.Consider the enzymatic reaction scheme: Asparagine + H20 Aspartate + NH3: a) calculate the specific activity of the enzyme, if in 30 seconds as a result of a reaction involving 3 mg of the enzyme under optimal conditions (pH 8.0, 37 °C) 75 umol aspartate is obtained; b) describe the reasons for the decrease in enzyme activity after incubation for 6. . 10 minutes at 70 °C (provide an appropriate graph).
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- Write a balanced equation for each of the following reactions or reaction sequences. (a) The reaction catalyzed by PFK-2 (b) The conversion of 2 moles of oxaloacetate to glucose (c) The conversion of glucose to UDP-Glc (d) The conversion of 2 moles of glycerol toglucose (e) The conversion of 2 moles of malate to glucose-6-phosphateCalculate the standard free-energy change of the reaction catalyzed by theenzyme phosphoglucomutase, given that, starting with 20 mM glucose 1-phosphate and no glucose 6- phosphate, the final equilibrium mixture at 25 °C and pH 7.0 contains 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate. Does the reaction in the direction of glucose 6-phosphate formation proceed with a loss or a gain of free energy?Let’s consider vmax. Recall that uncompetitive inhibitors bind to the enzyme only after thesubstate is bound.a) As the concentration of substrate increases, the fraction of enzyme bound to substrate (circleone) increases / decreases?b) Does an uncompetitive inhibitor bind more readily when the substrate concentration is low orhigh?
- Compare and contrast Bound Fraction equation in ligand binding and Michaelis-Menten equation in enzyme kinetics, including their double-reciprocal forms. Discuss what Km is important for and what Vmax (or kcat) is important for? Under what (substrate) conditions is Km more important than Vmax, and under what (substrate) conditions is Vmax more important than Km? Based on the discussions in question 2, explain what type of inhibitors works best under (a) high substrate concentration and (b) low substrate concentration.Examine the ActiveModel for alcohol dehydrogenase and describe the structure and function of the catalytic zinc center.Using the ActiveModel for aldose reductase, describe the structure of the TIM barrel motif and the structure and location of the active site.
- Figure 27.3 illustrates the response of R (ATP-regenerating) and U (ATP-utilizing) enzymes to energy charge. a. Would hexokinase be an R enzyme or a U enzyme? Would glutamine: PRPP amidotransferase, the second enzyme in purine biosynthesis, be an R enzyme or a U enzyme? b. If energy charge = 0.5: Is the activity of hexokinase high or low? Is ribose-5-P pyrophosphokinase activity high or low? c. If energy charge = 0.95: Is the activity of hexokinase high or low? Is ribose-5-P pyrophosphokinase activity high or low?Consider the following mechanism: What kind of reaction is occurring during this step of glycolysis? What metabolite is in Box 1? What metabolite is in Box 2? What metabolite is in Box 3? What enzyme catalyzes this reaction? Circle and label one place on the mechanism where covalent catalysis is occurring. Suggest an amino acid that could fill the role of residue “A” in the mechanism above and draw its structure at pH 7.4. Suggest an amino acid that could fill the role of residue “B” in the mechanism above and draw its structure at pH 7.4. Under standard conditions, this reaction is unfavorable (DG¢° = 23.8 kJ/mol). What conditions in the cell allow for the actual free energy change to be lower (DG » 0), making the reaction readily reversible? Explain your answer. *please help and explain as well as you can*Consider the following mechanism: Circle and label one place on the mechanism where covalent catalysis is occurring. Suggest an amino acid that could fill the role of residue “A” in the mechanism above and draw its structure at pH 7.4. Suggest an amino acid that could fill the role of residue “B” in the mechanism above and draw its structure at pH 7.4. Under standard conditions, this reaction is unfavorable (DG¢° = 23.8 kJ/mol). What conditions in the cell allow for the actual free energy change to be lower (DG » 0), making the reaction readily reversible? Explain your answer. *please help and explain as well as you can*
- If a reaction has a ΔG°′ value of at least −30.5 kJ · mol−1, suffi -cient to drive the synthesis of ATP (ΔG°′ = 30.5 kJ · mol−1), can it still drive the synthesis of ATP in vivo when its ΔG is only −10 kJ · mol−1? Explain.Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMthere are A-D questions to this picture set up. A) What enzyme catalyzes this reaction? B) What is Delta G, please answer in Joules, K=19 C) If concentration of Glucose-1_Phosphate is 48.82 uM at equalibrium, what is the concentration of Glucose-6-phosphate in uM? D) If the reaction is NOT at equalibrium, what is delta G at 25C if the concentration of Glucose-1-phosphate is 15.04 uM and concentration of Glucose -6-phosphate is 1.62 mM? please answer in Joules and in significant figures. *note, 10^3uM in 1 mM Thank you!!