v (mm/sec)5.04.03.02.01.00.00.01.00Estimate from the M-M plot below:Vmax, Km, Keat, and enzyme efficiency2.0[ET] =2.0 mM3.0[S](MM)SmartArt4.0 5.06.07.0Box

The rate (v) v/s substrate concentration ([S]) graph of enzymes that follow MM (Michaelis-Menten)…

Answered: v (mm/sec) 5.0 4.0 3.0 2.0 1.0 0.0 0.0…

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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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-
ver is given. A +B C+ D 1) Calculate AG for the above react un and indicáte whether the reaction is favorable or unfavorable [A] = 0.9 M 20°C AG° = 4 KJ/mol %3D [B] = 15mM [C] = 4mM [D] = 3 M R= 8.314 J/moleK %3D %3D 4.7 and explain how you know.
I need the answer as soon as possible
Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the above enzyme (lactase). (enzyme lactase has a Vo of 0.111111111111 mM per minute when [S] = 1.0 mM, and a Vo of 0.20 mM per minute when [S] = 5.0 mM) 0.125 per minute 0.25 per minute 0.50 per minute 1.25 per minute 5.0 per minute
You are presented with the following prescription: Rx Drug Y 175 mg Witepsol H15 q.s. Supply 6 suppositories (using 3 g nominal moulds) When using Witepsol H15, moulds will hold 1.06 times their nominal weight. The displacement value of Drug Y in Witepsol H15 is 1.1. Calculate the mass of Witepsol H15 required to prepare this product. Give your answer to two decimal places.
8 7- 6 2- 180 225 270 315 [Substrate] (nM) 45 90 135 360 405 450 495 540 58! What is the KM of the enzyme data shown above? rate (nM/sec) 3.
What percentage of max is obtained when the substrate is present at 80% of the Km? Use two digits in your answer.
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?
Calculate for rate constant(k)
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?
stock aliquot - 1mg/ml use the sequence below to determine the molarity using expasy prot program atgggcgaggagtataaagttgtcctcacatttggatccccaatgagccctaatgcaaataacaaacagacttgggttaataaacctcttgatgcgccttcgggccattacaatgtgaaaattgcaaaggatgttgaccactatctaaccatgcagggtttcacttctatagcatctgttgactggtacactatagattttcaaccatctgaggcgcctgccccgataaaaggcttgcaggtacttgtgaacatctcgaaaaaagctgatgtgtatgccgtcaaacaatttgtcacagcgcagaccaacaacaagcaccaggttacaagcctgttcctagtaaaagtaacaactggttttcaggtgaacaactacctgagttacttttacagggcgtctgctactggggatgccacaactaacctgttggttagaggagacacatacacagcagggataagttttacccagggtggatggtatttgttgacaaatacatctattgttgatggggctatgccacctggctgggtctggaataacgtggaacttaaaactaacacagcgtatcacatggacaaaggcttggtccatctaataatgcctttgcctgagtccacgcaaatgtgttatgagatgctgacatctattccagcggccgcagagctcgctctggtgccacgcggtagttccgctcatcaccaccatcatcaccatcaccaccact
What is the Vmax of an enzyme at 1 μM with -4G RT k = 6.2x 10¹2 s-1 (Ideal gas constant R = 8.314 J K-1 mol-1) cat 0.14 μM S-1 1.4 µM s-1 14 μM s-1 0.14 mM s-1 1.4 mM s-1 activation energy (AG) of 55000 J/mol at 298K?

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v (mm/sec) 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Estimate from the M-M plot below: Vmax, Km, Keat, and enzyme efficiency 1.0 [ET] =2.0 mM 2.0 3.0 [S](MM) 4.0 5.0 6.0 7.0