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- An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-If 287.9 umol of enzyme X has a Vmax = 47.8 mmol/sec, what is the value of kcat %3D sec-1? Please report answer with 1 decimal place. Please do not report units. Your Answer: Answer units MacBook Air 888 F5 F4 F3 F2 %23 %24 %24How many net molecules of nucleoside triphosphate (ATP and equivalent molecules) are produced by complete aerobic catabolism of a glucose going through glycolysis, the pyruvate dehydrogenase complex and the citric acid cycle (TCA cycle)? Do not count the ATP eventually generated by re-oxidation of reduced coenzymes, just the number of NTPs produced in reactions of these pathways. Choose the one best answer. 02 03 04 05 06 8
- ch Select all statements that are correct. Note there might be more than 1 correct statement. From the Lineweaver-Burk plot the equilibrium constant (Keq) can be obtained The Lineweaver-Burk plot gives a more accurate prediction for Vmax than the Michaelis- Menten plot The Lineweaver-Burk plot assumes that products and reactants are present at equal concentrations during the entire time of the reaction The Lineweaver-Burk plot shows velocity of reaction vs substrate concentration The Lineweaver-Burk plot shows 1/velocity of reaction vs 1/substrate concentration O 20°C D) // EThe following were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Reaction Velocity (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 20 652 1000 The Km for this enzyme is approximately _____________. (Round to the nearest integer)An enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a keat value of 181 s1. At an initial enzyme concentration of 0.0100 µM, the initial reaction velocity was found to be 1.07 x 10-0 µM/s. What was the initial concentration of the substrate, S, used in the reaction ? Express your answer in micromolar to three significant figures. > View Available Hint(s) ? [S] !! µM Submit
- 1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…Sample pH vs Initial Velocity (AA450/60 seconds) a bo16 0.0014 0.0012 o.001 0.0008 0.0006 0.0004 0 0002 pH Sample what does it mean? Does it agree with what we know about enzyme kinetics and why or why nol? If there were anomalies, what were they and what are some possible reasons for this occurring? Was an optimal pH identifiable? Why or why not? Initial Velocity (AA450/60seconds)8 7- 6- 2- 1- 45 90 135 180 225 270 315 360 405 450 495 540 58! [Substrate] (nM) What is the Vmax of the enzyme data shown above? rate (nM/sec) 3.
- You are working on an enzyme that obeys standard Michaelis-Menten kinetics. What variable is the V, dependant on if the concentration of the substrate is substantially higher than the concentration of the enzyme? [S] [E] [ES] O [P] O not enough information provided7.1 The following results were obtained for an enzyme-catalysed reaction Substrate concentration (mmolr'): Initial velocity (umol I' min"): 6.67 182 10.0 20.0 40.0 147 233 323 400 Calculate Km and Vmax8 7- 6 2- 180 225 270 315 [Substrate] (nM) 45 90 135 360 405 450 495 540 58! What is the KM of the enzyme data shown above? rate (nM/sec) 3.