Unformatted text preview: Explain. (2 points) The Km is a constant and would not be affected by the concentration of substrate. How would the Km of the reaction be affected by adding more enzyme? Explain. (2 points) The Km is a constant and would not be affected by the concentration of enzyme. 3. An enzyme has a K cat /K M of 10 6 M-1 s-1 . In a reaction with 1nM enzyme and 10 µ M substrate if the enzyme’s K M for substrate is 100 µ M, What is Vmax? (2 points) Vmax = [E] Kcat Kcat = 10 6 x 10-4 = 100 s-1 Vmax = 10-9 M x 100s-1 = 10-7 Ms-1 What is the reaction rate at this concentration of substrate? (2 points) V = Vmax [S]/Km + [S] = (10-7 Ms-1 x 10-5 M)/(10-4 M + 10-5 M) = 10-12 M 2 s-1 /1.1 x 10-4 M = 9 x 10-9 Ms-1 What fraction of the substrate is bound to enzyme under these conditions? (2 points) Concentration of ES = [E] x (Vo/Vmax) = 10-9 M x (9 x 10-9 Ms-1 /10-7 Ms-1 ) = 9 x 10-11 M Fraction of substrate = 0.9 x 10-11 M/10-5 M = 9 x 10-6...
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- Winter '10
- Reaction, Enzyme inhibitor, 4m, 1Nm