SM(2001) T MEXEGiven, Iso s ri Glucose + Pi→ Glucose-6-P + H₂O K'eq = 3.9 x 10-3 M-1 utoe MATP + H₂O → ADP + PibsK'eq = 2 x 105 Mrink ats1014 Glucose +DADetermine K 'eq for a coupled reactionenimenteGlucose + ATP →q8)otsdGlucose-6-P + ADP 100K'eq?Ollom\xA12800.0=8).noitoso stericeoria-lf)undili

Consider a reaction ; A B. Its biochemical equilibrium constant (K'eq 1) will be;Now consider a…

Answered: Given, isas ri Glucose + Pi→…

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…
The activity of the enzyme β-galactosidase produced bywild-type cells grown in media supplemented with different carbon sources is measured. In relative units, thefollowing levels of activity are found:Glucose Lactose Lactose + glucose0 100 1Predict the relative levels of β-galactosidase activity incells grown under similar conditions when the cells arelacI−, lacIS, lacO+, and crp−.
What is the Keq of Reaction #3? Show your work & clearly label your answer. Is Reaction #3 favorable under mammalian cellular conditions when the concentration of dGTP is 0.2 M, dCMP is 20 mM, dGMP-dCMP is 7 mM, and PPi is 10 mM? Use △G to support your answer.
-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?
Can you explain the answe?
A solution of [U 14C] glucose-1-phosphate (specific activity = 16,000 cpm/mmole) was incubated with glycogen and glycogen phosphorylase, an enzyme which adds glucose units on to glycogen. Radioactivity was incorporated into the glycogen primer at a rate = 2550 cpm/min. The rate of the enzymatic reaction in units of mmole glucose incorporated per minute is: (a) 0.016 mmol/min (b) 0.57 mmol/min (c) 0.16 mmol/min (d) 5.7 mmol/min
Calculcate Kcat for PNP substrate for both enzyme concentrations. enzyme volume: 20 ul Bovine Intensince Alkaline phosphatase molecular weight: 140,000 Bovine intenstine Alkaline phosphatase activity: 300 units/ml and 14 units/mg extinction coefficient PNP: 18.5 abs (mM-1 cm-1) Vmax: 0.332 moles/sec a) enzyme 1 concentration: undiluted b) enzyme 2 concentration: 1:1 dilution
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?
The table contains data about how different ASFV mutants affect enzyme activity-- for example the D29A mutant, where the D at position 29 was mutated to an A. Select three mutants at random, and for each mutant explain how that mutation could affect intermolecular force interactions (an example: hydrogen bonding could be disrupted by this mutation because…
3L6
14) Calculate K’eq and ΔG’0 for the following: A 0.1 M solution of glucose 1-phosphate at 25oC is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Glucose 1-phosphate (4.5 X 10-3 M) and Glucose 6-phosphate (9.6 X 10-2 M).
Please provide an explanation for how to work through this practice problem: Using table 12.1, calculate the free energy change for the synthesis of ATP from cAMP and inorganic phosphate. (Note: cAMP is hydolyzed to AMP, and the free engery of hydrolysis for ATP and ADP is approximately equal.) Table 12.1: (Compound: Change in Free Energy in kJ/mol) cAMP: -50.4 Creatine phosphate: -43.3 ATP: -30.5 Glucose 6-phosphate: -13.9 AMP: -9.2

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Given, isas ri Glucose + Pi→ Glucose-6-P ATP + H₂O → ADP + Pi Determine K 'eq for a coupled reaction Tot 24 h Glucose + ATP 18) (OUT) I MEX + H₂O K'eq = 3.9 x 10-3M-1uto beriasen K'eq=2 x 105 Marbl 93A otsdge Glucose-6-P + ADP K'eq?