Question 9: For the reaction catalyzed by the enzyme hexokinase, Glucose + ATP = glucose 6-phosphate+ ADP, the equilibrium constant is 7.8 x 10² at 298 K. Please answer the following independentquestions.A. Determine the standard Gibbs free energy of the reaction.B. In dying E. coli cells, the concentrations of metabolites were found to be: [ATP] = 0.5 mM, [ADP]= 100 mM, [glucose] =0.1 mM, and [glucose 6-phosphate] = 2 mM ay 298 K. Does the reactionproceed spontaneously?C. What would be the ATP:ADP ratio at equilibrium at 37 C if the concentration of glucose is 100-fold less than that of glucose-6-phosphate? The standard enthalpy of this reaction at 298 K is -23.8 kJ/mol.

Hexokinase is a initial glycolysis enzyme and rate limiting enzyme, catalyzes the phosphorylation of…

Answered: A. Determine the standard Gibbs free…

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

See similar textbooks

Similar questions

Question 23 An enzyme has a single active site at which it can bind and hydrolyze either X or Y but the enzyme cannot bind X and Y at the same time. Which of the following statements are TRUE? Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation... SHOW MORE V The Km for X will be affected if Y is present in the reaction mixture. a Y is a competitive inhibitor of X. The Km for X will increase. d. The V for X will be affected if Y is present in the reaction mixture. max pH dependence of Vmax reflects the ionization state of catalytic site e residues. Consider the following: X and Y are methanol (poisonous) and ethanol respectively. If the Km for X = 0.01 M and the Km for Y = 0.001 M then 0.01 M Y is 10 times the concentration of Y required for 0.5 Vmax. Addition of an enzyme to a chemical reaction increases the ratio of g products to reactants (Keg).
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
Which rate law has a molecularity of zero? Select one: O a. Rate = k[A]° O b. Rate = k[A][B] %3D O c. Rate = k[A][B][C] %3D O d. Rate k[A] %3D K[AJ? e. Rate = Determine the value of K. for the following reaction if the equilibriu
I need help with parts d-g a.) For Ado: catalytic efficiency = 4.18 x 106 M-1s-1 For ATP: catalytic efficiency = 12.1 x 107 M-1s-1 b.) Both specificity constants are below the diffusion limit so neither substrate is limiting the reaction rate due to diffusion c.) Line weaver Burk equation for AdoK enzyme with Ado substrate: y = 0.68x + (2.04 x 104), where y = 1/vo and x = 1/[Ado]
.When saturated with substrate, an enzyme has a maximum initial rate of 110 umoles of substrate converted to product per second. At a substrate concentration of 100 µM, the same enzyme converts substrate to product at a rate of 0.010 mmoles/sec. Assuming that Michaelis- Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2 x 10 M.
Can you explain the answe?
Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places
The free energy change of each step of glycolysis is given in the table below. ∆G°’ is the free energy under standard conditions (25°C, 1M each reactant, pH 7), while ∆G is the free energy change at presumed physiological conditions. Why must no step have a positive ∆G under physiological conditions?
Two enzyme catalyzed reactions, RA (red) and Rp (blue), have kinetic profiles shown below. Report which reaction has the higher maximal rate and which reaction has the smaller Km. в A 1/ [S] Higher Vm [ Select ] Lower Km [ Select ] A/T
In glycolysis, fructose 1,6-bisphosphate is converted to two products with a standard free- energy change (AG) of 23.8 kJ/mol. Under what cellular conditions will the free-energy change be negative, enabling the reaction to proceed spontaneously to form products? O In a cell where the concentrations of products and reactants yield a low value of the mass-action ratio Q ○ The reaction will not go to the right spontaneously under any conditions because the AG is positive ○ The reaction will proceed spontaneously to the right if there is a high concentration of products relative to the concentration of fructose 1,6-bisphosphate ○ The reaction will proceed spontaneously to the right if there is a high concentration of enzyme to catalyze the reaction O None of these conditions is sufficient
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 23.7, and the temperature is 37.0 °C? AG" for the reaction is -16.7 kJ/mol. The constant R = 8.3145 J/(mol-K) AG= Number kJ/mol
The Keq (25C) of the reaction below is 635.67. Fructose 1,6-biphosphate <-->fructose -6-phosphate + Pi. a) What is the standard Gibbs free energy change for this reaction? b) if the concentrationof fructose 1,6 biphosphate is adjusted to 0.85 M and that of fructose 6 phosphate and phosphate adjusted to 0.055 M, what is the actual free energy change

SEE MORE QUESTIONS

Recommended textbooks for you

Biochemistry

ISBN:

9781319114671

Author:

Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:

W. H. Freeman

Lehninger Principles of Biochemistry

Biochemistry

ISBN:

9781464126116

Author:

David L. Nelson, Michael M. Cox

Publisher:

W. H. Freeman

Fundamentals of Biochemistry: Life at the Molecul…

Biochemistry

ISBN:

9781118918401

Author:

Donald Voet, Judith G. Voet, Charlotte W. Pratt

Publisher:

WILEY

Biochemistry

ISBN:

9781319114671

Author:

Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:

W. H. Freeman

Lehninger Principles of Biochemistry

Biochemistry

ISBN:

9781464126116

Author:

David L. Nelson, Michael M. Cox

Publisher:

W. H. Freeman

Fundamentals of Biochemistry: Life at the Molecul…

Biochemistry

ISBN:

9781118918401

Author:

Donald Voet, Judith G. Voet, Charlotte W. Pratt

Publisher:

WILEY

Biochemistry

ISBN:

9781305961135

Author:

Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal

Publisher:

Cengage Learning

Biochemistry

ISBN:

9781305577206

Author:

Reginald H. Garrett, Charles M. Grisham

Publisher:

Cengage Learning

Fundamentals of General, Organic, and Biological …

Biochemistry

ISBN:

9780134015187

Author:

John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson

Publisher:

PEARSON

SEE MORE TEXTBOOKS