Suppose the data below are obtained for an enzyme catalyzed reaction with and without the inhibitor I. (s)( mM) 0.2 0.4 0.8 1.0 2.0 4.0 V without i (mM/min) 5.0 7.5 10.0 10.7 12.5 13.6 V with I (mM/min) 3.0 5.0 7.5 8.3 10.7 12.5   Make a Lineweaver Burke plot for this data using graph paper or a spreadsheet  Calculate KM and Vmax without inhibitor.  What type of inhibition is observed? show graph and work 2. Give the Lineweaver Burk equation and define all the parameters. 3. When substrate concentration is much greater than Km, the rate of catalysis is almost equal to   a. kcat b. none of these c. all of these d. Kd e. Vmax

Energetics of the Hexokinase Reaction The standard-state free energy change. Gfor the hexokinase reaction, is — 1 6.7 kJ/mol. Use the values in Table I to calculate the value of Gfor this reaction in the erythrocyte at 37°C.

Distinguishing the Mechanisms of Class I and Class I Aldolases Fructose bisphosphate aldolase in animal muscle is a class 1 aldolase, which forms a Schiff base intermediate between substrate (for example. fructose-1, 6-bisphosphate or dihydroxyacetone phosphate) and a lysine at the active site (see Figure I8.12). The chemical evidence for this intermediate conies from studies with aldolase and the reducing agent sodium borohydride, NaBH4. Incubation of the enzyme with dihydroxyacetone phosphate and NaBH4 inactivates the enzyme. Interestingly, no inactivation is observed if NabH4 is added to the enzyme in the absence of substrate. Write a mechanism that explains these observations and provides evidence for the formation of a Schiff base intermediate in the aldolase reaction.

MATHEMATICAL The enzyme -methylaspartase catalyzes the deamination of -methylaspartate [V. Williams and J. Selbin, J. Biol. Chem. 239, 1636 (1964)]. The rate of the reaction was determined by monitoring the absorbance of the product at 240 nm(A240). From the data in the following table, determine KM for the reaction. How does the method of calculation differ from that in Questions 26 and 27?