in a relatively depolarised state.
- The transition from 11-cis to all-trans retinal in the light causes a conformational change in the opsin, which, through a series of biochemical events leads to the closure of the Na channels in the outer segment.
-K continues to leave the cell through the leakage channels, driving the membrane potential towards the equilibrium potential for K (said to be hyperpolarising).
Under the relative depolarisation of the dark-adapted state, there is continuous release of neurotransmitter from the synapses on to the bipolar cells, and this is reduced on hyperpolarisation in the light.
