1/v (sec/μmol)Enzymes A and B catalyze different reactions but use the same reactant molecule as a substrate. The graphbelow presents the double-reciprocal plots observed when enzyme A and enzyme B are mixed together in asingle test tube containing molecule X.X12° 1>0.450.40Y0.350.30Z0.250.200.150.102.050.00-1.20 -1.00 -0.80 -0.60 -0.40 -0.20 0.00 0.20 0.40 0.60 0.80 1.00 1.201/[X] (1/μM)A-BWhat are the Vmax and the Km values for each enzymeunder these conditions?AVmax= 5.9μmolsecKM=1.00 μΜBVmax = 10.μmolsecKM = 3.0 μΜ

The Km and Vmax are two parameters we are looking into if we study enzymes. Therefore, we can use…

Answered: 1/v (sec/μmol) Enzymes A and B catalyze…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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The following data were collected for an enzyme that can be described by Michaelis Menton kinetics. Use graphical analysis to determine the Km and Vmax for this exzyme / substrate combination. [S] mM Product formed (umol/min) 0.21 1.5 2.0 0.24 3.0 0.28 4.0 0.33 8.0 0.40 16.0 0.45
If a data from an enzyme experiment is plotted as a Lineweaver - Burk plot, and the Vmax is 0.02 sec/mol, and x-intercept is -2.5 mM-1, then what is the KM value?
You are showing your dad graphs of lactase activity. In the graph below, what should the vertical axis be titled? 12 10 4 2 2 6. 8 10 12 14 16 Time (minutes) (Accessibility info: One line on graph, uses these coords: 0,0; 2, 025; 4,1; 6,4; 8,7; 10,10; 12,11; 14, 11.25) Curve is slow to rise, up on a linear slope, slowly flattens at top.) Lactase (units/ml) cross out а. O b. Lactose (µM) cross out С. Glucose (µM) cross out d. Galactose (µM) cross out e. Starch (uM) cross out f. None of these answers are correct. cross out O O O O
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Acetazolamide is a drug which inhibits carbonic anhydrase. Carbonic anhydrase participates in regulation of the pH and bicarbonate content of a number of body fluids. Figure 2 shows the experimental curve of initial reaction velocity (as percentage of Vma) versus [S] (concentration) for the carbonic anhydrase reaction. The graph also shows the curve in the presence of acetazolamide. 100 No inhibitor 50 Acetazolamide 0.2 0.4 0.6 0.8 [S] (mM) Figure 2 (i) Compare the maximal velocities and Michaelis Menten constants of the enzyme in the absence and the presence of the inhibitor acetazolamide. Determine the nature of inhibition by acetazolamide. Explain your answer. (ii) Name TWO (2) other types of inhibitions besides the inhibition shown by acetazolamide. Sketch a graph of V versus [S] showing curves in the absence of an inhibitor and in the presence of the types of inhibitors not shown by acetazolamide. ("AJO %) A
SUBSTRATE CONCENTRATION [S] µM INITIAL VELOCITY V0 s-1 10 0.13 25 0.27 50 0.45 100 0.65 150 0.77 200 0.85 300 0.94 500 1.03 (i) a) Construct an empty table with the following column headings: Substrate concentration [S] and initial velocity (Vi) where [S] has the unit µM, and Vi has the unit mM/s. (ii) The table provided is the enzyme kinetic data for your mutated enzyme, whereby Vi was expressed using the unit ∆A(405 nm)/s. Using the standard curve, express Vi with the unit mM/s rather than ∆A(405 nm)/s. Place your answer in the table above alongside the appropriate [S]. Hint: To answer this question you need to use this standard curve equation=0.0419x (The slope of the line is= 0.0419) (iii) The unmutated form of your protein has a Km of 25 µM and a Vmax of 43 mM/s. The enzyme kinetic data for your enzyme with the amino acid substitution should now be displayed in the table above. Based on these data, what is Vmax? Km? and…
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