BCCB2000 Lecture 14 Questions Flashcards
(36 cards)
The free energy for ATP hydrolysis in vivo is actually greater than the standard free energy change of -30 kJ mol-1 because: A. of the actual concentrations of ATP and its hydrolysis products in cells. B. it can participate in phosphoryl group transfers. C. All of the above D. it has stronger electrostatic repulsion. E. it can easily be formed from other nucleotide triphosphates.
A. of the actual concentrations of ATP and its hydrolysis products in cells.
Which of the following half-reactions has the greatest reduction potential (i.e. can act as the better reducing agent) compared with all the others in the list. A. O2 + 2H+ + 2e- ——> H2O2 E’o = 0.295V B. NAD+ + H+ + 2e- ——> NADH E’o = -0.320V C. NO3- + 2H+ + 2e- —> NO2- + H2O E’o = 0.421V D. Fumarate2- + 2H+ +2e- ——> succinate2- E’o = 0.031V
B. NAD+ + H+ + 2e- ——> NADH E’o = -0.320V
The chemical name for ATP is alanine triphosphate. True or False?
False
Calculate (with 2 significant figures) the standard Gibbs Free Energy available for the reaction: Pyruvate + NADH + H+ ——> lactate + NAD+ Given the following information: Pyruvate + 2H+ + 2e- ——> lactate E’o = -0.19V NAD+ + H+ + 2e- ——> NADH E’o = -0.32V Faradays constant (F) = 96485 J V-1 mol-1 A. -98 kJ/mol B. -98,000 kJ/mol C. -25 kJ/mol D. +25 kJ/mol E. -25,000 kJ/mol F. +98 kJ/mol
C. -25 kJ/mol The reaction: NAD+ + H+ + 2e- ——> NADH E’o = -0.32V occurs in the reverse direction in the overall reaction and so we must reverse the sign of the reduction potential: + 0.32V ΔE’o = -0.19 + 0.32 = 0.13V ΔG’o = -nFΔE’o Two electrons are transfered so n=2 F = Faradays constant = 96485 J V-1 mol-1 ΔG’o = -2 x 96485 x 0.13 = -250861 J/mol = -25 kJ/mol (2 sig fig)
If the standard Gibbs Free Energy for the hydrolysis of ATP is -30.5kJ/mol, then it would take +30 kJ/mol to synthesise ATP from the concentrations of ADP and Pi that exist in the cell. True or False?
False
Which of the following half-reactions has the greatest oxidising potential (i.e. can act as the better oxidising agent) compared with all the others in the list. A. Fumarate2- + 2H+ +2e- ——> succinate2- E’o = 0.031V B. NO3- + 2H+ + 2e- —> NO2- + H2O E’o = 0.421V C. O2 + 2H+ + 2e- ——> H2O2 E’o = 0.295V D. NAD+ + H+ + 2e- ——> NADH E’o = -0.320V
B. NO3- + 2H+ + 2e- —> NO2- + H2O E’o = 0.421V
The Standard Free energy change for ATP hydrolysis is large and negative ΔG’o = -30.5 kJ/mol However the actual free energy of hydrolysis (ΔG) of ATP in cells may be smaller or larger because it depends upon the cellular concentrations of ATP, ADP, and Pi. True or False?
True
Exergonic reactions can be used to ‘drive’ endergonic reactions. This statement means that: A. A reaction with positive Free Energy (+ΔG) can be coupled to provide energy for a reaction with negative Free Energy (-ΔG) B. A reaction with negative enthalpy (-ΔH) can be coupled to provide energy for a reaction with positive enthalpy (+ΔG) C. A reaction with negative Free Energy (-ΔG) can be coupled to provide energy for a reaction with positive Free Energy (+ΔG) D. A reaction with positive enthalpy (+ΔH) can be coupled to provide energy for a reaction with negative enthalpy (-ΔG)
C. A reaction with negative Free Energy (-ΔG) can be coupled to provide energy for a reaction with positive Free Energy (+ΔG)
The bacterium Pseudomonas saccharophilia contains phosphorylase an enzyme that catalyses the phosphorolytic cleavage of sucrose as follows: Sucrose + Pi —> glucose 1-phosphate + fructose (where Pi = phosphate). Calculate the standard free energy for the reaction above given the following information: sucrose + H2O —> glucose + fructose ΔG’o = -29 kJ/mol glucose 1-phosphate + H2O -> glucose + Pi ΔG’o = -21kJ/mol A. -8 kJ/mol B. +8 kJ/mol C. +50 kJ/mol D. -50 kJ/mol
A. -8 kJ/mol We recognise that we have a ‘coupled reaction’ and can get the overall reaction by adding the two other reactions given in the question. The common intermediate is glucose. Consequently, we can also add the standard free energies of the coupled reactions to give the standard free energy of the summed reaction. The reaction from sucrose to glucose and fructose proceeds in the direction written, so we don’t need to change the sign of the standard free energy. However, the hydrolysis of glucose 1-phosphate occurs in the the opposite direction to that written. Consequently, we need to reverse the sign of the standard free energy because we are saying the reaction goes in the opposite direction to that written. Thus, the free energy value is +21 kJ/mol for the reaction glucose + Pi —> H2O + glucose 1-phosphate Summing the two reactions: -29 + 21 = -8kJ/mol
Estimate the energy content of Acetyl CoA (CH3COS-CoA) using the energy value of -220kJ/mol for reduced bonds. A. ΔG’° = - 1100 kJ/mol B. ΔG’° = 1100 kJ/mol C. ΔG’° = 880 kJ/mol D. ΔG’° = -880 kJ/mol E. ΔG’° = -660 kJ/mol
D. ΔG’° = -880 kJ/mol Acetyl CoA (CH3COS-CoA) has: 3 x C-H bonds Hence 3 x -220 = -660 kJ/mol 1 x C-C bonds. Hence 1 x -220 = -220 kJ/mol Total estimated energy = -880 kJ/mol
The Free Energy contained in one mol of glucose (C6H12O6), assuming -220kJ/mol for each reduced bond, is: A. +2.640 kJ/mol B. +2640 J/mol C. -2200 kJ/mol D. -2640 kJ/mol E. -2.200 J/mol
D. -2640 kJ/mol Thus: 5 C-C bonds = 5 x -220 kJ/mol = -1100 kJ/mol 7 C-H bonds = 7 x -220 kJ/mol = -1540 kJ/mol Total estimated energy generated from one mole of glucose = -2640kJ/mol
The Gibbs free energy change for ATP hydrolysis is large and negative in part because: A. the products are relatively less stable B. the terminal anhydride bonds in ATP are ‘weaker’ compared with the bonds in the products. C. there is less electrostatic repulsion among four negative charges of phosphate after hydrolysis D. the products (phosphate and ADP) are less soluble in water than ATP
B. the terminal anhydride bonds in ATP are ‘weaker’ compared with the bonds in the products. C. there is less electrostatic repulsion among four negative charges of phosphate after hydrolysis
ATP is known as the ‘energy currency’ of the cell and needs to be relatively reactive for this role. However, it can remain in a bottle on the shelf (usually in a refrigerator) for many months unchanged. This is because ATP has a relatively high free energy of hydrolysis. True or False?
False
The tendency of a metabolic reaction to proceed is due to the free energy of both the reactants and products as well as the change in randomness of that reaction. True or False?
True
The major carrier of chemical energy in all cells is: A. adenosine monophosphate. B. adenosine diphosphate. C. adenosine triphosphate D. adenosine tetraphosphate. E. acetyl triphosphate.
C. adenosine triphosphate
Four reactions are given below with their reduction potentials. Indicate the flow of electrons that would occur from one reaction to the next if they could be coupled to each other. 1. NO3- + 2H+ + 2e- ——> NO2- + H2O E’o = 0.421V 2. Fumarate2- + 2H+ +2e- ——> succinate2- E’o = 0.031V 3. NAD+ + H+ + 2e- ——> NADH E’o = -0.320V 4. O2 + 2H+ + 2e- ——> H2O2 E’o = 0.295V A. Flow of electrons would occur from reaction 1 to 4 to 2 to 3 B. Flow of electrons would occur from reaction 3 to 2 to 4 to 1 C. Flow of electrons would occur from reaction 1 to 2 to 3 to 4 D. Flow of electrons would occur from reaction 4 to 3 to 2 to 1
B. Flow of electrons would occur from reaction 3 to 2 to 4 to 1 Electrons flow from low values of E’o (better reducing agents) to higher values of E’o (better oxidising agents)
Some electron carriers in the cell include: A. ubiqone B. ferrohydrolase C. cytochrome proteins D. coenzyme Q
C. cytochrome proteins D. coenzyme Q
The phosphorylation of glucose is an endergonic reaction as follows: Glucose + Pi = glucose 6-phosphate + H2O (where Pi = phosphate). ΔG’o = +13.8 kJ/mol The hydrolysis of ATP is an exergonic reaction as follows: ATP + H2O = ADP + Pi (where Pi = phosphate). ΔG’o = -30.5 kJ/mol These two reactions can be summed to give the overall reaction as follows: Glucose + ATP = glucose 6-phosphate + ADP Calculate the overall standard Gibbs Free Energy value for the summed (coupled) reactions. A. -16.7 kJ/mol B. -44.3 kJ/mol C. +44.3 kJ/mol D. +16.7 kJ/mol
A. -16.7 kJ/mol We can add the standard free energies of the coupled reactions. Both reactions proceed in the directions as written from left to right. Consequently, we do not need to change the sign of the standard free energy values. -30.5 + 13.8 = -16.7 kJ/mol
The conversion of ATP to ADP in the cell is able to drive thermodynamically unfavourable processes such as glucose to glucose-6-phosphate. A. ATP drives reactions because coversion of ATP to ADP makes the overall reaction favourable B. The breaking of bond A in the diagram releases energy C. The ΔG activation of ATP breakdown to ADP (hydrolysis) is high D. ATP is a relatively stable product compared with its products ADP and Pi E. The ATP molecule is a good source of energy due to the strong P-O bonds formed in the reaction F. All of the Free Energy in the hydrolysis reaction of ATP to ADP is released as heat if not coupled to another reaction G. ATP breakdown can be coupled to other reactions in a test tube with no other components except for the reactants.
A. ATP drives reactions because coversion of ATP to ADP makes the overall reaction favourable C. The ΔG activation of ATP breakdown to ADP (hydrolysis) is high E. The ATP molecule is a good source of energy due to the strong P-O bonds formed in the reaction
NAD+ transfers electrons by: A. accepting a hydride ion in its structure B. accepting two electrons and two hydrogen atoms in its structure C. accepting a hydroxyl group in its structure D. accepting one electron at a time in its structure
A. accepting a hydride ion in its structure
FAD transfers electrons by: A. accepting a hydride ion in its structure B. accepting two electrons and one hydrogen atom in its structure C. accepting one electron and one hydrogen atom C. accepting one electron and one hydrogen atom in its structure D. accepting a hydroxyl group in its structure
C. accepting one electron and one hydrogen atom in its structure
The Gibbs Free energy change for ATP hydrolysis is large and negative in part because: A. the products (ADP and phosphate) are more soluble in water than ATP B. the terminal anhydride bonds in ATP are ‘stronger’ compared with the bonds in the products. C. there is increased electrostatic repulsion among four negative charges of phosphate after hydrolysis D. the products (ADP and phosphate) are relatively more stable
A. the products (ADP and phosphate) are more soluble in water than ATP D. the products (ADP and phosphate) are relatively more stable
A carbon is reduced if: A. The number of bonds to more electronegative atoms (e.g. O, N, F, Cl, I or S) decreases B. The number of hydrogen atoms bonded to a carbon increases C. The number of hydrogen atoms bonded to a carbon decreases D. The number of bonds to more electronegative atoms increases
A. The number of bonds to more electronegative atoms (e.g. O, N, F, Cl, I or S) decreases B. The number of hydrogen atoms bonded to a carbon increases
All of the following contribute to the large, negative, free-energy change upon hydrolysis of “high-energy” compounds (e.g. ATP) except: A. electrostatic repulsion in the reactant. B. stabilization of products by ionization. C. low activation energy of forward reaction. D. stabilization of products by solvation. E. stabilization of products by extra resonance forms.
C. low activation energy of forward reaction.
