Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an ATP-requiring reaction, firefly luciferase can be used to test for the presence of ATP. In this way, luciferase can test for the presence of life. The coupled reactions are 1. luciferin + O2 = oxyluciferin + light 2. ATP = AMP + PP; AG°=-47.99 kJ/mol If the overall AG° of the coupled reaction is -5 kJ/mol , what is the standard free energy of the first reaction at 27°C? Express your answer numerically to 1 decimal place.

Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an ATP-requiring reaction, firefly luciferase can be used to test for the presence of ATP. In this way, luciferase can test for the presence of life. The coupled reactions are 1. luciferin + O2 = oxyluciferin + light 2. ATP = AMP + PP; AG°=-47.99 kJ/mol If the overall AG° of the coupled reaction is -5 kJ/mol , what is the standard free energy of the first reaction at 27°C? Express your answer numerically to 1 decimal place.

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter16: Spontaneity Of Reaction

Section: Chapter Questions

Problem 75QAP: When glucose, C6H12O11, is metabolized to CO2 and H2O in living systems, the first step is the...

See similar textbooks

Similar questions

In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation: Glu+ATPG6P+ADPG298=17kJ In this process, ATP becomes ADP summarized by the following equation: ATPADPG298=30kJ Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process: GluG6PG298=?
Glycolysis is the process by which glucose is metabolized to lactic acid according to the equation C6H12O6(aq)2C3H6O3(aq) G=198 kJ at pH 7.0 and 25°C Glycolysis is the source of energy in human red blood cells. In these cells, the concentration of glucose is 5.0103 M, while that of lactic acid is 2.9103 M. Calculate AG for glycolysis in human blood cells under these conditions. Use the equation G=G+RT In Q, where Q is the concentration quotient, analogous to K.
Aerobic respiration links the oxidation of glucose with the production of ATP form ADP.Given the following thermodynamic data, C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔrH° = −2803 kJ/mol-rxn ATP + H2O(l) → ADP + HPO42-(aq) ΔrH° ≈ −24 kJ/mol-rxn calculate the enthalpy of reaction for the aerobic production of 25 ATP shown below.C6H12O6(s) + 6O2(g) + 25ADP + 25HPO42-(aq)→ 6CO2(g) + 25ATP + 31H2O(l) ΔrH° = ?
The hydrolysis of ATP to form ADP requires energy to break the bonds initially, but releases moreenergy as bond are remade making the reaction exothermic overall. The Gibbs free energy for thisreaction is -30.5 kJ/mole, meaning that it is also exergonic (ΔG=-). This energy can be used by otherbiological systems if this reaction is coupled with another less favorable one. In this case, the creation ofglucose-6-phosphate, C6H10O9P, this is a pivotal intermediate in many biochemical pathways. PO43- + C6H12O6 ⇌ H2O + C6H10O9P ΔG° = +13.8 kJ/moleATP + H2O ⇌ ADP + PO43- ΔG° = -30.5 kJ/mole A) What is the overall reaction that takes place? B) Calculate the ΔG° for this reaction C) Will this be a successful coupled reaction?
The efficiency of striated muscle is typically 37%, that is, the muscle converts 37% of the chemical energy reaching it (in the form of glucose) into mechanical work, converts the remainder into waste heat. Your biceps muscle performs mechanical work at the rate of 50 W. (a) What is the rate at which it consumes chemical energy? (b) What is the rate at which it produces waste heat? (c) Given that the oxidation of glucose yields 3.70 103 kcal/kg, what is the rate at which this muscle consumes glucose?
The standard free energy variation, at 25 ºC, for equilibrium: Glucose-6-phosphate (G-6-P) Glucose-1-phosphate (G-1-P) is ΔGº '= + 7280 J / mol. Calculate a) The equilibrium constant of the reaction. b) The real change in free energy when one mole of G-6-P is transformed into G-1-P, both concentrations remaining constant and equal to 10mM and 2 mM respectively.
The oxidation of glucose to CO2 and water is a major source of energyin aerobic organisms. It is a reaction favored mainly by a large negativeenthalpy change. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(1) ΔH ° = -2816 kJ/mol ΔS ° = +181 J/mol • K(a) At 37 °C, what is the value for ΔG° ?(b) In the overall reaction of aerobic metabolism of glucose, 32 moles ofATP are produced from ADP for every mole of glucose oxidized. Calculatethe standard state free energy change for the overall reaction whenglucose oxidation is coupled to the formation of ATP at 37 °C.(c) What is the efficiency of the process in terms of the percentage of theavailable free energy change captured in ATP?
When nitric acid is produced industrially, nitrogen monoxide, NO, is first formed at high temperature. Bakefetr reacts NO on cooling further with oxygen to nitrogen dioxide: 2 NO(g) + O2 ⇌ 2 NO2 (g) Table 1: Thermodynamic data at 25°C. Bond ΔfHom Som Cop,m NO(g) 90.25 210.76 29.34 O2(g) 0.00 205.14 29.36 NO2(g) 33.18 240.06 37.20 1) Calculate (with all relevant intermediate calculations) the standard reaction Gibbs free energy, ΔrG25o, for reaction (1) at 25°C from the data in Table 1 2) Calculate (with all relevant intermediate calculations) the equilibrium constant K25, for reaction (1) at 25°C. 3) Industrially, however, the reaction does not proceed at 25°C but at 500°C. Therefore, calculate (with all relevant intermediate calculations) the standard reaction Gibbs free energy, ΔrG500o, for reaction (1) at 500°C under the assumption that the standard molar heat capacities, Cop, in Table 1 are independent of temperature in the interval [25°C, 500°C]
Combustion of glucose ( C6H12O6 ) is the main source of energy for animal cells: C6H12O6 ( s ) + 6O2 ( g ) → 6CO2 ( g ) + 6H2O ( l ) =ΔGrxn37°C−2872.kJ This energy is generally stored as ATP (adenosine triphosphate) molecules, which can release it when convenient by hydrolysis ("water-assisted decomposition") into ADP (adenosine diphosphate) molecules and a phosphate anion, often given the symbol Pi in biochemistry: ATP ( aq ) → ADP ( aq ) + Pi ( aq ) =ΔGrxn−35 to 70kJ The actual amount of free energy released by the hydrolysis of ATP varies, depending on the exact conditions inside the cell. Suppose under certain conditions hydrolysis of ATP actually releases −51.3/kJmol . Calculate the maximum number of ATP s that could be created from ADPs and Pi by the combustion of a molecule of glucose. Your answer must be a whole number.
Energy from ATP hydrolysis drives many nonspontaneouscell reactions:AT P⁴⁻(aq) +H₂O(l) ⇌ADP³⁻(aq) +HPO₄²⁻(aq) +H(aq) ΔG°'=-30.5 kJ Energy for the reverse process comes ultimately from glucose metabolism:C₆H₁₂O₆(s)+ 6O₂g) →6CO₂(g) +6H₂O(l)(a) Find K for the hydrolysis of ATP at 37°C.(b) Find ΔG'°ᵣₙₓ for metabolism of 1 mol of glucose. (c) How many moles of ATP can be produced by metabolism of1 mol of glucose? (d) If 36 mol of ATP is formed, what is the actual yield?
At 25 oC an antibody binds a protein with a ∆H = -87.9 kJ/mol and ∆S = -118.9 J mol-1 K-1 Which of the following is/are true? (Choose all that are correct) a. The process is exergonic b. The process is endothermic c. The entropy of the binding process increases d. The binding process is exothermic e. The entropy during the binding process decreases
Consider the reaction in the attached image: a. The standard Gibbs free energy (ΔfG˚) for this reaction (up to two decimal places) is ____________ kJ/mol. b. If the concentration of AMP is adjusted to 1.5M and that of adenosine and phosphate is adjusted to 0.50M, the actual free energy change (ΔG) (up to two decimal places) is equal to ___________ kJ/mol.