Given the following enzyme pathway, A→B→C→D The AGO' for each of the individual reactions is as follows: A→B is -10; for B→C is +5; for C→D is +1 Using this information, calculate the AGO' for the complete process of A→D. +5 +2 -4
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- In a bisubstrate reaction, a small amount of the fi rst product P is isotopically labeled (P*) and added to the enzyme and the fi rst substrate A. No B or Q is present. Will A (= P—X) become isotopically labeled (A*) if the reaction follows a Sequential mechanism?One of the examples that we have used to illustrate the concept of equilibrium is the isomerization of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P), which is the second step in g ycolysis. Draw a graph to show how the reaction Gibbs energy varies with the fraction fof F6P in solution.Label the regions of the graph that correspond to the formation of F6P and G6P being spontaneous, respectively.The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.
- for the following reactions: arrive at the Michaelis menten enzyme kinetics: E+2A <-----k1,k2----->A.E+A reversible A.E+A<------k3,k4------>A 2E reversible A2.E------k5--------> E+CD non-reverisble show all stepsThe enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose 6-phosphate. After the reactants and products were mixed and allowed to reach equilibrium at 25°C, the concentration of glucose 1-phosphate was 4.5 mM and that of glucose 6-phosphate was 86 mM. Calculate Keq' and AG for this reaction. The reaction coordinate diagram for an enzyme-catalyzed reaction is shown below. How many transition states and intermediates are in the reaction? Is the reaction thermodynamically favorable? Which step is the rate-determining step of the reaction? G Reaction coordinateIn enzyme kinetics, for the reversible with two central complexes mechanism, please provide complete proof that the rate equation is the equation below. The variables denoted with f indicate forward direction while the variables denoted with b indicate backward direction.
- The enzyme ATCase (Aspartate TransCarbamoylase) catalyzes an early step in the synthesis of the pyrimidine nucleotides. The activities of the ATCase reactions are shown in the following plots. Which one of the following statements is most correct description about this enzyme and/or this enzyme-catalyzed reaction? ATP No allosteric effectors Vo CTP ATCase 10 20 30 40 (Aspartate] (mM) O A. This enzyme is allosterically inhibited by ATP. B. This enzyme is composed of multiple subunits. C. "ATP" binds to catalytic subunits of ATCase. O D."CTP" binds to the R-state enzyme. O E. At high concentrations of ATP, ATP binds to the T-state enzyme, thus increase the affinity for the substrate [S].You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1 is much larger than ?2, which of the following statements are correct? The mutant version has a higher affinity for the substrate. The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. The reaction equilibrium is reached once there is no net change in the concentration of the substrate or the product. Based on the data table and your initial…In enzyme kinetics, for the reversible with one complex mechanism, please provide complete proof that the rate equation is the equation below. The variables denoted with f indicates forward direction while the variables denoted with b indicate backward direction.
- The active site of an enzyme that uses a general acid-base catalytic mechanism contains a Glu and an Asp residue (both of which are essential for catalysis) with pKa values of 5.9 and 4.5, respectively. If the enzyme is found in the lysosome (pH = 5.2), which residue will act as the general acid and which will act as the general base during the initial steps of the reaction?You are working on an enzyme that obeys standard Michaelis-Menten kinetics. Based on the following reaction expression, what is the Km value for this enzyme? E+SESE + P . . . k₁ = 880.8 M-¹5-1 k.₁ = 42.18 S-1 k₂ = 56.29 S-1Consider an enzyme that catalyzes the reaction S2 P, by the following simple reaction mechanism: k, E + S 2 E•S →E kcat + P Suppose the enzyme acquires a mutation that causes k1 to be 10-times smaller than for the wild-type (non-mutant) enzyme. Suppose you measure the initial reaction rate (vo) at several different [S] for the mutant and the wild-type enzymes. Under what conditions would the mutation have a greater effect on the reaction rate (vo) of the mutant enzyme compared to the wild-type enzyme - at very low [S], or at very high [S]? Explain briefly how you decided.