sol6 - process we studied in this problem This guarantees...

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Biophysics 11b, Spring 2006 – Solutions to Problem Set 6 1. Problem #1 Solution. (a) The Nernst potential is V Na Nernst = k B T e log c out c in = (1 . 38 × 10 - 23 J/K )(298 K ) 1 . 6 × 10 - 19 C log 145 12 = 64 mV . (1) According to Ohm’s law the current density is j = (64 mV + 90 mV )0 . 13 m Ω - 1 = 20 × 10 - 3 Am - 2 . (2) The current per unit length in the axon is then I/` = 2 πjR = (20 × 10 - 3 Am - 2 )2 π (0 . 5 × 10 - 3 m ) = 6 . 3 × 10 - 5 Am - 1 . (3) (b) The charge per unit length is Q/` = c in R 2 πe = (7 . 2 × 10 24 m - 3 )(0 . 5 × 10 - 3 m ) 2 π 1 . 6 × 10 - 19 C = 0 . 9 Cm - 1 . (4) If the interior concentration matched the exterior one (but leaving all else fixed) Q/` = c out R 2 πe = (8 . 7 × 10 25 m - 3 )(0 . 5 × 10 - 3 m ) 2 π 1 . 6 × 10 - 19 C = 11 Cm - 1 . (5) (c) Assuming the current does not change in the process, Q f - Q - i = It (6) t = ± Q f ` - Q i ` ² ` I 2 days . (7) 1
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(d) The nerve impulse depolarization happens on a time scale that is by eight orders of magnitudes smaller than the equilibration
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Unformatted text preview: process we studied in this problem. This guarantees that the equilibration cannot interfere with the information flow. 2. Problem #2 Solution . (a) The rate of ions passing through a single channel is n = D Δ C L A = (1 μm 2 /ms ) 100 mM 5 nm 1 nm 2 = 1 . 2 × 10 7 s-1 . (8) (b) The number of ions is the current divided by unit charge n = I/e = 4 . 5 pA/e = 2 . 8 × 10 7 s-1 . (9) This is consistent with the result of part a. (c) The channel is open approximately 4 out of 27 seconds. For a channel that is 50% of the time open the “square wave” pattern should be symmetrical with equal times open and closed within one period. 2...
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This note was uploaded on 09/21/2008 for the course BIPH 11 taught by Professor Kondev during the Spring '07 term at Brandeis.

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sol6 - process we studied in this problem This guarantees...

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