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Tymoczko’s Biochemistry A Short Course 3Rd Edition by John L. Tymoczko – 
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Chapter 3 Amino Acids

 

 

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. Answers may be used more than once.

  1. a) l amino acids
  2. b) water
  3. c) protons
  4. d) zwitterions
  5. e) arginine
  6. f) serine
  7. g) tyrosine
  8. h) cysteine
  9. i) glutamate
  10. j) histidine
  11. k) proline
  12. l) asparagine
  13. m) d amino acids

 

1. ____________: Chiral type of amino acids found in proteins.
Ans: a
Section: 3.1

 

2. ____________: Another name for dipolar molecules.
Ans: d
Section: 3.1

 

3. ____________: Disulfide bonds are formed by pairs of this amino acid.
Ans: h
Section: 3.2

 

4. ____________: The amino acid with a side-chain pKa just below neutral pH.
Ans: j
Section: 3.2

 

5. ____________: The amino acid with a side group that has a terminal carboxamide.
Ans: l
Section: 3.2

 

6. ____________: The amino acid with an imidazole side chain.
Ans: j
Section: 3.2

 

7. ____________: An amino acid that must be supplied by the diet.
Ans: j
Section: 3.3

 

8. ____________: The amino acid with a negatively charged side chain at neutral pH.
Ans: i
Section: 3.2

 

9. ____________: The amino acid with a sulfhydryl side chain.
Ans: h
Section: 3.2

 

10. ____________: The amino acid with the abbreviation Ser.
Ans: f
Section: 3.2

 

Fill-in-the-Blank Questions

 

11. The amino acid that contains a weakly acidic “phenolic” group is      .
Ans: tyrosine               Section 3.2

 

12.       are amino acids with neutral R groups containing an electronegative atom.
Ans: Polar amino acids            Section 3.2

 

13. The amino acid with the smallest-size side chain allowing greatest flexibility in a protein is      .
Ans: glycine                 Section 3.2

 

14. The charge of glycine when the pH is < 2.0 is      .
Ans: +1                        Section 3.1

 

15. Between the amino and the carboxyl functional group, the       has the lowest affinity for a proton.
Ans: carboxyl              Section: 3.1

 

16. The amino acid with an indol ring is      .
Ans: tryptophan                       Section: 3.2

 

17.        is an amino acid with a hydrophobic side chain containing a thioether.
Ans: Methionine                      Section: 3.2

 

18. The       group is the functional group that makes an amino acid more reactive than nonpolar amino acids such as valine, alanine, and phenylalanine.
Ans: hydroxyl              Section: 3.2

 

19.
Ans: nonessential         Section: 3.3

 

20.       is often seen in a child with a protein-deficient diet.
Ans: Edema                 Section: 3.3

 

Multiple-Choice Questions

 

21. What charged group(s) is/are present in glycine at a pH of 7?
A) –NH3+
B) –COO
C) –NH2+
D) A and B
E) A, B, and C
Ans: D             Section: 3.2

 

22. At a pH of 12, what charged group(s) is/are present in glycine?
A) –NH3+
B) –COO
C) –NH2+
D) A and B
E) A, B, and C
Ans:  B            Section:  3.2

 

23. In what pH range is zwitterionic alanine the predominate structure?
A) 0–2
B) 9–14
C) 8–10
D) 2–4
E) 2–9
Ans:  E     Section 3.2

 

24. Which amino acids contain reactive aliphatic hydroxyl groups?
A) serine and methionine
B) serine and threonine
C) methionine and threonine
D) cysteine and methionine
E) cysteine and threonine
Ans: B             Section: 3.2

 

25. Name three amino acids that are positively charged at a neutral pH.
A) lysine and arginine
B) histidine and arginine
C) cysteine and arginine
D) lysine and proline
E) glutamine and histidine
Ans: A             Section: 3.2

 

26. What would interactions between side chains of aspartate and arginine at neutral pH be?
A) hydrophobic
B) ionic
C) hydrogen bonding
D) steric
E) covalent
Ans: B             Section: 3.2

 

27. Which amino acid has a side chain with a hydroxyl group?
A) serine
B) alanine
C) tryptophan
D) histidine
E) glutamine
Ans: A             Section: 3.2

 

28. Which amino acid has a carboxyl group in its side chain?
A) glutamine
B) galanine
C) cysteine
D) glutamate
E) None of the above.
Ans: D             Section: 3.2

 

29. What would the overall charge of a peptide of the following peptide sequence at pH 1 be (Asp-Gly-Arg-His)?
A) −1
B) 0
C) 1
D) 2
E) 3
Ans: E              Section: 3.2

 

30. Which of the following amino acids would most likely be soluble in a nonpolar solvent such as benzene?
A) valine
B) histidine
C) glutamine
D) glycine
E) All of the above.
Ans: A             Section: 3.2

 

31. Below is a list of five tripeptides identified by their single letter codes. They are listed as A, B, C, D, and E. Which tripeptide contains an amino acid capable of forming covalent disulfide bonds?
A) FNC
B) RGK
C) VIL
D) MDE
E) SYT
Ans: A             Section: 3.2

 

32. Below is a list of five tripeptides identified by their single letter codes. They are listed as A, B, C, D, and E. Which tripeptide is negatively charged at physiological pH?
A) FNC
B) RGK
C) VIL
D) MDE
E) SYT
Ans: D             Section: 3.2

 

33. Below is a list of five tripeptides identified by their single letter codes. They are listed as A, B, C, D, and E. Which tripeptide has the most polar side chains?
A) FNC
B) RGK
C) VIL
D) MDE
E) SYT
Ans: E              Section: 3.2

 

34. Where are Trp and Phe found in a globular protein and why?
A)  exterior due to the hydrophilic effect
B) interior due to the hydrophobic effect
C) exterior forming polar H-bonds with water
D) interior forming ionic bonds with other amino acids
E) exterior forming ionic-polar bonds with water
Ans: B             Section: 3.2

 

35. Amino acids contain all of the following functional groups except:
A) indole.
B) thioester.
C) phenyl.
D) sulfhydryl.
E) amine.
Ans: B             Section: 3.2

 

Short-Answer Questions

 

36. What is the advantage of having multiple functional groups in proteins?
Ans: The rich diversity of functional groups in proteins can contribute independently to protein structure and accounts for the diversity in function as well.
Section: Introduction

 

37. What is the advantage of protein interaction and assembly with other proteins?
Ans: When proteins interact or assemble, new functions and specificity become available. These protein interactions provide multifunctional activity and specificity.
Section: Introduction

 

38. Draw the general structure of an amino acid at pH 7.0 with the side group shown as an “R.”
Ans: The figure should look like either one of the structures shown in the left margin on p. 38.
Section: 3.1

 

39. Why is the central carbon on an amino acid so important?
Ans: This is the chiral center of the molecule and is linked to each important functional group of an amino acid.
Section: 3.1
40. Draw the structure of alanine, aspartic acid, and histidine when the pH is 1.0, 7.0, and 12.0.
Ans: Use the figures in your book and the pKa for each functional group to determine the ionization state for each amino acid.
Section: 3.2

 

41. What is the net charge of each the following amino acid: alanine, aspartic acid, and histidine when the pH is 1.0, 7.0, and 12.0?
Ans: For alanine, the charges are: 1, 0, and −1. For aspartic acid, the charges are: 1, −1, and −2. For histidine, the charges are: 2, 0, and −1.
Section: 3.2

 

42. A gene is mutated so the amino acids glycine and glutamate are now alanine and leucine, respectively. What are the potential results of each of these mutations? Assume that the mutations are not near each other in the primary sequence and have no impact on the other.
Ans: The glycine-to-alanine mutations are similar and will have little or no effect. Glutamate and leucine have very different chemistries and will impact the function and structure of the protein, as one is charged and water soluble, and the other is hydrophobic and nonpolar.
Section: 3.2

 

43. What are the four ways amino acids can be classified?
Ans: hydrophobic, polar, positively charged, and negatively charged
Section: 3.2

 

44. What are the three aromatic amino acids?
Ans: phenylalanine, tyrosine, and tryptophan
Section: 3.2

 

45. Which amino acid side chains are capable of ionization?
Ans: The amino acids are aspartate, glutamate, histidine, cysteine, tyrosine, lysine, and arginine.
Section: 3.2

 

46. Which are the branched amino acids, and what impact do they have on protein shape?
Ans: These are the aliphatic, hydrophobic amino acids, valine, leucine, and isoleucine. They are hydrophobic, which drives the hydrophobic interactions in the interior of a protein. These are also bulky amino acids that will lend to steric strain if forced close to each other in a peptide.
Section: 3.2

 

47. Draw a titration curve for glycine.
Ans: Use the information from Section 2.5 and the graph from Figure 3.2.
Section: 3.2

 

48. What do serine, threonine, and tyrosine have in common?
Ans: Each has a hydroxyl (–OH) group, which makes the first two amino acids more water soluble and increases the reactivity of all three amino acids.
Section: 3.2

 

49. Which amino acid is responsible for stabilizing the structure of a protein by forming pairs of sulfhydryl groups?
Ans: cysteine
Section: 3.2

 

50. What functions make histadine an important amino acid?
Ans: The pKa of the imidazole ring is near physiological pH. This means that the side group may be charged and protonated or neutral and deprotonated. This results in an amino acid that can either lend or accept a proton or charge in the active site of an enzyme.
Section: 3.2

 

51. Which amino acids have a side chain that includes a modified carboxyl group, carboxaminde?
Ans: asparagine and glutamine
Section: 3.2

 

52. Which ionizable group has the lowest affinity for protons: the terminal a-carboxyl group, the aspartic acid side group, or the terminal a-amino group?
Ans: the terminal a-carboxyl group
Section: 3.2

 

53. Malnourished children with Kwashiorkor display a distended stomach, giving the illusion of being full. Why does this happen?
Ans: This is a nutritional state where there is an extremely low or poor protein intake in the diet. The osmolar shift of the blood, which is poor in protein content, causes water to flow into the tissues.
Section: 3.3

 

54. What is the difference between nonessential and essential amino acids?
Ans: The former are amino acids that humans can generate de novo, or from scratch. The latter cannot be made and must be ingested for the mature formation of proteins.
Section: 3.3

 

55. List the essential amino acids.
Ans: histadine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine
Section: 3.3

 

 

Chapter 13         Signal-Transduction Pathways

 

 

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) protein kinase A (PKA)
  2. b) calmodulin
  3. c) fatty acids
  4. d) proto-oncogene
  5. e) R2C2
  6. f) R1C2
  7. g) cAMP
  8. h) phospholipase C
  9. i) protein kinase C (PKC)
  10. j) epinephrine
  11. k) oncogene
  12. l) small G proteins

 

1.1. ____________ The primary messenger responsible for the “fight or flight” response.
Ans: j
Section: 13.2

 

2. The enzyme that becomes active when bound to cAMP is ____________.
Ans: a
Section: 13.2

 

3. ____________ The a and g subunits of heterotrimeric G proteins are anchored to the cell membrane by being covalently linked to these types of molecules.
Ans: c
Section: 13.2

 

4. The ____________ enzyme becomes active when bound to Ca2+ and diacylglycerol.
Ans: i
Section: 13.2

 

5. The inactive form of protein kinase A is ____________.
Ans: e
Section: 13.2

 

6. Ras is a member of the ____________ family of proteins.
Ans: l
Section: 13.3

 

7. ____________ A gene that leads to the transformation of susceptible cell types into cell types with cancer-like characteristics.
Ans: k
Section: 13.6

 

8. The ____________ protein binds to calcium ions and serves as a Ca2+ sensor in eukaryotic cells.
Ans: b
Section: 13.5

 

9. The ____________ molecule functions as a secondary messenger.
Ans: g
Section: 13.2

 

10. ____________ The enzyme that catalyzes the cleavage of PIP2.
Ans: h
Section: 13.2

 

 

Fill-in-the-Blank Questions

 

11. Protein kinase A phosphorylates serine and       residues.
Ans: threonine                         Section: 13.2

 

12.       is the membrane protein that catalyzes the conversion of ATP to cAMP.
Ans: Adenylate cyclase                       Section: 13.2

 

13. The cytosolic side, or β subunit, of the insulin receptor is a     kinase.
Ans: tyrosine                           Section: 13.4

 

14. The     receptor undergoes dimerization and cross-phosphorylation when activated.
Ans: EGF                                 Section: 13.3

 

15.       is a secondary messenger and is abbreviated IP3.
Ans: Inositol 1,4,5-trisphosphate         Section: 13.1

 

16. 7TM is an abbreviation for      receptors.
Ans: seven-transmembrane-helix         Section: 13.2

 

17.       binds to β-andrenergic receptors.
Ans: Epinephrine or adrenaline                       Section: 13.2

 

18. The binding of IP3 to the IP3 receptor results in the release of      from the endoplasmic reticulum.
Ans: calcium ions                    Section: 13.2

 

19. When activated, the insulin receptor results in the mobilization of      transporters to the cell surface.
Ans: glucose                            Section: 13.4

 

20. The      is a calcium-binding unit in many proteins and is characterized by a helix-loop-helix structure.
Ans: EF hand                           Section: 13.5

 

 

Multiple-Choice Questions

 

21. Most signal molecules:
A) easily diffuse through the membrane and bind to a receptor in the cytoplasm.
B) bind to membrane receptors and transmit information across a membrane without traversing the membrane.
C) carry out functions in the nucleus after binding to a receptor in the cell membrane.
D) A and C.
E) A, B, and C.
Ans: B             Section: 13.1

 

22. Examples of second messengers include:
A) cAMP.
B) calcium ion.
C) diacylglycerol.
D) A and B.
E) A, B, and C.
Ans: E              Section: 13.1

 

23. Advantages of second messengers include:
A) the signal can be amplified by making many second messengers.
B) can freely diffuse to other sites within the cell.
C) a few common second messengers can be used in multiple signaling pathways.
D) All of the above.
E) None of the above.
Ans: D             Section: 13.1

 

24. Which of the following amino acids can be phosphorylated?
A) tyrosine, serine, threonine
B) tyrosine, serine, tryptophan
C) serine, threonine, asparagine
D) histidine, serine, phenylalanine
E) tyrosine, methionine, tryptophan
Ans: A             Sections: 13.2 and 13.3

 

25. Which form of the guanyl nucleotide is bound in the unactivated state?
A) GTP
B) GDP
C) GMP
D) dGTP
E) None of the above.
Ans: B             Section: 13.2

 

26. The mechanism by which insulin-signaling processes might be terminated includes:
A) change in temperature.
B) the aggregation of all protein subunits.
C) protein dephosphorylation by phosphatases.
D) All of the above.
E) None of the above.
Ans: C             Section: 13.4

 

27. How does the binding of a hormone to a receptor activate a G-protein?
A) It causes an exchange of GTP for bound GDP.
B) It causes the γ subunit to be released from binding to the β subunit.
C) It causes an exchange of GDP for bound GTP.
D) A and B.
E) None of the above.
Ans: A             Section: 13.1

 

28. Why is bound GTP considered a “clock”?
A) It behaves in specific time intervals.
B) GTP is exchanged for GDP after binding to adenylate cyclase.
C) The Ga receptors have intrinsic GTPase activity, hydrolyzing GTP to GDP and Pi.
D) All of the above.
E) None of the above.
Ans: C             Section: 13.2

 

29. The enzyme responsible for induction of the phosphoinositide cascade is:
A) phospholipase C.
B) phospholipase A.
C) C-dependent protein (CDP).
D) All of the above.
E) None of the above.
Ans: A             Section: 13.2

 

30. What are the two messenger products formed by cleavage of PIP2?
A) diacylglyercol and inositol 1,4,5-triphosphate
B) diacylglyercol and inositol 1,3,5-triphosphate
C) diacylglyercol and inositol 1,3-diphosphate
D) diacylglyercol phosphate and inositol 1,4,5-trisphosphate
E) None of the above.
Ans: A             Section: 13.2

 

31. How is calmodulin activated?
A) by binding of both calcium and potassium
B) by binding Ca2+ when the cytosolic concentration is raised
C) by binding to a positively charged helix on another protein
D) All of the above.
E) None of the above.
Ans: B             Section: 13.5

 

32. Cross-phosphorylation is possible when two receptor proteins with kinase domains
A) are cleaved.
B) form dimers.
C) are internalized into organelles.
D) All of the above.
E) None of the above.
Ans: B             Section: 13.4

 

 

 

 

33. Example(s) of disease(s) caused by altered G-protein activity include
A) whooping cough.
B) cholera.
C) diabetes.
D) A and B.
E) B and C.
Ans: D             Sections: 13.2

 

34. ______________ may be effective anti-cancer drugs.
A) Monoclonal antibodies against offending receptors
B) EGF mimics
C) Protein kinase inhibitors
D) All of the above.
E) Both A and C.
Ans: C             Section: 13.6

 

35. When insulin binds to its receptor, which of the following occurs?
A) A PIP2-dependent kinase is activated.
B) Calmodulin binds to Ca2+.
C) Sos stimulates the exchange of GTP for GDP.
D) All of the above.
E) None of the above.
Ans: A             Section: 13.4

 

36. Which of the five steps in the generalized scheme of transduction pathways is defective in Cushing Syndrome?
A) termination
B) release of primary messenger
C) relay of information by second messenger
D) reception of primary messenger
E) activation of effectors
Ans:     E          Section: 13.2

 

 

37. That Gα subunits have intrinsic GTPase activity is important in which of the five steps in the generalized scheme of transduction pathways?
A) termination
B) release of primary messenger
C) relay of information by second messenger
D) reception of primary messenger
E) activation of effectors
Ans: A             Section: 13.2

 

38. The cleavage of PIP2 is important in which of the five steps in the generalized scheme of transduction pathways?
A) termination
B) release of primary messenger
C) relay of information by second messenger
D) reception of primary messenger
E) activation of effectors
Ans:     C          Section: 13.2

 

 

39. 7TM proteins action is important in which of the five steps in the generalized scheme of transduction pathways?
A) termination
B) release of primary messenger
C) relay of information by second messenger
D) reception of primary messenger
E) activation of effectors
Ans: D             Section: 13.3

 

Short-Answer Questions

 

40. What are some of the common structural features of the receptors to which signal molecules bind?
Ans: The molecule must have a signal-binding site on the extracellular side of the membrane, and must have an intracellular domain. Binding of the ligand to the receptor must induce change into another form (transduced) that affects the shape of the intracellular portion so the signal can be transmitted.
Section: 13.1

 

41. What is a disadvantage of using common molecules for signaling paths?
Ans: When second messengers are involved in more than one signaling pathway, fine tuning and sensitive responses can result. However, when the cross talk becomes inappropriate, the signaling paths and responses will be in error.
Section: 13.1

 

42. What happens when signaling paths are not terminated properly?
Ans: The cell will not be able to respond properly to new stimuli. The errant signals may lead to cancer, uncontrolled cell growth, or other problems.
Section: 13.1

 

43. How many 7TM membranes are there? What are some of their functions?
Ans: There are estimated to be thousands of 7TM receptors. Functions include sensory signals, physiological control, exocytosis, chemotaxis, neurotransmission, cell development and growth, and viral infection. (See Table 13.1 of the textbook for a complete list.)
Section: 13.2

 

44. What is the general mechanism for signal transmission by 7TM receptors?
Ans: The receptors “snake” through a membrane, with domains extending on the extracellular and cytoplasmic sides. A ligand binds to a site on the extracellular side, inducing a conformation change that is detectable on the cytoplasmic side of the cell.
Section: 13.2

 

45 How does binding of epinephrine initiate the cAMP production? Discuss briefly in terms of receptor structure and function.
Ans: Epinephrine binds to a 7TM receptor, which interacts with heterotrimeric G protein, causing it to exchange GDP with GTP. The binding of GTP causes the G protein to dissociate and the active Gα–GTP complex binds to the enzyme adenylate cyclase. Adenylate cyclase is a large membrane-embedded protein, with two large domains located on the inside of the cell. This interaction induces a conformational change in the enzyme, resulting in a more catalytically active form; thus, more cAMP is made. See Figure 13.4 of the textbook for more detail.
Section: 13.2

 

46. How is the hormone-bound activated receptor reset after activation?
Ans: The hormone dissociates, and the receptor returns to its initial, unactivated state. It may bind the hormone, depending on the concentration in the environment.
Section: 13.2

 

47. What are receptor tyrosine kinases? Provide an example.
Ans: These are proteins that bind ligands, such as EGF or insulin, on the extracellular domain. They also contain tyrosine kinase domains on the intracellular side.
Section: 13.3

 

48. Describe how phosphatidylinositol-4,5-diphosphate is converted into two secondary messengers.
Ans: Phosphatidylinositol 4,5-diphosphate (PIP2) is hydrolyzed by phospholipase C to produce diacylglycerol (DAG) and inositol 1,4,5-trisphosphate (IP3). The DAG binds to and activates protein kinase C, and the IP3 binds to a receptor that opens Ca2+ channels allowing Ca2+ to be released from the cellular stores.
Section: 13.2

 

49. What is the difference between a proto-oncogene and an oncogene?
Ans: A proto-oncogene is a normally expressed, unmutated form of a gene. Proto-oncogenes, under normal conditions, tend to regulate cell growth. When a proto-oncogene suffers a mutation that leads to unrestrained growth, it is referred to as an oncogene.
Section: 13.6

 

50. What is meant by an EF hand? Draw or describe the structure.
Ans: The EF hand is a protein that binds calcium ions and contains a calcium ion–binding domain that contains a helix-loop-helix arrangement. The structure resembles a hand, index finger, and thumb extended, with the ion held near the palm by the middle finger. Figure 13.15 of the textbook shows an example of what this structure might look like.
Section: 13.5

 

51. Give the reaction catalyzed by tyrosine kinase.
Ans:
Section: 13.3

 

52. What steps lead from the activation of a cross-phosphorylated receptor tyrosine kinase to an activated small G protein such as Ras?
Ans: The phosphorylated receptor binds an adapter protein called Grb-2, which in turn binds Sos. Sos stimulates the exchange of GTP for GDP in Ras.
Section: 13.3 and Figure 13.16

 

53. What is the difference between heterotrimeric G proteins and small G proteins?
Ans: Heterotrimeric G proteins are composed of αβγ subunits. The α subunit contains the guanyl nucleotide binding site. Upon activation by the signal-receptor event, the GDP is exchanged with a GTP, and the βγ subunits dissociate from the α bound with GTP, which is the form that activates adenyl cyclase. Small G proteins, such as Ras, are single-subunit proteins. They are activated by proteins such as Sos in the EGF signal pathway. The activation causes the exchange of GDP for GTP to convert it into an active kinase.
Sections: 13.2 and 13.3

 

54. Describe the role of phosphoinositol 4,5-bisphosphate (PIP2) in insulin signal transduction.
Ans: PIP2 is one of the signal molecules in the insulin pathway. It is converted into PIP3 by phoshoinositide 3-kinase. The binding of PIP3 activates PIP3-dependent protein kinase, which phosphorylates and activates other kinases in the pathway.
Section: 13.4

 

55. Why are mutated or overexpressed receptor tyrosine kinases frequently observed in tumors?
Ans: Because some small amount of the receptor can dimerize and activate the signaling pathway even with binding to the primary messenger, the overexpression of the receptor increases the likelihood that a “grow and “divide” signal will be inappropriately sent to the cell.
Section: 13.6

Chapter 23   The Calvin Cycle

 

 

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) rubisco
  2. b) transketolase
  3. c) Crassulacean acid metabolism
  4. d) autotrophs
  5. e) C6
  6. f) pyruvate-Pi dikinase
  7. g) hexose monophosphate pool
  8. h) heterotrophs
  9. i) glycolysis
  10. j) C4
  11. k) sucrose
  12. l) 3-phosphoglycerate

 

1. ____________ These organisms can synthesize glucose from carbon dioxide and water.

 

Ans:  d
Section:  Introduction

 

2. ____________ This is the product of CO2 fixation with ribulose-1,6-bisphosphate.

 

Ans:  l
Section:  23.1

 

3. ____________ This is another name for ribulose 1,5-bisphosphate carboxylase/oxygenase.

 

Ans:  a
Section:  23.1

 

4. ____________ This is the pathway that ensures that sufficient amounts of CO2 are available to minimize wasteful photorespiration.

 

Ans:  j
Section:  23.2

 

5. Glucose 1-phosphate, glucose 6-phosphate, and fructose 1-phosphate belong to the ____________.

 

Ans:  g
Section:  23.1

 

6. ____________ This enzyme is involved in both the Calvin cycle and the pentose phosphate pathway.

 

Ans:  b
Section:  23.1

 

7. ____________ This metabolic adaptation is employed by plants living in hot, dry climates.

 

Ans:  c
Section:  23.2

 

8. ____________ This sugar found in plants is readily transported and easily mobilized.

 

Ans:  k
Section:  23.1

 

9. ____________ These organisms obtain energy from chemical fuels.

 

Ans:  h
Section:  Introduction

 

10. ____________ is the final enzyme in the C4 pathway.

 

Ans:  f
Section:  23.2

 

 

Fill-in-the-Blank Questions

 

11. The biochemist who first described the pathway for fixing CO2 is _____.
Ans: Melvin Calvin   Section:  Introduction

 

12. The enzymes that catalyze the dark reactions of photosynthesis are located in the _____of the chloroplasts.
Ans:  stroma     Section:  23.1

 

13. The first step of the Calvin cycle involves the addition of CO2 to _____ to produce two molecules of 3-phosphoglycerate.
Ans:  ribulose-1,5-bisphosphate     Section:  23.1

 

14. The binding site of Mg2+ to rubisco involves the formation of a _____ group between lysine 221 and CO2.
Ans:  carbamate     Section:  23.1

 

15. In photorespiration oxygen is consumed and _____ is released.
Ans:  carbon dioxide     Section:  23.1

 

16.  The three stages of the Calvin cycle are _____.
Ans: carbon fixation, reduction of fixed carbon, and regeneration of ribulose 1.5 bisphosphate Section:  23.1
17. In C4 plants, carbon dioxide is added to _____ to form oxaloacetate, which is reduced to malate, which carries CO2 to the bundle-sheath cells.
Ans:  phosphoenolpyruvate     Section:  23.2

 

18. Volcanoes, which spew millions of tons of gasses and particles into the atmosphere, cause a(an) (decrease/increase) _____ in photosynthesis specifically during the daytime.
Ans: increase   Section:  23.1

 

19. _____is an example of a mechanism of accelerating photosynthesis by increasing carbon dioxide concentration.
Ans: C4 or hack-slack pathway   Section:  23.2

 

20. Plants adapt to arid ecosystems using _____.
Ans: crassuacean acid metabolism   Section:  23.2

 

 

Multiple-Choice Questions

 

21. ATP is called the energy currency. The currency of biosynthetic reducing power is
A) NADPH.     B) CoA.     C) AMP.     D) ADP.     E) None of the above.
Ans:  A     Section:  Introduction

 

22. What is the source of carbons for the Calvin cycle?
A) glucose D) glyoxylate
B) carbon dioxide E) None of the above.
C) glycogen
Ans:  B     Section:  23.1

 

23. Plants store glucose as ___________ and ____________.
A) starch; sucrose
B) fructose; sucrose
C) starch; fructose
D) All of the above.
E) None of the above.
Ans:  A     Section:  23.1

 

24. The most abundant protein on Earth is:
A)  ribulose 1.5-bisphosphate.
B)  aldolase.
C)  rubisco.
D) phosphopentose epimerase.
E) transketolase.
Ans:  C     Section:  23.1

 

 

25. In the Calvin cycle, 3-phosphoglycerate is converted into which hexose phosphate?
A) glucose 1-phosphate D) All of the above.
B) glucose 6-phosphate E) None of the above.
C) fructose 6-phosphate
Ans:  D     Section:  23.1

 

26. Which form of thioredoxin activates certain Calvin cycle enzymes?
A) reduced     B) oxidized     C) dimeric     D) A and C     E) B and C
Ans: A     Section:  23.2

 

27. Which coenzyme is required by glyceraldehyde 3-phosphate dehydrogenase in chloroplasts to convert 3-phosphoglycerate into glyceraldehyde-3-phosphate?
A) NADH D) NADP+
B) NADPH E) thiamine pyrophosphate
C) NAD+
Ans:  B     Section:  23.1

 

28. The C4 pathway is necessary in tropical plants because:
A) at high temperatures, the plants cannot maintain sufficient water levels within the cells.
B) tropical plants do not have proper day/light cycles to maintain the balance of CO2 necessary for carbohydrate storage.
C) at high temperatures, the oxygenase activity of rubisco is high.
D) All of the above.
E) None of the above.
Ans:  C     Section:  23.2

 

29. What is the energy cost of the C4 pathway?
A) Thirty molecules of ATP are used per hexose molecule made.
B) Eighteen molecules of ATP are used per hexose molecule made.
C) Thirty molecules of NADPH are used per hexose molecule made.
D) Eighteen molecules of ATP and 12 of NADPH are used per hexose molecule made.
E) None of the above.
Ans:  A     Section:  23.2

 

30. Rubisco is found:
A) in the meso sheath bundles.
B) in the stroma of the chloroplasts.
C) embedded in the plant cell wall.
D) in the lumen of the thylakoid disc.
E) None of the above.
Ans:  B     Section:  23.1

 

31. The enzyme that catalyzes the rate-limiting step in hexose synthesis is:
A) aldolase.
B) transketolase.
C) rubisco.
D) phosphosepentose isomerase.
E) None of the above.
Ans:  C     Section:  23.1

 

32. ________ rounds of the Calvin cycled are required for the production of hexose.
A)  One
B)  Three
C) Six
D) Twelve
E) Nine
Ans:  C     Section:  23.1

 

 

33. The immediate unwanted side reaction of rubisco is:
A)  phosphoglycolate .
B)  ribulose 1,5-bisphosphate.
C)  glycerol.
D) amylose.
E) phosphofuctose.
Ans:  A     Section:  23.2

 

34. The energetic equivalent of ______ ATP molecules is(are) consumed in transporting CO2 to the chloroplasts of the bundle-sheath cells.
A) 1
B) 2
C) 3
D) 4
E) 6
Ans:  B     Section:  23.2

 

35. The protein that regulates the Calvin cycle is:
A)  protein kinase C.
B)  rubisco .
C)  thioredoxin.
D) ferredoxin-thioredoxin reductase.
E) None of the above.
Ans:  C     Section:  23.2

 

36. On a molecular level, how might global warming affect carbon dioxide fixation?
A) Carbon dioxide, being a denser gas than oxygen, will cause more plants to sequester CO2 in mesophyll cells.
B) Increase in temperatures leads to increase in water loss in leafy plants compared to succulents, thus, the light reactions are inhibited.
C) Rubisco’s oxygenase activity decreases with an increase in temperature, requiring more plants to use the C3 cycle.
D) Ribusco’s oxygenase activity increases with an increase in temperature, requiring more plants to use the C4 pathway.
E) Ribusco’s carboxylase activity increases with an increase in temperature, requiring more plants to use the C3 pathway.
Ans:   D    Section:  23.2

 

37. Desert plants prevent loss of water vapor by closing stomata during the heat of the day and opening them at night. How does this affect the movement of CO2 and what are the implications for CO2 fixation?
A) Stomata, like cellular transporters can be selective and bent water molecules enter through different stomata than linear CO2 and O2; thus, desert plant CO2 fixation is not affected.
B) CO2, being a non-polar molecule moves easily thought cell membranes, thus, the opening and closing of stomata does not affect CO2 fixation.
C) Under normal conditions, the light reactions don’t work at maximum saturation, therefore, a diurnal pattern of H2O decreases CO2+ fixation only minimally.
D) CO2 entry into the plant is also inhibited by the closing of the stomata; however, the impact on CO2 fixation is minimal.
E) CO2 entry into the plant is also inhibited by the closing of the stomata; however, CO2 is sequestered in vacuoles in the form of malate.
Ans:  E     Section:  23.1

 

38. Knowing what you do about the distribution of the light reaction enzymes, where would you expect to find Calvin cycle enzymes and why?
A) lumen of the thylakoid membrane, as this is the location of the splitting of water
B) lumen of the thylakoid membrane, as this is the location of NADPH and ATP synthesis
C) thylakoid integral membrane proteins that derive energy from proton pumping
D) stromal side of thylakoid membrane, as this is the location of NADPH and ATP formation
E) stromal side of the thylakoid membrane, as this is the location of the splitting of water
Ans:   D    Section:  23.1

 

39. The mechanics of CO2 binding involves Mg2+ and Lys 201. What would you expect to be the pH optimum and [Mg2+] for this to occur?
A) high pH, high [Mg2+]
B) high pH, low [Mg2+]
C) low pH, high [Mg2+]
D) low pH, low [Mg2+]
E) neutral pH, low [Mg2+]
Ans:   A    Section:  23.2

 

Short-Answer Questions

 

40. Why are reactions of the Calvin cycle called the “dark reactions”?
Ans: Unlike photosynthetic reactions, the reactions of the Calvin cycle do not depend on light in order to proceed.
Section:  Introduction

 

41. Describe the stages of the Calvin cycle.
Ans: The stages of the Calvin cycle include:

1.      the fixation of carbon dioxide by reaction with ribulose 1,5-bisphosphate to form two molecules of 3-phosphoglycerate

2.      the formation of hexose sugars from 3-phosphoglycerate

3.      the regeneration of ribulose 1,5-bisphosphate

Section:  23.1

 

42. Which enzyme is cited as the most abundant enzyme in the biosphere? Why is this so?
Ans: Rubisco (for 1,5-bisphosphate carboxylase/oxygenase) makes up 22% of the chloroplast total protein, and is probably the most common protein in the biosphere. This is partially due to the slow catalytic activity of the enzyme, which means larger numbers of rubisco are necessary to carry out the reaction at sufficient levels.
Section:  23.1

 

43. What is photorespiration? Why is it wasteful?
Ans: Normal photosynthesis produces O2 and fixes CO2 into carbohydrate.  Photorespiration occurs when O2 is used and CO2 is produced. It is a result of oxygen combining with Ru-1,5-BP instead of CO2.  Phosphoglycolate is produced instead of one of the 3-phosphoglycerates.  Phosphoglycolate is not a versatile metabolite, and a salvage pathway recovers three carbons from two glyoxylates.  The fourth carbon is released as CO2.  The pathway is wasteful because there is no ATP produced, nor is any other molecule produced that can serve as an energy source.
Section:  23.2

 

44. Give the sequence of reactions involved in the synthesis of sucrose from two glucose 6-phosphate.
Ans:
Section:  23.1

 

45. Why is the chemistry of the “dark reactions” referred to as a cycle and not a pathway?
Ans: The Calvin cycle begins with the addition of CO2 to ribulose-1,5-bisphosphate and the last step of the process produces ribulose-1,5-bisphosphate.  The starting compound is regenerated and acts catalytically.  Thus, it is considered a cycle and not a pathway.
Section:  23.1

 

46. Explain/describe the stoichiometry of the Calvin cycle to convert three molecules of CO2 into one molecule of dihydroxyacetone phosphate.
Ans: Use Figure 23.5 to describe how CO2 is fixed to ribulose 1,5-bisphosphate to generate 3-phosphoglycerate. Then, 3-phosphoglycerate undergoes phosphorylation to 1,3-bisphosphoglycerate followed by reduction to glyceraldehyde-3-phosphate. Finally, using triosephosphate isomerase, glyceraldehyde-3-phosphate is converted to dihydroxyacetone phosphate.
Section:  23.1

 

47. What is the energy requirement for the formation of hexose from CO2 in terms of ATP equivalents and NADPH?
Ans: It requires six molecules of CO2 to make one hexose phosphate.  Each round of the Calvin cycle consumes two ATP and two NADPH.  So, 12 molecules each of ATP and NADPH are consumed.  Also, the regeneration of the 1,5-bisphosphate requires an additional six molecules of ATP.  Thus a total of 18 equivalents of ATP and 12 NADPH are required to convert six CO2 into a hexose.
Section:  23.1

 

48. How is rubisco activity controlled?
Ans: The presence of the carbamate at lysine 210 is critical for magnesium ion coordination and enzyme activity. The carbamate is most likely to be formed at high pH, which occurs during light induced proton pumping from the stroma into the thylakoid. This pumping also causes an increase in magnesium ion concentration in the stroma, which also stimulates formation of the carbamate.
Section:  23.1

 

49. Describe the role of thioredoxin in regulating the Calvin cycle.
Ans: The light reactions of photosynthesis result in the reduction of ferredoxin, one of the electron carriers of photosynthesis.  Then ferredoxin-thioredoxin reductase catalyzes the reduction of sulfhydryl groups of thioredoxin by reduced ferredoxin.  The reduced thioredoxin then transfers electrons to the component enzymes that contain regulatory disulfide bonds.  The inactive enzymes are activated by the reduction of the disulfide bonds.  The enzymes are converted to the inactive form by spontaneous oxidation by oxygen, a product of photosynthesis.
Section:  23.2

 

50. Write the balanced reaction for the Calvin cycle.
Ans: 6 CO2 + 18 ATP + 12 NADPH + 12 H2O → C6H12O6 + 18 ADP + 18 Pi + 12 NADP+ +  6H+
Section:  23.1

 

51. Describe, in biochemical terms, how bread becomes stale.
Ans: During the baking process, starches in bread are hydrated and become what is called “amorphous starch structure.” As bread cools and is left exposed to air, the water in the amorphous starch structure evaporates leaving crystalline amylose and amylopectin.
Section:  23.1

 

52. Explain how the relationship of the mesophyll cell and the bundle-sheath cell help to regulate the Calvin cycle.
Ans: The enzyme rubisco is inefficient and requires a high concentration of CO2 for fixing to occur. C4 plants produce Malate in mesophyll cells to be used as CO2 by the Calvin cycle in the bundle-sheath cells, thus building up the concentration of CO2 for fixation.
Section:  23.2

 

53. How does light alter the fixation of CO2?
Ans: Rubisco is activated when plants are exposed to light. In the dark, the magnesium is primarily located in the thylakoid and until light-driven pumping of protons, magnesium ions are not available in the stroma for the carbamation of lysine 201 of rubisco.
Section:  23.2

 

54. What is the basic difference between C4 and C3 plants?
Ans: C3 plants comprise the majority of plants on Earth and C4 plants are found in drier climes.  C3 is called C3 because carbon dioxide is fixed into a three-carbon sugar, where C4 pathways fix CO2 into a four-carbon sugar.  C3 plants use rubisco to directly fix the cells where C3 uses a second pathway (PEP carboxylate) to fix CO2.
Section:  23.1 and 23.2

 

55. Rubisco’s Km for CO2 is 50 times higher than for O2 but kcat values are similar. This suggests that at the level of oxygenase activity would be negligible. Why then is it not? [O2] vs. [CO2].
Ans: The answer lies in the oxygen concentration vs. carbon dioxide concentration in the atmosphere. Carbon dioxide is a trace gas with a concentration of 0.04% by volume. Oxygen, on the other hand, is 21% in air by volume.
Section:  23.

 

Chapter 33         The Structure of Informational Macromolecules: DNA and RNA

 

 

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) supercoiling
  2. b) DNA
  3. c) 5¢ → 3¢
  4. d) circular
  5. e) stem loop
  6. f) B-DNA
  7. g) retroviruses
  8. h) exon
  9. i) Erwin Chargaff
  10. j) histones
  11. k) RNA
  12. l) Rosalind Franklin
  13. m) A-DNA
  14. n) 3¢ → 5¢

 

1. The stable genetic information passed on from one generation to the next is _______________.
Ans: b
Section: Introduction

 

2. _______________ A transient copy of genetic information.
Ans: k
Section: Introduction

 

3. The scientist who provided data for Watson and Crick’s model of DNA is _______________.
Ans: l
Section: 33.2

 

4. The  ____________ form is more accessible for interactions with proteins.
Ans: f
Section: 33.3

 

5. The axis of a double helix can be twisted by a mechanism called _______________.
Ans: a
Section: 33.3

 

6. DNA can exist in both linear and _______________ forms.
Ans: d
Section: 33.3

 

7. _______________ Form of DNA that exists under dehydrating conditions.
Ans: m
Section: 33.3

 

8. The base sequence, as written by convention, is in the _______________ direction.
Ans: c
Section: 33.1

 

9. _______________ is the simple motif of RNA structure.
Ans: e
Section: 33.5

 

10. H2A, H2B, H3, and H4 are all major classes of _______________.
Ans: j
Section: 33.4

 

 

Fill-in-the-Blank Questions

 

11. A nitrogen containing an aromatic base attached to a ribose is a _____.
Ans: nucleoside                                                                       Section: 33.1

 

12. 5¢ dAMP refers to _____.
Ans: deoxyadenosine 5¢ monophosphate                                Section: 33.1

 

13. The distribution of parental atoms in newly synthesized DNA is called _____ replication.
Ans: semiconservative                                                 Section: 33.2

 

14. There are approximately _____ bases per turn of a B-DNA double helix.
Ans: 10                                                                                    Section: 33.2

 

15. The _____ effect stabilizes the structure of DNA, resulting in more polar surfaces being exposed to the aqueous media.
Ans: hydrophobic                                                        Section: 33.2

 

16. The nature of replication was determined using DNA labeled with nitrogen isotopes and _____.
Ans: density-gradient centrifugation                Section: 33.2

 

17. The nucleoside that is composed of a d-ribose linked to a cytosine base is called _____.
Ans: cytidine                                                               Section: 33.1

 

18. _____ is a left-handed double helix.
Ans: Z-DNA                                                                Section: 33.3

 

19. Unwinding of naturally occurring plasmid, circular DNA results in _____.
Ans: supercoiling                                                         Section: 33.3

 

20. Proteins that bind to DNA are rich the amino acids _____and _____.
Ans: arginine; lysine                                                    Section: 33.4

 

 

Multiple-Choice Questions

 

21. The difference in RNA bases compared to DNA bases is
A) RNA contains A instead of T.
B) RNA contains U instead of G.
C) RNA contains U instead of T.
D) RNA contains A instead of U.
E) None of the above.
Ans: C             Section: 33.1

 

22. How does a nucleotide differ from a nucleoside?
A) Nucleosides are found in DNA, whereas nucleotides are found in RNA.
B) Purines are only found in nucleotides.
C) Nucleosides contain only deoxyribose sugars.
D) A nucleotide is a nucleoside with a phosphate ester linked to the sugar.
E) None of the above.
Ans: D             Section: 33.1

 

23. Replication takes place in which manner?
A) conservative
B) random
C) semiconservative
D) N-linked
E) None of the above.
Ans: C             Section: 33.2

 

24. How many different sequence possibilities are there for a DNA polymer that is 10 bases long?
A) 262,13232
B) 320
C) 32,000
D) 1,048,576
E) 100,000
Ans: D             Section: 33.1

 

25. The shorthand notation pApCpG denotes that
A) a phosphate is attached to the 5¢ of the adenosine nucleotide unit.
B) a phosphate is attached to the 3¢ of the guanine nucleotide unit.
C) the cytosine nucleotide has a free hydroxyl group.
D) A and C.
E) All of the above.
Ans: A             Section: 33.1

 

26. The feature(s) of DNA deduced by Watson and Crick include
A) two antiparallel polynucleotide chains coiled in a helix around a common axis.
B) that the pyrimidine and purine bases lie on the inside of the helix.
C) that the bases are nearly perpendicular to the axis.
D) All of the above.
E) None of the above.
Ans: D             Section: 33.2

 

27. The chemical forces that contribute to the stability of the DNA due to the base stacking present in the DNA helix are
A) hydrogen bonds.
B) van der Waals interactions.
C) disulfide bonds.
D) C and B.
E) None of the above.
Ans: B             Section: 33.2

 

28. What is the approximate error rate in DNA replication?
A) 1 ´ 10−8
B) 1 ´ 10−6
C) 1 ´ 10−15
D) 1 ´ 10−10
E) None of the above.
Ans: A             Section: 33.3

 

29. The form of DNA that is the narrowest but having the longest pitch per turn of helix is the
A) A form.
B) B form.
C) C form.
D) Z form.
E) None of the above.
Ans: D             Section: 33.3

 

30. The temperature at which half of the DNA helical structure is lost is called the
A) denaturation temperature.
B) Tm.
C) annealing temperature.
D) dehybridization temperaure.
E) replication temperature.
Ans: B             Section: 33.2

 

31. After two generations of replication in the Meselson and Stahl experiment, what was the composition of the two bands?
A) One band was all 14N and one ba nd was all 15N.
B) One band was all 14N and one band was half 14N and half 15N.
C) One band was all 15N and one band was half 14N and half 15N.
D) One band was all 14N and one band was one quarter 14N and three quarters 15N.
E) One band was all 15N and one band was one quarter 14N and three quarters 15N.
Ans: B             Section: 33.3

 

32. The process of rehybridizing melted DNA is called
A) denaturation.
B) Tm.
C) annealing.
D) dehybridization.
E) replication.
Ans: C             Section: 33.2

 

33. Unwinding prior to ligation results in
A) relaxed DNA.
B) a wider major groove.
C) negative supercoiling.
D) positive supercoiling.
E) None of the above.
Ans: C             Section: 33.3

 

34. The anticancer drug cisplatin disrupts
A) histone binding to DNA.
B) supercoiling of DNA.
C) chromatin–DNA binding.
D) replication and transcription.
E) None of the above.
Ans: D             Section: 33.4

 

35. RNA
A) is oxygenated on the 2¢ of ribose.
B) can form non-Watson–Crick base pairings.
C) uses metals to stabilize folding.
D) may form triplet base pairings.
E) All of the above.
Ans: E              Section: 33.5

 

36. The mole-fraction composition of one of the strands of a double-helical DNA molecule is [T] = 0.22 and [C] = 0.28.  What can you say about [A], [G], [T], and [C] of the complementary strand?
A) [A] = 0.22, [G] = 0.28, [T] = 0.50 – [C], [C] = 0.50 – [T]
B) [A] = 0.22, [G] = 0.28, [T] = 0.28, [C] = 0.22
C) [A] = 0.22, [G] = 0.28, [T] = 0.22, [C] = 0.28
D) [A] = 0.50 – [G], [G] = 0.50 – [A], [T] = 0.22, [C] = 0.28
E) Not enough information is given to determine these concentrations.
Ans:     A         Section: 33.2

 

37. You perform melting experiments on double stranded DNA, starting at low salt concentrations (~0.2 M NaCl) and then increasing the salt concentration to about 0.6 M NaCl. How does salt concentration affect the melting temperature of the DNA?
A) Salt at these low concentrations have no effect on Tm.
B) Increasing salt causes a decrease in Tm due to the stabilizing actions of the salt.
C) Increasing salt causes an increase in Tm due to the destabilizing actions of the salt.
D) Increasing salt causes a decrease in Tm due to the destabilizing actions of the salt.
E) Increasing salt causes an increase in Tm due to the stabilizing actions of the salt.
Ans:     E          Section: 33.2

 

38. What makes DNA so much more stable than RNA in the presence of a basic solution?
A) The negative charge on the phosphate repels the negative charged base.
B) Histones block access to all but a few nucleotides that act as linkers.
C) The 2¢-H in DNA is not reactive, whereas the 2¢OH of RNA is under basic conditions.
D) DNA precipitates in basic solutions making it unreactive.
E) RNA forms elaborate structures that are susceptible to alkaline digestion.
Ans:     C          Section: 33.5

 

39. Which of the following explain why RNA forms unique structures not found in DNA?
A) RNA can wrap around itself to form supercoiled structures but not DNA.
B) RNA contains hydrogen-bond donors and acceptors that are not normal participants in Watson-Crick base pairs.
C) Histones can direct the folding of RNA into unique structures, as it does with DNA.
D) The uracil base in RNA allows for additional hydrogen bonding to stabilize unique structures.
E) Unique structures in RNA are due to mismatches because there is no proofreading in RNA synthesis.
Ans:     B          Section: 33.5

 

40. The mole-fraction composition of a strand of a RNA molecule is [U] = 0.19 and [C] = 0.33.  What can you say about the [A] and [G] of this RNA?
A) [A] = 0.19, [G] = 0.33
B) [A] = 0.33, [G] = 0.19
C) [A] + [G} = 0.52
D) [A] + [G} = 0.48
E) There is not enough information to determine these concentrations.
Ans: D              Section: 33.4

 

 

 

Short-Answer Questions

 

41. Draw two nucleotides in DNA, showing the linkage between the sugars.
Ans: The 3¢ oxygen of one nucleotide is linked to a phosphorous atom, which is linked to the oxygen on the 5¢ carbon on the next sugar. Two extra oxygens should be attached to the phosphorous atom, and the phosphate should be shown ionized at neutral pH. There should be no oxygen at the 2¢ position of the sugar.
Section: 33.1 and Figure 33.4

 

42. What advantage do phosphodiesters have compared to other esters?
Ans: The phosphodiester negative charge serves to repel nucleophilic species such as OH. Thus, the phosphodiester linkage is more stable because it is more resistant to hydrolytic cleavage.
Section: 33.1

 

43. Describe the DNA helix proposed by Watson and Crick.
Ans: The DNA model proposed by Watson and Crick contains two DNA polymer strands coiled around a common axis. The strands are oriented in opposite directions, from 5¢ to 3¢ (antiparallel), and the two strands are twisted in a right-handed coil. The sugar–phosphate backbone is located on the outside. The base pairs are stacked in the center of the helix and are stabilized by specific hydrogen bonds between AT and GC base pairs. The base pairs lie perpendicular to the axis, with 10 base pairs per turn. The helical structure repeats every 34 Å.
Section: 33.1

 

44. Draw the hydrogen-bonded base pairs of A to T and show why A does not hydrogen bond to C.
Ans:
Section: 33.1

 

45. Briefly describe the Meselson and Stahl experiment that indicated that DNA replication is semiconservative.
Ans: Meselson and Stahl grew bacteria in the presence of 15N, or “heavy” nitrogen. They rapidly replaced the 15N media with 14N, or “light” nitrogen. DNA was extracted at various intervals during the growth of the bacteria, representing different generations. Meselson and Stahl examined the DNA using density-gradient centrifugation and observed that no “heavy” DNA was present in the first generation, but that the DNA was intermediate between light and heavy. After the second generation, there were equal amounts of intermediate and light. This confirmed that one strand of the parent DNA is present in each daughter strand of replicated DNA.
Section: 33.2

 

46. Describe the structure of chromatin.
Ans: Chromatin is the packaged form of eukaryotic DNA. The DNA is tightly bound to a group of small basic proteins called histones. The DNA and all of its associated proteins are collectively called chromatin.
Section: 33.4

 

47. What are topoisomers of DNA?
Ans: Topoisomers, or topological isomers, are molecules of DNA that have the same sequence but different structures due to differences in coiling. A strand of unwound DNA is a topoisomer of supercoiled DNA of the same sequence.
Section: 33.3

 

48. What role does a nucleosome play in DNA structure?
Ans: A nucleosome consists of approximately 200 base pairs of DNA wound around eight histone proteins. Between nucleosomes is a stretch of DNA called linker DNA. Nucleosomes are the repeating units of DNA structure in chromatin.
Section: 33.4 and Figure 33.25

 

49. Why is there a major and a minor groove in DNA?
Ans: The major and minor grooves arise because of the orientation of the glycosidic bonds of each base pair. These glycosidic bonds are not diametrically opposite each other on the backbone, resulting in each base pair having a large side and a small side. The large side defines the major groove and the small side defines the minor groove.
Section: 33.3

 

50. Compare and contrast the three forms of DNA with respect to handedness and hydration state.
Ans: The Watson and Crick model of DNA is known as B-DNA and is a right-handed helix. A-DNA is also a right-handed helix, but is a less hydrated form. Z-DNA is a left-handed helix characterized by a zigzag arrangement of the backbone phosphoryl groups.
Section: 33.3

 

51. Describe, in simple terms, the hallmarks of DNA structure.
Ans: DNA consists of two chains of polynucleotides, paired via hydrogen bonds, but running in opposite directions in a right-handed helical form around a central axis. The bases are found on the inside of the helix, with the phosphates and sugars on the outside. The bases are held together by hydrogen bonds: adenine and thymine by two hydrogen bonds, and guanine and cytosine by three hydrogen bonds. The bases are perpendicular to the axis.
Section: 33.1