Q7. Kinetics were analysed for an enzyme in the absence and presence of inhibitor. In the absence of inhibitor, the Km was 10 mM and the Vmax Was 20 mM/min. In the presence of inhibitor, the apparent Km was 2 mM and the apparent Vmax Was 4 mM/min. Which statement is correct? Alpha and alpha' are both > 1 O Alpha equals 5 The inhibitor is uncompetitive In the presence of inhibitor, at very high (S). Vo approaches Vmax

Q7. Kinetics were analysed for an enzyme in the absence and presence of inhibitor. In the absence of inhibitor, the Km was 10 mM and the Vmax Was 20 mM/min. In the presence of inhibitor, the apparent Km was 2 mM and the apparent Vmax Was 4 mM/min. Which statement is correct? Alpha and alpha' are both > 1 O Alpha equals 5 The inhibitor is uncompetitive In the presence of inhibitor, at very high (S). Vo approaches Vmax

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter21: Biochemistry

Section: Chapter Questions

Problem 28QAP: . Describe the lock-and-key model for enzymes. Why are the shapes of the enzyme and its substrate...

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. Describe the lock-and-key model for enzymes. Why are the shapes of the enzyme and its substrate important in this model? What does it mean to say that an enzyme is inhibited by a particular molecule? What happens if this inhibition is irreversible? Can you think of a situation in which it might be advantageous to be able to inhibit an enzyme?
5 At a very low concentration of a certain substrate, we find that when the substrate concentration doubles, the rate of the enzyme-catalyzed reaction also doubles. Would you expect the same finding at a very high substrate concentration? Explain.
Compare the activation energy in uncatalyzed reactions and in enzyme-catalyzed reactions.
The figure below shows the dependence of the enzyme's rate, v (in uM/min), as a function of substrate concentration, S (in mM). Also shown is the dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. Based on this information, the Vmax of the enzyme is approximately: 1/v 1- 0.5 1/[S] O A. 0.25 pM/min O B. 0.5 pM/min O C. 1.0 uM/min D. 2 pM/min O E. 4 µM/min
Following data were obtained for an enzyme in the absence and presence of an inhibitor. What are the Km and Vmax values for the inhibited and uninhibited enzymes? What is the type of inhibitor?
An enzyme catalyzes the reaction M ⇌ N. The enzyme is present at a concentration of 1 nM, and the Vmax is 2 μM s−1. The Km for substrate M is 4 μM. (a) Calculate kcat. (b) What values of Vmax and Km would be observedin the presence of sufficient amounts of an uncompetitive inhibitor to generate an α′ of 2.0?
The following results were obtained for an enzyme-catalyzed reaction Substrate concentration (mmol 1-¹) 5.0 6.67 10.0 20.0 40.0 Initial velocity (μmol l-¹ min-¹) 147 182 233 323 400 Calculate K and Vmax
You performed a kinetic study of the enzyme lipase and you obtained the following double reciprocal (Lineweaver-Burk) plot: 4 y = 4.3511x + 1.8889 %3D 1 -0.5 -0.3 -0.1 0.1 0.3 0.5 0.7 0.9 1/[S] When you performed the same kinetic study in the presence of an inhibitor, you obtained a new line that had y-intercept 3 and x-intercept -0.3. Based on this information select all correct statements about this inhibitor. This is mixed inhibitor. When this inhibitor is present apparent Km of enzyme decreases. This inhibitor prevents substrate from binding to the enzyme. This is uncompetitive inhibitor. When this inhibitor is present apparent Vmax of enzyme decreases. When this inhibitor is present apparent Km of enzyme increases. This inhibitor can bind to either enzyme or enzyme-substrate complex.
Q1: the following table gives the results of enzyme-catalyzed reaction. The enzyme concentration [E] = 0.00012 mmol/L Use the information of substrate concentration [S] and rate Vo for both experiments, with inhibitor and without inhibitor to answer the following questions (NOTE: [E] = 0.00012 mmol/L) [S], mmol/L 0.710 0.400 0.310 0.098 0.066 0.040 Experiment 1 without inhibitor VO, mmol/L-min 0.200 0.180 0.160 0.120 0.100 0.070 Experiment 2 with inhibitor [1] = 0.033 mmol/L VO, mmol/L-min 0.180 0.150 0.110 0.070 0.050 0.040 1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive) Note:- show all your work to answer the above questions. Upload your answers…
:The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics Vo (mmol/min) S (mmol/L) 0.8 3 4 217 325 433 488 1,000 :The Km for this enzyme is approximately 647 mM 4 O mM 1 O mM 1,000 Ⓒ mM 3 O mm 6 O
3. (а) 0.0050 M operate at one-quarter of its maximum rate? At what substrate concentration would an enzyme with a kcat of 30.0 s-1 and a KM of (b) trations [So]: ½ Km, 2 KM, and 10 KM. Determine the fraction of Vmax that would be obtained at the following substrate concen- (c) 1 (HIV-1) has been the object of innumerable studies to develop effective chemotherapeutic agents. It has been shown that p6* known as the late assembly protein is an inhibitor of HIV protease. An assay was developed using an artificial polypeptide substrate containing a p-nitrophenylalanine residue at the cleavage point that undergoes a small change in absorption at 295 nm upon bond hydrolysis that could be followed spectrophotometrically. The cleavage of the peptide bond is shown schematically on the right. Results of the assay are given in the table below. The protease of the human immunodeficiency virus- Lys NH2 Ala Nle Arg Ala Val-Nle-NH-CH-Ċ–NH-Glu CH2 Lys NH2 Ala NO2 Nle H2ON HIV-1 protease Ala Vo…
Enzyme X can act on either substrate A or B; the Km on A < Km on B. Vmax for A > Vmax on B.Sketch 1 plot for 1/v vs. 1/[S] plot for Enzyme X in the presence of A and in the presence of B.