Synaptic transmission

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Synaptic transmission
2012-08-29 00:04:07

Neuroscience Exam 1
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  1. Neurotransmitter: Glutamate
    Most wide-spread excitatory transmitter in the CNS
  2. Neurotransmitter: Acetlcholine
    • Funciotn on the motor end plate, neuromuscular synapse, and also a major NT in the CNS.
    • In motoneurons between the pre- and post-synaptic 
  3. Neurotransmitter: Norepinephrine
    • Found in NT, CV system, also in CNS
    • When released at heart, increase heart rate and force of contraction, costrict blood vessels
  4. Neurotransmitter: alpha-aminobutyric acid(GABA)
    • Widespread inhibitory transmitter in CNS
    • Reflexes
  5. Why do stroke patients become rigid?
    • Damage to the inhibitory neurons so activity goes up
    • Normally, there is a balance/SUM between inhibitory and excitatory
  6. Temporal Summation
    • Repetitive firing of a single presynaptic cell.
    • Summing the effect on AP
    • Typical way to allow pre-synapse to affect the post-synaptic membrane
    • Concentration of NT goes up
  7. Spatial Summation
    • Pre-synaptic
    • AP goes down multiple pre-synptic neurons so the 2 AP will have affect on the post-synsptic membrane.
    • Added in space
    • Response is greater than either AP alone
    • Result--post-synaptic membrane can depolarize 
    • TIMING important: need to be added in space. If one arrives earlier than will not depolarize enough to reach threshold 
  8. Initial segment
    • Where the fast soidium channels are located
    • Get end plate potentials to summate to get propagation of AP to peripheral.
    • Even if membrane is partially depolarizes, if on post-synaptic membrane and not the initial segment, does NOT matter.
  9. How does Cl- inhibit without being hyperpolarized?
    • Tranference Equation
    • Excitationn when Cl- channels are open.
    • now if activation of inhibitory synapses causes a large number of  Cl-  channels to open, so that TCl-  =
    • 0.5
    • Calculate new membrane potential that must consider
    • all ions.
    • opening of the Na+ and  K+ channels will not cause as much depolarization, as shown by the transference equation:

    • Now have to split the transferences to 0.25 because all transferences must equal 1
    • Em = TK+ EK+  +  TNa+ENa+  + TCl-  ECl-

     Em = (0.25)(-90 mV) +(0.25)(60 mV) + (0.5)(-85 mV) = -50 mV