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ackman678
2018-10-18 18:20:35 -07:00
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neurophysiology2.md
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@@ -26,7 +26,7 @@ The solution was to build an electrophysiological recording apparatus with feedb
## Action potential summary video
<div><video height=400px controls src="figs/Animation02-03TheActionPotential.mp4"></video><figcaption>Neuroscience 5e Animation 2.3</figcaption></div>
<div><video height=400px controls src="figs/Animation02-03TheActionPotential_OC.mp4"></video><figcaption>Neuroscience 5e Animation 2.3</figcaption></div>
Note:
@@ -60,28 +60,28 @@ Note:
4. 0 mV:
* `Pk = 0.5; Pna = 0.5; Pcl = 0; kOut = 1; kIn = 10; naOut = 10; naIn = 1; clIn = 11; clOut = 11`
* `(58)*log10( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) ) = 0 mV`
* `(58)*Math.log10( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) ) = 0 mV`
* 0 mV:
* `Pk = 1; Pna = 1; Pcl = 0; kOut = 1; kIn = 10; naOut = 10; naIn = 1; clIn = 11; clOut = 11`
* `(58)*log10( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) )`
* `(58)*Math.log10( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) )`
* -59 mV (room temp and low Pna):
* `Pk = 1; Pna = 0.001; Pcl = 0.5; kOut = 1; kIn = 10; naOut = 10; naIn = 1; clIn = 1; clOut = 11`
* `(58)*log10( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut)`
* `(58)*Math.log10( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut)`
-62 mV (body temp and low Pna):
* R = 8.3; F = 9.6e4; T = (273+37)
* `Pk = 1; Pna = 0.001; Pcl = 0.5; kOut = 1; kIn = 10; naOut = 10; naIn = 1; clIn = 1; clOut = 11`
* `((R*T)/F)*log( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) )`
* `((R*T)/F)*Math.log( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) )`
-69 mV (body temp and low Pna and physiol concentrations):
* `R = 8.3; F = 9.6e4; T = (273+37)`
* `Pk = 1; Pna = 0.05; Pcl = 0.45; kOut = 5; kIn = 140; naOut = 145; naIn = 5; clIn = 5; clOut = 110`
* `((R*T)/F)*log( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) )`
* `((R*T)/F)*Math.log( (Pk*kOut + Pna*naOut + Pcl*clIn) / (Pk*kIn + Pna*naIn + Pcl*clOut) )`
Calculate the total concentration of all ions for these solutions. For every one NaCl that dissolves, two ions are produced (one Na⁺ and one Cl¯). Thus for 10 mmol/L NaCl outside there are (10 mmol/L)x(1 total Cl ions/NaCl) = 10mM. And for 1mM KCl outside there are (1 mmol/L)x(1 total Cl ions/KCl) = 1mM. Thus the total number of Cl⁻ ions per liter is 11mmol/L = 11mM
@@ -229,7 +229,7 @@ Note:
## Voltage clamp method summary
<div><video height=400px controls src="figs/Animation03-01TheVoltageClampMethod.mp4"></video><figcaption>Neuroscience 5e Animation 3.1</figcaption></div>
<div><video height=400px controls src="figs/Animation03-01TheVoltageClampMethod_OC.mp4"></video><figcaption>Neuroscience 5e Animation 3.1</figcaption></div>
Note: