HW3%20Chap8%20plus%20Matlab

HW3 Chap8 plus%2 - BME 100L Modeling Cellular and Molecular Systems Homework 3 Chapter 8 Problems 2 4 5 7 10 and 11 Due 5pm September 23rd(reversed

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BME 100L. Modeling Cellular and Molecular Systems Fall 2009 Homework 3 Chapter 8. Problems 2, 4, 5, 7, 10, and 11 Due: 5pm September 23 rd ( reversed due date ) Hints: 2.(a) Hydration reaction mechanism is CO 2 + E k 1 k -1 ECO 2 ECO 2 + H 2 O k 2 E + HCO 3 - + H + ] [ ] ][ [ ' 2 2 1 2 1 ECO E CO k k k K H M , V max = k 2 ’[E] 0 [H 2 O] is combined with k 2 to form a new constant k 2 ’ = k 2 [H 2 O] The curve fitting of the data will provide the values of K M and V max . (b) Dehydration reaction mechanism is CO 2 + E k -1 ECO 2 ECO 2 + H 2 OE + H C O 3 - + H + k 2 k -2 ] [ ] ][ [ ' ' 2 3 2 2 1 ECO E HCO k k k K DH M , V max = k -1 [E] 0 [H + ] can be absorbed into k -2 to form a new constant k -2 ’ = k -2 [H 2 O] (c) E + CO 2 + H 2 O complex E + HCO 3 - + H + k 1 k -1 k 2 k -2 When the system is at equilibrium, both reversible reactions are at equilibrium. Using the values of K M H , K M DH , k 2 ’ and k -1 obtained from (a) and (b) to calculate the overall equilibrium constant (K eq = [HCO 3 - ] e [H + ] e /[CO 2 ] e [H 2 O] e or K eq = [HCO 3 - ] e [H + ] e /[CO 2 ] e .
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This note was uploaded on 01/17/2010 for the course BME 100 taught by Professor Yuan during the Fall '07 term at Duke.

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HW3 Chap8 plus%2 - BME 100L Modeling Cellular and Molecular Systems Homework 3 Chapter 8 Problems 2 4 5 7 10 and 11 Due 5pm September 23rd(reversed

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