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BME314hw3 - BME 314 HW#3 Fall 2010(Roy BME 314 Fall 2010 HW...

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BME 314 HW #3 Fall 2010 (Roy) BME 314, Fall 2010 HW 3 Due November 23, 2010 Total points = 75 Always work problems in the units provided (if provided with English units, work in English, if provided with metric units, then work in metric). Define all equations used and provide explanatory comments to allow one to easily follow your work. Box final solutions. In problem A, B and C show all calculations and use the given values only to derive/obtain related parameters. A. SOLVE the following problems from the EXERCISES given at the end of CHAPTER 11 of your textbook (Introduction to Biomedical Engineering, Enderle, Blanchard, Bronzino, SECOND EDITION). 3. Assume that a membrane is permeable to Ca ++ and Cl _ . The initial concentrations on the inside are different from the outside, and these are the only ions in the solution. (a) Write an equation for J Ca and J Cl . J Ca = J Ca (drift) + J Ca (diffusion) J Ca = -2μ[Ca 2+ ] - Where the valence, Z, for Ca is 2+ J Cl = J Cl (drift) + J Cl (diffusion) J Cl = μ[Cl -1 ] - Where the valence, Z, for Cl is -1 J x = flow of ions due to drift in an Electric Field μ = mobility in m 2 /sV Z = ionic valence = voltage gradient across membrane [I] = ion concentration (b) Write an expression for the relationship between J Ca and J Cl . From space charge neutrality, 2J Ca = J Cl By plugging in values from part 1a we obtain: 2(-2μ[Ca 2+ ] - )= μ[Cl -1 ] -
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BME 314 HW #3 Fall 2010 (Roy) -4μ[Ca 2+ ] - = μ[Cl -1 ] - (c) Find the equilibrium voltage. J Ca = -2μ[Ca 2+ ] - ; begin with the flow rate for Calcium Under a constant electric field, =; Substitute into original equation J Ca = -2μ[Ca 2+ ] - Now lets obtain a relationship for the permeability of Calcium: P Ca = Using this relationship in our equation for J Ca we obtain: J Ca = Rearranging the terms yields: Taking the integral of both sides, while assuming J Ca is independent of x, gives: Resulting in:
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BME 314 HW #3 Fall 2010 (Roy) Removing from both sides and bringing the term to the other side of the equation, and then taking the exponential of both sides yields: Similarly, we can solve the derivation for: From space charge neutrality = J Ca , and this gives: = Solving for the exponential terms yields: Solving for V on both sides gives:
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BME 314 HW #3 Fall 2010 (Roy) V m = V i -V o = mV (d) Find the relationship between the voltage across the membrane and [CaCl 2 ] before equilibrium. From Problem 1b we know 2J Ca = J Cl or: -4μ[Ca 2+ ] - = μ[Cl -1 ] - Reducing the equation for similar variables on both sides we obtain: -4[Ca 2+ ] - = [Cl -1 ] - Separating the variables in terms of similarities on both sides: -4[Ca 2+ ] - [Cl -1 ] = - Factoring out similar constants: (-4[Ca 2+ ]- [Cl -1 ] )= (2-) Divide to get dv by itself: = Integrate both sides from inside to outside: = - - V i = voltage inside membrane V o = voltage outside membrane [R 3+ ] I = concentrations inside [R 3+ ] o
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