Biochemistry Exam 3 Iclicker Questions

The hydrolysis of phosphoenylpyruvate proceeds with a ∆G′˚ of about -62 kJ/mol. The greatest contributing factors to this reaction are the destabilization of the reactants by electrostatic repulsion & stabilization of the product pyruvate by? A. electrostatic attraction. B. ionization. C. polarization. D. resonance. E. tautomerization.

tautomerization

phosphofructokinase-1.

All of the following enzymes involved in the flow of carbon from glucose to lactate (glycolysis) are also involved in the reversal of this flow (gluconeogenesis) except: 3-phosphoglycerate kinase. aldolase. enolase. phosphofructokinase-1. phosphoglucoisomerase.

are irreversible

All of the bypass reactions used in gluconeogenesis require high-energy equivalents. involve the removal of phosphate groups. are irreversible. requires mitochondrial enzymes.

What is the ∆G′˚ for the following reaction at 25˚C & pH 7.0? Glutamate + oxaloacetate → aspartate + α-ketoglutarate Keq = 6.8 R = 8.315 J/mol · K -4.7 kJ/mol 0 +4.7 J/mol +4.7 kJ/mol

-4.7 kJ/mol dont need to use eq.we know that Keq is greater than 1, so delta G is negative.

Aldolase catalyzes the glycolytic reaction: Fructose 1,6-bisphosphate glyceraldehyde 3-phosphate + dihydroxyacetone phosphate The standard free-energy change for this reaction in the direction written is +23.8 kJ/mol. The concentrations of the 3 intermediates in the hepatocyte of a mammal are: fructose 1,6-bisphosphate, 1.4 x 10-5 M, glyceraldehyde 3-phosphate, 3 x 10-6 M & dihydroxyacetone phosphate, 1.6 x 10-5 M. At body temperature (37˚C), what is the actual free-energy change for the reaction? A. -8.6 kJ/mol B. -6.8 kJ/mol C. 0 kJ/mol D. +6.8 kJ/mol E. +8.6 kJ/mol

-8.6 kJ/mol ΔG = ΔG΄˚ + RT ln Q ΔG = 23.8 kJ/mol + (8.315 x 10-3 kJ/mol · K) (310K) ln Q ΔG = 23.8 kJ/mol + 2.578 kJ/mol ln [(3 x 10-6)(1.6 x 10-5)/(1.4 x 10-5)] ΔG = 23.8 kJ/mol + 2.578 kJ/mol ln (3.43 x 10-6) = -8.6 kJ/mol

The standard reduction potentials (E′˚) for the following half reactions are given: Fumarate + 2 H+ + 2 e- succinate E′˚ = +0.031 V FAD + 2 H+ + 2e- FADH2 E′˚ = -0.219 V If you mixed succinate, fumarate, FAD & FADH2 together, all at 1M concentrations & in the presence of succinate dehydrogenase, which of the following would happen initially? A. Fumarate & succinate would become oxidized; FAD & FADH2 would become reduced. B. Fumarate would become reduced, FADH2 would become oxidized. C. No reaction would occur because all the reactants & products are already at their standard concentrations. D. Succinate would become oxidized, FAD would become reduced. E. Succinate would become oxidized, FADH2 would be unchanged because it is a cofactor.

. Fumarate would become reduced, FADH2 would become oxidized.

Below are the standard reduction potentials (E′˚) for 2 conjugate redox pairs. Which is the following is true? Pyruvate/lactate E′˚ = -0.185 V NAD+/NADH E′˚ = -0.320 V A. The pyruvate/lactate conjugate redox pair has a greater tendency to lose electrons than the NAD+/NADH pair. B. Pyruvate has a greater affinity for electrons than NAD+. C. NAD+ is a reducing agent. D. NAD+ + oxaloacetate NADH + H+ + malate E. Under standard conditions, NAD+ is more likely to be converted to NADH, than pyruvate is to converted to lactate.

. Pyruvate has a greater affinity for electrons than NAD+.

Hydrolysis of 1M glucose 6-phosphate catalyzed by glucose 6-phosphatase is 99% complete at equilibrium (only 1% of substrate remains). Which of the following statements is correct? A. ∆G′˚ = -11 kJ/mol B. ∆G′˚ = -5 kJ/mol C. ∆G′˚ = 0 kJ/mol D. ∆G′˚ = 11 kJ/mol E. ∆G′˚ cannot be determined from the information given.

. ∆G′˚ = -11 kJ/mol Keq= 99/1, so it is large and positive, so dental g must be large and negative.

Which of the following does not have a large, negative energy of hydrolysis? A. 1,3-bisphosphoglycerate. B. 3-phosphoglycerate. C. ADP. D. Phosphoenolpyruvate. E. Thioesters (e.g. acetyl-CoA

3-phosphoglycerate.

When a mixture of 3-phosphoglycerate & 2-phosphoglycerate is incubated at 25˚C with phosphoglycerate mutase until equilibrium is reached, the final mixture is contains 6 times as much 2-phosphoglycerate as 3-phosphoglycerate. Which one of these statements is correct? 3-phosphoglycerate → 2-phosphoglycerate A. ∆G′˚ = -4.44 kJ/mol B. ∆G′˚ = 0 C. ∆G′˚ = 12.7 kJ/mol D. ∆G′˚ is incalculably large & positive. E. ∆G′˚ cannot be calculated from the information given.

: A. ∆G′˚ = -4.44 kJ/mol

will proceed spontaneously from left to right.

: If the ∆G′˚ of the reaction AB is -40 kJ/mol, under standard conditions, the reaction: is at equilibrium. will never reach equilibrium. will not occur spontaneously. will proceed at a rapid rate. will proceed spontaneously from left to right

It stimulates acetyl-CoA synthesis because it will lower the value of Q for the acetyl-CoA synthesis reaction.

: Synthesis of the activated form of acetate (acetyl-CoA) is carried out in an ATP-dependent process: Acetate + CoA + ATP acetyl-CoA + AMP + PPi Almost all cells contain the enzyme inorganic pyrophosphatase which catalyzes the hydrolysis of PPi to Pi. What effect does the presence of this enzyme have on the synthesis of acetyl-CoA? It inhibits acetyl-CoA synthesis because it will increase the value of Q for the acetyl-CoA synthesis reaction. It stimulates acetyl-CoA synthesis because it will lower the value of Q for the acetyl-CoA synthesis reaction. It inhibits acetyl-CoA synthesis because the hydrolysis of PPi is a thermodynamically unfavorable process. It stimulates acetyl-CoA synthesis because it will increase the value of Q for the acetyl-CoA synthesis reaction.

the products of the reaction are more disordered than the reactants

A chemical reaction is more likely to occur spontaneously if the products of the reaction are more complex than the reactants. the system takes up heat from its surroundings. the products of the reaction are more disordered than the reactants. the system gains free energy.

If Keq is greater than 1 & ∆G′˚ is negative.

A chemical reaction starts with 1M concentrations each of reactants A & B & products C & D, under what conditions of Keq & ∆G′˚ will the reaction proceed in the forward direction? If Keq is greater than 1 & ∆G′˚ is negative. If Keq is 0 & ∆G′˚ is negative. If Keq is less than 1 & ∆G′˚ is negative. If Keq is less than 1 & ∆G′˚ is positive.

The reaction ATP → ADP + Pi is an example of a __ reaction. homolytic cleavage internal rearrangement free radical group transfer oxidation/reduction

Group Transfer There is exergonic cleavage by the addition of H2), of the phosphate anhydrive bond in ATP

_____ is the energy currency that links catabolism and anabolism

ATP

What is the standard free-energy change for this reaction? Acetaldehyde + NADH + H+ ethanol + NAD+ Acetalydehyde + 2H+ + 2e- ethanol E′˚ = -0.197 V NAD+ + 2H+ + 2e- NADH + H+ E′˚ = -0.320 V A. -99.8 kJ/mol B. -23.7 kJ/mol C. 0 kJ/mol D. +23.7 kJ/mol E. +99.8 kJ/mol

Acetalydehyde + 2H+ + 2e- ethanol E′˚ = -0.197 V NADH + H+ NAD+ + 2H+ + 2e- E′˚ = +0.320 V Acetaldehyde + NADH + H+ ethanol + NAD+ E′˚ = +0.123 V ∆G' = -nFE' = -2(96.5 kJ/V · mol)(0.123 V) = -23.7 kJ/mol

3-phosphoglycerate kinase.

An enzyme used in both glycolysis & gluconeogenesis is: 3-phosphoglycerate kinase. glucose 6-phosphatase. hexokinase. phosphofructokinase-1. pyruvate kinase.

No net yield of ATP would occur during glycolysis.

Arsenate is structurally & chemically similar to Pi & many enzymes that require phosphate will also use arsenate. Organic compounds of arsenate are less stable than analogous phosphate compounds. For example, acyl arsenates decompose rapidly by hydrolysis: What would be the consequence to an organism if arsenate were substituted for phosphate? No net yield of ATP would occur during glycolysis. No net yield of NADH would occur during glycolysis. The GAPDH reaction would not occur. Pyruvate would not be formed.

If the carbon is bound to: 1. Hydrogen, then the carbon "owns" __ 2. Carbon, then the carbon "owns" ___ 3. Nitrogen or Oxygen, then the carbon "owns" ___

How to solve for oxidation states 2 e- 1 e- 0 e-

Conversion of NAD+ to NADH is accomplished by the addition of: A. a proton. B. an electron. C. a hydride ion. D. a hydrogen atom.

Hydride Ion: :H-

irreversible

Living systems are far from equilibrium AKA ?

When Delta E prime not ( standard reduction potential ) is positive, this makes Delta G prime not.......

Negative ( spontaneous)

When the concentrations of NAD+, NADH, oxaloacetate & malate are all 10-5 M, the "spontaneous" reaction is: Oxaloacetate2- + 2 H+ + 2 e- malate2- E′˚ = -0.166 V NAD+ + H+ + 2e- NADH E′˚ = -0.320 V A. Malate + NAD+ oxaloacetate + NADH + H+ B. Malate + NADH + H+ oxaloacetate + NAD+ C. NAD+ + NADH + H+ malate + oxaloacetate D. NAD+ + oxaloacetate NADH + H+ + malate E. Oxaloacetate + NADH + H+ malate + NAD+

Oxaloacetate + NADH + H+ malate + NAD+ Oxaloacetate2- + 2 H+ + 2 e- -> malate2- E′˚ = -0.166 V NADH NAD+ + H+ + 2e- E′˚ = +0.320 V Oxaloacetate + NADH + H+ malate + NAD+ E′˚ = +0.154 V It is telling us that it is a spontaneous rxn, so Dental G will be negative. We reverse one, and then add.

a measure of an atom's or molecule's affinity for electrons

Reduction Potential (E΄˚):

Formation of lactate. Pyruvate + H+ + 2 e- lactate E′˚ = -0.19 V NADH NAD+ + 2 e- E′˚ = +0.32 V Pyruvate + NADH + H+ lactate + NAD+ ∆E′˚ = +0.13 V ΔG΄˚ = -n F ΔE΄

The E′˚ values for the NAD+/NADH & pyruvate/lactate conjugate redox pairs are -0.32 V & -0.19 V, respectively. Beginning with 1 M concentrations of each reactant & product at pH 7.0 & 25˚C, which direction will the reaction proceed? Pyruvate + NADH + H+ lactate + NAD+ Formation of pyruvate. Formation of lactate. The reaction is at equilibrium. There is not enough information to determine.

NAD+/NADH Reduction potential is a measure of a molecule's affinity for electrons

The E′˚ values for the NAD+/NADH & pyruvate/lactate conjugate redox pairs are -0.32 V & -0.19 V, respectively. Which redox pair has a greater tendency to lose electrons? NAD+/NADH pyruvate/lactate They both have equal tendency to lose electrons Neither pair since, both reduction potentials are negative.

pyruvate/lactate Pyruvate is more likely to gain electrons (become reduced), so pyruvate is a better oxidizing agent.

The E′˚ values for the NAD+/NADH & pyruvate/lactate conjugate redox pairs are -0.32 V & -0.19 V, respectively. Which redox pair is a stronger oxidizing agent? NAD+/NADH pyruvate/lactate They are both equally strong oxidizing agents. Neither pair since, both reduction potentials are negative.

1,3-bisphosphoglycerate is rapidly converted to 3-phosphoglycerate. This makes Q < Keq, so ∆G will be negative

The oxidation of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate, catalyzed by GAPDH, proceeds with an unfavorable equilibrium constant (Keq = 0.08), yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell overcome the unfavorable equilibrium? 1,3-bisphosphoglycerate is rapidly converted to 3-phosphoglycerate. B. The ∆G′˚is negative. C. Redox reactions are always thermodynamically favorable. D. Phosphoryl transfer from ATP drives the reaction.

In glycolysis, fructose 1,6-bisphosphate is converted to 2 products with a standard free-energy change of (∆G′˚) of 23.8 kJ/mol. Under what conditions encountered in a normal cell will the free-energy change (∆G) be negative, enabling the reaction to proceed spontaneously to the right? A. Under standard conditions, enough energy is released to drive the reaction to the right. B. The reaction will not go to the right spontaneously under any conditions because ∆G′˚ is positive. C. The reaction will proceed spontaneously to the right if there is a high concentration of products relative to fructose 1,6-bisphosphate. D. The reaction will proceed spontaneously to the right if there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products. E. None of the above conditions are sufficient.

The reaction will proceed spontaneously to the right if there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products. ∆G = ∆G'-> + RT ln Q Q = [PRODUCTS]/[REACTANTS] = mass-action ratio

B formation is kinetically slow; equilibrium has not yet been reached after 24 hours. Keq = [B]/[A] Delta G Prime Not is negative in this example, and Keq is 2/8=.25, but Keq should be greater than because Delta G is negative, so we can conclude that equilibrium has not been reached

The ∆G′˚ of the reaction AB is -60 kJ/mol. The reaction is started with 10 mmol of A; no B is initially present. After 24 hours, analysis reveals the presence of 2 mmol of B & 8 mmol of A. Which is the most likely explanation? A & B have reached equilibrium concentrations. An enzyme has shifted the equilibrium toward A. B formation is kinetically slow; equilibrium has not yet been reached after 24 hours. Formation of B is thermodynamically unfavorable. The result described is impossible, given the fact that ∆G′˚ = -60kJ/mol.

Glucose + 2 ATP 2 glyceraldehyde 3-phosphate + 2 ADP + 2 Pi + 2 H+ FYI Glucose + 2 ATP 2 glyceraldehyde 3-phosphate + 2 ADP + 2 Pi + 2 H+: Energy investment phase of glycolysis (Steps 1-5) B. Glucose + 2 ATP glyceraldehyde 3-phosphate + dihydroxyacteone phosphate +2 ADP + + 2 Pi + 2H+: Glycolysis (Steps 1-4) C. Glucose + 2 NAD+ + 2 ADP + 2 Pi 2 pyruvate + 2 NADH + 2 ATP + 2 H+ + 2 H2O: Glycolysis (Steps 1-10) D. 2 pyruvate + 4 ATP + 2 GTP + 2 NADH + 2 H+ + 4 H2O glucose + 4 ADP + 2 GDP + 6 Pi + 2 NAD+: Gluconeogenesis (Steps 1-11)

What is the net equation for the Energy investment phase of glycolysis? Glucose + 2 ATP 2 glyceraldehyde 3-phosphate + 2 ADP + 2 Pi + 2 H+ B. Glucose + 2 ATP glyceraldehyde 3-phosphate + dihydroxyacteone phosphate +2 ADP + + 2 Pi + 2H+ C. Glucose + 2 NAD+ + 2 ADP + 2 Pi 2 pyruvate + 2 NADH + 2 ATP + 2 H+ + 2 H2O D. 2 pyruvate + 4 ATP + 2 GTP + 2 NADH + 2 H+ + 4 H2O glucose + 4 ADP + 2 GDP + 6 Pi + 2 NAD+

Which of the following has a thioester bond? A. phosphoenolpyruvate (PEP) B. ATP C. phosphocreatine D. acetyl-CoA

acetyl-CoA

: The reaction A + B C has a ∆G′˚ of -20 kJ/mol at 25˚C. Starting under standard conditions, one can predict that: A. at equilibrium, the [B] will exceed the [A]. B. at equilibrium, the [C] will be less than the [A]. C. at equilibrium, the [C] will exceed the [A] or the [B]. D. C will rapidly break down to A + B. E. when A & B are mixed, the reaction will proceed rapidly toward formation of C.

at equilibrium, the [C] will exceed the [A] or the [B]. Keq= (C)/(A)(B) This is greater than one, bc we know detla g is negative.(proceeds forward)

All of the following contribute to the large, negative free-energy change upon hydrolysis "high-energy" compounds except: A. electrostatic repulsion in the reactant. B. low activation energy of the forward reaction. C. stabilization of products by extra resonance forms. D. stabilization of products by ionization. E. stabilization of products by solvation.

low activation energy of the forward reaction.

The hydrolysis of ATP has a large, negative ∆G′˚; nevertheless, it is stable in solution due to: entropy stabilization. ionization of the phosphates. resonance stabilization. the hydrolysis reaction being endergonic. the hydrolysis reaction having a large activation energy.

the hydrolysis reaction having a large activation energy. Hydrolysis is spontaneous and stable b/c of activation energy and it is inversely proportional to the rate of the reaction.

iclicker: When a mixture of glucose 6-phosphate & fructose 6-phosphate is incubated with enzyme phosphohexose isomerase (which catalyzes the interconversion of these 2 compounds), until equilibrium is reached, the final mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate. Which of the following statements is true? glucose 6-phosphate → fructose 6-phosphate A. ∆G′˚ = -1.72 kJ/mol B. ∆G′˚ = 0 C. ∆G′˚ = 1.72 kJ/mol D. ∆G′˚ is incalculably large & negative. E. ∆G′˚ is incalculably large & positive.

∆G′˚ = 1.72 kJ/mol Keq is 1/2, less than one so delta G must be positive.