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Unformatted text preview: glutamate harder to protonate. 12. a. Deducing ~ 3 mM is good thinking but not technically correct. You should set the michaelis menton equation equal to 1⁄4 Vmax and solve for [S]. If you do this, you’ll get 2 mM. b. The kcat/Km ratio for Enzyme Y is 5 and the kcat/Km ratio for Enzyme X is 10,000. Therefore, Enzyme X is more catalytically efficient. 13. a. By inspection, you can estimate Vmax. The velocity changes very little as the substrate concentration increases 5 fold from 2 to 10 mM compared to the much larger velocity changes with much smaller increases in [S] at lower [S]. Therefore, Vmax = 140 μM/min. However this can be calculated using the data provided. Using the [S], Vo combinations of 4 x 10 5 M (40 μM), 112 μM/min and 1 x 10 4 M (100 μM), 128 μM/min, I will show a sample calculation: 1/[S] = x and 1/Vo = y (x1, y1) = (0.025, 8.9 x 10 3) (x2, y2) = (0.01, 7.8 x 10 3) slope = ∆y / ∆x = (y1 y2) / (x1 x2) = (8.9 x 10 3 – 7.8 x 10 3) / (0.025 0.01) = 0.0011 / 0.015 = 0.073 Therefore, Km/Vmax = 0.073, substitute this into 1/Vo = Km/ Vmax (1/[S]) + 1/ Vmax along with any other [S], Vo pair in reciprocal form. (I’m using the 2 x 10 5 M (20 μM), 95 μM/min points) 1.1 x 10 2 = 0.073(0.05) + 1/ Vmax 1.1x10 2 =3.7x10 3+1/Vmax 0.0071 = 1/ Vmax Vmax = 141 μM/min For Km calculations: Method 1: If you estimated Vmax by inspection, then the simplest...
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 Spring '14
 Potenza

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