Chemistry, 07.04.2020 17:54 conroyjoann22

The inhibitory effect of an uncompetitive inhibitor is greater at high [S] than at low [S].Explain this observation. Check all that apply. a. At [S] > KM the effect of the inhibitor on reducing Vmax is apparent because as [S] decreases, V approaches Vmax[S] / α′.b. At [S] > KM the effect of the inhibitor on reducing Vmax is apparent because as [S] increases, V approaches Vmax / α′c. At [S] < KM the effect of the inhibitor is minimal because as [S] increases, V approaches Vmax / [S]KM. d. At [S] < KM the effect of the inhibitor is minimal because as [S] decreases, V approaches Vmax[S] / KM. B. Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32 μmol/L. If the total enzyme concentration was 1 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute?Express your answer to three significant figures. kcat =Calculate kcat/KM for the enzyme reaction. Express your answer to two significant figures and include the appropriate units. kcat /KM =

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The inhibitory effect of an uncompetitive inhibitor is greater at high [S] than at low [S].Explain t...

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