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Unformatted text preview: Assuming steady state, this gives [EP] = {k2/(k2 + k3)}[ES] Then you get: [ES] = {(k2 + k3)/k2}[EP] Plug into eq 1 and get: {(k2 + k3)/k2}[EP] = (k1[E][S] + k2[EP])/k2 [EP] = (k1/k3)[E][S] [E] = {(k3/k1)[EP]}/[S] Mass balance: [E0] = [E] + [ES] + [EP] Plug in to get: [E0] = {(k3/k1)[EP]}/[S] + {(k2 + k3)/k2}[EP] + [EP] [E0] = [EP] {(k3/k1)/[S] + {(k2 + k3)/k2} + 1} [E0] = [EP] {(k3/k1) + {(k2 + k3)/k2}[S] + [S]}/[S] [EP] = [E0][S]/{k3/k1 + [S](1+(k2 + k3)/k2)} V = k3[EP] = k3[E0][S]/{k3/k1 + [S](1+(k2 + k3)/k2)} V = Vmax[S]/ {k3/k1 + [S](1+(k2 + k3)/k2)} 4. Identify apparent Km and Vmax for [I] = 1 if the enzyme concentration was half as what is shown on the plot. (0.5 pts each, 1 pt total) For [I] = 1 X intercept: 0.25 Y intercept: 2.5 Enzyme concentration halved: X intercept: 0.25 Y intercept: 5 Vmax: 1/(Y intercept) = 0.2 Km = 1/(0.5) = 2 X intercept = 1/[Km(1+[I]/K I )] = 0.25 apparent Km = Km(1++[I]/K I ) = 1/(X intercept) = 4...
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This document was uploaded on 02/23/2010.
 Winter '08

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