An enzyme-catalyzes the isomerization of substrate S to product P. The enzyme has a molecular weight of 120,000 g/mol. In assays using 1 μg of enzyme per assay the Km was 3 x 10^-3M and the Vmax was 2.75 μmole per minute. What would be the Kcat (turnover number or molecular activity) of the enzyme under these conditions? 2.75 min^-1? 3,300,000 min^-1? 330,000 s^-1? 19,800,000 min^-1? 5,500 s^-1?

An enzyme-catalyzes the isomerization of substrate S to product P. The enzyme has a molecular weight of 120,000 g/mol. In assays using 1 μg of enzyme per assay the Km was 3 x 10^-3M and the Vmax was 2.75 μmole per minute. What would be the Kcat (turnover number or molecular activity) of the enzyme under these conditions? 2.75 min^-1? 3,300,000 min^-1? 330,000 s^-1? 19,800,000 min^-1? 5,500 s^-1?

Biochemistry

9th Edition

ISBN:9781305961135

Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal

Publisher:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal

Chapter6: The Behavior Of Proteins: Enzymes

Section: Chapter Questions

Problem 31RE: MATHEMATICAL You do an enzyme kinetic experiment and calculate a Vmax of 100 mol of product per...

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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…
Part 1: Assess the following partial results section below by editing it for brevity by omitting any unnecessary parts (1 point), explain why you decided to remove certain sections (1 point): To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each molecule were prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized for the enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units of inorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture was incubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of the substrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes. Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of the molecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were further diluted to find the minimal…
A synthetic substrate, the para-nitrophenylacetate (PNPA), is used to monitor enzyme activity of protein P. The product of the hydrolysis reaction absorbs at 410 nm with a molar extinction coefficient of 4 000 M-¹.cm-¹. 1- Write the reaction catalyzed by the protease P using the pNPA substrate. 2- The enzyme extract is too concentrated and a 1/300 dilution is needed for enzyme tests. Considering that you would need at least 600 µL of diluted enzyme extract for activity tests, indicate which volume of buffer and enzyme extract you must use for the dilution.
KINETIC CONSTANT No Na2HPO4 25mM Na2HPO4 50mM Na2HPO4 Vmax nmol p-NP. Min- 20.3252 14.30615 17.30104 Km mM -0.819106 -0.46495 -0.352941 1. What does this suggest about the structure of the active side of the enzyme?
200 ml of a 2% protein solution containing an enzyme that you want to purify. Half of the sample is subjected to method A, consisting of fractionated precipitations and 5 ml of final solution are obtained, with a concentration protein equal to 5 mg / ml and enzymatic activity equal to 2000 U / ml. The other half is subjected to method B, consisting of ion exchange chromatography, and a final solution of 10 ml, with protein richness equal to 10 mg / ml and with an activity enzymatic also equal to 2000 U / ml. You want to know: a) Which of the two methods has provided the purest enzyme. b) By which of the methods the greatest amount of enzyme has been obtained.
CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the Km of the reaction represented by Line (B).
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the catalytic efficiency of the reaction represented by Line (A).
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Line (B) represents ____________. A. an enzyme-catalyzed reaction in the absence of any inhibitor. B. an enzyme-catalyzed reaction in the presence of a competitive inhibitor. C. an enzyme-catalyzed reaction in the presence of an uncompetitive inhibitor. D. an enzyme-catalyzed reaction in the presence of a noncompetitive, or mixed, inhibitor.
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the Vmax of the reaction represented by Line (C). Show all mathematical work, please.
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. QUESTION: Line (A) represents ____________. A. an enzyme-catalyzed reaction in the absence of any inhibitor. B. an enzyme-catalyzed reaction in the presence of a competitive inhibitor. C. an enzyme-catalyzed reaction in the presence of an uncompetitive inhibitor. D. an enzyme-catalyzed reaction in the presence of a noncompetitive, or mixed, inhibitor.
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Which of the following statements is true? Select any/all answers that apply. A. Both types of inhibitor mediate a slope effect on the Lineweaver-Burke plot. B. Both types of inhibitor decrease the apparent Vmax for this enzyme-catalyzed reaction. C. Both types of inhibitor alter the apparent Km of this enzyme-catalyzed reaction. D. Lines (A) and (C) share the same X-intercept, indicating that the noncompetitive inhibitor decreases the apparent Km of this enzyme-catalyzed reaction. E. Lines (A) and (C)…