Cellular Neuroscience 1) In part B, why is the slope of the open circle (O) line

Question

Cellular Neuroscience

1) In part B, why is the slope of the open circle (O) line considerably less that the slope of the open triangle line?

2) In part B, why does the open circle (O) line not represent the leakage current through non-gated channels? How would such a current look on in part B? Why isn't it shown?

3) Why do all four lines go through point (0,0)?

Peak current Late current NMDA-activated pA 1 100 Membrane component potential Late current 20 remaining after blockade with APV 50 150 100 t 50 mV APV -40 Late current APV AVA 100 APV -80 A. SCANSAKANDELACH11MK11-6, GIF 200 100 pA Peak (early) current 50 msec (kainate-quisqualate) 300

Explanation / Answer

Answer :

Here in this figure figure B

1) The slope of the open circle line is considerably less than the slope of the open triangle line, here

open circle represents the NMDA receptors and open triangle shows the non-NMDA receptor. As exitatory

postsynaptic post synaptic current has measured in presence of APV, which blocks the NMDA component but not the non-NMDA component

2) In open circle in part B, exccitatory post synaptic current mediated by NMDA receptor has been measured.

As the APV blocks the late NMDA receptor components, such things observed in part B. The drug APV selectively

binds to the NMDA receptor component and blocks the activity.

3) AS current and voltage have linear relationship, and non-NMDA receptors behave as simple resistors,

the reverse potential of both NMDA and non- NMDA receptors channels is at 0 mV.

Hence all four lines go through point (0,0)

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