BIOC 460 Introduction to Enzymes – Sample Problems p. 1 BIOC 460 Sample Problems on Introduction to Enzymes 1. Assume an uncatalyzed reaction. Suppose that kF = 5 x 10–4 s–1, and Keq = 2.5 x 102. What is kR? 2. Consider the same enzyme as in #1, with rate constants and equilibrium constant given or calculated in #1 for the uncatalyzed reaction. Suppose that in the presence of an enzyme, kF(catalyzed) = 5 x 105 s–1. A. What is the rate enhancement brought about by the enzyme? B. What is kR(catalyzed)? C. What is Keq for the catalyzed reaction?BIOC 460 Introduction to Enzymes – Sample Problems p. 2 BIOC 460 Sample Problems on Introduction to Enzymes -- ANSWERS 1. Assume an uncatalyzed reaction. Suppose that kF = 5 x 10–4 s–1, and Keq = 2.5 x 102. What is kR? ANSWER: kR(uncat) = kF(uncat)/Keq(uncat) = 2 x 10–6 s–1. 2. Consider the same enzyme as in #1, with rate constants and equilibrium constant given or calculated in #1 for the uncatalyzed reaction. Suppose that in the presence of an enzyme, kF(catalyzed) = 5 x 105 s–1. A. What is the rate enhancement brought about by the enzyme? ANSWER: rate enhancement = k(cat) / k(uncat) kF(cat) / kF(uncat) = 5 x 105 s–1/ 5 x 10–4 s–1 = 1 x 109 (Rate enhancement is always unitless.) B. What is kR(catalyzed)? ANSWER: Catalyst must increase kR by the same factor that it increased kF. kR(uncat) (calculated in #1) = 2 x 10–6 s–1. Rate enhancement = 109 = kR(cat) / kR(uncat) so kR(cat) = 109 x kR(uncat) = 2 x 103 s–1. C. What is Keq for the catalyzed reaction? ANSWER: Catalyst doesn’t change Keq, so Keq(cat) = Keq(uncat) = 2.5 x
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