Download_life the science of biology 11th edition sadava test bank

Life The Science of Biology 11th Edition Sadava Test Bank

Life The Science of Biology 11th Edition Sadava Test Bank To download the complete and accurate content document, go to: https://testbankbell.com/download/life-the-science-of-biology-11th-edition-sadava-test -bank/

Visit TestBankBell.com to get complete for all chapters

Test Bank to accompany

Life: The Science of Biology, Eleventh Edition Sadava • Hillis • Heller • Hacker

Chapter 8: Energy, Enzymes, and Metabolism

TEST BANK QUESTIONS Multiple Choice 1. A plant makes sugars via photosynthesis and is consumed by an herbivore. The herbivore in turn breaks down the sugars, producing energy it then uses to build a nest.

Which sequence represents the energy transformation of this pathway? a. Light → chemical → mechanical b. Chemical → kinetic → heat c. Heat → chemical → potential d. Light → kinetic → chemical e. Mechanical → heat → potential Answer: a Learning Outcome: 8.1.1.a Describe the forms of energy in biological systems.

Bloom’s Level: 4. Analyzing 2. In thermodynamic terms, water held back by a dam represents _______ energy. a. electrical b. irrigation c. potential d. kinetic e. heat Answer: c Learning Outcome: 8.1.1.b Differentiate between potential energy and kinetic energy.

Bloom’s Level: 3. Applying 3. Which of the following represents kinetic energy? a. Chemical bonds b. Concentration gradient c. Electric charge imbalance d. Muscle contraction e. Separation of opposite charges Answer: d Learning Outcome: 8.1.1.b Differentiate between potential energy and kinetic energy.

Bloom’s Level: 2. Understanding

© 2017 Sinauer Associates, Inc.

4. Refer to the table below.

A student constructed the table to summarize characteristics that distinguish anabolic reactions from catabolic reactions. Which statement best evaluates the accuracy of the student’s work? a. The table is correct as shown. b. The entries concerning molecular changes and role in metabolism are incorrect and should be reversed. c. The entries concerning direction of energy flow are incorrect and should be reversed. d. The entries concerning changes in potential energy are incorrect and should be reversed. e. The examples used are incorrect and should be reversed.

Answer: c Learning Outcome: 8.1.2.b Compare and contrast anabolic and catabolic reactions.

Bloom’s Level: 5. Evaluating 5. The sum total of all chemical reactions in a living organism is called its a. energetics. b. activity. c. digestive power. d. entropy. e. metabolism.

Answer: e Learning Outcome: 8.1.4.a Explain how the first law of thermodynamics relates to biological systems.

Bloom’s Level: 1. Remembering

© 2017 Sinauer Associates, Inc.

6. The first law of thermodynamics states that the total energy in the universe is a. decreasing. b. increasing. c. constant. d. being converted to free energy. e. being converted to matter.

Answer: c Learning Outcome: 8.1.4.a Explain how the first law of thermodynamics relates to biological systems.

Bloom’s Level: 2. Understanding 7. Which of the following is neither created nor destroyed during its conversion? a. Entropy b. Energy c. Free energy only d. Thermal energy only e. Potential energy only Answer: b Learning Outcome: 8.1.4.a Explain how the first law of thermodynamics relates to biological systems.

Bloom’s Level: 1. Remembering 8. During photosynthesis, plants use light energy to synthesize sugars, using the carbon in carbon dioxide. They do not make new energy in this process; they merely convert it from light energy to chemical energy. Photosynthesis is thus an illustration of a. increasing entropy. b. chemical equilibrium. c. the first law of thermodynamics. d. the second law of thermodynamics. e. a spontaneous reaction.

Answer: c Learning Outcome: 8.1.4.a Explain how the first law of thermodynamics relates to biological systems.

Bloom’s Level: 2. Understanding 9. Free energy is a. enthalpy. b. entropy. c. usable energy. d. always positive. e. always negative.

Answer: c Learning Outcome: 8.1.5.b Explain the concept of free energy.

Bloom’s Level: 2. Understanding 10. The G of a reaction tells us all of the following except

© 2017 Sinauer Associates, Inc.

a. the rate of the reaction. b. the direction of the reaction. c. whether the reaction is exergonic or endergonic. d. whether the reaction requires or releases energy. e. whether a higher free energy is in the product or in the reactants.

Answer: a Learning Outcome: 8.1.5.b Explain the concept of free energy.

Bloom’s Level: 2. Understanding 11. Which statement about the exergonic hydrolysis of maltose to glucose is true? a. The reaction requires the input of free energy. b. The free energy of glucose is larger than the free energy of maltose. c. The reaction is not spontaneous. d. The reaction releases free energy. e. At equilibrium, the concentration of maltose is higher than the concentration of glucose.

Answer: d Learning Outcome: 8.1.5.b Explain the concept of free energy.

Bloom’s Level: 3. Applying 12. If the temperature of a reaction with a G of ‒7.1 kJ/mol increases, what will happen to the G of the reaction? a. It will stay the same. b. It will equal zero. c. It will also increase. d. It will decrease. e. Not enough information is provided to make a determination.

Answer: e Learning Outcome: 8.1.5.c Use the equation ∆G = ∆H – T∆S to describe biological systems.

Bloom’s Level: 3. Applying 13. How does the second law of thermodynamics apply to living organisms? a. It does not apply to living organisms; the complexity of organisms contradicts the second law. b. As energy transformations occur, free energy increases and unusable energy decreases. c. The potential energy of ATP is converted to kinetic energy, as with muscle contractions. d. Living organisms require a constant input of energy to maintain complex structures and order. e. All of the reactions in an organism require an input of energy.

Answer: d Learning Outcome: 8.1.5.d Apply the second law of thermodynamics to biological systems.

Bloom’s Level: 2. Understanding

© 2017 Sinauer Associates, Inc.

14. When energy is converted from one form to another, only some of the energy is usable. The unusable energy is released to a form associated with disorder. A measure of this disorder is referred to as a. free energy. b. entropy. c. enthalpy. d. thermodynamics. e. equilibrium.

Answer: b Learning Outcome: 8.1.5.d Apply the second law of thermodynamics to biological systems.

Bloom’s Level: 1. Remembering 15. Suppose that, in a closed system, there is an input of energy as 1 kJ of light. The light energy is converted to glucose, then ATP, and finally to mechanical energy. Which statement about the amount of mechanical energy available is true? a. There is 1 kJ of mechanical energy available to do work. b. The amount of usable energy decreases with each conversion, so less than 1 kJ of energy is available for mechanical work. c. All of the reactions involved are anabolic, so there is 0.25 kJ of mechanical energy available. d. All of the reactions involved are exergonic, so the number of kilojoules of mechanical energy produced is greater than the 1 kJ input. e. All of the reactions involved are endergonic, so without constant input of energy, no mechanical energy is produced.

Answer: b Learning Outcome: 8.1.5.d Apply the second law of thermodynamics to biological systems.

Bloom’s Level: 3. Applying 16. The standard free energy change for the hydrolysis of GTP (guanosine triphosphate) to GDP + Pi is –7.3 kcal/mol, similar to that for ATP. From this information, one can conclude that the a. reaction will never reach equilibrium. b. free energy of GDP and phosphate is higher than the free energy of GTP. c. reaction requires energy. d. reaction is endergonic. e. reaction is exergonic.

Answer: e Learning Outcome: 8.1.6.a Differentiate between exergonic and endergonic reactions.

Bloom’s Level: 3. Applying 17. The G of ATP hydrolysis is negative, and the change in entropy is positive. The reaction, therefore, a. requires energy. b. is endergonic.

© 2017 Sinauer Associates, Inc.

c. is exergonic. d. will not reach equilibrium. e. decreases the disorder in the system.

Answer: c Learning Outcome: 8.1.6.a Differentiate between exergonic and endergonic reactions.

Bloom’s Level: 3. Applying 18. Refer to the figure below showing the change in free energy resulting from a chemical reaction.

Which statement about the reaction is true? a. It is an endergonic reaction. b. The reactants have less energy than the products. c. G is negative. d. It is an example of a condensation reaction. e. It is an anabolic reaction.

Answer: c Learning Outcome: 8.1.6.a Differentiate between exergonic and endergonic reactions.

Bloom’s Level: 4. Analyzing 19. Chemical equilibrium a. is a static state where all motion stops. b. represents a state of negative energy change. c. represents a state of positive energy change. d. is essential for normal cell functions. e. is a state in which G = 0.

Answer: e Learning Outcome: 8.1.7.a Use an example to explain the concept of chemical equilibrium.

Bloom’s Level: 2. Understanding 20. If the enzyme phosphohexoisomerase is added to a 0.3 M solution of fructose 6phosphate and the reaction is allowed to proceed to equilibrium, the final concentrations © 2017 Sinauer Associates, Inc.

are 0.2 M glucose 6-phosphate and 0.1 M fructose 6-phosphate. What would be the final concentration of glucose 6-phosphate if the initial concentration of fructose 6-phosphate were changed to 3 M? a. 2 M b. 3 M c. 1 M d. 0.5 M e. 0.2 M Answer: a Learning Outcome: 8.1.7.a Use an example to explain the concept of chemical equilibrium.

Bloom’s Level: 3. Applying 21. You make a solution of glucose 1-phosphate and allow it to isomerize to form glucose 6-phosphate. Eventually, you have an equilibrium mixture of the two isomers. What can you do to increase the concentration of glucose 6-phosphate in the solution? a. Add more glucose. b. Add more phosphate. c. Add a higher concentration of reactant and allow it to make more product. d. Add ATP as an energy source for the forward reaction (G = ‒7.1 kJ/mol) to make more product. e. Add an enzyme to lower the ΔG of the reaction so more product is made.

Answer: c Learning Outcome: 8.1.7.a Use an example to explain the concept of chemical equilibrium.

Bloom’s Level: 3. Applying 22. The reaction below takes place in cells and is catalyzed by the enzyme phosphoglucoisomerase.

At equilibrium, the ratio of fructose 6-phosphate to glucose 6-phosphate is 0.30:1. What will happen if the concentration of glucose 6-phosphate reaches 0.5 nM and the concentration of fructose 6-phosphate reaches 0.8 nM at a particular instant in the cell? a. Only the forward reaction will occur. b. Only the reverse reaction will occur. c. The rate of the forward reaction will be greater than the rate of the reverse reaction. d. The rate of the reverse reaction will be greater than the rate of the forward reaction. e. The rate of the forward reaction will be the same as the rate of the reverse reaction.

Answer: d Learning Outcome: 8.1.7.b Predict the direction of a reaction for a system at equilibrium when conditions are changed.

© 2017 Sinauer Associates, Inc.

Bloom’s Level: 4. Analyzing 23. Knowing the change in free energy (G) of a reaction tells us a. the equilibrium point of the reaction. b. the rate of the reaction. c. how quickly equilibrium will be reached. d. the optimum temperature for the reaction. e. the activation energy.

Answer: a Learning Outcome: 8.1.7.c Describe the change in free energy for a reaction given information about its point of equilibrium.

Bloom’s Level: 2. Understanding 24. A readily reversible reaction, in which reactants and products have almost equal free energies, is indicated by a a. slightly negative G. b. change in free energy. c. negative G. d. G near zero. e. large positive G. Answer: d Learning Outcome: 8.1.7.c Describe the change in free energy for a reaction given information about its point of equilibrium.

Bloom’s Level: 2. Understanding 25. Which reaction will go the furthest toward completion? a. A reaction with a G of zero b. A reaction with a G of +2 kcal/mol c. A reaction with a G of +12 kcal/mol d. A reaction with a G of ‒2 kcal/mol e. A reaction with a G of ‒12 kcal/mol Answer: e Learning Outcome: 8.1.7.c Describe the change in free energy for a reaction given information about its point of equilibrium.

Bloom’s Level: 3. Applying 26. When ADP gains a phosphate to form ATP, a. free energy is released by the loss of a phosphate. b. the reaction ends. c. energy is absorbed. d. chemical energy is converted to light energy. e. ribose loses an oxygen to become deoxyribose.

Answer: c Learning Outcome: 8.2.1.a Write a chemical equation describing the formation and hydrolysis of ATP and identifying the role of energy in the reaction.

Bloom’s Level: 2. Understanding

© 2017 Sinauer Associates, Inc.

27. What is the order in which the following molecules would be produced if phosphate ions were removed from ATP one at a time? a. ATP, ADP, AMP, adenosine b. AMP, ADP, ATP, adenosine c. Adenosine, ATP, AMP, ADP d. ATP, AMP, ADP, adenosine e. ADP, adenosine, AMP, ATP Answer: a Learning Outcome: 8.2.1.a Write a chemical equation describing the formation and hydrolysis of ATP and identifying the role of energy in the reaction.

Bloom’s Level: 2. Understanding 28. When ATP is hydrolyzed, the reaction produces all of the following except a. AMP. b. ADP. c. phosphate ions. d. pyrophosphate ions. e. water.

Answer: e Learning Outcome: 8.2.1.a Write a chemical equation describing the formation and hydrolysis of ATP and identifying the role of energy in the reaction.

Bloom’s Level: 2. Understanding 29. ATP can phosphorylate many different molecules. This means that ATP can a. receive phosphate groups. b. donate phosphate groups. c. convert molecules to nucleic acids. d. release a large amount of energy when hydrolyzed. e. waste usable energy available in its phosphate bonds.

Answer: b Learning Outcome: 8.2.1.c Analyze evidence that supports the claim that cells convert the chemical energy of ATP into other forms of energy.

Bloom’s Level: 2. Understanding 30. Which statement about ATP is true? a. The hydrolysis of ATP is endergonic. b. ATP consists of adenine bonded to deoxyribose. c. ATP releases a large amount of light energy when hydrolyzed. d. An active cell requires about 100 molecules of ATP per second. e. On average, ATP is consumed within 1 second of its formation.

Answer: e Learning Outcome: 8.2.1.c Analyze evidence that supports the claim that cells convert the chemical energy of ATP into other forms of energy.

Bloom’s Level: 1. Remembering

© 2017 Sinauer Associates, Inc.

31. Fireflies a. release a considerable amount of energy as heat. b. light up to signal danger. c. use ATP to drive luciferin oxidation. d. are constantly converting light energy into chemical energy. e. have a short life cycle because of rapid depletion of ATP. Answer: c Learning Outcome: 8.2.2.a Use diagrams to explain the role of ATP in the coupling of an energy-requiring biochemical transformation to an energy-releasing one.

Bloom’s Level: 2. Understanding 32. The phosphorylation of glucose to glucose 6-phosphate is a reaction with a G of +4.0 kcal/mol. What is required for this reaction to take place? a. The input of energy from hydrolysis of ATP b. An increase in glucose concentration to push the reaction toward the product c. The coupling of this reaction with an endergonic reaction d. The addition of a noncompetitive inhibitor e. Nothing is required; this is a spontaneous reaction.

Answer: a Learning Outcome: 8.2.2.a Use diagrams to explain the role of ATP in the coupling of an energy-requiring biochemical transformation to an energy-releasing one.

Bloom’s Level: 3. Applying 33. You are given a solution in which a biochemical reaction is occurring, and you are unsure whether G is positive or negative. You suspect that the reaction might make use of the ATP to ADP + Pi energy-coupling system. To test whether it uses this system, and to determine whether the reaction is exergonic or endergonic, you would add _______. If the reaction _______, you would conclude that it is _______. a. ATP; lit up; exergonic b. ATP; made more product; endergonic c. ADP; made more product; exergonic d. ADP; made phosphate ions; endergonic e. ATP; produced a decrease in temperature; exergonic Answer: b Learning Outcome: 8.2.2.b Analyze evidence to identify reactions as exergonic or endergonic.

Bloom’s Level: 3. Applying 34. In glycolysis, the exergonic reaction 1,3-diphosphoglycerate → 3-phosphoglycerate is coupled to the reaction ADP + Pi→ ATP. Based on this information, which statement is most likely to be true about the reaction ADP + Pi→ ATP? a. The reaction never reaches equilibrium. b. The reaction is spontaneous. c. There is a large decrease in free energy. d. The reaction is endergonic. e. Temperature will not affect the rate constant of the reaction.

© 2017 Sinauer Associates, Inc.

Answer: d Learning Outcome: 8.2.2.b Analyze evidence to identify reactions as exergonic or endergonic.

Bloom’s Level: 3. Applying 35. In order for a coupled reaction to have a net G of ‒3.3 kcal/mol, a. the exergonic reaction needs to release more energy than the endergonic reaction requires. b. both reactions need to have the same G. c. the G of the endergonic reaction needs to be more negative than the G of the exergonic reaction. d. both reactions need to be exergonic. e. both reactions need to be endergonic.

Answer: a Learning Outcome: 8.2.2.c Predict biochemical transformations that require coupling to ATP hydrolysis.

Bloom’s Level: 2. Understanding 36. Refer to the table below.

Predict which reactions are most likely to be coupled to ATP hydrolysis in a cell. a. Reactions 1 and 4 b. Reactions 2 and 3 c. Reactions 1 and 2 d. Reactions 2, 3, and 4 e. Reactions 1, 2, and 4 Answer: a Learning Outcome: 8.2.2.c Predict biochemical transformations that require coupling to ATP hydrolysis.

© 2017 Sinauer Associates, Inc.

Bloom’s Level: 3. Applying 37. The phosphorylation of glucose to glucose 6-phosphate has a G of +4.0 kcal/mol. If hexokinase is added to speed up the rate of this reaction, the G will a. increase. b. decrease. c. stay the same. d. increase, but only with an increase in enzyme amount. e. decrease, but only with a decrease in enzyme amount.

Answer: c Learning Outcome: 8.3.1.a Describe a catalyst.

Bloom’s Level: 3. Applying 38. Most enzymes are which type of macromolecule? a. Protein b. Carbohydrate c. Lipid d. Nucleic acid e. Hydrocarbon Answer: a Learning Outcome: 8.3.1.a Describe a catalyst.

Bloom’s Level: 1. Remembering 39. What is a transition state in a chemical reaction? a. The place where a substrate molecule binds to an enzyme b. A reactant with potential energy that is higher than that of the product c. The combination of a substrate and an enzyme d. The state at which the bonds of reactants are unstable e. The active site where reactants are oriented Answer: d Learning Outcome: 8.3.2.a Draw an energy diagram for a reaction, identifying the transition state and free energy changes.

Bloom’s Level: 2. Understanding 40. In a chemical reaction, transition-state intermediates have free energies that are a. lower than those of either the reactants or the products. b. lower than those of the reactants, but higher than those of the products. c. higher than those of either the reactants or the products. d. higher than those of the reactants, but lower than those of the products. e. lower than those of the reactants, but the same as those of the products.

Answer: c Learning Outcome: 8.3.2.a Draw an energy diagram for a reaction, identifying the transition state and free energy changes.

Bloom’s Level: 2. Understanding 41. Which of the following affects the rate of a reaction?

© 2017 Sinauer Associates, Inc.

a. S b. G c. H d. The activation energy e. The overall change in free energy Answer: d Learning Outcome: 8.3.2.b Explain the concept of activation energy.

Bloom’s Level: 2. Understanding 42. The hydrolysis of sucrose to glucose and fructose is exergonic. However, if sucrose is dissolved in water and the solution is kept overnight at room temperature, there is no detectable conversion to glucose and fructose. Why? a. The change in free energy of the reaction is positive. b. The activation energy of the reaction is high. c. The change in free energy of the reaction is negative. d. This is a hydrolysis reaction, so it requires an input of energy. e. The free energy of the products is higher than the free energy of the reactants.

Answer: b Learning Outcome: 8.3.2.b Explain the concept of activation energy.

Bloom’s Level: 3. Applying 43. Enzymes of the acid–base catalysis type contain a. a metal ion bound to a side chain. b. a prosthetic group. c. a coenzyme. d. acidic or basic amino acid side chains (R groups) in the active site. e. a covalently activated active site.

Answer: d Learning Outcome: 8.3.3.a Identify the category in which an enzyme belongs based on the type of reaction it catalyzes.

Bloom’s Level: 1. Remembering 44. The five reactions listed are catalyzed by enzymes in a cell. Which reaction is catalyzed by an oxidoreductase? a. Glucose + ATP → glucose 6-phosphate + ADP b. Sucrose + H2O → glucose + fructose c. Ethanol + NAD+ → acetaldehyde + NADH d. ATP → cAMP + Pi e. Maltose + H2O → 2 glucose Answer: c Learning Outcome: 8.3.3.a Identify the category in which an enzyme belongs based on the type of reaction it catalyzes.

Bloom’s Level: 4. Analyzing 45. Which will not increase the rate of conversion of reactant A to product B in the presence of enzyme C?

© 2017 Sinauer Associates, Inc.

a. An increase in enzyme concentration b. An increase in temperature to the optimum for the reaction c. An increase in substrate concentration d. The addition of a nonbiological molecule that can catalyze the same reaction e. The addition of D that can compete with A for binding at the active site of C Answer: e Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 4. Analyzing 46. Trypsin and elastase are both enzymes that catalyze hydrolysis of peptide bonds.

However, trypsin only hydrolyzes peptide bonds next to lysine, and elastase only hydrolyzes peptide bonds next to alanine. Why? a. Trypsin is a protein, and elastase is not. b. The Gs for the two reactions are different. c. The shapes of the active sites for the two enzymes are different. d. One of the reactions is endergonic and the other is exergonic. e. Hydrolysis of lysine bonds requires water; hydrolysis of alanine bonds does not.

Answer: c Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 3. Applying 47. The statement “enzymes are highly specific” means that specific a. enzymes are found in specific cells. b. reactions involving specific substrates are catalyzed by specific enzymes. c. enzymes require specific concentrations of substrates. d. reactions with specific activation energies are catalyzed by specific enzymes. e. concentrations of substrates work with specific enzymes.

Answer: b Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 2. Understanding 48. An active site is the a. part of the substrate that binds with an enzyme. b. site where enzymes are found in cells. c. site where energy is added to an enzyme catalyst. d. part of the enzyme that binds with a substrate. e. part of the enzyme that binds with an allosteric activator or inhibitor.

Answer: d Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 1. Remembering 49. In some cases, a substrate–enzyme complex is stabilized by a. peptide bonds. b. disulfide bonds. c. ester bonds. d. hydrogen bonds.

© 2017 Sinauer Associates, Inc.

e. phosphate bonds.

Answer: d Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 2. Understanding 50. The molecules that are acted on by an enzyme are called a. products. b. substrates. c. carriers. d. prosthetics. e. effectors.

Answer: b Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 1. Remembering 51. The enzyme glyceraldehyde 3-phosphate dehydrogenase catalyzes the reaction glyceraldehyde 3-phosphate → 1,3-diphosphoglycerate. The binding of the glyceraldehyde 3-phosphate to the enzyme creates a(n) a. transition state. b. activation groove. c. catalyst. d. enzyme–substrate complex. e. energy barrier.

Answer: d Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 3. Applying 52. The enzyme glucose oxidase binds the six-carbon sugar glucose and catalyzes its conversion to glucono-1,5-lactone. Mannose is also a six-carbon sugar, but glucose oxidase cannot bind mannose. The specificity of glucose oxidase is based on the a. free energy of the transition state. b. activation energy of the reaction. c. change in free energy of the reaction. d. three-dimensional shape and structure of the active site. e. rate constant of the reaction.

Answer: d Learning Outcome: 8.3.3.b Describe how an enzyme–substrate complex forms.

Bloom’s Level: 3. Applying 53. The enzyme -amylase increases the rate at which starch is broken down into smaller oligosaccharides by _______ of the reaction. a. decreasing the equilibrium constant b. increasing the change in free energy c. decreasing the change in free energy d. increasing the change in entropy e. lowering the activation energy

© 2017 Sinauer Associates, Inc.

Answer: e Learning Outcome: 8.3.4.a Explain how enzymes increase reaction rates.

Bloom’s Level: 2. Understanding 54. The enzyme sucrase increases the rate at which sucrose is broken down into glucose and fructose. Sucrase works by a. increasing the amount of free energy of the reaction. b. lowering the activation energy of the reaction. c. decreasing the equilibrium constant of the reaction. d. supplying energy to speed up the reaction. e. changing the shape of the active site.

Answer: b Learning Outcome: 8.3.4.a Explain how enzymes increase reaction rates.

Bloom’s Level: 3. Applying 55. Which statement about enzymes is false? a. Enzymes do not affect the equilibrium point of a reaction. b. Enzymes lower activation energy. c. Enzymes are highly specific. d. Enzymes can convert an endergonic to an exergonic reaction. e. Enzymes undergo no net chemical change during catalysis.

Answer: d Learning Outcome: 8.3.4.a Explain how enzymes increase reaction rates.

Bloom’s Level: 2. Understanding 56. The presence of a catalyst affects the a. position of equilibrium of the reaction. b. overall energy change, or G. c. energy of the reactants. d. energy of the products. e. free energy of the transition state.

Answer: e Learning Outcome: 8.3.4.a Explain how enzymes increase reaction rates.

Bloom’s Level: 2. Understanding 57. Refer to the figure below.

© 2017 Sinauer Associates, Inc.

Which type of mechanism used by an enzyme to catalyze a reaction is illustrated in the figure? a. Covalent catalysis b. Acid–base catalysis c. Metal ion catalysis d. Induced fit e. Induced strain Answer: d Learning Outcome: 8.4.1.c Give examples of mechanisms for enzyme catalysis.

Bloom’s Level: 3. Applying 58. Mechanisms used by enzymes to catalyze reactions are similar in that they all a. involve transient chemical change in the enzyme. b. involve permanent chemical change in the enzyme. c. lead to stabilization of the transition state of the substrate. d. require the involvement of cofactors or coenzymes. e. work individually in any one enzyme and not in conjunction with one another.

Answer: c Learning Outcome: 8.4.1.d Compare ways that enzymes catalyze biochemical reactions.

Bloom’s Level: 3. Applying 59. Refer to the table below.

© 2017 Sinauer Associates, Inc.

Hexokinase catalyzes the reaction between glucose and ATP to form glucose 6-phosphate and ADP. The table summarizes results of experiments using purified hexokinase. In the experiments, different sugars were substituted for glucose, and the rate of each catalytic reaction was measured and divided by the rate of reaction using glucose as substrate.

These data provide evidence for which general statement that has been made about enzymes? a. Enzymes are able to speed up a reaction by lowering its activation energy. b. Enzymes display high specificity with respect to their substrates. c. Enzymes often require cofactors and/or coenzymes in addition to their substrates. d. Enzymes bind substrates at their active sites using noncovalent forces of attraction. e. Enzymes lower activation energy of a reaction by stabilizing the substrate’s transition state.

Answer: b Learning Outcome: 8.4.2.a Explain the high degree of specificity of enzymes for their substrates.

Bloom’s Level: 5. Evaluating 60. An RNA molecule that has enzyme activity is called a. RNAse. b. ribonuclease. c. an allosteric enzyme. d. a regulatory enzyme. e. a ribozyme.

Answer: e Learning Outcome: 8.4.4.a Identify the types of biological molecules that function as enzymes.

Bloom’s Level: 1. Remembering

© 2017 Sinauer Associates, Inc.

61. Many enzymes require small nonprotein molecules for a reaction to occur. These molecules temporarily bind to and then are released from the substrate before participating in other reactions. These types of molecules are called a. side chains. b. coenzymes. c. substrates. d. prosthetic groups. e. cofactors.

Answer: b Learning Outcome: 8.4.5.a Describe coenzymes and cofactors and how they function.

Bloom’s Level: 1. Remembering 62. Refer to the expression below.

The enzyme alcohol dehydrogenase catalyzes this reaction in cells. The enzyme uses the coenzyme NAD+ in the reaction. Not shown are details about the active site, which contains a zinc ion that functions as a cofactor. How do the roles of cofactor and coenzyme compare in this particular case? a. Both function in the catalytic mechanism and do not undergo net chemical change themselves as a result of the reaction. b. Both function as substrates and undergo net chemical change as a result of the reaction. c. The coenzyme functions in the catalytic mechanism, changing only transiently, whereas the cofactor functions as a substrate and undergoes net chemical change as a result of the reaction. d. The cofactor functions in the catalytic mechanism, changing only transiently, whereas the coenzyme functions as a substrate and undergoes net chemical change as a result of the reaction. e. The cofactor functions as a substrate, changing only transiently, whereas the coenzyme functions in the catalytic mechanism and undergoes net chemical change as a result of the reaction.

Answer: d Learning Outcome: 8.4.5.b Differentiate between the roles coenzymes and cofactors play in catalysis.

Bloom’s Level: 3. Applying 63. Refer to the figure below.

© 2017 Sinauer Associates, Inc.

NAD+ functions as a coenzyme in many enzyme-catalyzed reactions. The changes that take place in this coenzyme are the same for all of these reactions and are illustrated in the figure. It is likely that, in these reactions, NAD+ a. functions as an electron donor in an oxidation–reduction reaction. b. functions as an electron acceptor in an oxidation–reduction reaction. c. forms a covalent intermediate during covalent catalysis. d. induces strain in other substrates bound at active sites. e. functions as a base in acid–base catalytic mechanisms.

Answer: b Learning Outcome: 8.4.5.c Analyze the role of a small molecule or ion in an enzymecatalyzed reaction.

Bloom’s Level: 4. Analyzing 64. In the presence of alcohol dehydrogenase, the rate of reduction of acetaldehyde to ethanol increases as the concentration of acetaldehyde increases. Eventually, the rate of the reaction reaches a maximum, at which point further increases in the concentration of acetaldehyde have no effect. Why? a. All the alcohol dehydrogenase molecules are bound to acetaldehyde molecules. b. At high concentrations of acetaldehyde, the activation energy of the reaction increases. c. At high concentrations of acetaldehyde, the activation energy of the reaction decreases. d. The enzyme is no longer specific for acetaldehyde. e. At high concentrations of acetaldehyde, the change in free energy of the reaction decreases.

Answer: a Learning Outcome: 8.4.6.a Describe how reaction rates change with differences in substrate concentration.

Bloom’s Level: 3. Applying 65. In order to determine whether a reaction is at “saturation,” you can add _______ and determine whether the reaction rate _______. a. ATP; increases b. more substrate; increases c. a competitive inhibitor; decreases d. a downstream product; decreases e. a noncompetitive inhibitor; decreases

© 2017 Sinauer Associates, Inc.

Answer: b Learning Outcome: 8.4.6.a Describe how reaction rates change with differences in substrate concentration.

Bloom’s Level: 3. Applying 66. Enzymatic reactions can become saturated as substrate concentration increases because a. enzymes have the highest possible number of hydrogen atoms attached to them. b. the concentration of substrate reaches a point at which it cannot increase any further. c. substrates are inhibitors of enzymes at high concentrations. d. the activation energy of the reaction reaches a point at which it cannot be lowered further. e. a limited number of enzyme molecules are present.

Answer: e Learning Outcome: 8.4.6.a Describe how reaction rates change with differences in substrate concentration.

Bloom’s Level: 2. Understanding 67. The enzyme hexokinase catalyzes the reaction between ATP and glucose to form glucose 6-phosphate and ADP. In order to determine the efficiency of hexokinase, you wish to conduct an experiment to calculate its turnover number. To do so, you would set up a. a reaction with a maximum amount of hexokinase at optimal pH and temperature. b. a reaction with a known amount of enzyme and saturating levels of glucose 6phosphate at optimal pH and temperature. c. a reaction with a known amount of enzyme and saturating levels of glucose and ATP at optimal pH and temperature. d. multiple reactions at varying pH levels but with a constant amount of glucose, a constant temperature, and a known amount of enzyme. e. multiple reactions with varying temperatures but a constant amount of glucose, constant pH levels, and a known amount of enzyme.

Answer: c Learning Outcome: 8.4.6.b Predict changes in the rate of an enzyme-catalyzed reaction as factors in the reaction are varied.

Bloom’s Level: 3. Applying 68. Which graph shows changes in the rate of an enzyme-catalyzed reaction as substrate concentration is varied? a.

© 2017 Sinauer Associates, Inc.

b.

c.

d.

e.

Answer: b Learning Outcome: 8.4.6.b Predict changes in the rate of an enzyme-catalyzed reaction as factors in the reaction are varied.

Bloom’s Level: 3. Applying 69. If you wanted to permanently block the function of an enzyme, you could add a. a molecule that can covalently bind to the active site. b. more product. c. a noncompetitive inhibitor. d. an allosteric inhibitor. © 2017 Sinauer Associates, Inc.

e. a competitive inhibitor.

Answer: a Learning Outcome: 8.5.2.a Differentiate between irreversible and reversible inhibition of enzyme catalysis.

Bloom’s Level: 2. Understanding 70. Which action would reverse the effects of a competitive inhibitor on an enzyme but not the effects of a noncompetitive or an irreversible inhibitor? a. Dilute the solution by adding more solvent. b. Add an allosteric activator. c. Heat the solution. d. Add substrate. e. Remove substrate.

Answer: d Learning Outcome: 8.5.2.b Compare and contrast the chemical natures, reversibility, and effects of competitive, noncompetitive, and covalent inhibitors.

Bloom’s Level: 5. Evaluating 71. The inhibition of enzyme activity by noncompetitive inhibitors can be reduced a. by decreasing the concentration of allosteric enzymes. b. by decreasing the concentration of substrate. c. by increasing the concentration of competitive inhibitor. d. by increasing the concentration of substrate. e. only when they become unbound.

Answer: e Learning Outcome: 8.5.2.b Compare and contrast the chemical natures, reversibility, and effects of competitive, noncompetitive, and covalent inhibitors.

Bloom’s Level: 2. Understanding 72. Competitive inhibitors of enzymes work by a. fitting into the active site. b. fitting into a site other than the active site. c. altering the shape of the enzyme. d. changing the enzyme into an inactive form. e. increasing the activation energy of the enzyme-catalyzed reaction.

Answer: a Learning Outcome: 8.5.2.c Distinguish between different types of inhibitors.

Bloom’s Level: 1. Remembering 73. How do competitive and noncompetitive enzyme inhibitors differ from each other? a. Competitive inhibitors bind to the active site, whereas noncompetitive inhibitors change the shape of the active site. b. Competitive inhibitors have a higher energy of activation than noncompetitive inhibitors have. c. Noncompetitive enzyme inhibitors contain magnesium, whereas competitive inhibitors contain iron.

© 2017 Sinauer Associates, Inc.

d. Noncompetitive enzyme inhibitors are reversible, whereas competitive inhibitors are irreversible. e. They function at different pH values.

Answer: a Learning Outcome: 8.5.2.c Distinguish between different types of inhibitors.

Bloom’s Level: 2. Understanding 74. A noncompetitive inhibitor hinders binding of a substrate to an enzyme by a. binding to the substrate. b. binding to the active site. c. lowering the activation energy. d. increasing the G of the reaction. e. changing the shape of the active site.

Answer: e Learning Outcome: 8.5.2.c Distinguish between different types of inhibitors.

Bloom’s Level: 1. Remembering 75. Which type of inhibitor can be overcome completely by the addition of more substrate? a. Irreversible b. Noncompetitive c. Competitive d. Prosthetic e. Feedback Answer: c Learning Outcome: 8.5.2.c Distinguish between different types of inhibitors.

Bloom’s Level: 2. Understanding 76. You observe an enzymatic reaction that exhibits a standard relationship between reaction rate and substrate concentration as it converts substrate B to product C. When you add compound A, the creation of product C comes to a complete stop. You add large quantities of substrate B but still are unable to create product C in the presence of compound A. Compound A is therefore most likely a a. noncompetitive inhibitor. b. competitive inhibitor. c. coenzyme. d. cofactor. e. transcription factor.

Answer: a Learning Outcome: 8.5.2.d Analyze data from a reaction to determine the mode of enzyme regulation.

Bloom’s Level: 4. Analyzing 77. The addition of the competitive inhibitor mevinolin slows the reaction HMG-CoA → mevalonate, which is catalyzed by the enzyme HMG-CoA reductase. In order to overcome the effects of mevinolin and increase the rate of the reaction, you would

© 2017 Sinauer Associates, Inc.

a. add more mevalonate. b. add more HMG-CoA. c. lower the temperature of the reaction. d. add a prosthetic group. e. lower the rate constant of the reaction.

Answer: b Learning Outcome: 8.5.2.e Explain how and why enzymes can be inactivated.

Bloom’s Level: 3. Applying 78. An allosteric inhibitor a. decreases the concentration of an inactive enzyme. b. changes the shape of an enzyme. c. increases the concentration of a product. d. changes the shape of a substrate. e. increases the concentration of an enzyme–substrate complex.

Answer: b Learning Outcome: 8.5.3.a Explain allosteric regulation.

Bloom’s Level: 1. Remembering 79. Which statement about allosteric regulators is true? a. The plot of reaction rate versus substrate concentration for allosteric regulators often displays a sigmoid curve. b. All enzymes can be allosterically regulated. c. Allosteric regulators affect substrate binding but do not affect enzyme structure. d. Binding of allosteric regulators is nonspecific. e. Allosteric regulators bind to the active site, blocking enzyme function.

Answer: a Learning Outcome: 8.5.3.a Explain allosteric regulation.

Bloom’s Level: 2. Understanding 80. You are working with an enzyme and wish to determine its saturation point. You note that the reaction rate increases slowly at first, then increases dramatically over a narrow range of substrate concentrations, and then levels off, creating a sigmoid reaction curve.

It is likely that this enzyme is regulated by a. temperature. b. pH. c. an allosteric mechanism. d. a competitive inhibitor. e. a coenzyme.

Answer: c Learning Outcome: 8.5.3.b Differentiate between an allosteric enzyme and a nonallosteric enzyme.

Bloom’s Level: 3. Applying

© 2017 Sinauer Associates, Inc.

81. In an experiment, a small amount of allosteric inhibitor is added to tube A and no inhibitor is added to tube B. The full reaction is completed in tube B twice as quickly as it is completed in tube A. The allosteric inhibitor acted by a. decreasing the number of enzyme molecules. b. increasing the amount of the active form of the enzyme. c. decreasing the amount of the inactive form of the enzyme. d. decreasing the amount of the active form of the enzyme. e. increasing the amount of substrate.

Answer: d Learning Outcome: 8.5.3.c Apply the concept of allosteric regulation to explain experimental results.

Bloom’s Level: 3. Applying 82. An enzyme is found to be inhibited by a molecule that has a very different shape and size than either of the enzyme’s two substrates. Moreover, the inhibitory molecule can bind to the enzyme in the absence of the substrates, and when it does bind, the molecule causes the enzyme to dissociate from a four-part quaternary structure to a two-part structure. What type of inhibitor is the molecule? a. Competitive b. Uncompetitive c. Feedback d. Irreversible e. Allosteric Answer: e Learning Outcome: 8.5.3.c Apply the concept of allosteric regulation to explain experimental results.

Bloom’s Level: 4. Analyzing 83. The process that involves an end product acting as an inhibitor of an earlier step in a metabolic pathway is called a. feedback activation. b. feedback inhibition. c. positive feedback. d. concerted activation. e. competitive inhibition.

Answer: b Learning Outcome: 8.5.4.a Describe feedback inhibition.

Bloom’s Level: 1. Remembering 84. End products of biosynthetic pathways often act to block the initial step in that pathway. This phenomenon is called a. posterior inhibition. b. denaturation. c. branch pathway inhibition. d. feedback inhibition. e. binary inhibition.

© 2017 Sinauer Associates, Inc.

Answer: d Learning Outcome: 8.5.4.a Describe feedback inhibition.

Bloom’s Level: 1. Remembering 85. Refer to the figure and table below.

The figure shows a metabolic pathway catalyzed by a series of enzymes labeled with numbers. The metabolite T was tested for inhibitory effects on each enzyme in the pathway, and the results are summarized in the table. The extent of inhibition is indicated by the number of plus signs. Lack of inhibition is indicated by a minus sign. Which statement gives a valid conclusion from these results? a. The production of T is not regulated by feedback inhibition. b. The inhibitory mechanism of T involves reversible phosphorylation. c. T is a competitive inhibitor of E4. d. The step from X to Y is the first commitment step in the production of T. e. The step from Q to R is the first commitment step in the production of T. Answer: e Learning Outcome: 8.5.4.b Analyze systems that involve feedback inhibition.

Bloom’s Level: 4. Analyzing 86. Denatured enzymes are a. ribozymes. b. synthesized in vitro. c. allosteric enzymes. d. nonfunctional enzymes. e. converted to coenzymes.

Answer: d Learning Outcome: 8.5.6.a Explain why enzymes are sensitive to environmental conditions.

Bloom’s Level: 2. Understanding 87. Enzymes are highly sensitive to pH and temperature because a. changes in the environment raise their activation energy. b. changes in temperature and pH readily break their peptide bonds. c. pH and temperature affect their three-dimensional structure and their side chain chemistry.

© 2017 Sinauer Associates, Inc.

d. at extreme temperatures and pH levels, coenzymes add chemical groups to the substrate. e. extremes of temperature and pH level change their ionization rate.

Answer: c Learning Outcome: 8.5.6.a Explain why enzymes are sensitive to environmental conditions.

Bloom’s Level: 2. Understanding 88. If the temperature of an enzyme-catalyzed reaction is increased from 20°C to 70°C, the rate of the reaction a. will increase indefinitely. b. will decrease immediately. c. will decrease to zero because the enzyme will become denatured. d. will decrease and then increase. e. cannot be determined without information regarding optimal temperature.

Answer: e Learning Outcome: 8.5.6.b Draw graphs of enzyme activity vs. temperature and enzyme activity vs. pH, and explain both.

Bloom’s Level: 3. Applying 89. Refer to the diagram below showing the activity of several enzymes at varying pH levels.

Based on the diagram, what is the pH at which pepsin would exhibit the most activity? a. 1.0 b. 1.5 c. 2.0 d. 2.8 e. 3.4 Answer: c Learning Outcome: 8.5.6.b Draw graphs of enzyme activity vs. temperature and enzyme © 2017 Sinauer Associates, Inc.

activity vs. pH, and explain both.

Bloom’s Level: 4. Analyzing 90. Refer to the diagram below showing the activity of several enzymes at varying pH levels.

Based on the diagram, achieving the maximum rate of both chymotrypsin and arginase activity in the same reaction would require a pH of about a. 6.8. b. 7.2. c. 7.5. d. 8.2. e. 9.0.

Answer: d Learning Outcome: 8.5.6.b Draw graphs of enzyme activity vs. temperature and enzyme activity vs. pH, and explain both.

Bloom’s Level: 4. Analyzing

Fill in the Blank 91. Heat, light, electricity, and motion are all examples of _______ energy.

Answer: kinetic Learning Outcome: 8.1.1.b Differentiate between potential energy and kinetic energy.

Bloom’s Level: 2. Understanding 92. Glycogen is a large molecule consisting of hundreds of molecules of glucose bonded together, and thus plays an energy storage role in animal cells. The chemical reaction that adds glucose units onto a glycogen molecule is an example of a(n) _______ reaction in the cell’s metabolism.

Answer: anabolic

© 2017 Sinauer Associates, Inc.

Learning Outcome: 8.1.2.a Explain how to determine whether a reaction is anabolic or catabolic.

Bloom’s Level: 3. Applying 93. The breaking down of molecules in a living system is known as _______.

Answer: catabolism Learning Outcome: 8.1.2.b Compare and contrast anabolic and catabolic reactions.

Bloom’s Level: 1. Remembering 94. Refer to the table below.

Of the four reactions listed in the table, the one that is spontaneous under all temperature conditions is reaction _______.

Answer: Y Learning Outcome: 8.1.5.c Use the equation ∆G = ∆H – T∆S to describe biological systems.

Bloom’s Level: 4. Analyzing 95. If the G of a spontaneous reaction is negative, indicating that the reaction releases free energy, the reaction is _______.

Answer: exergonic Learning Outcome: 8.1.6.a Differentiate between exergonic and endergonic reactions.

Bloom’s Level: 2. Understanding 96. A reaction with a G of ‒14.3 kcal/mol releases energy that is then used by a second reaction with a G of +10.2 kcal/mol. The net free energy change of the coupled reactions is _______.

Answer: ‒4.1 kcal/mol Learning Outcome: 8.2.2.b Analyze evidence to identify reactions as exergonic or endergonic.

Bloom’s Level: 3. Applying 97. Refer to the graph below.

© 2017 Sinauer Associates, Inc.

Ten sequential reactions take place in cells during glycolysis. The graph shows the relative free energies of each chemical intermediate relative to glucose, the starting reactant in glycolysis. Based on the information shown in the graph, reactions numbered _______ are all possibilities for reactions that might be coupled to the reaction in which ATP is hydrolyzed.

Answer: 2, 4, 5, 6, 8, and 9 Learning Outcome: 8.2.2.c Predict biochemical transformations that require coupling to ATP hydrolysis.

Bloom’s Level: 4. Analyzing 98. Enzymes function as biological _______.

Answer: catalysts Learning Outcome: 8.3.1.a Describe a catalyst.

Bloom’s Level: 1. Remembering 99. The amount of energy needed to start a reaction is known as its _______ energy.

Answer: activation Learning Outcome: 8.3.2.b Explain the concept of activation energy.

Bloom’s Level: 1. Remembering 100. Refer to the reaction below.

© 2017 Sinauer Associates, Inc.

The enzyme catalyzing this reaction would be classified along with other enzymes of the same type as a(n) _______.

Answer: isomerase Learning Outcome: 8.3.3.a Identify the category in which an enzyme belongs based on the type of reaction it catalyzes.

Bloom’s Level: 4. Analyzing 101. The greater the affinity of an enzyme for its substrate, the _______ the substrate dissociation constant (KD).

Answer: smaller Learning Outcome: 8.3.3.c Analyze the affinity of an enzyme for a substrate using KD. Bloom’s Level: 3. Applying 102. Refer to the table below.

The table shows data collected from three forms of the same enzyme—one normal form and two mutant forms. The data show that the _______ form has the greatest affinity for substrate.

Answer: mutant 2 Learning Outcome: 8.3.3.c Analyze the affinity of an enzyme for a substrate using KD. Bloom’s Level: 4. Analyzing 103. A chemical reaction is found to have an equilibrium constant of 0.95, which means that there are almost equal concentrations of products and reactants when equilibrium is achieved. If an enzyme is added that will catalyze this reaction, the reaction rate will _______ and the equilibrium point will not change.

Answer: increase Learning Outcome: 8.3.5.a Describe the relationship between an enzyme and the equilibrium point of the reaction it catalyzes.

Bloom’s Level: 3. Applying 104. Refer to the figure below.

© 2017 Sinauer Associates, Inc.

Hexokinase catalyzes the reaction between glucose and ATP to form glucose 6-phosphate and ADP. The figure shows how the active site of hexokinase orients ATP for attack by glucose during the reaction. In the absence of enzyme, this reaction would be much slower because there would be a _______ number of orientations of the two substrates compared with the number of orientations provided by the enzyme.

Answer: larger Learning Outcome: 8.4.1.a Contrast the orientations of reactant molecules with and without the presence of enzymes.

Bloom’s Level: 3. Applying 105. Two amino acids shown to be essential in the catalytic mechanism of phosphohexose isomerase are a histidine and a lysine. These two side chains work together in a reversible series of protonation and deprotonation steps to break one bond in the substrate and form a new one. This is an example of a(n) _______ mechanism of catalysis.

Answer: acid–base Learning Outcome: 8.4.1.c Give examples of mechanisms for enzyme catalysis.

Bloom’s Level: 3. Applying 106. Glyceraldehyde 3-phosphate dehydrogenase catalyzes the reaction between glyceraldehyde 3-phosphate, NAD+, and phosphate ion to form 1,3-bisphosphoglycerate and NADH. Under certain conditions, a permanently radiolabeled enzyme intermediate forms if 14C-glyceraldehyde 3-phosphate and NAD+ are used as substrates and phosphate is left out of the reaction mix. The radiolabeled molecule is linked to a cysteine side chain of the enzyme. This result suggests that this enzyme uses _______ as a catalytic mechanism.

Answer: covalent catalysis Learning Outcome: 8.4.1.c Give examples of mechanisms for enzyme catalysis.

© 2017 Sinauer Associates, Inc.

Bloom’s Level: 4. Analyzing 107. Refer to the figure below.

Enzymes use a variety of strategies for inducing substrates to assume their transition states. Of the three strategies shown in the figure, labeled A, B, and C, the two that operate using physical mechanisms are _______.

Answer: A and B Learning Outcome: 8.4.1.d Compare ways that enzymes catalyze biochemical reactions.

Bloom’s Level: 3. Applying 108. A change in enzyme shape caused by substrate binding is called _______.

Answer: induced fit

© 2017 Sinauer Associates, Inc.

Learning Outcome: 8.4.3.a Explain how induced fit relates to the enzyme’s role in catalysis.

Bloom’s Level: 1. Remembering 109. Zinc or iron ions that bind to enzymes and aid in their function are called _______.

Answer: cofactors Learning Outcome: 8.4.5.a Describe coenzymes and cofactors and how they function.

Bloom’s Level: 1. Remembering 110. Refer to the figure below.

The figure shows activities at the active site of the enzyme hexokinase as it catalyzes a reaction between ATP and glucose. Based on this figure, it can be inferred that the magnesium ion functions to _______ the ATP molecule so it can be attacked by glucose.

Answer: orient Learning Outcome: 8.4.5.c Analyze the role of a small molecule or ion in an enzymecatalyzed reaction.

Bloom’s Level: 4. Analyzing 111. Refer to the figure below.

The figure illustrates how _______ of a key enzyme in a metabolic pathway can shift the flow of molecules into a different pathway.

Answer: inhibition

© 2017 Sinauer Associates, Inc.

Learning Outcome: 8.5.1.a Explain how metabolic pathways can shift in different directions.

Bloom’s Level: 3. Applying 112. Refer to the figure and table below.

A student is constructing a table that lists poisons whose mechanisms of action involve enzyme inhibition. She will put a checkmark in any column that characterizes a compound’s mode of inhibition. Arsenate acts to inhibit many enzymes that use ATP as a coenzyme. To complete this row in the table the student should check the box(es) under _______.

Answer: reversible and competitive inhibition Learning Outcome: 8.5.2.b Compare and contrast the chemical natures, reversibility, and effects of competitive, noncompetitive, and covalent inhibitors.

Bloom’s Level: 3. Applying 113. Each day, an adult human uses approximately 2,000 kilocalories of metabolic energy. This quantity of energy can be supplied by about 250 moles of ATP as it hydrolyzes to ADP and Pi. At any one time, there is much less than this quantity of ATP present in the human body. Therefore, the production of ATP must be tightly regulated so that appropriate amounts are available, depending on need. For example, certain enzymes involved in ATP production will be _______ when a person shifts from high levels of activity to a period of sleep.

Answer: inhibited Learning Outcome: 8.5.2.e Explain how and why enzymes can be inactivated.

Bloom’s Level: 3. Applying 114. Refer to the figure below.

© 2017 Sinauer Associates, Inc.

Portions of three biosynthetic pathways regulated through feedback inhibition and activation are shown. Two of these pathways intersect at carbamoyl phosphate. The third pathway generates a product that activates production of carbamoyl phosphate through allosteric modification of the enzyme carbamoyl phosphate synthetase. In this set of pathways, arginine biosynthesis is shut down through feedback inhibition when _______ and arginine reach high levels of concentration.

Answer: UMP Learning Outcome: 8.5.4.b Analyze systems that involve feedback inhibition.

Bloom’s Level: 4. Analyzing 115. When an enzyme is heated until its three-dimensional structure is destroyed, the enzyme is said to be _______.

Answer: denatured Learning Outcome: 8.5.6.a Explain why enzymes are sensitive to environmental conditions.

Bloom’s Level: 1. Remembering

Short Answer/Essay 116. Discuss the meaning of the term “thermodynamics,” and explain how biological systems and thermodynamics are related.

© 2017 Sinauer Associates, Inc.

Answer: The term comes from the root words “therm” meaning “heat” and “dynamics” meaning “change.” It refers to the study of changes in energy in natural systems. Since biological systems are natural systems that undergo energy changes, thermodynamics is a study that can be applied to them. All of the laws of thermodynamics apply just as well to living systems as they do to nonliving systems in the universe.

Learning Outcome: 8.1.3.a Identify the meaning of the term “thermodynamics.”

Bloom’s Level: 3. Applying 117. It is estimated that approximately 90 percent of energy that passes between levels in a food web is lost at each level. Explain the first law of thermodynamics, and discuss why this apparent loss of energy does not contradict that law.

Answer: The first law of thermodynamics states that energy cannot be created or destroyed but that it may be converted from one form to another. In the transfer between levels of a food web, approximately 90 percent of the energy is converted to unusable heat energy. There is a net loss of usable energy during each conversion, but the total amount of energy (usable and unusable) remains the same.

Learning Outcome: 8.1.4.a Explain how the first law of thermodynamics relates to biological systems.

Bloom’s Level: 3. Applying 118. Refer to the figures below.

These figures can be used to illustrate the first and second laws of thermodynamics.

Assign each figure to one of these laws, then explain the difference between the figures.

Answer: Figure 1 illustrates the first law and Figure 2 illustrates the second law of thermodynamics. Figure 1 is much simpler than Figure 2 and shows only that the total amount of energy remains constant during energy transformations. Figure 2 contains elements of both laws, because it shows that the total amount of energy remains constant but the amount of that energy that is usable by a biological system decreases at each energy transformation step.

Learning Outcome: 8.1.5.a Compare and contrast the first and second laws of thermodynamics.

© 2017 Sinauer Associates, Inc.

Bloom’s Level: 3. Applying 119. Plot free energy versus the time course of an endergonic reaction, and do the same for an exergonic reaction. Include the activation energy on both plots. Label Ea and G on both graphs.

Answer:

Learning Outcome: 8.1.6.a Differentiate between exergonic and endergonic reactions.

Bloom’s Level: 3. Applying 120. Refer to the figure below.

This cellular reaction is catalyzed by the enzyme triosephosphate isomerase. At equilibrium, the ratio of dihydroxyacetone phosphate to glyceraldehyde phosphate is 24:1. What changes in conditions in the cell will cause changes in the direction of flux through this reaction? Give two changes, one that will cause a change in flux in the forward direction, and a second that will cause change in the reverse direction.

Answer: Increasing the glyceraldehyde phosphate concentration will cause a shift in flux

© 2017 Sinauer Associates, Inc.

in the forward direction of the reaction by increasing the rate of the forward reaction so that it is greater than the rate of the reverse reaction. In contrast, increasing the dihydroxyacetone phosphate concentration will cause a shift in flux in the reverse direction by increasing the rate of the reverse reaction so that it is greater than the rate of the forward reaction. In each case, once the ratio of dihydroxyacetone phosphate to glyceraldehyde phosphate reaches 24:1 again, the rates of the forward and reverse reactions will go back to being equal.

Learning Outcome: 8.1.7.b Predict the direction of a reaction for a system at equilibrium when conditions are changed.

Bloom’s Level: 3. Applying 121. What is the result of the hydrolysis of ATP? Draw the products using the reactants’ structures given in the figure below as a guide.

Answer: The hydrolysis of ATP to ADP breaks the bond between the phosphate groups and releases energy.

Learning Outcome: 8.2.1.b Explain the characteristics of ATP that account for the high free energy released during its hydrolysis to form ADP and Pi. Bloom’s Level: 2. Understanding 122. Refer to the table below.

© 2017 Sinauer Associates, Inc.

In a cell, these two reactions are coupled. Explain what that means and the benefit of having the reactions coupled. Include a diagram in your explanation.

Answer: When two reactions are coupled, the reactions occur together in time and place.

The benefit of having reactions coupled is that it allows an exergonic reaction to provide the energy needed for an endergonic reaction to proceed. In this case, the hydrolysis of ATP releases 7.3 kcal/mol of ATP hydrolyzed. Some of this energy can be used to drive the reaction between glutamic acid and ammonia to produce glutamine, which requires an input of 3.4 kcal/mol of glutamine synthesized. The overall free energy change is the sum of the free energy changes for the two reactions, or (3.4 kcal) + (−7.3 kcal) = −3.9 kcal.

Since the sum results in a negative value, the coupled reactions proceed spontaneously.

(Diagram should resemble Figure 8.7.)

Learning Outcome: 8.2.2.a Use diagrams to explain the role of ATP in the coupling of an energy-requiring biochemical transformation to an energy-releasing one.

Bloom’s Level: 3. Applying 123. Refer to the expression and table below.

© 2017 Sinauer Associates, Inc.

The enzyme tryptophanase catalyzes the reaction shown. However, research has revealed that compounds other than tryptophan can act as substrates for this enzyme. Some of these compounds are listed in the table along with their dissociation constants. What does the information in the table indicate about the affinity of the enzyme for compounds other than its natural substrate, tryptophan?

Answer: The results indicate that all but one of the compounds bind to the enzyme with lower affinity than the normal substrate tryptophan. Only nitrophenyl cysteine binds to the enzyme with greater affinity than tryptophan, since it has a smaller KD than tryptophan.

Learning Outcome: 8.3.3.c Analyze the affinity of an enzyme for a substrate using KD. Bloom’s Level: 4. Analyzing 124. -lactamase is an enzyme that catalyzes the hydrolysis of the antibiotic penicillin.

Infrared spectra taken of penicillin free in solution and penicillin bound to the active site of the enzyme have revealed that certain bonds are stretched to a greater extent when penicillin is bound to the enzyme. Explain how this observation relates to the concept of enzyme-induced strain.

Answer: It has been proposed that one way an enzyme functions to lower the activation energy of a chemical reaction is by inducing strain in the substrate. The enzyme achieves this by forcing the substrate to bind in such a way that the chemical bonds to be broken are stretched and strained as a result of the substrate binding to the active site. This physical strain makes it easier for the bonds to break, which speeds up the reaction. The finding that bonds in penicillin are stretched to a greater extent when penicillin is bound to the active site of -lactamase than when penicillin is free in solution is consistent with this idea. The bonds that are stretched are likely the bonds that are broken during the reaction, and the enzyme is facilitating this breakage by inducing strain in the bonds during the binding of penicillin to its active site.

Learning Outcome: 8.4.1.b Explain the concept of induced strain.

Bloom’s Level: 3. Applying 125. The enzyme threonine deaminase converts the amino acid threonine to 2-

© 2017 Sinauer Associates, Inc.

ketobutyrate. However, only L-threonine serves as a substrate. The enzyme does not recognize D-threonine as a substrate. Explain this.

Answer: Enzyme active sites are highly specific in terms of the molecules they recognize and bind. This recognition includes types of atoms and functional groups as well as their three-dimensional arrangements in space. In the case of D- and L-threonine, these molecules have identical functional groups, but these groups are arranged in threedimensional space in different ways. Because of this difference in spatial arrangement of groups, only one will fit into and bind to the active site of the enzyme. This site is complementary to only one arrangement.

Learning Outcome: 8.4.2.a Explain the high degree of specificity of enzymes for their substrates.

Bloom’s Level: 3. Applying 126. Refer to the figure below.

The graph was constructed from reaction-rate data collected using a constant amount of purified enzyme and increasing concentrations of substrate. Under conditions at point X, the graph indicates that the reaction rate changed dramatically when just a small increase in substrate concentration was made. Under conditions at point Y, the reaction rate changed very little with the same increase in substrate concentration. Explain why.

Answer: At point X, there is a large excess of enzyme with respect to substrate. When more substrate is added, it can bind to empty active sites on the enzyme molecules, which are abundant in the solution. Once bound, the enzyme catalyzes the reaction, and the reaction rate increases, as shown by the rise in the curve. At point Y, there is a large excess of substrate with respect to enzyme. Thus, there are no available enzyme active sites for the substrate molecules to bind. The active sites are saturated with substrate.

Therefore, no increase in reaction rate is observed, because no additional sites for reaction are available to generate product.

Learning Outcome: 8.4.6.b Predict changes in the rate of an enzyme-catalyzed reaction as factors in the reaction are varied.

Bloom’s Level: 3. Applying 127. Refer to the figure below.

© 2017 Sinauer Associates, Inc.

The figure shows the metabolic conversions between metabolite X and metabolite Y in a cell. Each direction of this conversion is catalyzed by a different enzyme. Explain how this arrangement enables shifts in the flux of molecules in different metabolic directions, depending on the needs of the cell.

Answer: Enzymes 1 and 2 can be regulated by different molecules produced farther downstream in pathways 1 and 2 to allow for specific flux in one direction or the other.

For example, when products of pathway 1 build up in concentration, they can act as feedback inhibitors of E2 and activators of E1 to signal a shift away from production of X and toward production of Y and flux through pathway 2. Later, when products of pathway 2 build up in concentration, they can act as feedback inhibitors of E1 and activators of E2 to signal a shift back in the other direction.

Learning Outcome: 8.5.1.a Explain how metabolic pathways can shift in different directions.

Bloom’s Level: 3. Applying 128. Construct a diagram to illustrate how an enzyme can be regulated in a reversible fashion via phosphorylation.

Answer:

Learning Outcome: 8.5.5.a Draw a diagram to illustrate how reversible phosphorylation regulates an enzyme.

Bloom’s Level: 6. Creating 129. Amylase is a digestive enzyme secreted in human saliva that breaks down starch.

Amylase functions well in the mouth but ceases to function in the stomach. Explain why.

Answer: The pH optimum of amylase is approximately 7.0. At that pH, the protein has the three-dimensional shape that allows starch to bind to its active site and catalyze its hydrolysis. When it is at stomach pH (approximately 2.0), the protein is denatured, and its three-dimensional shape and active site are lost; therefore, it can no longer catalyze the reaction.

Learning Outcome: 8.5.6.a Explain why enzymes are sensitive to environmental conditions.

Bloom’s Level: 3. Applying

© 2017 Sinauer Associates, Inc.