QUESTION 3In the presence of 0.000005 M competitive inhibitor, an enzyme exhibits an apparent Km of 0.002M. If the Kt of theinhibitor is 0.0000003 M, what is the true Km of the enzyme? Give your answer with 3 significant figures anscientific notation. For example, 0.000023434 is 2.34eQUESTION 4Approximately how many domains are in an 82.5 kDa protein?O a. 2O b.3O C. 4O d. 12.5QUESTION 5How long is an a-helix that contains 36 amino acid residues?6.0 nm5.4 nm78 Mostly clody a nClick Save and Submit to save and submit. Click Save All Answers to save all ansuersP Type here to search

A competitive inhibitor generally reduces the rate of catalysis by decreasing the proportion of…

Answered: QUESTION 3 In the presence of 0.000005…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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QUESTION 2 For the peptide: TREATS, what would be the y3 ion? EATS TRE TS ATS QUESTION 3 In the presence of 0.000005 M competitive inhibitor, an enzyme exhibits an apparent Km of 0.002M. If the Kq of the inhibitor is 0.0000003 M, what is the true Km of the enzyme? Give your answer with 3 significant figures an scientific notation. For example, 0.000023434 is 2.34e QUESTION 4 Approximately how many domains are in an 82.5 kDa protein? O a.2 70 Mostly cloudyA Click Save and Submit to save and submit. Click Save All Answers to save all answers. Tune here to search
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 %24
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
An enzyme has the following values: vmax's; 0.028 M s-' and 0.021 M s-' (I), and km's; 0.00198 M and 0.00197 M (I), when the total concentration of enzyme is 10-º M. The (I) values were determined with 1 microM inhibitor. Select the value closest to the k, for this enzyme. (hint 1 microM = 10-6 M).
A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.
columns) and absence (second column = control) of enzyme inhibitor. Both inhibitors were added in each reaction at a concentration of 2 mM. The enzyme concentration was similar in all and was approximately 0.001 Им. Calculate both Vmax and KM for the control using Lineweaver-Burk curve. b. Provide the type of inhibition for both? Find, KI, for the inhibitor binding to the enzyme, for experiments (2) and (3). d. Calculate the reaction Kcat for the Control in experiment (1). Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and Ky for the enzyme a. C. e. [S] (mM) (1) V. {(umol/(ml.s)] (2) V. [[umol/(ml.s)]| 4.4 (3) V. umol/{ml.s] 6.6 11.4 76 14.6 8.6 26.6 16.4 29 8 17.8 24.6 28.2 16 45.8 24 60 40.8
If a michaelis-type enzyme has an apparent Km of 2*10^-8 and apparent Vmax of 1*10^-4 moles/min in the presence of 1*10^-3 M uncompetitive inhibitor (Ki=1*10^-7 M), what is the true Km of the enzyme
At body temperature (37°C), k of an enzyme-catalyzed reaction is 2.3X10¹⁴ times greater than k of the uncatalyzed reaction. Assuming that the frequency factor A is the same for both reactions, by how much does the enzyme lower the Eₐ?
An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 μM and a kcat value of 206 s−1 . At an initial enzyme concentration of 0.0100 μM , the initial reaction velocity was found to be 1.07×10−6 μM/s . What was the initial concentration of the substrate, [S] , used in the reaction ? Express your answer in micromolar to three significant figures..
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…
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A research group discovers a new enzyme they decide to name happyase. This enzyme has a Michaelian behavior and catalyzes the chemical reaction below: HAPPY ↔ SAD The researchers begin to characterize the enzyme. 1-In the first experiment, the researchers use a total enzyme concentration ([Et]) of 4 nM. They find that the happyase has a Vmax of 1.6 µs-1 in these conditions. Calculate the k2 of happyase based on this finding. Show your work and include appropriate units. 2-In another experiment, with [Et]=1 nM and [HAPPY]=30 µM, the researchers find that V0 = 300 nM.s-1. Calculate the Km of happyase for its substrate HAPPY based on this finding. Show your work and include appropriate units. 3-Further research shows that the purified happyase enzyme used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. After the careful removal of ANGER, the measured Vmax is increased to 4.8 µM.s-1, and the calculated Km becomes equal to 15 µM. Compare…

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QUESTION 3 In the presence of 0.000005 M competitive inhibitor, an enzyme exhibits an apparent Km of 0.002M. If the Kt of the inhibitor is 0.0000003 M, what is the true Km of the enzyme? Give your answer with 3 significant figures an scientific notation. For example, 0.000023434 is 2.34e