Help? 1. The following data were recorded for the enzyme-catalyzedreaction S –→ P.[S]in MOmole/min|6.25 х 10%157.50 x 1056.251.00 х 10460|1.00 х 10374.9|1.00 x 10275 (b) What would v be at [S] = 2.5 x 10* M.

The relationship between velocity and substrate concentration of a reaction is represented by the…

Answered: 1. The following data were recorded for…

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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1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
For an enzymatic reaction, the following data were obtained for two different initial enzyme concentrations: [S](g/L) v ([E0] = 0,015 g/L)(g/L.min) v ([E0] = 0,00875 g/L) (g/L.min) 40 2,28 1,34 20 1,74 1,02 13,4 1,4 0,82 10 1,18 0,68 8 1 0,58 (a)Calculate Vmax (in g/(L.min)) for the initial enzyme concentration equal to 0.015 g/L. (b)Calculate Vmax (in g/(L.min)) for the initial enzyme concentration equal to 0.00875 g/L. (c)Calculate Km (in g/L) for the initial enzyme concentration equal to 0.015 g/L. (d)Calculate Km (in g/L) for the initial enzyme concentration equal to 0.00875 g/L.
You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. ?0=V0=
The Km value of an enzyme-catalyzed reaction and its Vmax is 70 mmol/min. What is the rate of the reaction of the reaction when the substrate concentration is 7 × 10−2 mmol/min? a. 35 mmol/min b. 50 mmol/min c. 60 mmol/min d. 70 mmol/min
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.
Q is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound Q
An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.
Suppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number
Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμM
1. Calculate the reaction Kcat for the Control in experiment (1). 2. Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and KM for the enzyme.
Calculate the rate enhancement for the enzyme pair: Enzyme 1: Kcat (sec-1)= 5.5 KM (µ M) = 1245 Kcat / KM = 0.00442 (µ M-1 sec-1) Enzyme 2: Kcat (sec-1)= 3.2 KM (µ M) = 1785 Kcat / KM = 0.00179 (µ M-1 sec-1)
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?

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1. The following data were recorded for the enzyme-catalyzed reaction S → P. [S] in M Omole/min 6.25 x 10 15 7.50 x 10 56.25 1.00 x 104 1.00 x 103 1.00 x 10 60 74.9 75