6.1 Multiple-Choice Questions
1) Which of the following best describes a metastable state?
- A) This state is composed of the difference in activation energy of a catalyzed versus an uncatalyzed reaction.
- B) The metastable state is formed by transient complexes with the substrate.
- C) The metastable state is created by the prosthetic group of the enzyme.
- D) This state changes the position of the equilibrium but not the rate.
- E) The metastable state is a state of the substrate in which the reaction can proceed but typically requires a catalyst.
Answer: E
Chapter Section: 6.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
2) Ribozymes were discovered ________ protein enzymes, even though they ________.
- A) before; only catalyze one type of cellular reaction
- B) before; are more heat labile than protein enzymes
- C) after; catalyze many important cellular reactions
- D) after; are the only catalysts found in bacteria and archaea
- E) at the same time as; only catalyze one type of cellular reaction
Answer: C
Chapter Section: 6.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.1
Global LO: G2
3) An enzyme
- A) decreases the rate of a reaction.
- B) binds substrates in a manner that facilitates the formation of product.
- C) changes the position of the equilibrium of the reaction.
- D) does not change the rate at which the equilibrium is achieved.
- E) is always a protein.
Answer: B
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
4) The site on an enzyme that will bind the substrate is called the
- A) prosthetic group.
- B) catalyst.
- C) active site.
- D) metastable site.
- E) activation site.
Answer: C
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
5) Which of the following is an example of a prosthetic group?
- A) a zinc ion
- B) a glycine residue
- C) a polypeptide chain
- D) a hydrogen ion
- E) carboxypeptidase A
Answer: A
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
6) Which of the following is a mechanism of substrate activation?
- A) bond distortion
- B) proton transfer
- C) electron transfer
- D) neutron transfer
- E) bond distortion, proton transfer, and electron transfer
Answer: E
Chapter Section: 6.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.1
Global LO: G2
7) Which of the following is an enzyme?
- A) iron
- B) histidine
- C) carboxypeptidase A
- D) ATP
- E) N-acetylmuramic acid
Answer: C
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
8) Irreversible enzyme inhibitors
- A) bind to the active site of an enzyme through noncovalent bonds.
- B) have only a temporary effect on enzyme activity.
- C) are often toxic to cells.
- D) are not active against ribozymes.
- E) are uncommon in nature.
Answer: C
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2, G7
9) As new enzymes are discovered, the EC system for naming enzymes is to be used. The names are to be based on which of the following criteria?
- A) the name of the substrate
- B) a description of substrate function
- C) an indication of the size of the substrate
- D) the six major classes of enzyme function
- E) the size of the enzyme
Answer: D
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
10) The active site for many enzymes
- A) contains amino acids that are contiguous to one another along the primary sequence of the protein.
- B) may require a prosthetic group such as a metal ion.
- C) involves amino acids that are brought into close proximity by extensive protein folding.
- D) usually depends on only one amino acid.
- E) involves amino acids that are brought into close proximity by extensive protein folding and may require a prosthetic group such as a metal ion.
Answer: E
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
11) According to the EC system, which is not one of the major groups of enzymes?
- A) proteases
- B) hydrolases
- C) oxidoreductases
- D) transferases
- E) ligases
Answer: A
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
12) An enzyme is active in the stomach of an animal but quickly loses its activity when it leaves the stomach. This example illustrates that enzymes are
- A) specific to the organs in which they are produced.
- B) inactivated by movement.
- C) sensitive to changes in pH.
- D) digested in the small intestine.
- E) consumed by the quantities of substrate in the small intestine.
Answer: C
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
13) A sick person often has a fever, which can inhibit the growth of bacteria because
- A) bacteria reproduce more rapidly at higher body temperature.
- B) enzymes do not function as well at temperatures other than the optimal temperature.
- C) the higher temperature increases the activity of lyases.
- D) sweating removes prosthetic groups from biological enzymes.
- E) fever blocks synthesis of proteins in the bacterial nucleus.
Answer: B
Chapter Section: 6.2
Bloom’s Taxonomy: Application
Learning Outcome: 6.1
Global LO: G2
14) The rate of a reaction catalyzed by an enzymes usually ________ with temperature; however, at temperatures above the optimum range ________.
- A) increases; protein denaturation occurs
- B) increases; the reaction proceeds without the enzyme
- C) decreases; the reaction proceeds without the enzyme
- D) increases; the direction of the reaction reverses
- E) decreases; prosthetic groups are necessary
Answer: A
Chapter Section: 6.2
Bloom’s Taxonomy: Application
Learning Outcome: 6.1
Global LO: G2, G7
15) The equation AB + H2O → A + B would be catalyzed by which of the following classes of enzymes?
- A) oxidoreductases
- B) transferases
- C) hydrolases
- D) ligases
- E) isomerases
Answer: C
Chapter Section: 6.2
Bloom’s Taxonomy: Application
Learning Outcome: 6.1
Global LO: G2
16) The equation A − PO4 + B → A + B − PO4 would be catalyzed by which of the following classes of enzymes?
- A) transferases
- B) oxidoreductases
- C) hydrolases
- D) ligases
- E) isomerases
Answer: A
Chapter Section: 6.2
Bloom’s Taxonomy: Application
Learning Outcome: 6.1
Global LO: G2
17) Substrate activation may involve
- A) a change in enzyme conformation induced by substrate binding.
- B) accepting protons from the enzyme.
- C) formation of temporary covalent bonds.
- D) donation of protons to the enzyme.
- E) All of these are correct.
Answer: E
Chapter Section: 6.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.1
Global LO: G2
18) An enzyme reduces the free energy of which of the following?
- A) substrate
- B) product
- C) transition state
- D) cofactor
- E) intermediate product
Answer: C
Chapter Section: 6.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.1
Global LO: G2
19) A competitive inhibitor will affect the ________ of an enzymatic reaction.
- A) Km
- B) Vmax
- C) S
- D) P
- E) both Km and Vmax
Answer: A
Chapter Section: 6.3
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G4
20) A noncompetitive inhibitor will affect the ________ of an enzymatic reaction.
- A) Km
- B) Vmax
- C) S
- D) P
- E) both Km and Vmax
Answer: B
Chapter Section: 6.3
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G4
21) Mixed-type inhibitors will affect the ________ of an enzymatic reaction.
- A) Km
- B) Vmax
- C) S
- D) P
- E) both Km and Vmax
Answer: E
Chapter Section: 6.3
Bloom’s Taxonomy: Application
Learning Outcome: 6.2
Global LO: G2, G4
22) The type of inhibitor that binds to the enzyme active site is a(n) ________ inhibitor.
- A) competitive
- B) noncompetitive
- C) uncompetitive
- D) coenzyme
- E) mixed-type
Answer: A
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2, 6.3
Global LO: G2
23) Which of the following is not true of the enzyme–substrate interaction?
- A) Many enzymes are extremely specific regarding a substrate.
- B) Many enzymes cannot recognize a stereoisomer of their substrate.
- C) Some enzymes accept any of a whole group of substrates.
- D) Carboxypeptidase recognizes any of the amino acids from the carboxyl end of a polypeptide.
- E) Cells are often able to carry out metabolic activity with only a handful of enzymes because many are nonspecific.
Answer: E
Chapter Section: 6.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2
24) The induced-fit model
- A) was proposed by Hans Buchner.
- B) distorts the shape of the enzyme and the substrate.
- C) is also called the lock-and-key model.
- D) states that enzyme–substrate interactions are rigid.
- E) proposes that very strong covalent bonds are formed upon substrate binding.
Answer: B
Chapter Section: 6.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2
25) Why is the Lineweaver–Burk plot important in enzyme kinetics?
- A) It is a single-reciprocal plot.
- B) It illustrates enzyme specificity.
- C) It reveals the presence of prosthetic groups in enzymes.
- D) It makes it easier to determine Vmaxand K
- E) It is nonlinear.
Answer: D
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2, G4
26) The Michaelis constant
- A) can be determined using the Lineweaver–Burk plot.
- B) is equal to twice the Vmax.
- C) is equal to the substrate concentration at Vmax/2.
- D) can be determined using the Lineweaver–Burk plot and is equal to the substrate concentration at Vmax/2.
- E) can be determined using the Lineweaver–Burk plot and is equal to twice the Vmax.
Answer: D
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2
27) Feedback inhibition prevents cells from
- A) the harmful effects of enzyme activation by covalent modification of unneeded enzymes.
- B) making products that are not needed by inhibiting the activity of enzymes in biosynthetic pathways allosterically.
- C) destroying enzymes by proteolytic cleavage when they are needed in biosynthetic pathways.
- D) accumulating unnecessary proteins.
- E) irreversibly inhibiting critical enzymes.
Answer: B
Chapter Section: 6.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.3
Global LO: G2
28) Which of the following variables is not part of the Michaelis–Menten equation?
- A) kcat
- B) Km
- C) Vmax
- D) [S]
- E) v
Answer: A
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2, G4
29) Saturation can be defined as
- A) denaturation of an enzyme.
- B) the inability to increase reaction velocity beyond a finite upper limit.
- C) inhibition of enzyme function by blocking the active site.
- D) the substrate concentration at which velocity reaches one-half maximum velocity.
- E) a characteristic of all uncatalyzed reactions.
Answer: B
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2, G4
30) The enzyme carbonic anhydrase catalyzes the formation of carbonic acid from carbon dioxide and water. Imagine that you have two sealed tubes with carbon dioxide gas above a buffer containing water and carbon dioxide. One tube has carbonic anhydrase and the other does not. After 24 hours, both tubes are at equilibrium. Which statement below accurately describes the conditions in the tubes?
- A) The tube with carbonic anhydrase will have a lower concentration of carbon dioxide and a higher concentration of carbonic acid than the tube with any enzyme.
- B) The tube with carbonic anhydrase will have a higher concentration of carbon dioxide and a lower concentration of carbonic acid than the tube with any enzyme.
- C) The tubes will have equal concentrations of carbon dioxide and carbonic acid.
- D) The tube with carbonic anhydrase will have a higher concentration of carbonic acid than the tube with any enzyme, but the concentration of carbon dioxide will be the same in both tubes.
- E) It is not possible to determine the relative concentrations of the products and reactants without more information about carbonic anhydrase, such as the K
Answer: C
Chapter Section: 6.3
Bloom’s Taxonomy: Evaluation
Learning Outcome: 6.2
Global LO: G2, G4
31) A linear relationship between Vmax and enzyme concentration would be expected when
- A) [S] << Km.
- B) [S] >> Km.
- C) [S] = Km.
- D) both [S] << Km and [S] = Km.
- E) both [S] >> Km and [S] = Km.
Answer: C
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2, G4
32) An allosteric inhibitor
- A) increases the rate of substrate binding.
- B) binds and activates the high-affinity state of the enzymes.
- C) is identical to the active site.
- D) binds at the regulatory site.
- E) is converted to an activator by the enzyme.
Answer: D
Chapter Section: 6.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.3
Global LO: G2, G7
33) A competitive inhibitor
- A) binds at a site other than the active site.
- B) irreversibly binds and inactivates the enzyme.
- C) cannot be processed by the enzyme.
- D) does not inhibit enzyme activity but does lower substrate concentration.
- E) binds to and inactivates the substrate.
Answer: C
Chapter Section: 6.3, 6.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.3
Global LO: G2, G7
34) An example of an irreversible inhibitor is
- A) a competitive inhibitor.
- B) angiotensin converting enzyme (ACE) inhibiting drugs.
- C) a noncompetitive inhibitor.
- D) penicillin.
- E) isoleucine.
Answer: D
Chapter Section: 6.3
Bloom’s Taxonomy: Application
Learning Outcome: 6.2
Global LO: G2, G7
35) Enzyme regulation may occur by several methods. Which of the following is not a means of enzyme regulation?
- A) substrate-level phosphorylation
- B) feedback inhibition
- C) allosteric regulation
- D) covalent modification
- E) saturation
Answer: E
Chapter Section: 6.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.3
Global LO: G2
36) Covalent modification
- A) can activate an enzyme.
- B) affects the activity of an enzyme by adding or removing a chemical group.
- C) can involve the addition of phosphate groups.
- D) produces modifications that can sometimes be reversed.
- E) All of these are correct.
Answer: E
Chapter Section: 6.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.3
Global LO: G2, G7
37) All of the following are examples of ribozymes or ribozyme activity except
- A) peptidyl transferase activity.
- B) autocatalytic RNAs.
- C) ribonuclease P.
- D) intron removal from pre-rRNA.
- E) zymogen.
Answer: E
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2, G7
38) Which of the following is/are means whereby a catalyst can lower the activation energy of a reaction?
- A) decreasing the number of reactive molecules
- B) altering the temperature within the cell to one appropriate for reactions to proceed
- C) quantum tunneling
- D) inefficient collisions
- E) permanently binding substrates
Answer: C
Chapter Section: 6.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
39) Of the following, which is used to inhibit specific enzymes in the treatment of many bacterial and viral diseases?
- A) substrate analogues
- B) noncompetitive inhibitors
- C) denaturing agents
- D) allosteric regulation
- E) hydrogenases
Answer: A
Chapter Section: 6.3
Bloom’s Taxonomy: Application
Learning Outcome: 6.2
Global LO: G2, G5, G7
40) A noncompetitive inhibitor will
- A) bind to free enzyme.
- B) bind to free product.
- C) decrease Vmax.
- D) decrease Km.
- E) bind to free enzyme and decrease Vmax.
Answer: C
Chapter Section: 6.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 6.2
Global LO: G2
6.2 Matching Questions
Match the symbol on the left with its description on the right.
- A) y-intercept of a Lineweaver–Burk plot
- B) turnover number
- C) Michaelis–Menten constant
- D) maximum velocity
- E) substrate concentration
- F) slope of a Michaelis–Menten plot
1) Km
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
2) kcat
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
3) Vmax
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
4) [S]
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
5) Km/Vmax
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
6) 1/Vmax
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
Answers: 1) C 2) B 3) D 4) E 5) F 6) A
Match the choice on the left with the correct choice on the right.
- A) end-product inhibition
- B) trypsin
- C) Lineweaver–Burk plot
- D) RNA
- E) zinc
- F) protein kinases
- G) distortion of substrate and enzyme
7) double-reciprocal
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
8) feedback inhibition
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
9) covalent modification
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
10) proteolytic cleavage
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
11) ribozymes
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
12) prosthetic group
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
13) induced-fit model
Chapter Section: 6.2, 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.1, 6.2, 6.3
Global LO: G2, G7
Answers: 7) C 8) A 9) F 10) B 11) D 12) E 13) G
6.3 Short Answer Questions
1) The ________ is the minimum energy required before two molecules can be successful in producing a reaction.
Answer: activation energy
Chapter Section: 6.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
2) The site on an enzyme that the substrate binds to is called the ________.
Answer: active site
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
3) A nonprotein component of an enzyme that is usually a metal ion or small organic molecule is called a(n) ________.
Answer: prosthetic group
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
4) ________ are a class of enzymes responsible for the movement of functional groups from one molecule to another.
Answer: Transferases
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
5) The ________ velocity of an enzymatic reaction is the velocity at very high substrate concentration.
Answer: maximum
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
6) ________ inhibitors bind the enzyme at a location other than the active site but still interfere with product formation.
Answer: Noncompetitive
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
7) ________ inhibitors bind reversibly at the active site of an enzyme.
Answer: Competitive
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
8) ________ are enzymes able to convert a substrate to its mirror image.
Answer: Isomerases
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
9) Nonprotein catalysts are known as ________.
Answer: ribozymes
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
10) ________ is the turnover number for a given enzyme.
Answer: kcat
Chapter Section: 6.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.2
Global LO: G2
11) Two specific coenzymes we need to obtain from our diet are niacin and ________, as our bodies cannot synthesize them.
Answer: riboflavin
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
12) Azidothymidine (AZT) is a(n) ________ used in the treatment of AIDS and targets the enzyme reverse transcriptase.
Answer: substrate analogue
Chapter Section: 6.3
Bloom’s Taxonomy: Application
Learning Outcome: 6.2
Global LO: G2
13) ________ is a means of overcoming the activation energy barrier that depends on quantum effects.
Answer: Quantum tunneling
Chapter Section: 6.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 6.1
Global LO: G2
6.4 Inquiry
Inhibitors of enzymes can be either reversible or irreversible. In addition, most reversible inhibitors are either competitive or noncompetitive. Based on what you know about enzyme inhibition, classify the following examples as irreversible, competitive, or noncompetitive enzyme inhibition.
- A) competitive
- B) noncompetitive
- C) irreversible
1) Diisopropyl fluorophosphate binds to acetylcholinesterase and permanently inactivates the enzyme. Paralysis results.
Chapter Section: 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G5, G7
2) A drug binds to the active site of an enzyme but disassociates and leaves the enzyme active.
Chapter Section: 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G5, G7
3) A toxin binds to the surface of an enzyme. The enzyme then binds the substrate, but no product is produced. The toxin may disassociate and the enzyme will become active again.
Chapter Section: 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G5, G7
4) Vitamin K is a coenzyme involved in blood clotting. An anticoagulant drug binds at the site of vitamin K bonding, blocking vitamin K binding and preventing clotting. Clotting resumes after the patient stops taking the drug.
Chapter Section: 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G5, G7
5) Aspirin binds to prostaglandin synthetase and permanently stops its ability to produce prostaglandin.
Chapter Section: 6.3, 6.4
Bloom’s Taxonomy: Application
Learning Outcome: 6.2, 6.3
Global LO: G2, G5, G7
Answers: 1) C 2) A 3) B 4) A 5) C
Enzyme activity may be regulated by several means. What is the mode of regulation in each of the following examples?
- A) allosteric regulation or feedback inhibition
- B) substrate-level regulation
- C) inhibition
- D) proteolytic cleavage
6) When glucose is converted to glucose-6-phosphate by hexokinase, the accumulation of glucose-6-phosphate inhibits the reaction.
Chapter Section: 6.4
Bloom’s Taxonomy: Analysis
Learning Outcome: 6.3
Global LO: G2, G5, G7
7) A foreign substance is added to the reaction in part a above. This substance binds to hexokinase and prevents its ability to catalyze the reaction. E) covalent modification
Chapter Section: 6.4
Bloom’s Taxonomy: Analysis
Learning Outcome: 6.3
Global LO: G2, G5, G7
8) An inactive form of an enzyme becomes active after being phosphorylated.
Chapter Section: 6.4
Bloom’s Taxonomy: Analysis
Learning Outcome: 6.3
Global LO: G2, G5, G7
9) Pepsinogen becomes pepsin after being released into the lumen of the stomach.
Chapter Section: 6.4
Bloom’s Taxonomy: Analysis
Learning Outcome: 6.3
Global LO: G2, G5, G7
10) In a series of enzymatic reactions, it is found that the product of the last reaction inhibits the first enzyme in the series of reactions.
Chapter Section: 6.4
Bloom’s Taxonomy: Analysis
Learning Outcome: 6.3
Global LO: G2, G5, G7
Answers: 6) B 7) C 8) E 9) D 10) A
11) Two algal species utilize different transport proteins to take up carbon dioxide for use in photosynthesis. The kinetics of carbon dioxide uptake by the transport proteins in species A and B are different:
Species A Species B
Km 1000 mM 10 mM
Vmax 1000 mmol/min 100 mmol/min
Assuming that that rate of carbon dioxide uptake is the rate-limiting step in photosynthetic growth, which algal species will grow faster if the concentration of carbon dioxide is: 5 mM? 100 mM? 2000 mM?
Answer: Students can calculate the v at each carbon dioxide concentration for both algal species. Some students may also be able to logically deduce which species will grow faster at low and high carbon dioxide concentrations based on the very different kinetic characteristics. Species B will grow faster (uptake carbon dioxide faster) at 5 mM, the two species have an equal rates of carbon dioxide uptake at 100 mM, and species A will grow faster at 2000 mM.
Chapter Section: 6.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 6.2
Global LO: G2, G4, G7, G8
Becker’s The World of the Cell, 9e (Hardin/Bertoni/Kleinsmith)
Chapter 7 Membranes: Their Structure, Function, and Chemistry
7.1 Multiple-Choice Questions
1) Each of the following is a function of membranes except
- A) defining cell and organelle boundaries.
- B) sites for specific biochemical functions.
- C) information storage.
- D) regulation of transport.
- E) cell–cell communication.
Answer: C
Chapter Section: 7.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
2) Each of the following is a model for membrane structure except
- A) Overton lipid coat.
- B) Langmuir monolayer.
- C) Gorter and Grendel bilayer.
- D) Singer and Nicholson fluid mosaic.
- E) Watson and Crick double helix.
Answer: E
Chapter Section: 7.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
3) In response to temperature changes, cell membranes change state to become more solid or more fluid by undergoing
- A) lipid raft formation.
- B) membrane folding.
- C) a phase transition.
- D) transverse diffusion.
- E) differential scanning calorimetry.
Answer: C
Chapter Section: 7.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 7.2
Global LO: G2
4) When a mouse cell and a human cell with different cell-surface protein markers are fused using polyethylene glycol (PEG) and immediately placed at 0°C, what would you expect happens to the mouse and human marker proteins?
- A) Both the mouse and human marker proteins will rapidly disperse evenly throughout the fused membrane.
- B) Only the mouse cell marker proteins will disperse throughout the fused membrane; the human marker proteins will remain confined to the original human region of the fused membrane.
- C) The mouse and human markers will migrate to opposite poles of the fused membrane.
- D) The mouse and human markers will migrate little and remain confined to their original membrane regions.
- E) Both mouse and human markers will be endocytosed by the fused cell and destroyed in the fused cell.
Answer: D
Chapter Section: 7.4
Bloom’s Taxonomy: Analysis
Learning Outcome: 7.2
Global LO: G2
5) Of the following molecules, which would you predict diffuses most readily across membranes?
- A) water
- B) glucose
- C) oxygen
- D) serine
- E) hydrogen ions
Answer: C
Chapter Section: 7.1
Bloom’s Taxonomy: Application
Learning Outcome: 7.1
Global LO: G7
6) Which of the following molecules enters kidney cells via a specific transporter?
- A) water
- B) carbon dioxide gas
- C) cholesterol
- D) ethanol
- E) oxygen
Answer: A
Chapter Section: 7.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
7) Each of the following is a type of cell–cell junctions except
- A) adhesive.
- B) tight.
- C) gap.
- D) plasmodesmata.
- E) All are cell–cell junctions.
Answer: E
Chapter Section: 7.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
8) When examining an electron micrograph of cells obtained from a new deep-sea life-form, you notice that the plasma membranes appear as two dark lines separated by a lightly stained region. Which of the following investigator(s) used a similar observation as the basis for a model of membrane structure?
- A) Robertson
- B) Gorter and Grendel
- C) Unwin and Henderson
- D) Overton
- E) Singer and Nicolson
Answer: A
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G1
9) Which of the following proposed the “sandwich” model of membranes, in which lipid bilayers are coated on both sides with thin sheets of proteins?
- A) Overton
- B) Langmuir
- C) Gorter and Grendel
- D) Davson and Danielli
- E) Robertson
Answer: D
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
10) The composition of lipids in the outer and inner monolayers of cell membrane lipid bilayers is
- A) asymmetrical; i.e., different in each monolayer.
- B) identical in each monolayer.
- C) twice as concentrated in the inner monolayer as in the outer monolayer.
- D) highly random for each monolayer.
- E) the same for all cell plasma membranes but different from the composition in mitochondrial and chloroplast membranes.
Answer: A
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
11) The most prominent lipids in animal cell membranes are
- A) phospholipids.
- B) glycolipids.
- C) cholesterol.
- D) phytosterol.
- E) cerebrosides.
Answer: A
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
12) Which of the following types of organisms contain sphingomyelin in cell membranes?
- A) prokaryotes
- B) algae
- C) animals
- D) yeast
- E) plants
Answer: C
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
13) Which of the following lipids would you expect to find associated with chloroplast membranes?
- A) phosphatidylserine
- B) glycosphingolipid
- C) monogalactosyldiacylglycerol
- D) galactocerebroside
- E) hopanoids
Answer: C
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
14) Which of the following lipids is found concentrated in lipid rafts in animal cell plasma membranes?
- A) cholesterol
- B) phosphatidylcholine
- C) phosphatidylserine
- D) phophatidylethanolamine
- E) phosphatidylinositol
Answer: A
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
15) Which of the following would you expect to find predominating in the plasma membrane of a unicellular eukaryotic organism thriving in glacier ice?
- A) 20 carbon long saturated fatty acids.
- B) 18 carbon long saturated fatty acids.
- C) 20 carbon long fatty acids with 1 double bond.
- D) 18 carbon long fatty acids with 1 double bond.
- E) 16 carbon long fatty acids with 3 double bonds
Answer: E
Chapter Section: 7.3
Bloom’s Taxonomy: Evaluation
Learning Outcome: 7.2
Global LO: G2
16) Cholesterol
- A) is found in abundance in prokaryotic cell membranes.
- B) buffers membrane fluidity by increasing fluidity at low temperature and decreasing fluidity at high temperature.
- C) increases permeability for Ca2+.
- D) increases permeability for small polar molecules.
- E) All of the above are correct.
Answer: B
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
17) Which of the following would be the most thermodynamically unfavorable membrane lipid activity in a membrane?
- A) lateral diffusion
- B) transverse diffusion
- C) association with a neighboring lipid
- D) association with cholesterol
- E) rotation
Answer: B
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
18) Each of the following is typical characteristics of an integral membrane protein transmembrane segment except
- A) α-helical structure.
- B) 20–30 amino acids long.
- C) composed primarily of hydrophobic amino acids.
- D) amphipathic.
- E) All are characteristics of a typical transmembrane segment.
Answer: D
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
19) Each of the following is true about homeoviscous adaptation in membranes except
- A) maintains membrane viscosity despite changes in environmental temperature.
- B) may decrease the average length of membrane lipid fatty acid tails.
- C) may increase the unsaturation of membrane lipid fatty acid tails.
- D) occurs primarily in homeothermic (warm-blooded) organisms and only rarely in poikilothermic (cold-blooded) organisms.
- E) may increase the proportion of cholesterol in animal cell membranes.
Answer: D
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
20) Using a laser beam to inactivate fluorescence dye molecules connected to proteins or lipids on a cell membrane is called
- A) liposome formation.
- B) electrophoresis.
- C) photobleaching.
- D) freeze-fracturing.
- E) affinity labeling.
Answer: C
Chapter Section: 7.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 7.2
Global LO: G2
21) Predict which of the following lipid characteristics would be most important to form the best liposomes for delivering a drug into a cell.
- A) It contains a carbon ring structure similar to that in cholesterol.
- B) It is linked to mannose
- C) It is short and polar like glycerol.
- D) It is amphipathic.
- E) It has three fatty acid chains like a triglyceride.
Answer: D
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
22) Hopanoids are
- A) found exclusively in yeast.
- B) related to cholesterol.
- C) a type of glycolipid.
- D) a type of phospholipid.
- E) All of the above are correct.
Answer: B
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G7
23) The most common number of carbons in fatty acid hydrocarbon chains of membrane phospholipids is
- A) 7.
- B) 10.
- C) 16.
- D) 19.
- E) 24.
Answer: C
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
24) Which of the following is a sterol-like lipid associated with plant cell membranes?
- A) stearate
- B) phytosterol
- C) linoleate
- D) palmitate
- E) hopanoids
Answer: B
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
25) E. coli uses which of the following enzymes to regulate membrane fluidity?
- A) glucose-6-phosphatase
- B) desaturase
- C) fatty acid convertase
- D) gangliosidase
- E) kinase
Answer: B
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
26) Lipid rafts are associated with which of the following activities?
- A) responding to extracellular signals
- B) transport of nutrients across cell membranes
- C) immune responses
- D) transport of cholera toxin into cells
- E) All of the above are correct.
Answer: E
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
27) Which of the following diseases results from impaired glycosphingolipid degradation?
- A) heart disease
- B) glycosphingolipid anemia
- C) metabolic syndrome
- D) erythrocyte spherocytosis
- E) Tay-Sachs disease
Answer: E
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
28) The technique that separates denatured proteins based on size is
- A) SDS-PAGE.
- B) FRAP.
- C) thin-layer chromatography.
- D) Western blotting.
- E) freeze-fracture microscopy.
Answer: A
Chapter Section: Key Technique
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
29) Starting with shredded spinach leaves, you follow a procedure that separates cellular organelles into different fractions. To identify the fraction that contains the mitochondria, you should test for the presence of
- A) aquaporin.
- B) insulin receptors.
- C) phospholipids present only in the mitochondrial outer membrane.
- D) mitochondrial wall polysaccharides.
- E) electron transport proteins.
Answer: E
Chapter Section: 7.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G7
30) Which of the following types of protein would be most easily removed from a membrane by changing pH or ionic strength?
- A) GPI-anchored protein
- B) fatty acid-anchored protein
- C) integral protein
- D) peripheral protein
- E) glycosylated integral protein
Answer: D
Chapter Section: 7.4
Bloom’s Taxonomy: Application
Learning Outcome: 7.2
Global LO: G2
31) A colleague gives you two tubes containing membrane fractions from an animal cell lysate. One tube contains the plasma membrane fraction, and the other tube contains the mitochondrial inner membrane fraction, but the tubes are not labeled. When you analyze the macromolecule composition of the samples, you are confident that the second tube contains the mitochondrial fraction, because the sample has
- A) a higher ratio of cholesterol to phospholipids.
- B) a higher protein to lipid ratio.
- C) a lower protein to lipid ratio.
- D) more carbohydrate in glycoproteins.
- E) more GPI-anchored proteins.
Answer: B
Chapter Section: 7.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 7.2
Global LO: G7
32) Which of the following would best detect a specific carbohydrate sugar on a glycoprotein?
- A) thin-layer chromatography
- B) freeze-fracture microscopy
- C) photobleaching
- D) ferritin-conjugated lectins
- E) SDS-PAGE
Answer: D
Chapter Section: 7.4
Bloom’s Taxonomy: Application
Learning Outcome: 7.2
Global LO: G7
33) Localized regions of plasma membranes that contain cholesterol and proteins involved in cell signaling are known as
- A) gap junctions.
- B) arterial plaques.
- C) cadherin junctions.
- D) lipid rafts.
- E) plasmodesmata.
Answer: D
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
34) The number and location of transmembrane segments in an integral membrane protein can be inferred from the amino acid sequence of the protein by a computer-generated
- A) Western blot.
- B) DNA sequence.
- C) freeze-fracture replica.
- D) hydropathy plot.
- E) prediction of glycosylation sites.
Answer: D
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
35) Cell–cell communication in plants takes place via specialized structures called
- A) plasmodesmata.
- B) chloroplast outer membrane pores.
- C) gap junctions.
- D) lipid rafts.
- E) connexons.
Answer: A
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
36) Which of the following would best be used to determine the lipid content of an isolated membrane fraction?
- A) freeze-fracture analysis
- B) thin-layer chromatography
- C) SDS-PAGE
- D) ferretin-linked antibodies
- E) Western blot analysis
Answer: B
Chapter Section: 7.3
Bloom’s Taxonomy: Application
Learning Outcome: 7.2
Global LO: G1
37) FRAP has revealed that some proteins move in cell membranes much slower than they move in reconstituted liposomes. Which of the following could account for limited mobility of proteins in cell plasma membranes?
- A) association with other proteins in a large complex.
- B) association with lipid rafts.
- C) anchorage to the extracellular matrix.
- D) anchorage to the cell cytoskeleton.
- E) All could limit protein mobility.
Answer: E
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
38) Naturally occurring unsaturated fatty acids typically
- A) are highly branched.
- B) are omega-3 fatty acids.
- C) contain double bonds primarily in the trans
- D) contain double bonds primarily in the cis
- E) contain an odd number of carbon atoms.
Answer: D
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
39) Each of the following contributes to the system providing structural support to the erythrocyte plasma membrane except
- A) the peripheral protein spectrin.
- B) the integral protein glycophorin.
- C) the peripheral protein glyceraldehyde-3-phosphate dehydrogenase (GAPDH).
- D) the peripheral protein ankyrin.
- E) short actin filaments.
Answer: C
Chapter Section: 7.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 7.1
Global LO: G2
40) Which of the following is true about glycosylated plasma membrane proteins?
- A) Carbohydrate is added only after the protein is in the plasma membrane.
- B) The carbohydrate usually is one sugar and rarely more than 5 sugars long.
- C) Only one specific site is glycosylated on each protein.
- D) N-linked carbohydrates are linked to hydroxyl groups in protein R groups.
- E) The carbohydrates can be detected experimentally with ferretin-linked lectins.
Answer: E
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
41) Individuals with the O blood group are known as universal donors because the Gal and GlcNAc groups that are recognized by the antibodies in A, B, and AB in other individuals are missing. Which of which of the following types of erythrocyte plasma membrane components lacks these GlcNAc groups?
- A) a glycoprotein
- B) a glycolipid
- C) glycophorin
- D) a GPI-anchored protein
- E) a glycocalyx protein
Answer: B
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
7.2 Matching Questions
Match the scientist(s) on the left with the statement on the right.
- A) proposed the protein-lipid-protein “sandwich” model of the cell membrane
- B) proposed the “unit membrane” model based on electron microscopy images of the “railroad track” pattern of membranes
- C) proposed the fluid mosaic model of the cell membrane
- D) described the chemical nature of the cell membrane
- E) created lipid monolayers from solubilized cell membrane lipids
- F) determined the 3D structure of a membrane protein demonstrating transmembrane segments
- G) created lipid bilayers from solubilized cell membrane lipids
- H) demonstrated that a biological membrane is a lipid bilayer
1) Davson and Danielli
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
2) Gorter and Grendel
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
3) J. David Robertson
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
4) Irving Langmuir
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
5) Unwin and Henderson
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
6) Charles Overton
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
7) Singer and Nicolson
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
Answers: 1) A 2) H 3) B 4) E 5) F 6) D 7) C
Match the membrane lipid group on the left with the correct type of membrane lipid on the right.
- A) sphingomyelin
- B) Monogalactosyldiacylglycerol (MGDG)
- C) cerebrosides
- D) phosphatidylserine
- E) cholesterol
8) phosphosphingolipids
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
9) glycosphingolipids
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
10) sterols
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
11) phosphoglycerides
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
12) glycoglycerolipid
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
Answers: 8) A 9) C 10) E 11) D 12) B
7.3 Short Answer Questions
1) The E. coli enzyme responsible for creating unsaturated bonds from saturated bonds in the cell membrane fatty acids at a lower environmental temperature is ________.
Answer: desaturase
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
2) Substances that can readily pass through the cell membrane are chemically ________ in nature.
Answer: hydrophobic or nonpolar
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
3) The shift in membrane fluidity from a gel-like state to a more liquid state as temperature rises is known as a ________.
Answer: phase transition
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
4) ________ organisms are unable to regulate their body temperature and must rely on ________ to maintain optimal membrane fluidity.
Answer: Poikilothermic; homeoviscous adaption
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
5) Transmembrane proteins involved in detecting external chemical signals and relaying that information to the cell are known as ________ proteins.
Answer: receptor
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G2
6) The current understanding of cell membrane structure is based on the ________ model.
Answer: fluid mosaic
Chapter Section: 7.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
7) The ability of some organisms to alter their membrane lipid composition in response to changes in temperature is known as ________.
Answer: homeoviscous adaptation
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
8) Using a radioactive molecule to identify which membrane proteins bind the molecule is called ________.
Answer: affinity labeling
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.1
Global LO: G1
9) Membrane lipids can be separated by ________, based on their ________.
Answer: thin-layer chromatography; hydrophobicity
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G1
10) ________ is a sterol associated with fungal cell membranes.
Answer: Ergosterol
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
11) Most carbon double bonds in naturally occurring unsaturated fatty acids are in the ________ configuration, but some are in the ________ configuration, making the shape and packing of those fatty acids more like that of saturated fatty acids.
Answer: cis; trans
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
12) The presence of trans fats in membranes ________ (increases or decreases) membrane fluidity and ________ (increases or decreases) the membrane transition temperature.
Answer: decreases; increases
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
13) The technique of transferring proteins from SDS-PAGE gels onto a membrane and using labeled antibodies to identify particular proteins is known as ________.
Answer: Western blotting
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G1
14) ________ are small peptide molecules (10–50 amino acids in length) that can disrupt membrane structure.
Answer: Antimicrobial peptides or AMPs
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
15) The structures of porin proteins in bacterial membranes and mitochondrial and chloroplast outer membranes is atypical for multipass integral proteins, because the porin transmembrane region is composed of a ________ instead of ________ transmembrane segments as in bacteriorhodopsin.
Answer: β-barrel (cylindrical β-sheet); α-helical
Chapter Section: 7.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
7.4 Inquiry
1) Hydropathy analysis of a membrane protein’s amino acid sequence predicts the protein’s transmembrane segments and orientation in the membrane. It is necessary to confirm those predictions with experimental analysis. One experimental approach for plasma membrane proteins is to use the protease trypsin to digest extracellular domains of these proteins. (Trypsin is hydrophilic and cannot cross the plasma membrane to enter the cell.) When added to cells, trypsin digests the hydrophilic portions of plasma membrane proteins exposed outside the cell into small fragments but will not have access to transmembrane or cytosolic regions of proteins. Design a general experiment using trypsin and SDS-PAGE to address membrane protein association and orientation, and describe specific expected results that would be consistent with the following predicted membrane protein associations and orientations:
- a singlepass integral membrane protein with only its N-terminus exposed on the extracellular surface
- a singlepass integral membrane protein with only its C-terminus exposed on the extracellular surface
- a multipass integral membrane protein with only three hydrophilic loops exposed on the extracellular surface
- a fatty acylated protein
- a GPI-anchored protein
Answer: Experiment: Treat one set of intact cells with trypsin and use a second set of untreated cells as a control source of the full-length protein. Use SDS-PAGE to separate the total cell proteins or proteins in an isolated plasma membrane fraction from both the treated and untreated cells. Identify and compare the protein of interest and any of it proteolysis fragments in the treated and untreated control samples.
Expected results:
- The undigested region of this protein (transmembrane segment and cytosolic C-terminus) would be shorter and therefore migrate further in the gel in the treated sample than in the control sample, because the trypsin would digest the extracellular N-terminal end.
- The undigested region of this protein (transmembrane segment and cytosolic N-terminus) would be shorter and therefore migrate further in the gel in the treated sample than in the control sample, because the trypsin would digest the extracellular C-terminal end.
- Trypsin would cut the three extracellular loops of this protein, leaving four smaller fragments (migrating further in the gel) in the treated sample.
- The entire fatty acylated protein anchored to the outer monolayer of the plasma membrane is cytosolic and would be completely protected from trypsin proteolysis. This protein would be the same size in the treated and control samples.
- The entire GPI-anchored protein anchored to the outer monolayer of the plasma membrane is exposed extracellularly and would be completely digested into rapidly migrating small fragments in the treated sample but would appear full length in the untreated sample.
Chapter Section: 7.4 and Key Technique
Bloom’s Taxonomy: Evaluation
Learning Outcome: 7.2
Global LO: G1, G8
2) You are a microbiologist working with a new species of bacteria thriving in soil at 15°C. In the lab, you find this bacterium adjusts well to growth at 37°C. If this bacterium were “typical,” what changes would you expect to find in the fatty acid composition of the membrane as the bacteria undergo homeoviscous adaption?
Answer: Expected would be an increase in the average length of the membrane lipid fatty acid chains from 16 to 18 carbons and an increase in their overall saturation (decrease in C=C double bonds), both of which would increase the melting temperature of the membrane. Stearate is an example of a fatty acid that fits both of these criteria and might be found at higher levels following the temperature change.
Chapter Section: 7.3
Bloom’s Taxonomy: Synthesis
Learning Outcome: 7.2
Global LO: G1, G8
3) You are interested in the mobility of two different unknown integral plasma membrane proteins—protein X and protein Y—in a cell. In one cell, you fluorescently label all the X proteins. In another cell, you label all the Y proteins. When you perform FRAP analysis on both cells, you find that most of the protein X fluorescence has recovered within 3 minutes, but recovery of the protein Y fluorescence is significantly slower. Even after 10 minutes, only 50% of the protein Y fluorescence has recovered. Describe the FRAP experimental protocol, explain what the results tell you about mobility of the X and Y proteins, and propose a possible explanation for why the mobilities of the two proteins differ.
Answer: For Fluorescence Recovery After Photobleaching (FRAP), the fluorescence of proteins in a spot on the membrane is bleached with a laser beam and recovery of fluorescence in that spot is measured over time. Recovery occurs when the proteins with photobleached labels diffuse out of the spot and proteins with unbleached labels diffuse in. Rapid recovery of protein X fluorescence indicates that protein X readily diffuses in the plasma membrane with little to no constraint. The slower and incomplete recovery of protein Y fluorescence indicates protein Y diffusion is constrained. Protein Y diffusion could be constrained, because the protein is associated associated with other membrane constituents in large complexes or lipid rafts, or it could be associated with extracellular or intracellular peripheral structures such as the extracellular matrix or the cytoskeleton.
Chapter Section: 7.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 7.2
Global LO: G2
4) Amphipathic cationic antimicrobial peptides (AMPs) have potential as novel therapeutic agents in the treatment of skin bacterial diseases such as leprosy (caused by Mycobacterium leprae). Describe the nature of AMPs, where they can be found in nature, the mechanism by which amphipathic cationic AMPs kill bacteria, and how such AMPS could be developed into a treatment for leprosy.
Answer: AMPs are biologically active 10–50 amino-acid-long peptides, which are naturally produced by a variety of cell and tissue types in invertebrates, plants, and animals as part of the organism’s innate immune system. Amphipathic cationic AMPs are structurally and chemically similar to some detergents. Like those detergents, they dissolve negatively charged membrane lipids, generating holes in bacterial cell membranes that destroy the bacterial cell’s permeability barrier. Because leprosy is associated with the skin, AMPs that disrupt M. leprae membranes could be incorporated into a salve and applied topically.
Chapter Section: 7.3
Bloom’s Taxonomy: Synthesis
Learning Outcome: 7.2
Global LO: G1, G8
5) Recently, consumption of trans fatty acids has been linked to high blood cholesterol levels and increased risk of heart disease. The food industry has responded by decreasing use of “trans fats.” Consider the following questions: (a) How do the structures and transition temperatures of trans, cis, and saturated fatty acids differ, and what are the effects on membrane fluidity? (b) Are trans fatty acids abundant in nature? (c) The major source of trans fats is hydrogenated oils. (The trans fats are produced by a side reaction during the hydrogenation process.) Why has the food industry used synthetically produced hydrogenated oils?
Answer:
- The single carbon bonds in saturated fatty acids are saturated with hydrogen bonds and the fatty acid structure can be straight. The terms trans and cis describe the structures of unsaturated carbon double bonds in the fatty acid. In trans carbon double bonds, the hydrogen atoms are on opposite sides of the C=C bond, resulting in a fatty acid chain that is much straighter than one with cis carbon double bonds, where the hydrogen atoms are on the same side of the C=C bond, forming a rigid kink or bend in the fatty acid. The straighter saturated and trans fatty acids can pack closer together and have higher transition temperatures than do cis fatty acids. At a given temperature, membranes with more cis fatty acids are more fluid than are membranes with more trans or saturated fatty acids.
- Most unsaturated fatty acids found in nature contain cis double bonds.
- The food industry converted some unsaturated fats to saturated fats by converting carbon-carbon double bonds to single bonds in a process called hydrogenation. This process converts oils (liquid at room temperature) to products that are solid at room temperature. These products (such as Crisco® shortening and margarine) are inexpensive to manufacture, have increased shelf life, and have lower levels of saturated fats than butter or lard. Unfortunately, a side reaction in this process also converts some cis double bonds to the trans double bonds that have been linked to heart disease.
Chapter Section: 7.3
Bloom’s Taxonomy: Application
Learning Outcome: 7.2
Global LO: G2, G5, G8
