Handout 1 Sol

Handout 1 Sol - TA: Anna Gruzman E-mail: agruzman@ucsd.edu...

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TA: Anna Gruzman Section 1 E-mail: agruzman@ucsd.edu 12/10/10 1. a. Write out the Nernst equation and explain what each variable means. A: Equilibrium potential of an ion (Eion) = RT/FZ ln ( [Cion out]/ [Cion in] ) o RT/F is a conversion factor o Z is the charge (as in. .monovalent or divalent, etc.) o C is the concentration of the ion o 2.303: convert to log base ( easier to do calculations) o T = 37 deg C (mammalian body temperature) Need to convert this to Kelvin ( 273.15 K + 37) o F = 96500 Coulombs/ mol o R= 8.314 J.K -1 .mol - o Volt = J/C and then 1 volt = 1000 millivolts o If you plug these constants into the equation you will get 61 ( Can just remember this number for the test) o Eion = 61mV/z log ( [Cion out]/ [Cion in] ) o Every 61 mV indicates a 10 fold gradient b. What does the Nernst equation tell you? A: Tells you what electrical potential will balance the concentration gradient for that particular ion. In other words the membrane potential that exactly opposes the concentration gradient if the cell was only permeable tot the ion you are looking at. c. Fill in the values for the following table: (It would be a good idea to memorize/ understand this for the exams) Conc. Inside Cell Conc. Outside Cell Down Gradient Flow Effect on Membrane Potential K+ 3.5-5 mmol Na+ 140 mmol Cl- 5-10 mmol Ca++ 1 mmol A: Conc. Inside Cell Conc. Outside Cell Down Gradient Flow Effect on Membrane Potential K+ 150 mmol 3.5-5 mmol Out of cell More neg. Na+ 14 mmol 140 mmol Into cell More pos.
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Cl- 5-10 mmol 110 mmol Into cell More neg. Ca++ .0001 mmol 1 mmol Into cell More pos. d. Fill in the following table [Ion]in: (mM) [Ion]out: (mM) Eion: (mV) Driving force (mv): Vm= - 70 mV Driving force (mV) Vm = + 100 mv Direction Vm= -70 Direction Vm = + 100 K= 150 1.5 Na= 14 140 Cl = 100 10 Ca= 0.0001 1 A: [Ion]in: (mM) [Ion]out: (mM) Eion: (mV) Driving force (mv): Vm= - 70 mV Driving force (mV) Vm = + 100 mv Direction Vm= -70 Direction Vm = + 100 K= 150 1.5 -122 52 222 out out Na= 14 140 61 - 131 39 in out Cl = 100 10 61 - 131 39 out in Ca= 0.0001 1 122 - 192 - 22 in in Brief Review of Logs: (Can look at this later if it’s been a while since you did math)
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This note was uploaded on 03/10/2011 for the course BIPN 100 taught by Professor French during the Winter '07 term at UCSD.

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Handout 1 Sol - TA: Anna Gruzman E-mail: agruzman@ucsd.edu...

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