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<html><body>CHAPTER 2 Protein Synthesis 9

In Chapter 29 the mechanism of protein synthesis, a process called translation, is ex

amined in detail. Translation is a complicated process in which the four-letter alphabet of nucleic acids is translated into the 20-letter alphabet of proteins. The

chapter begins with an introduction to the major components of translation—mRNA, tRNA, ribosomes, and aminoacyl-tRNA synthetases. The detailed structures and conformations of tRNAs, the adaptor molecules that recognize both the codons on the mRNAs and the enzymes that attach the corresponding amino acids, are discussed first.Next the authors explain how amino acids are activated for the subsequent formation of peptide bonds through their attachment to tRNAs by the two classes of aminoacyl-tRNA synthetases. The exquisite specificity of these reactions is explored, in terms of correct binding of amino acids and tRNAs to a given synthetase. ThreonyltRNA synthetase is used as an example of specificity at the level of amino acid selection. This enzyme discriminates between threonine and the isosteric valine and the isoelectronic serine, using a combination of selective binding at the active site and proofreading after aminoacylation. Several aminoacyl-tRNA synthetases are used as examples of ways in which the correct tRNA is chosen, ranging from those which require multiple contact points (glutaminyl-tRNA synthetase) to alanyl-tRNA synthetase, which will recognize a “microhelix” containing only the acceptor stem and a hairpin loop.

The authors next turn to the structure and composition of the ribosome, a mo

lecular machine that coordinates charged tRNAs, mRNA, and proteins, leading to protein synthesis. The fact that the ribosome is now recognized to be a ribozyme, with the RNA components playing the major role in catalysis, is introduced. The experiments that showed the polarities of polypeptide formation and the translation of mRNA are presented next. Then initiation is described, and the roles of a specialized 517 518

CHAPTER 29

initiator tRNA, the mRNA start codon, and 16S rRNA sequences are outlined. The spatial and functional relationships of the sites on the ribosome that bind aminoacyl-tRNAs and peptidyltRNAs, the peptide-bond–forming reaction, the role of GTP, and the mechanism of the translocation of the peptidyl-tRNA from site to site on the ribosome are presented in the description of the elongation stage of protein synthesis. The wobble hypothesis is then presented to explain the lack of strict one-to-one Watson-Crick base-pairing interactions among the three nucleotides of the tRNA anticodons and the mRNA codons.

The critical role that protein factors play in translation is discussed next, including ini

tiation, elongation, and release factors. The termination of translation is outlined, and the role of release factors that recognize translation stop codons is described. The chapter closes with a brief overview of translation in eukaryotes, emphasizing the major contrasting features with respect to translation in prokaryotes. Differences in the initiator tRNA, the selection mechanism of the initiator codon, the ribosomes, and the overall complexity of the process are highlighted. Last, the mechanisms of several potent inhibitors of translation and the mechanism of the bacterial toxin that causes diphtheria is presented. LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Introduction

1. Provide an overview of protein synthesis that includes the roles of the amino acids, the tRNAs, the amino acid activating enzymes, mRNA, and the ribosome. Protein Synthesis Requires the Translation of Nucleotide Sequences into Amino Acid Sequences (Text Section 29.1)

2. Draw the cloverleaf structure of a tRNA and identify the regions containing the anticodon

and the amino acid attachment site.

3. List the features common to all tRNAs.

4. Relate the two-dimensional cloverleaf representation of the tRNA structure to its three

dimensional configuration. Aminoacyl-Transfer RNA Synthetases Read the Genetic Code (Text Section 29.2)

5. Write the two-step reaction sequence of the aminoacyl-tRNA synthetases. Enumerate the

high-energy phosphate bonds that are consumed in the overall reaction.

6. Describe the mechanisms of amino acid selection and proofreading that contribute to

the accuracy of the attachment of the appropriate amino acid to the correct tRNA.

7. Describe the different modes of recognition of the correct tRNA molecule by aminoacyl

tRNA synthetases.

8. Recount an experiment that showed that the tRNA rather than the amino acid in an

aminoacyl-tRNA recognizes an mRNA codon.

9. Outline the distinguishing properties of class I and class II aminoacyl-tRNA synthetases.

PROTEIN SYNTHESIS 519 A Ribosome Is a Ribonucleoprotein Particle (70S) Made of a Small (30S) and a Large (50S) Subunit (Text Section 29.3)

10. List the kinds and numbers of macromolecular components of the prokaryotic ribosome.

Give the mass, sedimentation coefficient, and dimensions of the ribosome of E. coli.

11. Outline the three-dimensional structure of a ribosome.

12. List the evidence that suggests that the RNA components of ribosomes have active roles

in protein synthesis.

13. Recount the experiments that established the direction of translation, both in terms of

protein synthesis and the reading of mRNA.

14. Name the major initiator codon and the amino acid it encodes. Explain the roles of the

nucleotide sequences in 16S rRNA, mRNA, and tRNA in selecting the initiation codon rather than the identical codon that encodes an internal amino acid.

15. Distinguish among the initiator tRNA, tRNAf, and tRNAm and outline the conversion of

methionine into formylmethionyl-tRNAf.

16. Explain how some codons are recognized by more than one anticodon, that is, how they in

teract with more than one species of aminoacyl-tRNA. List the base-pairing interactions allowed according to the wobble hypothesis.

17. Define the polysome. Correlate the polarity of ribosome movement with the polarity of

the growing polypeptide chain. Protein Factors Play Key Roles in Protein Synthesis (Text Section 29.4)

18. List the components of the 70S initiation complex and indicate the roles of the initiation factors (IF) and GTP in its formation.

19 Outline the elongation stage of protein synthesis and describe the roles of the elongation factors (EFs) and GTP in the process. Locate the aminoacyl-tRNAs and peptidyl-tRNAs in the A or P sites of the ribosome during one cycle of elongation.

20. Describe how the GTP–GDP cycle of EF-Tu controls its affinity for its reaction partners.

21. Explain the role of EF-Tu in determining the accuracy and timing of protein synthesis.

22. Outline the translocation steps that occur after the formation of a peptide bond and de

scribe the roles of EF-G and GTP.

23. Name the translation stop codons, describe the termination of translation, and explain the

roles of the release factors (RFs) in the process. Eukaryotic Protein Synthesis Differs from Prokaryotic Protein Synthesis Primarily in Translation Initiation (Text Section 29.5)

24. Contrast eukaryotic and prokaryotic ribosomes with respect to composition and size.

25. Contrast the mechanisms of translation initiation in prokaryotes and eukaryotes. Note the

different initiator tRNAs, AUG codon selection mechanisms, and numbers of IFs and RFs.

26. Describe the mechanism by which the diphtheria toxin inhibits protein synthesis in

eukaryotes.

27. Provide examples of antibiotics that inhibit translation, and describe their mechanisms

of action. 520

CHAPTER 29 SELF-TEST Protein Synthesis Requires the Translation of Nucleotide Sequences into Amino Acid Sequences

1. Which of the following statements about functional tRNAs are correct?

(a) They contain many modified nucleosides. (b) About half their nucleosides are in base-paired helical regions. (c) They contain fewer than 100 ribonucleosides. (d) Their anticodons and amino acid accepting regions are within 5 Å of each other. (e) They consist of two helical stems that are joined by loops to form a U-shaped

structure.

(f)

They have a terminal AAC sequence at their amino acid accepting end.

2. Explain why tRNA molecules must have both unique and common structural features. Aminoacyl-Transfer RNA Synthetases Read the Genetic Code

3. Which of the following statements about the aminoacyl-tRNA synthetase reaction are

correct?

(a) ATP is a cofactor. (b) GTP is a cofactor. (c) The amino acid is attached to the 2′- or 3′-hydroxyl of the nucleotide cofactor (ATP). (d) The amino group of the amino acid is activated. (e) A mixed anhydride bond is formed. (f)

An acyl ester bond is formed.

(g) An acyl thioester bond is formed. (h) A phosphoamide (P–N) bond is formed.

4. The DG°′ of the reaction catalyzed by the aminoacyl-tRNA synthetases is

(a) ~0 kcal/mol. (b) <0 kcal/mol. (c) >0 kcal/mol.

5. Considering the correct answer to question 4, explain how aminoacyl-tRNAs can be pro

duced in the cell.

6. Match the class I and class II aminoacyl-tRNA synthetases with the appropriate items in

the right column.

(a) class I

(1) are generally dimeric

(b) class II

(2) are generally monomeric (3) acylate the 2′-hydroxyl of the tRNA (4) generally acylate the 3′ hydroxyl of the tRNA (5) contain b strands at the activation domain

7. Indicate the possible ways in which different aminoacyl-tRNA synthetases may recog

nize their corresponding tRNAs.

(a) by recognizing the anticodon (b) by recognizing specific base pairs in the acceptor stem (c) by recognizing the 3′ CCA sequence of the tRNA (d) by recognizing both the anticodon and acceptor stem region (e) by recognizing extended regions of the L-shaped molecules

8. In an experiment, it was found that Cys-tRNACys can be converted to Ala-tRNACys and

used in an in vitro system that is capable of synthesizing proteins.

(a) If the Ala-tRNACys were labeled with 14C in the amino acid, would the labeled Ala

be incorporated in the protein in the places where Ala residues are expected to occur? Explain.

(b) What does the experiment indicate about the importance of the accuracy of the

aminoacyl-tRNA synthetase reaction to the overall accuracy of protein synthesis?

9. Which of the following answers completes the sentence correctly? The wobble hypothesis

(a) accounts for the conformational looseness of the amino acid acceptor stem of tRNAs

that allows sufficient flexibility for the peptidyl-tRNA and aminoacyl-tRNA to be brought together for peptide-bond formation.

(b) accounts for the ability of some anticodons to recognize more than one codon. (c) explains the occasional errors made by the aminoacyl-tRNA synthetases. (d) explains the oscillation of the peptidyl-tRNAs between the A and P sites on the

ribosome.

(e) assumes steric freedom in the pairing of the first (5′) nucleotide of the codon and

the third (3′) nucleotide of the anticodon.

10. Assuming that each nucleoside in the left column is in the first position of an anticodon,

with which nucleoside or nucleosides in the right column could it pair during a codon– anticodon interaction if each of the nucleosides on the right is in the third position (3′ position) of a codon?

(a) adenosine

(1) adenosine

(b) cytidine

(2) cytidine

(3) guanosine

(d) inosine

(4) uridine

(e) uridine A Ribosome Is a Ribonucleoprotein Particle (70S) Made of a Small (30S) and a Large (50S) Subunit

11. Which of the following statements about an E. coli ribosome are correct?

(a) It is composed of two spherically symmetrical subunits. (b) It has a large subunit comprising 34 kinds of proteins and two different rRNA mol

ecules.

(c) It has a sedimentation coefficient of 70S. (d) It has two small subunits, one housing the A site and the other the P site. (e) It has an average diameter of approximately 200 Å. (f)

It has a mass of approximately 270 kd, one-third of which is RNA.

12. What is the significance of the reconstitution of a functional ribosome from its separated

components?

13. Which of the following statements about translation are correct?

(a) Amino acids are added to the amino terminus of the growing polypeptide chain. (b) Amino acids are activated by attachment to tRNA molecules. (c) A specific initiator tRNA along with specific sequences of the mRNA ensures that

translation begins at the correct codon.

(d) Peptide bonds form between an aminoacyl-tRNA and a peptidyl-tRNA positioned

in the A and P sites, respectively, of the ribosome.

(e) Termination involves the binding of a terminator tRNA to a stop codon on the mRNA. 522

CHAPTER 29 Protein Factors Play Key Roles in Protein Synthesis

14. An experiment is carried out in which labeled amino acids are added to an in vitro trans

lation system under the direction of a single mRNA species. Samples are withdrawn at different times, and the labeling patterns below are observed in the completed polypeptide chains. The dashes (-) represent unlabeled amino acids, X represents labeled amino acids, and A and B represent the ends of the intact protein.

Time 1 (early)

A - - - - - - - - - - - - - - XXB

Time 2

A - - - - - - - - - - - X X X X B

Time 3

A - - - - - - - X X X X X X B

Time 4 (late)

A - - - - X X X X X X X X B

Which of the following statements about these proteins are correct?

(a) The labeled amino acids are added in the B-to-A direction. (b) The labeled amino acids are added in the A-to-B direction. (c) A is the amino terminus of the protein. (d) B is the amino terminus of the protein.

15. Given an in vitro system that allows protein synthesis to start and stop at the ends of any

RNA sequence, answer the following questions:

(a) What peptide would be produced by the polyribonucleotide 5′-UUUGUUUUUGUU-3′?

(See the table with the genetic code on the inside back cover of the textbook.)

(b) For this peptide, which is the N-terminal amino acid and which is the C-terminal

amino acid?

16. What is the role of the vitamin folic acid in prokaryotic translation?

17. Match the functions or characteristics of prokaryotic translation in the right column with

the appropriate translation components in the left column.

(a) IF1

(1) moves the peptidyl-tRNA from the A to

(b) IF2

the P site

(2) delivers aminoacyl-tRNA to the A site

(d) EF-Tu

(3) binds to the 30S ribosomal subunit

(e) EF-Ts

(4) recognizes stop codons

(f)

EF-G

(5) forms the peptide bond

(g) peptidyl transferase

(6) delivers fMet-tRNA Met

f

to the P site

(h) RF1

(7) cycles on and off the ribosome

(i)

RF2

(8) binds GTP (9) prevents the combination of the 50S and

30S subunits

(10) is involved in the hydrolysis of GTP to GDP (11) associates with EF-Tu to release a bound

nucleoside diphosphate

(12) hydrolyzes peptidyl-tRNA (13) modifies the peptidyl transferase reaction

18. Which of the following statements about occurrences during translation are correct?

(a) The carboxyl group of the growing polypeptide chain is transferred to the amino

group of an aminoacyl-tRNA.

(b) The carboxyl group of the amino acid on the aminoacyl-tRNA is transferred to the

amino group of a peptidyl-tRNA.

PROTEIN SYNTHESIS 523

(c) Peptidyl-tRNA may reside in either the A or the P site. (d) Aminoacyl-tRNAs are shuttled from the A to the P site by EF-G.

19. About 5% of the total bacterial protein is EF-Tu. Explain why this protein is so abundant.

20. For each of the following steps of translation, give the nucleotide cofactor involved and

the number of high-energy phosphate bonds consumed. (a) amino acid activation (b) formation of the 70S initiation complex (c) delivery of aminoacyl-tRNA to the ribosome (d) formation of a peptide bond (e) translocation

21. Which of the following statements about release factors are correct?

(a) They recognize terminator tRNAs. (b) They recognize translation stop codons. (c) They cause peptidyl transferase to use H2O as a substrate. (d) They are two proteins in E. coli, each of which recognizes two mRNA triplet sequences. Eukaryotic Protein Synthesis Differs from Prokaryotic Protein Synthesis Primarily in Translation Initiation

22. Many antibiotics act by inhibiting protein synthesis. How can some of these be used in

humans to counteract microbial infections without causing toxic side effects due to the inhibition of eukaryotic protein synthesis?

23. Which of the following statements about eukaryotic translation are correct?

(a) A formylmethionyl-tRNA initiates each protein chain. (b) It occurs on ribosomes containing one copy each of the 5S, 5.8S, 18S, and 28S

rRNA molecules.

the rRNA of the small ribosomal subunit with an mRNA sequence upstream from the translation start site.

(d) It is terminated by a release factor that recognizes stop codons and hydrolyzes GTP. (e) It involves proteins that bind to the 5′ ends of mRNAs. (f)

It can be regulated by protein kinases.

24. Increasing the concentration of which of the following would most effectively antago

nize the inhibition of protein synthesis by puromycin?

(a) ATP

(d) peptidyl-tRNAs

(b) GTP

(e) eIF3

25. Which of the following statements about the diphtheria toxin are correct?

(a) It is cleaved on the surface of susceptible eukaryotic cells into two fragments, one

of which enters the cytosol.

(b) It binds to peptidyl transferase and inhibits protein synthesis. (c) It reacts with ATP to phosphorylate eIF2 and prevent the insertion of the Met-tRNAi

into the P site.

(d) It reacts with NAD+ to add ADP-ribose to eEF2 and prevent movement of the peptidyl

tRNA from the A to the P site.

(e) One toxin molecule is required for each translation factor inactivated, that is, it acts

stoichiometrically. 524

CHAPTER 29 ANSWERS TO SELF-TEST

1. a, b, c. The molecules consist of two helical stems, each of which is made of two stacked

helical segments. However, the molecules are L-shaped, and the anticodon and amino acid accepting regions are some 80 Å from each other. Functional tRNAs have a CCA sequence, not an AAC sequence at their 3′ termini.

2. Transfer RNAs need common features for their interactions with ribosomes and elonga

tion factors but unique features for their interactions with the activating enzymes.

3. a, e, f. The carboxyl group of the amino acid is activated in a two-step reaction via the

formation of an intermediate containing a mixed anhydride linkage to AMP. The amino acid is ultimately linked by an ester bond to the 2′- or 3′-hydroxyl of the tRNA.

4. a. Since the standard free energy of the hydrolysis of an aminoacyl-tRNA is nearly equal

to that of the hydrolysis of ATP, the reaction has a DG°′ ~ 0; that is, it has an equilibrium constant near 1.

5. In the cell, the hydrolysis of PPi by pyrophosphatase shifts the equilibrium toward the

formation of aminoacyl-tRNA.

6. (a) 2, 3, 5 (b) 1, 4, 5. Both classes of enzymes contain b sheet domains at their activa

tion domains. Class I enzymes have a Rossmann fold (p. 440 in text) and class II enzymes consist of b strands.

7. a, b, d, e. There are many different ways in which aminoacyl-tRNA synthetases recog

nize their specific tRNAs. Answer (c) is incorrect because the 3′ CCA sequence is common to all tRNAs and cannot be used to distinguish among them.

8. (a) No. The tRNA, acting as an adaptor between the amino acid and mRNA, would as

sociate with Cys codons in the mRNA through base-pairing between codon and anticodon. The labeled alanine would incorporate only at sites encoded by the Cys codons and not at those encoded by Ala codons. The experiment demonstrates that the tRNA and not the amino acid reads the mRNA. Thus, if the activating enzyme mistakenly attaches an incorrect amino acid to a tRNA, that amino acid will be incorporated erroneously into the protein.

(b) Answer (e) is incorrect because the ambiguities in base-pairing occur between the

third nucleotide of the codon and the first nucleotide of the anticodon.

10. (a) 4 (b) 3 (c) 2, 4 (d) 1, 2, 4 (e) 1, 3. (See Table 29.3 on p. 832 of the text.)

11. b, c, e. Answer (f) is incorrect because two-thirds of the 2700-kd mass of a ribosome is rRNA.

12. It shows that the components themselves contain all the information necessary to form

the structure and that neither a template nor any other factors are involved. Thus, the ribosome serves as model from which we might learn the general principles involved in self-assembly. Reassembly allows systematic study of the roles of the individual components through the determination of the effects of substitutions of mutant or altered individual proteins or rRNAs.

13. b, c, d. Answer (a) is incorrect because the incoming activated aminoacyl-tRNA, in the

A site of the ribosome, adds its free amino group to the activated carboxyl of the growing polypeptide on a peptidyl-tRNA in the P site. Answer (e) is incorrect because termination does not involve tRNAs that recognize translation stop codons but rather protein release factors that recognize these signals and cause peptidyl transferase to donate the growing polypeptide chain to H2O rather than to another aminoacyl-tRNA.

PROTEIN SYNTHESIS 525

14. b, c. Although longer incubation times result in completed proteins that have labeled

polypeptides closer to their amino terminus, the chains actually grow in the amino-tocarboxyl direction. When the labeled amino acid is introduced into the system, it begins adding to the carboxyl ends of the growing chains that are already present in all stages of completion. The completed chains in samples withdrawn after a short time will have labeled polypeptides only near their carboxyl end. As time passes, more and more polypeptides that began adding labeled polypeptides near their amino terminals will become complete chains.

15. (a) Phenylalanylvalylphenylalanylvaline

(b) Phe is the N-terminal amino acid and Val is the C-terminal amino acid.

16. After folic acid is converted to N10-formyltetrahydrofolate, it acts as a carrier of formyl groups

and is a substrate for a transformylase reaction that converts Met-tRNAf to fMet-tRNAf—the initiator tRNA. (See p. 828 in the text.)

17. (a) 3, 7, 9 (b) 3, 6, 7, 8, 10 (c) 3, 7, 9 (d) 2, 7, 8, 10 (e) 11 (f) 1, 7, 8, 10 (g) 5, 12 (h)

4, 7, 13 (i) 4, 7, 13

18. a, c. The aminoacyl-tRNA in the A site becomes a peptidyl-tRNA when it receives the

carboxyl group of the growing polypeptide chain from the peptidyl-tRNA in the P site. After the free tRNA leaves, the extended polypeptide on its new tRNA is then moved to the P site by EF-G. Answer (d) is incorrect because transfer RNAs bearing aminoacyl derivatives with free amino groups are never found in the P site.

19. The large amounts of EF-Tu in the cell bind essentially all of the aminoacyl-tRNAs and

protect these activated complexes from hydrolysis.

20. (a) ATP, 2 (b) GTP, 1 (c) GTP, 1 (d) none (e) GTP, 1. With regard to the answer for (d),

the formation of a peptide bond per se does not require a cofactor. The energy for the exergonic reaction is supplied by the activated aminoacyl-tRNA.

21. b, c, d. Each of the two release factors of E. coli recognizes two of the three translation

stop codons and interacts with the synthesis machinery such that peptidyl transferase donates the polypeptide chain to H2O and thus terminates synthesis by hydrolyzing the ester linkage of the protein to the tRNA.

22. The inhibition of the prokaryotic translation and not that of the eukaryote can result

from differences between their respective ribosomes. Some antibiotics interact with the RNA components that are unique to bacterial ribosomes and, consequently, can inhibit bacterial growth without affecting the human cells.

23. b, d, e, f. Eukaryotic ribosomes usually scan the mRNA from the 5′ end for the first

AUG codon, which then serves to initiate the synthesis. Answer (e) is correct because proteins that bind to the cap of the mRNA are involved in the association of the ribosome with the mRNA.

24. c. Puromycin is an analog of aminoacyl-tRNA. It inhibits protein synthesis by binding

to the A site of the ribosome and accepting the growing polypeptide chain from the peptidyl-tRNA in the P site and thus terminating polymer growth. Because aminoacyltRNAs compete with the puromycin for the A site, increasing their concentration would lessen the extent of inhibition.

25. d. Answer (e) is incorrect because the toxin acts catalytically and is thus extremely

deadly; one toxin molecule can inactivate many translocase molecules by modifying them covalently. 526

CHAPTER 29 PROBLEMS

1. (a) The template strand of DNA known to encode the N-terminal region of an E. coli

protein has the following nucleotide sequence: GTAGCGTTCCATCAGATTT. Give the sequence for the first four amino acids of the protein.

(b) Suppose that the sense strand of the DNA known to encode the amino acid se

quence of the N-terminal region of a mammalian protein has the following nucleotide sequence: CCTGTGGATGCTCATGTTT. Give the amino acid sequence that would result.

2. The nucleotide sequence on the sense strand of the DNA that is known to encode the

carboxy terminus of a long protein of E. coli has the following nucleotide sequence: CCATGCAAAGTAATAGGT. Give the resulting amino acid sequence.

3. Suppose that a particular aminoacyl-tRNA synthetase has a 10% error rate in the for

mation of aminoacyl-adenylates and a 99% success rate in the hydrolysis of incorrect aminoacyl-adenylates. What percentage of the tRNAs produced by this aminoacyl-tRNA synthetase will be faulty?

4. Students of biochemistry are frequently distressed by “Svedberg arithmetic,” that is, for

instance, by the fact that the 30S and 50S ribosomal subunits form a 70S particle rather than an 80S particle. Why don’t the numbers add up to 80? (See p. 88 of the text.)

5. The possible codons for valine are GUU, GUC, GUA, and GUG.

(a) For each of these codons write down all the possible anticodons with which it might

pair (use the wobble rules in Table 29.3 in the text).

(b) How many codons could pair with anticodons having I as the first base? How many

could pair with anticodons having U or G as the first base? How many could pair with anticodons beginning with A or C?

6. What amino acid will be specified by a tRNA whose anticodon sequence is IGG?

7. According to the wobble principle, what is the minimum number of tRNAs required to

decode the six leucine codons—UUA, UUG, CUU, CUC, CUA, and CUG? Explain.

8. Coordination of the threonine hydroxyl by an active site Zn in the threonyl-tRNA syn

thetase allows discrimination between threonine and the isosteric valine (Sankaranarayanan et al., Nat. Struct. Biol. 7[2000]:461–465). Given the similarity of serine and threonine (Ser lacks only the methyl group of Thr), if this is the only mechanism for amino acid discrimination available, threonyl-tRNA synthetase mistakenly couples Ser to threonyl-tRNA at a rate several-fold higher than it does threonine. Since this would lead to unacceptably high error rates in translation, how it is it avoided?

9. Mutations from codons specifying amino acid incorporation to one of the chain

terminating codons, UAA, UAG, or UGA, so-called nonsense mutations, result in the synthesis of shorter, usually nonfunctional, polypeptide chains. It was discovered that some strains of bacteria can protect themselves against such mutations by having mutant tRNAs that can recognize a chain-terminating codon and insert an amino acid instead. The result would be a protein of normal length that may be functional, even though it may contain an altered amino acid residue. How can bacteria with such mutant tRNA molecules ever manage to terminate their polypeptide chains successfully?

PROTEIN SYNTHESIS 527

10. Change of one base pair to another in a sense codon frequently results in an amino acid

substitution. Change of a C–G to a G–U base pair at the 3:70 position of tRNACys causes that tRNA to be recognized by alanyl-tRNA synthetase.

(a) What amino acid substitution or substitutions would result with the mutated

tRNACys present?

(b) How does the pattern differ from that resulting from base substitutions within a codon?

11. The methionine codon AUG functions both to initiate a polypeptide chain and to direct

methionine incorporation into internal positions in a protein. By what mechanisms are the AUG start codons selected in prokaryotes?

12. Laboratory studies of protein synthesis usually involve the addition of a radioactively

labeled amino acid and either natural or synthetic mRNAs to systems containing the other components. To observe the formation of protein, advantage is taken of the fact that proteins, but not amino acids, can be precipitated by solutions of trichloroacetic acid. Thus, one can observe the extent to which radioactivity has been incorporated into “acid-precipitable material” as a function of time to estimate the rate of formation of protein. In one such experiment, poly(U) is used as a synthetic mRNA in an in vitro system derived from wheat germ (a eukaryote).

(a) What labeled amino acid would you add to the reaction mixture? (b) What product will be formed?

For each of the following procedures, explain the results observed. Assume that in a

complete system 3000 cpm (counts per minute) are found in acid-precipitable material at the end of 30 minutes and that values below 150 cpm are not significantly above the background level.

(c) 85 cpm is recovered when RNase A is added to the complete system. (d) 2900 cpm is recovered when chloramphenicol is added to the complete system. (e) 300 cpm is recovered when cyclohexamide is added to the complete system. (f)

640 cpm is recovered when puromycin is added to the complete system.

(g) 1518 cpm is recovered when puromycin and extra wheat germ tRNA are added to

the complete system.

(h) 120 cpm is recovered when poly(A) is used instead of poly(U). ANSWERS TO PROBLEMS

1. (a) The sequence of the first four amino acids of the protein is (formyl)Met-Glu-Arg-Tyr.

As the name implies, the template (antisense) strand of DNA serves as the template for the synthesis of a complementary mRNA molecule. (Remember that by convention nucleotide sequences are always written in the 5′ to 3′ direction unless otherwise specified.) The template strand of DNA and the mRNA synthesized are as follows:

DNA template strand:

5′-GTAGCGTTCCATCAGATTT-3′

mRNA:

3′-CAUCGCAAGGUAGUCUAAA-5′

Remember that the codons of an mRNA molecule are read in the 5′-to-3′ direction. Because this particular nucleotide sequence specifies the N-terminal region

528

CHAPTER 29

of an E. coli protein, the first amino acid must be (formyl)-methionine, which may be encoded by either AUG or GUG. Because there is no GUG and only a single AUG in the mRNA sequence, the location of the initiation codon can be established unambiguously. The portion of the mRNA sequence encoding protein and the first four amino acids it encodes are

mRNA:

5′-AUG-GAA-CGC-UAC-3′

Amino acid sequence:

(Formyl)Met-Glu-Arg-Tyr

(b) The expected amino acid sequence is Met-Leu-Met-Phe. The nucleotide sequences

on DNA and mRNA are

Sense strand of DNA:

5′-CCTGTGGATGCTCATGTTT-3′

mRNA:

5′-CCUGUGGAUGCUCAUGUUU-3′

In eukaryotes the first triplet specifying an amino acid is almost always the AUG that is closest to the 5′ end of the mRNA molecule. In this example there are two AUGs, so there will be two Met residues in the polypeptide that is produced. The reading frame and the resulting amino acids are as follows:

mRNA:

5′-CCUGUGG-AUG-CUC-AUG-UUU-3′

Amino acid sequence:

Met-Leu-Met-Phe

2. The sequence is His-Ala-Lys. The DNA and mRNA sequences are

Sense strand of DNA:

5′-CCATGCAAAGTAATAGGT-3′

mRNA:

5′-CCAUGCAAAGUAAUAGGU-3′

Since this sequence specifies the carboxyl end of the peptide chain, it must contain one or more of the chain-termination codons: UAA, UAG, or UGA. UAA and UAG occur in tandem in the sequence, so we can infer the reading frame. The mapping of the amino acid residues to the mRNA is as follows:

mRNA:

5′-C-CAU-GCA-AAG-UAA-UAG-GU-3′

Amino acid sequence:

His-Ala-Lys

3. The percentage of tRNAs that will be faulty is 0.11%. For every 1000 aminoacyl-adenylates

that are produced, 100 are faulty and 900 are correct. The 900 correct intermediates will be converted to correct aminoacyl tRNAs because the intermediates are tightly bound to the active site of the aminoacyl-tRNA synthetase. Of the 100 incorrect aminoacyladenylates, 99 will be hydrolyzed and will therefore not form aminoacyl tRNAs. Only one will survive to become an incorrect aminoacyl tRNA. The fraction of incorrect aminoacyl tRNAs is therefore 1/901, or 0.11%.

4. The Svedberg unit (S) is a sedimentation coefficient, which is a measure of the veloc

ity with which a particle moves in a centrifugal field. It represents a hydrodynamic property of a particle, a property that depends on, among other factors, the size and shape of the particle. When two particles come together, the sedimentation coefficient of the resulting particle should be less than the sum of the individual coefficients because there is no frictional resistance between the contact surfaces of the particles and the centrifugal medium.

PROTEIN SYNTHESIS 529

5. (a) The possible anticodons with which the codons might pair are as follows: Codon Possible anticodon

GUU

AAC, GAC, IAC

GUC

GAC, IAC

GUA

UAC, IAC

GUG

CAC, UAC

(b) Three codons could pair with the anticodon beginning with I; two codons could

pair with an anticodon beginning with U or G; only one codon could pair with an anticodon beginning with A or C.

6. Proline. The three codons that will pair with IGG—CCU, CCC, and CCA—all spec

ify proline.

7. A minimum of three tRNAs would be required. One tRNA having the anticodon UAA

could decode both UUA and UUG. For the other four codons, which have C in the first position and U in the second, there are two different combinations of two tRNAs each that could decode them. The first combination would be two tRNAs that have anticodons with A in the second position and G in the third, one with I in the first position to decode CUU, CUC, and CUA, and the other with U or C in the first position to decode CUG. The second combination would be two tRNAs that have anticodons with A in the second position and G in the third, one with G in the first position to decode CUU and CUC, and the other with U in the first position to decode CUA and CUG.

8. Threonyl-tRNA synthetase has a proofreading mechanism. Any Ser-tRNAThr that is mis

takenly formed is hydrolyzed by an editing site 20 Å from the activation site. The “decision” to hydrolyze the aminoacyl-tRNA appears to depend on the size of the amino acid substituent. If it is smaller than the correct amino acid, the amino acid fits into the hydrolytic site and is cleaved. If it is the same size as the correct amino acid, it does not fit and is not destroyed. Discrimination between amino acids that are larger than the correct one or are not isoelectronic with it occurs at the aminoacylation step.

9. If two different legitimate stop codons are present in tandem, it would be extremely im

probable that mutant tRNAs would exist for both and would simultaneously bind to each of them and thereby prevent proper chain termination.

10. (a) The tRNACys will become loaded with Ala rather than Cys, and as a result will in

sert Ala rather than Cys into polypeptide chains.

(b) In the case of a base change within a codon, only a single amino acid of a single

polypeptide is changed. In the case of a tRNA recognition mutation, amino acid substitutions at many positions of many polypeptides would occur.

11. A purine-rich mRNA sequence, three to nine nucleotides long (called the Shine-Dalgarno

sequence), which is centered about 10 nucleotides upstream of (to the 5′ side of) the start codon, base-pairs with a sequence of complementary nucleotides near the 3′ end of the 16S rRNA of the 30S ribosomal subunit. This interaction plus the association of fMet-tRNAf with the AUG in the P site of the ribosome sets the mRNA reading frame.

12. (a) Poly(U) codes for the incorporation of phenylalanine. Therefore, labeled pheny

lalanine must be added to the reaction mixture.

(b) Polyphenylalanine will be formed. 530

CHAPTER 29

plate for polyphenylalanine synthesis. Also the tRNA will be digested and the ribosomes damaged. No protein synthesis will take place.

(d) Chloramphenicol inhibits the peptidyl transferase activity of the 50S ribosomal sub

unit in prokaryotes but has no effect on eukaryotes so synthesis in a eukaryote is unaffected.

(e) Cyclohexamide inhibits the peptidyl transferase activity of the 60S ribosomal sub

unit in eukaryotes so synthesis is largely blocked.

(f)

Puromycin mimics an aminoacyl-tRNA and causes premature polypeptide chain termination leading to a low level of protein synthesis.

(g) The addition of extra wheat germ tRNA reduces the inhibiting effect of puromycin,

since they both compete for the A site on ribosomes. Therefore synthesis is increased over that in experiment (f ).

(h) Poly(A) directs the synthesis of polylysine; since there is no lysine (either labeled

or unlabeled) in the system, no product can be detected. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. The enzyme-bound Ile-AMP intermediate is necessary for the 32PPi exchange into ATP. Since

isoleucine is a requirement, labeled ATP will be formed only in (c).

2. Four bands: light, heavy, a hybrid of light 30S and heavy 50S, and a hybrid of heavy 30S

and light 50S. Recall that “ribosomes dissociate into 30S and 50S subunits after the polypeptide product is released.” As protein synthesis continues, 70S ribosomes are reformed from the various heavy and light subunits.

3. About 799 high-energy phosphate bonds are consumed—400 to activate the 200 amino

acids, 1 for initiation, and 398 to form 199 pepitide bonds.

4. Type I: b, c, and f; type 2: a, d, and e.

5. A mutation caused by the insertion of an extra base can be suppressed by a tRNA that

contains a fourth base in its anticodon. For example, UUUC rather than UUU is read as the codon for phenylalanine by a tRNA that contains 3′-AAAG-5′ as its anticodon.

6. One approach is to synthesize a tRNA that is acylated with a reactive amino acid analog.

For example, bromoacetylphenylalanyl-tRNA is an affinity-labeling reagent for the P site of E. coli ribosomes. See H. Oen, M. Pellegrini, D. Eilat, and C. R. Cantor. Proc. Nat. Acad. Sci. 70(1973):2799.

7. The sequence GAGGU is complementary to a sequence of five bases at the 3′ end of 16S

rRNA and is located several bases on the 5′ side of an AUG codon. Hence this region is a start signal for protein synthesis. The replacement of G by A would be expected to weaken the interaction of this mRNA with the 16S rRNA and thereby diminish its effectiveness as an initiation signal. In fact, this mutation results in a tenfold decrease in the rate of synthesis of the protein specified by this mRNA. For a discussion of this informative mutant, see J. J. Dunn, E. Buzash-Pollert, and F.W. Studier. Proc. Nat. Acad. Sci. 75(1978):2741.

8. Proteins are synthesized from the amino to the carboxyl end on ribosomes, and in the

reverse direction in the solid-phase method. The activated intermediate in ribosomal synthesis is an aminoacyl-tRNA; in the solid-phase method, it is the adduct of the amino acid and dicyclohexylcarbodiimide.

PROTEIN SYNTHESIS 531

9. The error rates of DNA, RNA, and protein synthesis are of the order of 10−10, 10−5, and 10−4

per nucleotide (or amino acid) incorporated. The fidelity of all three processes depends on the precision of base-pairing to the DNA or mRNA template. No error correction occurs in RNA synthesis. In contrast, the fidelity of DNA synthesis is markedly increased by the 3′

5′ proofreading nuclease activity and by postreplicative repair. In protein synthe

sis, the mischarging of some tRNAs is corrected by the hydrolytic action of the aminoacyltRNA synthetase. Proofreading also takes place when aminoacyl-tRNA occupies the A site on the ribosome; the GTPase activity of EF-Tu sets the pace of this final stage of editing.

10. GTP is not hydrolyzed until aminoacyl-tRNA is delivered to the A site of the ribosome.

An earlier hydrolysis of GTP would be wasteful because EF-Tu–GDP has little affinity for aminoacyl-tRNA.

11. The translation of an mRNA molecule can be blocked by antisense RNA, an RNA mol

ecule with the complementary sequence. The antisense–sense RNA duplex is degraded by nucleases. Antisense RNA added to the external medium is spontaneously taken up by many cells. A precise quantity can be delivered by microinjection. Alternatively, a plasmid encoding the antisense RNA can be introduced into target cells. For an interesting discussion of antisense RNA and DNA as research tools and drug candidates, see H. M. Weintraub. Sci. Amer. 262(January 1990):40.

12. (a) Intact protein isolated after only one minute will have been started with unlabeled

amino acids. (Only during the last minute of synthesis will label have been incorporated into the protein.) Therefore, the carboxyl-terminal peptide A5, the last segment to be synthesized, will be most heavily labeled.

(b) Due to the continuation of previously initiated chains, the order, from most labeled

to least, will reflect the reverse order of synthesis:

A >

5

A4

A3

A2

13. Aminoacyl-tRNA synthetases are the only components that actually match a nucleotide

sequence (the three-base RNA anticodon) with a particular amino acid to define the genetic code. All the other interactions of genetic code components involve simply “Watson-Crick” pairing between complementary bases.

14. The rate of protein synthesis would be slower because the cycling of EF-Tu between its

GTP-bound and GDP-bound forms would be slowed.

15. The nitrogen atom of the deprotonated a-amino group of aminoacyl-tRNA makes a nu

cleophilic attack on the ester bond of peptidyl-tRNA to form the new peptide bond. As a result, the growing peptide chain is transferred to the tRNA that bears the new amino acid. The tRNA that formerly held the peptide is released:

(f)-Met

(f)-Met

J

J (HNJ (HNJ CHR

CHR

J

KO)

KO) n n + 1

J

J

NH

:

NH

NH

J

RJ

R*J

R*J

H

J

OK

OK

J

J

OH

J

O

J

tRNA

tRNA„

tRNA„ 532

CHAPTER 29

16. The ornithinyl-tRNA is unstable because the nitrogen of the side chain will serve as a nu

cleophile to hydrolyze the ester bond to tRNA. The hydrolysis is facile because the transition state involves a six-membered ring. Lysyl-tRNA, by contrast, is more stable because a similar internal reaction would require a less favorable transition state with a sevenmembered ring.

J

NH

J

2

H

K

Self-hydrolysis of putative

N :

O

H

O

ornithinyl-tRNA

H

tRNA

17. EF-Ts catalyzes the exchange of GTP for GDP bound to EF-Tu. In G protein cascades,

an activated seven-helix receptor catalyzes GTP–GDP exchange in a G protein. For example, photoexcited rhodopsin triggers GTP–GDP exchange in transducin (see Section 32.3 in the text for further discussion).

18. Many G proteins are sensitive to ADP ribosylation by cholera toxin or pertussis toxin

(See Chapter 15). In each case, an ADP-ribose unit is transferred from NAD+, but the acceptor residue varies.For the modification of EF2 by diphtheria toxin, the acceptor is diphthamide (a derivative of histidine; see Figure 29.35 in the text), whereas the acceptors for the cholera toxin or pertussis toxin modifications of G proteins are arginine or cysteine.

19. (a) In Graph A, eIF4H exhibits two effects: (1) The higher slope observed at early re

action times shows that the rate of helix unwinding increases. (2) The extent of helix unwinding in the plateau region at late reaction times also increases.

(b) To establish that eIF4H by itself does not have inherent helicase activity. (c) Half-maximal activity was achieved with about 0.11 mM eIF4H, that is, about half

of the concentration of eIF4. Depending on the relative kinetics of association and dissociation, this result may suggest a stoichiometric 1:1 binding of the helper to the initiation factor.

(d) The upward displacement of the straight line indicates that eIF4H enhances the rate

of unwinding of all helices. The smaller slope when eIF4H is present indicates that the helper effect is greater for the more stable helices.

(e) Several answers are possible. Graph A shows that the helper enhances both the rate

and extent of helix unwinding. Both of these effects would result if the helper would slow the dissociation of eIF4 from the RNA helix. Such a mechanism would increase the processivity and also would be consistent with the energetics shown in Graph C. CHAPTER 3 The Integration of Metabolism 0

This chapter, which concludes the two major sections of the text devoted to me

tabolism, provides an integrated view of mammalian metabolism and a review of the principal themes of metabolism. The chapter starts with a recapitulation

of the roles of ATP, NADPH, and the building-block molecules derived from fuels in biosynthesis and cellular processes. The regulatory mechanisms that control metabolism, such as allostery, covalent modification, and compartmentation, are reviewed. (Chapters 10 and 17 contain important background material for this section.) The authors then review the major metabolic pathways (glycolysis, citric acid cycle and oxidative phosphorylation, pentose phosphate pathway, gluconeogenesis, and glycogen and fatty acid metabolism) as well as their principal sites of control (Chapters 16 through 22). The roles of glucose 6-phosphate, pyruvate, and acetyl CoA as key intermediates at junctions between the various metabolic pathways is discussed. The metabolic characteristics of the major organs are presented next, emphasizing the metabolic interchanges between the organs.

The authors then consider the ways in which the body responds to a series of

physiological conditions, such as the well-fed state, the early fasting state, and the refed state. The metabolic consequences of prolonged starvation are discussed, with attention to the priorities of metabolism in starvation, such as maintaining a blood-glucose level above 2.2 mM, and preserving protein by shifting the fuel being used from glucose to fatty acids and ketone bodies. The metabolic derangements in diabetes and the role of the hormones insulin and leptin in caloric homeostasis are discussed. The authors then turn to the fuel choices that the body makes during exercise and how those choices differ between aerobic and anaerobic activity. The chapter concludes with the ways in which ethanol alters energy metabolism in the liver and the adverse consequences of excess consumption. 533 534

CHAPTER 30 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Metabolism Consists of Highly Interconnected Pathways (Text Section 30.1)

1. Review the sources of ATP, NADPH, and the building-block molecules and their roles in

biosyntheses.

2. List the general mechanisms for the regulation of metabolism. Give examples of metabolic

regulation by each mechanism.

3. Locate the major metabolic pathways in the compartments of a eukaryotic cell.

4. Describe the roles of glycolysis, name its products, and describe the regeneration of NAD+

under aerobic and anaerobic conditions. Outline the regulation of phosphofructokinase in the liver.

5. Discuss the roles of the citric acid cycle, the electron transport chain, and oxidative phos- phorylation in the oxidative degradation of fuels. Explain the control of these pathways by the availability of ADP.

6. Describe the roles and the regulation of the pentose phosphate pathway.

7. Discuss the physiological role of gluconeogenesis and the reciprocal regulation of gluco

neogenesis and glycolysis.

8. Outline the synthesis and degradation of glycogen and their coordinated control by phos- phorylation and dephosphorylation and allosteric effectors.

9. Summarize the metabolism of fatty acids. Describe the regulation of fatty acid b-oxidation

and fatty acid synthesis. Explain how compartmentation is involved in these processes.

10. Describe the fates of glucose 6-phosphate in cells. Outline the pathways that give rise to

glucose 6-phosphate, and discuss the release of glucose into the blood.

11. Discuss the sources and fates of acetyl CoA and pyruvate. Each Organ Has a Unique Metabolic Profile (Text Section 30.2)

12. Summarize the fuel requirements of the brain and note the relationship of blood to brain

glucose levels. Describe the glucose transporter in the brain.

13. Explain the use of fuels by resting and active muscle and list the approximate fuel reserves

of a 70-kg man in kilocalories.

14. Describe the synthesis and turnover of triacylglycerols by adipose tissue. Indicate the role

of glucose in this tissue.

15. Discuss the role of the liver in providing glucose to other tissues and in regulating lipid

metabolism. List the fuels used by the liver for its own needs. Food Intake and Starvation Induce Metabolic Change (Text Section 30.3)

16. Explain and contrast the metabolic effects of insulin and glucagon.

17. Describe how the blood-glucose level is controlled by the liver in response to glucagon and

insulin in well-fed, early fasting, and refed states. Discuss the contributions of muscle and adipose tissue to the regulation of the blood-glucose level.

THE INTEGRATION OF METABOLISM 535

18. Describe the metabolic changes that occur after one and three days of starvation. Discuss

the metabolic adaptations that occur after prolonged starvation; note especially the shift in brain fuels and the decreased rate of protein degradation.

19. Describe the metabolic derangements in diabetes mellitus resulting from relative insulin

insufficiency and glucagon excess.

20. Explain the role of the signaling molecule leptin in maintaining caloric homeostasis and

appetite control. Fuel Choice During Exercise Is Determined by Intensity and Duration of Activity (Text Section 30.4)

21. Discuss the different patterns of fuel use in short- and long-distance running. Ethanol Alters Energy Metabolism in Liver (Text Section 30.5)

22. Describe the two pathways by which ethanol is metabolized and note the deleterious

effects of large levels of ethanol metabolism. SELF-TEST Metabolism Consists of Highly Interconnected Pathways

1. Consider the following examples of metabolic regulation:

(1) Fatty acid oxidation in mitochondria is diminished when fatty acid biosynthesis in the

cytosol is active due to the inhibition of carnitine acyltransferase I by malonyl CoA.

(2) The synthesis of HMG CoA reductase in various cells is inhibited by low-density

lipoproteins.

(3) Glucose 6-phosphatase is present in the liver and kidneys but not in muscle. (4) Amidophosphoribosyl transferase, the enzyme that catalyzes the committed step in

the biosynthesis of purine nucleotides, is inhibited by all purine nucleotides.

(5) The enzyme that catalyzes the synthesis and degradation of fructose 2,6-bisphos

phate is phosphorylated and dephosphorylated in response to hormonal signals.

Indicate which of these examples apply to each of the following modes of metabolic regulation:

(a) allosteric interactions (b) covalent modifications (c) enzyme levels (d) compartmentation (e) metabolic specialization of organs

2. Match each metabolic pathway in the left column with its major role in metabolism from

the right column.

(a) glycolysis

(1) control of glucose levels in blood

(b) gluconeogenesis

(2) formation of NADH and FADH2

(3) storage of fuel

(d) glycogen synthesis

(4) synthesis of NADPH and ribose

(e) fatty acid degradation

5-phosphate

(5) production of ATP and building blocks

of biomolecules

3. The control of phosphofructokinase in the liver and in muscle is different. Both epi

nephrine and glucagon initiate responses to low glucose levels, yet epinephrine stimulates glycolysis in muscle, whereas glucagon inhibits glycolysis in the liver. Explain this fact.

4. Which of the following answers completes the sentence correctly? Regulation of fatty

acid biosynthesis occurs at the enzymatic step catalyzed by

(a) carnitine acyltransferase I. (b) acetyl CoA carboxylase. (c) pyruvate carboxylase. (d) citrate synthase. (e) citrate-malate translocase.

5. Match the three key metabolic intermediates in the left column with their major prod

ucts from the right column. Indicate the most direct relationships, that is, those not separated by other key intermediates.

(a) glucose 6-phosphate

(1) ketone bodies

(b) pyruvate

(2) oxaloacetate

(3) pyruvate (4) glycogen (5) CO2 (6) lactate (7) ribose 5-phosphate (8) fatty acids (9) alanine

(10) cholesterol (11) acetyl CoA Each Organ Has a Unique Metabolic Profile

6. Which of the following statements about the metabolism of the brain are INCORRECT?

(a) It uses fatty acids as fuel in the fasting state. (b) It uses about 60% of the glucose consumed by the whole body in the resting state. (c) It lacks fuel reserves. (d) It can use acetoacetate and 3-hydroxybutyrate under starvation conditions. (e) It releases lactate during periods of intense activity.

7. Adipose cells constantly break down and resynthesize triacylglycerols, but synthesis

cannot proceed without an external supply of glucose. Explain why.

8. Which of the following statements about the metabolism of adipose tissue are correct?

(a) It has an active pentose phosphate pathway. (b) It contains a hormone-sensitive lipase that hydrolyzes triacylglycerols. (c) It uses ketone bodies as its preferred fuel. (d) It releases fatty acids to the blood as triacylglycerols that are packaged in VLDL. (e) It is the most abundant source of stored fuel.

9. In adipose tissue, glucose 6-phosphate is not converted into which of the following?

(a) pyruvate

(b) glycogen

(d) ribose 5-phosphate

THE INTEGRATION OF METABOLISM 537

10. In the liver, the major fates of pyruvate include the formation of which of the following?

(a) acetyl CoA

(b) lactate

(d) alanine

11. Which of the following tissues converts pyruvate to lactate most effectively?

(a) liver (b) muscle (c) adipose tissue (d) brain (e) kidney

12. Select the statements from the right column that best describe the metabolism of each

organ, tissue, or cell in the left column.

(a) brain

(1) releases glycerol and fatty acids into the

(b) muscle

blood during fasting periods

(2) in a normal nutritional state, utilizes

(d) liver

glucose as the exclusive fuel

(3) synthesizes ketone bodies when the

supply of acetyl CoA is high

(4) can release lactate into the blood (5) utilizes a-keto acids from amino acid

degradation as an important fuel

(6) can store glycogen but cannot release

glucose into the blood

(7) can synthesize fatty acids, triacyl

glycerols, and VLDL when fuels are abundant Food Intake and Starvation Induce Metabolic Change

13. When fuels are abundant, the liver does not degrade fatty acids; rather, it converts them

into triacylglycerols for export as very low-density lipoproteins (VLDL). Explain how boxidation of fatty acids and the formation of ketone bodies from fatty acids are prevented under these conditions.

14. Use an “S” to indicate the following metabolic processes that are stimulated by and an

“I” to indicate those that are inhibited by the action of insulin.

(a) gluconeogenesis in liver (b) entry of glucose into muscle and adipose cells (c) glycolysis in the liver (d) intracellular protein degradation (e) glycogen synthesis in liver and muscle (f)

uptake of branched-chain amino acids by muscle

(g) synthesis of triacylglycerols in adipose tissue

15. Explain the allosteric effects of glucose on glycogen metabolism.

16. The blood-glucose level of a normal person, measured after an overnight fast, is approxi

mately 80 mg/100 ml. After a meal rich in carbohydrate, it rises to about 120 mg/100 ml and then declines to the fasting level. The approximate time course of these changes and

538

CHAPTER 30

the inflection points is shown in Figure 30.1. After examining the figure, complete the following sentences:

(a) The increase in the glucose level from A to B is due to (b) The decrease in the glucose level from B to C is due to (c) The leveling off of the glucose level from C to D is due to (d) The slight overshoot that is sometimes observed at C can be explained by FIGURE 30.1 Blood-glucose levels after a meal rich in carbohydrate.

160

B

120

el (mg/100ml)

D

v

80 A

C

40

Blood glucose le

0

1

2

3

4

Time after a meal (hr)

17. Match the fuel storage forms in the left column with the most appropriate characteris

tics from the right column.

(a) glycogen

(1) largest storage form of calories

(b) triacylglycerols

(2) most readily available fuel during

muscular activity

(3) major source of precursors for glucose

synthesis during starvation

(4) depleted most rapidly during starvation (5) not normally used as a storage form

of fuel

18. Relative to the well-fed state, fuel utilization after three days of starvation shifts in which

of the following ways?

(a) More glucose is consumed by the brain. (b) Adipose tissue triacylglycerols are degraded to provide fatty acids to most tissues. (c) The brain begins to use ketone bodies as fuels. (d) Proteins are degraded in order to provide three-carbon precursors of glucose. (e) Glycogen is stored as a reserve fuel.

19. Metabolic adaptations to prolonged starvation include which of the following changes

relative to the metabolic picture after three days of starvation?

(a) The rate of lipolysis (mobilization of triacylglycerols) in the adipose tissue increases. (b) The glucose output by the liver decreases. (c) The ketone body output by the liver decreases. (d) The utilization of glucose by the brain decreases as the utilization of ketone bodies

increases.

(e) The rate of degradation of muscle protein decreases.

THE INTEGRATION OF METABOLISM 539

20. Show the changes in blood-glucose levels you would expect to see for an insulin

dependent diabetic patient after a meal rich in carbohydrate by plotting their time course on Figure 30.1 (see question 16). Explain your answer.

21. Which of the following occur in people with untreated diabetes?

(a) Fatty acids become the main fuel for most tissues. (b) Glycolysis is stimulated and gluconeogenesis is inhibited in the liver. (c) Ketone body formation is stimulated. (d) Excess glucose is stored as glycogen. (e) Triacylglycerol breakdown is stimulated.

22. Is it true or false that in diabetes the brain shifts to ketone bodies as its major fuel?

Explain. Fuel Choice During Exercise Is Determined by Intensity and Duration of Activity

23. List the following metabolic pathways or sources in the order of decreasing ATP pro

duction rate during strenuous exercise.

(a) muscle glycogen to CO2 (b) liver glycogen to CO2 (c) muscle glycogen to lactate (d) adipose tissue fatty acids to CO2 (e) muscle creatine phosphate

24. From the energy sources given in question 23, select the ones that provide most of the

energy in

(1) a 100-meter race (2) a 1000-meter race (3) a marathon race Ethanol Alters Energy Metabolism in Liver

25. Which of the following are consequences of ethanol consumption?

(a) accumulation of NADH (b) accumulation of NADPH (c) generation of acetaldehyde (d) generation of lactate (e) metabolism of triacylglycerols in the liver (f)

regeneration of glutathione ANSWERS TO SELF-TEST

1. (a) 1, 4 (b) 5 (c) 2 (d) 1 (e) 3

2. (a) 5 (b) 1 (c) 4 (d) 3 (e) 2, 5

3. The different effects of glucagon and epinephrine in the liver and in muscle are due to

the different properties of the kinase and phosphatase enzymes that catalyze the synthesis and degradation of fructose 2,6-bisphosphate in these organs. In the liver, the

540

CHAPTER 30

cAMP cascade leads to inhibition of the kinase and activation of the phosphatase. This decreases fructose 2,6-bisphosphate levels and inhibits glycolysis. In muscle, the phosphorylation of a homologous enzyme activates the kinase, thus stimulating the formation of fructose 2,6-bisphosphate, the activity of phosphofructokinase, and glycolysis.

4. b

5. (a) 3, 4, 7 (b) 2, 6, 9, 11 (c) 1, 5, 8, 10

6. a, e

7. The synthesis of triacylglycerols requires glycerol 3-phosphate, which is derived from

glucose. The glycerol that is released during triacylglycerol hydrolysis cannot be reutilized in adipose cells because they lack glycerol kinase. Thus, externally supplied glucose is required.

8. a, b, e

9. b, c

10. a, c. The conversions of pyruvate into lactate or alanine mainly occur in the muscles and

red cells. In the liver, pyruvate is mostly used in gluconeogenesis or for lipid synthesis.

11. b

12. (a) 2 (b) 4, 6 (c) 1 (d) 3, 5, 7

13. The selection of the pathway depends on whether or not the fatty acids enter the mito

chondrial matrix, the compartment of b-oxidation and ketone body formation. When citrate and ATP concentrations are high, as in the fed state, the activity of acetyl CoA carboxylase is stimulated. The resulting malonyl CoA, which is a precursor for fatty acid synthesis, inhibits carnitine acyltransferase I, which translocates fatty acids from the cytosol into the mitochondria for oxidation.

14. (a) I (b) S (c) S (d) I (e) S (f) S (g) S

15. Phosphorylase a binds glucose, which makes this enzyme susceptible to the action of

phosphatase. The resulting phosphorylase b is inactive; therefore, glycogen degradation is decreased. Since phosphorylase b does not bind phosphatase, phosphatase is released and activates glycogen synthase by dephosphorylating it, leading to the production of glycogen.

16. (a) The increase in the glucose level from A to B is due to the absorption of dietary

glucose.

(b) The decrease in the glucose level from B to C is due to the effects of insulin, which

is secreted in response to increased blood glucose. Glucose is removed from the blood by the liver, which synthesizes glycogen, and by muscle and adipose tissue, which store glycogen and triacylglycerols.

of glucagon and the diminished concentration of insulin. Glucagon maintains blood-glucose levels by promoting gluconeogenesis and glycogen degradation in the liver and by promoting the release of fatty acids, which partially replace glucose as the fuel for many organs.

(d) The slight overshoot that is sometimes observed at C can be explained by the contin

ued effects of insulin, which are not yet balanced by the metabolic effects of glucagon.

17. (a) 2, 4 (b) 1 (c) 3, 5

18. b, c, d

19. b, d, e. Answer (a) is incorrect because lipolysis remains essentially constant.

THE INTEGRATION OF METABOLISM 541

20. See Figure 30.2. In diabetic patients, the level of insulin is too low and that of glucagon

is too high, so after a meal the glucose levels will reach higher values than in a normal person. Also the removal of glucose from blood will be slower, and the fasting glucose levels in the blood may remain higher. FIGURE 30.2 Blood-glucose levels for a diabetic patient and a normal person after

a meal rich in carbohydrate.

200

160

120

Diabetic

el (mg/100ml) v

80

Normal

40

Blood glucose le

0

1

2

3

4

Time after a meal (hr)

21. a, c, e

22. False. Although ketone body concentrations in blood may become high in diabetes, glu

cose is even more plentiful. Therefore, the brain continues to use glucose as its major fuel.

23. e, c, a, d, b. Liver glycogen and adipose tissue fatty acids as fuels for active muscle are

the slowest and are about equivalent in terms of the maximal rate of ATP production (6.2–6.7 mmol/s). This rate is probably limited by the slow transport of the fuels from the storage sites to the muscle.

24. (1) c, e (2) a, c, e (3) a, b, d

25. a, c, and d. Ethanol consumption leads to production of NADH through the oxidation

of ethanol to acetaldehyde and acetaldehyde to acetate. The increase in NADH inhibits gluconeogenesis and leads to accumulation of lactate. NADPH is used by liver cytochrome P450-dependent pathways and inhibits the regeneration of glutathione. PROBLEMS

1. Cardiac muscle exhibits a high demand for oxygen, and its functioning is severely im

paired when coronary circulation is blocked.

(a) Considering the energy-generating substrates available to and used by the heart

under normal circumstances, why is oxygen required by heart muscle?

(b) Suppose that in an intact animal you can measure the concentrations of biochem

ical metabolites in the arteries leading to the cardiac muscle and in the veins carrying blood away from heart tissue. If the supply of oxygen to heart tissue is reduced, what differences in arterial and venous glucose concentrations will you

2. An infant suffering from a particular type of organic acidemia has frequent attacks of

vomiting and lethargy, which are exacerbated by infections, fasting, and the consumption of protein or fat. During these episodes, the patient suffers from hypoglycemia, which can be alleviated by injections of D-3-hydroxybutyrate. In addition, concentrations of ketone bodies in the blood are extremely low. The patient also has elevated concentrations of a number of organic acids in both blood and urine. Among these acids are 3-hydroxy-3-methylglutarate, b-methylglutaconate, and isovalerate. From the evidence of the buildup of these compounds, the enzyme that converts 3-hydroxy-3-methylglutaryl CoA (HMG CoA) to acetoacetate and acetyl CoA (HMG CoA cleavage enzyme) is probably deficient.

(a) How could this enzyme deficiency lead to a reduction in the concentration of ke

tone bodies?

(b) How does fasting exacerbate the symptoms of the disorder? (c) The consumption of fat causes a noticeable increase in the concentration of 3-hy

droxy-3-methylglutarate. Why?

(d) How can the administration of D-3-hydroxybutyrate relieve hypoglycemia?

3. Patients who remain unconscious after a serious surgical operation are given 100 to 150 g

of glucose daily through the intravenous administration of a 5% solution. This amount of glucose falls far short of the daily caloric needs of the patient. What is the benefit of the administration of glucose?

4. A biochemist in the Antarctic is cut off from his normal food supplies and is forced to

subsist on a diet that consists almost entirely of animal fats. He decides to measure his own levels of urinary ketone bodies, beginning on the day he starts the high-fat diet. What changes in urinary ketone body levels will he find?

5. In liver tissue, insulin stimulates the synthesis of glucokinase. What implications does

this have for a person who has an insulin deficiency?

6. Within a few days after a fast begins, nitrogen excretion accelerates to a relatively high level.

After several weeks, the rate of nitrogen excretion falls to a lower level. The excretion of nitrogen then continues at a relatively constant rate until the body is depleted of triacylglycerol stores; then the rate of urea and ammonia excretion again rises to a very high level.

(a) What events trigger the initial surge of nitrogen excretion? (b) Why does the nitrogen excretion rate decrease after several weeks of starvation? (c) Explain the increase in nitrogen excretion that occurs when lipid stores are exhausted.

7. Among the difficulties caused by prolonged fasting are metabolic disorders caused by vi

tamin deficiencies. What vitamins are needed during starvation to ensure that cells can continue to carry out the metabolic adaptations discussed in Section 30.3.1 of the text?

8. Describe the general fate of each of the following compounds in the mitochondria and

in the cytosol of a liver cell:

(a) palmitoyl CoA

(d) NADH

(b) acetyl CoA

(e) glutamate

(f)

malate

9. Young men who are championship marathon runners have levels of body fat as low as

4%, whereas most casual runners have levels ranging from 12% to 15%. Why would marathoners be at greater risk during prolonged fasting?

10. Assume that a typical 70-kg man expends about 2000 kcal of energy per day. If the

energy for his activities were all derived from ATP, how many grams of ATP would

have to be generated on a daily basis? How many grams of glucose would be required to drive the formation of the needed amount of ATP? The molecular weight of ATP is 500, and that of glucose is 180. One mole of glucose generates 686 kcal of energy when completely oxidized, and 7.3 kcal are required to drive the synthesis of one mole of ATP. Assume that 40% of the energy from the oxidation of glucose can be used for ATP synthesis.

11. Suppose that the concentration of mitochondrial oxaloacetate increases dramatically in

a liver cell. Briefly describe how the increase would affect each of the pathways below, and justify your answer.

(a) gluconeogenesis

(b) biosynthesis of palmitoyl CoA

(d) degradation of acetyl CoA

12. In chronically malnourished people or in healthy people who have missed one or two

meals, hypoglycemia develops rapidly with the ingestion of moderate amounts of ethanol. A reduced rate of hepatic glucose synthesis is observed, along with increases in intracellular ratios of lactate to pyruvate, of glycerol 3-phosphate to dihydroxyacetone phosphate, of glutamate to a-ketoglutarate, and of D-3-hydroxybutyrate to acetoacetate. In a well-fed person whose liver contains normal amounts of glycogen, ethanol infusion is less likely to induce hypoglycemia. The rate of hepatic glucose production is relatively normal. Increases in intracellular ratios of the pairs of compounds named above do occur, however.

(a) How do elevated intracellular ratios of NADH to NAD+ in response to ethanol in

fusion lead to an increase in ratios of the pairs of compounds named above?

(b) Briefly describe how an increase in the ratios of any of the pairs of compounds could

impair glucose synthesis in the liver of a malnourished person.

glycogen content is normal?

13. During starvation, the rate of liver-cell lipolysis accelerates, and the concentration of

acetyl CoA increases. These changes are accompanied by an increase in the activity of pyruvate carboxylase, which converts pyruvate to oxaloacetate. Give two reasons why it is desirable to increase oxaloacetate concentrations in starved cells. Also name a source of pyruvate during starvation, when glucose availability in the liver is low.

14. For several hours after birth, premature infants are particularly susceptible to hypo

glycemia and are also unable to rapidly generate ketone bodies. Describe how each of the characteristics below would contribute to hypoglycemia, low circulating levels of ketone bodies, or both.

(a) a large brain-to-body-weight ratio (b) a small store of liver glycogen (c) Low specific activity of cytosolic carnitine long-chain acyl-CoA transferase in liver (d) Very low levels of liver phosphoenolpyruvate carboxykinase

15. The Cori cycle (see Section 16.4.2 of the text) is especially important during early phases

of starvation, in which lactate molecules generated in the peripheral tissues are sent to the liver for use in gluconeogenesis.

(a) Why is it important for fatty acid oxidation to occur in the liver while the Cori cycle

is operating?

(b) Suppose that lactate from muscle were completely oxidized to carbon dioxide and

water in the liver. How would this make it more difficult for that organ to maintain glucose homeostasis during the early phases of starvation?

the Cori cycle. 544

CHAPTER 30 ANSWERS TO PROBLEMS

1. (a) Under normal conditions, heart muscle consumes acetoacetate and D-3-hydroxy

butyrate, both of which are converted to two molecules of acetyl CoA. Oxygen is required for terminal oxidation of acetyl CoA in the citric acid cycle.

(b) Cardiac cells deprived of oxygen are unable to generate metabolic energy by oxi

dizing acetyl CoA. An alternative source of energy is glucose, which can be converted to lactate under anaerobic conditions, with the subsequent generation of ATP. Thus, under anaerobic conditions, glucose uptake by heart muscle increases; the concentration of glucose in cardiac veins will decrease relative to glucose levels in coronary arteries. The concentration of lactate in the coronary veins is also elevated, compared with the levels in the coronary arteries.

2. (a) The production of the ketone bodies acetoacetate and D-3-hydroxybutyrate is de

pendent on the activity of HMG CoA cleavage enzyme, which is deficient in the infant.

(b) Fasting causes an increase in acetyl CoA production through the increased rates of

lipolysis that occur in the attempt to generate sources of metabolic energy. Any increase in acetyl CoA concentration stimulates HMG CoA synthesis, and the inability to convert HMG CoA to acetoacetate leads to an increase in concentrations of 3-hydroxy-3-methylglutarate, which is excreted by the cells into the plasma.

acetyl CoA, from the b-oxidation of fatty acids. As noted in the answer to (b), elevation in the level of acetyl CoA stimulates the production of HMG CoA, leading to an increase in the concentration of 3-hydroxy-3-methylglutarate.

(d) Because the liver is unable to generate normal levels of ketone bodies, those tissues

that normally utilize them are required to use other substrates as metabolic fuels. Since glucose is the substrate of choice in such situations, increased demand for blood glucose results in hypoglycemia. D-3-Hydroxybutyrate can be oxidized into the ketone body acetoacetate, and both can be used as fuel sources for the brain and heart. The administration of D-3-hydroxybutyrate therefore provides an alternative source of metabolic energy, conserving glucose and reducing hypoglycemia

3. The limited amount of glucose is given to prevent the hydrolysis of muscle protein dur

ing fasting. It serves as a source of energy for the brain and blood cells; otherwise, during fasting, body proteins are hydrolyzed to provide carbon atoms for the generation of glucose by gluconeogenesis in the liver and kidney.

4. A high-fat diet will stimulate ketone body formation because the oxidation of fatty acids causes

an increase in acetyl CoA concentration, which in turn stimulates the production of ketone bodies. The overall profile of ketone body production may resemble that found during starvation because a source of carbon for glucose production will be lacking. However, dietary fats will serve as a source of energy instead of the triacylglycerols stored in body tissue.

5. Under normal conditions, glucokinase acts to phosphorylate glucose when concentra

tions of the hexose are high. The failure to synthesize sufficient quantities of glucokinase means that the liver cannot control the levels of glucose in blood, compounding the other difficulties caused by insulin deficiency described in the text.

6. (a) During the first few days of starvation, the brain continues to utilize glucose.

Glycogen stores are exhausted, so the primary source of carbon atoms for gluconeogenesis is amino acids. Because the concentration of free amino acids in the tissues is limited, body proteins are broken down to provide the amino acids to

support gluconeogenesis. Nitrogen excretion increases because the amino groups of those amino acids are eliminated as urea.

(b) After several weeks of fasting, the brain adapts to the utilization of ketone bodies

as a source of energy, so less glucose is required. The resulting reduction in gluconeogenesis means a reduction in the rate of oxidation of amino acids and in the production of ammonia and urea.

as a source of glucose for the brain but also as a source of energy for all other tissues. These requirements cause a great increase in the rate of body protein catabolism, with corresponding increases in amino acid oxidation and nitrogen excretion. Often more than a kilogram per day in weight is lost, indeed causing a threat to life.

7. Most of the vitamins and cofactors discussed in previous chapters of the text would be

needed during starvation because many of the essential metabolic pathways must continue to operate. Among the most obvious vitamins needed for those pathways are pyridoxal phosphate (for the transamination of amino acids), niacin and riboflavin (for electron transport), thiamin (for the oxidative decarboxylation of pyruvate, a-ketoglutarate, and the branched-chain amino acids), biotin (for the carboxylation of pyruvate), and cobalamin (for the conversion of methylmalonyl CoA to succinyl CoA).

8. (a) In the cytoplasm, the acyl chain of palmitoyl CoA can be esterfied to glycerol as

part of the process of triacylglycerol formation in the liver cytosol, or the palmitoyl chain can be transferred to carnitine for subsequent transport to the mitochondria. In mitochondria, palmitoyl CoA is oxidized to CO2 and H2O.

(b) In mitochondria, acetyl CoA is oxidized to carbon dioxide and water, or it can un

dergo carboxylation to form oxaloacetate. It can also be used for the synthesis of ketone bodies. In the cytosol acetyl CoA serves as a precursor of malonyl CoA, as well as a precursor of HMG CoA and cholesterol.

rulline in the urea cycle, whereas in the cytosol, it serves as a precursor of pyrimidines when it condenses with aspartate to yield carbamoyl aspartate.

(d) In mitochondria, NADH is reoxidized to NAD+ in the electron transport chain. In

the cytosol, it can be reoxidized to NAD+ through the action of lactate dehydrogenase or NADPH dehydrogenase or by means of the glycerol–phosphate or malate– aspartate shuttles.

(e) In mitochondria, glutamate undergoes oxidative deamination, yielding ammonia

and a-ketoglutarate. It can also serve as a source of amino groups for a number of aminotransferase enzymes. In the cytosol, it can also serve as an amino donor for aminotransferases. Glutamate can also be used in the cytosol as a precursor of glutamine, and it can of course be incorporated into newly synthesized protein.

(f)

Malate is converted to oxaloacetate in the citric acid cycle, which takes place in the mitochondria. In the cytoplasm, as a component of the malate–aspartate shuttle, it serves as an electron carrier to transfer electrons from NADH to the inner mitochondrial membrane. Malate can also be used as a source of electrons for the generation of NADPH in the reaction catalyzed by malic enzyme.

9. The greater the percentage of body fat, the larger the reserves of triacylglycerols, which

are the primary source of metabolic energy during fasting. Once these reserves are depleted, the body accelerates the breakdown of muscle protein as an energy source. Extensive hydrolysis of muscle tissue threatens many vital body functions and can lead to death. 546

CHAPTER 30

10. Since 7.3 kcal are required to drive the synthesis of 1 mole of ATP, the caloric value of

the energy in each mole of ATP is 7.3 kcal. Therefore, the amount of ATP needed per day is

2000 kcal/day

= 274 mol ATP per day

7.3 kcal/ mol ATP

274 mol ATP/day × 500 g/mol = 137 kg ATP per day

If the oxidation of 1 mole of glucose yields 686 kcal of energy, and if 40% can be used to drive the synthesis of ATP, glucose yields (0.4)(686 kcal/mol) = 274 kcal/mol of usable energy. Therefore, the amount of glucose needed per day is

2000 kcal/day

= 7 4

. mol glucose per day

274 kcal/ mol glucose

7.3 mol glucose/day × 180 g/mol = 1.31 kg glucose per day

Normal fuel stores available in the blood to the typical 70-kg man include about 250 g of glycogen and approximately 60 g of glucose. These figures make it evident that humans can rely on stores of carbohydrate for only a short time.

11. (a) Oxaloacetate is a source of carbon atoms for the synthesis of phosphoenolpyruvate,

an intermediate in gluconeogenesis. An increase in oxaloacetate concentration will therefore stimulate gluconeogenesis.

(b) In mitochondria, oxaloacetate combines with acetyl CoA to form citrate. When cit

rate levels increase, the citrate is shuttled across the mitochondrial membrane where it is a source of cytosolic acetyl CoA. Formation of malonyl CoA from acetyl CoA allows synthesis of palmitoyl CoA through the action of the fatty acyl CoA synthase complex. An increase in oxaloacetate concentration will therefore stimulate synthesis of palmitoyl CoA.

CoA. As discussed in (b), an increase in oxaloacetate concentration results in an increase in cytosolic concentrations of acetyl CoA, which could drive the formation of additional cholesterol.

(d) The synthesis of citrate in the mitochondrion is stimulated by higher concentrations

of oxaloacetate, which serves as an acceptor of acetyl units from acetyl CoA. An increase in mitochondrial citrate concentration accelerates the rate of the reactions of the citric acid cycle, the function of which includes the oxidation of acetyl CoA. Therefore an increase in oxaloacetate concentration will stimulate oxidation or degradation of acetyl CoA.

12. (a) Each pair of compounds referred to in the problem is interconvertible through

oxidation–reduction reactions linked to the NADH/NAD+ redox pair. An increased NADH/NAD+ ratio will therefore suppress net oxidation reactions because the NAD+ needed to serve as an electron acceptor (e.g., for conversion of lactate to pyruvate) is limited in concentration. Thus an ethanol-induced increase in NADH will suppress conversion of lactate to pyruvate, of glycerol 3-phosphate to dihydroxyacetone phosphate, of glutamate to a-ketoglutarate, and of D-3-hydroxybutyrate to acetoacetate.

(b) During starvation the liver carries out gluconeogenesis using lactate, amino acids,

and a-glycerol phosphate (from triacylglycerols) as initial substrates. These compounds are converted via NAD+-linked oxidation to compounds like pyruvate,

a-ketoglutarate, and dihydroxyacetone phosphate, all directly in the pathway to glucose synthesis. The inability of the cell to produce these and other compounds (like oxaloacetate) because of the lack of NAD+ would result in a low level of glucose production. The acetoacetate/D-3-hydroxybutyrate pair is not involved in gluconeogenesis; these are ketone bodies. However, the inability to convert D-3-hydroxybutyrate to acetoacetate would interfere with the terminal oxidation of these compounds as well.

and gluconeogenesis through the liver’s utilization of amino acids and other compounds will be somewhat impaired. However, stored hepatic glycogen serves as a source of glucose in response to any drop in blood sugar levels. Because liver glycogen levels are virtually depleted after 24 to 36 hours of starvation, ethanol-induced hypoglycemia can develop rapidly in malnourished people.

13. Increased concentrations of oxaloacetate are needed to provide more acceptors of acetyl

groups from acetyl CoA to form citrate, ensuring that the citric acid cycle can operate at higher capacity. This makes it able to oxidize the increasing amounts of acetyl CoA present in the cell. Elevated levels of oxaloacetate are also required to provide more molecules that can serve as precursors for gluconeogenesis. During starvation, liver cells increase their rate of glucose formation through glycogenolysis and gluconeogenesis, in order to provide more glucose to peripheral tissues. Gluconeogenesis depends primarily on the availability of oxaloacetate molecules, which are converted first to phosphoenolpyruvate and then ultimately to glucose. One source of pyruvate during starvation is alanine, produced by the degradation of muscle tissue proteins. Alanine, along with glutamine, serves as a carrier of carbon atoms and nitrogen from muscle to liver. Alanine is converted to pyruvate by aminotransferase enzymes, which use another a-keto acid as an acceptor of the amino group from alanine.

14. (a) Because brain tissue preferentially uses glucose as a fuel, the demand for glucose in

neonates is disproportionately high compared with that of an older person with a lower brain-to-body weight ratio.

(b) Low reserves of liver glycogen mean that the ability of the liver to export glucose

synthesized from glycogen is limited.

fuel during hypoglycemia because the low specific activity of carnitine acyl transferase limits transport of long-chain fatty acids across the mitochondrial membrane. The depletion of fatty acids in the mitochondrion means that only limited amounts of acetyl CoA from fatty chain oxidation are available for synthesis of acetoacetyl CoA and b-hydroxybutyrate.

(d) The liver cannot effectively carry out gluconeogenesis because the activity of phos

phoenolpyruvate carboxykinase (which carries out synthesis of PEP from oxaloacetate), a key enzyme in glucose synthesis, is present only at a very low level.

15. (a) Because most of the ATP used for gluconeogenesis in the liver is generated by

b-oxidation of fatty acids.

(b) Compounds of the Cori cycle, such as lactate and pyruvate, provide a readily avail

able source of carbon for gluconeogenesis in the liver. If those molecules were unavailable, additional carbon atoms from proteolysis would be required to sustain the level of glucose required through production by gluconeogenesis.

to the liver, where it can be disposed of as urea. Conversion of pyruvate to alanine means that the electrons in NADH normally consumed in the conversion of pyruvate to lactate can now be sent to the mitochondrion to be used to drive ATP synthesis. 548

CHAPTER 30 EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. The liver contains glucose 6-phosphatase, whereas muscle and the brain do not. Hence,

muscle and the brain, in contrast to the liver, do not release glucose. Another key enzymatic difference is that the liver has little of the transferase needed to activate acetoacetate to acetoacetyl CoA. Consequently, acetoacetate and 3-hydroxybutyrate are exported by the liver for use by heart muscle, skeletal muscle, and the brain.

2. (a) Adipose cells normally convert glucose to glycerol 3-phosphate for the formation

of triacylglycerols. A deficiency of hexokinase would interfere with the synthesis of triacylglycerols.

(b) A deficiency of glucose 6-phosphatase would block the export of glucose from the

liver following glycogenolysis. This disorder (called von Gierke disease) is characterized by an abnormally high content of glycogen in the liver and a low bloodglucose level.

acids. Fasting and exercise precipitate muscle cramps in these people.

(d) Glucokinase enables the liver to phosphorylate glucose even in the presence of a

high level of glucose 6-phosphate. A deficiency of glucokinase would interfere with the synthesis of glycogen.

(e) Thiolase catalyzes the formation of two molecules of acetyl CoA from acetoacetyl

CoA and CoA. A deficiency of thiolase would interfere with the utilization of acetoacetate as a fuel when the blood sugar level is low.

(f)

Phosphofructokinase will be less active than normal because of the lowered level of F-2,6-BP. Hence, glycolysis will be much slower than normal.

3. (a) A high proportion of fatty acids in the blood are bound to albumin. Cerebrospinal

fluid has a low content of fatty acids because it has little albumin.

(b) Glucose is highly hydrophilic and soluble in aqueous media, in contrast to fatty

acids, which must be carried by transport proteins such as albumin. Micelles of fatty acids would disrupt membrane structure.

4. (a) A watt is equal to 1 joule (J) per second (0.239 calorie per second). Hence, 70 W

is equivalent to 0.07 kJ/s or 0.017 kcal/s.

(b) A watt is a current of 1 ampere (A) across a potential of 1 volt (V). For simplicity,

let us assume that all the electron flow is from NADH to O2 (a potential drop of 1.14 V). Hence, the current is 61.4 A, which corresponds to 3.86 × 1020 electrons per second (1 A = 1 coulomb/s = 6.28 × 1018 charges/s).

formed per 0.80 electron transferred. A flow of 3.86 × 1020 electrons per second therefore leads to the generation of 4.83 × 1020 ATP per second or 0.8 mmol per second.

(d) The molecular weight of ATP is 507. The total body content of ATP of 50 g is equal

to 0.099 mol. Hence, ATP turns over about once per 125 seconds when the body is at rest.

5. (a) The stoichiometry of complete oxidation of glucose is

C

+

6H12O6

6 O2

6 CO2

6 H2O

and that of tripalmitoylglycerol is

C

+

51H98O6

72.5 O2

51 CO2

49 H2O

Hence, the RQ values are 1.0 (6/6) and 0.703 (51/72.5), respectively.

THE INTEGRATION OF METABOLISM 549

(b) An RQ value reveals the relative usage of carbohydrate and fats as fuels. The RQ of

a marathon runner typically decreases from 0.97 to 0.77 during the race. The lowering of the RQ reflects the shift in fuel from carbohydrate to fat.

6. One gram of glucose (molecular weight 180.2) is equal to 5.55 mmol, and one gram of

tripalmitoylglycerol (molecular weight 807.3) is equal to 1.24 mmol. The reaction stoichiometries (see problem 5) indicate that 6 mol of H2O are produced per mole of glucose oxidized, and 49 mol of H2O per mole of tripalmitoylglycerol oxidized. Hence, the H2O yields per gram of fuel are 33.3 mmol (0.6 g) for glucose and 60.8 mmol (1.09 g) for tripalmitoylglycerol. Thus, complete oxidation of this fat gives 1.82 times as much water as does glucose. Another advantage of triacylglycerols is that they can be stored in essentially anhydrous form, whereas glucose is stored as glycogen, a highly hydrated polymer (see Section 22.1 in the text). A hump consisting mainly of glycogen would be an intolerable burden—far more than the straw that broke the camel’s back!

7. A typical macadamia nut has a mass of about 2 g. Because it consists mainly of fats (~9

kcal/g), a nut has a value of about 18 kcal. The ingestion of 10 nuts results in an intake of about 180 kcal. As was discussed in problem 4, a power consumption of 1 W corresponds to 0.239 calorie per second, and so 400-W running requires 95.6 cal/s, or .0956 kcal/s. Hence, one would have to run 1882 s, or about 31 min, to spend the calories provided by 10 nuts.

8. A high blood-glucose level would trigger the secretion of insulin, which would stimu

late the synthesis of glycogen and triacylglycerols. A high insulin level would impede the mobilization of fuel reserves during the marathon

9. Insulin-dependent diabetes is characterized by high levels of blood glucose due to poor

entry of glucose into cells. The impaired carbohydrate utilization leads to uncontrolled breakdown of lipids to acetyl-CoA. However, much of the acetyl-CoA cannot enter the citric acid cycle because of a shortage of oxaloacetate; furthermore, the acetyl-CoA cannot be converted to pyruvate or glucose. Acetyl-CoA therefore will be converted back to triacylglycerides, some of which will accumulate in the bloodstream.

10. Glycolysis is inhibited in the liver so that available glucose can be saved for use by the

brain. Meanwhile, the liver supplies its energy needs by oxidizing fatty acids.

11. Electron transfer pathways depend on reactions in both compartments. For example,

NADH is produced in both the cytoplasm and the mitochondria. NADH equivalents from glycolysis must be transported into the mitochondria by the glycerol–phosphate shuttle or malate–aspartate shuttle. Furthermore, ATP that is produced in the mitochondria must be transported specifically to the cytoplasm to support the energy needs of many reactions.

12. (a) Insulin inhibits lipolysis. An abundance of glucose and fatty acids in adipose tissue

will lead to synthesis and storage of triacylglycerols.

(b) Insulin promotes uptake of branched amino acids and has a general stimulating ef

fect on protein synthesis and inhibitory effect on protein degradation. Nevertheless, the individual will continue to be protein deficient due to the poor diet.

fluid to leak into extracellular spaces.

13. The oxygen will serve as the ultimate acceptor of electrons from NADH, as important re

covery reactions take place. During the recovery, lactate that was produced during exercise will be converted back into pyruvate (primarily in the liver), with the concomitant production of NADH from NAD+ by lactate dehydrogenase. Electrons from NADH will pass through the electron-transport chain to oxygen, producing NAD+ in addition to ATP. The NAD+ will be used to oxidize more lactate and will be available as an electron

550

CHAPTER 30

acceptor for future glycolysis, when needed. Some of the ATP will be used to regenerate phosphocreatine and some for gluconeogenesis, to replenish the expended supplies of glucose and glycogen, in the liver and muscle.

14. Excess oxygen is needed because thermodynamic machines, including mammalian bod

ies, are less than 100% efficient. Some of the energy is lost as heat, and additional energy is expended because gluconeogenesis (to replenish muscle glycogen) is not the thermodynamic or chemical reverse of glycolysis. Rather, the resynthesis of glucose from lactate requires more ATP than is produced by anaerobic glycolysis. The amount of excess oxygen consumed typically is about 15% of the total oxygen consumed during exercise (see J. Appl. Physiol., 62[1987]:485–490).

15. Many brain functions depend on a balance between excitatory and inhibitory neuro

transmission. It is likely that the diverse effects of ethanol result from alterations in this balance. Although ethanol interacts with several receptor and channel systems, the detailed mechanisms are not yet understood.

16. One possible approach would be to attempt to fix samples for microscopy under aero

bic and anaerobic conditions. Perhaps differences in fiber morphology or crossbridge formation could be observed (particularly for type I fibers) in the presence and absence of oxygen. Alternatively, the subsclasses of myosin differ in type I and type II fibers and can be distinguished using specific antibodies as labels that can be viewed bound to the respective fiber types in the electron microscope (see J. Cell. Biol., 90[1981]:128–144).

17. (a)

(

,

200 000 kcal)(503 gmol 1

− ) = .84 × 106 g

1

12 kcalmol−

(b) Multiply the answer from part (a) by the cost per gram to obtain a total cost of

$1.2 billion! CHAPTER 3 The Control of Gene Expression 1

In this chapter, the authors describe the biochemistry underlying several mechanisms

that control gene expression in prokaryotes and eukaryotes. The lactose (lac) operon in bacteria is an example of a mechanism in which the initiation of transcription is

regulated. Negative control is exerted through the binding of a repressor protein to DNA carrying the lac operator. Positive control of the lac operon is accomplished by a complex of catabolite activator protein (CAP) and cyclic AMP (cAMP). The complex binds near the lac promoter in the absence of the repressor and through proteinprotein interactions stimulates the activity of RNA polymerase at the lac promoter.

The text then explains how, in eukaryotic cells, the abundance of DNA, chro

mosome structure, and the existence of the cellular differentiation and the cell cycle complicate the control of gene expression. The authors describe the structure of nucleosomes, which are the lowest-level, repeating units of chromatin. Each nucleosome consists of DNA wrapped around an octameric protein core composed of histones. Individual nucleosomes are linked together by intervening stretches of DNA. When genes are not being expressed, the nucleosomes are packed, along with other proteins, to form a highly condensed chromosome.

The text describes the modifications of chromatin structure that occur during tran

scription as an introduction to the topic of the regulation of eukaryotic gene expression. The authors describe how enhancers and combinations of regulatory proteins, DNA modification by methylation, and histone modifications by acetylation and deacetylation effect gene regulation. The text describes the steroid-hormone estrogen receptor that uses the zinc-finger motif to bind specifically estrogen response elements in DNA and thereby directly activate transcription. The mechanisms of coactivators and corepressors are also described. The cyclic AMP-response element binding protein (CREB) and the coactivator CREB-binding protein (CBP) are described as an example of how membrane-associated receptors acting through phosphorylation cascades can also regulate transcription. 551 552

CHAPTER 31

The phenomenon of attenuation of translation of the trp operon in bacteria is provided

as an example of posttranscriptional gene regulation. This mechanism, which is used by several amino acid biosynthetic operons, relies on alternative RNA secondary structures and on the coupling of transcription and translation in prokaryotes. The regulation of iron metabolism in animals is presented to show how RNA secondary structures can by bound specifically by proteins and thereby regulate translation.

In preparation for studying this chapter, you should review Chapter 5, Sections 5.4 and

5.6, and Chapter 28. Material on page 748 of the text describes the functional groups on DNA that can serve as determinants for specific interactions with proteins. Section 9.3.3 describes the binding of EcoRV endonuclease to DNA. These examples provide the principles by which the proteins described in this chapter could interact with specific sequences of DNA. LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Introduction

1. Define gene expression and indicate the primary level of its regulation during expression

of the genetic information. Prokaryotic DNA-Binding Proteins Bind Specifically to Regulatory Sites in Operons (Text Section 31.1)

2. Outline the metabolism of lactose in E. coli. Draw the structure of lactose, describe its

entry into the cell, and write the equations for the reactions catalyzed by b-galactosidase, providing both the substrates and the products.

3. Recount the observations by Jacob and Monod that led to the concept of the lac operon

and its regulation. Define “operon.”

4. Draw the genetic map of the lac operon and outline the functions of the promoter, repres- sor, operator, and inducer in controlling the production of polycistronic lac mRNA. Distinguish between regulatory and structural genes.

5. Describe the role of the lac repressor in determining whether a bacterium is inducible or constitutive (continuously expressed) for b-galactosidase.

6. Describe the subunit structure of the lac repressor and relate it to the symmetrical sequence

of the lac operator. Compare the affinities of the lac repressor for lac operator DNA and nonspecific DNA. Describe the effect of the binding of allolactose or isopropylthiogalactoside by the repressor on its affinity for the operator.

7. Appreciate that many other gene-regulatory networks in prokaryotes function like the lac operon.

8. Explain the functions of cAMP and the CAP in modulating the expression of the lac

operon.

9. Diagram the relative positions of cAMP–CAP complex, RNA polymerase, and lac repres

sor on the DNA template. Explain their effects on one another and on the DNA structure.

10. Describe the helix-turn-helix motif and relate it to DNA binding. Note the existence of

other DNA-binding motifs.

THE CONTROL OF GENE EXPRESSION 553 The Greater Complexity of Eukaryotic Genomes Requires Elaborate Mechanisms for Gene Regulation (Text Section 31.2)

11. State the relative amounts of genomic DNA in the haploid genomes of E. coli (bacterium), S. cerevisiae (yeast), and H. sapiens (human).

12. Describe the composition and structure of the nucleosome, and relate the nucleosome to

the proposed structure of the chromatin fiber.

13. Describe the composition of chromatin. List the types of histones and describe their gen

eral characteristics. Note the evolutionary stability of the sequences of the H3 and H4 histones and assign the types of histones to their locations within the nucleosome.

14. Distinguish between the DNA associated with the nucleosome core and that in the inter- nucleosome linker.

15. Relate hypersensitivity to DNase I to chromatin structure. Discuss the implications of

the chromatin immunoprecipitation experiments with yeast protein GAL4 concerning chromatin structure.

16. Describe enhancers and outline their mechanism of action.

17. Relate the covalent modification of DNA by methylation to gene expression. Describe CpG islands. Transcriptional Activation and Repression Are Mediated by Protein–Protein Interactions (Text Section 31.3)

18. Outline how the combination of transcription factors gives rise to cell-specific transcription.

19. Contrast the mechanism of the steroid hormones with hormones initiating their actions

through interactions with a transmembrane receptor.

20. Describe the zinc finger structure of nuclear hormone receptors and describe how it is in

volved in the interaction of the estrogen–estrogen receptor complex interaction with estrogen response elements.

21. Outline the roles of coactivators and corepressors in transcription complexes. Contrast mech

anisms of positive and negative gene regulation in eukaryotes.

22. Explain how Tamoxifen serves as an anticancer agent.

23. Outline the covalent modification and demodification of histone tails.

24. Provide examples of phosphorylation cascades that regulate transcription. Gene Expression Can be Controlled at Posttranscriptional Levels (Text Section 31.4)

25. Provide an overview of the regulation of the tryptophan (trp) operon by attenuation.

26. Sketch the RNA secondary structures of the trp attenuator and antiterminator and relate

them to control of the trp operon.

27. Explain the function of leader peptides and the consequences of the coupling of transcrip- tion and translation in the regulation of several biosynthetic operons.

28. Outline the role of RNA secondary structure in the regulation of iron metabolism in animals.

Describe the roles of transferrin, transferrin receptor, ferritin, the iron-response element, and IRE-binding protein. Relate the IRE-binding protein to aconitase and iron sensing. 554

CHAPTER 31 SELF-TEST Prokaryotic DNA-Binding Proteins Bind Specifically to Regulatory Sites in Operons

1. Which of the following are common mechanisms used by bacteria to regulate their meta

bolic pathways?

(a) control of the expression of genes (b) control of enzyme activities through allosteric activators and inhibitors (c) formation of altered enzymes by the alternative splicing of mRNAs (d) deletion and elimination of genes that specify enzymes (e) control of enzyme activities through covalent modifications

2. Which of the following statements about b-galactosidase in E. coli are correct?

(a) It is present in varying concentrations, depending on the carbon source used

for growth.

(b) It is a product of a unit of gene expression called an operon. (c) It hydrolyzes the b-1,4-linked disaccharide lactose to produce galactose and glucose. (d) It forms the b-1,6-linked disaccharide allolactose. (e) It is activated allosterically by the nonmetabolizable compound isopropylthio

galactoside (IPTG).

(f)

Its levels rise coordinately with those of galactoside permease and thiogalactoside transacetylase.

3. Match each feature or function in the right column with the appropriate DNA sequence

element of the lac operon in the left column.

(a) i

(1) contains a specific binding sequence

(b) p

for the lac repressor

(2) encodes a galactoside permease

(d) z

(3) contains a binding sequence for the

(e) y and a

cAMP–CAP complex

(f)

CAP binding site

(4) encodes a protein that interferes with

the activation of RNA polymerase

(5) encodes a protein that binds allolactose (6) contains a specific binding sequence

for RNA polymerase

(7) encodes b-galactosidase (8) is a regulatory gene (9) is the lac promoter

(10) encodes thiogalactoside transacetylase (11) is the lac operator (12) encodes the lac repressor

4. Explain the stoichiometric relationship among the concentrations of b-galactosidase,

galactoside permease, and thiogalactoside transacetylase in E. coli.

5. When E. coli is added to a culture containing both lactose and glucose, which of the sug

ars is metabolized preferentially? What is the mechanism underlying this selectivity?

6. What happens after the first-used sugar is depleted during the experiment described in

question 5?

7. Which of the following statements about the cAMP–CAP complex are correct?

(a) It protects the :87 to :49 sequence of the lac operon from nuclease digestion. (b) It protects the :48 to ;5 sequence of the lac operon from nuclease digestion.

THE CONTROL OF GENE EXPRESSION 555

(c) It protects the :3 to ;21 sequence of the lac operon from nuclease digestion. (d) It affects RNA polymerase activity in a number of operons. (e) Upon binding to the lac operon, it contacts RNA polymerase.

8. The helix-turn-helix motif

(a) is a protein-folding pattern. (b) is observed in a variety of prokaryotic DNA-binding proteins. (c) contains a recognition helix that inserts itself into the minor groove of DNA. (d) is often observed in proteins that bind DNA as dimers. The Greater Complexity of Eukaryotic Genomes Requires Elaborate Mechanisms for Gene Regulation

9. Which of the following statements about DNA that has been isolated from a eukaryotic

chromosome are correct?

(a) It is resistant to breakage by shearing forces because of its large size. (b) It is linear and unbranched. (c) It can be sized by a viscoelastic technique, which measures the time a stretched and

elongated molecule takes to relax to its normal conformation.

(d) It is a single molecule. (e) It can be more than 100 Mb long.

10. Match the organism listed in the left column with the amount of DNA in its haploid

genome from the right column.

(a) E. coli (bacterium)

(1) 4600 Kb

(b) S. cerevisiae (yeast)

(2) 3000 Mb

(3) 17 Mb

11. Which of the following statements about histones are correct?

(a) They are highly basic because they contain many positively charged amino acid

side chains.

(b) They are extensively modified after their translation. (c) In combination with DNA, they are the primary constituents of chromatin. (d) They account for approximately one-fifth of the mass of a chromosome.

12. Which of the following statements about nucleosomes are correct?

(a) They constitute the repeating units of a chromatin fiber. (b) Each contains a core of eight histones. (c) They contain DNA that is surrounded by a coating of histones. (d) They occur in chromatin in association with approximately 200 base pairs of DNA,

on average.

13. Which of the following statements about eukaryotic genes that are actively being tran

scribed are correct?

(a) They are cell-type specific. (b) They are highly condensed. (c) They are more susceptible to hydrolysis by DNAase I than are silent genes. (d) They are developmentally regulated. (e) They can be detected by chromatin immunoprecipitation.

14. Describe the structure of the nucleosome.

15. Does the formation of nucleosomes account for the observed packing ratio of human

metaphase chromosomes? Explain. 556

CHAPTER 31

16. The DNA methylation involved in gene regulation

(a) requires S-adenosylmethionine as a source of methyl groups. (b) occurs at 5„-CpG-3„ sequences. (c) uses N5, N10-methylenetetrahydrofolate to form the 5-methyl group of thymine. (d) converts cytosine in DNA to 5-methylcytosine. (e) is less frequent at sites adjacent to actively transcribed genes. Transcriptional Activation and Repression Are Mediated by Protein–Protein Interactions

17. Specific combinatorial control of transcription

(a) is enabled by specific interactions between transcription factors and specific DNA

sequences.

(b) allows a given regulatory protein to have different effects depending upon the neigh

boring proteins with which it is associated.

(c) is effected by transcription factors, some of which do not themselves interact with DNA. (d) depends upon the assembly of multicomponent nucleoprotein complexes. (e) results from the ability of one protein to recruit another to a complex.

18. Which of the following statements about steroid hormones are correct?

(a) They bind to a seven-helix transmembrane receptor to initiate a series of phospho

rylations that culminate in gene transcription.

(b) Upon binding to their specific receptor proteins, they enable the receptors to bind

specific DNA sequences.

(c) They activate specific protein kinases and protein phosphatases. (d) They are recognized by members of the nuclear receptor superfamily of proteins. (e) They require plasma membrane transporters to go from the blood to the cytosol.

19. Nuclear hormone receptors

(a) are dimers. (b) bind to response elements, which are specific DNA sequences at or near the genes

the hormones control.

(c) undergo conformational changes when they bind their ligand. (d) contain zinc finger domains. (e) interact with coactivators and corepressors in the presence of their ligands.

20. The tails of histones

(a) when acetylated have lower affinity for DNA. (b) are involved in recruiting chromatin-remodeling engines that move nucleosomes. (c) when acetylated, serve as substrates for histone deacetylases. (d) have their positive charges reduced by acetylation. (e) when acetylated interact with the bromodomain of many eukaryotic transcription

factors when that domain is brominated.

21. What is a primary consequence on gene regulation of the folding of chromatin? Gene Expression Can be Controlled at Posttranscriptional Levels

22. Regulation of the trp operon involves which of the following?

(a) controlling the amount of polycistronic mRNA formed at the level of transcrip

tion initiation

(b) controlling the amount of polycistronic mRNA at the level of transcription termination

THE CONTROL OF GENE EXPRESSION 557

lism from a single mRNA

(d) the sequential and coordinate production of five enzymes of tryptophan metabo

lism from five different mRNAs produced in equal concentrations

(e) the production of transcripts of different sizes, depending on the level of trypto

phan in the cell

23. Which of the following statements concerning the trp operon leader RNA, which has

162 nucleotides preceding the initiation codon of the first structural gene of the operon, are correct?

(a) A deletion mutation in the DNA encoding the 3„ region of the leader RNA gives rise

to increased levels of the biosynthetic enzymes forming Trp.

(b) A short open reading frame, containing Trp codons among others, exists within the

leader RNA.

the level of Trp-tRNA in the cell.

(d) The leader RNA may form two alternative and mutually exclusive secondary

structures.

(e) The structure of the leader RNA in vivo depends on the position of the ribosomes

translating it.

24. What are the biochemical similarities and differences between an iron-response element (IRE) and an estrogen-response element (ERE)? ANSWERS TO SELF-TEST

1. a, b, e. Answer (c) is incorrect because the splicing of mRNA is rare in bacteria.

2. a, b, c, d, f. Although IPTG is an inducer for the synthesis of b-galactosidase, it is nei

ther a substrate nor an allosteric activator of the enzyme, so answer (e) is incorrect.

3. (a) 4, 5, 8, 12 (b) 6, 9 (c) 1, 11 (d) 5, 7 (e) 2, 10 (f) 3. For (d), 5 is correct because al

lolactose is the product of a reaction catalyzed by b-galactosidase and, as a product, it binds to the enzyme.

4. The genes encoding these three enzymes are transcribed as a polycistronic mRNA, so

there are approximately equal numbers of copies of the mRNA sequences specifying each of the enzymes. See problem 6.

5. Glucose is metabolized preferentially because it results in a decrease in the synthesis of

cAMP by adenylate cyclase. The lack of cAMP prevents the formation of the cAMP–CAP complex, which is necessary for the efficient transcription of the lac operon and other catabolite-repressible operons.

6. When glucose is depleted, the concentration of cAMP rises. The cAMP–CAP complex

forms and binds to the CAP binding site just upstream of the RNA polymerase binding site in the lac promoter. At the same time, some lactose has entered the cell, has been converted to allolactose by b-galactosidase, and is bound by the lac repressor so that it no longer binds to the lac operator. RNA polymerase now binds to the lac promoter even more effectively because of protein–protein interactions with the cAMP–CAP complex. The enzymes and permease of the lac operon are expressed fully; consequently, lactose readily enters the cell and is efficiently metabolized.

7. a, d, e. Answers (b) and (c) are incorrect because they correspond to the binding se

quences for RNA polymerase and lac repressor, respectively. 558

CHAPTER 31

8. a, b, d. Answer (c) is incorrect because the recognition helix inserts into the wider major

groove rather than into the narrow minor groove.

9. b, c, d, e. Answer (a) is incorrect because DNA molecules more than a few kilobases long

are sensitive to fragmentation by the shearing forces developed when solutions are stirred. Answer (c) is correct because the longer a DNA molecule, the longer it takes to reassume its normal solution conformation after it has been stretched by being in a flowing solution.

10. (a) 1, (b) 3, and (c) 2. Note that the length of the bacterial chromosome is expressed

here in kilobases, not megabases.

11. a, b, c. Answer (d) is incorrect because histones make up nearly half the mass of a

chromosome.

12. a, b, d. A nucleosome core consists of ~145 base pairs of DNA wrapped around a his

tone octamer. The nucleosome cores are connected by linker DNA, which contains from fewer than 20 to more than 100 base pairs (bp), the exact length depending on the organism and the tissue. The average total length is ~200 bp.

13. a, c, d, e. Answer (b) is incorrect, because actively transcribed genes are less compact

than those that are transcriptionally silent.

14. The nucleosome core has a disk shape and is composed of eight histone molecules. The

octameric core of histones has ~145 base pairs of DNA wound about it in approximately 13⁄4 turns of a left-handed torroidal supercoil. Two copies each of histones H2A, H2B, H3, and H4 are on the inside of the toroidal coil, whereas histone H1 is associated with the DNA where it emerges from the core. Each core histone has a basic tail that protrudes from the core structure. In total, ~200 bp of DNA is present per nucleosome.

15. No. As mentioned, each nucleosome is associated with approximately 200 base pairs of

DNA. If this DNA were coiled into a sphere with a diameter of approximately 100 Å, it would be condensed from 200 base pairs¥3.4 Å per base pair=680 Å of linear DNA to 100 Å, which is a packing ratio of about 7. The chromatin fiber, which is composed a helical array of nucleosomes, must be formed, and the resulting 360-Å coils must themselves be looped and folded. Scaffolding proteins, topoisomerases, and small basic molecules, such as the polyamines, also contribute to the ultimate compaction of 104 that is observed in metaphase chromosomes.

16. a, d, e. Answer (b) is wrong because the methylation takes place at 5„-CpG-3„ sequences.

Answer (c) is wrong because SAM is the methyl donor for postsynthetic DNA methylation.

17. a, b, c, d, e. Although a critical feature of combinatorial control is mediated by protein–

protein interactions, some components of the transcription complex must interact specifically with DNA in order to locate the transcriptional assemblage to the proper region on the DNA. For instance, some transcription factors bind to enhancers far from the site of transcription initiation.

18. b, d. Answers (a) and (c) are wrong because they are properties of a more numerous class

of hormones that act by initiating phosphorylation cascades within cells after binding outside the cell to a transmembrane receptor.

19. a, b, c, d, e.

20. a, b, c, d. Answer (e) is incorrect because, although named “bromodomain,” bromination

has nothing to do with the action of this acetyllysine-binding protein structure. The name derives from the brahma gene in Drosophila, where the archetype bromodomain was found.

21. The tight folding of chromatin renders many of the sites on DNA inaccessible to the pro

teins that must be assembled to form an active transcription complex. Chromatin structure decreases the amount of DNA available to nonchromatin proteins. Remodeling of chromatin makes some of these sites accessible.

THE CONTROL OF GENE EXPRESSION 559

22. a, b, c, e. Answer (a) is correct. Although not mentioned in the text, the trp operon con

tains an operator, and interaction with a repressor, in addition to attenuation, is involved in transcription regulation. Attenuation provides a rapid, sensitive fine-tuning mechanism on top of the control exerted by the repressor–operator interaction. When the RNA is not terminated at the attenuator, a single polycistronic mRNA, which encodes five enzymes, is produced, so answer (d) is incorrect.

23. a, b, c, d, e. The discovery of a deletion mutation in front of the first structural gene of

the operon, and not in the operator, was the first clue that a control mechanism operating at the level of transcription termination was involved in regulating the expression of the trp biosynthetic enzymes. This deletion changed the potential mRNA structures so that the rho-independent transcription-termination structure could no longer form.

24. The IRE is a sequence in the 5„-untranslated region of the mRNA that encodes the fer

ritin molecule. The IRE-binding protein (IRE-BP) binds to the IRE and blocks translation. The ERE is a DNA sequence to which the estrogen–estrogen receptor complex binds to facilitate trancription. PROBLEMS

1. What property of enzymes makes them more suitable than, say, structural proteins for

studies of the genetic regulation of protein synthesis? Explain.

2. When lactose is used as an inducer a lag occurs before the enzymes of the lactose operon

are synthesized. With IPTG synthesis starts without a lag. Explain this observation. (See Figure 31.1.) FIGURE 31.1 Kinetics of b-galactosidase induction by lactose and IPTG. (Assume that each

inducer has been removed after an appropriate period.)

IPTG

Inducer added

Lactose

-Galactosidase activity b

Time D

3. Some of the known constitutive mutations of the lactose operon occur in the operator

sequence rather than the regulator gene.

(a) Would you expect such an oc mutant to be dominant or recessive to its wild-type o; allele? Explain.

(b) Is a constitutive mutation in an operator cis-acting or trans-acting in its effects?

Explain.

(c) Design an experiment involving the genes i;, oc, o;, and z; that would confirm your

answer to part (b). Assume that it is possible to detect whether enzymes are produced in diploid (;;) or haploid (;) amounts.

4. Since the permease required for the entry of lactose into E. coli cells is itself a product of

the lactose operon, how might the first lactose molecules enter uninduced cells? Explain. 560

CHAPTER 31

5. Design an experiment to show that lactose stimulates the synthesis of new enzyme mol

ecules in E. coli rather than fostering the activation of preexisting enzyme molecules, for example, by zymogen activation.

6. The three enzymes of the lactose operon in E. coli are not produced in precisely equimo

lar amounts following induction. Rather, more galactosidase than permease is produced, and more permease than transacetylase is produced. Propose a mechanism to account for this that is consistent with known facts about the lactose operon.

7. Assume that the following allelic possibilities exist for the i genes and o sequence of the

lactose operon of E. coli: i;=wild-type regulator gene ic=regulator constitutive mutation, makes inactive repressor is=repressor insensitive to inducer o;=wild-type operator oc=operator consitutive mutation

In addition, assume that the mutations z:, y:, and a: lead to nonfunctional enzymes Z, Y, and A, respectively.

For each of the following, predict whether active enzymes Z, Y, and A will or will not be

produced. For partially diploid cells, assume semidominance; that is, the enzyme activity in a diploid cell will be twice that found in a haploid cell. Use the following answer code:

0 =active enzyme absent ;

=active enzyme present in haploid amounts

;; =active enzyme present in diploid amounts Without IPTG With IPTG Z Y A Z Y A

(a) i;o;z;y;a;

_____ _____ _____

(b) ic o ; z ; y ; a ;

_____ _____ _____

(d) i ; ocz;y;a;

_____ _____ _____

+ + + + +

(e) i o z y a

_____ _____ _____

_____ _____ _____ ico+z+ y+a+

+ + + + +

(f) i o z y a

_____ _____ _____

_____ _____ _____ ico+z− y+a+

+ + + + +

(g) i o z y a

_____ _____ _____

_____ _____ _____ i+ocz+ y+a+ s + + + +

(h) i o z y a

_____ _____ _____

_____ _____ _____ i+o+z+ y−a− s + + + +

(i) i o z y a

_____ _____ _____

_____ _____ _____ i+ocz− y+a+

+ + + + +

(j) i o z y a

_____ _____ _____

_____ _____ _____ icocz+ y+a−

(a) Are enzymes X and Y likely to be biosynthetic or degradative? Justify your an

swer briefly.

(b) Is substance S active or inactive in the presence of T? Explain. (c) For each of the following, predict whether active enzymes X and Y will or will not

be produced under the specified conditions. For partially diploid cells, assume semidominance; that is, the enzyme activity in a diploid cell will be twice that found in a haploid cell. Use the following answer code:

0

=active enzyme absent

;

=active enzyme present in haploid amounts

; ; =active enzyme present in diploid amounts Without T With T X Y X Y

(1) r ; o ; x ; y ;

_____ _____

_____ _____

(2) r ; o ; x : y ;

_____ _____

_____ _____

(3) r s o ; x ; y ;

_____ _____

_____ _____

(4) r s o o x ; y ;

_____ _____

_____ _____

(5) r ; o o x ; y ;

_____ _____

_____ _____

(6) r o o ; x ; y ;

_____ _____

+ + + +

(7) r o x y

_____ _____

_____ _____ r+o+ x− y+

+ + + +

(8) r o x y

_____ _____

_____ _____ r o

o + x+ y+

s + − −

(9) r o x y

_____ _____

_____ _____ roo+ x+ y+

+ o + −

(10) r o x y

_____ _____

_____ _____ r o

o + x+ y+

9. The kinetics of induction of enzyme X are shown in Figure 31.2. What percentage of

total cellular protein is due to enzyme X in induced cells when 60 mg of total bacterial protein has been synthesized? 562

CHAPTER 31 FIGURE 31.2 Kinetics of induction of enzyme X.

Inducer

removed

8

g) m

4

Inducer

added

Enzyme X (

2

30

60

90

Total bacterial protein (mg)

10. Assume that the dissociation constant K for the repressor–operator complex is 10:13 M

and that the rate constant for association of operator and repressor is 1010 M:1 s:1. Calculate the rate constant kdiss for the dissociation of the repressor–operator complex. What is the t1/2 (half-time of dissociation, or half-life) of the repressor–operator complex?

11. In systems of genetic regulation involving positive control, a regulatory gene produces a

substance that enhances rather than inhibits transcription. Are there elements of positive control in the lactose operon of E. coli? Explain. What positive regulatory proteins often participate in eukaryotic transcription regulation?

12. An operon for the biosynthesis of amino acid X in a certain bacterium is known to be

regulated by a mechanism involving attenuation. What can one confidently predict about the amino acid sequence in the leader peptide for that operon? Explain.

13. In order to prove that regulation by attenuation occurs in vivo, Charles Yanofsky and

others studied tryptophan synthesis regulation in a series of E. coli mutants. For each mutant described below predict the expression of tryptophan synthesis genes in the presence or absence of tryptophan. Mutant A:

mutations with decreased, but detectable, tryptophan-tRNA synthetase activity

Mutant B:

mutations in AUG or Shine-Dalgarno sequence of leader peptide sequence

Mutant C:

same as mutant B but with the leader peptide fully expressed on a plasmid

Mutant D:

mutations replacing the two Trp codons with Leu codons

Mutant E:

mutant E with a mutation in the Leu-tRNA synthetase gene

Mutant F:

mutant E that constitutively synthesizes leucine

14. Would you expect the interaction between protamines and DNA to be enhanced or

diminished in solutions that have highly ionic strength, that is, high salt concentrations? Protamines, which are found in very high concentrations in sperm where they participate in condensation of the DNA, are low-molecular-weight compounds rich in groups with high pKa values, that is, they are basic compounds. Explain the basis for your answer. Does the action of histone acetyltransferases (HATs) act using a similar principle? Explain.

15. One measure of the evolutionary divergence between two proteins is the number of amino

acid differences between them. It can be argued that a better measure would be the minimum number of mutational events that must have occurred to result in those differences.

THE CONTROL OF GENE EXPRESSION 563

(a) The differences in the amino acid sequences of histone H4 between calf thymus and

pea seedlings are as follows: AA position Pea seedlings Calf thymus

60

Ile

Val

77

Arg

Lys

What minimum number of mutational events accounts for these differences? Give the changes that must occur in both mRNA and DNA. (Refer to the genetic code inside the back cover of the text.)

(b) From your knowledge of amino acid chemistry, comment on the nature of the

changes. What conclusion follows about the function of H4?

16. Chromatin that is transcriptionally active (euchromatin) is disperse in structure, whereas

chromatin that is transcriptionally inactive (heterochromatin) is compact. When the nuclei of chicken globin-producing cells were treated briefly with pancreatic DNase, the adult globin genes were selectively destroyed, but the genes for embryonic globins and ovalbumin remained intact. In contrast, when the nuclei of oviduct cells were treated with DNase, the ovalbumin genes were destroyed. Explain these results.

17. The chromatin of globin-producing cells can be treated with micrococcal nuclease under

conditions that cleave DNA almost exclusively in the linker regions between intact nucleosomes. When the resulting nucleosomes are isolated and their DNA is examined, it is found to contain DNA sequences for the synthesis of globin. Are these results consistent or inconsistent with the explanation for the results in problem 16? Explain.

18. Suppose that a system regulating the expression of a single copy DNA leads to the syn

thesis of an enzyme having a turnover number (kcat) of 104 s:1. Each DNA copy is transcribed into 103 molecules of mRNA and each of the mRNA molecules is translated into 105 molecules of enzyme protein. How many molecules of substrate are converted into product per second for each wave of transcription that sweeps over the DNA?

19. Would you expect a zinc deficiency in eukaryotes to be associated with any sort of de

velopmental abnormality? Explain.

20. You have isolated the M gene, which is involved in muscle cell differentiation, and, after

introducing it into cultured mammalian cells, you wish to study its expression. In sequencing the region upstream from the open reading frame you notice that there are a number of CG-rich regions, from 100 to 1000 bp upstream.

(a) When you grow the cells in presence of 5-azacytidine, the M gene is expressed,

whereas cells grown in the absence of the analogue do not express it. Provide an explanation for these observations, relating them to the CG-rich regions and how they would be affected by 5-azacytidine. (Hint: How might changing the 5 position in the heterocyclic ring of cytidine from carbon to nitrogen affect its chemistry?)

(b) You decide to carry out some experiments to verify your explanation of the re

sults with 5-azacytidine. Isolating the M gene upstream sequence that contains the CG-rich regions, you place it into a bacterial plasmid immediately next to the gene for the enzyme chloramphenicol acetyl transferase (CAT), which is not found in mammalian cells and can be easily assayed. You then amplify this plasmid construct in, and isolate it from, bacteria. When the isolated plasmid molecules are used to infect transiently the muscle cells in culture medium lacking 5-azacytosine, CAT is formed. You then isolate the upstream DNA sequence from cell-derived chromosomes as well as from the plasmid, and digest samples from both

564

CHAPTER 31

isolates with the restriction endonucleases HpaII and MspI. The DNA from the digests is separated using gel electrophoresis. What data would you expect to obtain from these experiments? You will need to look up the cutting specificities of these restriction endonucleases at http://rebase.neb.com/rebase/rebase.html.

21. Proteins that interact with a specific sequence of DNA usually remain bound under con

ditions of low ionic strength during electrophoresis in gels formed with low percentages of acrylamide. (Low percentages of acrylamide form gels with pores sufficiently large to allow entry of large macromolecular complexes.) Under these conditions, protein–DNA complexes usually migrate more slowly than does the free DNA. Suppose you have a sample of a DNA fragment radioactively labeled with 32P that contains an entire promoter sequence. Describe how you could use the sample and gel electrophoresis to isolate transcription factors from extracts of protein from eukaryotic cells. How might you compensate for the possibly competing binding of histones to the DNA? How could you use unlabeled samples of your promoter fragment to demonstrate the specificity of the interaction between the fragment and a putative transcriptional factor? Might you also detect proteins that do not themselves bind directly to the DNA?

22. Would you be surprised if your analysis of the gene regulatory machinery of a eukary

otic cell indicated that DNA sequences far removed (1 or more kbp) from the site of transcription initiation were involved? Explain.

23. In sex determination in humans, female is the default state. To become male, genes must be

activated that lead to the development of the testes and external male genitalia and suppression of the development of what would ultimately become the female sex organs. Many of these genes are on the Y chromosome, which is absent in genotypic (XX) females. The steroid hormone testosterone, an androgen, is involved in this process. What would you predict would happen to a genetic male (XY) fetus whose testosterone receptor had a mutation in its C-terminal domain that rendered that domain resistant to binding the androgen?

24. A group of molecular biologists showed that the three eukaryotic DNA sequences below

can activate transcription of a reporter gene, such as the gene encoding chloramphenicol transacetylase. Sequence X: 5„-TAATTGCGCAATTA-3„

3„-ATTAACGCGTTAAT-5„

Sequence Y: 5„-TAATTGCTACGGTA-3„

3„-ATTAACGATGCCAT-5„

Sequence Z: 5„-TACCGTATACGGTA-3„

3„-ATGGCATATGCCAT-5„

Gel-shift assays, described in problem 21, were used to purify three dimeric protein activators, each of which bound to one of the three DNA sequences shown above. Further investigation revealed that although three dimeric proteins could be purified, only two polypeptide monomers could be isolated. How is this possible?

25. Unlike bacterial RNA polymerases, eukaryotic polymerases have relatively low affinity

for their promoters and therefore often depend on several activator proteins for initiation of transcription. Thus, while many bacterial genes are subject to negative regulation by repressor proteins, eukaryotic genes are more likely to be under positive regulatory control. The reasons for this difference in the mode of regulation may be related to the great difference in genome sizes. For example, the human genome is ~650 times larger than that of E. coli and may contain over 40,000 genes. What are the advantages of positive regulation in the control of gene expression in eukaryotes?

THE CONTROL OF GENE EXPRESSION 565 ANSWERS TO PROBLEMS

1. Enzymes are catalysts, and thus small amounts can be readily detected. An enzyme hav

ing a turnover number of 300,000 s:1 will provide an assay that is 300,000 times as sensitive as that for a structural protein, which must be assayed stoichiometrically, that is, as a single molecule. Many cellular proteins are produced in amounts that are too small to be detected by direct chemical methods.

2. The actual inducer of the lactose operon in vivo is 1,6-allolactose (see text, p. 871). The

lag represents the time it takes for lactose to be converted into 1,6-allolactose by residual b-galactosidase. IPTG itself directly acts as an inducer. Therefore, no lag is observed.

3. (a) Imagine a partial diploid that has one o; and one oc sequence. The o; gene will bind

a repressor, so the structural genes on its chromosome will not be expressed. The oc sequence will not bind a repressor, so the structural genes on its chromosome will always be expressed. Thus, an oc mutant would be dominant to its wild-type o; allele.

(b) Repressors do not bind oc sequences. Only the structural genes on the same chro

mosome as the oc mutant will be affected, a cis-acting effect (see p. 794 in the text for a discussion of cis-acting and trans-acting effects).

(c) One could prepare a partial diploid with the following genotype i+o+z+ i+ocz+

If the effect of the mutation is cis, the haploid amount of enzyme Z will be produced in the absence of inducer. (Its synthesis will be specified by the chromosome containing i;ocz;.) If the effect is trans, the diploid amount of Z will be produced in the absence of inducer.

4. Very low levels of lactose operon enzymes are synthesized even in the absence of an inducer

(see text, p. 872, which indicates that few enzymes are produced in the absence of inducer).

5. Add IPTG to E. coli cells growing in a medium containing a carbon source other than

lactose in both the presence and the absence of an inhibitor of prokaryotic protein synthesis, like chloramphenicol. If zymogen activation is involved, chloramphenicol will not inhibit induction. If the synthesis of new protein is involved (as it is), induction will not be observed in the presence of chloramphenicol.

6. Differential expression of the three structural genes in the lactose operon must be at the

level of translation and not transcription since a single, polycistronic mRNA molecule is formed. Following induction, mRNA transcripts containing genetic information for all three genes are produced. Some ribosomes might drop off the messenger at the end of the structural genes, with a smaller number reading through the more distal genes.

7. (a) 0

0

;

(b) ;

;

0

(d) ;

;

(e) 0

0

;;

;;

(f)

0

;

;;

;;

(g) ;

;

;

;;

;;

(h) 0

0

(i)

0

;

0

;

(j)

;

0

;;

; 566

CHAPTER 31

8. (a) The enzymes will be biosynthetic. The clue is provided by the statement that the

enzymes are repressible. (Degradative enzymes are often inducible, whereas biosynthetic enzymes are often repressible.)

(b) Because this operon is under negative control, substance S must act as a repres

sor. Substance S is active as a repressor in the presence of T, so T functions as a corepressor.

(c)

(1) ;

;

0

(2) 0

;

0

(3) 0

0

(4) ;

;

(5) ;

;

(6) ;

;

(7) ;

;;

0

(8) ;

;;

0

(9) 0

0

(10) ;;

;

0

9. From the graph in Figure 31.2, we see that 2 mg of enzyme X is present when the total

bacterial protein present is equal to 60 mg. Thus, the percentage of enzyme X is 100%¥2/60=3.3%.

10. Remembering that the dissociation constant K is equal to the ratio of the off rate to the

on rate for a reaction,

[RO]

[

G R ] + [O]

[R ][O]

−13 kdiss

= K = 10

M

[RO]

1010 M−1s−1 k

=

−

10 3 s−1

diss

.

0 693

.

t =

= 0 693 = 693 s ≅ 11.6 min

1

−3 − k

1

diss

10 s

2

11. CAP is a positive control element. When the level of glucose in cells is low, the level of

cAMP is high, leading to the formation of cAMP–CAP complex. The cAMP–CAP complex binds to DNA in the promoter region, creating an entry site for RNA polymerase. The result is the transcription of the lactose operon (providing that no repressor is present) (see Figure 31.10 on p. 873 of the text). Coactivators act as positive control elements in eukaryotic transcription (see pp. 881–882 in the text).

12. The leader sequence should contain codons for amino acid X. If sufficient X is present

in the cell, there will be sufficient X-tRNAX for the synthesis of the leader peptide (as well as for the synthesis of other X-containing proteins in the cell). Therefore, there will be no need to biosynthesize the enzymes needed to produce X.

13. Mutant A: There will be increased transcription of the trp synthesis genes in both the

presence and absence of tryptophan. The low levels of Trp-tRNA synthetase will slow down the rate of translation regardless of the levels of tryptophan. It is the stalling or slowing of the ribosome in segment 1 that is important for regulation, not just the inability to synthesize the leader peptide. Mutant B: Since the ribosome will never start translation, transcription will always terminate regardless of the levels of tryptophan.

THE CONTROL OF GENE EXPRESSION 567

Mutant C: The results will be the same as in mutant B. Attentuation relies on the coupling of transcription and translation. Providing the leader peptide in trans would have no effect on the level of attenuation.

Mutant D: Removing the Trp codons would lose all regulation by tryptophan, and Trp synthesis would be regulated by the levels of leucine. In this mutation, the Trp synthesis genes would not be transcribed even in the absence of tryptophan, and this strain, like mutants B–D, would always require tryptophan to grow.

Mutant E: There would be constitutive expression of the Trp synthesis genes, even in the presence of tryptophan or leucine.

Mutant F: This mutant would behave like mutants B–E and never express the genes for Trp synthesis.

14. Just as histones can be dissociated from DNA with salt, the interaction between prota

mines and DNA is diminished in highly ionic solutions because the salt in solution disrupts the ionic interactions between the ligands and the DNA. Protamines are arginine-rich proteins whose positively charged guanidinium groups can associate with the negatively charged phophodiester bonds in a DNA helix. These electrostatic interactions bind the protamines or histones tightly to the polynucleotide. The positively and negatively charged ions that result from the addition of a salt to an aqueous solution compete with DNA-ligand interactions and, hence, weaken them. By acetylating the primary amino groups on the side chains of lysine, HATs remove their positive charges and thereby weaken the interaction between the histone and the DNA (see p. 884 in the text). Thus, the physicochemical principle, reduction of charge-charge interactions is the same whether accomplished through the action of HATs or the addition of salts.

15. (a) There would be a minimum of two mutations involved. The possible codons for Ile

and Val are as follows: Ile Val

AUU

GUU

AUC

GUC

AUA

GUA GUG

A single change from A to G in the first position of the codon would give a substitution of Val for Ile. This corresponds to a change from an A–T to a G–C base pair on DNA.

The possible codons for Arg and Lys are as follows: Arg Lys

CGU

AAA

CGA

AAG

CGG CGC AGA AGG

Again, a single change, from G to A in the second position of the codon, would account for the amino acid difference. This corresponds to a change from a G–C to an A–T base pair on DNA. 568

CHAPTER 31

(b) These are conservative changes. Both Lys and Arg have positively charged side

chains, and both Ile and Val are hydrophobic. Therefore, we would expect virtually no structural or functional difference in the H4 of calf thymus and pea seedlings. However, the two organisms are clearly different in other respects.

16. In cells that are actively synthesizing adult globins, chromatin is dispersed so that the

globin genes may be transcribed into mRNA. Accordingly, this region is sensitive to DNase. In an adult cell specialized for the production of globins, neither embryonic globins nor ovalbumin is produced to a significant extent. Accordingly, the regions of DNA carrying the information for these genes are compact and are therefore not sensitive to DNase. Conversely, in those oviduct cells making ovalbumin, the ovalbumin gene is destroyed by DNase but not the globin genes.

17. The results are consistent. The genes for globin synthesis are contained within the nu

cleosomes, many of which are required to cover the globin genes. When these genes become transcriptionally active, the chromatin becomes dispersed and the linker regions between the nucleosomes become susceptible to cleavage by micrococcal nuclease.

18. For each transcription, the number of molecules of substrate that are converted into

product per second is given by 104 s:1¥103¥105=1012 s:1

19. The transcription of many eukaryotic genes is activated by proteins containing from one

or more zinc fingers, each of which is an ~30-residue-long amino acid sequence containing (usually) two cysteines and two histidines coordinated to a zinc ion. For example, zinc is involved in the structure of DNA-binding domains of the nuclear hormone receptors (see the text, p. 880). It is also involved in a large number of other enzymecatalyzed reactions, including the conversion of acetaldehyde to ethanol, the formation of bicarbonate ion, and the cleavage of peptides by chymotrypsin. Because of its essential roles in gene expression and in cellular metabolism, it is likely that zinc deficiencies could lead to significant developmental abnormalities.

20. (a) The presence of CG-rich regions (islands) in the upstream region of the M gene in

dicates that these are sequences that might be subject to methylation and could thereby control transcription by influencing promoter activity. In many cells, about three-fourths of CG sequences are methylated, whereas those sequences in transcriptionally active regions are less methylated. In your experiments, cells grown in the presence of 5-azacytidine incorporate the analog into DNA, and the azacytosine residue cannot be methylated at N-5 by specific DNA methyltransferases because the atom at the 5 position is N, not the normally present C. The undermethylated region is therefore more susceptible to transcriptional activation. Transcription of the M gene leads to expression of the gene in those cells grown in 5-azacytosine. Cells grown in the absence of the analog are more likely to have methylated CG-rich sequences in the promoter region and are less likely to express the gene you are studying.

(b) If your explanation in (a) is correct, you would expect the CG sequences in the up

stream island to be methylated in samples taken from cells, but not methylated in samples taken from the plasmid grown in bacteria (methylation of C residues occurs frequently in vertebrates but is rare in prokaryotes). Thus, expression of the gene from the CAT gene must be due to the activity of its promoter. You would expect that both restriction enzymes would cleave your upstream sequence isolated from the bacterial plasmid, while only MspI would cleave the sequence isolated from chromosomes. The fragment would be resistant to cleavage by HpaII, which does not cleave CmCGG sequences, that is, sequences methylated at the 5 position of the interior C.

21. Assays that exploit the differential electrophoretic mobility of protein–DNA complexes

and free DNA are called gel-shift or electrophoretic-mobility-shift assays. In these experi

ments, specific DNA sequences are allowed to associate with putative DNA-binding proteins from cell extracts or from fractionated samples from the extracts. They are then subjected to gel electrophoresis along with control samples of labeled fragment by itself. The radioactivity on the gels is visualized by autoradiography on film or in a phosphoreimaging machine. Labeled samples that are retarded on the gel are candidates for sequence-specific protein–DNA complexes. To minimize retardation caused by nonspecific binding of proteins, such as histones, to the labeled fragment, large amounts of unlabeled, random sequences of DNA can be added to the sample before electrophoresis. This DNA binds the nonspecific proteins, but is not observed on the gel because it is not labeled. Because sequence-specific proteins usually have a much higher affinity for a specific sequence than those proteins binding nonspecifically, the addition of the extra DNA does not usually interfere with the gel-shift assay. Experiments that assess specificity of the DNA–protein interaction can be conducted by including an excess amount of unlabeled DNA which contains the same specific sequence in an analyzed sample. If the interaction is truly sequence-specific, the unlabeled DNA should bind the protein and by specific competition abolish the shift of the labeled sequence on the gel. Gel-shift analysis is a widely employed technique. Although it is not a true equilibrium technique and therefore cannot easily provide true thermodynamic equilibrium binding constants, it reliably indicates the relative affinities of proteins for specific DNA sequences. You might also detect proteins that are associated tightly through protein–protein interactions with the DNA-binding proteins; for example, you might detect coactivators or corepressors.

22. You would not be surprised because the distant sequences might well be enhancers that

bind transcription factors that themselves associate with the core transcription machinery by looping the DNA to achieve proximity to the transcription start site.

23. The testosterone-nuclear hormone receptor complex could not form, and the genes nec

essary to promote virilization and suppress feminization would not function properly. The outcome would be a genetic male who developed into a phenotypic female. Recall (p. 880 in the text) that the ligand-binding domain of nuclear hormone receptors is near their C termini. The inability to form the testosterone–receptor complex results in a male genotype expressing a phenotype similar to one arising from a missing Y chromosome, that is, being a female—in neither case is a functional androgen receptor formed. The disorder arising from the situation described is called testicular feminization, and many other biochemical and developmental factors beyond those mentioned here are involved.

24. The most likely possibility is that each of the two monomers can form homodimers, each

of which could bind to one of the sequences above. In addition to forming homodimers, the two different monomers associate to form a heterodimer, which binds to a third sequence. You would expect an activator that is a homodimer to bind to a DNA sequence, which has dyad symmetry (is an inverted repeat), whereas a heterodimer would bind to an asymmetrical sequence, which contains sequences common to each of the individual symmetric sequences. Inspection of the three sequences shows that sequence X and Z have dyad symmetry, whereas sequence Y is asymmetrical, containing sequences that represent half of X and half of Z. Thus one homodimer binds to and activates sequence X, whereas another homodimer associates with the symmetric sequence Z. A heterodimer, composed of each of the two monomers, binds to and activates the asymmetric sequence Y, which contains half of each sequence X and Z. This problem illustrates the important principle that heterodimer formation can allow recognition of DNA sequences that do not have dyad symmetry, thereby increasing the potential for regulation of expression. The CREB protein is an example of a homodimer DNA binding protein. The oncogenes fos and jun form transcription-regulatory proteins that can associate with themselves to form homodimers or with each other to form a heterodimer—the situation described in this problem. 570

CHAPTER 31

25. In bacteria, negative regulation requires synthesis of a specific repressor that blocks tran

scription of a gene or an operon. To carry out negative regulation of genes in a human genome, 40,000 repressor proteins would need to be synthesized, which would be an inefficient means of controlling transcription. Because most eukaryotic genes are not in operons and are normally inactive with regard to transcription, selective activation through synthesis of a small group of activator proteins is used to promote transcription of a particular array of genes needed by the cell at a certain time. Another reason for positive control may be related to the fact that a larger genome presents the possibility that a relatively short DNA sequence for a regulatory protein would be present in multiple and possibly wrong locations, bringing about inappropriate or unneeded gene activation. This can be avoided by requiring that several positive regulatory proteins form a complex that can specifically activate a gene and promote its transcription, thereby reducing the possibility of incorrect gene activation. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. (a) Without the lac repressor gene, the repressor protein will not be produced. Without

a repressor, the lac z, y, and a proteins will be produced constitutively (independently of the presence or absence of lactose, albeit at low levels when glucose is present, due to catabolite repression).

(b) Provided that the lac promoter remains intact, the effect would be the same as in

(a): no repression, and constitutive production of the lac z, y, and a proteins.

are produced will remain low, even in the presence of lactose.

2. A liter contains 1000 cm3, so a cell volume of about 10:12 cm3 is 10:15 liter. One mole

cule divided by Avogadro’s number in 10:15 liter corresponds to a concentration of about 1.7 * 10:9 M.

(1 molecule)

= .

1 7 ∗10 9

− M

23

1

(6.02 ∗10 molecules mol− )(10 1

− 5 liter)

Since the repressor concentration is much higher than the dissociation constant for the repressor/operator complex (10:13 M), the single molecule will be bound to (operator) DNA.

3. The probability of having a particular chosen nucleotide sequence at a given site is (1⁄4)

for a single base, (1⁄4)2 for a two-base sequence, and (1⁄4)n for a sequence of n bases. The number of statistically expected occurrences of a particular sequence of length n in a genome of L base pairs is L times (1⁄4)n. The table below summarizes the results for the E. coli genome, which contains about 4.8 * 106 base pairs. Length of sequence, n (1⁄4)n Predicted number of sites in E. coli genome

8

1.5 * 10:5

73

10

9.5 * 10:7

4.6

12

6.0 * 10:8

0.03

4. In the table below, a charge of :1 is assigned to each Asp and each Glu, and a charge of ;1

is assigned to each Lys and each Arg in each of the histones. These charges are then summed

to give estimated net charges of ;17 for H2A, ;18 for H2B, ;20 for H3 and ;18 for H4. The histone octamer net charge then would equal 2*(17+18+20+18)=;146. A 150-base pair sequence of DNA, with one negative charge per phosphodiester linkage, has a net charge of about 2*(:149)=:298. H2A H2B H3 H4

Asp (D)

:3

:4

:3

Glu (E)

:6

:7

:4

Lys (K)

14

21

13

11

Arg (R)

12

7

18

14

Sum

;17

;18

;20

;18

5. The isolated mixture of DNA fragments could be tested for fragments that would hybridize

to a single-stranded probe corresponding to a portion of the known sequence of interest. To prepare for the analysis, the known probe DNA could be attached to a filter. The DNA fragments isolated from the immunoprecipitation could be amplified by the polymerase chain reaction, if necessary, labeled with 32P using 5„-polynucleotide kinase and g-32PATP, and heated to separate the strands of the DNA double helix. Labeled fragments would then be incubated with the filter-attached probe (under “stringent” hybridization conditions). After washing the filter, the extent of binding could be determined by counting the specific radioactivity that the filter acquired during the hybridization reaction. For a lac repressor immunoprecipation experiment, only one unique DNA fragment is expected to be protected by the binding of the repressor. The pur repressor, by contrast, should protect about 20 or more different sites on the E. coli chromosome (see Figure 31.9 in the text).

6. Transcriptionally inactive regions of DNA have a high content of 5-methylcytosine. Incor

porating 5-azacytidine into DNA will prevent methylation. The lack of methylation will lead to the activation of some normally inactive genes.

7. Because 5-methylcytosine often is a signal for gene inactivity, the protein domain might

play a role in gene inactivation. The domain could perhaps block transcription by binding to regulatory regions of double-stranded DNA that contain 5-methylcytosine. The protein domain would bind in the major groove. The 5-methyl group will be on the “outside” of a GC base pair (see diagram below) and will protrude into the major groove of double-stranded DNA. (To view an example, examine the C5 positions on the cytosines in structure 1D64 in the Protein Data Bank.)

H C

N J H

O

3

N

H

J

H J

N

H J N

N

deoxyribose

N

K

J

O

H J N

deoxyribose

base pair between 5-methyl-C and G

8. Whereas the lac repressor is released from DNA by binding to a small molecule, the pur

repressor is induced to associate with DNA by the binding of a small corepressor molecule, either guanine or hypoxanthine. An additional difference is the number of respective binding sites in the E. coli genome, which contains only one binding site for the lac repressor but about 20 sites for the pur repressor (see Figure 31.9 in the text). 572

CHAPTER 31

9. The anti-inducer could be a competitive inhibitor of the inducer. As such, the anti-inducer

would bind to the repressor at a similar or overlapping site to that of the inducer, but would not cause the conformational change necessary to release the repressor from the operator DNA. Higher concentrations of inducer would then be needed to displace the competitively bound anti-inducer from its site on the repressor.

10. Because symmetry is a recurring theme for protein–DNA interactions, the DNA se

quence may have functional importance. One possibility is that the DNA sequence could be a binding site for a dimeric regulatory protein. Alternatively, inverted repeat sequences sometimes serve as hot spots for genetic rearrangements because they may form hairpin secondary structures that block DNA polymerases or are processed by structure-specific endonucleases.

11. The lysine amino group can make a nucleophilic attack on the carbonyl carbon of the

thioester of acetyl-CoA to give a tetrahedral intermediate. The tetrahedral intermediate then could eliminate CoASH as a leaving group to yield acetyl-lysine.

O

K

O:

J

O

K

J

J

CoA

H C

N

3

3

3

D

HN

D

H

:

+

H N

2

CoASH

12. The injected DNA fragments may bind competitively to CREB and thereby prevent CREB

from binding to its true physiological target sites. For this reason, CREB would be unable to perform its role in stimulating the synthesis of new proteins. A proposed pathway for the stimulation of long-term memory by serotonin would be: (1) Serotonin binds to a receptor on the surface of a neuron cell and activates a G protein. (2) The G protein activates adenylate cyclase, which increases the intracellular concentration of cAMP. (3) cAMP activates protein kinase A. (4) Protein kinase A phosphorylates CREB. (5) Phosphorylated CREB binds the coactivator CBP. (6) The CREB/CBP complex activates the transcription of new proteins for long-term memory. (Step 6 would be inhibited by the DNA fragments that contain binding sites for CREB.)

13. A large percentage of the cytosine residues in mouse DNA are methylated, whereas very

few C’s in Drosophila or E. coli DNA are methylated. Therefore, the Drosophila and E. coli DNA are cut by HpaII into pieces of average size about 256 base pairs, while the mouse DNA is cut into pieces of average size about 50,000 base pairs. CHAPTER 3 Sensory Systems 2

This chapter describes the functioning of the five major senses—smell, taste, vision,

hearing, and touch—on a molecular level. All are shown to rely on mechanisms involved in transduction of other sorts of signals (hormones, neurotransmit

ters, etc.). Olfaction, taste, and vision utilize G-protein-linked 7TM receptors. Hearing and touch have different receptors but appear to share ankyrin repeats as part of their structures.

The human genome contains sequences for hundreds of different odorant re

ceptors (OR), each of which is a 7TM receptor. When an odorant arrives with the proper shape to bind, a G protein binds to GTP and triggers adenylate cyclase. The stimuli send signals to areas of the brain, and the perception appears to be decoded by a combinatorial mechanism. In other words a familiar scent may be the result of a dozen ORs firing at once. The bitter receptors on the tongue form a very similar family of 7TM receptors, but it appears that all of the signals converge on a single area of the brain, so that many different molecules can trigger the same bitter signal. Glucose and other sugars trigger a similar process for sweetness and glutamate binds to its own 7TM receptor for the savory umami flavor. Ion channels account for salty (Na+) and sour (H+) tastes.

Vision is mediated by rhodopsin, a 7TM receptor with retinal bound at the cen

ter. The change in shape when light isomerizes retinal is exactly the same as when other 7TM receptors bind to their appropriate ligands. And the result inside the cell is parallel, but in this case the G protein (transducin) triggers the cleavage of cyclic GMP. Color is perceived by cone photoreceptors very similar to rhodopsin, but modified to absorb maximally in the red, green, and blue regions. Color blindness is due to homologous recombination of the photoreceptor genes.

Hearing and touch use related systems to sense mechanical stimuli. Displacement

of hair cells in the cochlea causes direct stimulation of nerves by opening ion channels. 573 574

CHAPTER 32

A similar receptor in Drosophila, known as NompC, appears to be an ion channel with 29 ankyrin repeats at the amino terminus, inside the cell. Touch appears to utilize similar receptors for touch, plus receptors for temperature and pain. Capsaicin is the active compound in hot peppers, and the capsaicin receptor (VR1) has been studied. It also appears to be an ion channel with ankyrin repeats, and is involved in sensing pain. The authors finish with a discussion of the magnetic sense, pheromones, and circadian rhythms. LEARNING OBJECTIVES A Wide Variety of Organic Compounds Are Detected by Olfaction (Text Section 32.1)

1. Describe the properties of a typical odorant, and explain the relationship between shape

and smell.

2. Define the term specific anosmia.

3. Explain how scientists grew to suspect that G proteins are involved with olfaction.

Furthermore, explain how we know there are hundreds of different odorant receptors (OR) in humans.

4. Interpret what happens when an OR binds to an odorant. Include details about G(olf),

cAMP, and calcium.

5. Describe what is meant by a combinatorial mechanism, and how it applies to decoding of

olfactory signals.

6. Explain the mechanism by which functional magnetic resonance imaging (fMRI) allows vi

sualization of active regions of the brain. Taste Is a Combination of Senses That Function by Different Mechanisms (Text Section 32.2)

7. List the five primary tastes. Describe the different receptor types for each taste, and name

compounds that stimulate each receptor.

8. Identify gustducin, PROP, and T2R-1.

9. Explain why a large family of bitter receptors does not produce a broad spectrum of dif

ferent bitter flavors.

10. Present the evidence that suggests that the family of sweet receptors is closely related and

parallel to the family of bitter receptors.

11. Describe the structure of an amiloride sensitive sodium channel. Compare this sodium chan

nel to the potassium channel shown in figures on page 360 of the text.

12. Explain how protons interact with various ion channels to send the sour sensation to the

brain.

13. Describe the glutamate receptor, which is responsible for sensing the umami taste. Photoreceptor Molecules in the Eye Detect Visible Light (Text Section 32.3)

14. Distinguish between cone and rod photoreceptor cells.

15. Describe the structure of a rod cell.

16. Define rhodopsin, opsin, and 11-cis-retinal.

SENSORY SYSTEMS 575

17. Explain how light absorption affects the structure of retinal.

18. Identify transducin, and describe the reaction that is stimulated in response to light in

rod cells.

19. Briefly describe how rod cells recover after being stimulated by light.

20. Explain how the red, blue, and green cone receptors resemble one another, and how

they differ.

21. Describe the relationship between DNA recombination and color blindness. Hearing Depends on the Speedy Detection of Mechanical Stimuli (Text Section 32.4)

22. Define cochlea, hair cell, stereocilia, and tip link.

23. Relate the evidence that systems found in Drosophila may be homologous to auditory

sensors in vertebrates. Explain the role ankyrin appears to play. Touch Includes the Sensing of Pressure, Temperature, and Other Factors (Text Section 32.5)

24. Identify nociceptor, capsaicin, and VR1.

25. Explain how some organisms sense Earth’s magnetic field.

26. Define circadian rhythm. SELF-TEST A Wide Variety of Organic Compounds Are Detected by Olfaction

1. If we have 750 odorant receptor (OR) genes in the human genome, but only 30% of

them are functional, then:

(a) How many functional genes would there be? (b) What percentage of the complete human genome is taken up by OR genes?

2. At the start of Section 32.1.1 the structures of R- and S-carvone are presented. One of

these enantiomers smells like spearmint and the other like caraway seeds. How can two “identical” compounds have such different sensory qualities?

3. Which would NOT be an indicator that a certain gene codes for an OR?

(a) codes for 7TM protein (b) genes found in cells of nasal epithelium (c) part of a large and diverse family of similar genes

4. Is it likely that there is a single OR responsible for a smell like orange peel or chocolate?

5. Functional magnetic resonance focuses on what in the brain?

(a) electron flow in nerves (b) temperature of active brain regions (c) speed of blood flow (d) hemoglobin versus oxyhemoglobin in the brain (e) turnover of neurotransmitters in the brain

576

CHAPTER 32 Taste Is a Combination of Senses That Function by Different Mechanisms

6. Which of the following is INCORRECT?

(a) Like OR, bitter receptors form a large family of 7TM proteins. (b) There are many subtle shades of bitter linked with the different receptors. (c) Bitter and sweet taste receptors are closely related. (d) Bitter receptors are found mostly on the back of the tongue.

7. Compare the structures of tetrodotoxin (in Section 13.5.4, p. 358) and amiloride (in

Section 32.2.3). What structural similarity do you see? Are there similarities in the actions of the two compounds?

8. How does the umami receptor differ from receptors in the brain that detect glutamate as

a neurotransmitter? Photoreceptor Molecules in the Eye Detect Visible Light

9. Match the two main regions of the rod cell in the left column with the appropriate func

tions or properties from the right column.

(a) outer segment

(1) contains discs

(b) inner segment

(2) carries out normal cellular processes (3) contains the photoreceptors that

absorb light

10. Which of the following statements about retinal, the chromophore of rhodopsin, is

INCORRECT?

(a) The unprotonated Schiff base absorbs maximally at 440 nm and higher. (b) In the dark, it is covalently bound to opsin through a Schiff base linkage. (c) In the dark, it is present as the 11-cis-retinal isomer. (d) When bound to rhodopsin, it absorbs light maximally at 500 nm. (e) It becomes the all-cis isomer after absorbing light.

11. Place the following events in the excitation of rhodopsin by light in their correct sequence.

(a) metarhodopsin II (b) conversion of all-trans-retinal to 11-cis-retinal (c) triggering enzyme cascade (d) conversion of 11-cis-retinal to all-trans-retinal (e) activation of transducin G protein (f)

bathorhodopsin

Which of the following statements about G(olf), gustducin, and transducin are true?

(a) All are G proteins. (b) All are associated with 7TM sensory receptors. (c) All bind calcium ions. (d) All bind GTP, which is hydrolyzed to GDP.

13. Explain briefly the major roles of the following participants in the enzymatic cascade that

is triggered by the photoexcitation of rhodopsin.

(a) transducin (b) cyclic GMP (cGMP) (c) activated phosphodiesterase

14. With its seven transmembrane helices, rhodopsin has a structure similar to that of which

of the following integral membrane proteins?

(a) nicotinic acetylcholine receptor channel (b) Na+-K+ pump (c) sarcoplasmic Ca2+-ATPase (d) hexokinase (e) bacteriorhodopsin light-driven pump (f)

photoreceptors of retinal cones

15. How does the photoexcited system return to the dark state?

16. Which of the following statements about human color vision are correct?

(a) It is mediated by three different chromophores. (b) It is mediated by three different photoreceptors. (c) It involves only the cone cells. (d) It involves only the rod cells. (e) It involves seven-transmembrane-helix proteins.

17. Which colors do not correspond to human visual pigments? What would be different if

this question were about color vision in chickens?

(a) red (b) yellow (c) green (d) blue (e) violet

18. Why is the proportion of color-blind males so much higher than that of color-blind females? Hearing Depends on the Speedy Detection of Mechanical Stimuli

19. The text states (in Section 32.4.2) that a likely ortholog of the transduction channel used

in human hearing has been found in fruit flies. What is an ortholog? How similar is its use in flies and people? Touch Includes the Sensing of Pressure, Temperature, and Other Factors

20. Birds appear to make use of the Earth’s magnetic field while migrating. How could this

be proven? ANSWERS TO SELF-TEST

1. There would be some 260 functional genes. If the human genome has 40,000 genes, then

one could either calculate the percentage of functional genes (260/40,000 = 0.65%) or calculate the percentage of “apparent” genes—which is justifiable because the figure of 40,000 includes all “apparent” genes. This yields 750/40,000 = 1.875 %. Either way, this is a surprisingly large chunk of the genome, and it shows how important the sense of smell must be in higher animals. 578

CHAPTER 32

2. Even though they look very much the same on paper, in three dimensions they would

bind to an active site very differently. Thus they interact with different OR and smell completely different.

3. (b) is wrong because all cells in an individual have the same germ-line DNA. Thus OR

receptors are found everywhere. If cDNA can be made from cells in the nasal epithelium, that means that mRNA is being expressed there, and thus that would be an indication that the gene might be an OR.

4. No. Even when a scent is known to be triggered by a single chemical (like vanilla) there

will be an array of receptors that perceive it, and the sensation will be decoded by a combinatorial process.

5. d

6. b

7. Both contain the guanidinium moiety. This cation would be attracted to sodium chan

nels of all types, and then the rest of the molecule will block the channel. So both compounds function as the “cork” in the “bottle.”

8. See Figure 32.18. The large yellow portion (high affinity Glu-binding domain) is pres

ent in brain receptors but missing in umami receptors, resulting in a lower affinity for Glu in the umami receptors.

9. (a) 1, 3 (b) 2

10. e

11. d, f, a, e, c, b

12. c.

13. (a) Photoexcited rhodopsin binds inactive transducin (T–GDP), catalyzes the exchange

of GTP for GDP, and releases Ta–GTP. This form of transducin then activates cGMP phosphodiesterase. Hydrolysis of GTP bound to Ta deactivates phosphodiesterase and allows the binding of Tbg, regenerating T–GDP.

(b) In the dark, cGMP keeps the cation-specific channels open. Activated phosphodi

esterase hydrolyzes cGMP to 5′-GMP, leading to the closing of the channels.

leads to the closing of the cation-specific channels and to hyperpolarization of the plasma membrane.

14. a, f

15. The return to the dark state requires the deactivation of both cGMP phosphodiesterase

and photoexcited rhodopsin and the formation of cGMP. Phosphodiesterase is deactivated by the hydrolysis of GTP bound to Ta to return transducin to the inactive state. Photoexcited rhodopsin is deactivated by rhodopsin kinase, which catalyzes the phosphorylation of photoexcited rhodopsin at multiple sites. The phosphorylated rhodopsin binds arrestin, which blocks the binding of transducin. Guanylate cyclase catalyzes the synthesis of cGMP from GTP. Guanylate cyclase is inhibited by high Ca2+ levels, and therefore light-induced lowering of Ca2+ reactivates cGMP formation.

16. b, c, e

17. b, e. Chickens have a violet pigment so the answer would be (b) only.

SENSORY SYSTEMS 579

18. The red and green visual pigment genes are both on the X chromosome. Women get two

chances to “get it right” because they have two copies of X. Men get only one chance because they are XY. Homologous recombination ensures that many X chromosomes have a missing visual pigment gene.

19. Review Chapter 7, page 173. Orthologs arise by gene duplication and generally diverge

in function. Paralogs arise by a speciation event and generally retain similar function in different species. The use of the protein in flies and humans appears remarkably similar; both are sensing vibrations or disturbances in the air.

20. One way would be to make magnetic “hats” and see if the birds can be made to fly per

pendicularly to their normal routes. This may not work, because birds appear to pay attention to the stars, and use the magnetic field as a kind of calibration. One recent article suggests that some species use the magnetic field to tell them when to fatten up for a long flight with little available food (Nature 414[2001]:35). PROBLEMS

1. A very common specific anosmia is the inability to smell musk. When exposed to the

pure compound, some 10% of the population smell nothing, and another 20% find the smell unpleasant. What is happening on the molecular level to explain this?

2. Wine drinkers sometimes find that a bottle is “corked.” This means that a defective cork

has leached trichloro-anisole, or TCA, into the wine. This compound ruins the wine, making it smell like wet cardboard. Some wine collectors report that as much as 20% of the bottles they open are corked, while others hardly ever encounter the problem. How would you account for the individual differences? How could the problem of corked wines be solved?

3. Why are specific anosmias so common, but the lack of some particular taste (would we

call it specific “dis-gustia”?) is quite rare? Consider the bitter receptor in your answer.

4. The fruit of the African Miracle Berry bush (Synsepalum dulcificum) has no flavor. But after

it has been chewed, acidic foods taste sweet and not sour. A lemon tastes delicious, and a tomato is sweeter than an apple. The active substance appears to be a protein, and the mechanism has not been discovered. What are some possible mechanisms?

5. One close relative of the umami receptor is the NMDA receptor found in the brain. The

NMDA receptor is one of several glutamate receptors in the central nervous system. NMDA stands for N-methyl-D-aspartate, a compound that does not exist in nature. Why do you think a glutamate receptor would be named for a compound not found in the brain?

6. Late at night, police investigated a car pulled over to the side of the highway. When they

interviewed the driver, he stated that he had seen streams of blue light coming out of his dashboard. At first the officers thought that the man had taken hallucinogenic drugs, but then he told them he had taken sildenafil (Viagra) earlier in the evening. Why does Viagra sometimes have visual side effects? Why are they most likely to occur at night? Sildenafil works by inhibiting a phosphodiesterase known as PDE5, an enzyme that breaks down cGMP. 580

CHAPTER 32

7. Which of the compounds in Figure 32.1 would be expected to have the smallest ab

sorbance per mole in the visible light range? Give the reason for your choice. FIGURE 32.1 Four light-absorbing compounds.

H C

CH

H C

CH

3

3

CH

3

3

J

CHO

A

B

CH

3

CHO

H C

CH

H C

CH

3

3

CH

3

3

J

CHO

J

CH OH

2

C

D

CH

3

8. Figure 32.2 shows the absorption spectrum of a light-absorbing pigment that is involved

in color vision.

(a) What color light is absorbed by the pigment? (b) When the pigment is extracted into an organic solvent, what is the color of the re

sulting solution? FIGURE 32.2 Absorption spectrum of a color vision pigment.

e absorbance

Relativ

400

500

600

700

Wavelength (nm)

9. 11-cis-Retinal is covalently linked to a lysine side chain of opsin by the formation of a

Schiff base:

K O

H;

≈

J

RJ

C +H NJ(CH ) JOpsinGRJCKNJ(CH ) JOpsin+H O

J

2

2 4

2

H

J

H

Such linkages can typically be stabilized in the laboratory by reduction with borohydride:

H

:

J

≈

BH

RJCKNJ(CH ) JOpsinD

4

RJCH JNJ(CH ) JOpsin

2 4

2

2 4

J

H

Would you expect a rod cell preparation that has been treated with borohydride to be active in the cycle of visual excitation? Why or why not?

SENSORY SYSTEMS 581

10. A color vision pigment that absorbs red light is chemically cleaved to separate the reti

nal from the protein. The same is done for a pigment that absorbs blue light. Then a new pigment is constituted using the retinal from the red-absorbing pigment and the opsin from the blue-absorbing pigment. What color of light will be absorbed by the new pigment?

11. Many people who are hard of hearing or completely deaf have been helped by cochlear

implants. How is it possible to mimic the natural hearing process?

12. Zostrix is one of several skin creams that contain capsaicin. Rubbing the cream on the

less sensitive parts of the body (knee, elbow, neck) normally does not produce a burning sensation, and can provide relief from neuralgia or arthritis pain. One must avoid contacting eyes, nose, or mouth because capsaicin is the hot principle of the chili pepper and can cause painful burning of mucus membranes. Why would capsaicin work as an analgesic, when it directly stimulates pain receptors?

13. Section 32.5.2 of the text implies that there are human pheromones. Are there? What sort

of evidence can be found for and against the idea? You will need to try searching the internet or the literature to answer this question, as the answer is not in the textbook. ANSWERS TO PROBLEMS

1. The 10% who are anosmic for musk either lack the receptor or have a mutation that

changes its specificity. People who can smell musk normally describe the scent as sweet and floral. But for the 20% who dislike it, it smells like old wet newspapers or something stored in the attic too long. It seems that a different receptor must have mutated and developed the ability to bind to musk. The structure of natural musk is interesting; it is the cyclic lactone of w-hydroxypalmitate. If one appreciates the sensory differences these genetic variations cause in humans, it brings home the truth of the old saying “de gustibus non disputandem est” (to each his own taste).

2. Just as with musk, there are many people with specific anosmia to TCA. This can ei

ther lead to an inability to smell it, or a diminished ability to smell it. The unlucky people who smell it very intensely often find that they can’t drink their expensive wine because of a cheap little cork. The obvious solution is to switch to plastic corks, or screw-on caps. Some people argue that the gentle oxidation that occurs through a cork allows the wine to age better. Others say that the disadvantages of real corks greatly outweigh the advantages.

3. Loss of an OR leads to the inability to smell that odor, or the diminished ability to smell

it. But because of the way bitter receptors are “wired,” loss of one kind of bitter receptor has essentially no effect on the perception of bitter tastes. All of the many receptors send their signals to the same area of the brain, so nothing is really lost when one or two bitter receptors mutate and lose their functionality.

4. Some people think that the protein binds directly to sweet receptors and stimulates them

when exposed to acid. It also might be possible that it is a glycoprotein, and some sugar hydrolyzes off in the presence of acid. The compound has been proposed as an aid to dieters, but artificial sweeteners solve the problem more directly. Artichokes contain a substance that produces similar changes in the taste of other foods.

5. NMDA isn’t the preferred transmitter for the NMDA receptor. But that receptor is

distinguished by its ability to bind and respond to the artificial compound, NMDA. Joe Z. Tsien, a researcher at Princeton, attracted much attention in the popular press

582

CHAPTER 32

when he found a way to enhance the functioning of NMDA receptors and produced “smart” mice, which could solve problems better and faster than normal mice (Nature 401[1999]:63).

6. The mechanism of action of Viagra is based on its ability to inhibit PDE-5. In other

words, it prevents cGMP from being cleaved to yield GMP. If levels of the drug are high enough, then cGMP levels in the retina are propped up at a high level because the local enzyme, PDE-6, becomes inhibited, so that normal cycling cannot occur. In a small percentage of the population this yields blue-tinted or blurry vision, sometimes with more extreme manifestations. See problem 9 in the text for more discussion. Scientists are working on second-generation Viagra(tm)-like drugs which will be more specific for PDE-5, and not stimulate the retina as much.

7. Compounds that absorb visible light significantly have long sequences of alternating sin

gle and double bonds, that is, they are conjugated. Compound B is unconjugated, so it would have negligible absorbance in the visible range of the spectrum. Compound A is 11-cis-retinal, C is all-trans-retinal, and D is all-trans-retinol. All of these would have significant absorbance in the visible range.

8. (a) Red light is absorbed by the pigment.

(b) A solution of the pigment would be blue or blue-green. The pigment absorbs red

light, but it transmits light at the blue end of the spectrum.

9. The photoexcitation of rhodopsin leads initially to the isomerization of 11-cis-retinal to

all-trans-retinal and ultimately to the cleavage of all-trans-retinal from the protein. In a rod cell preparation that has been treated with borohydride, the Schiff base would be stabilized by reduction, so the removal of the all-trans-retinal would be impaired. As a result, the cycle of visual excitation could not occur. (See Figures 32.22 and 32.23 and Section 32.3.2 of the text for a discussion of these events.)

10. The new pigment will absorb blue light. The retinal is the same in all color vision pig

ments. The protein component, however, varies among the pigments, giving absorption maxima at different wavelengths.

11. Physiologists learned enough about the sort of nerve signals generated by the cochlea that

it was possible to simulate them. As long as enough nerves are present in the inner ear to convey the signals, an external device will perceive auditory vibrations and transduce them into electrical signals just like the hair cells in the cochlea. So the electrical impulses received by the brain are the same as would be experienced by a hearing person.

12. Even though the pain is not perceived when capsaicin is rubbed on a knee or elbow, the

receptors are being stimulated. And enough capsaicin applied regularly appears to flatten the response of the pain receptors on the surface of the skin. This means that there is less response to real pain, which occurs under the skin.

13. The most compelling piece of evidence against human pheromones is the fact that the

vomeronasal organ (VNO), a pheromone receptor found in many higher animals including mammals, is vestigial in humans and disappears before birth. Contradicting this is the fact that humans do have one apparently functional receptor, V1R, which is closely homologous to known pheromone receptors in other mammals. For a review see Nature 413(2001):211; pheromones are discussed near the end of the article. Other evidence is more circumstantial, like the fact that women living in the same dormitory (and breathing the same air) will have synchronized menstrual cycles. Human pheromones have yet to be demonstrated decisively.

SENSORY SYSTEMS 583 EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. Isoleucine has the same geometry as valine and is one carbon larger. Likewise, heptanal

has the same geometry as octanal and is one carbon smaller. It is plausible that these two features may compensate at the active site of the mouse receptor. The extra carbon on Ile-206 could therefore interfere with octanal binding but still nicely accommodate the shorter heptanal.

2. The specificity would be switched, and the “attractant” would become a “repellant.” The

AWB neurons induce avoidance behavior when their receptors encounter the corresponding ligands. This general behavior would be expected to remain true in the transgenic nematode.

3. At least two compounds must be present: C5-COOH (receptor 5) and HOOC-C7

COOH (receptor 9, as well as receptors 3, 12, and 13). Other compounds that activate receptor 5 cannot be present because of the pattern of non-activated receptors. Each of the following additional compounds could be present, but is not necessarily present: Br-C3-COOH, Br-C4-COOH, HOOC-C4-COOH, HOOC-C5-COOH (all activating receptor 13 only), and HOOC-C6-COOH (receptors 3, 12, and 13).

4. Sour and salty taste responses result from the direct action of hydrogen ions or sodium

ions on channels; these responses therefore have the potential for very rapid time resolution. Taste responses (bitter and sweet) that are likely to require 7TM receptors and second messengers will exhibit slower time resolution.

5. The estimated time difference is (0.15 m) / (350 m s−1) = 4.3 × 10−4 s. The time res

olution of the human hearing system is slightly faster than this, namely about 2 × 10−5 s, so that indeed the difference in arrival time for sound at the two ears can be easily detected. A system that would use 7TM receptors and G proteins, however, would be too slow because it would require a response time of at least several milliseconds.

6. The olfactory response would be eliminated entirely, and the sensitivity of the visual re

sponse would be greatly reduced. Within the olfactory system, if adenylate cyclase were always fully active, then there would be no opportunity for increasing the intracellular concentration of cAMP in response to the binding of an odorant to its receptor. No signals could be processed. In the visual system, if guanylate cyclase were always fully active, then it would be more difficult for phosphodiesterase to reduce the level of cGMP, and consequently more difficult to close the cGMP-gated channels and initiate neuronal signalling. The signal amplification would therefore be impaired, and the visual sensitivity would be reduced. The kinetics of visual recovery would also be altered because the rate of cGMP synthesis would no longer be sensitive to calcium.

7. The tastant would probably be sweet because the rodent showed a preference for drink

ing the water that contained the added tastant. (If the tastant had been bitter, then the rodent would have preferred the bottle with the plain water.)

8. There exist significant examples of mimicry in the competitive biological world. By im

itating a toxic plant with a bitter taste, the nontoxic plant may be able to decrease the extent to which it is eaten by animals.

9. The restriction is a precaution for airline safety. The inhibition of a cGMP phosphodi

esterase that is prevalent in smooth muscle could have an unknown side effect that potentially might affect the pilot’s response (e.g., to an emergency situation) while flying the airplane. 584

CHAPTER 32

10. For all senses, ATP hydrolysis is required to generate and maintain ion gradients and

membrane potential. Olfaction: ATP is required for the synthesis of cAMP. Gustation: ATP is required for the synthesis of cyclic nucleotides, and GTP is required for the action of gustducin in the detection of bitter and sweet tastes. Vision: GTP is required for the synthesis of cGMP and for the action of transducin. Hearing and touch: ATP hydrolysis is required to generate and maintain ion gradients and membrane potential and may be required for other roles as well.

11.

HO

R

H;

O

H

NH

+

K

NH

H

R Lysine Retinal

HO:

R

H

H;

R

H

K

NH

N

2 Schiff base CHAPTER 3 The Immune System 3

Chapter 33 deals with the immune system. The cells and proteins of this system co

operate to detect and inactivate foreign (nonself) molecules, microorganisms, and viruses. The humoral immune response acts through soluble antibodies, secreted

into the circulatory system, that bind antigens with high specificity and affinity. The cellular immune response acts through receptor proteins similar to antibodies on the surface of specialized cells (T lymphocytes) that bind to peptides presented on another kind of cell-surface protein, the major-histocompatibility-complex (MHC) proteins. The immune system affords a good example of a system governed by the basic principles of evolution.

After defining essential immunological terms, the authors discuss the structure of

immunoglobulin G, including the molecular details of antibody–antigen recognition. Next they relate the variable and constant regions of immunoglobulins to the organization of the genes that encode them. They explain how somatic recombination of a large variety of V-, D-, and J-segment genes with a few C genes, plus imprecise joining, can generate the enormous diversity of antibody molecules. The five classes of immunoglobulins with their characteristic polypeptide compositions and functions are introduced. These five classes are generated by gene rearrangements, in a process called class switching. Alternative mRNA splicing leads to the formation of membrane-bound or soluble immunoglobulins. Finally the authors turn to the cellular immune response. They describe the functions of cytotoxic T cells and helper T cells, and discuss the diversity, polypeptide composition, and structures of MHC proteins and T-cell receptors. They conclude the chapter with a discussion of the human immunodeficiency virus (HIV) infection of helper T cells, and a discussion of autoimmune disease and how the system normally avoids it.

Preceding chapters in the text that deal with protein structure (Chapters 3 and 4),

evolution (2 and 7), molecular recognition (Chapters 9 and 10), and flow of genetic information (Chapters 5 and 6) constitute important background for this chapter. Note 585 586

CHAPTER 33

that some basic information about the immune system, and a discussion of the production of monoclonal antibodies, are covered in Section 4.3.

When you have mastered this chapter, you should be able to complete the following

objectives. LEARNING OBJECTIVES Adaptation of the Immune System Relies on the Power of Evolution (Text Section 33.0.1)

1. Contrast the usual targets of the humoral immune response and the cellular immune response.

2. Distinguish between cytoxic T cells (“killer T cells”) and helper T cells and describe their

functions.

3. Describe the basic evolutionary principles that govern the immune system. Antibodies Possess Distinct Antigen-Binding and Effector Units (Text Section 33.1)

4. Relate the intact structure of an immunoglobulin G (IgG) molecule to the Fab and Fc

fragments produced by proteolysis. Describe the functions performed by the different regions of IgG.

5. Sketch the polypeptide chains of an IgG molecule, and relate the heavy chain (H)–light chain (L) subunit composition (H2L2) to the Fab and Fc fragments of the molecule. Understand the function of the hinge region of IgG.

6. Describe the features of the constant (C), variable (V), and hypervariable amino acid

sequences of the L and H chains of IgG molecules.

7. List the different classes of immunoglobulins and give their functions. Note the common

occurrence of (k) and (l) L chains in all classes and the (a, m, d, g, or ¯) H chains that provide the structural bases for the function of each class. The Immunoglobulin Fold Consists of a Beta-Sandwich Framework with Hypervariable Loops (Text Section 33.2)

8. Discuss the biological distribution of proteins that contain the immunoglobulin-fold

structural motif.

9. Relate the hypervariable sites of the H and L chains of IgG to the complementarity- determining regions (CDRs) of the immunoglobulin. Describe the function of the constant regions of the H and L chains of IgG. Antibodies Bind Specific Molecules Through Their Hypervariable Loops (Text Section 33.3)

10. Describe the domain structure of IgG as revealed by crystallography, and note the presence

of the immunoglobulin fold as a common structural feature. Locate the antigen-binding sites.

11. Summarize the types of bonds that form complexes between immunoglobulins and

antigens. Note the similarities of combining sites of immunoglobulins with the active sites of enzymes.

THE IMMUNE SYSTEM 587 Diversity Is Generated by Gene Rearrangements (Text Section 33.4)

12. Describe the contributions of the number of immunoglobulin genes (the germ-line repertoire), somatic recombination, and somatic mutation in the generation of antibody diversity. Describe the roles of the V, C, D, and J segment genes in these processes.

13. Calculate the diversity of immunoglobulin structures that arise from the combinatorial association of different genes and from somatic mutation.

14. Describe the process of affinity maturation and how it can lead to a 1000-fold increase in

binding affinity during the course of a disease.

15. Define ITAM and explain its significance in linking oligomerization of B-lymphocyte sur

face antibodies to secretion of soluble antibodies.

16. Discuss the mechanism and use of the drug cyclosporin.

17. Define hapten, epitope, antigen, and immunogen.

18. Describe the phenomenon of class switching, and note its significance in maintaining con

stant recognition specificity among the immunoglobulin classes. Major-Histocompatibility-Complex Proteins Present Peptide Antigens on Cell Surfaces for Recognition by T-Cell Receptors (Text Section 33.5)

19. Describe the formation and recognition of foreign peptides displayed on cell surfaces in

complexes with MHC proteins.

20. Describe the interaction of a peptide with the peptide-binding site of HLA-A2. Mention

the typical length of the peptide and the anchor residues for this site.

21. Compare the structures and features of T-cell receptors with those of the immunoglobulins.

Note the sizes and conformations of the epitopes that are recognized by each kind of protein.

22. Explain the origins of the diversity of T-cell receptors and why these receptors can be

even more diverse than immunoglobins.

23. Outline a model that accounts for the recognition of a combined epitope by a T-cell receptor.

24. Explain the role of CD8 coreceptors in the activation of cytotoxic T cells, and describe the

functions of perforin and granzymes in leading to the cell’s death from apoptosis.

25. Explain the role of CD4 coreceptors in the activation of helper T cells, and describe the

functions of cytokines such as interleukin-2 and interferon-g.

26. Describe the subunit structures of the class I and class II MHC proteins. Locate the

peptide-binding sites.

27. Describe the three classes of proteins (class I, class II, and class III) encoded by the MHC genes. Note the diversity of the class I and class II MHC proteins and their significance in transplantation rejection.

28. Describe HIV and the disease it causes—AIDS.

29. Outline the mechanism of infection and lysis of helper T cells by HIV. Immune Responses Against Self-Antigens Are Suppressed (Text Section 33.6)

30. Explain the selection process, both positive and negative, that is applied to T cells in

the thymus.

31. List three autoimmune diseases. Explain the consequences when the immune system fails

to distinguish between self and nonself.

32. Describe the role that the immune system plays in cancer prevention. 588

CHAPTER 33 SELF-TEST Antibodies Possess Distinct Antigen-Binding and Effector Units

1. Match the structure or feature listed in the right column with the appropriate IgG frag

ment on the left.

(a) Fab

(1) contains an antigen combining site

(b) Fc

(2) one is formed per IgG molecule (3) contains an H-chain fragment (4) contains an intact L chain (5) mediates effector functions such as

complement fixation in the intact IgG

(6) two are formed from an IgG molecule (7) forms a precipitate upon binding

an antigen

2. Which of the following statements about the L and H chains of IgG are correct?

(a) The H chains of IgG molecules have variable and constant regions of amino acid

sequences.

(b) The H chains are responsible for segmental flexibility. (c) The constant region of L chains exists in two forms (k and l). (d) The variable region of the L chain has a counterpart of the same length and amino

acid sequence in the variable region of the H chain.

3. Match the immunoglobulin class listed in the left column with its property or function

from the right column.

(a) IgA

(1) most prevalent soluble antibodies

(b) IgD

(2) unknown function

(3) first soluble antibodies to appear in

(d) IgG

serum after immunization

(e) IgM

(4) protect against parasites (5) major antibodies in tears, saliva,

and mucus The Immunoglobulin Fold Consists of a Beta-Sandwich Framework with Hypervariable Loops

4. Fill in the blanks: A molecule of IgG contains _____ immunoglobulin domains. Each

heavy chain has _____ of these sandwiches, and each light chain has _____.

5. The immunoglobulin fold is made up of

(a) seven alpha-helical segments. (b) a beta-barrel. (c) a sandwich of two parallel beta sheets. (d) a sandwich of two antiparallel beta sheets. (e) a beta saddle domain. Antibodies Bind Specific Molecules Through Their Hypervariable Loops

6. Which of the following statements about IgG structure are correct?

(a) Each of the two antigen-combining sites on an IgG molecule can bind to a struc

turally distinct epitope.

(b) Both interchain and intrachain disulfide bonds stabilize IgG structure.

THE IMMUNE SYSTEM 589

(c) Both the L and the H chains of IgG contain domains with similar structures. (d) The hypervariable regions of the L chain are the sole determinants for the binding

of the IgG to the specific antigen.

7. Which of the following are properties of antigen-binding sites of IgG?

(a) They are located between the two sheets of b strands of the V domains. (b) They are made up of loops formed by the complementarity-determining regions of

both the VL and the VH domains.

(c) They may undergo conformational changes upon binding of the hapten or antigen. (d) They contain a specific amino acid residue that covalently binds to the antigen. (e) They form numerous weak electrostatic, hydrogen-bond, van der Waals, and

hydrophobic interactions with the antigen surface.

8. X-ray analysis has revealed that when small antigens bind to antibodies

(a) they usually fit into a cleft. (b) they contact all six CDRs. (c) they can bind to the Fc end of the antibody. (d) they are attracted by the same noncovalent bonds found in enzyme/substrate

interactions.

9. Explain why the antibodies produced in an animal in response to a given antigen display

a range of binding constants for the antigen eliciting them. Diversity Is Generated by Gene Rearrangements

10. If the mRNA encoding the L chain of an IgG molecule were isolated, radiolabeled, and

hybridized to genomic DNA that has been isolated from either the plasma cell producing the antibody or from germ-line cells from the same organism, what would you observe with respect to the relative locations of the L-chain gene sequences on the two DNAs?

11. Some antibody diversity arises from the combination of one V gene and one C gene from

pools containing numerous different copies of each. Why doesn’t this mechanism account completely for the observed diversity?

12. Match the following gene segments with their approximate number in the germ line.

(a) Vk

(1) 5

(b) VH

(2) 6

(3) 27

(d) JH

(4) 40

(e) D

(5) 51

13. Small foreign molecules do not usually elicit the formation of soluble antibodies, and the

cellular immune system also ordinarily responds only to macromolecules. Explain how an antibody can sometimes be directed against a small foreign molecule.

14. Match the immunological term in the left column with its description or definition in

the right column.

(a) antigen

(1) particular site on an immunogen to

(b) antibody

which an antibody binds

(2) protein synthesized in response to

(d) hapten

an immunogen

(3) macromolecule that elicits antibody

formation

(4) small foreign molecule that elicits anti

body formation

(a) RNA splicing joins the sequences that encode VHDJH regions to sequences that en

code different class CH regions.

(b) Plasma cells that initially synthesize IgM switch to form IgG with the same antigen

specificity.

(c) Class switching doesn’t affect the variable region of the H chains. (d) Class switching allows a given recognition specificity of an antibody to be coupled

with different effector functions. Major-Histocompatibility-Complex Proteins Present Peptide Antigens on Cell Surfaces for Recognition by T-Cell Receptors

16. Which are properties of the peptide-binding site of HLA-A2, a MHC class I protein?

(a) consists of a deep groove with a b sheet floor and a-helical walls (b) can bind all peptides of 7 to 11 amino acid residues with equal affinity (c) interacts specifically with two anchor residues of the peptide and nonspecifically

with the rest of the amino acid residues

(d) binds peptides that retain their a-helical conformation (e) has a very high affinity for peptides

17. Which of the following statements about T-cell receptors are correct?

(a) T-cell receptors recognize soluble foreign molecules in the extracellular fluid. (b) T-cell receptors recognize T cells. (c) For a T-cell receptor to recognize a foreign molecule, the molecule must be bound

by proteins encoded by the genes of the major histocompatibility complex.

(d) The T-cell receptor is structurally similar to an IgG immunoglobulin in that it has

two H and two L chains.

(e) The T-cell receptor is encoded by genes that arise through the recombination of a

repertoire of V, J, D, and C DNA sequences.

(f)

T-cell receptors primarily recognize fragments derived from foreign macromolecules.

18. Match the receptor proteins in the left column with appropriate structural features in the

right column.

(a) class I MHC proteins

(1) contain an immunoglobulin fold motif

(b) class II MHC proteins

(2) contain domains homologous to the

V and C domains of immunoglobulins

(3) contain a b-microglobulin chain (4) contain two transmembrane polypep

tide chains

(5) contain one transmembrane chain (6) are encoded by six different genes

that are highly polymorphic

(7) are encoded by V, D, and J gene

segments

19. Match the type of T cell with the corresponding feature listed in the right column.

(a) helper T cell

(1) detects foreign peptides presented

(b) cytotoxic T cell

on cell surfaces

(2) recognizes class I MHC protein plus

peptide

(a) MHC proteins play a role in the rejection of transplanted tissues. (b) One class of MHC proteins is present on the surfaces of nearly all cells, binds frag

ments of antigens, and presents them to T-cell receptors.

(c) MHC proteins are encoded by multiple genes. (d) One class of MHC proteins provides components of the complement system. (e) The genes encoding MHC proteins produce three classes of soluble proteins.

21. Which one of the following is NOT a property of the human immunodeficiency virus (HIV)?

(a) It is a retrovirus, that is, it has an RNA genome that produces viral DNA in the

host cell.

(b) It contains a bilayer membrane with two kinds of glycoproteins. (c) It interacts with the CD4 coreceptor of helper cells through the gp120 glycoprotein. (d) It injects its RNA into the cell and releases the coat into the medium surrounding

the cell.

(e) It impairs and destroys the host cell by increasing its permeability. Immune Responses Against Self-Antigens Are Suppressed

22. T cells are subject to both positive and negative selection during fetal development in

vertebrates. Why are both needed?

23. Human cancer cells are very much like normal human cells. How then can the immune

system play a role in cancer prevention? ANSWERS TO SELF-TEST

1. (a) 1, 3, 4, 6 (b) 2, 3, 5. Answer (7) is inappropriate for either fragment because the Fc

fragment lacks an antigen-binding site and the Fab fragment contains only one. The insoluble lattice of antigen–antibody molecules forms because intact IgG molecules each have two antigen-binding sites and can therefore link several antigens together.

2. a, b, c. Answer (d) is incorrect because the variable sequences of the L and H chains in

a given IgG are different from one another.

3. (a) 5 (b) 2 (c) 4 (d) 1 (e) 3

4. 12, 4, 2

5. (d), but (b) isn’t entirely wrong. Some people describe the beta-sandwich as a “col

lapsed barrel.”

6. b, c. Answer (a) is incorrect because the two antigen-combining sites on an IgG mol

ecule are directed toward the same epitope. Thus, an IgG can bind only one kind of

592

CHAPTER 33

antigen. Answer (d) is incorrect because the hypervariable regions of both H and L chains form the antigen-combining site.

7. b, c, e

8. a, d

9. The antigen stimulates several B lymphocytes that bear different surface antibodies that rec

ognize it to differentiate into plasma cells and secrete antibodies. Each plasma cell secretes a different kind of antibody that binds the antigen through a unique array of noncovalent interactions between the hypervariable regions of the antibody and the epitope bound.

10. The mRNA probe would hybridize to one region on the DNA from the plasma cell, but to

widely separated regions on the germ-line DNA. The gene on the plasma cell DNA encodes the intact gene sequence for the V, J, and L regions of the L chain, including introns. The genes encoding the V, J, and L regions of the L chain are in distant locations in the germline DNA because the DNA has not yet rearranged to bring these regions into proximity.

11. There aren’t enough unique V and C genes to provide a sufficient number of different

sequences when they are recombined in all the possible combinations. Joining (J) and diversity (D) genes increase the number of possible combinations.

12. (a) 4 (b) 5 (c) 1 (d) 2 (e) 3, see Section 33.4.2 in the text.

13. If the small molecule (hapten) becomes attached to a macromolecule (carrier), it can act

as an immunogen and serve as an epitope (haptenic determinant) to which an antibody can be selected to bind. See page 933, Section 33.4.3 in the text.

14. (a) 3 (b) 2 (c) 1 (d) 4

15. b, c, d. Answer (a) is incorrect because the sequence rearrangements of class switching

take place through DNA recombination.

16. a, c, e. Peptides are bound in “extended” conformation.

17. c, e, f. Answer (a) is incorrect because T-cell receptors recognize fragments of foreign

macromolecules only when the antigen is bound on the surface of a cell. Answer (d) is incorrect because the T-cell receptor is composed of one a and one b chain, each having sequences that are homologous to the V regions of the chains of immunoglobulins.

18. (a) 1, 3, 5, 6 (b) 1, 4, 6 (c) 1, 2, 4, 7

19. (a) 1, 3, 5, 6 (b) 1, 2, 4, 7

20. a, b, c, d. Answer (e) is incorrect because MHC proteins are all bound to the cell surface

and are not soluble. Answer (d) is correct because class III MHC proteins contribute to the complement cascade; they are mentioned in Section 33.5.6.

21. d. This answer is incorrect because the HIV RNA and other core components enter the

helper T cells as the viral membrane fuses with the cell membrane.

22. The positive selection winnows out the T-lymphocytes that do not bind to any of the

available MHC-peptide complexes. This step explains the Nobel-prize–winning observations, by Peter C. Doherty and Rolf M. Zinkernagel (1996), who observed that cytotoxic T-cells, which would kill virus infected cells from the mouse they came from, would not kill mouse cells infected with the same virus, but from an unrelated mouse (Nature 251[1974]:547). Most developing T-cells are discarded during the “positive selection” phase including cells that would respond to MHC proteins from other individuals, but not those present in “self ” cells. The negative selection process then removes T-cells that bind too tightly to “self ” peptides complexed with MHC proteins.

23. As the text points out, cancer cells will sometimes produce proteins that are inappropri

ate for the developmental stage of the individual, like the CEA associated with colorectal

cancer. And sometimes unique abnormal proteins can be produced. A widely used test for prostate cancer is the blood test for prostate specific antigen (PSA). The fact that human cancer cells are basically human cells poses a serious challenge for cancer treatment, and explains the fact that chemotherapy is generally rather difficult and uncomfortable. Antibiotics, directed against prokaryotic organisms, can be quite safe. But chemotherapeutic drugs generally have to inhibit processes that occur both in cancer cells and normal cells, so the therapeutic ratio is less favorable. PROBLEMS

1. The human genome contains only some 40,000 genes, but millions of antibodies are

produced by gene rearrangements as described in the text. This diversity generating system is confined to jawed vertebrates, but all higher animals (down to the sponge) have some kind of self versus nonself recognition system. What happens when a crab or an octopus is infected? How would its immune system cope?

2. Assuming that antigen–antibody precipitates have lattice-like structures (see Fig. 33.4 in the

text), draw simple sketches showing possible arrangements of antigen and antibody molecules in a precipitate in which the ratio of antibodies to antigens is (a) 1.14 and (b) 2.83.

3. When polyacrylamide gel electrophoresis (text Section 4.1.4) of a monoclonal antibody

preparation is conducted, a single sharp band appears. When the antibody preparation is treated with b-mercaptoethanol, two bands appear. Why is this the case?

4. Pepsin cleaves IgG molecules on the carboxyl-terminal side of the interchain disulfide

bonds between heavy chains. How many physical pieces would result from the cleavage of IgG by pepsin? How many of the pieces derive from the Fc region of IgG?

5. Suppose that antibody is prepared against the extracellular portion of a particular hormone

receptor known to have intracellular tyrosine kinase activity, and known to carry out autophosphorylation of its tyrosine kinase. Suppose further that addition of this antibody to target cells elicits intracellular responses similar to those obtained by addition of hormone itself. Propose a mechanism by which addition of antibody may mimic the effects of adding hormone. (You may wish to review Section 15.4, pp. 411–413 in the text, before tackling this question. Hint: Fab fragments do not elicit the hormone-like response.)

6. DNP or 2,4-dinitrophenyl is used as a hapten (artificial epitope) in many experiments

on the immune system. The addition of a bifunctional DNP affinity-labeling reagent (one with two affinity-labeling groups) to myeloma protein (making the protein an artificial antigen) produces light and heavy chains that are cross-linked through Tyr 34 and Lys 54, respectively. What conclusion is suggested by this observation?

7. Most antigens are polyvalent, that is, they have more than one antibody-binding site. In

the case of macromolecules that contain regular, repeating sequences, like polysaccharides, it is easy to understand how a molecule might have multiple binding sites. In the case of proteins with nonrepeating sequences, it is more difficult to envision how polyvalence might be accounted for. Yet proteins with single polypeptide chains are polyvalent as antigens. What feature of antibody production accounts for this behavior?

8. Quantitative measures of the interactions between antigens and antibodies are frequently

given as association constants, the reciprocal of dissociation constants. (See problem 1 on p. 949 of the text.) The association constant for the binding of a given hapten to an antibody is 109 M−1, and the second-order rate constant for its binding is 108 M−1 s−1. Calculate the rate constant for the dissociation of the hapten from the antibody. 594

CHAPTER 33

9. Propose a method using the technique of affinity chromatography (p. 82 of the text) that

would allow one to select lymphocytes producing antibody to one particular antigen from a heterogeneous population of immature lymphocytes. Explain the rationale behind your proposal.

10. Suppose that dinitrophenol is attached to a protein with many potential DNP binding

sites and that the resulting antigen is used to stimulate antibody production in rabbits. When serum is harvested and the immunoglobulin fraction is purified and mixed together with antigen, no precipitate forms, yet fluorescence measurements reveal the presence of antigen–antibody complexes. Explain this paradox.

11. The system that supplies MHC proteins with peptides is called “cut and display” (text

p. 935). Distinguish between what is cut and where it is cut in cells with MHC class I and MHC class II proteins.

12. In a 1995 experiment, researchers had male college students sleep in plain white T-shirts

for two nights. They then had female volunteers sniff the shirts and rate them as to how attractive they smelled. Because the researchers had tissue-typed all participants, they were able to determine that the women preferred the scent of men whose MHC proteins were the most different from their own. Why would this be logical, on an evolutionary basis? In a 2002 experiment, along parallel lines, other researchers found that women favored men who smelled like their fathers. Can this also be considered reasonable?

13. The text (p. 943) states that the mutation rate of HIV is more than 65 times higher than

that of the influenza virus. What are the medical implications of this? What about the evolutionary implications?

14. Although the immune system is designed so that antibodies are not ordinarily directed

toward one’s own tissue components, sometimes that process goes awry, leading to socalled autoimmune diseases. One human disease thought to involve such an autoimmune mechanism is myasthenia gravis, a relatively common disorder in which antibodies directed toward acetylcholine receptors lead to a decrease in receptor numbers. Those suffering from myasthenia gravis show weakness and fatigability of skeletal muscle. Eventually death results from loss of function of breathing muscles. Medical therapy for people suffering from myasthenia gravis may include two types of drugs, immunosuppressive agents and inhibitors of acetylcholinesterase. Give a brief rationale for the use of each. ANSWERS TO PROBLEMS

1. Even without a diversity-generating system, higher invertebrates will have a rather large

number of variable sequences in the germ-line DNA. These will allow for an immune response although without the power and flexibility of the vertebrate system. Some invertebrates live for many years, but most have a fairly short life-span, which probably diminishes the importance of the immune system somewhat.

2. The sketches for the two precipitates are shown in Figure 33.1. In (A) the Ab/Ag ratio is

8/7 = 1.14. In (B) the ratio is 17/6 = 2.83.

3. Polyacrylamide gel electrophoresis separates proteins on the basis of size. b-mercap

toethanol reduces the disulfide bonds that link the light and heavy antibody chains. Thus, the light and heavy chains are separated from one another on the gel.

4. Three pieces result from the cleavage of IgG by pepsin, as shown in Figure 33.2. One

piece contains both Fab units. The other two pieces derive from the bisection of the Fc region.

THE IMMUNE SYSTEM 595 FIGURE 33.1 Possible lattice structures for Ab/Ag ratios of (A) 1.14 and (B) 2.83.

•

A

B

•=Ag

=Ab FIGURE 33.2 Cleavage of IgG by pepsin.

Fab

SJS

Fc

5. Addition of hormone to receptors in some instances causes the hormone-receptor com

plex to dimerize. The tyrosine kinase domains of two such approximated receptor monomers will then become capable of cross-phosphorylation, with a consequent increase in tyrosine kinase activity. The addition of bivalent antibodies would also have the effect of drawing monomeric receptor units together in such a way that they could crossphosphorylate and thus trigger intracellular effects.

6. The observation suggests that both Tyr 34 on the light chain and Lys 54 on the heavy

chain are involved in binding of the DNP hapten.

7. A given antigenic protein will stimulate the production of a mixed population of anti

bodies, with each type of antibody being specific for a different region in the tertiary structure of the antigenic protein.

8. Since K = a kon/koff, substituting the given values yields

8

1

−

1

−

9

1

−

10 M s

10 M

= koff koff = 10 1

−

1

−

s 596

CHAPTER 33

9. One could attach antigen to insoluble beads, and then pass the lymphocyte preparation

over a column containing the beads. Lymphocytes capable of producing antibody toward the particular antigen would have a receptor complementary to that antigen on their cell surface and hence would be bound to the beads. Other lymphocytes would pass through the column without binding. Bound lymphocytes could then be released from the column by the addition of free antigen, which would compete with bound antigen for combination with the cell-surface receptors. The fact that such a purification method works is a consequence of the fact that many types of lymphocytes are produced by the immune system, each having a specific receptor on the cell surface capable of reacting with some (previously unencountered) antigen. Combination of antigen with that cell-surface receptor causes the lymphocyte to divide and stimulates antibody production.

10. The results would occur if the haptens were clustered so densely on the antigen that bi

valent antibodies combined preferentially with two neighboring haptens on a given antigen molecule. Such behavior is actually found in some systems and is termed monogamous bivalency.

11. Nearly all cells have MHC class I proteins. Protein degradation by proteasomes occurs

constantly in the cytoplasm. Both self and nonself proteins will be degraded, and the peptides will be translocated across the ER membrane where they will encounter nascent MHC class I proteins. The combination of peptide and MHC will eventually be displayed on the cell’s exterior, and nonself peptides will attract cytotoxic T-cells.

In contrast, MHC class II proteins are only found in B-cells from the immune system.

The peptides do not originate in the cytoplasm but from foreign proteins bound by antibodies outside the cell, and brought into the cell in endosomes where acidic hydrolysis occurs. The nonself peptides attract helper T-cells, which stimulate the B-cell to reproduce, thus bolstering the immune response.

12. It is surprising that the pattern of MHC proteins is detectable in the scent of an individual,

but not surprising that females would prefer males with maximally different MHCs. This would ensure that the offspring would have widely varying MHCs and thus perhaps improved survival chances in an epidemic of some virulent disease. The data in the study are not overwhelmingly convincing, and there was some variation in the results according to whether the women were taking birth control pills. The work was done by Claus Wedekind (Proc. Roy. Soc. Lond. B 260 [1995]:245). The more recent paper by Suma Jacob (Nature Gen. 30[2002]:175) contradicted Wedekind’s conclusions, but corresponded to the part of the study on women taking birth control pills. It seems possible that “dating” women would be interested in adventure (maximum difference) while “settled” or pregnant women (birth control pills simulate pregnancy) would prefer the familiar scents of home.

13. Each year, there is a different flu shot because we know that the influenza virus has

changed, and the previous year’s shot wouldn’t work any more. The HIV virus changes so rapidly that after an individual has been infected for several years, there are numerous strains within the body. Obviously it is difficult to find a way to design a vaccine when the target is constantly moving. AIDS is used as an example by some writers on evolution, because it seems to adapt to the life-style of the infected population. It certainly fits the model of reproduction with variation and subsequent selection. This would allow it to change in whatever ways it can to keep spreading the infection from person to person. These facts, plus the fact that it is a retrovirus that actually becomes part of the genes of infected cells, makes it terribly hard to eradicate or even cure.

14. Use of immunosuppressive agents will retard the synthesis of antibodies against acetylcholine

receptors (as well, of course, as inhibiting the synthesis of many other useful and protective antibodies). Administration of acetylcholinesterase inhibitors will lead to an increase in

acetylcholine concentrations at motor end plates. This increase will lead in turn to an increased number of acetylcholine–acetylcholine receptor complexes by mass action, and will therefore compensate somewhat for the decrease in receptor concentration. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. (a) DG°′ = −2.303 RT log

= −

10 K′dis

1.36 × log10 (3 × 10−7) = 8.9 kcal/mol. Note

that this positive value is the DG°′ for dissociation of the Fab-hapten complex. Since binding is the reverse of dissociation, the DG°′ for binding = −8.9 kcal/mol.

(b) K =

1/dissociation constant = 1/3 × 10−7 M = 3.3 × 106 M. Note—see part

(a)—that the DG°′ of binding = −2.303 RT log10 Ka.

ward (off) reaction to the rate constant of the reverse (on) reaction. Therefore, k

=

off/kon

120 s−1/kon

3 × 10−7 M. Solving this equation gives kon

4 × 108 M−1s−1.

This value is close to the diffusion-controlled limit for combination of a small molecule with a protein. Hence, the extent of structural change is likely to be small because extensive conformational transitions take time.

2. The size of the dextran-binding site is typical for an antigen-recognizing surface on an

antibody. It is also noteworthy that the active site in lysozyme accommodates six sugar residues. The range of sizes of the combining sites of antibodies is similar to that of active sites of enzymes.

3. The fluorescence enhancement and shift to the blue indicate that water is largely ex

cluded from the combining site when the hapten is bound. Hydrophobic interactions contribute significantly to the formation of most antigen-antibody complexes.

4. (a) Using K′ =

=

eq

10−DG°′/1.36 (text, p. 187), Ka

107/136 = 105.15. Since Kdis is the reciprocal

of K

= −

= −

=

a, the log of Kdis

log of Ka

5.15 and Kdis

7.1 × 10−6 M.

(b) DG°′ = 2 × −7 + 3 = −11 kcal/mol. Using the same logic as in (a), K =

=

1011/1.36 = 108.1 and Kdis

10−8.1 = 8 × 10−9 M.

A comparison of bivalent binding with univalent binding is given by Ka2/Ka1 = Kdis1/Kdis2 = 7.1 × 10−6/8 × 10−9 = 888, meaning that the bivalent binding is 888 times tighter than the univalent binding.

5. (a) An antibody-combining site is formed by CDRs from both the H and L chains. The

VH and VL domains are essential. A small portion of Fab fragments can be further digested to produce FV, a fragment that contains just these two domains. CH1 and CL contribute to the stability of Fab but not to antigen binding.

(b) A synthetic FV analog 248 residues long was prepared by expressing a synthetic

gene consisting of a VH gene joined to a VL gene through a linker. See J. S. Huston, et al., Proc. Nat. Acad. Sci. 85(1988):5879.

6. (a) Multivalent antigens lead to the dimerization or oligomerization of transmembrane

immunoglobulins, an essential step in their activation. This mode of activation is reminiscent of that of receptor tyrosine kinases.

(b) An antibody specific for a transmembrane immunoglobulin will activate a B cell by

cross-linking these receptors. This experiment can be carried out using, for example, a goat antibody to cross-link receptors on a mouse B cell.

7. B cells do not express T-cell receptors. Hybridization of T-cell cDNAs with B-cell mRNAs

removes cDNAs that are expressed in both cells. Hence, the mixture of cDNAs following

598

CHAPTER 33

this hybridization are enriched in those encoding T-cell receptors. This procedure, called subtractive hybridization, is generally useful in isolating low-abundance cDNAs. Hybridization should be carried out using mRNAs from a closely related cell that does not express the gene of interest. See S. M. Hedrick, M. M. Davis, D. I. Cohen, E. A. Nielsen, and M. M. Davis, Nature 308(1984):149, for an interesting account of how this method was used to obtain genes for T-cell receptors.

8. The model could be tested by an unfolding/refolding experiment. An antibody would be

reversibly unfolded using high temperature or a chemical denaturant such as guanidine hydrochloride (see Chapter 3, Section 3.6). Then a slow refolding would be attempted by gradually lowering the temperature or removing the denaturant in the presence of different putative small-molecule haptens. The model would suggest that different folding should be produced by refolding the presence of different antigens.

9. The gene rearrangements (Section 33.4) that generate antibody diversity inevitably may

introduce some premature termination codons. If expressed, the resulting truncated proteins or peptides could adversely affect the immune response. Nonsense-mediated RNA decay would provide a way to prevent or down-regulate the expression of the mistakenly terminated genes (see also J. Immunology 167[2001]:6901).

10. The Fc fragments lack the regions of variable sequence. Because their (nearly constant)

sequences are quite similar to each other, the properties of their surfaces also are quite similar, and they are able to interact with each other to form a regular lattice. The group of Fab fragments, by contrast, display a heterogeneous variety of binding surfaces.

11. The peptide should have 8–10 residues, with leucine at position 2 and valine at the C

terminal (Section 33.5.1). The given sequence contains only one valine, so this V should be the C-terminal of the presented peptide. The most likely peptide therefore is LLQATYSAV (or GLLQATYSAV).

12. The transition-state analog matches the geometry of the tetrahedral transition state that is

formed during the hydrolysis of amides or esters by serine proteases (or cysteine proteases). For catalytic activity, therefore, one would expect an antibody to have an appropriate nucleophile such as serine (or cysteine) in its binding site. (A neighboring general base such as histidine would be useful to enhance the nucleophilicity of the Ser or Cys.)

13. A distinct intramolecular domain (e.g., a domain homologous to SH2; see Section 15.4)

may bind to the critical phosphotyrosine residue and maintain the enzyme in an inactive conformation. If a phosphatase such as CD45 would remove the phosphate from the phosphotyrosine, then the second domain may not bind to the tyrosine and the inhibition could be relieved. (See also J. Biol. Chem. 276[2001]:23173, and references therein.)

14. (a) Kd, the antibody concentration needed to achieve 50% binding, is about 10−7 M for

antibody A.

(b) From the graph, Kd is about 10−9 M for antibody B. (c) Antibody B results from repeated immunizations, binds more tightly to the antigen

and is therefore improved over antibody A. We recall that antibody B has been produced by a process known as “affinity maturation.” Somatic mutation is a likely mechanism for this process because a single codon change has led to the selection of antibodies that more precisely fit the antigen (Section 33.4.2). CHAPTER 3 Molecular Motors 4

This chapter describes the transduction of chemical energy into mechanical en

ergy—for example, the use of ATP hydrolysis to drive the contraction of muscles or to move cells, or the exploitation of transmembrane proton-motive force

to rotate bacterial flagella. The authors describe how nanometer motions of proteins can be converted into the coordinated movements of cellular organelles, bacteria, and even animals themselves.

The authors begin with the fact that most molecular-motor proteins are based on

the P-loop NTPase structure. Myosin, kinesin, and dynein are described briefly and compared. Then the subunit structures of all three are shown in detail, and the mechanism of myosin’s dramatic flexing in response to ATP hydrolysis is explained. The authors then describe the structure of vertebrate skeletal muscle by showing how thick and thin protein filaments give rise to the striated appearance revealed in micrographs. Muscle contraction occurs by the oriented sliding of these filaments past one another. The thick filaments are primarily myosin and the thin filaments are principally an actin polymer with associated troponin and tropomyosin molecules. They next show the structure of the F-actin polymer, which is composed of a linear coiled array of G-actin monomers. The opposite polarities of the thick and thin filaments within a sarcomere indicate how coordinated molecular movement can result in the shortening of myofibrils. Reconstituted, moving assemblies of myosin-coated beads traversing along actin filaments reveal that myosin is the motor driving movement along the track. The repeated association and dissociation of the S1 heads of myosin with actin and the conformational changes in myosin that are effected by the binding of ATP, its hydrolysis to ADP and Pi, and the release of the hydrolysis products suggest how the power stroke occurs.

Microtubules are found in nearly every cell and serve multiple structural and

functional roles. They are composed primarily of a-tubulin and b-tubulin monomers that form tubular structures which have large diameters (300 Å). Microtubules also 599 600

CHAPTER 34

CHAPTER 3

contribute the basic macromolecular assembly of the axoneme, which is the fundamental structural component of the cilia and flagella of eukaryotic cells. Dynein and kinesin interact with microtubules to bend cilia and flagella and to move vesicles along microtubules, respectively. The rapid association of GTP-tubulin with microtubules and the rapid dissociation of GDP-tubulin from the ends of tubulin polymers in conjunction with the GTPase activity of the tubulins explains the dynamic instability of microtubules. Vesicle transport in neurons is explained as their ATP-dependent, kinesin-driven movement along microtubule tracks, which is analogous to the myosin–actin interactions that slide muscle filaments past one another.

The chapter ends with a description of the flagellar motor of bacteria and a discussion

of how the interplay of two kinds of protein components forming the motor give rise to directional rotation as a consequence of protons moving from outside the membrane into the cytosol. The system is related to bacterial chemotaxis, which can reverse the rotation of the flagella by phosphorylation of CheY. The examples given in this chapter demonstrate how energy in one form (chemical) can be converted into another form (kinetic) by the regulated activities of proteins.

When you have mastered this chapter, you should be able to complete the following

objectives. LEARNING OBJECTIVES Most Molecular-Motor Proteins Are Members of the P-Loop NTPase Superfamily (Text Section 34.1)

1. List the two sources of energy that power coordinated molecular movement.

2. Give a general description of P-loop NTPases, and (using the text’s index) list several

examples.

3. Sketch the general structure of the polypeptide backbone of a myosin molecule. Describe

the subunit composition of myosin, and note its a-helical coiled-coil structure and its dual globular head.

4. Describe the fragmentation of myosin into light meromyosin (LMM) and heavy meromyosin (HMM) by proteolysis and the further fragmentation of HMM into its S1 and S2 subfragments. Sketch the polypeptide structures of the fragments, and associate them with the ATPase activity, the actin-binding sites, and the thick-filament-forming domains of the intact molecule.

5. Compare kinesin to dynein and myosin.

6. Describe the large conformational difference between myosin-ADP and myosin-ATP

(same as myosin-ADP-vanadate). Locate switch I, switch II, and the relay helix in relation to the P-loop, and explain how they cause the protein to flex.

7. Describe the differences in protein affinity between myosin, kinesin, and the a subunit

of heterotrimeric G-protein. Myosins Move Along Actin Filaments (Text Section 34.2)

8. Define the terms sarcomere and myofibril.

MOLECULAR MOTORS 601

9. Identify the A band, H zone, M line, I band, and Z line of a sarcomere in an electron mi

crograph of a myofibril.

10. Relate the locations of the thick filaments and thin filaments to the I band, the A band, and

the H zone of a sarcomere.

11. Relate the structures of the thick and thin filaments to their compositions of actin, myosin, tropomyosin, and the troponin complex. Note the cross-bridges between the thick and thin filaments.

12. Describe muscle contraction in terms of the sliding-filament model, and relate contraction

to changes in the sizes of the A band, I band, and H zone.

13. Distinguish between G-actin and F-actin. Define critical concentration as it relates to the

polymerization of actin. Contrast the roles of the ATPase activities of actin and myosin.

14. Recount the experiment that displayed the unidirectional movement of myosin mole

cules along an actin cable. Describe the experiment indicating that the force of muscle contraction is generated by the S1 head of myosin.

15. Provide an overview of a model for the mechanism of the power stroke during muscle

contraction. Explain the roles of actin and the ATPase activity of myosin in the model. Kinesin and Dynein Move Along Microtubules (Text Section 34.3)

16. Sketch the general structure of a microtubule, noting its diameter and showing its alter

nating subunits of a-tubulin and b-tubulin.

17. Describe the structure of the axoneme, and sketch the 9 + 2 array of the microtubule

doublets and singlets that form its basic ring motif.

18. Note the role of dynamic instability in growing microtubules. Describe the effects of taxol

on the polymerization and depolymerization of microtubules.

19. Describe the kinesin-dependent movement of vesicles and organelles along microtubules.

Distinguish between the plus and minus ends of a microtubule.

20. Explain how small structural differences can cause ncd protein to have reversed polarity

compared with normal kinesins. A Rotary Motor Drives Bacterial Motion (Text Section 34.4)

21. Outline the structure and function of bacterial flagella. Distinguish between the mecha

nism by which eukaryotic and prokaryotic flagella generate motile force.

22. Explain the role of proton-motive force in flagellar rotation. Present a model that explains

the production of rotary motion from the effects of a proton gradient on the transmembrane flagellar motor.

23. Define chemotaxis and outline the sequence of events constituting it.

24. Describe the flagella and the motors of the chemotactic apparatus of E. coli. Relate their

actions to the smooth swimming and tumbling motions of a bacterium in response to a temporal gradient (change over time) of attractant or repellent substances.

25. Describe the effect of phosphorylation of the che gene product CheY on flagellar rotation. 602

CHAPTER 34 SELF-TEST Most Molecular-Motor Proteins Are Members of the P-Loop NTPase Superfamily

1. Match the proteins in the left column with their descriptions in the right column.

(a) myosin

(1) enormous protein

(b) kinesin

(2) helps separate chromosomes

(3) important in muscle contraction

2. Which of the following statements about HMM and LMM are true?

(a) HMM and LMM are formed from myosin by tryptic cleavage. (b) HMM can be cleaved by papain to yield two globular proteins that polymerize to

form the thin filaments.

filaments in vitro.

(d) HMM contains the two globular heads of the myosin molecule, hydrolyzes ATP, and

binds actin in vitro.

3. Energy is required to drive the contraction of striated muscle, the beating of flagella or

cilia, and the intracellular transport of vesicles along microtubules. Which of the following statements about energy transduction in these systems are correct? (a) The proton-motive force across the plasma membrane surrounding the cell provides

the energy for these movements.

(b) The binding of ATP to proteins, which then undergo a conformational change, pro

vides the energy for these processes.

proteins of these systems to drive them.

(d) The hydrolysis of protein-bound ATP and the release of ADP + Pi lead to confor

mational transitions that complete a movement cycle.

(e) The binding of GTP to oriented proteins at the interface of the moving assemblies

drives these processes. Myosins Move Along Actin Filaments

4. Which of the following answers complete the sentence correctly? Actin

(a) is formed from a 42-kd monomer (G-actin). (b) monomers can exist with either ATP or ADP bound to themselves. (c) exhibits an ATPase activity that helps power muscle contraction. (d) in its F form binds myosin in an oriented manner. (e) in its F form with myosin has a barbed and a pointed end.

5. Figure 34.1 is a schematic diagram of a longitudinal segment of a skeletal muscle myo

fibril. Label the structures indicated in the figure by matching them with the listed choices. (1) I band (2) A band (3) thin filaments (4) M line (5) H zone (6) thick filaments (7) Z line

6. Which of the following statements about myosin are true?

(a) Myosin binds to polymerized actin. (b) In vitro, myosin assembles spontaneously into the thin filaments. (c) Myosin is an ATPase. (d) Myosin has domains that interact with one another to effect its physiological

functions.

(e) Myosin is composed of two polypeptide chains, one of which forms an a-helical

coiled coil and the other a globular head.

7. Assign the proteins in the right column to the appropriate myofibrillar component in the

left column. (a) thin filament

(1) tropomyosin

(b) thick filament

(2) myosin (3) actin (4) troponin complex

8. Which proteins are homologs of actin?

(a) hexokinase (b) MreB (c) Hsp-70 (d) myosin

9. Which of the following statements concerning events related to the power stroke of mus

cle contraction are correct? (a) The hydrolysis of ATP to ADP and Pi by myosin is fast relative to the release of the

ADP and Pi from the protein.

(b) The binding of actin to myosin stimulates the ATPase activity of myosin by facili

tating the release of ADP and Pi.

ATP to the myosin head domains.

(d) Repeated cycles of ATP binding, ATP hydrolysis, and the resulting association and

dissociation of cross-bridges and conformational changes in myosin contribute to the contractile process.

(e) In the region of overlapping thick and thin filaments of a sarcomere, the cross

bridges will either all be formed or all be dissociated, depending on the phase of the power stroke. 604

CHAPTER 34

10. Considering only the power stroke of skeletal muscle contraction and the events that

precede and follow it, place the following states or processes that occur in going from the resting state to the contracted state and back again in their proper order. (a) The thick filament moves with respect to the thin filament. (b) The S1 heads of myosin interact with actin, Pi is released, and myosin changes its

conformation.

ADP and Pi.

(e) ATP is hydrolyzed to ADP and Pi, and myosin undergoes a conformational change. (f)

ADP dissociates from the myosin to complete the power stroke.

11. Which of the domains of myosin is primarily responsible for generating the force of skele

tal muscle contraction? (a) the hinge between the S1 and S2 domains (b) the hinge between the S2 and the LMM domains (c) the LMM domain (d) the S1 globular head (e) the S2 domain that connects the S1 domain to the a-helical coiled coil (f)

the a-helical coiled coil Kinesin and Dynein Move Along Microtubules

12. Which of the following statements concerning microtubules are correct?

(a) Microtubules are filaments composed of a-helical coiled-coil polypeptide chains. (b) Microtubules contain a-tubulin and b-tubulin protomers that are disposed in a hel

ical array around a hollow core to form a cylindrical structure.

round a pair of microtubule singlets.

(d) The outer microtubules in an axoneme are linked together and to an ATPase called dynein.

(e) The powered movements of dynein in an axoneme shorten the structure.

13. A neuron can move a vesicle approximately a meter from the central cell body to the end

of an axon in a day. How are microtubules involved in this process, what provides the energy for the movement, and what protein is directly involved in the movement? A Rotary Motor Drives Bacterial Motion

14. Contrast the protein compositions and molecular movements of the flagella of bacteria

and of eukaryotic cells.

15. How would the movements of a starved bacterium (one whose energy stores had been

depleted) be affected by having the pH value of the media in which it resides lowered below that of its cytoplasm?

16. Match the major components of the chemotaxis system of E. coli in the left column with

the appropriate descriptions from the right column.

(a) chemoreceptor

(1) contains several rings, a hook, and a rod

(b) processing system

(2) has binding sites for attractants or repellents

(3) includes cytosolic peripheral membrane

(d) flagellum proteins

(4) contains flagellin and adopts a helical

configuration (b) Tumbling will be more frequent. (c) The counterclockwise rotation of flagella will occur more frequently. (d) The clockwise rotation of flagella will occur more frequently.

18. The proposed model for the transduction of chemotactic signals via CheY, the tumble

regulator, includes which of the following?

(a) Phosphorylated CheY promotes the counterclockwise rotation of the motor. (b) The activation of CheY requires ATP. (c) Attractants block the CheY pathway, and smooth swimming results. ANSWERS TO SELF-TEST

1. (a) 3 (b) 2 (c) 1

2. a, c, d. Answer (b) is incorrect because the two globular heads arising from papain

digestion cannot polymerize, as they lack a-helical coiled-coil structures.

3. b, d

4. a, b, d, e. Answer (c) is incorrect because the ATPase activity of actin is involved in the

formation and disassembly of F-actin.

5. (a) 2 (b) 1 (c) 5 (d) 7 (e) 4 (f) 6 (g) 3. The M lines lie halfway between the Z lines and

correspond to the middle of the bare zone separating regions where oppositely oriented myosin molecules point toward each other (see text, Figure 34.13).

6. a, c, d. Answer (b) is incorrect because myosin assembles to form the thick filaments.

Answer (e) is incorrect because myosin is composed of six polypeptide chains. Two heavy chains intertwine their C-terminal portions to form the a-helical coiled coil, with their N-terminal segments forming two globular heads. Four more polypeptide chains of two kinds associate with the globular heads.

7. (a) 1, 3, 4 (b) 2

8. a, b, and c. Heat shock protein (Hsp-70) is not mentioned in the current chapter, but its

homology with actin is discussed in Chapter 7 of the text (Figure 7.14, p.180).

9. a, b, d. Answer (c) is incorrect because the binding of ATP to actomyosin dissociates the

cross-bridges. Answer (e) is incorrect because, at any instant, numerous cross-bridges will be in all stages of forming and breaking because the process is dynamic and asynchronous.

10. d, b, a, f, c, e, d

11. d. Although the whole molecule is required for muscle contraction, the activities of the

S1 head domains provide the biochemical activity for the power generation. The cyclic changes in the conformation of the head domains and in their affinities for actin, ATP, ADP, and Pi are the basis of the energy transduction.

12. b, c, d. Answer (a) is incorrect because microtubules are assembled from relatively glob

ular tubulin subunits and lack the a-helical coiled-coil structure of myosin and the intermediate filaments. Answer (e) is incorrect because dynein movements lead to the bending, not the contraction, of the axoneme.

13. The microtubules, which form a network of fibers that traverse the cell, provide tracts along

which vesicles and organelles can move. ATP hydrolysis by the protein kinesin acts as a molecular engine, in a manner analogous to the way myosin acts, to power the movements. 606

CHAPTER 34

14. Bacterial flagella are formed principally from flagellin subunits, whereas eukaryotic fla

gella contain microtubules, among other components. Eukaryotic flagella bend owing to an internally generated force. Bacterial flagella are rotated by a motor attached to one of their ends.

15. The bacterium would commence swimming because protons would move from the

higher concentration outside the cell through the intramembrane motors of the flagella, causing their counterclockwise rotation and thereby coordinated movement.

16. (a) 2 (b) 3 (c) 1 (d) 4. The flagellar motor is shown in Figure 34.30 of the text.

17. a, c

18. b, c. Answer (a) is incorrect because phosphorylated CheY induces clockwise rotation of

the flagella, leading to tumbling. PROBLEMS

1. In the text, compare Figure 34.10 on page 955 to Figure 9.51 on page 255. We can as

sume that there is some relationship between adenylate kinase and myosin. Which of these proteins is likely to be older? Why?

2. Apply your knowledge of the a helix and protein folding to answer this question: What

features of the tails of myosin molecules contribute to their ability to interact with one another to form thick filaments?

3. Design an experiment involving ATP labeled in the g phosphoryl group with 32P that

would show that actin stimulates the hydrolysis of ATP by myosin.

4. Decide whether each of the following will remain unchanged or will decrease upon

muscle contraction. Assume that the sliding-filament model applies. Refer to page 957 of the text.

(a) the distance between adjacent Z lines (b) the length of the A band (c) the length of the I band (d) the length of the H zone

5. The symmetry of thick and thin filaments in a sarcomere is such that six thin filaments

ordinarily surround each thick filament in a hexagonal array. (See Figure 34.13 on p. 957 of the text.) In electron micrographs of cross-sections of fully contracted muscle, the ratio between thin and thick filaments has been found to be double that of resting muscle.

(a) Propose an explanation for this observation based on the sliding-filament model. (b) How might the explanation you have given in part (a) also account for the long

appreciated fact that a fully contracted muscle is, paradoxically, “weaker” than a resting muscle?

6. Some people have defective dynein, which causes an “immotile-cilia syndrome.” This

leads to chronic respiratory disorders and also infertility in males. Explain.

7. Colchicine, and the mold products vincristine and vinblastine, interfere with the poly

merization of microtubules. Vincristine and vinblastin are widely used in the treatment of rapidly growing cancers. Explain the basis for their effects, remembering that the mitotic spindles of dividing cells are composed of microtubules.

8. The processive nature of kinesin motion is shown in Figure 34.25 of the text. There are

two possible interpretations of how the two “feet” of kinesin walk down the “tightrope”

of the microtubule. One is called the “inchworm” model, and the other the “hand-overhand” model. The first one can be visualized as walking with one foot, say the left foot, always in front, and the other as normal walking, left foot and then right foot in front. How do you think scientists were able to show which model was correct?

9. A single kinesin molecule can move a single vesicle from the nucleus of a nerve cell to

the end of its axon. Each “step” in the kinesin cycle is 8 nm, and each cycle uses one ATP.

(a) How many cycles of kinesin binding and ATP hydrolysis will it take for a vesicle to

reach the little toe from the upper cervical region of a tall human? Assume that the length of the nerve cell is 6 ft.

(b) Given that the rate of vesicle transport is 400 mm per day, how long will it take the

vesicle to reach the little toe?

10. Myosin, kinesin, and dynein all contain a globular head which harbors ATPase activity

and a tail region which can be thought of as a protein-binding region. The specificity of these protein-binding domains changes the role of each of these three proteins in tissues. What is the specificity of each of these three protein-binding domains, and how does it relate to the roles of myosin, kinesin, and dynein in tissues?

11. Figure 34.2 depicts the response of bacteria to two chemotactic agents, X and Y. Which

agent is likely to be an attractant, and which is likely to be a repellent? Give the reasons for your choices. FIGURE 34.2 Bacterial response to two chemotactic agents.

1.0

kwise rotation

0.5

EPulse of Y

EPulse of X

Probability of cloc

0

Probability of cloc

0

5

10

15

20

5

10

15

20

Time (s)

Time (s) ANSWERS TO PROBLEMS

1. Both proteins flex dramatically when ATP binds to the P-loop area. The change in myosin

is amplified by nearby structures like the relay helix, and by the length of the lever arm, but the change is generally parallel. Adenylate kinase maintains an equilibrium between ATP and ADP and must be extremely ancient. Myosin is confined to eukaryotes, and hence should be somewhat younger.

2. The absence of proline, which would interfere with a-helix formation (p. 66), and the

abundance of regularly spaced leucine, alanine, and glutamate residues in seven-long repeating motifs that form long a helices with hydrophobic pockets and knobs on one

608

CHAPTER 34

face and charged residues on the opposite face allow two such helices in two myosin molecules to form a long coiled-coil rod. Because one turn of the a helix occupies 3.6 residues (p. 56) seven residues represents approximately two turns, with hydrophobic R-groups on one side and the more polar glutamate on the other.

3. Incubate the labeled ATP with myosin and measure the amount of labeled inorganic

phosphate liberated as a function of time without actin. Then add actin. The result should be a burst in the amount of labeled Pi released. (See Figure 34.18 in the text for the ATP/ADP cycle.)

4. (a) The distance between adjacent Z lines will decrease.

(b) The length of the A band will remain unchanged. (c) The length of the I band will decrease. (d) The length of the H zone will decrease. See Figure 34.3. FIGURE 34.3 Schematic illustration of (A) uncontracted and (B) partially

contracted sarcomeres.

A Band

I Band

(A)

H Zone

A Band

I Band

(B)

H Zone

Z line

5. (a) The number of thin filaments per thick filament doubles because the thin filaments

override one another when the sarcomere is fully contracted. See Figure 34.4. FIGURE 34.4 Schematic illustration of a fully contracted sarcomere.

(b) Thin filaments and thick filaments must have the same polarity to interact with one

another to form cross-bridges, but the polarity of the thick filaments reverses halfway between the Z lines. (See Figure 34.19 on p. 961 of the text.) Thus, in the

region where the thin filaments override one another in a fully contracted sarcomere, the ends of the thin filaments have the wrong polarity to interact with an adjacent thick filament. Hence, a slightly smaller number of cross-bridges are formed, so the fully contracted muscle develops less tension; that is, it is “weaker.”

6. Defective dynein molecules not only immobilize the cilia of the respiratory tract, prevent

ing inhaled particles from being swept out of the lungs, but also render sperm immotile.

7. Cancer cells have a greater than normal rate of cell division. In the presence of vincristine

or vinblastine, mitotic spindle fibers fail to form, and cell division is retarded. Since cancer cells grow more rapidly, they are more sensitive to these drugs.

8. Normal human walking makes use of hip and knee joints to allow the body to remain

aimed straight ahead. Scientists realized that the “hand-over-hand” model would require kinesin to rotate about its own axis, turning 180° with each step, whereas the “inchworm” model would allow kinesin to move with very little rotation. So scientists at Brandeis University attached a length of microtubule to the top of kinesin and took photographs that showed that as the microtubule was carried along, it stayed aimed in more or less the same direction. Thus it appears that the “inchworm” model is the correct one. (Science 295[2002]:844). Look at Figure 34.25 on page 965 of the text—Which version is shown there? Early editions showed the “hand-over-hand” version, which is apparently incorrect.

6

ft × 1

2 in /

ft × 2 54

.

cm/

in × 0 0

. 1 m/cm

9. (a)

= 2 3

. × 1

08 steps

8 × 1

0 9

− m/step

and 2.3 × 108 molecules of ATP hydrolyzed.

6 ft × 1

2 in / ft × 2

.

cm / in × 0 0

. 1 m / cm

(b)

= 4 6

. days

0 4

. m

/ day

4 6

. days × 24 h/

day × 60 min/

h × 0

6 s/min

(c)

= 1 7

. × 1

0 3

− s/steps

2

× 108

.3 steps

and the inverse of 1.7 × 10−3 seconds/step gives 588 ATPs hydrolyzed per second.

10. The tail of myosin binds other myosin molecules to form the thick filaments of striated

muscle. Without this self-aggregating property of myosin, striated muscle would not form. The tail regions of kinesin or dynein bind the surface of a vesicle that is being transported. Changing their tail regions would change the vesicles or molecules that are being transported. Finally, cellular and ciliary dynein must have different tail regions so that the former will transport vesicles and the latter will polymerize with tubulin in cilia.

11. X is likely to be an attractant and Y is likely to be a repellent. The addition of X causes

a decrease in clockwise rotation (an increase in counterclockwise rotation), which leads to smooth swimming. The addition of Y causes an increase in clockwise rotation (a decrease in counterclockwise rotation), which leads to tumbling. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. (a) ATP is the energy source for skeletal muscle and eukaryotic cilia, and proton-motive

force is the energy source for bacterial flagella.

(b) There are two essential components in each system: myosin–actin, dynein–microtubule,

and motA–motB, respectively. 610

CHAPTER 34

2. 6400 = 80 “body lengths” per second

80

For a 10-foot automobile, 80 body lengths per second would correspond to 800 ft s−1, or:

(800

1

ft s− ) × (3600

1

s hr− ) = 545 miles per hour!

(5280

1

ft mile− )

3. The force generated, in grams, is

(0.22 lb) × (4 × 10−12) × (454 g lb−1) = 4 × 10−10 g.

The protein mass, in grams, is

(

,

100 000

1

g mole− )

= .

1 66 × 1

0 18

−

1

g molecule− .

( .

6 02 × 1

023

1

molecules mole− )

The ratio, force lifted per myosin molecule, obtained by dividing the above two numbers, is: 2.4 × 108 molecular “bodyweights” lifted. Heavy lifting, indeed!

4. The rapid decrease in the level of ATP following death has two consequences. First, the

cytosolic level of calcium rises rapidly because the Ca2+-ATPase pumps in the plasma membrane and sarcoplasmic reticulum membrane no longer operate. High Ca2+, through troponin and tropomyosin, enables myosin to interact with actin. Second, a large proportion of S1 heads will be associated with actin. Recall that ATP is required to dissociate the actomyosin complex. In the absence of ATP, skeletal muscle is locked in the contracted (rigor) state.

5. The critical concentration for polymerization is 20-fold lower for actin–ATP than for

actin–ADP (Section 34.2.2). For monomer concentrations in between the critical concentrations for actin–ATP and actin–ADP, therefore, the actin–ATP will polymerize and later will gradually depolymerize, as the bound ATP becomes hydrolyzed to ADP.

6. The dynamic polymerization and depolymerization of actin filaments involve energy and

are linked to ATP hydrolysis (see problem 5, above). These processes therefore could play active roles.

7. Figure 34.27 suggests that the choice of direction is determined by the neck and

stalk domains and not by the motor domains. Therefore, it is reasonable to predict that (a) conventional kinesin with a motor domain from ncd will continue to move in the plus direction (characteristic of kinesin), whereas (b) ncd protein with a motor domain from kinesin will move in the minus direction (characteristic of ncd). (Nevertheless, single mutations within the chimeric constructs are able to reverse the direction, suggesting that the regulatory mechanisms are complex; see Science 281[1998]:1200.)

8. For single-stranded DNA in an extended conformation, the total distance between suc

cessive bases is about 12 Å (measured along the backbone). (This is different from the translation per base along a double-helical axis!) The hydrolysis of 50 ATP molecules per second therefore would correspond to a movement along 50 bases over a distance of (12 × 50) = 600 Å, at a rate of 600 Å/s, or 0.06 micrometer per second. The rate of movement is about one-tenth that of kinesin, which moves at about 6400 Å/s (Section 34.3.2).

MOLECULAR MOTORS 611

9. Proton flow from the acidic solution can drive the bacterial flagellar rotation.

10. (a) Substituting into the equation given for force (F), we get F = 6p × 10−2 g cm−1 s−1

× 10−4 cm × 0.5 × 10−4 cm s−1 = 9.4 × 10−10 g cm s−2 = 9.4 × 10−10 dyne.

(b) If one erg = dyne cm and the bead moved 0.5 mm (5 × 10−5 cm), the work per

second = 9.4 × 10−10 dyne × 5 × 10−5 cm = 4.7 × 10−14 erg.

4.19 × 1010 ergs. Then for ATP hydrolysis within the cell DG = −12 kcal/mol =

−50.2 × 1010 ergs per mole. Dividing ergs per mole by Avogadro’s number (6.02

× 1023), we get 8.3 × 10−13 ergs/molecule. The hydrolysis of 40 ATP molecules could yield 3.3 × 10−11 ergs, much more energy than the actual work performed in moving the 2-mm-diameter bead. Thus, the hydrolysis of ATP by a single kinesin motor provides more than enough free energy to power the transport of micrometersize cargos at micrometer-per-second velocities.

11. A step size of only 6 nm would be inconsistent with the actual distance of 8 nm between

equivalent binding sites on tubulin subunits.

12. One or more additional tether domains might allow the KIF1A protein to remain bound

to a microtubule during times when the motor domain needed to detach. An alternation between tether and motor attachement could enable the protein to be processive. (For a discussion see Proc. Natl. Acad. Sci., USA, 97[2000]:640.)

13. The direction of proton flow would reverse when the direction of flagellar rotation re

verses. Following the converse of Figure 34.32, protons would need to flow from the inner half-channel to the MS ring, and then following clockwise rotation to the outer half-channel.

14. The effect is mediated through Ca2+-calmodulin, which stimulates myosin light-chain ki

nase (MLCK). Phosphoryl groups introduced by MLCK are removed from myosin by a Ca2+-independent phosphatase.

15. (a) From the graph, the maximum velocity is about 13 molecules of ATP hydrolyzed per

myosin molecule per second, that is, k

cat

13 s−1. KM is the ATP concentration re

quired to yield half maximal activity (6.5 s−1), that is, K

M

15 micromolar ATP.

(b) It appears that the average step size in the figure is about 50 nm. (c) It is plausible that the two heads of myosin may “walk” along the actin filament as

two feet alternately cycle past each other, and alternately exchange leading and trailing positions, when a person walks. The respective cycles of binding, ATP hydrolysis, ADP release, and movement would alternate between the two myosin heads. Processivity could be achieved by one head always remaining attached when the other was released or moving, and vice versa. </body></html>

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