Given the active site diagram below, which circled component coordinates the cofactor? 1 05 02 5 NH 2 *H₂N. Mn²+ HN H₂ Mn²+ -3
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- 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 \muμMGive a complete and well descriptive definition of the following:1.1 Enzyme catalysis1.2 Co-enzyme1.3 Negative heterotropic co-cooperativitComment on Figure 2a and Figure 2b regarding stabilization of the transition state and destabilization of ES complex during catalysis. (b) AG AG EX EX E+S E+P ES EP AG AG,+ AG- TAS ES EP Figure 2
- The protein catalase is an enzyme that catalyzes the decomposition of hydrogen peroxide:2 H2O2 (aq) → 2 H2O (l) + O2 (g)and has a Michaelis-Menten constant of 25 × 10-3 mol·dm-3 and a turnover number of 4.0×107s-1.The total enzyme concentration is 0.016×10-6 mol·dm-3 and the initial substrate concentration is4.32×10-6 mol·dm-3 Calculate the maximum reaction rate (????) for this enzyme, and the initial rateof this reaction. Note that catalase has a single active site.Why is the overall coupled reaction exergonic?The equilibrium constant for the hydrolysis of the peptide alanylglycine (Gly-Ala in the reaction from Part B) by a peptidase is K = 9.0 × 10² at 310 K. Calculate AG for this reaction. Express the Gibbs free energy to three significant figures. AG = Submit ΠΑΠΙ ΑΣΦ Request Answer ? kJ/mol Keq [Gly] [Ala] [Gly-Ala]
- Consider the mechanism of enolase, as indicated below. Which of the following correctly describes the roles of the Mg2+ as illustrated in the figure? (This is a multi- select question). Mg2+ Mg2 Enolase PO3- OH -C-C-H H OH HO H-N-H Lys 345 Glu211 2-Phosphoglycerate bound to enzyme Mg2+ Mg2 PO3- OH C-C-H OH HO H H-N*-H Lys 345 O Glu211 Enolic intermediate HOH PO3- H Phosphoenolpyruvate The metal ion (Mg2+) is helping to stabilize the extra negative charge that developed on the carboxyl group in the enolic intermediate. The metal ion (Mg2+) is serving as a general base, removing a proton in order to improve the quality of the nucleophile. The metal ion (Mg2+) is assisting in the oxidation of the carboxyl carbon through metal ion catalysis. The metal ion (Mg2+) is helping to orient the substrate properly in the active site. The metal ion (Mg2+) is accepting a proton in order to improve the quality of the leaving group.Acid phosphatases are an important group of enzymes that can be detected in human blood serum. Under slightly acidic conditions (pH 5.0), this group of enzymes can hydrolyze biological phosphate esters as follows: R-O-P-O;² + H 20 → R-OH + HO-P-O;² Acid phosphatases are produced and can be detected in erythrocytes, kidney, spleen, the liver, and prostrate gland. The enzyme from the prostrate gland is clinically important because an increased activity in the blood is frequently an indication of cancer of the prostrate gland. Tartrate ion can strongly inhibit the phosphatase from the prostrate gland, but not acid phosphatases from other tissues. How can you use the information above to develop a specificHuman blood serum contains a class of enzymes known as acid phosphatases, which hydrolyze biological phosphate esters under slightly acidic conditions (pH 5.0): R-O-P-O3-2 + H2O --> R-OH + HO-P-O3-2. Acid phosphatases are produced by erythrocytes, the liver, kidney, spleen, and prostate gland. The enzyme from the prostate gland is clinically important because an increased activity in the blood is frequently an indication of cancer of the prostate gland. The phosphatase from the prostate gland is strongly inhibited by tartrate ion, but acid phosphatases from other tissues are not. How can this information be used to develop a commercial specific procedure for measuring the activity of the acid phosphatase of the prostate gland in human blood serum? * 1. Prostate cancer cannot be diagnosed biochemically. 2. Use tartrate to inhibit phosphatase from prostate gland and then subtract the results from the total serum enzyme activities to get an…
- b) Why might the compound shown below act as a transition state analog of phosphoglucose isomerase? A drawing of the normal transition state for this enzyme is needed. HO- OH T .N -OH -OH CH₂OPO₂²-The 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 coordinateThe ΔG°′ for hydrolytically removing a phosphoryl group from ATP is about twice as large as the ΔG°′ for hydrolytically removing a phosphoryl group from AMP (−14 kJ · mol−1). Explain the discrepancy.