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- a. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Draw the pathway leading to the production of protein E.The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of β-Dglucose to D-glucono-δ-lactone. This enzyme is highly specific for the β anomer of glucose and does not affect the α anomer. In spite of this specificity, the reaction catalyzed byglucose oxidase is commonly used in a clinical assay for total blood glucose—that is, for solutions consisting of a mixture of β- and α-D-glucose. What are the circumstances required to make this possible? Aside from allowing the detection of smaller quantities of glucose, what advantage does glucose oxidase offer over Fehling’s reagent for measuring blood glucose?The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of B-D-glucose to D-glucono-6-lactone. The enzyme is highly specific for the B anomer of glucose and does not affect the a anomer. In spite of this specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose -that is, for solutions consisting of a mixture of B- and a-D-glucose, as well as other sugars present in blood. The oxidation proceeds in the presence of oxygen and forms hydrogen peroxide, in addition to the lactone. A second enzyme, called peroxidase, catalyzes the reaction of hydrogen peroxide with colorless compounds to create a colored product, which is quantified with a simple spectrophotometer. What are the circumstances required to make this possible? Aside from allowing the detection of smaller quantities of glucose, what advantage does glucose oxidase offer over the Fehling's reagent for measuring blood glucose?
- *The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of 3-D-glucose to D-glucono-6- lactose. This enzyme is highly specific for the ß anomer of In spite of this glucose and does not affect the a anomer. specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose that is, for solutions consisting of a mixture of 3- and a-D- glucose. What are the circumstances required to make this possible? Aside from allowing the detection of smaller quan- tities of glucose, what advantage does glucose oxidase offer over non-enzymatic oxidizing agents like Tollens reagent? *Is B-D-glucosamine a reducing sugar?The enzymatic activity of PFK1 is generally measured by set- ting up a coupled enzyme assay system whereby aldolase, triose phos- phate isomerase, and glycerol-3-phosphate dehydrogenase are added to the assay mixture. For the latter enzyme, NADH is added and its change in concentration is readily monitored at 340 nm. Write the chain of reactions catalyzed by these enzymes using structural formulas, label substrates and products, and explain why the coupled en- zyme assay system leads to oxidation of NADH. While the chain of reac- tions is similar to those in glycolysis, there is a critical difference because of the dehydrogenase enzyme. Describe how this enzyme causes the chain of reactions to differ from those in glycolysis.The kcat and KM for chymotrypsin-catalysed cleavage of a synthetic substrate, S, were determined to be 60 s-1 and 0.5 mM, respectively, using steady-state kinetics. The concentration of enzyme in the assay was 5 x 10-8 M. ii) Draw a graph on the axes shown in Figure 1 below showing how the initial rate of the reaction varies with [S], making use of the values for KM and Vmax. Label the axes with appropriate titles and units.
- For the following aspartate reaction in the presence of inhibitor, Km = 0.00065 M. Determine Vmax in both reactions and in the reaction without inhibitor, the Km. Identify whether the inhibition is competitive, non-competitive or uncompetitive. ( see attached picture ) how I and S bind to the E as shown by the Lineweaver Burk plot. the significance of the following obtained values for Km and Vmax. effect in slope and x-interceptEquation 7a of the text, (Umax/KM1) [S] (v'max/KM2) [P] 1+ [S]/KM1 + [P]/KM2 - gives the expression for the rate of formation of product by a modified version of the Michaelis-Menten mechanism in which the second step is also reversible. Derive the expression and find its limiting behaviour for high and low concentrations of substrate.6-Mercaptopurine , after its conversion to the corresponding nucleotide through salvage reactions, is a potent competitive inhibitor of IMP in the pathways for AMP and GMP biosynthesis. It is therefore a clinically useful anticancer agent. The chemotherapeutic effectiveness of 6 mercaptopurine is enhanced when it is administered with allopurinol. Explain the mechanism of this enhancement.
- The enzymatic activity of lysozyme is optimal at pH 5.2 and decreases above and below this pH value. Lysozyme contains two amino acid residues in the active site essential for catalysis: Glu35 and Asp52. The pK value for the carboxyl side chains of these two residues are 5.9 and 4.5 respectively. What is the ionization state of each residue at the pH optimum of lysozyme? How can the ionization states of these 2 amino acid residues explain the pH-activity profile of lysozyme?Staphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)The Y150F (Tyr 150 to Phe) mutant of ß-lactamase catalyzes the reaction with a keat of 3.1 * 10³ s¹ at pH = 7.0 and 37 °C. Assuming this decrease in keat is due to the loss of transition state stabilization by the Y150F mutant, calculate the AAG* between the wild-type and Y150F enzymes. Show your work and state any assumptions you make in solving this problem.