11 XFor the hydrolysis ofglucose-6-phosphate,glucose-6-P + H₂O =glucose + PiAG'°= -13.8 kJ/mol andK'eq = 3.81x10-³ at 25°C.Therefore the [glucose-6-phosphate]/[glucose] ratiois about 1/260 atequilibrium.What is the [glucose-6-phosphate]/[glucose] ratioat equilibrium when thisreaction is coupled to ATPhydrolysis in the coupledreactionglucose + ATP =glucose-6-phosphate +ADPand [ATP] = ADP]? For ATPhydrolysis AG'°= -30.5 kJ/mol at 25°C.0/1 Point Earned9/10 Attempts Remaining

In this coupled reaction, the hydrolysis of glucose-6-phosphate is coupled with the hydrolysis of…

Answered: × For the hydrolysis of…

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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If a 0.1 M solution of glucose 1- phosphate at 25 °C 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 Calculate Keq and ΔG′° for this reaction.
Consider the following equilibrium at 25ºC :Glucose-1-Phosphate Glucose-6-PhophateUsing the equilibrium concentrations of [Glucose-1-Phosphate] = 0.35 M and [Glucose-6-Phosphate] = 1.65 M, calculate BOTH K′eqand Gº′ for this reaction. Is this reaction exergonicor endergonic? R = 8.314 J/K·mol
ΔG°′ for the isomerization reaction glucose-1-phosphate (G1P) ⇌ glucose-6-phosphate (G6P) is −7.1 kJ · mol−1. Calculate the equilibrium ratio of [G1P] to [G6P] at 25°C.
In the cytosol of rat hepatocytes, the temperature is 37 °C and the mass-action ratio, Q, is Calculate the free energy required to synthesize ATP in a rat hepatocyte.
Given that the AG" values for the hydrolysis of glucose 1-phosphate and glucose 6-phosphate are approximately -21 kJ/mol and –14 kJ/mol, respectively, which statement is true regarding the isomerization shown? glucose 1-phosphate → glucose 6-phosphate O The reaction can likely proceed in either direction and will therefore depend on the concentrations of glucose 1-phosphate and glucose 6-phosphate. O Only the anabolic direction will occur because the free energy change is positive. O The reaction will proceed in both directions simultaneously. This reaction can only proceed in one direction and must be the rate limiting reaction for this point in metabolism.
The effect of temperature on the hydrolysis of lactose by a ß-galactosidase is shown below in Table 1. The temperature coefficient, Q10 is the factor by which the rate increases by raising the temperature 10°C. The universal gas constant, R is 8.314 J/mol.K. (a) (b) Table 1: Data of Vmax over temperature T (°C) 20 30 35 40 45 Vmax (umoles/min.mg protein) 4.50 8.65 11.80 15.96 21.36 Plot the graph of In Vm vs 1/T using any spreadsheet software (include all appropriate labels and equation). Calculate the activation energy Ea and temperature coefficient Q10.
The equilibrium constant for the hydrolysis of the peptide alanylglycine (Gly-Ala in the reaction from Part B) by a peptidase is K = 9.0 × 10² at 310 K. Calculate AG for this reaction. Express the Gibbs free energy to three significant figures. AG = Submit ΠΑΠΙ ΑΣΦ Request Answer ? kJ/mol Keq [Gly] [Ala] [Gly-Ala]
Calculate the actual, physiological AG for the reaction Phosphocreatine + ADP = creatine + ATP at 37 °C, as it occurs in the cytosol of neurons, where [phosphocreatine] = 4.7 mM, [creatine] = 1.0 mM, [ADP] = 0.73 mM, and [ATP] = 2.6 mM. Standard Free Energies of Hydrolysis of Some Phosphorylated Compounds Phosphorylated compound AGʻ° (kJ/mol) phosphoenolpyruvate -61.9 phosphocreatine -43.0 ADP (→ AMP + P;) -32.8 ATP (→ ADP + P;) -30.5
The ΔG°′ for hydrolytically removing a phosphoryl group from ATP is about twice as large as the ΔG°′ for hydrolytically removing a phosphoryl group from AMP (−14 kJ · mol−1). Explain the discrepancy.
ATP can be hydrolyzed into ADP + inorganic phosphate (P.) with a AG of -30500 J/mol. If, galactose-1-phosphate can be hydrolyzed into galactose + P, with a AG of -15.3 kJ/mol what is the Gibbs free energy change for the reaction of galactose + ATP = galactose-1-phospate + ADP? O a. -45.8 kJ/mol O b. 15.2 kJ/mol Oc. -30484.7 J/mol O d. -15.2 kJ/mol Oe. -1520 J/mol
In muscle cells, the AG for glucose+ATP2 glucose-6-P+ADP is –33.5 kJ•mol-'. In contrast, the AG for glucose – 6 – P fructose – 6 - P is –2.5 kJ · mol-1. (a) In a physiological context, which reaction is faster? How do you know? (b) The enzyme that catalyzes glucose + ATP 2 glucose important point of regulation in glycolysis, while the enzyme that catalyzes glucose – 6 – PZ fructose – 6 – P (phosphoglucose isomerase) is not. Why is hexokinase a good step at which to regulate glycolysis relative to phosphoglucose isomerase? 6 – P + ADP (hexokinase) is an | - |
13. Calculate the equilibrium constant K'eg, for each of the following reactions at pH 7.0 and 25°C. glucose + Pi a. Glucose 6-phosphate + H20 enz. Glucose 6-phosphatase; AG'O=-13.8kJ/mol

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