Tymoczko’s Biochemistry A Short Course THIRD EDITION (Six Month Access) John L. Tymoczko -Test Bank




Tymoczko’s Biochemistry A Short Course THIRD EDITION (Six Month Access) John L. Tymoczko -Test Bank

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


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 __________.
C) DGº′
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


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


Ans: c
Section:  Introduction


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


Ans: j
Section:  7.1


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


Ans: a
Section:  7.1


58. A reaction with two substrates is considered a ____________ reaction.
Ans: e


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


Ans: h
Section:  7.2


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


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


Ans: f
Section:  7.2


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


Ans: g
Section:  7.2


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


Ans: l
Section:  7.3


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


Ans: d
Section:  7.4



Fill-in-the-Blank Questions


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


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


67. 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


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


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


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


71. 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


72. 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


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


74. 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


75. 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


76. 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


77. 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



78.   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


79. 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


80. 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


81.   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


82. 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


83. 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


84. 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


85. 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


86. 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


87. 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


88.   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


89.  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


90.  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


91.  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


92.  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


93.  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


94. 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


95. 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



96. 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


97. 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


98. 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


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


Initial velocity



Maximum velocity



Substrate concentration



Michaelis constant



Section:  7.2


100. 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


101. 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


102. 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


103. 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


104. 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


105. 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


106. 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


107. 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


108. 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