You were asked to determine the mode of inhibition exerted by Inhibitor DEDS to a newly discovered enzyme known as BILISTASE, Thekinetics data are shown below[Substrate, uM]Vo with DEDS M/min)Vo without DEDS (uM/min)M3.33310.774,11.19614,0000.8771.3165.0001.012#1.46210.0001.462088'120.0001.88012198Using linear regression analysis. deterane tievaluesof Vun K orthe ercymehine ABSENCE of inhibitoe Express your answerIn 3 decimal places, do NOT include the units) Using linear regression analysis, determine the values of V and Ky of the enzyme in the PRESENCE ofinhibitor: (5in 3 decimal places, do NOT incude the units)VmaxKM=What is the MODE of inhibition exerted by DEDS?

Lineweaver-Burk plot, also known as double-reciprocal plot, is the graph plotted for Lineweaver-Burk…

Answered: You were asked to determine the mode of…

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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Similar questions

Hello can someone please help me graph 1 and 2 using the 3 data sets listed, and help with the following questions 3 and 4. Thank you.
In the same kinetics experiment above without the inhibitor, what is the fraction of occupied enzyme at substrate concentration 1microM ? How about when it's 1mM ?
The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077
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-
-Inhibitor +Inhibitor [S] (mM) V0&νβσπ;(μmol/sec). νο&νβσπ: &ν βσπ (μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITH iinhibitor?
Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the catalytic efficiency for this reaction. Report your answer in scientific notation to three significant figures in units of M-1s-1.
Km value should be more than 10% apart, use non-competitive inhibitor
For an enzyme-catalyzed reaction, the velocity was determined at two different concentrations of the substrate. Estimate the value of Vmax. [S] (MM) 10 20 88 nmol/min 79 nmol/min 67 nmol/min V(nmol/min) 51 nmol/min 27 Not enough information is given to form a reasonable estimation. 48
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (uM) 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Km of the inhibited enzyme reaction?
Consider the Michaelis-Menten equation, below: Vmaz (S V. k + [S] %3D What is the relationship between changes in substrate concentration and velocity when the concentration of substrate, [S), is well below k7 The S terms cancel out in this equation, so there is no effect of changing substrate concentration. The S] term in the numerator is negligible, so there is no impact of changing substrate concentration. Because the enzyme has reached Vr there is no effect of changing substrate concentrations on enzyme velocity. OThe [S term in the denominator is negligible compared to k There is a linear relationship between substrate concentration and velocity.
Please don't provide handwritten solution ....
The hydrolysis of a substrate, S, by an enzyme has been studied in the lab. The following initial rates, vo, were recorded at different concentrations. [S] (M) Vo (M/min) 2.10-10-4 1.20.10-6 4.20-10-4 3.10.10-6 9.30-10-4 6.30-10-6 1.42-10-3 9.10-10-6 A. Determine the rate constants for degradation of the substrate B. What is the rate of reaction at [S] = 1.1.104 M? C. Explain, why enzymes can make reactions go faster? Does enzymes also catalyse the reverse reaction from product to substrate?

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