Tymoczko Biochemistry A Short Course 3rd Edition – Test Bank

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INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS

 

Tymoczko’s Biochemistry A Short Course THIRD EDITION (Six Month Access – Test Bank

 

Sample  Questions

 

Chapter 6   Basic Concepts of Enzyme Action

 

 

Matching Questions

Use the following to answer questions 1-10:

 

Choose the correct answer from the list below. Not all of the answers will be used.

  1. a) apoenzymes
  2. b) hydrolyases
  3. c) active site
  4. d) transition state
  5. e) spontaneous
  6. f) induced fit
  7. g) energy
  8. h) prosthetic group
  9. i) lock and key
  10. j) substrate(s)
  11. k) oxidoreductases
  12. l) equilibria

 

1. ____________ The site on the enzyme where the reaction occurs.

 

Ans: c
Section:  6.4

 

2. ____________ The substance that the enzyme binds and converts to product.

 

Ans: j
Section:  6.1

 

3. Enzymes that do not have the required cofactor bound are called ____________.

 

Ans: a
Section:  6.2

 

4. A tightly bound cofactor might be called a(n) ____________.

 

Ans: h
Section:  6.2

 

5. Enzymes will decrease the energy of activation but do not change the ____________ of a chemical reaction.

 

Ans: l
Section:  6.3

 

 

6. A reaction that is exergonic will be ____________.

 

Ans: e
Section:  6.3

 

7. An endergonic reaction requires an input of ____________ to proceed.

 

Ans: g
Section:  6.3

 

8. Enzymes that transfer electrons are called ___________.

 

Ans: k
Section:  6.1

 

9. Enzymes that cleave molecules by addition of water are called ____________.

 

Ans: b
Section:  6.4

 

10. Which model is more appropriate to explain an enzyme binding to its substrate?

 

Ans: f
Section:  6.4

 

Fill-in-the-Blank Questions

 

11. Enzymes accelerate the rate of a chemical reaction by       the free energy of activation of the reaction.
Ans:  lowering     Section:  6.3

 

12. The difference between the standard-state free energy, ΔGº, and the biochemical standard-state free energy is that ΔGº refers to the standard free-energy change at      .
Ans:  pH 7     Section:  6.3

 

13.  An enzyme that loosely binds substrate will have a      level of specificity.
Ans:  low      Section:   6.1

 

14. Organic cofactors are referred to as      .
Ans:  coenzymes     Section:  6.2

 

15. A reaction can occur spontaneously only if ΔG is      .
Ans:  negative     Section:  6.3

 

16. When ΔG for a system is zero, the system is at      .
Ans:  equilibrium     Section:  6.3

 

17.  An enzyme that has been stripped of small molecules needed for activity is called      .
Ans:   an apoenzyme    Section:  6.2

 

18.  The total change of free energy in a reaction depends on      and      .
Ans:  the substrate DG; the DG  of the product     Section:  6.3

 

19.  The difference in values for DG and DGo′ is in the      .
Ans:    concentration of reactants and products    Section:  6.3

 

20. Competitive inhibitors that mimic the substrate while in the transition state are called

      inhibitors.

Ans:   transition-state analog     Section:  6.4

 

 

Multiple-Choice Questions

 

21. What is the common strategy by which catalysis occurs?
A) increasing the probability of product formation
B) shifting the reaction equilibrium
C) stabilization the transition state
D) All of the above.
E) None of the above.
Ans:  C     Section:  6.4

 

22.  An enzyme will specifically bind its substrate because of____________
A) a tight lock and key binding mechanism.
B) a high number of hydrophobic amino acids in the center of the protein.
C) a large number of weak interactions at the active site.
D) additional nonprotein cofactors.
E) None of the above.
Ans:  C     Section:  6.4

 

23. Examples of cofactors include:
A) Zn+2, Mg+2, and Ni+2.
B) biotin and thiamine pyrophosphate.
C) pyridoxal phosphate and coenzyme A.
D) B and C.
E) All of the above.
Ans:  E     Section:  6.2

 

24. A cofactor is best defined as ______________.
A) another protein
B) a covalently bound inorganic molecule
C) a small molecule that holds the substrate in the active site
D) a molecule responsible for most of the catalytic activity of the enzyme
E) None of the above.
Ans:  E     Section:  6.2

 

25. Which of the following is true?
A) Enzymes force reactions to proceed in only one direction.
B) Enzymes alter the equilibrium of the reaction.
C) Enzymes alter the standard free energy of the reaction.
D) All of the above.
E) None of the above.
Ans:  E     Section:  6.3

 

26. The Gibbs free energy of activation is:
A) the difference between the substrate and the transition state.
B) the difference between the substrate and the product.
C) the difference between the product and the transition state.
D) All of the above.
E) None of the above.
Ans:  A     Section:  6.4

 

27. At equilibrium, the Gibb’s free energy is ___________.
A) a positive value
B) neutral
C) a negative value
D) zero
E) one
Ans:  D     Section:  6.3

 

28. The rate of a reaction, or how fast a reaction will proceed, is best determined by __________.
A) DR
B) DG
C) DGº′
D) DH
E) None of the above.
Ans:  B     Section:  6.3

 

29. The relationship between DGo′ and DG is best described as ______________.
A) determined by the temperature
B) described by changes in Keq
C) differ from standard state to physiological or actual concentrations of reactants and products
D) dependent on the reaction mechanism of the reaction
E) differ only in terms of the types of reactions used for each value
Ans:  C     Section:  6.3

 

30. For the two reactions a)  A→B  DGo′ = 2 kJmol-1  and   b) X→Y  DGo′ = –3.5  kJmol-1, which of the following statements is correct?
A) Reaction a is not spontaneous at cellular concentrations.
B) Reaction b will react very quickly.
C) Reaction a is a more thermodynamically favorable reaction than b.
D) Neither reaction is reversible.
E) None of the above.
Ans:  E     Section:  6.3

 

 

31. A graph of product versus time (as in Fig. 6.2 in your textbook) for an enzyme is determined to be hyperbolic. Why does the amount of product level off as time increases?
A) The reaction has reached equilibrium, that is, the forward and reverse reactions are occurring at a fixed rate.
B) There is a product inhibition of the enzyme.
C) The reaction runs out of reaction materials.
D) The enzyme has finished accelerating the reaction.
E) None of the above.
Ans:  A     Section:  6.3

 

 

32. The free energy of activation is _______________.
A) the amount of chemical energy available in the transition state
B) the difference in free energy between the substrate and product
C) the free energy gained by adding a catalyst
D) the difference in free energy between the transition state and the substrate
E) All of the above.
Ans:  D     Section:  6.4

 

 

33. The molecular structure that is short-lived and neither substrate nor product is known as_______.
A) substrate analog
B) transition state
C) free energy stabilization state
D) catalysis state
E) equilibrium intermediate
Ans:  B     Section:  6.4

 

34. Riboflavin is a water-soluble organic substance that is not synthesized by humans.  Metabolically, it is chemically converted into a substance called flavin adenine dinucleotide, which is required by succinate dehydrogenase.  Which of the following statements is most correct?
A) Riboflavin is a coenzyme.
B) Flavin adenine dinucleotide is a vitamin.
C) Succiniate dehydrogenase is a coenzyme.
D) Flavin adenine dinucleotide is a coenzyme.
Ans:  D     Section:  6.2

 

35. The active site of an enzyme_____________.
A) is a series of amino acids that bind the enzyme
B) is a linear sequence of amino acids that react with each other
C) binds covalently to the substrate
D) allows water to enter into the solvate the substrate
E) None of the above.
Ans:  E     Section:  6.4

 

36. The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reaction was allowed to come to equilibrium. If the Keq′ is 0.50 and the equilibrium [glucose-6-phosphate] is 1.43 M, what is the equilibrium [fructose-6-phosphate]?
A) 1.00 M
B) 1.33 M
C) 0.667 M
D) 0.250 M
E) 0.150 M
Ans:  C      Section:  6.3

 

 

37. The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reaction was allowed to come to equilibrium. If the Keq′ is 0.50, what is the DG°′ in kJ/mol?
A) +0.99
B) +1.71
C) 0, as defined by equilibrium conditions
D) –0.99
E) –2.27
Ans:  B     Section:  6.3

 

 

38. The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Under cellular conditions (37oC), the glucose-6-phosphate is 6.6 μM and the fructose-6-phosphate is 1.3 μM. If the Keq′ is 0.50, what is the ΔG in kJ/mol? (Hint: Use the DG°′ from the previous question.)
A) +4.19
B) –1.81
C) –4.03
D) –2.50
E) –1.75
Ans:  D     Section:  6.4

 

 

39. That many transition-state analogs bind more tightly than the native substrate reinforces the concept that:
A) transition-state analogs are planar structures.
B) transition-state analogs are highly charged at physiological pH.
C) binding to the transition state is through a lock-and-key-mechanism.
D) transition-state analogs are hydrophobic.
E) binding to the transition state is through an induced-fit mechanism.
Ans:  E     Section:  6.4

 

 

Short-Answer Questions

40. What is the relation between an enzyme-catalyzed reaction and the transition state of a reaction?
Ans: Enzymes only catalyze reactions that are thermodynamically favorable.  To facilitate the catalysis, enzymes reduce the transition state (activation) energy of the reaction.  Enzymes do this by binding the substrate with several weak interactions and forming a temporary transition state intermediate.
Sections:  6.3 and 6.4

 

41. What is the difference between prosthetic groups and coenzymes?
Ans: Coenzymes are small molecules that are not tightly bound to the enzyme, while prosthetic groups are either covalently bound to the enzyme or nearly irreversibly associated with the protein.
Section:  6.2

 

42. How do enzymes facilitate the formation of the transition state?
Ans: When enzymes bind substrate, free energy is released by the formation of a large number of weak interactions. Only the correct substrate can participate in the most or all possible interactions with the enzyme. The full complement of interactions occurs when the transition state is achieved. This causes maximal release of free energy.
Section:  6.4

 

43. How is the substrate bound to the active site?
Ans: The active site is a small part of the total enzyme structure. It is usually a three-dimensional cleft or crevice, which is formed by amino acid residues from different regions of the polypeptide chain. The substrate is bound by multiple noncovalent attractions such as electrostatic interactions, hydrogen bonds, van der Waals forces, and hydrophobic interactions. The specificity is dependent on the precise arrangement of the various functional groups in the binding site.
Section:  6.4

 

44. You believe a substrate fits into a cleft like a key into a lock, but your roommate does not. Who is right?
Ans: You are both partially correct. Like a lock and key, the substrate fits precisely into the enzyme. However, the site is not a rigid cleft, but is flexible. Thus, it is possible for the substrate to actually modify the shape of the site a bit, a hypothesis known as induced fit. See textbook Figs. 6.5 and 6.6 for further detail.
Section:  6.4

 

45. In an enzymatic reaction in a test tube, the reaction will eventually reach equilibrium. Why does this not happen in living organisms?
Ans: In a cell, the product may be utilized for a subsequent reaction, thus the reaction may not reach equilibrium.
Section:  6.3

 

46. How is free energy useful for understanding enzyme function?
Ans: Free energy (DG) is the key thermodynamic parameter in determining if an enzyme catalyzed-reaction will occur.  Understanding if a reaction is thermodynamically favorable is the first step in our knowledge the basic function of an enzyme.
Section:  6.3

 

47. While some enzymes have very specific substrates, others are more promiscuous.  What would you suspect is the reason for this?
Ans: Specificity of binding is separate from catalysis.  The specificity of the enzyme for its substrate is due to many weak interactions between the substrate and the amino acids of the protein.  Thus, for the less specific binding protein, there must be less required interactions for binding.
Section:  6.4

 

48. Multiple dilution and dialysis of a purified protein results in a loss of enzymatic activity.  What might be the cause for this?  Assume the structure of the protein is retained.
Ans: The dilution and dialysis must have separated small molecules from the enzyme.  These are likely cofactors (cosubstrates) required for the activity.  One could test this hypothesis by adding back potential cofactors and observing a reconstituted activity.
Section:  6.2

 

 

49. If Keq = 1, what is the DG°′? If Keq >1, what is the DG°′? If Keq <1, what is the DG°′?
Ans: DG°′ = 0, negative value, positive value
Section:  6.3

 

50. The free energy change (ΔG′) for the oxidation of the sugar molecules in a sheet of paper into CO2 and H2O is large and negative (the = DG°′ – 2833 kJ/mol).  Explain why paper is stable at room temperature in the presence of oxygen (O2).
Ans: Activation energy!  While the reaction is highly thermodynamically favorable, it is not going to happen without a catalyst or sufficient added energy due to the energy barrier.
Section:  6.4

 

51. The DG°′ for the hydrolysis of ATP to ADP + Pi is approximately –31kJ/mole. Calculate the equilibrium constant for this reaction (R = 8.314J/°mole) at the cellular temperature of 37°C.  If the cellular concentrations of ATP, ADP, and Pi are 8, 1, and 8mM, respectively, is the above reaction at equilibrium in the cell?

 

Ans: Keq­ ­= 1.7 ´ 105. Note: the cell is not at equilibrium.
Section:  6.3

 

52. How does a rigid, lock and key model for substrate binding not fit with the formation of the transition state?
Ans: The amino acids in the active site must combine with any cofactors and the substrates in an orientation that promotes the active site.  This intermediate will not be like the substrate in shape and coordination.  Thus, the binding and reaction must allow for a flexible and dynamic binding and active site.
Section:  6.4

 

53.  A mutation of a proteolytic enzyme described in Section 6.1 results in a stable covalent bond between one of the catalytic amino acids of the protease with its protein substrate.  What would be the most likely outcome of enzyme function?
Ans: The enzyme, after one round of reaction, would be catalytically dead.  The stable transition state would render the active site amino acid unavailable for further reactions and would not drive forward the rest of the reaction.  This is much like the transition state analog discussed in Section 6.4.
Section:  6.4

 

54. What are transition state analogs?
Ans: These potent inhibitors mimic the structure of the transition state involved in the catalytic process. They bind very tightly to the catalytic site and are useful in determining the structure and catalytic mechanism of the enzyme.
Section:  6.4

 

 

Chapter 7 Kinetics and Regulation

 

 

Matching Questions

Use the following to answer questions 1-10:

 

Choose the correct answer from the list below. Not all of the answers will be used.

  1. a) first-order reaction
  2. b) second-order reaction
  3. c) metabolism
  4. d) ensemble
  5. e) biomolecular
  6. f) turnover number
  7. g) Michaelis
  8. h) equilibrium
  9. i) sequential
  10. j) kinetics
  11. k) initial reaction velocity
  12. l) allosteric
  13. m) ping-pong

 

1. ____________is a complex array of enzyme catalyzed reactions organized in multiple pathways.

 

Ans: c
Section:  Introduction

 

2. _______________ is the study of rates of chemical reactions.

 

Ans: j
Section:  7.1

 

3. A reaction that is directly proportional to the concentration of reactant is a ____________.

 

Ans: a
Section:  7.1

 

4. A reaction with two substrates is considered a ____________ reaction.
Ans: e
Section7.1

 

5. At ____________ there will be no net change in the concentration of substrate or product.

 

Ans: h
Section:  7.2

 

6. The value Vo is called the ____________.
Ans: k
Section:  7.2

 

7. The kcat is often referred to as the ____________.

 

Ans: f
Section:  7.2

 

8. The property that describes the enzyme-substrate interaction is measured by what constant?

 

Ans: g
Section:  7.2

 

9. ____________ Enzymes that do not obey Michaelis–Menten kinetics.

 

Ans: l
Section:  7.3

 

10. ____________ Experiments that determine the kinetics of a population of enzyme molecules.

 

Ans: d
Section:  7.4

 

 

Fill-in-the-Blank Questions

 

11. One way to measure the rate of an enzymatic reaction is to measure the loss of       over time.
Ans: substrate    Section:  7.1

 

12. Reactions that have more than two reactants or substrates are considered       reactions.
Ans: second-order     Section:  7.1

 

13. The      rule states that all subunits in an allosteric enzyme must be in either the R or the R state; no hybrids.
Ans: symmetry      Section:  7.3

 

14. The Michaelis–Menten model assumes that       is the rate constant ignored because P has not accumulated.
Ans: k2     Section:  Appendix

 

15.       is directly dependent on enzyme concentration.
Ans: Vmax    Section:  7.2

 

16. An enzyme will be most sensitive to changes in cellular substrate concentration when the concentration is     .
Ans: near the KM   Section:  7.2

 

17. The type of inhibition where the  product of one enzyme inhibits another enzyme that acts earlier in a metabolic pathway is considered a(an)       inhibitor.
Ans:  feedback     Section:  7.3

 

18. Allosteric enzymes can be identified because the plot of initial velocity, V0, versus substrate concentration, S, is not hyperbolic but      -shaped.
Ans:  sigmoidal     Section:  7.3

 

19. Negative allosteric      stabilize the T-state of the enzyme.
Ans:  effectors     Section:  7.3

 

20. The straight-line kinetic plot of 1/ V0 versus 1/S is called a      .
Ans:  Lineweaver–Burk plot, or double-reciprocal plot     Section:  7.2

 

 

Multiple-Choice Questions

 

21. A critical feature of the Michaelis–Menten model of enzyme catalysis is
A) increasing the probability of product formation.
B) shifting the reaction equilibrium.
C) formation of an ES complex.
D) All of the above.
E) None of the above.
Ans:  C     Section:  7.2

 

22. What value of [S], as a fraction of KM is required to obtain 20% Vmax? [S] =
A) 0.2 KM
B) 0.25 KM
C) 0.5 KM
D) 0.75 KM
E) 0.8 KM
Ans:  B     Section:  7.2

 

23. Allosteric proteins:
A) contain distinct regulatory sites and have multiple functional sites.
B) display cooperativity.
C) always consist of several identical subunits.
D) A and B.
E) A, B, and C.
Ans:  D     Section:  7.3

 

 

24.   Allosteric effectors alter the equilibria between:
A) the ES state.
B) the R and T forms of a protein.
C) the forward and reverse reaction rate.
D)  the formation of product and it’s reverse reaction.
E)  All of the above.
Ans:   B   Section:  7.3

 

25. The formula V0 = Vmax           [S]   , indicates the relationship between

[S] + KM

 

A) the enzyme activity and the equilibrium constant.
B) the rate of a catalyzed reaction and the equilibrium constant.
C) enzyme activity as a function of substrate concentration.
D) All of the above.
E) None of the above.
Ans:  C     Section:  7.2

 

26. The model describing allosteric regulation that requires all subunits to be in the same state is called the ________.
A) concerted model
B) syncopated model
C) cooperative model
D) equilibrium model
E) None of the above.
Ans:    A   Section:  7.3

 

27.   Loss of allosteric regulation in the production of purine nucleotides results in ___________.
A) excess nucleotides for DNA
B) loss of RNA due to ribose phosphate synthetase
C) decreased urate degradation
D) loss in urate concentration
E) None of the above.
Ans:   E    Section:  7.3

 

28. The KM is:
A) equal to the product concentration at initial reaction conditions.
B) equal to the substrate concentration when the reaction rate is half its maximal value.
C) proportional to the standard free energy.
D) All of the above.
E) None of the above.
Ans:  B     Section:  7.2

 

29. Given are five KM values for the binding of substrates to a particular enzyme. Which has the strongest affinity when k1 is greater than k2?
A) 150 mM     B) 0.15 mM     C) 150 mM     D) 1.5 nM     E) 15,000 pM
Ans:  D     Section:  7.2

 

30. When substrate concentration is much greater than KM, the rate of catalysis is almost equal to
A) Kd.     B) kcat.     C) Vmax.     D) All of the above.     E) None of the above.
Ans:  C     Section:  7.2

 

31. Which of the following is true under the following conditions: The enzyme concentration is 5 nM, the substrate concentration is 5 mM, and the KM is 5 mM.
A) The enzyme is saturated with substrate.
B) Most of the enzyme does not have substrate bound.
C) There is more enzyme than substrate.
D) All of the above.
E) None of the above.
Ans:  A     Section:  7.2

 

32. Homotrophic effects of allosteric enzymes:
A) are due to the effects of substrates. D) shift the kinetics curve to the right.
B) are due to the effects of allosteric activators. E) None of the above.
C) shift the kinetics curve to the left.
Ans:  A     Section:  7.3

 

33. Multiple substrate enzyme reactions are divided into two classes:
A) sequential reactions and double displacement reactions.
B) double displacement reactions and concerted reactions.
C) sequential reactions and concerted reactions.
D) A and C.
E) None of the above.
Ans:  A     Section:  7.2

 

34.   When [S] << KM, the enzymatic velocity depends on__________.
A)  the values of kcat/KM, [S], and [E]t
B)  the Vmax of the reaction
C)  the affinity of the substrate for the catalytic site
D)  kcat
E)  the formation of the ES complex
Ans:   A    Section:  7.2

 

35.  Allosteric effectors:
A) can cause large changes in enzymatic activity.
B) can lead to a decrease in the availability of a protein.
C) do not alter the sensitivity of a metabolic pathway.
D) decrease the sensitivity of the enzyme at nearly all concentrations of substrate.
E) alter enzyme activity by binding to the active site of an enzyme.
Ans:  A      Section:  7.3

 

36.  For decades, enzymes have been studied using ensemble methods, but technology now allows them to be studied in singulo. Which of the statements below states one of the significant outcomes of this new technology?
A) New methods better demonstrate cooperativity of allosteric enzymes.
B) New methods allow for better determination of kcat.
C) New methods reveal a distribution of enzyme characteristics.
D) New methods validate the steady-state assumption of Michaelis–Menten kinetics.
E) New methods provide understanding of average enzyme kinetic data.
Ans:  C      Section:  7.4

 

37.  When reaction conditions are such that the amount of substrate is far greater than the amount of enzyme present, then the following conditions are also met.
A) The [substrate] is much less than KM.
B) The V0 is half Vmax.
C) The enzyme is displaying second-order kinetics.
D) The enzyme is displaying first-order kinetics.
E) The enzyme is displaying zero-order kinetics.
Ans:  E     Sections:  7.1 and 7.2

 

38.  During the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity versus substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results.
A) This is an enzyme that displays Michaelis–Menten kinetics, and you purify away a homotrophic inhibitor.
B) This is an enzyme that displays Michaelis–Menten kinetics, but you must use a Lineweaver–Burk plot to determine KM and Vmax correctly.
C) This is an allosteric enzyme, but you must use a Lineweaver–Burk plot to determine KM and Vmax correctly.
D) This is an allosteric enzyme, and during purification you purify away a heterotrophic activator.
E) This is an allosteric enzyme displaying a double-displacement mechanism, and during purification you purify away one of the substrates.
Ans: D     Sections:  7.2 and 7.3

 

39.  After purifying the enzyme in the previous question, you determine the Mr to be 75,000. By assaying 5 μg of the enzyme under saturating [S] concentrations, you determine the Vmax to be 1.68 μmol/sec. Calculate the turnover number for this enzyme.
A)  2.25 ´ 106 sec-1
B) 1.50 ´ 105 sec-1
C) 2.50 ´ 104 sec-1
D) 1.79 ´ 105 sec-1
E) You need to also know the KM for this enzyme to calculate turnover number.
Ans: D     Section:  7.2

 

 

Short-Answer Questions

 

40. In many enzyme assays, the natural substrate and product are not used. Why?
Ans: Many products are difficult to measure accurately. Some are simply difficult to measure, while others are difficult to discern against the background of other molecules present in the reaction. Instead, substrates are chosen that the enzyme can still process but that result in products that can be easily measured. For example, substrates are chosen that result in products that are colored and can be detected spectrophotometrically.
Section:  7.1

 

41. A protease hydrolyzes the peptide backbone.  What is the substrate(s) and product for this reaction?  Assuming that the concentration of water is so high (~55M) that it does not appreciably change, to what kind of reaction order would one assign this reaction?
Ans: The reaction would be Protein + H2O ® peptide-1 + peptide-2.  As water doesn’t significantly change it’s concentration in an aqueous reaction, the concentration change is zero and can be ignored, thus the rate of the reaction is directly proportionate to the concentration of the protein and is a first-order reaction.
Section:  7.1

 

 

42. The rate of a reaction is dependent on [ES].  Using an enzyme catalyzed reaction scheme, (6), describe the kinetic model for [ES].
Ans:  The concentration of an ES complex is described as the enzyme binding to substrate and can be measured as one kinetic rate constant forming the ES complex.  Loss of the ES can be described as the separation of the two components without reacting (k-1) and the resulting reaction where ES ® EP ® E + P (k2­­).
Section:  7.2

 

43. Figure 7.8 is a simplified version of a common set of converging metabolic pathways.  Describe the type of regulation necessary if each of the reactions was reversed and a product, A or G, were preferred.
Ans: A feed-forward inhibition where a product, G or H, inhibits e1, e2, e3, or e5. Another possibility is that one or more of the products A through F inhibits e10 or e11.
Section:  7.3

 

44. Draw a Cleland notation for a sequential reaction and for a double-displacement reaction.
Ans: See Figures 7.6 A and 7.6B.
Section:  7.2

 

45. What is the Michaelis–Menten equation? Define all parameters.
Ans: V0 = Vmax(S/(S + KM))

 

Initial velocity

V0

 

Maximum velocity

Vmax

 

Substrate concentration

S

 

Michaelis constant

KM

 

Section:  7.2

 

46. What does Vmax indicate?
Ans: The maximum velocity or rate of reaction as catalyzed by a specific amount of enzyme when it is saturated with substrate.
Section:  7.2

 

47. What is the upper limit of kcat / KM?
Ans: The diffusion-controlled interaction of the substrate and enzyme determines the upper limit of the rate. The upper limit is 108 – 109 s-1M-1.
Section:  7.2

 

48. How do the intermediate steps in multi-substrate enzyme mechanisms differ?
Ans: In a sequential displacement reaction, both substrates bind and a ternary complex of all three is formed. In a double displacement (ping-pong), one or more products are released prior to binding of all substrates. Thus, a substituted enzyme intermediate is formed.
Section:  7.2

 

49. Describe the difference between the concerted and the sequential model of allosteric regulation.
Ans: The concerted model describes where a multi-subunit enzyme can only assume an R or T conformation, whereas the sequential model assumes that the subunits can assume its conformation different from the neighboring subunits.  In the former, substrate influences the equilibria between each subunit’s R-T conformation and in the latter, substrate can have an intermediate impact on affinity and conformation.
Section:  7.3

 

50. Would you expect the order of substrate binding to be critical for enzyme catalysis?
Ans: Yes, in some cases. For example, in ping-pong reactions, the proper substrate would have to bind to form the right substituted enzyme intermediate form. In sequential displacement, both conditions are observed. Substrates may need to bind in a particular order (lactate dehydrogenase) or the enzyme may bind substrates and release products in random order (creatine kinase).
Section:  7.2

 

51. What is the turnover number for an enzyme and what does this value tell us about the enzyme?
Ans: The rate of reaction and dissociation of the ES complex to E + P is k2.  That is the rate at which an enzyme saturated with substrate converts substrate to product.  This is basically the measure of the reaction without an impact on substrate binding and is dependent on the concentration of enzyme and describes the relative speed of a reaction.
Section:  7.2

 

52. When designing a drug to inhibit the formation of a product, which requires several enzymes in a metabolic pathway, what should be the first piece of information a biochemist needs in order to develop the drug?
Ans: Find the committed step.  In a metabolic pathway there will be a rate-limiting, committed enzyme step that is often the target of physiological regulation.  This protein would be the best target for a new drug.
Section:  7.2

 

53. How does the sequential model differ from the concerted model for allosteric enzymes?
Ans: The concerted model does not allow for anything other than an “all-or-none” complete tense- or relaxed-form protein. In contrast, the sequential model allows for a mixed type of protein, containing some tense and some relaxed subunits. The form is in response to the ligand binding by a particular subunit.
Section:  7.3

 

54. Draw a sketch of a Michaelis–Menten plot and a Lineweaver–Burk plot. Identify how you would determine KM and Vmax from each of these plots. Explain why the Michaelis–Menten is used more widely than the Lineweaver–Burk plot even though, in general, straight-line plots are easier to interpret.
Ans: Sketches should look like Fig 7.3 and 7.5 in the textbook.  KM and Vmax should be identified on the plots. The reason why Lineweaver–Burk plots are rarely used in enzyme studies is because the data points at high and low concentrations are weighted differently, making them sensitive to errors. In addition, computer software has advanced to the point where hyperbolic plots like Michaelis–Menten plots are much more readily analyzed by computers than they were originally.
Section:  7.2

Chapter 9   Hemoglobin: An Allosteric Protein

 

 

Matching Questions

Use the following to answer questions 1-10:

 

Choose the correct answer from the list below. Not all of the answers will be used.

  1. a) cooperative
  2. b) oxygen
  3. c) fMRI
  4. d) carbamate
  5. e) histidine
  6. f) hyperbolic
  7. g) myoglobin
  8. h) bicarbonate ion
  9. i) sickle-cell anemia
  10. j) protoporphyrin
  11. k) fetal
  12. l) carbonic acid

 

1. ____________ The shape of the myoglobin binding curve that shows that it is not regulated allosterically.

 

Ans:  f
Section:  9.1

 

2. ____________ This is the organic portion of the heme group in hemoglobin.

 

Ans:  j
Section:  9.2

 

3. ____________ This method of studying hemoglobin monitors changes in magnetic fields during the binding of oxygen.

 

Ans:  c
Section:  9.2

 

4. ____________ This is the chemical form in which most of the carbon dioxide is transported in the blood.

 

Ans:  h
Section:  9.5

 

5. ____________ This substance is produced when carbon dioxide reacts with water.

 

Ans:  l
Section:  9.5

 

6. ____________ This type of hemoglobin is composed of two α chains and two γ chains.

 

Ans:  k
Section:  9.4

 

7. ____________ This is the molecule whose function is to facilitate diffusion of oxygen in muscle cells.

 

Ans:  g
Section:  Introduction, 9.1

 

8. ____________ The iron atom in heme is bound to the fifth coordination site of this molecule.

 

Ans:  e
Section:  9.3

 

9. ____________ This type of binding is indicated by a sigmoidal-shaped binding curve.

 

Ans:  a
Section:  9.1

 

10. ____________ This condition is a result of a single-point mutation in the β chain of hemoglobin.

 

Ans:  i
Section:  9.6

 

 

Fill-in-the Blank Questions

 

11. Under normal conditions, the heme iron in myoglobin and hemoglobin is in the       oxidation state.
Ans:  ferrous, or Fe+2     Section:  9.2

 

12. The ability of myoglobin to bind oxygen depends on the presence of a bound prosthetic group called      .
Ans:  heme     Section:  9.2

 

13. The form of hemoglobin found in the R state is called      .
Ans:  oxyhemoglobin    Section:  9.3

 

14. The binding of 2,3-bisphosphogycerate to hemoglobin      (increases; decreases) its affinity of oxygen binding.
Ans:  decreases     Section:  9.4

 

15. The effect of pH on oxygen binding of hemoglobin is referred to as the      .
Ans:  Bohr effect     Section:  9.5

 

16. Carbon dioxide reacts with the amino terminal groups of hemoglobin to form carbamate groups, which carry a      charge.
Ans:  negative     Section:  9.5

 

17. The T state of hemoglobin is stabilized by a salt bridge between β1 Asp 94 and the C-terminal

     of the β1 chain.

Ans:  histidine     Section:  9.5

 

18. In normal adult hemoglobin, HbA, the β6 position is a glutamate residue, whereas in sickle-cell hemoglobin, HbS, it is a      residue.
Ans:  valine     Section:  9.6

 

19. As the partial pressure of carbon dioxide increases, the affinity of oxygen binding to hemoglobin      .
Ans:  decreases     Section:  9.5

 

20. 2,3-Bisphosphoglycerate binds only to the      form of hemoglobin.
Ans:  T-, or deoxy     Section:  9.4

 

 

Multiple-Choice Questions

 

21. What factor(s) influence(s) the binding of oxygen to myoglobin?
A) the concentration of bicarbonate ion, HCO3
B) the partial pressure of oxygen, pO2
C) the concentration of hemoglobin present
D) the concentration of 2,3-BPG
E) B and D.
Ans:  B     Section:  9.2

 

22. Which of the following is correct concerning the differences between hemoglobin and myoglobin?
A) Both hemoglobin and myoglobin are tetrameric proteins.
B) Hemoglobin exhibits a hyperbolic O2 saturation curve while myoglobin exhibits a sigmoid-shaped curve.
C) Hemoglobin exhibits cooperative binding of O2 while myoglobin does not.
D) Hemoglobin exhibits a higher degree of O2 saturation at all physiologically relevant partial pressures of O2 than does myoglobin.
E) All of the above.
Ans:  C     Section:  9.1

 

23. Which of the following is NOT correct concerning myoglobin?
A) The globin chain contains an extensive α-helix structure.
B) The heme group is bound to the globin chain by two disulfide bonds to cysteine residues.
C) The iron of the heme group is in the Fe2+ oxidation state.
D) The diameter of the iron ion decreases upon binding to oxygen.
E) The function of myoglobin is oxygen storage in muscle.
Ans:  B     Section:  9.2

 

24. The structure of normal adult hemoglobin can be described as
A) a tetramer composed of four myoglobin molecules.
B) a tetramer composed of two αβ dimers.
C) a tetramer composed of two α2 and two β2 dimers.
D) a tetramer composed of two α2 and two γ2 dimers.
E) None of these accurately describe hemoglobin.
Ans:  B     Section:  9.3

 

25. Which of the following is correct concerning fetal hemoglobin?
A) Fetal hemoglobin is composed of two α and two γ subunits.
B) Fetal hemoglobin binds 2,3-BPG more tightly than normal adult hemoglobin.
C) Fetal hemoglobin binds oxygen less than HbA at all pO2.
D) Fetal hemoglobin does not exist in the T-form.
E) None of the above.
Ans:  A     Section:  9.4

 

26. Hemoglobin binding of oxygen is best described as a:
A) concerted model.
B) Michaelis–Menten model.
C) sequential model.
D) combination of sequential and concerted models.
E) None of the above.
Ans:  D     Section:  9.3

 

27. 2,3-Bisphosphoglycerate
A) binds in the central cavity in the T-form of hemoglobin.
B) preferentially binds to deoxyhemoglobin and stabilizes it.
C) is present in the red blood cells.
D) All of the above.
E) None of the above.
Ans:  D     Section:  9.4

 

28. What is the Bohr effect?
A) the ability of hemoglobin to retain oxygen when in competition with myoglobin
B) the regulation of hemoglobin binding by hydrogen ions and carbon dioxide
C) the alteration of hemoglobin conformation during low oxygen stress
D) All of the above.
E) None of the above.
Ans:  B     Section:  9.5

 

29. Which of the following statements is correct for hemoglobin and oxygen transport?
A) The oxygen binds to the proximal histidine residue of the globin chain.
B) Bonding of carbon dioxide to hemoglobin molecules increases the binding of oxygen.
C) Hemoglobin binds more oxygen as the pH is lowered.
D) Hemoglobin binds more oxygen at higher [BPG] concentrations.
E) The binding of each O2 molecule to hemoglobin increases its affinity for the next O2.
Ans:  E     Section:  9.3

 

30. Which of the following describes the Bohr effect?
A) Lowering the pH results in the release of O2 from oxyhemoglobin.
B) Increasing the pressure of CO2 results in the release of O2 from oxyhemoglobin.
C) Increasing the pH increases the T-form of hemoglobin.
D) All of the above.
E) A and B.
Ans:  E     Section:  9.5

 

31. Which of the following is correct concerning the following equilibria?

CO2  +  H2O    H2CO3

A) An increase in the pressure of CO2 will result in a decrease of pH.
B) This reaction is catalyzed by carbonic anhydrase.
C) The H2CO3 dissociates to H+ and bicarbonate ion, HCO3.
D) The majority of CO2 is transported to the lungs in the form of HCO3.
E) All of the above.
Ans:  E     Section:  9.5

 

32. Carbon dioxide forms carbamate groups in proteins by reaction with:
A) aspartate residues.
B) cysteine residues.
C) N-terminal amino groups.
D) tyrosine residues.
E) heme groups.
Ans:  C     Section:  9.5

 

33. Sickle-cell anemia is caused by:
A) a decreased production of α chains of hemoglobin.
B) a substitution of a Glu residue for a Phe residue at the β6 position.
C) the loss of the heme group because the proximal His is oxidized.
D) a substitution of a Val residue for a Glu residue at the β6 position.
E) a substitution of Glu residue for His at the C-terminal of the α chain.
Ans:  D     Section:  9.6

 

 

34. Which of the following is correct concerning the oxygenation plot of proteins X and Y shown in the graph below?
A) Protein Y exhibits tighter oxygen binding than protein X.
B) Protein Y corresponds to fetal hemoglobin, and protein X corresponds to normal adult hemoglobin.
C) Protein X corresponds to fetal hemoglobin, and protein Y corresponds to normal adult hemoglobin.
D) Protein X corresponds to myoglobin, and protein Y corresponds to hemoglobin.
E) None of the above.
Ans:  C     Section:  9.4

 

35. Which of the following is NOT correct concerning the oxygenation plot of proteins X and Y shown in the graph below?
A) Protein X exhibits tighter oxygen binding than protein Y.
B) Protein Y would function as a better transport protein than protein X.
C) Protein X exhibits cooperative binding, whereas Y does not.
D) Protein X corresponds to myoglobin, and protein Y corresponds to hemoglobin.
E) Protein Y contains multiple-binding sites.
Ans:  D     Section:  9.1

 

 

36. Consider the oxygen-binding profile at three different pH values of 7.6, 7.4, and 7.2.  Which statement is most correct?

 

A) Curve X most likely corresponds to pH 7.2.
B) Curve Z most likely corresponds to pH 7.6.
C) Hb has a higher affinity for oxygen at the pH of curve Z.
D) Curve Y most likely corresponds to pH 7.4.
E) pH has no effect on the oxygenation of hemoglobin.
Ans:  D     Section:  9.5

 

37. What would be the expected result of a Lys residue being substituted with a Ser residue in the BPG binding site of hemoglobin?
A) BPG would bind tighter because of the loss of a positive charge.
B) BPG would bind tighter because of the gain of a positive charge.
C) BPG would bind less tightly because of the loss of a positive charge
D) BPG would bind less tightly because of the gain of a positive charge.
E) This substitution would have no effect on the binding of BPG.
Ans:  C     Section:  9.4

 

38. Why are blood transfusions a successful treatment for thalassemia but not sickle-cell anemia?
A) Thallasemia results from a reduced solubility of the deoxygenated form of hemoglobin.
B) Thallasemia results in high concentrations of deoxygenated hemoglobin.
C) Thallasemia results in low levels of functional hemoglobin leading to decreased production of RBCs.
D) Sickle-cell anemia has a single amino acid substitution of valine for glutamate.
E) Sickle-cell anemia results in tetramers that contain only the β chain leading to decreased production of RBCs.
Ans:  C     Section:  9.6

 

39. What does fMRI measure on the molecular level and what does this mean at the tissue level?
A) fMRI measures the changes in the fifth coordination site in binding iron revealing the amount of carbon monoxide bound in carbon monoxide poisoning.
B) fMRI measures the changes in the fifth coordination site in binding iron revealing sensory brain activity.
C) fMRI measures the changes in magnetic properties of γ-chain hemoglobin and is a noninvasive way of measuring fetal hemoglobin levels during pregnancy.
D) fMRI measures the changes in magnetic properties of hemoglobin revealing the relative amounts of deoxy- and oxyhemoglobin in the circulation of a specific organ
E) fMRI measures pO2  levels in the circulatory system in any organ.
Ans:  D     Section:  9.2

 

 

Short-Answer Questions

 

40. Why is it advantageous for hemoglobin to have allosteric properties?
Ans: Hemoglobin binds oxygen in a positive cooperative manner. This allows it to become saturated in the lungs, where oxygen pressure is high. When the hemoglobin moves to tissues, the lower oxygen pressure induces it to release oxygen and thus deliver oxygen where it is needed.
Section:  9.1

 

41. What is fetal hemoglobin? How does it differ from adult hemoglobin?
Ans: Fetal hemoglobin contains two a and two g chains, in contrast to adult hemoglobin with two a and two b chains. The fetal hemoglobin g chain is probably a result of gene duplication and divergence. The difference in the chains results in a lower binding affinity of 2-3 BPG to fetal hemoglobin. Thus, the fetal hemoglobin has a higher affinity for oxygen, and the oxygen is effectively transferred from the mother’s hemoglobin to fetal hemoglobin.
Section:  9.4

 

42. Describe the octahedral coordination sphere of the iron ion in hemoglobin and myoglobin.
Ans: The Fe2+ ion is coordinated to the four nitrogens in the center of the protoporphyrin of the heme.  The fifth coordination site is occupied by the “proximal histidine” of the globin chain.  The oxygen is bound to the sixth coordination site of the iron.
Section:  9.2

 

43. What functional role does the “distal histidine” play in the function of myoglobin and hemoglobin?
Ans: The bonding between the iron and oxygen can be described as a combination of resonance structures, one with Fe2+ and dioxygen and another with Fe3+ and superoxide.  The “distal histidine” donates a hydrogen bond to this complex, stabilizing the complex and inhibiting the oxidation of the iron to the ferric state.
Section:  9.2

 

44. Draw the oxygen-binding curve of myoglobin and that of hemoglobin.  Indicate the partial pressure of oxygen in the lungs and the range of pressure in tissue.
    ↑

Lungs

 

Ans:

20 – 40 torr
Section:  9.1

 

45. Describe the structure of normal adult hemoglobin.
Ans: Normal adult hemoglobin, HbA, is a tetramer.  It is composed of two α subunits and two β subunits.  Each subunit has a structure very similar to myoglobin.  It can be best described as a pair of identical αβ dimers.  Each subunit contains a heme group.  So, each molecule of hemoglobin can bind up to four molecules of oxygen.
Section:  9.1

 

46. Briefly describe cooperative binding.
Ans: Cooperative binding occurs in multi-subunit proteins that possess multiple-binding sites.  The binding of a ligand to one site causes a conformational change that influences the binding of the ligand to the next site.  The binding sites are not independent, but each binding event affects the affinity of the next binding event.
Section:  9.3

 

47. Describe the concerted model to explain allosteric cooperative binding.
Ans: The protein exists in two conformations, a T state (for tense) that has a lower affinity for the ligand and an R state (for relaxed) that has a higher affinity for the ligand.  In the concerted model, all of the molecules exist either in the T state or in the R state.  At each ligand concentration, there is an equilibrium between the two states.  An increase in the ligand concentration shifts the equilibrium from the T  to the R state.
Section:  9.3

 

48. Describe the role of 2,3-bisphosphoglycerate in the function of hemoglobin.
Ans: 2,3-bisphosphoglycerate, 2,3-BPG, is a relatively small, highly anionic molecule found in the RBC.  2,3-BPG only binds to the center cavity of deoxyhemoglobin (T state).  The size of the center cavity decreases upon the change to the R-form so that it cannot bind to the R state.  Thus, the presence of 2,3-BPG shifts the equilibrium toward the T state.  T state is unstable, and without BPG, the equilibrium shifts so far toward the R state that little oxygen would be released under physiological conditions.
Section:  9.3

 

49. Describe the chemical basis of the Bohr effect.
Ans: The effect observed by Christian Bohr is that hemoglobin becomes deoxygenated as the pH decreases.  In deoxyhemoglobin, three amino acid residues form two salt bridges that stabilize the T state.  One of these is formed between the C-terminal His β146 and an Asp residue (β94).  As the pH increases, this stabilizing salt bridge is broken because His becomes deprotonated and loses its positive charge.  At lower pH values, this His is positively charged.  The formation of the salt bridge shifts the equilibrium from the R state to the T state, thus releasing oxygen.
Section:  9.5

 

50. Describe how carbon dioxide affects the oxygenation of hemoglobin.
Ans: Increased levels of CO2.cause hemoglobin to release oxygen.  The more active the tissue, the more fuel is burned and the more CO2 is produced.  These active tissue cells have the greatest need for oxygen to produce more energy.  The CO2 combines with the N-terminal amino groups to form negatively charge carbamate groups.  The negatively charge carbamate groups form salt bridges that stabilize the T state.  Thus, the increase of carbon dioxide causes the conversion of the R state to the T state, releasing the bound oxygen to the tissues producing the most CO2.
Section:  9.5

 

51. Briefly describe the cause of sickle-cell anemia.
Ans: Sickle-cell anemia is a genetic disorder that is the result of a single substitution of β6 Glu with a Val.  This changes a negatively charged side chain to a nonpolar, hydrophobic side chain.  This Val binds into a hydrophobic pocket on the β chain of an adjacent molecule whose β6 Val binds to another molecule, thus hemoglobin aggregates.  These aggregates form long fibers that strain the RBC and force it into a sickled shape.  The distorted red blood cells clog capillaries and impair blood flow, resulting in the sickle-cell crisis.  The sickled cells are then destroyed, resulting in the anemia.
Section:  9.6

 

52. How does the binding of oxygen to hemoglobin result in the T-to-R-state transition?
Ans: When oxygen binds to hemoglobin, the iron ion moves into the plane of the porphyrin. The histidine residue bound in the fifth coordination site moves with the iron. Because this histidine is part of an α helix, the α helix moves as well. The carboxyl terminal of the α helix lies between the two αβ dimers, thus the movement of the iron ion causes changes in quaternary structure that corresponds to T-to-R-state transition.
Section:  9.3

 

53. Describe the molecular and physiological adaptations for the high-altitude flying bar-headed goose.
Ans: It is thought that the amino acid substitutions in the α and β chains of bar-headed geese hemoglobin, particularly, the change of an alanine for a proline in the α chain, disrupts a van der Waals contact and facilitates the formation of the R state. Thus, oxygen binds more readily, even at the pO2 of 30% at high altitudes found as these birds migrate over the Himalayas.
Section:  9.4

 

54. What is the driving force for moving CO2 out of body tissues and into red blood cells?
Ans: CO2 is formed by the oxidation of glucose in reactions that are not reversible. Thus, a nonpolar molecule such as CO2 readily crosses cell membranes moving down a concentration gradient from high, inside the cell, to low, outside the cell. CO2 then diffuses again into the RBC, where once again the concentration of CO2 is kept low by the activity of carbonic anhydrase. CO2 is converted to carbonic acid that ionizes to hydrogen ions and bicarbonate ions. Thus, there is a steady downhill diffusion gradient for CO2 from the actively metabolizing tissue to the RBC.
Section:  9.5

 

Chapter 11   Lipids

 

 

Matching Questions

Use the following to answer questions 1-10:

 

Choose the correct answer from the list below. Not all of the answers will be used.

  1. a) ether-linked lipid
  2. b) triacylglycerol
  3. c) sterol
  4. d) amphipathic
  5. e) 16 or 18
  6. f) organic solvent
  7. g) cholesterol
  8. h) cysteine
  9. i) phospholipid
  10. j) prokaryotes
  11. k) glycolipid
  12. l) 20 or 24
  13. m) sphingosine
  14. n) serine

 

1. ____________ The storage form of fatty acids.

 

Ans:  b
Section:  Introduction

 

2. ____________ This is the number of carbons in most common fatty acids.

 

Ans:  e
Section:  11.1

 

3. ____________ In addition to phospholipids and glycolipids, this is a major type of membrane lipid.

 

Ans:  g
Section:  11.3

 

4. ____________ This is a term applied to molecules that have both hydrophilic and hydrophobic moieties.

 

Ans:  d
Section:  11.3

 

5. ____________ A lipid is defined as a compound soluble in ________________.

 

Ans:  f
Section:  Introduction

 

6. ____________ Lipids that are bound to carbohydrates.
Ans:  k
Section:  Introduction

 

7. ____________Type of lipid with two acyl chains, a glycerol backbone, and a polar head group.
Ans:  i
Section:  11.3

 

8. ____________ Flat polycylic molecule absent in prokaryotic membranes.

 

Ans:  g
Section:  11.3

 

9. ____________ These lipids are less resistant to hydrolysis, potentially due to the way the acyl chain is linked to the glycerol backbone.

 

Ans:  a
Section:  11.3

 

10. ____________ A complex amino alcohol backbone for membrane lipids.

 

Ans:  m
Section:  11.3

 

 

Fill-in-the-Blank Questions

 

11.      is a membrane lipid composed of sphingosine, fatty acid, and a simple sugar.
Ans:  Cerebroside     Section:  11.3

 

12. The common name of hexadecanoic acid is      .
Ans:  palmitic acid     Section:  11.1

 

13. In phosphoglycerides, the fatty acids are linked to the glycerol backbone by the       linkages.
Ans:  ester     Section:  11.3

 

14. The configuration of most fatty acids in biological systems is      .
Ans:  cis     Section:  11.1

 

15. Fatty acids are ionized at physiological pH and so are referred to in their      form.
Ans:  carboxylate     Section:  11.1

 

16.      The short-hand notation indicating that there are two cis double bonds between carbons 9 and 10 and again between 12 and 13.
Ans:  cis, cis-912   Section:  11.1

 

17. The presence of double bonds in fatty acids limits tight packaging and the number of

      interactions.

Ans:  van der Waals     Section:  11.1

 

18.       is the type of glycolipid that contains a branched chain of as many as seven sugar residues.
Ans: Ganglioside   Section:  11.3

 

19. The reduction in tight packing due to cis double bonds       the melting temperature of a fatty acid.
Ans: lowers   Section:  11.1

 

20. One important       is EPA (eicosapentoenoate) and is found in fatty fish and shellfish.
Ans:  ω-3 fatty acid   Section:   11.1

 

 

Multiple-Choice Questions

 

21. Membrane lipids are primarily comprised of:
A) phospholipids.     B) glycolipids.     C) cholesterol.     D) A and B.     E) A, B, and C.
Ans:  E     Section:  11.3

 

22.  Which of the following is NOT a main function of lipids?
A) cell signaling
B) fuel source
C) structural rigidity of the cytoskeleton
D) membrane component
E) All of the above.
Ans:   C    Section:  Introduction

 

23.  Octadecatrienoic acid has how many double bonds?
A) 0
B) 1
C) 2
D) 3
E) 4
Ans:    D   Section:   11.1

 

24.  An w-3 fatty acid ____________.
A) has a methyl group at the carboxyl end of the fatty acid
B) has a methyl group on the third carbon of the chain
C) has a double bond the third carbon in from the carboxyl group
D) has a triple bond on the third carbon from the methyl end of the fatty acid
E) None of the above.
Ans:   E    Section:   11.1

 

25.  The notation 12:2 indicates which of the following about a fatty acid?
A) There are 12 carbons in the chain with two double bonds.
B) There are two 12-carbon chains for this fatty acid.
C) The second carbon has a fatty acid double bond.
D) The  12th carbon has a double bond.
E) There are two trans-double bonds on this 12-carbon fatty acid.
Ans:   A    Section:   11.1

 

26. Which of the following is NOT correct concerning the structure given?

 

A) It is a component of biological membranes.
B) It is amphipathic.
C) It is a sphingolipid.
D) It is a phosphoglyceride.
E) It is phosphatidyl choline.
Ans:  C     Section: 11.3

 

27.  The longer the fatty acid the ________ the fatty acid.
A) more oxidized
B) lower the melting point of
C) higher the melting point of
D) more reduced
E) more double-bond containing
Ans:   C    Section: 11.1

 

28.  Palmitate has how many carbons in its chain?
A) 12
B) 14
C) 16
D) 20
E) 24
Ans:      C    Section:   11.1

 

29.  Unsaturations ________ melting points of fatty acids and their derivatives.
A) maintain
B) decrease
C) increase
D) are unrelated to
E) None of the above.
Ans:  B     Section:   11.1

 

30.  Eating ____________ increases the w-3 fatty acids decreasing ___________.
A) arachidonic acid,  cardiovascular disease
B) a low fat diet, cardiovascular disease
C) fatty fish, cardiovascular disease
D) shellfish, lung cancer
E) vegetable oils, blood pressure
Ans:     C  Section:   11.1

 

31.  The backbone of a phospholipid is which of the following?
A) glucose
B) cholesterol
C) fatty acid chain
D) triacylglycerol
E) glycerol
Ans:   E    Section:   11.3

 

 

32.  The polar head group of phospholipids is found at which carbon of glycerol?
A) C1
B) C2
C) C3
D) C1-OH
E) C2-OH
Ans:    C   Section:   11.3

 

 

33.  Polar-head groups of phospholipids are esterified to what functional group?
A) methyl
B) phosphate
C) ketone
D) thiol
E) aldehyde
Ans:  B     Section:   11.3

 

34.  A phosphatidate lipid (phosphatidic acid) has which of the following components?
A) phosphate
B) glycerol
C) ester linkage
D) acyl chain
E) All of the above.
Ans:   E    Section:  11.3

 

35.  Which phospholipid is enriched in neural sheath membranes?
A) phosphatidic acid
B) phosphatidylcholine
C) sphingomyelin
D) diphophatidylglycerol (cardiolipin)
E) phosphatidylinositol
Ans:    C   Section:   11.3

 

36.  Identify the differences in archaea membrane lipids compared to those of eukaryotes or bacteria and how these differences help them withstand extreme environmental conditions.
A) The ether linkages are more, readily hydrolyzed by enzymes allowing the membrane to serve as an energy reserve.
B) The glycerol moiety is esterified to multiple complex carbohydrate chains making them more soluble in low pH environments.
C) The fatty acid chains are branched, allowing them to pack more tightly, thereby protecting membrane integrity.
D) There are two phosphate esters instead of only one giving the archaea better solubility in high salt environments.
E) Omega ω-3 fatty acids are common in membranes and may act as important precursors as they do in eukaryotes.
Ans:    C   Section:   11.3

 

37.  You are studying a protein known to be localized to the membrane surface. What protein modifications might you look for to determine how the protein is attached to the membrane?
A) Determine whether a farnesyl group is attached to a carboxy terminal cysteine residue.
B) Determine whether a fatty alcohol is attached to a serine residue on the surface of the protein.
C) Determine whether a glycosylphosphatidylinositol anchor is attached to the carbosy terminus.
D) A and C only.
E) All of the above.
Ans:    D   Section:   11.3

 

38. Cholesterol and other steroids are not soluble in blood, and therefore must be transported. Predict what chemical modifications must occur for cholesterol to move through the circulatory system.
A) Cholesterol forms micells in blood, the surface of which is hydrophilic and the interior is hydrophobic.
B) Cholesterol cannot be transported in blood and so it broken down and resynthesized in all cells.
C) Cholesterol moves through cell membranes from tissue to tissue.
D) Cholesterol forms glycolipids with large carbohydrate complexes in order to increase solubility.
E) Cholesterol is esterified to a fatty acid for transport by lipoprotein particles, the surface of which is hydrophilic and the interior is hydrophobic.
Ans:    E   Section:   11.3

 

Short-Answer Questions

 

39. What does the notation 18:2 for fatty acids imply?
Ans: In this fatty acid there are 18 carbons, with two double bonds.
Section:  11.1

 

40. What are the two systems for naming the positions of the double bonds? Provide examples.
Ans: Two systems are used. One system refers to the double bond relative to the last, or omega (w), carbon. (An example would be w-3 fatty acids.) The other system uses notation that indicates the position of the double bond relative to the carboxyl carbon, and indicates if the bond is cis or trans. (An example would be cis-D9.)
Section:  11.1

 

41. What are some molecules that form the polar-head group of phospholipids? Provide several examples.
Ans: Examples of head groups include serine, ethanolamine, choline, glycerol, and inositol.
Section:  11.3

 

42.  Explain the biochemical nature of why trans bonds do not have the same effect as cis bonds on the melting point of fatty acids.
Ans: This is due to the order or cis-order when a double bond is introduced.  A trans-double bond maintains the linear shape of the fatty acid, while a cis-double bond creates a kinked or bent shape.  The latter reduces the number of contact points, thus reducing the amount of energy needed to melt the fatty acid lattice.
Section:  11.1

 

43.  Explain why fats are an efficient way to store biochemical energy.
Ans: This has to do with the anhydrous nature of fat (less water means more mass of fat per gram) and with the fact that fatty acid tails are more reduced than carbohydrates or amino acids.  We will learn later, that the transfer of electrons through redox reactions leads to ATP production.
Section:  11.2

 

44. Draw and label a typical phospholipid.
Ans: The phospholipid drawn should resemble Figure 12.5 in the textbook. It should contain a central glycerol molecule, to which two fatty acids are attached in ester linkages at the center and an end. The other end should be linked to a phosphor alcohol group. The phosphate should be shown negatively charged at pH 7.
Section:  11.3

 

45.  How are birds that migrate across the Gulf of Mexico able to sustain flight over long distances?
Ans: The energy source for these migrations are fatty acids, stored as triacylglycerols (TAGs). TAGs are stored in a nearly anhydrous form, and as a result, a gram of fat stores more than six times as much energy as a gram of hydrated glycogen.
Section:  11.2

 

46.  What is the difference between a sphingolipid and a glycerolipid?
Ans: A sphingolipid has an alcohol such as a serine for the backbone and only one fatty acid tail, where as a glycerolipid uses glycerol as the backbone and has two acyl chains esterified to the hydroxyl groups of the glycerol.
Section:  11.3

 

47.  Steroid hormones come from what lipid?
Ans: Cholesterol is the main source of steroid hormones.
Section:  11.3

 

48.  Define the different chemical characteristics for the phospholipids, phosphatidylserine, phosphatidylcholine, and phosphadidylinositol.
Ans: See the figure for each phospholipid and focus on the head group.  Look for charge and hydrophobicity.
Section:  11.3

 

49.  How does the structure of cholesterol give it a unique structural quality among the lipids?
Ans: The flat double-bonded fused ring creates a bulky hydrophobic moiety for several potential modifications.
Section:  11.3

 

50.  Based on your knowledge of lipids, guess why some fats from plants are oils (liquid at room temperature) and animal fats are solid at room temperature.
Ans: Knowing that the longer the fatty acid chain the higher the melting point and that more double bonds lead to lower melting points, plant fats are likely to be shorter and primarily unsaturated or monounsaturated, while animal fats are more often polyunsaturated.
Section:  11.2

 

 

51.  Phosphoglycerides have common but varying structural features. Describe which structural features are common to all phosphoglycerides and which ones vary. Draw correlations between structural variability and membrane function.
Ans: All phosphoglycerides contain a glycerol backbone with two fatty acid chains attached through ester bonds to two of the hydroxyls. The third hydroxyl is attached by a phosphoester bond to phosphate, which is then attached to one of several alcohols. This amphipathic structure forms a bilayer in aqueous solutions, so that even with the differences identified as follows, phosphoglycerides form a stable hydrophobic barrier between the cell’s interior and its surroundings. The differences are seen in the length and saturation of the fatty acids, and the chemical nature of the alcohol attached to the phosphate. Fatty acid differences affect the fluidity of the membrane and the chemical nature of the alcohol provides multiple functional sites on the membrane surfaces.
Section:  11.2

 

52. Draw the structure of sphingomyelin and label the linkages in this structure.
Ans: See Figure 12.6
Section:  12.3

Chapter 15   Metabolism: Basic Concepts and Design

 

 

Matching Questions

Use the following to answer questions 1-10:

 

Choose the correct answer from the list below. Not all of the answers will be used.

  1. a) O2
  2. b) niacin
  3. c) phototrophs
  4. d) ATP
  5. e) CO2
  6. f) coenzyme A
  7. g) vitamins
  8. h) amphibolic
  9. i) ADP
  10. j) NADPH
  11. k) chemotrophs
  12. l) FAD

 

1. ____________ These organisms use energy from sunlight and convert it to chemical energy.

 

Ans:  c
Section:  15.1

 

2. ____________ These organisms obtain chemical energy from oxidation of foodstuffs.

 

Ans:  k
Section:  15.1

 

3. ____________ Pathways that can be either anabolic or catabolic depending on the energy conditions of the cell.

 

Ans:  h
Section:  15.2

 

4. ____________ In aerobic organisms, this is the ultimate acceptor of electrons.

 

Ans:  a
Section:  15.4

 

5. ____________ In aerobic metabolism, this is the product of oxidation of carbon containing fuels.

 

Ans:  e
Section:  Introduction

 

6. ____________ The electron carrier, NADH, is derived from this vitamin.

 

Ans:  b
Section:  15.5

 

7. ____________ This substance is the electron donor in most reductive biosyntheses.

 

Ans:  j
Section:  15.5

 

8. ____________ This compound serves as an acyl carrier in metabolism.

 

Ans:  f
Section:  15.5

 

9. ____________ This is the “chemical currency” of metabolism.

 

Ans:  d
Section:  15.2

 

10. ____________ These small organic compounds are required in the diet of higher organisms and are components of coenzymes.

 

Ans:  g
Section:  15.5

 

 

Fill-in-the-Blank Questions

 

11.       is the type of metabolism where useful energy is harvested.
Ans:  Catabolism     Section:  Introduction

 

12. A thermodynamically unfavorable reaction can be driven by a thermodynamically favorable reaction to which it is      .
Ans:  coupled     Section:  15.2

 

13. ATP is considered an “energy rich” compound because it contains two       bonds.
Ans:  phosphoanhydride     Section:  15.3

 

14. In the cell, the hydrolysis of an ATP molecule in a coupled reaction changes the equilibrium ratio of products to reactants by a factor of      .
Ans:  108     Section:  15.3

 

15. In vertebrate muscle,      serves as a reservoir of high-potential phosphoryl groups that can be readily transferred to ADP to regenerate ATP.
Ans:  creatine phosphate     Section:  15.3

 

16.       is the process of building larger molecules from smaller ones.
Ans:  Anabolism     Section:  15.2

 

17. FAD is an electron carrier that is derived from the vitamin      .
Ans:  riboflavin     Section:15.5

 

18. The acetyl group is attached to coenzyme A by a      bond.
Ans:  thioester     Section:15.5

 

19. ATP-generating (catabolic) pathways are inhibited by a      (high, low) energy charge.
Ans:  high     Section:15.6

 

20. One way that metabolism is regulated is through control of the accessibility of      .
Ans:  substrates     Section:  15.6

 

 

Multiple-Choice Questions

 

21. The major purpose(s) for which organisms require energy is/are:
A) performance of mechanical work. D) A and C.
B) active transport. E) A, B, and C.
C) synthesis of biomolecules.
Ans:  E     Section:  15.1

 

22. Reaction pathways that transform fuels into cellular energy are:
A) anabolic. D) All of the above.
B) catabolic. E) None of the above.
C) allobolic.
Ans:  B     Section:  15.2

 

23. Metabolic pathways that require energy and are often biosynthetic processes are:
A) anabolic. D) All of the above.
B) catabolic. E) None of the above.
C) allobolic.
Ans:  A     Section:  15.2

 

24. Electron carrier(s) that include ATP are:
A) NAD+     B) FAD     C) FMN     D) A and B.     E) A, B, and C.
Ans:  D     Section:  15.5

 

25. What is the standard-state free energy (DG°′) for the hydrolysis of ATP to ADP?
A) +45.6 kJ/mol D) -14.6 kJ/mol
B) -45.6 kJ/mol E) +30.5 kJ/mol
C) -30.5 kJ/mol
Ans:  C     Section:  15.3

 

26. Which of the following molecule(s) have a higher phosphoryl-transfer potential than ATP?
A) phosphoenolpyruvate D) A and B.
B) creatine phosphate E) C, B, and C.
C) 1,3-bisphosphoglycerate
Ans:  E     Section:  15.3

 

27. This energy source is used to regenerate ATP from ADP and Pi.
A) oxidation of carbon to CO2
B) electrochemical potential of stored glycogen
C) reduction of pyruvate to lactate
D) All of the above.
E) None of the above.
Ans:  A     Section:  15.2

 

28. The reduced form of flavin adenine dinucleotide is:
A) FADH.     B) FAD.     C) FADH++.     D) FADH2.     E) None of the above.
Ans:  D     Section:  15.5

 

29. Which of the following is the electron donor used for reductive biosynthesis?
A) NADH D) CoASH
B) NADPH E) ATP
C) FADH2
Ans:  B     Section:  15.5

 

30. Pantothenate kinase associated degeneration:
A) is a predominantly neurological disorder. D) A and C.
B) can cause anemia. E) A, B, and C.
C) affects tissues that are dependent on aerobic metabolism.
Ans:  D     Section:  15.5

 

31. Which is the correct coenzyme: carrier pair?
A) NADH: acyl
B) tetrahydrofolate: electrons
C) coenzyme A: acyl
D) lipoamide: aldehyde.
E) thiamine pyrophosphate: glucose
Ans:  C     Section:  15.5

 

32. Which activated carriers contain adenosine phosphate units?
A) NADH     B) FADH2     C) coenzyme A     D) A and B.     E) A, B, and C.
Ans:  E     Section:  15.5

 

33. Which of the following is an example of an oxidation reaction?
A)
B)
C)
D) Ala-Ser  +  H2O   →   Ala  +  Ser
E) None of the above.
Ans:  A     Section:  15.4

 

34. Metabolic processes are regulated by:
A) transcriptional regulation of the amount of enzyme.
B) allosteric control of enzyme activity.
C) accessibility of substrates by compartmentalization.
D) A and B.
E) A, B, and C.
Ans:  E     Section:  15.6

 

35. Some of the mechanisms by which enzyme catalytic activity is controlled are:
A) allosteric control. D) A and C.
B) feedback inhibition. E) A, B, and C.
C) covalent modification .
Ans:  E     Section:  15.6

 

36.  The phosphorylation of fructose-6-phosphate is an endergonic reaction with a ΔGoˊ of 16.3 kJ/mol. How do cells overcome this thermodynamic barrier for this reaction under standard conditions?
A) The enzyme that catalyzes this reaction couples it with the condensation of ADP and inorganic phosphate, resulting in an overall ΔGoˊ of –46.8 kJ/mol.
B) The enzyme that catalyzes this reaction couples it with the hydrolysis of ATP to ADP and inorganic phosphate, resulting in an overall ΔGoˊ of –14.2 kJ/mol.
C) This reaction will proceed to the right because the Keq is small.
D) This reaction will proceed to the right because the Keq is negative.
E) By uncoupling the reaction to the hydrolysis of ATP, the reaction can be driven forward.
Ans:  B     Section:  15.2

 

37.  You are interested in studying bacteria found in peat swamps and you identify a new bacterium that you believe is a chemotroph. Which of the following would you use to verify your belief?
A) screen for ATP synthesis
B) screen for enzymes that oxidize carbon
C) screen for light gathering structures
D) screen for digestive enzymes
E) screen for linked reactions
Ans:    C   Section:  15.1

 

38. The formation of ATP by creatine kinase is shown in the reaction below:

creatine phosphate  +  ADP  ↔  ATP  +  creatine

Using the Table of Standard Free Energies (Table 15.1 in text), determine if this reaction is thermodynamically favored under standard conditions.

A) No, it is not thermodynamically favored because the Kˊeq is 12.6.
B) No, it is not thermodynamically favored because the ΔGoˊis 12.6 kJ/mol.
C) No, it is not thermodynamically favored because the ΔGoˊis is -73.6 kJ/mol.
D) Yes, it is thermodynamically favored because the Kˊeq is 73.6.
E) Yes, it is thermodynamically favored because the ΔGoˊis is -12.6 kJ/mol.
Ans:    E   Section:  15.3

 

39. The hydrolysis of a phosphate group from ATP releases 30.5 kj/mol, whereas the hydrolysis of a phosphate from glucose 6-phosphate releases only 13.82 kJ/mol. In that the product is the same, what accounts for the difference?
A) ATP has greater resonance stabilization than the product orthophosphate.
B) There is a greater increase in entropy when ATP is hydrolyzed.
C) Water hydrates ATP greater than glucose 6-phosphate.
D) ATP has a larger phosphoryl-transfer potential.
E) The phosphate ester in ATP is more thermodynamically stable than in glucose 6-phosphate.
Ans:   D    Section:  15.3

 

 

 

Short-Answer Questions

 

40. Explain how a metabolic pathway can contain an energetically unfavorable reaction yet still occur.
Ans: The free-energy changes of the individual steps in a pathway are summed to determine the overall free-energy change. Thus, a step that might not normally occur can be driven if it is coupled to a thermodynamically stable reaction.
Section:  15.2

 

41. What are the two criteria that must be satisfied by a biochemical pathway?
Ans: 1) The reactions must be specific, in that only one set of products are formed from a specific set of reagents. 2)  As a whole, the pathway must be spontaneous, that is, must be thermodynamically favored.
Section:  15.2

 

42. Draw the structure of ATP and identify the phosphoanhydride bond(s).
Ans:  An example can be found in Figure 15.3 in the textbook.
Section:  15.2

 

43. What general factors contribute to the high phosphoryl-group transfer of ATP?
Ans: Resonance stabilization, electrostatic repulsion, and stabilization due to hydration are important.
Section:  15.3

 

44. Draw the resonance structures of orthophosphate and explain why these structures are not significant in ATP.
Ans: The answer is as observed in Figure 15.5 and 15.6 in the textbook. Each resonance structure has an overall –2 charge.  These structures do not contribute to the stabilization of ATP because the positively charged oxygen is next to positively charged phosphorus.
Section:  15.3

 

45. How much ATP is used daily by a typical human? How is it regenerated?
Ans: A human uses 40 kg of ATP per day. There is only about 100 g ATP available, thus the ATP is used and regenerated rapidly. ATP is regenerated from ADP and Pi, using the energy from the catabolic processes.
Section:  15.4

 

46. What is oxidative phosphorylation?
Ans: It is the process by which ATP is formed by the phosphorylation of ADP using the energy of the proton gradient that was generated by the transfer of electrons from reduced coenzymes to oxygen.
Section:  Introduction

 

47. Why are fats a more efficient fuel source than carbohydrates?
Ans: When any fuel molecule is oxidized, the free energy released is used to generate ATP. In that carbohydrates are already more oxidized than fats, less energy is released in the complete oxidation to CO2.
Section:  15.4

 

48. What is an activated carrier? Provide two examples.
Ans: Activated carriers are molecules that are used as the carrier molecules of a particular molecule, atom, electron, or proton. One example is ATP, which is the activated carrier of phosphoryl groups. Flavin derivatives (FAD) and nicotinamide derivatives (NAD+) are examples of activated carriers of electrons.
Section:  15.5

 

49. Compare ATP to acetyl CoA.
Ans: Both are activated carriers: Acetyl CoA carries acetyl groups with high acetyl group-transfer potential, whereas ATP carries phosphate groups with high phosphoryl-group transfer. Both molecules are common to several pathways.
Section:  15.4 and 15.5

 

50. How are metabolic processes unified? How can you use this to help learn and understand biochemistry?
Ans: Common molecules and mechanisms are evident in motifs and patterns throughout metabolic pathways. Understanding the logic of catabolic and anabolic paths, and knowing common molecules (such as ATP) and mechanisms (oxidation-reduction), makes it simpler to understand the myriad paths of metabolism.
Section:  15.5

 

51. List five activated carriers in metabolism and give the vitamins that are the precursors of these carriers.
Ans: Activated carrier Vitamin
NADH and NADPH niacin
FADH2 riboflavin
acetyl-coenzyme A pantothenate
biotin biotin
tetrahydrofolate folic acid
Section:  15.5 and Table 15.2

 

52. What is the relationship between the energy charge of a cell and control of the ATP-generating pathway?
Ans: When the energy charge is high, ATP-generating pathways are inhibited and visa versa.
Section:  15.6

 

53. How is metabolism controlled?
Ans: The amounts of enzymes and their catalytic activity are two controllable aspects of metabolism. Substrate accessibility is also important.
Section:  15.6

 

54. If many compounds are common to both anabolic and catabolic paths, how can metabolism be controlled?
Ans: The enzymes and their activities can be controlled by the energy charge in the cell.  The biosynthetic and catabolic paths are different from each other and may even be located in different compartments in the cell.  Thus, the two opposing processes can be controlled independently.
Section:  15.6

 

55. Why do we call ATP a carrier of phosphoryl groups and not a storage molecule for phosphoryl groups?
Ans:  ATP has a phosphoryl-transfer potential that is intermediate among the biologically important phosphorylated molecules. This intermediate position enables ATP to function efficiently as a carrier of phosphoryl groups. High phosphoryl-transfer-potential compounds derived from the metabolism of fuel molecules are used to power ATP synthesis. In turn, ATP donates a phosphoryl group to other biomolecules to facilitate their metabolism.
Section:  15.3