Biochemistry 7Th Ed By Berg – Test Bank

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Biochemistry 7Th Ed By Berg – Test Bank

Chapter 6   Exploring Evolution and Bioinformatics

 

 

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) BLAST
  2. b) glutamine to asparagine
  3. c) orthologs
  4. d) alignment
  5. e) convergent evolution
  6. f) paralogs
  7. g) gap
  8. h) evolutionary tree
  9. i) sequence templates
  10. j) combinatorial chemistry
  11. k) glycine to tryptophan
  12. l) conservative

 

1. ____________ Homologs that perform similar or identical functions.

 

Ans: c
Section:  6.1

 

2. ____________ Homologs from the same organism that perform different functions.

 

Ans: f
Section:  6.1

 

3. An amino acid substitution to another of similar size and charge is called a ____________ substitution.

 

Ans: l
Section:  6.2

 

4. ____________ Regions of a protein, critical to three-dimensional structure, that are conserved.

 

Ans: i
Section:  6.3

 

5. ____________ A database search procedure used to find homologous sequences in proteins.

 

Ans: a
Section:  6.2

 

6. ____________ The processes by which different evolutionary paths end up at a similar structure or mechanism.

 

Ans: e
Section:  6.3

 

7. ____________ Used to present a graphical depiction of how sequences are related in evolutionary terms.

 

Ans: h
Section:  6.4

 

8. ____________ is a technique that produces large populations of molecules and selects for a biochemical property.

 

Ans: j
Section:  6.5

 

9. ____________ An example of a conservative amino acid change.

 

Ans: b
Section:  6.2

 

10. In some sequence alignments it is necessary to include a ____________.

 

Ans: g
Section:  6.2

 

 

Fill-in-the-Blank Questions

 

11. __________________ are homologous molecules that are found in different species and have similar or identical functions.
Ans:  Orthologs     Section:  6.1

 

12. __________________ are homologous molecules that are found in a given species but possess different biological functions.
Ans:  Paralogs     Section:  6.1

 

13. ________________ amplification of well-preserved samples allows the determination of nucleotide sequences from extinct organisms.
Ans:  PCR     Section:  Introduction and 6.5

 

14. Two molecules are said to be _________________ if they have been derived from a common ancestor.
Ans:  homologous     Section:  6.1

 

15. In the Blosum-62 substitution matrix, a large ________________ score corresponds to substitution that occurs only rarely.
Ans:  negative     Section:  6.2

 

16. Analysis of substitution matrices indicates that cysteine (C) and __________________ residues tend to be conserved more than other amino acid residues.
Ans:  tryptophan (W)     Section:  6.2

 

17. The process by which very different evolutionary pathways lead to a protein with similar function is referred to as _________________ evolution.
Ans:  convergent     Section:  6.3

 

18. ___________________ can be detected in proteins by attempting to align a given sequence with itself, internally.
Ans: Repeated motifs or Domains     Section:  6.3

 

19. Evolutionary trees are constructed with the assumption that the number of sequence differences is related to the time since the two sequences ________________________.
Ans:  diverged     Section:  6.4

 

20. Comparison of three-dimensional structures indicated that heat shock protein 70 is a paralog of ____________, despite only 15.6% sequence identity.
Ans:  actin     Section:  6.3

 

 

Multiple-Choice Questions

 

21. Protein sequence comparisons can provide estimates of
A) protein function. D) All of the above.
B) protein shape. E) b and c only.
C) pathways of evolutionary descent.
Ans:  E     Section:  Introduction

 

22. The definition of a homolog is
A) molecules related by sequence similarity.
B) two molecules derived from a common ancestral protein.
C) molecules with similar domains.
D) All of the above.
E) None of the above.
Ans:  B     Section:  6.1

 

23. Paralogs differ in
A) detailed biochemical functions. D) All of the above.
B) ancestral evolution. E) None of the above.
C) ornithology.
Ans:  A     Section:  6.1

 

24. An example of a conservative substitution would be
A) Ala to Trp. D) All of the above.
B) Gly to Ser. E) None of the above.
C) Asp to Glu.
Ans:  C     Section:  6.2

 

25. An open reading frame is necessary to compare
A) DNA promoter sequences.
B) alignments of potential protein coding regions.
C) cDNA alignments.
D) All of the above.
E) None of the above.
Ans:  B     Section:  6.2

 

26. Based on their three-dimensional structures, actin and Hsp-70 are considered
A) homologs. D) All of the above.
B) orthologs. E) None of the above.
C) paralogs.
Ans:  C     Section:  6.3

 

27. What percentage of proteins contain two or more similar domains?
A) 10%     B) 36%     C) 8%     D) All of the above.     E) None of the above.
Ans:  A     Section:  6.3

 

28. An example of proteins that evolved similar mechanisms by convergent evolution is
A) DNA polymerase and elastase. D) b and c.
B) chymotrypsin and subtilisin. E) None of the above.
C) neuramidase and glycolase.
Ans:  B     Section:  6.3

 

29. Which of the following sequences would retain the base-pairing necessary to form a hairpin loop? The original sequence is UUGCUCAGUAAGAGCAA.
A) UAGCUCAGUAAGAGCUA D) All of the above.
B) UACCUCAGAGAGCUA E) None of the above.
C) UUGUUUCAGAGAGCAA
Ans:  A     Section:  6.3

 

30. Evolutionary trees provide relative scales of divergence. How can these be reconciled with real times?
A) by comparing the amount of sequence divergence
B) by comparing proteins from convergent evolution
C) by using fossil records when possible
D) All of the above.
E) None of the above.
Ans:  C     Section:  6.4

 

31. Which of the following molecules is the most stable?
A) mRNA D) rRNA
B) hemoglobin E) mitochondrial DNA
C) carbohydrates
Ans:  E     Section:  6.5

 

32. An example of paralogs would include
A) hemoglobin and myoglobin. D) All of the above.
B) actin and keratin. E) None of the above.
C) DNA polymerase and trypsin.
Ans:  A     Section:  6.1

 

33. In a combinatorial study of RNA described in the text, what functional feature was of interest?
A) ATP binding D) All of the above.
B) loop formation E) None of the above.
C) replication ability
Ans:  A     Section:  6.5

 

34. A population of molecules in which it is easy to study evolutionary processes using combinatorial chemistry.
A) proteins D) All of the above.
B) nucleotides E) None of the above.
C) lipids
Ans:  B     Section:  6.5

 

35. A BLAST search provides the following information.
A) three-dimensional motifs

B) a list of sequence alignments

C) an estimate that an alignment occurred by chance

D) All of the above.

E) b and c only

Ans:  E     Section:  6.2

 

 

Short-Answer Questions

 

36. Why are protein comparisons of three-dimensional shape more revealing than primary sequences?
Ans: The amino acid sequence can provide clues to a protein’s function and its relationship to other proteins. However, the shape can provide more information about the arrangement of the amino acids in space. This information is critical to understanding protein function. Comparison of structure can reveal other relationships that may not be apparent from the sequence alone.
Section:  Introduction

 

37. How does one determine if two homolog proteins are paralogs or orthologs?
Ans: From functional studies, paralogs have similar sequences, but differ in function. Orthologs have similar or identical functions.
Section:  6.1

 

38. Why is it more effective to compare protein sequences, rather than DNA sequences, in evolutionary studies?
Ans: More effective statistical comparisons can be made using amino acids as there are 20 amino acids versus 4 bases. Furthermore, base alterations may result in meaningless silent mutations.
Section:  6.2

 

39. How are sequence alignments made?
Ans: Two sequences are aligned and the best matches determined for all possible juxtapositions. In some cases, gaps are introduced to maximize the number of possible matches. Statistical formulas are used to determine the best fit according to defined parameters.
Section:  6.2

 

40. What is a substitution matrix?
Ans: A substitution matrix is deduced from aligned sequences, and scores correlate to how often an amino acid is substituted.
Section:  6.2

 

41. What is the difference between a simple scoring system for alignment and the
Blosum-62 matrix?
Ans: The Blosum-62 allows an examination of substitutions that considers mutations that are conservative. Thus, conservative substitutions can be considered in the scoring used to determine the significance of the change in the alignment and allow more significant conclusions about evolutionary relationships.
Section:  6.2

 

42. How are three-dimensional structures useful in evolutionary comparisons?
Ans: The structural details can be correlated with specific functions. Enough of the structure must be maintained to retain the function, thus sequence changes occur that do not disrupt the structure. It is difficult to determine which amino acids are critical to the structure without evaluation of the three-dimensional structure. Thus, comparisons of proteins with similar structures reveal more critical information than sequence alignments.
Section:  6.3

 

43. If a protein contains a repetitive region, what might be assumed, and what should be done next to test the hypothesis?
Ans: Similar sequences imply a gene duplication event, which may indicate that that region of the protein has an important functional or structural role. The next steps would be to determine if the region of repetition is statistically significant and to examine the three-dimensional structure of the region.
Section:  6.5

 

44. In addition to examining the base sequence, what other features of RNA are useful in determining evolutionary patterns?
Ans: Changes to bases that retain the ability to form base-pairs with partner bases in other sections of the sequence imply that the base-pair is important. These conservative changes will retain the three-dimensional shape of the RNA molecule.
Section:  6.3

 

45. What evidence exists for a duplication event that led to the a and b-hemoglobin?
Ans: Lamprey is a fish that diverged from bony fish long ago. It has hemoglobin that contains a single type of subunit.
Section:  6.4

 

46. What are some of the inherent difficulties of using DNA from ancient samples?
Ans: Contamination of the DNA source is a major problem. In addition, old samples are often too degraded to be of real utility.
Section:  6.5

 

47. What are the processes by which evolution occurs?
Ans: There are three distinct steps: 1) Generation of a diverse population, 2) selection of members based on some criteria, and 3) reproduction of the selected members.
Section:  6.5

 

48. If using protein structure is a better method to determine evolutionary relationships, why is most analysis done using DNA sequences?
Ans: Relatively few protein structures have been determined. It is faster, easier, and cheaper to determine DNA sequences.
Section:  Entire Chapter

 

49. Briefly describe the ATP – RNA binding experiment that showed that a structure had evolved that was capable of specific binding to ATP.
Ans: RNA sequences, synthesized by combinatorial chemistry, were selected for the ability to bind ATP using ATP affinity columns. The ATP-binding RNA molecules are released from the column and replicated, a process that occurs several times. The final RNA products with significant ATP-binding ability are isolated and characterized.
Section:  6.5

 

50. In an evolutionary tree, what is the relationship between the length of the branch connecting each pair and the sequence differences?
Ans: The length of the branch connecting each pair of proteins is proportional to the number of amino acid differences between sequences.
Section:  6.4

Chapter 7   Hemoglobin: A Portrait of a Protein in Action

 

 

Matching Questions

Use the following to answer questions 1-10:

 

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

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

 

51. Carbon dioxide reacts with the amino terminal groups of deoxyhemoglobin to form __________ groups.

 

Ans:  d
Section:  7.3

 

 

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

 

Ans:  j
Section:  7.1

 

53. ____________ This is a genetic disease due to the decreased production of one of the subunits of hemoglobin.

 

Ans:  c
Section:  7.4

 

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

 

Ans:  h
Section:  7.3

 

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

 

Ans:  l
Section:  7.3

 

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

 

Ans:  k
Section:  7.2

 

57. ____________ This is the molecule whose function is to store oxygen in muscle cells.

 

Ans:  g
Section:  Introduction

 

58. ____________ This oxidized hemeprotein does not reversibly bind oxygen.

 

Ans:  e
Section:  7.1

 

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

 

Ans:  a
Section:  7.2

 

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

 

Ans:  i
Section:  7.4

 

 

Fill-in-the-Blank Questions

 

61. Under normal conditions, the heme iron in myoglobin and hemoglobin is in the ____________ oxidation state.
Ans:  ferrous, or Fe+2     Section:  7.1

 

62. The ability of myoglobin to bind oxygen depends on the presence of a bound prosthetic group called _____________.
Ans:  heme     Section:  7.1

 

63. In hemoglobin, the iron of the heme is bonded to the four nitrogens of porphyrin and to the proximal ______________ residue of the globin chain.
Ans:  histidine     Section:  7.1

 

64. The binding of 2-3-bisphosphogycerate to hemoglobin ____________ (increases, decreases) its affinity of oxygen binding.
Ans:  decreases     Section:  7.2

 

65. The effect of pH on oxygen-binding of hemoglobin is referred to as the _____________.
Ans:  Bohr effect     Section:  7.3

 

66. Deoxyhemoglobin is stabilized through ___________________ interactions between the carbamates and positively charged amino acids at the interface between αβ dimmers.
Ans:  salt-bridge     Section:  7.3

 

67. The T-state of hemoglobin is stabilized by a salt bridge between β1 Asp 94 and the C-terminal ___________________ of the β1 chain.
Ans:  histidine     Section:  7.3

 

68. In normal adult hemoglobin, HbA, the β6 position is a glutamate residue, whereas in sickle-cell hemoglobin, HbS, it is a ____________ residue.
Ans:  valine     Section:  7.4

 

69. As the partial pressure of carbon dioxide increases, the affinity of oxygen binding to hemoglobin ______________.
Ans:  decreases     Section:  7.3

 

70. 2,3-Bisphosphoglycerate binds only to the __________________ form of hemoglobin.
Ans:  T-, or deoxy     Section:  7.2

 

 

Multiple-Choice Questions

 

71. What factor(s) influence(s) the binding of oxygen to myoglobin?
A) The concentration of bicarbonate ion, HCO3
B) The partial pressure of oxygen, pO2
C) The concentration of hemoglobin present
D) The concentration of 2,3-BPG
E) Both b and d
Ans:  B     Section:  7.2

 

72. Which of the following is correct concerning the differences between hemoglobin and myoglobin?
A) Both hemoglobin and myoglobin are tetrameric proteins.
B) Hemoglobin exhibits a hyperbolic O2 saturation curve while myoglobin exhibits a sigmoid shaped curve.
C) Hemoglobin exhibits cooperative binding of O2 while myoglobin does not.
D) Hemoglobin exhibits a higher degree of O2 saturation at all physiologically relevant partial pressures of O2 than does myoglobin.
E) All of the above.
Ans:  C     Section:  7.2

 

73. Which of the following is not correct concerning myoglobin?
A) The globin chain contains an extensive α-helix structure.
B) The heme group is bound to the globin chain by two disulfide bonds to cysteine residues.
C) The iron of the heme group is in the Fe+2 oxidation state.
D) The diameter of the iron ion decreases upon binding to oxygen.
E) The function of myoglobin is oxygen storage in muscle.
Ans:  B     Section:  7.1

 

74. The structure of normal adult hemoglobin can be described as
A) a tetramer composed of four myoglobin molecules.
B) a tetramer composed of two αβ dimers.
C) a tetramer composed of two α2 and two β2 dimers.
D) a tetramer composed of two α2 and two γ2 dimers.
E) None of these accurately describe hemoglobin.
Ans:  B     Section:  7.1

 

75. Which of the following is correct concerning fetal hemoglobin?
A) Fetal hemoglobin is composed of two α and two γ subunits.
B) Fetal hemoglobin binds 2,3-BPG more tightly than normal adult hemoglobin.
C) Fetal hemoglobin binds oxygen less than HbA at all pO2.
D) Fetal hemoglobin does not exist in the T-form.
E) None of the above.
Ans:  A     Section:  7.2

 

26. Hemoglobin-binding of oxygen is best described as a
A) concerted model.
B) Michaelis-Menten model.
C) sequential model.
D) combination of sequential and concerted models.
E) None of the above.
Ans:  D     Section:  7.2

 

27. 2-3 Bisphosphoglycerate
A) binds in the central cavity in the T-form of hemoglobin.
B) preferentially binds to deoxyhemoglobin and stabilizes it.
C) is present in the red blood cells.
D) All of the above.
E) None of the above.
Ans:  D     Section:  7.2

 

28. What is the Bohr effect?
A) the ability of hemoglobin to retain oxygen when in competition with myoglobin
B) the regulation of hemoglobin-binding by hydrogen ions and carbon dioxide
C) the alteration of hemoglobin conformation during low oxygen stress
D) All of the above.
E) None of the above.
Ans:  B     Section:  7.3

 

29. Why is the HbS mutation so prevalent in Africa and other tropical regions?
A) The oxygen binds with greater affinity to the proximal histidine residue of HbS.
B) Bonding of carbon dioxide to HbS molecules increases the binding of oxygen.
C) Hemoglobin binds more oxygen with the reduction of the hemoglobin S chain.
D) Hemoglobin binds more oxygen with aggregations of α chains found in sickle-cell hemoglobin.
E) People with sickle-cell trait are resistant to malaria, increasing the prevalence of the HbS allele.
Ans:  E     Section:  7.4

 

30. Which of the following describes the Bohr effect?
A) Lowering the pH results in the release of O2 from oxyhemoglobin.
B) Increasing the pressure of CO2 results in the release of O2 from oxyhemoglobin.
C) Increasing the pH increases the T-form of hemoglobin.
D) All of the above.
E) a and b.
Ans:  E     Section:  7.3

 

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

CO2  +  H2O    H2CO3

A) An increase in the pressure of CO2 will result in a decrease of pH.
B) This reaction is catalyzed by carbonic anhydrase.
C) The H2CO3 dissociates to H+ and bicarbonate ion, HCO3.
D) The majority of CO2 is transported to the lungs in the form of HCO3.
E) All of the above.
Ans:  E     Section:  7.3

 

32. Carbon dioxide forms carbamate groups in proteins by reaction with
A) aspartate residues.
B) cysteine residues.
C) N-terminal amino groups.
D) tyrosine residues.
E) heme groups.
Ans:  C     Section:  7.3

 

33. Sickle-cell anemia is caused by
A) a decreased production of α chains of hemoglobin.
B) a substitution of a Glu residue for a Phe residue at the β6 position.
C) the loss of the heme group because the proximal His is oxidized.
D) a substitution of a Val residue for a Glu residue at the β6 position.
E) a substitution of Glu residue for His at the C-terminal of the α chain.
Ans:  D     Section:  7.4

 

 

34. Which of the following is correct concerning the oxygenation plot of proteins X and Y shown below?
A) Protein Y exhibits tighter oxygen-binding than protein X.
B) Protein Y corresponds to fetal hemoglobin, and protein X corresponds to normal adult hemoglobin.
C) Protein X corresponds to fetal hemoglobin, and protein Y corresponds to normal adult hemoglobin.
D) Protein X corresponds to myoglobin, and protein Y corresponds to hemoglobin.
E) None of the above.
Ans:  C     Section:  7.2

 

35. Which of the following is not correct concerning the oxygenation plot of proteins X and Y shown below?
A) Protein X exhibits tighter oxygen binding than protein Y.
B) Protein Y would function as a better transport protein than protein X.
C) Protein X exhibits cooperative binding, whereas Y does not.
D) Protein X corresponds to myoglobin, and protein Y corresponds to hemoglobin.
E) Protein Y contains multiple binding sites.
Ans:  C     Section:  7.2

 

36. Which is not correct concerning the models that are accepted to describe cooperative binding?
A) In the sequential model, the binding of a ligand changes the conformation of the subunit to which it binds, which in turn induces a change in neighboring subunits.
B) All known allosteric proteins exhibit either the concerted or sequential model exclusive of the other.
C) Both models incorporate a low affinity T-state and a higher affinity R-state.
D) Both models explain the sigmoid-shaped binding curve.
E) In the concerted model, all molecules exist either in the T-state or the R-state.
Ans:  B     Section:  7.2

 

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

 

A) Curve X most likely corresponds to pH 7.2.
B) Curve Z most likely corresponds to pH 7.6.
C) Hb has a higher affinity for oxygen at the pH of curve Z.
D) Curve Y most likely corresponds to pH 7.4.
E) pH has no effect on the oxygenation of hemoglobin.
Ans:  D     Section:  7.3

 

38. What would be the expected result of a Lys residue being substituted with a Ser residue in the BPG binding site of hemoglobin?
A) BPG would bind tighter because of the loss of a positive charge.
B) BPG would bind tighter because of the gain of a positive charge.
C) BPG would bind less tightly because of the loss of a positive charge
D) BPG would bind less tightly because of the gain of a positive charge.
E) This substitution would have no effect on the binding of BPG.
Ans:  C     Section:  7.2

 

 

Short-Answer Questions

 

39. Why is it advantageous for hemoglobin to have allosteric properties?
Ans: Hemoglobin binds oxygen in a positive cooperative manner. This allows it to become saturated in the lungs, where oxygen pressure is high. When the hemoglobin moves to tissues, the lower oxygen pressure induces it to release oxygen and thus deliver oxygen where it is needed.
Section:  7.2

 

40. What is fetal hemoglobin? How does it differ from adult hemoglobin?
Ans: Fetal hemoglobin contains two a and two g chains, in contrast to adult hemoglobin with two a and two b chains. The fetal hemoglobin g chain is probably a result of gene duplication and divergence. The difference in the chains results in a lower binding affinity of 2-3 BPG to fetal hemoglobin. Thus, the fetal hemoglobin has a higher affinity for oxygen, and the oxygen is effectively transferred from the mother’s hemoglobin to fetal hemoglobin.
Section:  7.2

 

41. What is metmyoglobin?
Ans: Metmyoglobin is formed when the heme iron ion, which is normally in the +2 oxidation state, is oxidized to the +3 oxidation state.  This oxidized form of myoglobin does not bind molecular oxygen and is not functional.
Section:  7.1

 

42. Describe the recurring structure called the globin fold.
Ans: Each of the four subunits of hemoglobin consists of a set of α helices in the same arrangement as the α helices of myoglobin. The arrangement is known as the globin fold.
Section:  7.1

 

43. What functional role does the “distal histidine” play in the function of myoglobin and hemoglobin?
Ans: The bonding between the iron and oxygen can be described as a combination of resonance structures, one with Fe2+ and dioxygen and another with Fe3+ and superoxide.  The “distal histidine” donates a hydrogen bond to this complex, stabilizing the complex, and inhibits the oxidation of the iron to the ferric state.
Section:  7.1

 

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

Lungs

 

Ans:

20 – 40 torr

 

 

Section:  7.2 and Figure 7.8

 

45. Describe the structure of normal adult hemoglobin.
Ans: Normal adult hemoglobin, HbA, is a tetramer.  It is composed of two α subunits and two β subunits.  Each subunit has a structure very similar to myoglobin.  It can be best described as a pair of identical αβ dimers.  Each subunit contains a heme group.  So, each molecule of hemoglobin can bind up to four molecules of oxygen.
Section:  7.2

 

46. What is neuroglobin and what is its suspected role?
Ans: Neuroglobin is expressed in the brain, and at especially high levels in the retina. It may play a role in protecting neural tissues from hypoxia.
Section:  7.4

 

47. Describe the concerted model to explain allosteric cooperative binding.
Ans: The protein exists in two conformations: a T-state (for tense) that has a lower affinity for the ligand and an R-state (for relaxed) that has a higher affinity for the ligand.  In the concerted model, all of the molecules exist either in the T-state or in the R-state.  At each ligand concentration, there is an equilibrium between the two states.  An increase in the ligand concentration shifts the equilibrium from the T- to the R-state.
Section:  7.2

 

48. Describe the role of 2,3-bisphosphoglycerate in the function of hemoglobin.
Ans: 2,3-bisphosphoglycerate, 2,3-BPG, is a relatively small, highly anionic molecule found in the RBC.  2,3-BPG only binds to the center cavity of deoxyhemoglobin (T-state).  The size of the center cavity decreases upon the change to the R-form so that it cannot bind to the R-state.  Thus, the presence of 2,3-BPG shifts the equilibrium toward the T-state. The  T-state is unstable, and without BPG, the equilibrium shifts so far toward the R-state that little oxygen would be released under physiological conditions.
Section:  7.2

 

49. Describe the chemical basis of the Bohr effect.
Ans: The effect observed by Christian Bohr is that hemoglobin becomes deoxygenated as the pH decreases.  In deoxyhemoglobin, three amino acid residues form two salt bridges that stabilize the T-state.  One of these is formed between the C-terminal His β146 and an Asp residue (β94).  As the pH increases, this stabilizing salt bridge is broken because His becomes deprotonated and loses its positive charge.  At lower pH values, this His is positively charged.  The formation of the salt bridge shifts the equilibrium from the R-state to the T-state, thus releasing oxygen.
Section:  7.3

 

50. Describe how carbon dioxide affects the oxygenation of hemoglobin.
Ans: Increased levels of carbon dioxide cause hemoglobin to release oxygen.  The more active the tissue, the more fuel is burned and the more CO2 is produced.  These active tissue cells have the greatest need for oxygen to produce more energy.  The CO2 combines with the N-terminal amino groups to form negatively charged carbamate groups.  The negatively charged carbamate groups form salt bridges that stabilize the T-state.  Thus, the increase of carbon dioxide causes the conversion of the R-state to the T-state, releasing the bound oxygen to the tissues producing the most CO2.
Section:  7.3

 

51. Briefly describe the cause of sickle-cell anemia.
Ans: Sickle-cell anemia is a genetic disorder that is the result of a single substitution of β6 Glu with a Val.  This changes a negatively charged side chain to a nonpolar, hydrophobic side chain.  This Val binds into a hydrophobic pocket on the β chain of an adjacent molecule whose β6 Val binds to another molecule, thus hemoglobin aggregates.  These aggregates form long fibers that strain the RBC and force into a sickled shape.  The distorted red blood cells clog capillaries and impair blood flow, resulting in the sickle-cell crisis.  The sickled cells are then destroyed, resulting in the anemia.
Section:  7.4

 

52. What is thalassemia?
Ans: Thalassemia is caused by the substantial decreased production of one of the subunits of hemoglobin.  In α-thalassemia, the decreased production of the α chain results in the formation of tetramers of only the β chain.  This β4 binds oxygen more tightly than HbA and does not exhibit cooperative binding.  In β-thalassemia, the α chains form insoluble aggregates in the immature red blood cells.
Section:  7.4

 

53. What is the role of α-hemoglobin stabilizing protein?
Ans: Four genes express the α chains, and only two genes express the β chain.  Thus, there is an excess of α chains, which if allowed, would aggregate and become insoluble.  Red blood cells produce α-hemoglobin stabilizing protein (AHSP), which binds to the α chain monomers to from a soluble complex.  This prevents the aggregation of the α subunits.
Section:  7.4