87-6-2-1-4590135180225 27031536040545049554058![Substrate] (nM)What is the Vmax of the enzyme data shown above?rate (nM/sec)3.

Michaelis Menton kinetics: This is well known model for Enzyme kinetics. This model relates rate of…

Answered: 8- 2- 315 405 [Substrate] (nM) 45 90…

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 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…
For some Enzyme, the Vmax is 18 micromols/min, Km is 400 microM. If the substrate concentration is 100 microM, what is the velocity of the reaction?
5/ n SO a- 6.46 The following data show the binding of Mg2+ ions with a protein containing n equivalent sites: [Mg2+ ] total/μM [Mg2+ ]free/μM 108 Com 35 180 65 288 501 115 248 752 446 Apply the Scatchard plot to determine n and Ka. The protein concentration is 98 µM.
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-
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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.
Materials - 2021FA-CHM-1O X Bb 4427577 d-fleet02-xythos.content.blackboardcdn.com/6086c260d7e8f/4427577?X-Blackboard-Expiration=1633143600000&X-Blackboard-Sig 6 / 9 100% Exercise 1: Standard Curve for Protein Measurements: A standard curve for protein concentration is often created using known concentrations of bovine serum albumin (protein). This process is called the Bradford Assay; it is a colorimetric assay. A special reagent turns blue when it binds to amino acids present in protein. The intensity of the color is best measured with a spectrophotometer (a device for comparing two light radiations, wavelength by wavelength). In the case of the Bradford Assay the greater the absorbance, the higher the protein concentration. A series of tests were performed on some samples and spectrophotometer: following measurements were obtained using a Protein Concentration (mg/ml) Absorbance (A) 0.26 0.098 0.56 0.213 0.383 0.84 1.12 0.473 1.40 0.527 TASKS: 1. Enter the data into Excel - the…
KINETIC CONSTANT No Na2HPO4 25mM Na2HPO4 50mM Na2HPO4 Vmax nmol p-NP. Min- 20.3252 14.30615 17.30104 Km mM -0.819106 -0.46495 -0.352941 1. What does this suggest about the structure of the active side of the enzyme?
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?
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8- 2- 315 405 [Substrate] (nM) 45 90 135 180 225 270 360 450 495 540 58! What is the Vmax of the enzyme data shown above? 7. 6. 4. rate (nM/sec)