Unformatted text preview: key here is that the substrate is rate limiting. So your graph should indicate a decrease in rate as a function of time. 7. From top to bottom: A, D, B, D, B, A, B or C, B 8. b. If α is greater than α’, then KI is less than K’I. Since KI = [E] [I]/[EI] and K’I = [ES] [I]/[ESI], more inhibitor is bound to E than ES. 9. a. Since the enzyme catalyzes the reaction at 0.5 times its maximal rate of catalysis when the substrate is present at 3.3 mM (given) then the Km is 3.3 mM. Using the M M equation Vo =Vmax [S]/(Km + [S]) or some other equation, you can calculate the Vmax by plugging in a rate, [S] data point and the given Km. The Vmax is 11 μM/s. b. Now solve the same equation for Vo using the Km and Vmax from part a. Vo is 0.32 μM/s. c. Vmaxapp =Vmax/ α’ ; α’= 1 + 0.01/1.7 x 10 3; So Vmaxapp = 1.6 μM/s. The Kmapp =Km/ α’; So the Kmapp = 480 μM. Now use the M M equation to solve for Vo with the given [S]. Vo = 1.5 μM/s. 10. Same as PS4, Q8. a. Vmax = kcat [ET], Vmax is 110 μM/s. b. Since 55 μM per second is 1⁄2 Vmax , you are simply being asked to calculate the Kmapp. The Vmaxapp =Vmax for competitive inhibitors. α is 5.76 and Kmapp =αKm. Km was given as 32 μM. Therefore the [S] is 5.76(32) = 184.4 μM. 11. The pK of lysine will increase making lysine harder to deprotonate and the pK of glutamate will decrease making...
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This homework help was uploaded on 03/02/2014 for the course BIOCHEMIST 395 taught by Professor Potenza during the Spring '14 term at NMSU.
 Spring '14
 Potenza

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