1aa3-2010-tut12-answerkey

1aa3-2010-tut12-answerkey - CHEMISTRY 1AA3 April 5 8 2010...

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CHEMISTRY 1AA3 April 5 – 8, 2010 TUTORIAL PROBLEM SET 12 – Full SOLUTIONS ______________________________________________________________________________ 1 Questions 1 - 4: Some practical enzyme kinetics Our diet contains many complex polysaccharides (polymers of different sugars), which must be broken down to monosaccharides in order to be converted to energy. Polysaccharides are broken down through a variety of routes, which includes significant amounts of disaccharide formation. Let's consider three disaccharides, cellobiose, gentiobiose and lactose, that must be hydrolyzed to the corresponding monosaccharides, glucose & galactose. A hypothetical enzyme, glycosidase X (gX), which catalyzes this reaction. gX will be digest each disaccharide with different steady state rate constants. 1. (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: 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 can plug the values into the Michaelis-Menten equation: v 0 = (10 × 10 -7 × 2 × 10 -5 ) / (10 -4 + 2 × 10 -5 ) = 1.7 × 10 -7 M/s
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CHEMISTRY 1AA3 April 5 – 8, 2010 TUTORIAL PROBLEM SET 12 – Full SOLUTIONS ______________________________________________________________________________ 2 (c) Why does the rate roughly double when [gentiobiose] doubles? When [S] << K M , the reaction is first order w.r.t. [S], which is the case in this question. 2. (a) If v 0 = 2 × 10 -6 M/s with 4 × 10 -4 M cellobiose and 1.8 × 10 -6 M/s with 2 × 10 -4 M cellobiose, what can we conclude about K M (cellobiose)? There is almost no change in v 0 when [S] doubles, so the reaction is roughly zero order w.r.t. [S]. This occurs when [S] >> K M , so we know that K M (cellobiose) << 2 × 10 -4 M. (b) If [gX] = 10 -7 M, make an estimate of k cat . When [S] >> K M , ( K M + [S]) [S]. The M-M equation reduces to v 0 = k cat [E] 0 , or k cat = v 0 /[E] 0 k cat = (2 × 10 -6 M/s) / 10 -7 M = 20 s -1 3. (a) With lactose: k cat = 2 s -1 and K M = 3 × 10 -5 M. What would the v 0 vs [lactose] profile look like in comparison to the gentiobiose profile? The enzyme reaches 1/2 maximal rate at a lower value of [S] with lactose because its K M value is lower. Its k cat value is also lower, so the maximal rate is lower than gentiobiose. The maximal rate for gentiobiose is off the top of the graph shown. (b)
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1aa3-2010-tut12-answerkey - CHEMISTRY 1AA3 April 5 8 2010...

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