1aa3-2012-tut5-a

1aa3-2012-tut5-a - CHEMISTRY 1AA3 Week of FEBRUARY 6, 2012...

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CHEMISTRY 1AA3 Week of FEBRUARY 6, 2012 TUTORIAL PROBLEM SET 5 – QUESTIONS ______________________________________________________________________________ Page 1 of 8 Conceptual Questions about catalyzed reactions 1. How does a catalyst change the rate of a chemical reaction? SOLUTION: A catalyst provides a different pathway for the process, and lowers the activation energy, thus increasing the overall rate of the reaction. 2. Does adding a catalyst change the ratio of starting material(s) to product(s) in an equilibrium reaction? SOLUTION : adding a catalyst changes the pathway for the reaction, but not the potential energies of the starting material(s) or product(s). Adding a catalyst will speed establishment of the equilibrium, but will not change the position of the equilibrium. 3. Explain how catalyzed reactions can be approximately zero order as well as approximately first order with regards to substrate, depending on conditions. SOLUTION: This easiest way to explain this feature is with the help of a rate versus [S] graph: At low [S], the fraction of catalytic sites occupied by substrate ([E . S]) is low, and hence increases nearly linearly with increasing [S]. Conversely, at high [S], most active sites are occupied by substrate, and further increases in [S] hence do not lead to any significant increase in [E . S], or rate. The transition between first order and zero order conditions is defined as [S] = K M Enzyme kinetics Our diet contains many complex polysaccharides (polymers of different sugars) which must be broken down to monosaccharides to be converted to energy. This process often involves disaccharide intermediates. Assume that cellobiose, gentiobiose and lactose are three disaccharides that are hydrolyzed to the two constituting monosaccharides, glucose & galactose, by a hypothetical enzyme, glycosidase X, or gX, with different steady state rate constants.
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O O O HO HO OH HO OH OH cellobiose glycosidase X O O O HO HO OH HO OH OH lactose 2 x monosaccharides (2 x glucose or glucose + galactose) HO O HO OH OH gentiobiose O O HO HO OH OH OH OH OH OH 4. (a) With gentiobiose: k cat = 10 s -1 and K M = 10 -4 M. If: [gX] = 10 -7 M, and [gentiobiose] = 10 -5 M, what is v 0 ? SOLUTION: We can plug the values into the Michaelis-Menten equation: [S] [S] [E] 0 M cat K k v 0 v 0 = (10 × 10 -7 × 10 -5 ) / (10 -4 + 10 -5 ) = 9 × 10 -8 M/s (b) If [gentiobiose] increases to 2 × 10 -5 M, what is the new v 0 ? We again plug these values into the Michaelis-Menten equation:
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1aa3-2012-tut5-a - CHEMISTRY 1AA3 Week of FEBRUARY 6, 2012...

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