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Biochemistry Problems and Solutions

By Roger-Binny Nov 08, 2014 32747 Words
BIOCHEMISTRY PROBLEMS AND SOLUTIONS (FOR ADVANCED STUDENTS)

Chapter 01

Protein Structure and Function

1. Shape and dimension (形状与大小)
(a) Tropomyosin (原肌球蛋白), a 70-kd muscle protein, is a two-stranded α-helical coiled coil. What is the length of the molecule?
(b) Suppose that a 40-residue segment of a protein folded into a two-stranded anti-parallel β structure with a 4-residue hairpin turn. What is the longest dimension of this motif? Ans: (a) Each strand is 35 kd and hence has about 318 residues (the mean molecular mass of amino acid residue is about 110 daltons). Because the rise per residue in an α helix is 1.5Å, the length is 477 Å (318×1.5).

(b) Eighteen residues in each strand (40 minus 4 divided by 2) are in a β sheet conformation. Because the rise per residue is 3.5 Å, the length is 63 Å. 2. Contrasting isomers(差别悬殊的异构体). Poly-L-leucine in an organic solvent such as dioxane is α-helical, whereas poly-L-isoleucine is not. Why do these amino acids with the same number and kinds of atoms have different helix-forming tendencies? Ans: The methyl group attached to the β carbon of isoleucine sterically interferes with α helix formation. In leucine, this methyl group is attached to the γ carbon atom, which is farther from the main chain and hence does not interfere.

3. Active again(活性恢复). A mutation that changes an alanine residue in the interior of a protein to a valine is found to lead to a loss of activity. However, activity is regained when a second mutation at a different position changes an isoleucine residue to a glycine. How might this second mutation lead to a restoration of activity?

Ans: The first mutation destroys activity because valine occupies more space than alanine, and so the protein must take a different shape. The second mutation restores activity because of a compensatory reduction of volume; glycine is smaller than isoleucine. 4. Shuffle test(改组试验). An enzyme that catalyzes disulfide-sulfhydryl exchange reactions, called protein disulfide isomerase (PDI, 蛋 白 二 硫 键 异 构 酶 ) has been isolated. Inactive scrambled ribonuclease is rapidly converted into enzymatically active ribonuclease by PDI. In contrast, insulin is rapidly inactivated by PDI. What does this important observation imply about the relation between the amino acid sequence of insulin and its three-dimensional structure? Ans: The native conformation of insulin is not the thermodynamically most stable form. Indeed, insulin is formed from pro-insulin, a single-chain precursor containing 33 additional residues. In pro-insulin, residue 30 of the future B-chain of insulin is linked to the residue 1 of the future A-chain.

5. Stretching a target(目标的伸长). A protease is an enzyme that catalyzes the hydrolysis of peptide bonds of target proteins. How might a protease bind a target protein so that its main chain becomes fully extended in the vicinity of the vulnerable peptide bond? Ans: A segment of the main chain of the protease could hydrogen-bonded to the main chain of the substrate to form an extended parallel or anti-parallel pair of β strands. 6. Often irreplaceable(经常不能替换). Glycine is a highly conserved amino acid residue in the evolution of proteins. Why?

Ans: Glycine has the smallest side chain of any amino acid. Its smallness often is critical in allowing polypeptide chains to make tight turns or to approach one another closely. 7. Potential partners(潜在伙伴). Identify the groups in a protein that can form hydrogen bonds or electrostatic bonds with an arginine side chain at pH 7.

Ans: Glutamate, aspartate. The terminal carboxyl group can form salt bridges with the guanidinium group of arginine. In addition, this group can be a hydrogen bond donor to the side chains of glutamine, asparagine, serine, threonine, aspartate, and glutamate, and the main chain carbonyl.

8. Permanent waves(永久烫发). The shape of hair is determined in part by the patterns of disulfide bonds in keratin, its major protein. How can curls be induced? Ans: Disulfide bonds in hair are broken by adding a thiol and applying gentle heat. The hair is curled, and an oxidizing agent is added to re-form disulfide bonds to stabilize the desired shape. Chapter 02

Exploring Proteins

1. Valuable reagents(有用试剂). The following reagents are often used in protein chemistry: CNBr
Trypsin

Dabsyl chloride

Urea

Phenyl isothiocyanate

6N HCl
Performic acid

β-mercaptoethanol

Ninhydrin

Chymotrypsin

Which one is the best suited for accomplishing each of the following tasks? (a) Determination of the amino acid sequence of a small peptide. (b) Identification of the amino-terminal residue of a peptide (of which you have less than 0.1μg). (c) Reversible denaturation of a protein devoid of disulfide bonds. Which additional reagent would you need if disulfide bonds were present?

(d) Hydrolysis of peptide bonds on the carboxyl side of lysine and aromatic residues. (e) Cleavage of peptide bonds on the carboxyl side of methionine. (f) Hydrolysis of peptide bonds on the carboxyl side of lysine and arginine residues. Ans: (a) Phenyl isothiocyanate. (b) Dansyl chloride or dabsyl chloride. (c) Urea; β-mercaptoethanol to reduce disulfides. (d) Chymotrypsin. (e) CNBr. (f) Trypsin 2. Acid-base relations(酸硷关系). What is the ratio of base to acid at pH 4, 5, 6, 7, and 8 for an acid with a pKa of 6?

Ans: 0.01, 0.1, 1. 10, and 100.
3. Finding an end(寻找末端). Anhydrous hydrazine(肼) has been used to cleave peptide bonds

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in proteins. What are the reaction products? How might this technique be used to identify the carboxyl-terminal amino acid?
Ans: Each amino acid residue, except the carboxyl-terminal one, gives rise to a hydrazide on reacting with hydrazine. The carboxyl-terminal residue can be identified because it yields a free amino acid.

4. Crafting a new breakthrough(制作一个新的突破). Ethyleneimine reacts with cysteine side chains in proteins to for S-aminoethyl derivatives. The peptide bonds on the carboxyl side of these modified cysteine residues are susceptible to hydrolysis by trypsin. Why? Ans: The S-aminoethylcysteine side chain resembles that of lysine. The only difference is a sulfur atom in place of a methylene group.

5. Spectrometry(分光光度法). The absorbance A of a solution is defined as A = log10 (I0/I)
In which I0 is the incident light intensity and I is the transmitted light intensity. The absorbance is related to the molar absorption coefficient (extinction coefficient ε (in cm-1 M-1), concentration c (in M), and path length l (in cm) by

A =εlc
The absorption coefficient of myoglobin at 580nm is 15000 cm-1 M-1. What is the absorbance of a 1mg/ml solution across a 1-cm path? What percentage of the incident light is transmitted by this solution?

Ans: A 1mg/ml solution of myoglobin (17.8kd) corresponds to 5.62×10-5 M. The absorbance of a 1-cm path length is 0.84, which corresponds to an I0/I ratio of 6.96. Hence 14.4% of the incident light is transmitted.

6. A slow mover(缓慢的移动者). Tropomyosin, a 93-kd muscle protein, sediments more slowly than does hemoglobin (65kd). Their sedimentation coefficients are 2.6S and 4.3S respectively. Which structural features of tropomyosin accounts for its slow sedimentation? Ans: Tropomyosin is rod shaped, whereas hemoglobin is approximately spherical. 7. Sedimenting spheres(沉降球体). What is the dependence of the sedimentation coefficient S of a spherical protein on its mass? How much more rapidly does an 80-kd protein sediment than does a 40-kd protein?

Ans: The frictional coefficient f as well as the mass m determines S. Specifically, f is proportional to r. Hence, f is proportional to m1/3, and so S is proportional to m2/3. An 80-kd spherical protein sediments 1.59 times as rapidly as a 40-kd spherical protein. 8. Size estimate(大小估算). The relative electrophoretic mobilities of a 30-kd protein and a 92-kd protein used as standards on an SDS-polyacrylamide gel are 0.80 and 0.41, respectively. What is the apparent mass of a protein having a mobility of 0.62 on this gel? Ans: 50kd.(plotting the mobility against the value of log MW of the two known proteins, and then find the apparent masss of the unkown protein)

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9. A new partnership?(一个新的伙伴关系?) The gene encoding a protein with a single disulfide bond difference undergoes a mutation that changes a serine residue into a cysteine residue. You want to find out whether the disulfide pairing in this mutant is the same as in the original protein. Propose an experiment to directly answer this question. Ans: The positions of disulfide bonds can be determined by diagonal electrophoresis. The disulfide pairing is unaltered by the mutation if the off-diagonal peptides formed from the native and mutant proteins are the same.

10. Helix-coil transitions(螺旋与随机卷曲的转换).(a) Circular dichroism measurements have shown that poly-L-lysine is a random coil at pH7 but becomes α-helical as the pH is raised above 10. Account for this pH-dependent conformational transition. (b) Predict the pH dependence of the helix-coil transition of poly-L-glutamate.

Ans: (a) Electrostatic repulsion between positively charged ε-amino groups prevents α-helix formation at pH 7. At pH 10, the side chains become deprotonated, allowing α-helix formation. (b)Poly-L-glutamate is a random coil at pH 7 and becomes α-helical below pH 4.5 because the γ-carboxylate groups become protonated.

11. Sorting cells(分拣细胞). Fluorescence-activated cell sorting (FACS, 荧光激活的细胞分拣 法) is a powerful technique for separating cells according to their content of particular molecules.

For example, a fluorescent-labeled antibody specific for a cell-surface protein can be used to deect cells containing such molecules. Suppose that you want to isolate cells that possess a receptor enabling them to detect bacterial degradation products. However, you do not yet have an antibody directed against this receptor. Which fluorescent-labeled molecule would you prepare to identify such cells?

Ans: A fluorescent-labeled derivative of a bacterial degradation product (e.g. a formylmethionyl peptide) would bind to cells containing the receptor of interest. 12. Mirror images(镜像结构). Suppose that a protease is synthesized by the solid-phase method from D rather than L amino acids. How would the sedimentation, electrophoretic and circular dichroism properties of this enzyme compare with those of the native form? What prediction can you make about the relation of peptide substrates of the D and L enzymes? Ans: The sedimentation and electrophoretic properties of the L enzyme and the mirror-image D form would be the same. The circular dichromism spectra would have the same magnitude but of opposite sign because the two structures have opposite srew-sense. Peptide substrates that are mirror images of one another would be cleaved at the same rate by the L and D enzymes. 13. Peptides on a chip(芯片上的多肽). Large numbers of different peptides can be synthesized in a small area on a solid support. This high-density array can then be probed with a fluorescent-labeled protein to find out which peptides are recognized. The binding of an antibody to an array of 1024 different peptides occupying a total area of a thumbnail. How would you synthesize such a peptide array?(Hint: use light instead of acid to deprotect the terminal amino group in each round of synthesis).

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Ans: Light was used to direct the synthesis of these peptides. Each amino acid added to the solid support contained a photolabile protecting group instead of a t-Boc protecting group at its α-amino group. Illumination of selected regions of the solid support led to the release of the protecting group, which exposed the amino groups in these sites to make them reactive. The pattern of masks used in these illuminations and the sequence of reagents define the ultimate products and their locations.

Chapter 03 DNA and RNA: Molecules of Heredity
1.

Complements(互补物). Write the complementary sequence ( in the standard 5′→3′ notation)

for
(a) GATCAA
Ans: (a) TTGATC

( b) TCGAAC
(b) GTTCGA

(c) ACGCGT
(c) ACGCGT

(d) TACCAT
and

(d) ATGGTA

2. Compositional constraint(组成限制). The composition (in mole fraction units) of one of the strands of a double-helical DNA is [A] = 0.30 and [G] =0.24
(a) What can you say about [T] and [C] for the same strand?
(b) What can you say about [A], [G], [T], and [C] of the complementary strand? Ans: (a) [A]+[C] =0.46

(b) [T]=0.30 [C] =0.24, and [A]=[G] =0.46

3. Lost DNA(失去的 DNA). The DNA of a deletion mutant of λ bacteriophage has a length of 15μm instead of 17 μm. How many bas pairs are missing from this mutant? Ans: (17-15)×104/3.4= 5882 base pairs
4. An unseen pattern (看不见的模式). What result would Meselson and Stahl have obtained if the replication of DNA were conservative (i.e., the parental double helix stayed together)? Give the expected distribution of DNA molecules after 1.0 and 2.0 generations for comservative replication.

Ans: In conservative replication, after 1.0 generation, one half of the molecules would be 15N-15N, the other half 14N-14N. After 2.0 generations, one quarter of the molecules would be 15N-15N, the other three quarters 14N-14N. Hybrid

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N-15N molecules would not be observed in conservative

replication.
5. A fortunate circumstance(一个幸运情况). Griffith used heat-killed S pneumococci to transform R. mutants. Studies years later showed that the double-stranded DNA is needed for efficient transformation and that high temperatures melt the DNA double helix. Why were Griffith′s experiments nevertheless successful?

Ans: The DNA renatured when the heat-killed pneumococci were cooled before they were injected into mice.
6. A matter of competence(感受态问题).

Strains of Bacillus substilis that can be transformed

by foreign DNA are termed competent. Others, termed noncompetent, are insusceptible to transformation. How might theses strains differ from each other? Ans: Noncompetent strains may not be able to take up DNA Alternatively, they may have potent deoxyribonucleases, or they may not be able to integrate fragments of DNA into their genome. 5

7. A propitious choice(吉兆的选择). Bacteriophage M13 infects E.coli differently from the way bacteriophage T2 does. The M13 protein coat is removed in the inner membrane of the bacterial cell, where it is sequestered and subsequently used for the development of progeny DNA. Why would M13 have been much less suitable than T2 was for the experiments carried out by Hershey and Chase?

Ans: In the Hershey-Chase experiment, 35S-labeled T2 viral proteins did not become incorporated into infected cells. The labeled viral proteins were found in the supernatant when infected cells were centrifuged. In contrast, M13 proteins become imbedded in the inner membrane of infected cells; they would appear in the pellet rather than the supernatant after centrifugation. Hershey and Chase would not have been able to separate M13 into genetic and nongenetic parts, as they did for T2.

8. Tagging DNA(标签 DNA).
(a) Suppose that you want to radioactively label DNA but not RNA in dividing and growing bacterial cells. Which radioactive molecule would you add to the culture medium? (b)Suppose that you want to prepare DNA in which the backbone phosphorus atoms are uniformly labeled with 32P. which precursors should be added to a solution containing DNA polymerase I and primed template DNA? Specify the position of the radioactive atoms in these precursors. Ans: (a) tritiated thymine or tritiated thymidine. (b) dATP, dGTP, dCTP, and dTTP labeled with 32P in the innermost (α) phosphorus atom.

9. Finding a template (找模板). A solution contains polymerase I and the Mg2+ salts of DATP, dGTP, dCTP and dTTP. The DNA molecules listed below are added in aliquots of this solution. Which of them would lead to DNA synthesis?

(a) A single-stranded closed circle containing 1000 nucleotide units. (b) A double stranded closed circle containing 1000 nucleotide pairs. (c) A single-stranded closed circle of 1000 nucleotide base paired to a linear strand of 500 nucleotide with a free 3′-OH terminus.

(d) A double-stranded linear molecule of 1000 nucleotide pairs with a free 3′-OH at each end. Ans: Molecules (a) and (b) would not lead to DNA synthesis because they lack a 3′-OH group (a primer). Molecule (d) has a free 3′-OH at one end of each strand but no template strand beyond. Only (c) would lead to DNA synthesis.

10. The right start(正确的开始). Suppose that you want to assay reverse transcriptase activity. If polyriboadenylate is the template in the assay, what should you use as the primer? Which radioactive nucleotide should you use to follow chain elongation? Ans: A deoxythymidylate oligonucleotide should be used as the primer. The poly (A) template specifies the incorporation of dT; hence, radioactive dTTP should be used in the assay. 11. Essential degradation(必需的降解反应). Reverse transcriptase has ribonuclease activity as well as polymerase activity. What is the role of its ribonuclease activity? Ans: The ribonuclease serves to degrade the RNA strand, a necessary step in forming duplex DNA 6

from the RNA-DNA hybrid.
12. Virus hunting(寻找病毒). You have purified a virus that infects turnip leaves. Treatment of a sample with phenol removes viral proteins. Application of the residual material to scraped leaves results in the formation of progeny virus particles. You infer that the infectious substance is a nucleic acid. Propose a simple and highly sensitive means of determining whether the infectious nucleic acid is DNA or RNA.

Ans: treat one aliquot of the sample with ribonuclease and another with deoxyribonuclease. Test these nuclease-treated samples for infectivity.
13. Mutagenic consequences(突变剂的结果). Spontaneous deamination of cytosine bases in DNA occurs at low but measurable frequency. Cytosine is converted to uracil by loss of its amino group. After this conversion, which base pair occupies this position in each of the daughter strands resulting from one round of replication? Two rounds of replication? Ans: Deamination changes the original GC base pair into a GU pair. After one round of replication, one daughter duplex will contain a GC pair, and the other duplex an A-U pair. After two rounds of replication, there would be two GC pairs, one AU pair, and one A-T pair. 14. Eons ago(久远时代). The atmosphere of the primitive earth before the emergence of life contained N2, NH3, H2, HCN, CO, and H2O. Which of these compounds is the most likely precursor of most of the atoms in adenine? Why?

Ans: Hydrogen cyanide. Adenine can be viewed as a pentamer of HCN. 15. Information content(信息含量).
(a) How many different 8-mer sequences of DNA are there?
( b) How many bits of information are stored in an 8-mer DNA sequence? In the E.coli genome? In the human genome?
( c) Compare each of these values with the amount of informationb that can be stored on a personal computer diskette. A byte is equal to 8 bits.
Ans: (a) 48=65536. In computer terminology, there are 64K 8-mers of DNA. (b) A bit specifies two bases (say, A and C) and a second bit specifies the other two (G and T). Hence, two bits are needed to specify a single nucleotide (or base pair) in DNA. For example, 00, 01, 10, and 11, could encode A, C, G, and T. An 8-mer stores 16-bits (216=65536), the E.coli genome (4×106bp) stores 8×106 bits, and the human genome (2.9×109 bases) stores 5.8× 109 bits of genetic information. (c) A high-density diskette stores about 1.5 megabytes, which is equal to 1.2×107 bits. A large number of 8-mer sequences could be stored on such a diskette. The DNA sequence of E. coli, once known, could be written on a single diskette. Nearly 500 diskettes would be needed to record the human DNA sequence.

Chapter 04 Flow of Genetic Information
1. Key polymerases(重要的聚合酶). Compare DNA polymerase I and RNA polymerase from E. coli in regard to each of the following features:
(a) Activated precursors.
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(b) Direction of chain elongation.
(c) Conservation of the template.
(d) Need for a primer.
Ans: (a) Deoxyribonuleoside triphosphates versus ribonucleoside triphosphates.(b) 5′→3′ for both. (c) Semiconserved for DNA polymerase I, conserved for RNA polymerase. (d) DNA polymerase I needs a primer, whereas RNA polymerase does not.

2. Encoded sequences(被编码的序列).
(a) Write the sequence of the mRNA molecule synthesized from a DNA template strand having the sequence
5′-ATCGTACCGTTA-3′
(b) What amino acid sequence is encoded by the following base sequence of an mRNA molecule?Assume that the reading frame starts at the 5′ end. 5′-UUGCCUAGUGAUUGGAUG-3′
(c) What is the sequence of the polypeptide formed on addition of poly (UUAC) to a cell free protein synthesizing system?
Ans: (a) 5′-UAACGGUACGAU-3′; (b) Leu-Pro-Ser-Asp-Trp-Met. (c) Poly (Leu-Leu-Thr-Tyr) 3. A tougher chain(更坚韧的链). RNA is readily hydrolyzed by alkali, whereas DNA is not. Why?
Ans:The 2′-OH group in RNA acts as an intramolecular catalyst. In the alkaline hydrolysis of RNA, it forms a 2′, 3′- cyclic intermediate.
4. A potent blocker(强力阻断器). How does cordycepin(虫草素) (3΄-deoxyadenosine, 3΄-脱氧 腺苷) block the synthesis of RNA?
Ans: Cordycepin terminates RNA synthesis. An RNA chain containing cordycepin lacks a 3′-OH group.
5. Silent RNA(沉默 RNA). The code word GGG could not be deciphered in the same way as was UUU, CCC, and AAA because poly (G) does not act as a template. Poly (G) forms a triple-stranded helical structure. Why is it an ineffective template? Ans: Only single-stranded RNA can serve as a template for protein synthesis. 6. Two from one ( 一 分 为 二 ). Khorana synthesized by organic-chemical methods two complementary deoxyribonucleotides, each with nine residues: d (TAC)3 and d (GTA)3. Partially overlapping duplexes that formed on mixing these oligonucleotides then served as templates for the synthesis by DNA polymerase of long, repeating double helical DNA chains. The next step was to obtain long polyribonucleotide chains with a sequence complementary to only one of the two DNA strands. How did he obtain only poly (UAG) Only poly (GUA)?

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Ans: Incubation with RNA polymerase and only UTP, ATP, and CTP led to the synthesis of only poly (UAC). Only poly (GUA) was formed when GTP was used in place of CTP. 7. Back to the bench(回到实验桌). A protein chemist told a molecular geneticist that he had found a new mutant hemoglobin in which aspartate replaced lysine. The molecular geneticist expressed surprise and sent his friend scurrying back to the laboratory. (a) Why was the the molecular geneticist dubious about the reported amino acid substitutions? (b) Which amino acid substitution would have been more palatable to the molecular geneticist? Ans: (a) A codon for lysine cannot be changed to one for aspartate by the mutation of a single nucleotide. (b) Arg, Asn, Gln, Ile, Met, or Thr.

8. Triple entendre(三联体协议). The RNA transcript of a region of G4 phage DNA contains the sequence 5′-AAAUGAGGA-3′. This sequence encodes three different polypeptides. What are they?
Ans: A peptide terminating with Lys (UGA is a stop codon), -Asn-Glu-, and –Met-Arg9. Valuable synonyms(有用的同义词). Proteins generally have low contents of Met and Trp, intermediate ones of His and Cys, and high ones of Leu and Ser. What is the relationship between the number of codons of an amino acid and its frequency of occurrence in proteins? What might be the selective advantage of the relation?

Ans: Highly abundant amino acid residues have the most codons (e.g. Leu and Ser each have six), whereas the least abundant ones have the fewest (Met and Trp each have only one). Degeneracy allows (a) variation in base composition and (b) decreases the likelihood that a substitution of a base will change the encoded amino acid. If the degeneracy were equally distributed, each of the 20 amino acids would have three codons. Benefits (a) and (b) are maximized by assigning more codons to prevalent amino acids than to less frequently used ones. 10. A new translation(一个新的翻译方式). A transfer RNA with a UGU anticodon is enzymatically conjugated to 14C-labeled cysteine. The cysteine unit is then chemically modified to alanine. The altered aminoacyl-tRNA is added to a protein-synthesizing system containing normal components except for this tRNA. The mRNA added to this mixture contains the following sequence:

5′-UUUUGCCAUGUUUGUGCU-3′
What is the sequence of the corresponding radiolabeled peptide? Ans: Phe-Cys-His-Val-Ala-Ala
11. Fire and ice(冰与火). Valine is specified by four codons. How might the relative frequencies of their usage in an alga isolated from a volcanic hot spring differ from those of an alga isolated from an Antarctic bay?

Ans: GUG and GUC are likely to be used more by the algae from the hot springs to increase the melting temperature of its DNA.
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12. Eons apart(相距久远). The amino acid sequences of a yeast protein and a human protein carrying out the same function are found to be 60% identical. However, the corresponding DNA sequences are only 45% identical. Account for this differing degree of identity. Ans: The genetic code is degenerate. Eighteen of the twenty amino acids are specified by more than one codon. Hence, many nucleotide changes (especially in the third base of a codon) do not alter the nature of the encoded amino acid. Mutations leading to an altered amino acid are usually more deleterious than those that do not and hence are subject to more stringent selection.

Chapter 05 Exploring Genes
1. Reading Sequences(序列阅读).
(a) an autoradiogram of a gel containing four lanes of DNA fragments produced by ideal chemical cleavage at each of the four bases is shown in the following figure. The DNA contained a 32P label at its 5′ end. What is the sequence?

(b) Suppose that the Sanger dideoxy method shows that the template strand sequence is 5′-TGCAATGGC-3′. Sketch the gel pattern that would lead to this conclusion. Ans: (a) 5′-GGCATAC-3′. (b) The Sanger dideoxy method of sequencing would give the gel pattern shown below.

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2. The right template(正确的模板). Ovalbumin is the major protein of egg white. The chicken ovalbumin gene contains eight exons separated by seven introns. Should one use ovalbumin cDNA or ovalbumin genomic DNA to form the protein in E.coli? Why? Ans: Ovalbumin cDNA shoud be used. E.coli lacks the machinery to splice the primary transcript arising from genomic DNA.

3. The right cuts(正确切割). Suppose that a human genomic library is prepared by exhaustive digestion of human DNA with the EcoR I restriction enzyme. Fragments averaging about 4kb in length would be generated.

(a) Is this procedure suitable for cloning large genes? Why? (b) Is this procedure suitable for mapping extensive stretches of the genome by chromosome walking? Why?
Ans: (a) No, because most human genes are much longer than 4kb. One would obtain fragments containing only a small part of a complete gene. (b) No, chromosome walking depends on having overlapping fragments. Exhaustive digestion with a restriction enzyme produces nonoverlapping, short fragments.

4. A revealing cleavage(很说明问题的切割). Sickle-cell anemia arises from a mutation in the gene for the β chain of human hemoglobin. The change from GAG to GTG in the mutant elimilates a cleavage site for the restriction enzyme MstII, which recognizes the target sequence CCTGAGG. These findings from the basis of a diagnostic test for the sickle-cell gene. Propose a rapid procedure for distinguishing between the normal and the mutant gene. Would a positive result prove that the mutant contains GTG in place of GAG?

Ans: Southern blotting of an MstII digest would distinguish between the normal and mutant genes. The loss of a restriction site would lead to the replacement of two fragments on the Southern blot by a single longer fragment. Such a finding would not prove that GTG replaced GAG; other sequence changes at the restriction site could yield the same result. 5. Autocatalysis(自我催化). Thomas Cech showed that the ribosomal RNA (rRNA) precursor in the ciliated protozoa Tetrahymena (四膜虫) can self-splice without binding any Tetrahymena protein. He cloned in a plasmid a region of Tetrahymena DNA containing of the intron and flanking sequences present in the precursor RNA. Suggest how this plasmid was used to establish that Tetrahymena proteins are not required for the splicing of the ribosomal RNA precursor. Ans: Cech replicated the recombinant DNA plasmid in E.coli, and then transcribed the DNA in vitro using bacterial RNA polymerase. He then found that this RNA underwent self-splicing in vitro in the complete absence of any proteins from Tetrahymena. 6. Many melodies from cassette(录音带中的不同旋律). Suppose that you have isolated an enzyme that digests paper pulp and have obtained its cDNA. The goal is to produce a mutant that is effective at high temperature. You have engineered a pair of unique restriction sites in the cDNA that flank a 30-bp coding region. Propose a rapid technique for generating many different mutants in this region.

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Ans: A simple strategy for generating many mutants is to synthesize a degenerate set of cassettes by using a mixture of activated nucleosides in particular rounds of oligonucleotide synthesis. Suppose that the 30-mer begins with GTT, which encodes Val. If a mixture of all four nucleotides is used in the first and second rounds of synthesis, the resulting oligonucleotides will begin with the sequence XYT (where X and Y denote A, C, G, or T). These 16 different versions of the cassette will encode proteins containing either Phe, Leu, Ile, Val, Ser, Pro, Thr, Ala, Tyr, His, Asn, Asp, Cys, Arg, or Gly at the first position. Likewise, one can make degenerate cassettes in which two or more codons are simultaneously varied.

7. Terra incognita (未知领域). PCR is typically used to amplify DNA that lies between two known sequences. Suppose that you want to explore DNA on both sides of a single known sequence. Devise a variation of the usual PCR protocol that would enable you to amplify entirely new genomic terrain.

Ans: Digest genomic DNA with a restriction enzyme and select the fragment that contains the known sequence. Circularize the fragment. Then carry out PCR using a pair of primers that serve as templates for the synthesis of DNA away from the known sequence. 8. A puzzling ladder (令人困惑的梯状带). A gel pattern displaying PCR products shows four strong bands. The four pecies of DNA have lengths that are approximately in the ratio of 1:2:3:4. The largest band is cut out of the gel, and PCR is repeated with the same primers. Again, a ladder of four bands is evident in the gel. What does this reveal about the structure of the encoded protein.

Ans: The encoded protein contains four repeats of a specific sequence. 9. Landmarks in the genome(基因组中的界标). Many laboratories around the world are mapping the human genome. It is essential that the results be merged at an early stage to provide a working physical map of each chromosome. In particular, we need to know whether a YAC studied in one laboratory overlaps a YAC studied in another when only a small proportion of each (less than 5%) has been sequenced. Propose a simple test for overlap based on the two laboratories. Ans: Sequence-tagged sites (STSs) can serve as a common framework for establishing the relation between clones. An STS is a sequence 200 to 500 bp long that occurs only once in the entire genome. These sequences and those of a pair of PCR primers that can generate the STS are stored in a database. Laboratory 1 states its YAC contains STS-34, STS-102, and STS-860. Laboratory 2 can then synthesize the corresponding PCR primers and learn whether their YAC contains any of these STSs.

Chapter 06 Portrait of An Allosteric Protein
1. Hemoglobin content(血红蛋白含量). The average volume of a red blood cell is 87μm3. The mean concentration of hemoglobin in red cells is 34g/100ml.
(a) What is the weight of the hemoglobin contained in a red cell? (b) How many hemoglobin molecules are there in a red cell?
12

(c) Could the hemoglobin concentration in red cells be much higher than the observed value? (Hint: suppose that a red cell contained a crystalline array of hemoglobin molecules in a cubic lattice with 65 Ǻ sides.)

Ans: (a) 2.96×10-11 g; (b) 2.71×108 molecules; (c) No. There would be 3.22×108 hemoglobin molecules in a red blood cell if they were packed in a cubic crystalline array. Hence, the actual packing density is about 84% of the maximum possible.

2. Iron content(铁含量). How much iron is there in the hemoglobin of a 70-kg adult? Assume that the blood volume is 70ml/kg of body weight and that the hemoglobin content of blood is 16g/100ml.
Ans: 2.65g (or 4.75×10-2 moles) of Fe
3. Oxygenating myoglobin(肌红蛋白的氧合). The myoglobin content of some human muscles is about 8g/kg. In sperm whale, the myoglobin content of muscle is about 80g/kg. (a) How much O2 is bound to myoglobin in human muscle and in that of sperm whale? Assume that the myoglobin is saturated with O2.

(b) The aount of oxygen dissolved in tissue water (in equilibrium with vanous blood) at 37℃ is about 3.5×10-5M. What is the ratio of oxygen bound to myoglobin to that directly dissolved in the water of sperm whale muscle?

Ans: (a) In humans, 1.44×10-2 g (4.49 ×10-4 moles) of O2 per kilogram of muscle. In sperm whale, 0.144g (4.49×10-3 moles) of O2 per kilogram; (b) 128.
4. Release kinetics(释放动力学). The equilibrium constant K for the binding of oxygen to muoglobin is 10-6, where K is defined as K = [Mb] [O2]/[MbO2]. The rate constant for the combination of O2 with myoglobin is 2×107M-1S-1.

(a) What is the rate constant for the dissociation of O2 from oxymyoglobin? (b) What is the mean duration of the oxymyoglobin complex?
Ans: (a) koff =konK =20 S-1; (b) mean duration is 0.05s ( the reciprocal of koff ). 5. Tuning oxygen affinity(细调氧亲和力). What is the effect of each of the following treatments on the oxygen affinity of hemoglobin A in vitro?

(a) Increase in pH from 7.2 to 7.4.
(b) Increase in pO2 from 10 to 40 torrs.
(c) Increase in [BPG] from 2×10-4 to 8×10-4 M.
(d) Dissociation of α2β2 into monomer subunits.
Ans: (a) increased; (b) decreased; (c) Decreased, and (d) increased oxygen affinity. 6. Avian and reptilian counterparts(鸟类和爬行类的相应物). The erythrocytes of birds and turtles contain a regulatory molecule different from BPG. This substance is also effective in reducing the oxygen affinity of human hemoglobin stripped of BPG. Which of the following substances would you predict to be the most effective in this regard? (a) Glucose 6-phosphate. (b) Inositol hexaphosphate. (c) HPO42-. (d) Malonate. (e) Arginine. (f) Lactate.

13

Ans: Inositol hexaphosphate.
7. Tuning proton affinity(调节质子亲和力). The pK of an acid depends partly on its environment. Predict the effect of the following environmental changes on the Pk of a glutamic acid side chain. (a) A lysine side chain is brought into close proximity.

(b) The terminal carboxyl group of the protein is brought into close proximity. (c) The glutamic acid side chain is shifted from the outside of the protein to a nonpolar site inside.
Ans: The pK is (a) lowered, (b) raised, and (c) raised.
8. Linkage relations(连接关系). The concept of linkage is crucial for the understanding of many biochemical processes. Consider a protein molecule P that can bind A or B or both:

The dissociation contants for these equilibra are defined as KA=[P][A]/[PA], KB=[P][B]/[PB], B

KBA=[PB][A]/[PAB], KAB=[PA][B]/[PAB].
(a) Suppose KA=5×10-4M, KB=10-3M, and KBA=10-5M. Is the value of the fourth dissociation B

constant KAB defined? If so, what is it?
(b) What is the effect of [A] on the binding of B? What is the effect of [B] on the binding of A? Ans: (a) Yes. KAB +KBA (KB/KA) = 2 ×10-5 M; (b) the presence of A enhances the binding of B; hence, the presence of B enhances the binding of A.

9. Poison puzzle(毒物之谜). Carbon monoxide, an odorless gas, combines with hemoglobin to form CO-hemoglobin. Crystals of CO-hemoglobin are isomorphous with those of oxyhemoglobin. Each heme in hemoglobin can bind one carbon monoxide molecule, but O2 and CO can not simultaneouly bind to the same heme. The binding affinity for CO is about 200 times as great as that for oxygen. Exposure for 1 hour to a CO concentration of 0.1% in inspired air leads to the occupancy by CO of about half the heme siyes in hemoglobin, aproportion that is frequently fatal. An interesting problem was posed (and partly solved) by J.S. Haldane and J.G.Priestley in 1935: If the action of CO were simply to diminish the oxygen-carrying power of the hemoglobin, without other modifications of its properties, the symptoms of CO poisoning would be very difficult to understand in the light of other knoeledge. Thus, a person whose blood is half-saturated with CO is particularlly helpless, as we have just seen; but a person whose hemoglobin percentage is simply diminished to half by anemia may be going about his work as usual. What is the key to this seeming paradox?

Ans: Carbon monoxide bound to one heme alters the oxygen affinity of other hemes in the same hemoglobin molecule. Specifically, CO increases the oxygen affinity of hemoglobin and threby 14

decreases the amount of O2 released in actively metabolizing tissues. Carbon monoxide stabilizes the quaternary structure characteristic of oxyhemoglobin. In other words, CO mimics O2 as an allosteric effector.

9. Optimal transport(最佳转运). A protein molecule P reversibly binds a small molecule L. The dissociation constant K for the equilibrium: P + L + PL is defined as K = [P][L]/[PL]. The protein transports the small molecule from a region of high concentration [LB]. Assume that the B

concentrations of unbound small molecules remain constant. The protein goes back and forth between regions A and B.
(a) Suppose [LA] = 10-4 M and [LB] = 10-6 M. What value of K yields maximal transport? One B

way of solving this problem is to write an expression for ΔY, the change in saturation of the ligand-binding site in going from A to B. Then, take the derivative of ΔY with respect to K. (b) Treat oxygen transport by hemoglobin in a similar way. What value of P50 would give a maximal ΔY? Assume that the P in the lungs is 100 torrs, whereas P in the tissue capillaries is 20 torrs. Compare your calculated value of P50 with the physiological value of 26 torrs. Ans: (a) For maximal transport, K = 10-5M. In general, maximal transport is achieved when K = ([LA][LB])0.5; (b) For maximal transport, P50 =44.7 torrs, which is considerably higher than the physiologic value of 26 torrs. However, it must be stressed that this calculation ignores cooperative binding and the Bohr effect.

11. On the mutant trail(突变体寻找). A hemoglobin with an abnormal electrophoretic mobility is detected in a screening program. Fingerpringting after tryptic digestion reveals that the amino acid substitition

is

in

the

β

chain.

The

mormal

amino

terminal

tryptic

peptide

(Val-His-Leu-Thr-Pro-Glu-Glu-Lys) is missing. A new tryptic peptide consisting of six amino acid residues is found. Valine is the amino-terminal residue of this peptide. (a) Which amino acid sunstitutions are consistent with these data? (b) Which single-base changes in DNA sequence could give these amino acid substitutions? The DNA

sequence

encoding

the

normal

amino-terminal

region

is:

GTGCACCTGACTCCTGAGGAGAAG. (c) How should the electrophoretic mobility of this hemoglobin compare with those of HbA and HbS at pH8?
Ans: (a) Lys or Arg at position 6; (b) GAG (Glu) to AAG (Lys); (c) This mutant hemoglobin moves more rapidly toward the anode than does HbA and HbS because it is more positively charged.
12. Allosteric transition(变构转换). Consider an allosteric protein that obeys the concerted model. Suppose that the ratio of T to R formed in the absence of ligand if 105, KT=2mM, and KR=5uM. The protein contains four binding sites for ligand. What is the fraction of molecules in the R form when 0, 1, 2, 3, and 4 ligands are bound?

Ans: The fraction of molecules in the R form is 10-5, 0.004, 0.615, 0.998, and 1 when 0, 1, 2, 3, 15

and 4 ligands, respectively, are bound.
13. Triple hit(命中三个). Some mutations in a hemoglobin gene affect all three of the hemoglobins A, A2, and F, whereas others affect only one of them. Why? Ans: Mutations in the α gene affect all three hemoglobins because their subunit structures are α2β2, α2δ2, andα2γ2. Mutations in the β, δ or γ genes affect only one of them. 14. Saving grace. Hemoglobin A inhibits the formation of long fibers of hemoglobin S and the subsequent sickling of the red cell upon deoxygenation. Why does hemoglobin A have this effect? Ans: Deoxy HbA contains a complementary site, and so it can add on to a fiber of deoxy HbS. The fiber cannot then grow further because the terminal deoxy HbA molecule lacks a sticky patch. 15. Carbamoylation(氨甲酰化). Cyanate was a promising antisickling drug until clinical trials uncovered its toxic side effects, such as damage to peripherial nerves. Cyanate carbamoylates the terminal amino groups of hemoglobin. It behaves as a reactive analog of CO2. Predict the resulting changes in oxygen affinity. Why does carbamoylation have an antisickling effect? Ans: Carbamoylation of hemoglobin increases its oxygen affinity. Oxygenated HbS does not sickle.

16. A changing slope(变化的斜率). The Hill plot for hemoglobin has a slope of about 2.8 at the midpoint (Y=0.5). However, it is important to note that the slope of the Hill plot is not constant. At both very low and very high oxygen pressure, the slope approaches 1. Why? Ans: At very low fractional saturation, O2 binds to hemoglobin molecules that are almost entirely in the T state. At very high fractional saturation, O2 binds to molecules that are almost entirely in the R state. The slopes is ~1 (nearly noncooperative) when binding is not accompanied by a significant switch from T to R.

Chapter 07 Enzymes: Concepts and Kinetics
1. Hydrolytic driving force(水解动力). The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions such as the synthesis od DNA. This hydrolytic reaction is catalyzed in E. coli by a pyrophosphatase that has a mass of 120kd and consists of six identicla sununits. For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes 10umoles of pyrophosphate in 15 minutes at 37℃ under standard assay conditions. The purified enzyme has a Vmax of 2800 units per milligram of enzyme. (a) How many moles of substrate are hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than Km?

(b) How many moles of active site are there in 1 mg of enzyme? Assume that each subunit has one active site.
(c) What is the turnover number of the enzyme? Compare this value with others mentioned in this chapter.
Ans: (a) 31.3μmoles; (b) 0.05μmoles; (c) 622 s-1.
2. Destroying the Trojan horse(破坏特洛伊木马). Penicillin is hydrolyzed and thereby rendered 16

inactive by penicillinase (also known as β-lactamase), an enzyme present in some resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6kd. The amount of penicillin hydrolyzed in 1 minute in a 10-ml solution containing 10-9g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay.

(a) Plot 1/V versus 1/[S] for these data. Does penicilinase appear to obey Michaelis-Menten kinetics? If so, what is the value of Km?
(b) What is the value of Vmax?
(c ) What is the turnover number of penicillinase under these experimental conditions? Assume one active site per enzyme molecule.
Ans: (a) Yes. KM = 5.2×10-6M; (b) Vmax = 6.84×10-10 moles/min; (c) 337 s-1. 3. Counterpoint(对照). Penicillinase (β-lactamase) hydrolyzes penicillin. Compare penicillinase with glycopeptide transpeptidase.

Ans: Penicillinase, like glycopeptides transpeptidase, forms an acyl-enzyme intermediate with its substrate but transfers it to water rather than to the terminal glycine of the penta-glycine bridge. 4. Mode of inhibition(抑制模式). The kinetics of an enzyme are measured as a function of substrate concentration in the presence and absence of 2mM inhibitor (I)

(a) What are the values of Vmax and Km in the absence of inhibitors? In its presence? (b) What type of inhibition is this?
(c) What is the binding constant of this inhibitor?
(d) If [S] =30uM, and [I] = 2mM, what fraction of the enzyme molecules have a bound 17

substrate? A bound inhibitor?
(e) If {S} =30μM, what fraction of the enzyme molecules have a bound substrate in the presence and absence of 2mM inhibitors? Compare this ratio with the ratio of the reaction velocities under the same conditions.

Ans: (a) In the absence of the inhibitor, Vmax is 47.6 μmole/min and Km is 1.1× 10-5M. In the presence of inhibitor, Vmax is the same, and the apparent KM is 3.1×10-5M; (b) Competitive; (c) 1.1× 10-3M; (d) ƒES is 0.243 and ƒEI is 0.488; (e) ƒES is 0.73 in the absence of inhibitor and 0.49 in the presence of 2×10-3M inhibitor. The ratio of these values, 1.49, is the same as the ratio of the reaction velocities under these conditions.

5. A different mode(一种不同模式).The kinetics of the enzyme discussed in problem 4 are measured in the presence of a different inhibitor. The concentration of this inhibitor is 100uM. (a) What are the values of Vmax and Km in the presence of this inhibitor? Compare them with those obtained in problem 4.

(b) What type of inhibition is this?
(c) What is the dissociation constant of this inhibitor?
(d) If [s] =30μM, what fraction of the enzyme molecules have a bound sunstrate in the presence and absence of 100uM inhibitor?

Ans: Vmax is 9.5μmole/min. KM is 1.1×10-5M, the same as without inhibitor; (b) Noncompetitive; (c) 2.5× 10-5M; (d) 0.73, in the presence or absence of this noncompetitive inhibitor. 6. A fresh view(新颖观点). The plot of 1/V versus 1/[S] is sometimes called a lineweaver-Burk plot. Another way of expressing the kinetic data is to plot V versus V/[S], which is known as an Eadie-Hofstee plot.

(a) Rearrange the Michaelis-Menten equation to give V as the function of V/[S]. (b) What is the significance of the slope, the vertical intercept, and the horizontal intercept in a plot of V versus V[S]?

(c) Make a sketch of a plot of V versus V/[S] in the absence of a competitive inhibitor, and in the presence of a noncompetitive inhibitor.
Ans: (a) V= Vmax-(V/[S]) KM; (b) Slope =-KM, y-intercept = Vmax, x-intercept =Vmax/KM; (c) An Eadie-Hofstee plot is shown below.
18

7. Potential donors and acceptors(潜在供体与受体). The hormone progesterone contains two ketone groups. At pH 7, which side chains of the receptor might form hydrogen bonds with progesterone? (Assume that the side chains in the receptor protein have the same pKs as in the amino acids in aqueous solution.)

Ans: Potential hydrogen bond donors at pH 7 are the side chains of the following residues: arginine, asparagine, glutamine, histidine, lysine, serine, threonine, tryptophan, and tyrosine. 8. Competing substrates(竞争底物). Suppose that two substrates A and B compete for an enzyme. Derive an expression relating the ratio of the rates of utilization of A and B, VA/VB, to the B

concentrations of these substrates and their values of k3 and Km.(Hint: Express VA as a function of k3/Km for substrate A, and do the same

for VB.) Is specificity determined by Km alone?
B

Ans: The rates of utilization of A and B are given by

Thus, an enzyme discriminates between competing substrates on the basis of their values of K3/KM rather than that of KM alone.
9. A tenacious mutant(顽强的突变体). Suppose that a mutant enzyme binds a substrate 100-fold as tightly as does the native enzyme. What is the effect of this mutation on catalytic rate if the binding of the transition state is unaffected?

Ans: The mutation slows the reaction by a factor of 100 because the activation free energy is increased by +2.72 kcal/mol (2.303RTlog100). Strong binding of the substrate relative to the transition state slows catalysis.

Chapter 08 Catalytic Strategies
1. Kinetic forecast(动力学预测). Predict the relative rates of hydrolysis by lysozyme of these 19

oligosaccharides (G stands for an N-acetylglucosamine residue, and M for N-acetylmuramic acid): (a) MMMMM

(b) GMGMGM

(c) MGMGMG

Ans: The fastest is (b) and the slowest is (a).
2. Occupancy forecast(占位预测). Predict on the basis of the data given in the following figure, which of the sugar binding sites, A to F, on lysozyme will be occupied in the most prevalent complex with each of these oligosaccharides (same abbreviations as in problem 1).

(a) GG

(b) GM

(c) GGGG

Ans: (a) B-C, (b) A-B or E-F, and (c) A-B-C (one sugar residue does not interact with the enzyme, thus avoiding site D, which is energetically unfavorable.)
3. A revealing tag(很能说明问题的标签). Suppose that hexa-NAG is synthesized sothat the glycosidic oxygen between D and E sugar residues is labeled with

18

O. Where will this isotope

appear in the products formed by hydrolysis with lysozyme?
Ans: The 18O emerges in the C-4 hydroxyl of di-NAG (residues E-F) 4. Affinity accounting(亲和力解释). An analog of tetra-NAG containing –H in place of –CH2OH at C-5 of residue D binds to lysozyme much more strongly than does tetra-NAG. Propose a structural basis for this difference in binding affinity.

Ans: This analog lacks a bulky substituent at C-5, and so it can probably bind to site D without being strained. Consequently, the binding of residue D of this analog is likely to be energetically favorable, whereas the binding of residue D of tetra-NAG costs free energy. 5. Metal coordination ( 金 属 配 位 ) . Compare the coordination of the zinc atom in carboxypeptidase A with that of the iron atom in oxymyoglobin and oxyhemoglobin. (a) Which atoms are directly bonded to these metal ions?

(b) Which side chains contribute these metal-binding groups? (c) Which other side chains in proteins can bind Zn2+or Fe2+. Ans: (a) In oxymyoglobin, Fe is bonded to five nitrogens and one oxygen. In carboxypeptidase A, Zn is bonded to two nitrogens and two oxygens; (b) In oxymyoglobin, one of the nitrogens bonded to Fe comes from the proximal histidine residue, whereas the other four come from the heme. The 20

oxygen atom linked to Fe is that of O2. In carboxypeptidase A, the two nitrogen atoms coordinated to Zn come from histidine residues. One of the oxygen ligands is from a glutamate side chain, the other from a water molecule; (c) Aspartate, cycteine, and methionine. 6. Detective work(侦探工作). In Agatha Christie′s Murder on the Orient Express, Hercule Poirot′s task as a detective was compounded by the presence of so many plausible suspects. Likewise, the catalytic mechanism of carboxypepidase A was difficult tpo unravel because of the presence of several potential catalytic groups at the active site. For example, the hydroxyl group of tyrosine 248 was postulated to be the proton donor to the NH group of the substrate in the cleavage step. Design an experiment that critically tests this hypothesis. Ans: If tyrosine 248 were essential for catalysis, a mutant of this enzyme containing phenylalanine in its place would be inactive. The tyrosine 248 codon (TAT) was converted to that of Phe (TTT) by oligonucleotide-directed mutagenesis. The recombinant plasmid containing this gene was inserted into yeast and expressed. The striking finding was that the mutant enzyme had the same Kcat value as did the native enzyme but a KM value that was sixfold higher. This experiment showed that tyrosine 248 participates in the binding of substrate but is not essential for catalysis. Moreover, it graphically illustrated the power of site-specific mutagenesis in delineating the function of a particular residue in a protein.

7. On the thiol trail(巯基信息). What is a simple means of determining whether a recently discovered proteolytic enzyme is a thiol protease?
Ans: The essential sulfhydryl in its active site should be highly susceptible to chemical modification.

The

enzyme

should

be

inactivated

by

reacting

with

iodoacetamide,

N-ethyl-maleimide, or another sulfhydryl-specific reagent. Furthermore, bound substrate is likely to protect the key sulfhydryl from alkylation.
8. A criticla donation(重要捐献).The transfer of a proton from an enzyme to its substrate is often a key step in catalysis.
(a) Does this occur in the proposed catalytic mechanism for ribonuclease A, Chymotrypsin, lysozyme, and carboxypeptidase A?
(b) If so, in each case, identify the proton donor.
Ans: (a) Yes; (b) Histidine 119 in ribonuclease A, histidine 57 in chymotrypsin, glutamic acid 35 in lysozyme, and a zinc-bound water molecule in carboxypeptidase A. 9. Variation on an inhibitory theme(抑制主题的变种). Recall that TPCK, an affinity labeling reagent, inactivates chymotrypsin by alkylating histidine 57. (a) Design an affinity-labeling reagent for trypsin that resembles TPCK. (b) How would you test its specificity?

Ans: (a) Tosyl-L-lysine chloromethyl ketone (TLCK); (b) First, determine whether substrates 21

protect trypsin from inactivation by TLCK and, second, ascertain whether the D isomer of TLCK inactivates trypsin.
10. An aldehyde inhibitor(一个醛的抑制剂). Elastase is specifically inhibited by an aldehyde derivative of one of its substrates:

In fact, this aldehyde is an analog of the transition state for catalysis by elastase. (a)

Which residue at the active site of the elastase is most likely to a form a covalent bond with this aldehyde?

(b) What type of covalent link would be formed?
Ans: (a) Serine; (b) A hemiacetal between the aldehyde of the inhibitor and the hydroxyl group of the active site serine.
11. A perfect dimer(完美的二聚体). The HIV-1 protease, like other retroviral proteases, is a dimer of identical subunits rather than a single chain twice as large. What is the selective advantage to the virus of a dimeric arragement?

Ans: A dimmer of identical subunits is advantageous for two reasons. First, less RNA is needed to encode the protease. A small genome can be more readily replicated and packaged than a large one. Second, the protease does not become active until a critical concentration is attained. Premature excision of viral subunits from the poly-protein is thereby prevented. 12. Designing anti-AIDS drugs(抗艾滋病药物设计). Symmetric inhibitors of the HIV-1 protease are attractive candidates for anti-AIDS drugs because their symmetry matches that of the target.. propose another mechanism of selectively inhibiting HIV-1 protease. Ans: Inhibitors of protease subunit dimerization should be effective in blocking protease action. 13. An antihypertensive drug(抗高血压药). Angiotensin II, an octapeptide formed by the kidneys, raises blood pressure. This pressor hormone is generated by the action of two proteolytic enzymes. First, angiotensinogen, a 14-residue peptide, is cleaved by renin to yield angiotensin I, a decapeptide. Then, the last two residues of angiotensin I are removed in a single step by the hydrolytic action of angiotensin converting enzyme (ACE, 血 管 紧 张 肽 转 换 酶 ) to form angipotensin II. This hormone rapidly elevates blood pressure by constricting arterioles. ACE is a zinc-containing protease that is mechanistically similar to carboxypeptidase A except that it cleaves the penultimate rather than the last peptide bond. Captopril(卡托普利), a potent inhibitor (Ki =0.2nM) of ACE, is widely used to treat hypertension. How might captopril inhibit ACE?

22

Ans: Its sulfhydryl group probably binds to the zinc ion at the active site, and its carboxylated group very likely occupies the recognition site for the C terminus of the substrate. 14. Recapitulation(简要说明). List the factors contributing to the catalytic power of enzymes. Ans: Precise positioning of catalytic residues and substrates, geometrical strain (distortion of the substrate), electronic polarization, and desolvation of the substrate.

Chapter 09 Regulatory Strategies
1. Acitivity profile(活性图). A histidine residue in the active site of aspartate transcarbamoylase is thought to be important in stabilizing the transition state of the bound substrates. Predict the pH dependence of the catalytic rate, assuming that this interaction is essential and dorminates the pH-activity profile of the enzyme.

Ans: The protonated form of histidine probably stabilizes the negatively charged carbonyl oxygen atom of the scissile bond in the transition state. Deprotonation would lead to a loss of activity. Hence, the rate is expected to be half-maximal at a pH of about 6.5 (the pK of an unperturbed histidine side chain in a protein) and decrease as the pH is raised. 2. Allosteric switching(变构转换). A substrate binds 100 times as tightly to the R state of an allosteric enzyme as to its T state. Assume that the concerted(MWC) model applies to this enzyme.

(a) By what factor does the binding of one substrate molecule per enzyme molecule alter the ratio of the concentrations of enzyme molecules in the R and T states? (b) Suppose that L, the ratio of [T] to [R] in the absence of substrate, is 107 and that the enzyme contains four binding sites for substrate. What is the ratio of enzyme molecules in the R state to that in the T state in the presence of saturating amounts of substrate, assuming that the concerted model is obeyed?

Ans: (a) 100. The change in the [R] / [T] ratio on binding one substrate molecule must be the same as the ratio of the substrate affinities of the two forms.; (b) 10. The binding of four substrate molecules changes the [R]/[T] ratio by a factor of 1004 =108. The ratio in the absence of substrate is 10-7. Hence, the ratio in the fully liganded molecule is 108×10-7 =10. 3. Negative cooperativity(负协同效应). You have isolated a dimeric enzyme that contains two identical active sites. The binding of substrate to one active site decreases the substrate affinity of the other active site. Which allosteric model best accounts for this negative cooperativity?

23

Ans: The sequential model can account for negative cooperativity, whereas the concerted model cannot. Homotropical allosteric interactions must be cooperative if the concerted model holds. 4. Paradoxical at first glance(咋一看有点奇怪). Recal that phosphonacetyl-L-aspartate (PALA) is a potent inhibitor of ATCase because it mimics the two physiologic substrates. However, low concentrations of this unreactive bisubstrate analog increases the reaction velocity. This stimulatory action is particularly striking in the ATCase-catalyzed breakdown of N-carbamoyl phosphate by arsenate, a nonphysiologic reaction. On addition of PALA the reaction rate increases until an average of three PALA are bound per enzyme molecule. This maximal velocity is 17-fold greater than in the absence of PALA. The reaction rate then decreases to nearly zero on adding three molecules of PALA per enzyme. Why do low concentrations of PALA activate ATCase?

Ans: The binding of LALA switches ATCase from T to the R state because it acts as a substrate analog.. An enzyme molecule containing bound PALA has fewer free catalytic sites than does an unoccupied enzyme molecule. However, the PALA-containing enzyme will be in the R state and hence have higher affinity for the substrates. The dependence of the degree of activation on the concentration of PALA is a complex function of the allosteric constant L0, and of the binding affinities of the R and T states for the analog and the substrates. 5. Zymogen activation(酶原激活). When very low concentrations of pepsinogen are added to acidic media, how does the half-time for activation depend on zymogen concentration? Ans: Activation is independent of zymogen concentration because the reaction is intra-molecular. 6. A revealing assay(说明问题的检验).

Suppose that you have just examined a young boy

with a bleeding disorder highly suggestive of a classic hemophilia (Factor VIII deficiency). Because of the late hour, the laboratory that carries out speciaized coaggulation assays is closed. However, you happen to have a sample of blood from a classic hemophiliac you admitted to the hospital an hour earlier. What is the simplest and most rapid test you can perform to determine whether your present patient is also deficient in factor VIII activity? Ans: Add blood from the second patient to a sample from the first. If the mixture clots, the second patient has a defect different from that of the first. This type of assay is called a complementation 24

test.
7. Counterpoint(对比). The synthesis of Factor X, like that of prothrombin. Requires vitamin K. Factor X also contains γ-carboxyglutamate residues in its amino-terminal region. However, activated factor X, in contrast with thrombin, retains this region of the molecule. What is a likely functional consequence of this difference between the two activated species? Ans: Activated factor X remains bound to blood platelet membranes, which accelerates its activation of prothrombin.

8. A discerning inhibitor(能辨别的抑制剂). Antithrombin III forms an irreversible complex with thrombin but not with prothrombin. What is the most likely reason for this difference in reactivity?
Ans: Antithrombin III is a very slowly hydrolyzed substrate of thrombin. Hence, its interaction with thrombin requires a fully formed active site on the enzyme. 9. Repeating heptads(重复的七聚体). Each of the three types of chains of fibrin contains repeating heptapeptide units (abcdefg) in which residues a and d are hydrophobic. Propose a reason for this regularity.

Ans: Residues a and d are located in the interior of an α-helical coiled coil, near the axis of the superhelix. Hydrophobic interactions between these side chains contribute to the stability of the coiled coil.

10. Drug design(药物设计). A drug company has decided to prepare by recombinant DNA methods a modified α1-antitrypsin that will be more resistant to oxidation than is the naturally occurring inhibitor. Which single amino acid substitution would you recommend to them? Ans: Leucine would be a good choice. It is much more stable than methionine but has nearly the same volume and also is very hydrophobic.

Chapter 10 Membrane Structure and Dynamics
1. Population density(群体密度). How many phospholipid molecules are there in a 1-μm2 region of a phospholipid bilayer membrane? Assume that a phospholipid molecule occupies 70 Å2 of the surface area.

Ans: 2.86×106 molecules, because each leaflet of the bilayer contains 1.43× 106 molecules. 2. Lipid diffusion(脂质扩散). What is the average distance traversed by a membrane lipid in 1μs, 1ms ansd 1s? Assume a diffusion coefficient of 10-8cm2/s.

Ans: 2×10-7cm, 6.32×10-6cm, and 2×10-4cm
3. Protein diffusion(蛋白质扩散). The diffusion coefficient D of a rigid spherical molecule is

25

given by
D =kT/6πηr in which, η is the viscosity of the solvent, r is the radius of the sphere, k is the Boltzman constant (1.38×10-16 erg/deg), and T is the absolute temperature. What is the diffusion coefficient at 37C of a 100-kd protein in a membrane that has an effective viscosity of 1 poise (1 poise =1 erg s/cm3)? What is the average distance traversed by this protein in 1μs, 1ms ansd 1s? Assume that this protein is an unhydrated, rigid sphere of density 1.35g/ cm3. Ans: The radius of this molecule is 3.08×10-7 cm and its diffusion coefficient is 7.37×10-9 cm2/s. The average distances traversed are 1.72×10-7 cm in 1μs, 5.42×10-6 in 1ms, and 1.72×10-4cm in 1 s.

4. Cold sensitivity(冷敏感). The conductance of a lipid bilayer membrane containing a carrier antibiotic decreased abruptly when the temperature was lowered from 40℃ to 36℃. In contrast, there was little change in conductance of the same bilayer membrane when it contained achannel-forming antoibiotic. Why?

Ans: The membrane underwent a phase transition from a highly fluid to a nearly frozen state when the temperature was lowered. A carrier can shuttle ions across a membrane only when the bilayer is highly fluid. A channel former, by contrast, allows ions to traverse its pore even when the bilayer is quite rigid.

5. Flip-flop(翻转). The transverse diffusion of phospholipids in a bilayer membrane was investigated using a paramagnetic analog of phosphatidyl choline, called spin-labeled phosphatidyl choline.

The nitroxide (NO) group in spin-labeled phosphatidyl choline gives a distinctive paramagnetic resonance spectrum. This spectrum disappears when nitroxides are converted into amines by reducing agents such as ascorbate.

Lipid vesicles containing phosphatidyl choline (95%) and the spin-labeled analog (5%) were prepared by sonication and purified by gel-filtration chromatography. The outside diameter of these liposomes was about 25nm. The amplitude of the paramagnetic resonance spectrum (顺磁共 振谱) decreased to 35% of its initial value within a few minutes after the addition of a second aliquot of ascorbate. However, the amplitude of the residual spectrum decayed exponentially with a half-time of 6.5 hours. How would you interpret thses changes in the amplitude of the paramagnetic spectrum?

Ans: The initial decrease in the amptitude of the paramagnetic resonance spectrum results from the reduction of spin-labeled phosphatidyl cholines in the outer leaflet of the bilayer. Ascorbate does not traverse the membrane under these experimental conditions; hence, it does not reduce the 26

phospholipid in the inner leaflet. The slow decay of the residual spectrum is due to the reduction of phospholipids that have flipped over to the outer leaflet of the bilayer.

Chapter 11

Membrane Channels and Pumps

1. Concerted opening(协同开启). Suppose that a channel obeys the concerted allosteric model (MWC model). The binding of ligand to the R state (the open form) is 20 times as tight as to the T state (the closed form). In the absence of ligand, the ratio of closed to open channels is 105. If the channel is a tetramer, what is the fraction of open channels when 1, 2, 3, and 4 ligands are bound? Ans: The ratio of closed to open forms of the channel is 105, 5000, 250, 12.5, and 0.625 when 0, 1, 2, 3, and 4 ligands, respectively, are bound. Hence, the fraction of open channel is 10-5, 2×10-4, 3.98×10-3, 7.41×10-2, and 0.615.

2. Respiratory paralysis(呼吸瘫痪). Tabun (塔崩; N,N′-二甲氨基氰磷酸乙酯;N,N′-二甲氨基 氰基磷酸乙酯) and sarin (沙林) have been used as chemical warfare agents, and parathion has been employed as an insecticide. What is the molecular basis of their lethal actions?

Ans: These organic phosphates inhibit acetylcholinesterase by reacting with the active site serine to form a stable phosphorylated derivative. They cause respiratory paralysis by blocking synaptic transmission at cholinergic synapses.

3. Ligand-induced channel opening(配体诱导的通道开放). The ratio of open to closed forms of the acetylcholine receptor channel containing 0, 1, and 2 bound acetylcholines is 5×10-6, 1.2×10-3, and 14, respectively.

(a) By what factor is the open/closed ratio increased by the binding of the first acetylcholine? And the second acetylcholine?
(b) What are the corresponding free energy contributions to channel opening at 25℃? (c) Can the allosteric transition be accounted for by the MWC concerted model? Ans: (a) The binding of the first acetylcholine increases the open/closed ration by a factor of 240, and binding of the second by a factor of 11,700; (b) Hence, the free energy contributions are 3.3kcal/mol and 5.6 kcal/mol, respectively; (c) No. The MWC model predicts that the binding of each ligand will have the same effect on the open/closed ratio. The acetycholine receptor channel is not perfectly symmetric. The two α chains are not in identical environments. Also, the presence

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of desensitized states in addition to the open and closed ones indicates that a more complex model is required.
4. Voltage-induced channel opening(电压诱导的通道开放). The fraction of open channels at 5mV steps beginning at -45mV and ending at +5mV at 20C is 0.02, 0.04, 0.09, 0.19, 0.37, 0.59, 0.78, 0.89, 0.95, 0.98, and 0.99. (a) At what voltage are half the channels open? (b) What is the value of the gating charge? (c) How much free energy is contributed by the movement of the gating charge in the transition from -45mV to +5mV?

Ans: (a) -22mV; (b) +4.5; (c) 5.2kcal/mol.
5. Frog poison(蛙毒). Batrachotoxin (BTX, 箭毒蛙毒素) is a steroidal alkaloid from the skin of a poisonous Colombian frog. In the presence of BTX, sodium channels in an excised patch stay persistently open when the membrane is depolarized. They close when the membrane is depolarized. Which transition is blocked by BTX?

Ans: Batrachotoxin blocks the transition from the open to the inactivated state. 6. Valium Target [(安定(商标名称)作用目标]. γ-Amino butyric acid (GABA) opens channels that are specific for chloride ions. The GABA receptor channel is pharmacologically important because it is the target of Valium, which is used to diminish anxiety. The extracellular concentration of Cl- is 123 mM and the intracellular concentration is 4 mM. In which direction does Cl- flow through an open channel when the membrane potential is in the -60mV to +30mV range?

What is the effect of chloride channel opening on the excitability of a neuron? The hydropathy profile of the GABAA receptor resembles that of the acetylcholine receptor. Predict the number of subunits in this chloride channel.

Ans: (a) Chloride ions flow into the cell; (b) Chloride flux is inhibitory because it hyperpolarizes the membrane; (c) The channel consists of five subunits.
7. The price of extrusion(压榨的代价). What is the free-energy cost of pumping Ca2+ out of a cell when the cytosolic concentration is 0.4μM, the extracellular concentration is 1.5mM, and the membrane potential is -60mV?

Ans: The free energy cost is 7.6kcal/mol (4.88kcal/mol for the chemical work performed and 2.76kcal/mol for the electric work performed.)
8. Rapid transit(快速过渡). A channel exhibits current steps of 5 pA at a membrane potential of -50mV. The mean open time is 1ms, and the membrane potential is omV. (a) What is the conductance of this channel?

(b) How many univalent ions flow through the channel during its mean open time? 28

(c) How long does it take an ion to pass through the channel? Ans: (a) The conductance g of the channel is 100 pS; (b) 3.12×104 ions flow through the channel during its mean open time of 1 ms; (c) The mean transit time for an ion is 32 ns. 9. Pumping protons(泵动质子). Design an experiment to show that lactose permease can be reversed in vitro to pump protons.

Ans: Membrane vesicles containing a high concentration of lactose in their inner volume are formed. The binding of lactose to the inner face of the permease will be followed by the binding of a proton. Both sites will then evert. Because the lactose concentration on the outside is low, lactose and the proton will dissociate from the permease. The downhill flux of lactose will drive the uphill flux of protons in this in vitro system.

Chapter 12 Signal Transduction Cascades
1. The richness of phosphorylation(磷酸化作用的丰富性). Compare and contrast the role of phosphorylation in bacterial chemotaxis, vision, and growth-control cascade. Ans: Phosphorylation mediates the excitatory response in chemotaxis and receptor tyrosine kinase cascades, whereas it participates in recovery and adaptation in vision. Autophosphorylation of che A is essential in chemotaxis. Likewise, autophosphorylation of receptor tyrosine kinase is crucial in growth-control cascades. By contrast, phosphorylation of photoexcited rhodopsin leads to its deactivation.

2. Attractant response(吸引物反映). What is the sequence of molecular changes in the bacterial chemotaxis system following a step increase in the concentration of an attractant such as aspartate?
Ans: The binding of an attractant rapidly leads to the formation of less phosphorylated cheY, and hence the probability of tumbling decreases. On a slower time scale, the binding of attractant leads to the formation of less phosphorylated cheB (the more potent form of the methylesterase). Also, the rate of methylation is increased. Consequently, the methylation level of the receptor increases, which speeds the production of phosphorylated to bring the tumbling probability back to the initial value. Hence, excitation is followed by adaptation.

3. Eeyore and Candide(屹耳与憨第德). Bacterial chemotaxis mutants have distinctive personalities.
(a) A phenotype reminiscent of Eeyore, the persistent pessimist in Milne′s Winnie the Pooh, arises from the loss of which che gene?
(b) The loss of a different che gene gives a phenotype resembling Voltaire′s Candide, eternally optimistic, convinced that “this is the best of all possible worlds.” Which gene? Ans: (a) Loss of che B (the gene for the methylesterase) would give receptors that are fully methylated. The bacterium would persistently tumble, always convinced that it is heading in the wrong direction. (b) Loss of che R (the gene for methyltransferase) would give receptors that are 29

completely demethylated. The bacterium would swim smoothly all the time, confident that it is heading for greener pastures.
4. Truncated rods(切短的视杆细胞). A retinal rod outer segment can be severed near its base to produce a fragment that carries out phototransduction if provided with GTP and ATP. Small molecules and proteins can be diffused into a truncated rod because its base is not sealed. Predict the effect of each of the following manipulations on the response to a light flash that excites a few rhodopsin molecules:

(a) Addition of an inhibitor of rhodopsin kinase.
(b) Addition of an inhibitor of the cGMP phosphodiesterase.
(c) Addition of EGTA.
(d) Addition of GTPγS (a hydrolysis-resistent analog of GTP). Ans: (a) The photoresponse would be prolonged because deactivation of R* would be inhibited. (b) The amplitude of the photoresponse would be decreased because less cGMP would be hydrolyzed. (c) EGTA would lower the calcium level, which would increase guanylate cyclase activity. The photoresponse would be diminished by the higher cGMP level. (d) The photoresponse would be markedly prolonged because GTPγS bound to tranducin cannot be hydrolyzed. The phosphodiesterase would be persistently activated.

5. Color blindness (色盲). In 1794, John Dalton presented a lecture entitled“”Extraordinary Facts Relating to the Vision of Colours” to the Manchester Literary and Philosophical Society. Dalton, who later developed the atomic theory, emphasized how his own perception of color differed from that of most people: “… that part of the image which others call red appears to me little more than a shade or defect of light; after that the orange, yellow, and green seem one color which descends pretty uniformly from intense to a rare yellow, …” Dalton, like 8% of males and 0.6% of females generally, was red-green color blind. This form of color blindness is rarer in females because it is caused by a recessive sex-linked mutation. Recent molecular genetic studies have shown that genes for the red and green receptors are next to each other on the X chromosome. In some color blind men, the green gene is found in place of the red one. Predict the absorption spectrum of such a hybrid gene. Why would it lead to altered color vision?

Ans: The absorption maximum is likely to be intermediate between that of the red and green receptors. The capacity to distinguish hues between red and green depends on having receptors with well-separated absorption maxima. The separation between the hybrid and green receptors is less than in the normal case. Hence, color vision is impaired. 6. Inhibitory constraints(抑制性限制). Many proteins in signal transduction pathways are activated by the removal of an inhibitory constraint. Give three examples of this recurring mechanism.

Ans: cGMP phosphodiesterase, protein kinase A, and protein kinase C are activated by removal of an inhibitory constraint.
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7. Fulminating diarrhea(爆发性腹泻). Cholera toxin catalyzes the transfer of an adenosine diphosphate ribose unit to an arginnine residue of the stimulatory G protein. This modification, termed ADP-ribosylation, blocks the hydrolysis of GTP bound to the α subunit of Gs. How does cholera toxin induce a diarrhea that can be lethal if body fluids are not quickly replaced?( Hint: phosphorylation of chloride channels in intesyinal epithelial cells increases the efflux of Cl- and water.)

Ans: ADP-ribosylation by cholera toxin leads to persistent activation of adenylate cyclase because Gs-GTP is very slowly hydrolyzed to Gs-GDP. The abnormally high cAMP level leads to complete activation of protein kinase A, which then phosphorylates chloride channels. The proportion of open channels is much higher than normal, which leads to excessive loss of salt and water.

8. Calcium pump regulation(钙泵调节). The Ca2+-ATPase pump in the plasma membrane of eukaryotic cells is activated by Ca2+-calmudulin. The C-terminal region of this 138-kd pump contains a sequence similar to one in calmodulin.

Ca2+-ATPase

Glu-Glu-Glu-Ile-Phe-Glu

Calmodulin

Glu-Glu-Glu-Ile-Arg-Glu

How might this sequence make the pump responsive to activation by Ca2+-calmodulin? Ans: The Ca2+-ATPase pump contains a region (D) that binds this internal calmodulinlike sequence (C). The catalytic site is blocked by the formation of the C-D complex. Ca2+-calmodulin (C) disrupts this internal complex and activates the pump by binding D. Thus, calcium-calmodulin reverses the effect of an autoinhibitory region of its target. Once again we see that stimulation of one protein by another can be mediated by relief of a built-in inhibitory constraint. 9. Reversing oncogenesis (逆转癌发生). Recepter tyrosine kinases are potential drug targets in 31

cancer therapy. How might the effect of an oncogenic mutation be reversed by a small molecule? Propose a search strategy for such a drug.
Ans: One strategy is to search for compounds that block the dimerization of a receptor tyrosine kinase. Inhibitors of dimerization should block autophosphorylation.

Chapter 13 Antibodies and T-cell Receptors
1. Energetics and kinetics(能学与动力学). Suppose that the dissociation constant of an Fab-hapten complex is 3×10-7 M at 25℃.
(a) What is the standard free energy of binding?
(b) Immunologists often speak of affinity (Ka), the reciprocal of the dissociation constant, in comparing antibodies. What is the affinity of this Fab?
(c) The rate constant of release of hapten from the complex is 120 s-1. What is the the rate constant for association? What does the magnitude of this value imply about the extent of structural change in the antibody on binding hapten?

Ans: (a) ∆G0′ = -8.9kcal/mol. (b) Ka = 3.3×106 M-1. (c) kon = 4 ×108 M-1 s-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; extensive conformational transitions take time. 2. sugar niche(糖的位置). An antibody specific for dextran, a polysaccharide of glucose residues, was tested for its binding of glucose oligomers. The full binding affinity was obtained when the oligomer contained six glucose residues. How does the size of this site compare with that of the active site of lysozyme?

Ans: The active site in lysozyme also accommodates six sugar residues. The range of sizes of the combining sites of antibodies is like that of active sites of enzymes. 3. A brilliant emitter(聪明的发光者). Certain naphthalene derivatives exhibit a weak yellow fluorescence when they are in a highly polar environment (such as water) and an intense blue fluorescence when they are in a markedly nonpolar environment (such as hexane). The binding of ε–dansyl-lysine to specific antibody is accompanied by a marked increase in its fluorescence intensity and a shift in color from yellow to blue. What does this finding reveal about the hapten-anyibody complex?

Ans: The fluorescence enhancement and shift to the blue indicate that water is largely excluded from the combining site when the hapten is bound. Hydrophobic interactions contribute significantly to the formation of most antigen-antibody complexes. 4. Avidity versus affinity(亲和与亲和力). The standard free energy of binding of Fab derived from an antiviral Ig G is -7kcal/mol at 25℃.

(a) Calculate the dissociation constant of this interaction. (b) Predict the dissociation constant of the intact Ig G assuming that both conbining sites of the 32

antibody can interact with viral epitopes and that the free energy cost of assuming a favorable hinge angle is +3kcal/mol.
Ans: (a) 7.1μM. (b) ∆G0′ is equal to 2× (-7) +3 kcal/mol or -11 kcal/mol, which corresponds to an apparent dissociation constant of 8nM. The avidity (apparent affinity) of bivalent binding in this case is 888 times as much as the affinity of the univalent interaction. 5. Miniantibody(微型抗体). The Fab fragment of an antibody molecule has essentially the same affinity for a monovalent hapten as does intact Ig G.

(a) What is the smallest unit of an antibody that could retain the specificity and binding affinity of the whole protein?
(b) Design a compact single-chin protein that is likely to specifically bind antigen with high affinity.
Ans: (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 proportion 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. 6. Turning on B cells(激活 B 细胞). B lymphocytes, the precursors of plasma cells, are triggered to proliferate by the binding of multivalent antigens to receptors on their surface. The cell-surface receptors are transmembrane immunoglobulins. Univalent antigens, by contrast, do not activate B cells.

(a) What do these findings reveal about the mechanism of B-cell activation? (b) How might antibodies be used to activate B cells?
Ans: (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. Switching to a soluble antibody(转变为可溶性抗体). An unactivated B cell contains membrane-bound immunoglobulin (the μm form of Ig M), whereas an activated cell secretes soluble antibody of the same specificity (μs). The H chains of μm differ from those of the μs in possessing a tranmembrane segment and a short cytosolic tail. The L chains of μm and μs are identical. Furthermore, μm and μs are encoded by the same genes. Propose a mechanism for generating membrane-bound and soluble antibodies of the same specificity. Ans: μs and μm are formed by alternative splicing of the same primary transcript. The mRNA for μm encodes a C-terminal transmembrane anchor that is not present in μs.

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8. An ingenious cloning strategy(睿智的克隆策略). In the cloning of the gene for the α chain of the T-cell receptor. T-cell cDNAs were hybridized eith B-cell mRNAs. What was the purpose of this hybridization step? Can the principle be applied generally? Ans: B cells do not express T-cell receptors. Hybridization of T-cell cDNA with B-cell mRNAs removes cDNA that are expressed in both cells. Hence, the mixture of cDNAs following 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.

Chapter 14

Molecular Motors

1. Diverse motors(多样的马达). Skeletal muscles, eukaryotic cilia, and bacterial flagella use different molecular strategies for the conversion of free energy into coherent motion. Compare and contrast these motility systems with respect to (a) the free energy source, (b) the number of essential components and their identity, and (c) celluar localization. Ans: (a) ATP for skeletal muscle and eukaryotic cilia, and proton-motive force for bacterial flagella. (b) two in each system: myosin-actin, dynein-microtubule, and motA-motB, respectively. (c) Thick and thin filaments of muscle are located in the cytosol. The axosome of cilia is enveloped by the plasma membrane; the contents are continuous with the cytosol. The rotatory motor that drives bacterial flagella is located in the cytoplasmic membrane of the cell; the flagella are extra-cellular appendages.

2. Load-dependent ATPase activity(依赖于卸栽的 ATP 酶活性). Experiments carried out more than a half century ago showed that the ATPase rate of skeletal muscle is slowed when a large load is applied. Propose a molecular explanation for this observation. Ans: A large load resists the movement of the thin filament relative to the thick filament. ADP release is slower with a large load than a small one because the power stroke takes longer to complete.

3. Rigor mortis(死后僵直). Why does the body stiffen after death? Ans: 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. 4. Theme and variation(主题与变化). Calmodulin is typically activated by Ca2+ with an apparent K of about 1µM. Troponin C, however, has a K of about 10µM. What is a likely advantage of troponins′s lower calcium affinity? What price must be paid for this gain? 34

Ans: The dissociation constant K sets an upper limit on koff, the rate constant for calcium release from the binding protein. Because kon<109 M-1s-1 (the diffusion-controlled limit), koff<K 109M-1s-1. For calmodulin, koff <103 s-1, whereas for troponin C, koff <104s-1. Indeed, as predicted, troponin C is a faster switch than calmodulin, which enables skeletal muscle to rapidly contract and relax. The price paid is that a higher concentration of Ca2+ is needed to activate troponin C, and hence more ATP must be spent to pump Ca2+ out of the cytosol to return to the basal state. 5. Smooth muscle(平滑肌). Smooth muscle, in contrast with skeletal muscle, is not regulated by a troponin-tropomyosin mechanism. Instead, vertebrate smooth muscle contraction is controlled by the degree of phosphorylation of its light chains. Phosphorylation induces contraction, and dephosphorylation leads to relaxation. Smooth muscle contraction is triggered by an increase in the cytosolic Ca2+ level. Propose a mechanism for this action of Ca2+ based on your knoeledge of other signal transduction processes.

Ans: The effect is mediated through Ca2+-calmodulin, which stimulates myosin light-chain kinase (MLCK). Phosphoryl groups introduced by MLCK are removed from myosin by a Ca2+-independent phosphatase.

6. Mushroom poison(蘑菇毒素). The importance of rapid turnover of microfilaments in cell motility is highlighted by the action of phalloidin, a toxic cyclic peptide from Amanita phalloides, a highly poisonous mushroom. Propose four ways in which phalloidin could, in principle, interfere with microfilament function.

Ans: In principle, phalloidin could (a) block the assembly of microfilaments by capping the plus (growing) end, (b) sever already formed microfilaments, (c) prevent the disassembly of microfilaments, and (d) interfere with the binding of microfilaments to membranes and other cellular structures. In fact, phalloidin binds to F-actin and blocks depolymerization. 7. Vesicle hauling(囊泡移动). Consider the action of a single kinesin molecule in moving a vesicle along a microtubule track. The force F required to drag a spherical particle of radius α at a velocity υ in a medium having a viscosity η is

F = 6πηαυ
Suppose that a 2-µm bead is carried at a velocity of 0.5µm/s in an aqueous medium (η =0.01 poise =0.01g cm-1s-1)
(a) What is the magnitude of the force exerted by the kinesin molecule? Express the value in dynes(1 dynes =1 g cm s-2).
(b) How much work is performed in 1s? Express the value in ergs (an erg is a dyne cm). (c) A kinesin motor hydrolyzes 40 ATP per second. What is the energy content of ATP, expressed in ergs? Compare this value with the actual work done.

Ans: (a) F= 9.4×10-10 dyne. (b) The work performed is 4.7 × 10-14 erg. (c) The energy content of 35

ATP (based on a ∆G of -12 kcal/mol under typical cellular conditions) is 8.3 × 10-13 erg per molecule. Hence, 40 ATP could yield 3.3 × 10-11 erg, far more than the actual work performed in moving the 2-μm-diameter bead. Thus, the hydrolysis of ATP by a single kinesin motor provides more than enough free energy to power the transport of micrometer-size cargoes at micrometer-per-second velocities.

Chapter 15 Protein Folding and Design
Motif hunt(基序寻找). Camodulin contains four EF hands. They bind calcium ion as in the figure

Find them in the amino acid sequence shown below.

Ans: EF hands can be detected by scanning amino acid sequences for the 29-residue motif. The sequences of the four EF hands in calmodulin are

1. Multiple alanine mutagenesis(多个丙氨酸突变). A solvent-exposed α helix of bacteriophage T4 lysozyme contain three alanine residues. A variant with alanine at 10 consecutive positions of this helix folds normally and has nearly the same enzymatic activity as the wild type. Also, the backbone comformation of this multiple mutant is essentally identical with that of the wild type. (a) What does this mutant reveal about the information content of the amino acid sequence of the wild type enzyme?

(b) Substitution of a buried leucine of this helix with alanine lowers the melting temperature of the protein by 8℃. Why?
(c)

In contrast, replacement of an exposed serine with alanine raises the melting temperature by 1℃. Why?
36

(d) How might multiple alanine mutagenesis be used to simplify the protein folding problem? Ans: (a) This experiment shows that there is considerable redundancy of information in the amino acid sequence of phage lysozyme. (b) Alanine is much smaller than leucine. The absence of three carbon atoms creates a cavity in the hydrophobic core and diminishes the strength of van der Waals interactions. (c) Serine is a helical breaker, whereas is a helix former. (d) The substitution of non-critical residues with alanine focuses attention on positions that are important for folding and stability.

2. Circularly permuted sequences(环形排列的序列). Phosphoribosyl anthranilate isomerase, an enzyme that participates in the synthesis of tryptophan, folds as an αβ barrel in which the amino and carboxy termini are in close proximity. A mutant was constructed in which these termini were joined and new ones were generated at a different site. This mutant has essentially the same sequence as the wild type except that the order of secondary structure lelments with respect to the ends of the chain is different. This circularly permuted variant was enzymatically active. What does this finding reveal about the folding process?

Ans: The order of synthesis of the polypeptide chain is not important in determining the final folded structure.
3. Bispecific antibodies(双重特异性的抗体). The rearrangement of V, D, and J segments to form a functional immunoglobulin gene occurs on only one chromosome because the protein product suppresses the rearrangement of the other chromosome. Hence, all antigen-binding sites made by a single cell are the same. However, antibodies containing two binding sites with different specificities can be produced in the laboratory.

(a) How might two Fab units be joined to produce a bispecific antibody? (b) How might a bispecific antibody be used therapeutically? Ans: (a) The gene for each heavy chain is altered so that it encodes a coiled-coil sequence following the CH1 domain. The Fab units will then associate by forming an α-helical coiled coil. (b) A bispecific antibody has been used to increase the destruction of target cells by killer T cells. One of the Fab units of this antibody was specific for a cell-surface protein on killer T cells, and the other was specific for an interleukin receptor on a target cell. 4. Immunosuppression(免疫抑制). Cyclosporin A, a cyclic 11-residue peptide from fungi, is widely used to inhibit graft rejection following organ transplantation. This immunosuppressant acts by blocking the activation of cytotoxic T lymphocytes. The target is calcineurin, a calcium-sensitive protein phosphatase,. Hoever, cyclosporin must be activated by cyclophilin before it can inhibit the phosphatase. How might cyclophilin activate cyclosporin? Ans: One possibility, a priori is that cyxclophilin catalyzes the cis-trans isomerization of a peptide bond in CsA. The newly formed isomer would then dissociates from cyclophilin, bind to 37

calcineurin, and block its phosphatase activity. Alternatively. Calcineurin might be inhibited by CsA only when the cyclic peptide is bound to cyclophilin. In fact, the inhibitory species is a complex of CsA and cyclophilin. We see here that the simplest mechanism is not always the one utilized by nature.

5. Molecular design(分子设计). Propose a design for a membrane channel. Ans: A four-helix bundle in which the pattern of hydrophobic and hydrophilic residues is shown in the following figure might fold in a membrane and conduct ions. The length of each helical segment should be about 21 residues to span the bilayer.

f
Chapter 16

Metabolism: Basic Concepts and Design

1. Flow of ~P (磷酸基团的流动). What is the direction of each of the following reactions when the reactants are initially present in equimolar amounts? Use the data given in Table 17-1. (a) ATP + creatine

creatine phosphate + ADP

(b) ATP + glycerol

glycerol 3-phosphate +ADP

(c) ATP + pyruvate

phosphoenolpyruvate + ADP

(d) ATP +glucose

glucose 6-phosphate + ADP

Ans: Reactions (a) and (c), to the left; reactions (b) and (d), to the right. 2. A proper inference(恰当的推论). What information do the ΔG0

‫׳‬

data given in Table 17-1

provide about the relative rates of hydrolysis of pyrophosphate and acetyl phosphate? Ans: None whatsoever.
3. A potent donor(强大的供体). Cosider the reaction ATP + pyruvate

Phosphoenolphosphate + ADP
0 ‫׳‬

(a) Calculate ΔG

and K′eq at 25 0C for this reaction, using the data given in Table 17-1.

(b) What is the equilibrium ratio of pyruvate to phosphoenolpyruvate if the ratio of ATP to ADP is 10?

38

Ans: (a) ∆G0′ = +7.5kcal/mol and K′eq =3.06×10-6. (b) 3.28×104. 4. Isomeric equlibrium. Calculate ΔG0 ‫׳‬for the isomerization of glucose-6phosphate to glucose 1-phosphate. What is the equlibrium ratio of glucose 6-phosphate to glucose 1-phosphate at 25 ℃?

Ans: ∆G0′= 1.7kcal/mol. The equilibrium ratio is 17.8.
5. ~CH3(甲基). The formation of acetyl CoA from acetate is an ATP-driven reaction: Acetate +ATP +CoA
0 ‫׳‬

(a)

Calculate ΔG

acetyl CoA +AMP +PPi
for this reaction, using data given in this chapter.

(b) The PPi formed in the above reaction is rapidly hydrolyzed in vivo because of the ubuquity of inorganic pyrophosphatase. The ΔG0

‫׳‬

for the hydrolisis of PPi is -8kcal/mol. Calculate the ΔG0

‫׳‬

for the overall reaction. What effect does the the hydrolysis of PPi have on the formation of acetyl CoA?
Ans: (a) + 0.2 kcal/mol. (b) -7.8 kcal/mol. The hydrolysis of PPi drives the reaction toward the formation of acetyl CoA.
6. Acid strength (酸强度). The pK of an acid is a measure of its proton group-transfer potential. (a)

Derive a relation between ΔG0 and pK.

(b)

What is the ΔG0 for the ionization of acetic acid, which has a pK of 4.8?

Ans: (a) ∆G0 = 2.303 RTpK. (b) 6.53 kcal/mol at 25℃.
7. Activated sulfate(活化硫酸). Fibrinogen contains tyrosine-O-sulfate. Propose an activated form of sulfate that could react in vivo with the aromatic hydroxyl group of a tyrosine residue in a protein to form tyrosine-O-sulfate.

Ans: The activated form of sulfate in most organisms is 3′-phosphoadenosine 5′-phosphosulfate. 8. Dynamics revealed(动力学的阐明). Two chemical species give rise to a single peak in an NMR spectrum if their rate of interconversion is fast compared with the difference between their resonance frequences. For example, ortho phosphate gives a single peak at pH values at which both H2PO4- and HPO42- are present. The chemical shifts of these forms of orthop[hosphate differ by 2.4ppm relative to a standard compound that resonates at 129 Mhz. (a) Calculate the difference between the resonance frequencies of these species. (b) What is the minimum interconversion rate of these species? (c) What is the minimum rate constant for the association of H+ with HPO42-? Ans: (a) 310 Hz. (b) Both the protonation and deprotonation rates must be faster than 310 s-1. (c) The pK′ for the equilibrium of H2PO4- and HPO42- is 7.21. Hence, the dissociation constant K is 6.16×10-8 M. The rate constant for association kon is equal to koff/K. Because koff is greater than 310 s-1, kon must be greater than 5× 109 M-1 s-1.

39

9. Raison d′ĕtre(存在的理由). The muscle of some invertebrates are rich in arginine phosphate (phosphoarginine). Propose a function for this amino acid derivative. How would you test your hypothesis?
Ans: Arginine phosphate in invertebrate muscle, like cratine phosphate in vertebrate muscle, serves as the reservoir of high-potential phosphoryl groups. Arginine phosphate maintains a high level of ATP in muscular exertion. NMR is a choice technique for monitoring the levels of arginine phosphate and ATP in contrasting muscle.

10. Recurring motif(重复出现的主题) . What is the structural feature common to ATP, FAD, NAD+, and CoA?
Ans: An ADP unit (or a closely related derivative, in the case of CoA.

Chapter 17 Carbohydrates
1. Word origin(词汇起源). Account for the origin of the term carbohydrate. Ans: Carbohydrates were originally regarded as hydrates of carbon because the empirical formula of many of them is (CH2O)n.
2. Couples(成对关系). Indicate whether each of the following pairs of sugars consists of anomers, epimers, or an aldoseketose pair:
(a) D-glyceraldehyde and dihydroxyacetone.
(b) D-glucose and D-mannose.
(c) D-glucose and D-fructose.
(d) α-D-glucose and β-D-glucose.
(e) D-ribose and D-ribulose.
(f) D-galactose and D-glucose.
Ans: (a) Aldose-ketose; (b) epimers; (c) aldose-ketose; (d) anomers; (e) aldose-ketose; and (f) epimers.
3. Tollen′s test(托伦实验). Glucose and other aldoses are oxidized by an aqueous solution of a silver-ammonia complex. What are the reaction products?
Ans: Aldose are converted into aldonic acids; the aldehyde group of the sugar is oxidized to a carboxylate.
4. Mutarotation(变旋现象). The specific rotations ([α]D) of the α and β anomers of D-glucose are +112 degrees and +18.7 degrees. [α]D is defined as the observed rotation of light of wavelength 589nm (the D line of a sodium lamp) passing through 10 cm of a 1g/ml solution of a sample. 40

When a crystalline sample of α-D-glucopyranose is dissolved in water, the specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. At equilibrium, what are the proportions of α and β anomers? Assume that the concentration of the open-chain form is negligible.

Ans: The proportion of α anomer is 0.36, and that of the β anomer is 0.64. 5. Telltale adduct(说明问题的加成物). Glucose reacts slowly with hemoglobin and other proteins to form covalent compounds. Why is glucose reactive? What is the nature of the adduct formed.

Ans: Glucose is reactive because of the presence of an aldehyde group in its open-chain form. The aldehyde group slowly condenses with amino groups to form Shiff base adducts. 6. Periodate cleavage(高碘酸切割). Compounds containing hydroxyl groups on adjacent carbon atoms undergo carbon-carbon cleavage when treated with periodate ion(IO4-). How can this reaction be used to distinguish between pyranosides and furanosides? Ans: A pyranoside reacts with two molecules of periodate; formate is one of the products. A furanoside reacts with only one molecule of periodate; formate is not formed. 7. Oxygen source ( 氧 的 来 源 ) . Does the oxygen atom attached to C-1 in methyl α-D-glucopyranoside come from glucose or methanol?

Ans: From methanol.
8. Sugar lineup(糖排列). Identify the four sugars shown below.

Ans: (a) β–D-Mannose; (b) β–D-galactose;

(c) β–D-fructose; (d) β–D-glucosamine.

9. Cellular glue(细胞胶水). A trisaccharide unit of a cell-surface glycoprotein is postulated to play a critical role in mediating cell-cell adhesion in a particular tissue. Desin a simple experiment to test this hypothesis.

Ans: The trisaccharide itself should be a competitive inhibitor of cell adhesion if the trisaccharide unit of the glycoprotein is critical for the interaction.

Chapter 18 Glycolysis
1. Kitchen chemistry(厨房化学). Sucrose is commonly used to preserve fruits. Why is glucose

41

not well suited for preserving foods?
Ans: Glucose is reactive because its open chain form contains an aldehyde group. 2. Tracing carbon atoms(追踪碳原子).

Glucose labeled with

14

C at C-1 is incubated with

glycolytic enzymes and necessary cofactors.
(a) What is the distribution of

14

C in the pyruvate that is formed? (Assume that the

interconversion of glyceraldehyde 3-phosphate and dihydroxyacetone phosphate is very rapid compared with the subsequent step.)
(b) If the specific activity of the glucose substrate if 10 mCi/mM, what is the the specific activity of the pyruvate that is formed?
Ans: (a) The label is in the methyl carbon of pyruvate. (b) 5mCi/mM. The specific activity is halved because the number of moles of product (pyruvate) is twice that of the labeled substrate (glucose).

3. Lactic fermentation(乳酸发酵). Write a balanced equation for the conversion of glucose into lactate.
(a) Calculate the standard free-energy change of this reaction using the data given in Table 19-2 (p491) and the fact that

ΔG0

‫׳‬

+

is -6kcal for the reaction
lactate +NAD+

Pyruvate + NADH +H

(b) What is the free-energy change (ΔG0) of this reaction when the concentrations of reactants are : glucose,5mM; lactate,0.05mM; ATP, 2mM; ADP,0.2mM; and Pi, 1mM? Ans: (a) Glucose + 2Pi + 2ADP→ 2 lactate + 2 ATP. (b) ∆G′ =-27.2 kcal/mol. 4. High potential(高势能).

What is the equilibrium ratio of phosphoenolpyruvate to pyruvate

under standard conditions when [ATP]/[ADP] = 10?
Ans: 3.06×10-5.
5. Hexose-triose equlibrium(己糖与丙糖平衡). What are the equilibrium concentrations of fructose 1,6-bisphosphate, dihydroxyacetone phosphate, and glyceraldehyde 3-phosphate when 1mM fructose 1,6-bisphosphate is incubated with aldoase under standard conditions? Ans: The equilibrium concentration of fructose 1, 6-bisphosphate, dihydroxyacetone phosphate, and glyceraldehyde 3-phosphate are 7.76×10-4 M, 2.24×10-4M, and 2.24×10-4M, respectively. 6. Double labeling(双重标记). 3-phosphoglycerate labeled uniformly with 14C is incubated with 1, 3-BPG labeled with

32

P at C-1. What is the radioisotope distribution of 2, 3-BPG that is

formed on addition of BPG mutase?
Ans: All three carbon atoms of 2, 3-BPG are

14

C-labeled. The phosphorus atom attached to the

C-2 hydroxyl is 32P-labeled.

42

7. An informative analog(能给出信息的类似物). Xylose has the same structure as glucose except that it has a hydrogen atom at C-6 in place of a hydroxymethyl group. The rate of ATP hydrolysis by hexokinase is markedly enhanced by addition of xylose. Why? Ans: Hexokinase has a low ATPase activity in the absence of a sugar because it is in a catalytically inactive conformation. The addition of xylose closes the cleft between the two lobes of the enzyme. However, xylose lacks a hydroxymethyl group, and so cannot be phosphorylated. Instead, a water molecule at the site normally occupied by the C-6 hydroxymethyl group acts as the phosphoryl acceptor from ATP.

8. The far reach of glycolysis (糖酵解的远端). Oxygen transport can be affected in genetic disorders of glycolysis in red cells.
(a) How are glycolysis and oxygen transport linked?
(b) How is oxygen affinity altered by a deficiency of hexokinase? (c) How is oxygen affinity altered by a deficiency of pyruvate kinase? Ans: (a) 2, 3-Bisphosphoglycerate (BPG) lowers the oxygen affinity of hemoglobin. The rate of synthesis of 2, 3-BPG is controlled by the level of 1, 3-BPG, a glycolytic intermediate. (b) The lowered level of glycolytic intermediates leads to less 2, 3-BPG and hence, a higher oxygen affinity. (c) Glycolytic intermediates are present at a higher than normal level. The level of 2, 3-BPG is increased, which makes the oxygen affinity lower than normal. 9. Distinctive sugars(可区别的糖). The intrvenous infusion of fructose into healthy volunteers leads to a two- to fivefold increase in the level of lactate in the blood, a far greater increase than that observed following the infusion of the same amount of glucose. (a) Why is glycolysis more rapid following the infusion of fructose? (b) Fructose has been used in place of glucose for intravenous feeding. Why is this use of fructose unwise?

Ans:

(a)

The

fructose

1-phosphate

pathway

forms

glyceraldehyde

3-phosphate.

Phosphofructokinase, a key control enzyme, is bypassed. Furthermore, fructose 1-phosphate stimulates pyruvate kinase. (b) The rapid, unregulated production of lactate can lead to metabolic acidosis.

10. Catalytic metal(催化性金属). Aldolases in prokaryotes contain a tightly bound divalent metal ion that is essential for catalysis. Propose a catalytic function for this metal ion. Ans: The metal ion serves as an electron sink, as does the protonated Schiff base in animal aldolases.

11. Contrasting inactivators(对比鲜明的钝化剂). Prokaryotic aldolases are inactivated by EDTA, whereas animal aldolases are inactivated by sodium borohydride. Account for this difference.
Ans: EDTA removes the metal ion from the catalytic site of prokaryotic aldolases, whereas sodium 43

borohydride reduces the Schiff base intermediate in catalysis by animal aldolases. 12. Metabolic mutants(代谢突变体). Predict the effect of each of the following on the pace of glycolysis in liver cells.

(a) Loss of the allosteric site for ATP in phosphofructokinase. (b) Loss of the binding site for citrate in phosphofructokinase. (c) Loss of the phosphatase domain of the bifunctional enzyme that controls the level of fructose 2, 6-bisphosphate.

(d) Loss of the binding site for fructose 1, 6-bisphosphate in pyruvate kinase. Ans: (a) Increased; (b) increased; (c0 increased; (d) decreased.

Chapter 19 Citric Acid Cycle
1. Flow of carbon atoms(碳原子的流动). What is the fate of the radioactive label when each of the following compounds is added to a cell extract containing the enzymes and cofactors of the glycolytic pathway, the citric acid cycle, and the pyruvate dehydrogenase complex? (a) H3 14C -CO-COO-

(b) H3C-14C O-COO-

(c) H3C-CO-14C OO-

(d) H314C -CO-S-CoA

(e) Glucose 6-phosphate labeled at C-1.
Ans: (a) After one round of the citric acid cycle the label emerges in C-2 and C-3 of oxaloacetate. (b) The label emerges in CO2 in the formation of acetyl CoA from pyruvate. (c) After one round of the citric acid cycle, the label emerges in C-1 and C-4 of oxaloacetate. (d) and (e) Same fate as in (a).

2. C2 +C2 →C4.
(a) Which enzymes are required to get net synthesis of oxaloacetate from acetyl CoA? (b) Write a balanced equation for the net synthesis.
(c) Do mammalian cells contain the requisite enzymes?
Ans: (a) Isocitrate lyase and malate synthase are required in addition to the enzymes of the citric acid cycle. (b) 2 Acetyl CoA + 2NAD+ + FAD + 3H2O→ oxaloacetate + 2CoA + 2NADH + FADH2 + 3H+. (c) No. Hence, mammals cannot carry out the net synthesis of oxaloacetate from acetyl CoA.

3. Driving force(驱动力). What is the ΔG0 ‫׳‬for the complete oxidation of the acetyl unit of acetyl CoA by the citric acid cycle?
Ans: -9.8 kcal/mol.
4. Probing stereospecificity(探测立体特异性). A sample of deuterated reduced NAD was prepared by incubating H3C-CD2-OH and NAD+ with alcohol dehydrogenase. This reduced coenzyme was added to a solution of 1,3-BPG and glyceraldehyde 3-phosphate dehydrogenase. 44

The NAD+ formed by this second reaction contained one atom of deuterium, whereas glyceraldehyde 3-phosphate, the other product, contained none. What does this experiment reveal about the stereospecificity of glyceraldehyde 3-phosphate dehydrogenase? Ans: The coenzyme stereospecificity of glyceraldehde 3-phosphate dehydrogenase is the opposite of that of alcohol dehydrogenase (Type B versus type A, respectively). 5. A potent inhibitor(强抑制剂). Thiamine thiazolone pyrophosphate binds to pyruvate dehydrogenase about 20,000 times as strongly as does thiamine pyrophosphate, and it conpetitively inhibits the enzyme. Why?

Ans: Thiamine thiazolone pyrophosphate is a transition state analog. The sulfur-containing ring of this analog is uncharged, and so it closely resembles the transition state of the normal coenzyme in thiamine0catalyzed reactions (e.g., the uncharged resonance form of hydroxyethyl –TPP). 6. Making oxaloacetate(制造草酰乙酸). The oxidation of malate by NAD+ to form oxaloacetate is a highly endergonic reaction under standard conditions (ΔG0

‫׳‬

= +7 kcal/mol). The reaction

proceeds readily under physiological conditions.
(a) Why?
(b) Assuming an [NAD+]/[NADH] ratio of 8 and a pH of 7, what is the lowest [malate]/[oxaloacetate] ratio at which oxaloacetate can be formed from malate? Ans: (a) The steady-state concentrations of the products are low compared with those of the substrates. (b) The ratio of malate to oxaloacetate must be greater than 1.75×104 for oxaloacetate to be formed.

7. Theme and variation(主题与变化). Propose a reaction mechanism for the condensation of acetyl CoA and glyoxylate in the glyoxylate cycle of plants and bacteria. Ans: The enol intermediate of acetyl CoA attacks the carbonyl carbon atom of glyoxylate to form a C-C bond. This reaction is like the condensation of oxaloacetate with the enol intermediate of acetyl CoA in the reaction catalyzed by citrate synthase. Glyoxylate contains a hydrogen atom in place of the –CH2COO- of oxaloacetate; the reactions are otherwise nearly identical. 8. Phosphatase action(磷酸酶作用). Hydrolysis of the phosphoryl group attached to Ser 113 in the inactivated form of isocitrate dehydrogenase proceeds very slowly if the phosphatase is added

45

without ADP. Propose a mechanism for phosphatase action that accounts for the accelerating effect of ADP. (Hint: recall that the kinase and phosphatase activities of the modifying enzymes are mediated by the same active site.)

Ans: The phosphoryl group on Ser 113 is transferred to bound ADP and then to water, rather than directly to water.
10. Déjà vu(记忆错误). Which other regulatory enzyme in metabolism contains kinase and phosphatase activities on the same polypeptide chain? How does it differ from the one that acts on bacteria isocitrate dehydrogenase?

Ans: The kinase that synthesizes fructose 2, 6-bisphosphate (F-2, 6-BP) and the phosphatase that hydrolyzes F-2,6-BP to fructose 6-phosphate are present on a single polypeptide chain. However, in contrast with the bifunctional enzyme that modifies bacterial isocitrate dehydrogenase, the kinase and phosphatase active sites are on different domains.

Chapter 20 Oxidative Phosphorylation
1. Energy harvest(能量收获). What is the yield of ATP when each of the following substrates is completely oxidized to CO2 by a mammalian cell homogenate? Assume that glycolysis, the citric acid cycle, and oxidative phosphorylation are fully active.

(a)

Pyruvate

(b) Lactate

(d) Phosphoenolpyruvate

(c) Fructose 1, 6-bisphosphate
(e) Galactose

(f) Dihydroxyacetone phosphate.

Ans: (a) 12.5; (b) 14; (c) 32; (d) 13.5; (e) 30; (f) 16.
2. reference state(参照状态). The standard oxidation-reduction potential for the reduction of O2 to H2O is given as +0.82 V in biochemistry textbooks. However, the value given in textbooks of chemistry is 1.23V. Accounts for this difference.

Ans: Biochemists use E0′, the value at pH 7, whereas chemists use E0, the value in 1 M H+, the prime denotes that pH 7 is the standard state.
3. Potent poisons(强烈的毒物).

What is the effect of each of the following inhibitors on

electron transport and ATP formation by the respiratory chain? (a) Azide

(b) Atractyloside

(c) Rotenone

(d) DNP (e) Carbon monoxide (f) Antimycin A

Ans: (a) Blocks electron transport and proton pumping at site 3. (b) Blocks electron transport and ATP synthesis by inhibiting the exchange of ATP and ADP across the inner mitochondrial membrane. (c) Blocks electron transport and proton pumping at site 1. (d) Blocks ATP synthesis without inhibiting electron transport by dissipating the proton gradient. (e) Blocks electron transport and proton pumping at site 3. (f) Blocks electron transport and proton pumping at site 2. 4. P:O ratios(磷:氧比). The number of molecules of inorganic phosphate incorporated into organic form per atom of oxygen consumed, termed the P:O ratio, was frequently used as an index 46

of oxidative phosphorylation.
(a) What is the relation of P:O ratio to the ratio of the number of protons translocated per electron pair(H+/2e-) and the ratio of the number of protons needed to synthesize an ATP and transport it to the cytosol(~P/H+)?

(b) What are the P:O ratios for electrons donated by matrix NADH and by succinate? Ans: (a) The P:O ratio is equal to the product of (H+/2e-) and (~P/H+). Note that the P:O ratio is identical with (P:2e-) ratio. (b) 2.5 and 1.5, respectively. 5. Thermodynamic constraint(热力学限制). Compare the ΔG0 ‫׳‬for the oxidation of succinate by NAD+ and by FAD. Use the data given in Table 21-1, and assume that E0′ for the FAD/FADH2 redox couple is nearly 0V. Why is FAD rather than NAD+ the electron acceptor in the reaction catalyzed by succinate dehydrogenase?

Ans: ∆G0′ is + 16.1 kcal/mol for oxidation by NAD+ and + 1.4 kcal/mol for oxidation by FAD. The oxidation of succinate by NAD+ is not thermodynamically feasible. 6. Cyanide antidote(氰化物解毒剂). The immediate administration of nitrate is a highly effective treatment for cyanide poisoning. What is the basis for the action of this antidote? (Hint: Nitrite oxidizes ferro-hemoglobin to ferri-hemoglobin).

Ans: Cyanide can be lethal because it binds to the ferric form of cytochrome oxidase anfd thereby inhibits oxidative phosphorylation. Nitrite converts ferrohemoglobin to ferrihemoglobin, which also binds cyanide. Thus, ferrihemoglobin competes with cytochrome oxidase for cyanide. This competition is therapeutically effective because the amount of ferrihemoglobin that can be formed without impairing oxygen transport is much greater than the amount of cytochrome oxidase. 7. Chiral clue(手性线索). ATPγS, a slowly hydrolyzed analog of ATP, can be used to probe the mechanism of phosphoryl transfer reactions. Chiral ATPγS has been synthesized containing 18O in a specific γ position and ordinary 16O elswhere in the molecule. Hydrolysis of this chiral molecule by ATP synthase in 17O-enriched water yields inorganic [16O, 17O, 18O] thiophosphate having the absolute configuration shown as the following. In contrast, hydrolysis of this chiral ATPγS by a calcium-pumping ATPase from muscle gives thiophosphate of the opposite configuration. What is the simplest interpretation of these data?

Ans: The absolute configuration of thiophosphate is opposite to that of ATP in the reaction catalyzed by ATP synthase. This result is consistent with an in-line phosphoryl transfer reaction 47

occurring in a single step. The retention of configuration in the Ca2+-ATPase reaction points to two phosphoryl transfer reactions—inversion by the first, and a return to the starting configuration by the second. The Ca2+-ATPase reaction proceeds by a phosphorylated enzyme intermediate. 8. Currency exchange(货币交换). For a proton-motive force of 0.2V (matrix negative), what is the maximum [ATP]/[ADP][Pi] ratio compatible with ATP synthesis? Calculate this ratio three times, assuming that the number of protons translocated per ATP formed is 2, 3 and 4 and that the temperature is 25℃.

Ans: The available free energy from the translocation of 2, 3 and 4 protons is -9.23, -13.8, and -18.5 kcal, respectively. The free energy consumed in synthesizing a mole of ATP under standard conditions is 7.3 kcal. Hence, the residual free energy of 1.93, -6.5, and -11.2 kcal can drive the synthesis of ATP until the [ATP] / [ADP] [Pi] ratio is 26.2, 6.51×104, and 1.62×108, respectively. Suspensions of isolated mitochondria synthesize ATP until this ratio is greater than 104, which shows that the number of protons translocated per ATP synthesized is at least 3. 9. Runaway mitochondria(衰竭的线粒体). Suppose that the mitochondria of a patient oxidized NADH irrespective of whether ADP was present. The P:O ratio for oxidative phosphorylation by these mitochondria was less than normal. Predict the likely symptoms of this disorder. Ans: Such a defect (called Luft′s syndrome) was found in a 38-year-old woman who was incapable of performing prolonged physical work. Her basal metabolic rate was more than twice normal, but her thyroid function was normal. A muscle biopsy showed that her mitochondria were highly variable and atypical in structure. Biochemical studies then revealed that oxidation and phosphorylation were not tightly coupled in these mitochondria. In this patient, much of the energy of fuel molecules was converted into heat rather than ATP. 10. An essential residue(一个必需残基). Conduction of protons by the F0 unit of ATP synthase is blocked by modification of a single side chain by dicyclohexylcarbodiimide. What are the most likely targets of action of this reagent? How might you use site-specific mutagenesis to determine whether this residue is essential for proton conduction?

Ans: Dicyclohexylcarboiimide reacts readily with carboxyl groups, as was

discussed earlier in

regard to its use in peptide synthesis. Hence, the most likely targets are aspartate and glutamate side chains. In fact, aspartate 61 of subunit c of E. coli F0 is specifically modified by this reagent. Conversion of this aspartate by site-specific mutagenesis also eliminated proton conduction. 11. Recycling device(废物利用设施). The cytochrome b component of cytochrome reductase enables both electrons of QH2 to be effectively utilized in generating a proton-motive force. Cite another recycling device in metabolism that brings a potentially dead-end reaction product back into the mainstream.

48

Ans: Triose phosphate isomerase converts dihydroxyacetone phosphate (a potential dead end) into glyceraldehyde 3-phosphate ( a mainstream glycolytic intermediate). 12. Crossover point(交叉点). The precise site of action of a respiratory-chain inhibitor can be revealed by the crossover technique. Britton Chance devised elegant spectroscopic methods for determining the proportions of the oxidized and reduced forms of each carrier. This is feasible because they have distinctive absorption spectra, as illustrated for cytochrome c (in the following figure). You are given a new inhibitor and find that its addition to respiring mitochondria causes the carriers between NADH and QH2 to become more reduced, ant those between cytochrome c and O2 to become more oxidized. Where does your inhibitor act?

Ans: This inhibitor (like antimycin A) blocks the reduction of c1 by QH2, the crossover point. Chapter 21 Pentose Phosphate pathway and Gluconeogenesis
1. tracing glucose(追踪葡萄糖). Glucose labeled with 14C at C-6 is added to a solution containing the enzymes and cofactors of the oxidative branch of the pentose phosphate pathway. What is the fate of the radioactive label?

Ans: The label emerges at C-5 of ribulose 5-phosphate.
2. Recurring decarboxylations(反复出现的脱羧作用). Which reaction in the citric acid cycle is most analogous to the oxidative decarboxylation of 6-phosphogluconate to ribulose 5-phosphate? What kind of enzyme-bound intermediate is formed in both ractions? Ans: Oxidative decarboxylation of isocitrate to α-ketoglutarate. A β-keto acid intermediate is formed in both reactions.

3. Carbon shuffling(碳重排). Ribose 5-phosphate labeled with 14C at C-1 is added to a solution containing transketolase, transaldolase, phosphopentose, epimerase, phosphopentose isomerase, and glyceraldehyde 3-phosphate. What is the distribution of the radioactive label in the erythrose 4-phosphate and fructose 6-phosphate that are formed in this reaction mixture? Ans: C-1 and C-3 of fructose 6-phosphate are labeled, whereas erythrose 4-phosphate is not labeled.

49

4. Synthetic stoichiometries(合成反应的计量关系). What is the stoichiometry of synthesis of (a) Ribose 5-phosphate from glucose 6-phosphate without the concommitant generation of NADPH?
(b) NADPH from glucose 6-phosphate without the concommitant formation of pentose sugars? Ans: (a) 5 Glucose 6-phosphate + ATP→6 ribose 5-phosphate + ADP + H+. (b)

Glucose 6-phosphate + 12 NADP+ + 7 H2O→ 6 CO2 + 12 NADPH + 12H+ +Pi

5. Developmental transition(发育过渡). During fetal development, the proportion of H chains in lactate dehydrogenase increases and that of M chains decreases. Propose a selective advantage of this shift in isozyme pattern during development.

Ans: The lactate level in the maternal circulation, and hence in the fetal circulation, increases during pregnancy because the mother is carrying a growing fetus with its own metabolic demands. The shift to H4 in the fetal heart enables the fetus to use lactate as a fuel. Consequently, there is less need for gluconeogenesis by the mother.

6. Trapping a reactive lysine(捕获反应性赖氨酸). Design a chemical experiment to identify the lysine residue that forms a Shiff base at the active site of transaldolase. Ans: Form a Schiff base between a ketose substrate and transaldolase, reduce it with tritiated NaBH4, and fingerprint the labeled enzyme.

7. Reductive power(还原力). What ratio of NADPH to NADP+ is required to sustain [GSH] =10 mM and [GSSG]=1mM? Use the redox potentials given on page 532. Ans: ∆E0′ for the reduction of glutathione by NADPH is +0.09V. Hence, ∆G0′ is -4.15 kcal/mol, which corresponds to an equilibrium constant of 1126. The required [NADPH] / [NADP+] ratio is 8.9×10-2.

8. Metabolic mutant(代谢突变体). What are the likely consequences of a genetic disorder rendering fructose 2, 6-bisphosphatase in liver less sensitive ti regulation by fructose 2,6-bisphosphate?
Ans: Fructose 2,6-bisphosphate, present at high concentration when glucose is abundant, normally inhibits gluconeogenesis by blocking fructose 1, 6-bisphosphatase. In this genetic disorder, the phosphatase is active irrespective of the glucose level. Hence, substrate cycling is increased, which generates heat. The level of fructose 1,6-bisphosphate is consequently lower than normal. Less pyruvate is formed, resulting in less acetyl CoA. In turn, less ATP is formed by the citric acid cycle and oxidative phosphorylation.

9. Carriers on strings(串联在一起的载体). A lysine side chain serves as a flexible linker that enables the activated carboxyl group of carboxybiotin to rotate from one active site to another. Name another activated carrier on a molecular string and cite its function. 50

Ans: The lipoic acid is covalently attached to a lysine side chain in the dihydrolipoyl dehydrogenase component (E2) of pyruvate dehydrogenase. This flexible unit carries acetyl groups from bound thamine pyrophosphate to coenzyme A.

10. Biotin snatcher(生物素抢夺者). Avidin(亲和素), a 70-kd protein in egg-white, has very high affinity for biotin. In fact, it is a highly specific inhibitor of biotin enzymes. Which of the following conversions would be blocked by the addition of avidin to a cell homogenate? (a) glucose to pyruvatate

(b) Pyruvate to glucose

(d) Glucose to ribose 5-phosphate

(c) Oxaloacetate to glucose

(e) Pyruvate to oxaloacetate (f) Ribose 5-phosphate to

glucose
Ans: Reactions (b) and (e) would be blocked.

Chapter 22

Glycogen Metabolism

1. Carbohydrate conversion(碳水化合物转变). Write a balanced equation for the formation of glycogen from galactose.
Ans: Galactose +ATP +UTP +H2O+ glycogen n→glycogenn+1 + ADP + UDP +2Pi + H+. 2. Telltale products(能说明问题的产物). A sample of glycogen from a patient with liver disease is incubated with orthophosphate, phosphorylase, the transferase, and the debranching enzyme (α-1, 6-glucosidase). The ratio of glucose 1-phosphate to glucose formed in this mixture is 100. What is the most likely enzymatic deficiency in this patient? Ans: There is a deficiency of the branching enzyme.

3. Excessive storage(过量储藏). Suggest an explanation for the fact that the amount of glycogen in type I glycogen-storage disease (von Gierke′s disease) is increased. Ans: The high level of glucose 6-phosphate in von Gierke′s disease, resulting from the absence of glucose 6-phosphatase or the transporter, shifts the allosteric equilibrium of phosphorylated glycogen synthase toward the active form.

4. A shattering experience(崩解经历). Crystals of phosphorylase a grown in the presence of glucose shatter when a substrate such as glucose 1-phosphate is added. Why? Ans: Glucose is an allosteric inhibitor of phosphorylase a. Hence, crystals grown in its presence are in the T state. The addition of glucose 1-phosphate, a substrate, shifts the R→T equilibrium toward the R state. The conformational differences between these states are sufficiently large that the crystal shatters unless it is stabilized by chemical cross-links. The shattering of a crystal caused by an allosteric transition was first observed by Haurowitz in the oxygenation of crystals of deoxyhemoglobin.

5. Recouping an essential phosphoryl(重获一个必需磷酸基团). The phosphoryl group on 51

phosphoglucomutase is slowly lost by hydrolysis. Propose a mechanism for restoring this essential phosphoryl group that utilizes a known catalytic intermediate. How might this phosphoryl donor be formed?

Ans: The phosphoryl donor is glucose 1, 6-bisphosphate, which is formed from glucose 1-phosphate in a reaction catalyzed phosphoglucokinase.
6. Hydrophobia(疏水). Why is water excluded from the active site of phosphorylase? Predict the effect of a mutation that allows a water molecule to occasionally enter. Ans: Water is excluded from the active site to prevent hydrolysis. The entry of water could lead to the formation of glucose rather than glucose 1-phosphate. A site-specific mutagenesis experiment is revealing in this regard. In phosphorylase, Tyr 573 is hydrogen-bonded to the 2′-OH of a glucose residue. The ratio of glucose 1-phosphate to glucose product is 9000:1 for the wild type enzyme, and 500:1 for the Phe573 mutant. Model building suggests that a water molecule occupies the site normally filled by the phenolic OH of tyrosine and occasionally attacks the oxocarbonium ion intermediate to form glucose.

7. Two in one(二为一体). A single polypeptide chain houses the transferase and debranching enzyme. Cite a potential advantage of this arrangement.
Ans: The substrate can be handed directly from the transferase site to the debranching site. 8. Metabolic mutants(代谢突变体). Predict the major consequence of each of the following mutations:
(a) Loss of the AMP binding site in muscle phosphorylase.
(b) Mutation of Ser 14 to Ala 14 in liver phosphorylase.
(c) Overexpression of phosphorylase kinase in liver.
(d) Loss of the gene for inhibitor 1 of the protein phosphatase 1. (e) Loss of the gene for the glycogen-targeting subunit of protein phosphatase 1. (f) Loss of the gene for glycogenin.
Ans: (a) Muscle phosphorylase b will be inactive even when the AMP level is high. Hence, glycogen will not be degraded unless phosphorylase is converted into the a form by hormone-induced or Ca2+-induced phosphorylation. (b) Phosphorylase b cannot be converted into the much more active a form. Hence, the mobilization of liver glycogen will be markedly impaired. (c) The elevated level of the kinase will lead to the phosphorylation and activation of glycogen phosphorylase. Little glycogen will be present in the liver because it will be persistently degraded. (d) Protein phosphatase 1 will be continually active. Hence, the level of phosphorylase b will be higher than normal, and glycogen will be less readily degraded. (e) Protein phosphatase 1 will be much less effective in dephosphorylating glycogen synthase and glycogen phosphorylase. Consequently, the synthase will stay in the less active b form, and the phosphorylase will stay in 52

the more active a form. Both changes will lead to increased degradation of glycogen. (f) The absence of glycogenin will block the initiation of glycogen synthesis. Very little glycogen will be synthesized in its absence.

9. Multiple phosphorylation(多重磷酸化). Protein kinase A activates muscle phosphorylase kinase by rapidly phosphorylating its β subunits. The α subunits of phosphorylase kinase are then slowly phosphorylated, which makes α and β susceptible to the action of protein phosphatase 1. What is the functional significance of the slow phosphorylation of α? Ans: The slow phosphorylation of the α subunts of phosphorylase kinase serves to prolong the degradation of glycogen. The kinase cannot be deactivated until its α subunits are phosphorylated. The slow phosphorylation of α assures that the kinase and, in turn, phosphorylase stay active for a defined interval.

Chapter 23 Fatty Acid Metabolism
1. After lipolysis(脂分解后). Write a balanced equation for the conversionb of glycerol into pyruvate. Which enzymes are required in addition to those of the glycolytic pathway? Ans: (a) Glycerol + 2NAD+ + Pi + ADP→ pyruvate + ATP + H2O + 2NADH + H+. (b) Glucerol kinase and glycerol phosphate dehydrogenase.

2. From fatty acid to ketone body(从脂肪酸到酮体). Write a balanced equation for the conversion of stearate into acetoacetate.
Ans: Stearate + ATP +131/2 H2O + 8FAD+8NAD+→41/2 acetoacetate + 141/2 H++8FADH2 + 8NADH + AMP +2Pi.
3.Counterpoint(对比). Compare and contrast fatty acid oxidation and the synthesis with respect to (a) Site of the process.
(b) Acyl carrier.
(c) Reductants and oxidants.
(d) Stereochemistry of the intermediates.
(e) Direction of synthesis or degradation.
(f) Organization of the enzyme system.
Ans: (a) Oxidation in mitochondria, synthesis in the cytosol. (b) Acetyl CoA in oxidation, acyl carrier protein for synthesis. (c) FAD and NADP+ in oxidation, NADPH for synthesis. (d) L-isomer of 3-hydroxyacyl CoA in oxidation, D-isomer in synthesis. (e) Carboxyl to methyl in oxidation, methyl to carboxyl in synthesis. (f) The enzymes of fatty acid synthesis, but not those of oxidation, are organized in a multi-enzyme complex.

4. Sources(来源). For each of the following unsaturated fatty acids, indicate whether the biosynthetic precursors in animals is palmitoleate, oleate, linoleate, or linolenate. (a) 18:1 cis-Δ11

(b) 18:3 cis-Δ6, 9,12

(c) 20:2 cis-Δ11,14
53

(d) 20:3 cis-Δ5,8,11

(e) 22:1 cis-Δ13

(f) 22:6 cis-Δ4,7,10, 13,16,19

Ans: (a) palmitoleate; (b) linoleate; (c) linoleate; (d) oleate; (e) oleate; and (f) linoleate. 5. Tracing carbons(追踪碳原子). Consider a cell extract that actively synthesizes palmitate. Suppose that a fatty acid synthase in this preparation forms one palmitate in about five minutes. A large amount of malonyl CoA labeled with 14C in each carbon of its malonyl unit is suddenly added to this system, and fatty acid synthesis is stopped a minute later by altering the pH. The fatty acids in the supernatant are analyzed for radioactivity. Which carbon atom of the palmitate formed by this system is more radioactive---C-1 or C-14?

Ans: C-1 is more radioactive.
6. Two plus three to make four(二加三得四). Propose a reaction mechanism for the condensation of an acetyl unit with a malonyl unit to form an acetoacetyl unit in fatty acid synthesis. Ans: The enolate anion of one thioester attacks the carbonyl carbon atom of the other thioester to form a C-C bond.

7. Driven by decarboxylation(脱羧反应驱动). What is the the role of decarboxylation in fatty acid synthesis? Name another key reaction in the metabolic pathway that employs this mechanistic motif.
Ans: Decarboxylation drives the condensation of malonyl-ACP and acetyl-ACP. In contrast, the condensation of two molecules of acetyl-ACP is energetically unfavorable. In gluconeogenesis, decarboxylation drives the formation of phosphoenolpyruvate from oxaloacetate. 8. Kinase surfeit(激酶过量). Suppose that a promoter mutation leads to the overproduction of protein kinase A in adipose cells. How would fatty acid metabolism be altered by this mutation? Ans: Adipose–cell lipase is activated by phosphorylation. Hence, overproduction of the cAMP-activated kinase will lead to accelerated breakdown of triacylglycerols and depletion of fat stores.

9. An unaccepting mutant(不能接受的突变体). The serine in acetyl CoA carboxylase that is the target of the AMP-dependent protein kinase is mutated to alanine. What is a likely consequence of this mutation?

Ans: The mutant enzyme would be persistently active because it could not be inhibited by phosphorylation. Fatty acid synthesis would be abnormally active. Such a mutation might lead to obesity.
10. Blocked assets(阻断的资产). The presence of a fuel molecule in the cytosol does not assure that it can be effectively used. Give two examples of how impaired transport of metabolites between compartments leads to disease.

Ans: Carnitine translocase deficiency and glucose 6-phosphate transporter deficiency. 11. Elegant inversion(漂亮的翻转). Peroxisomes have an alternative pathway for oxidizing 54

polyunsaturated fatty acids. They possess a hydratase that converts D-3-hydroxyacyl CoA into trans-Δ2-enoyl CoA. How can this enzyme be used to oxidize CoAs containing a cis double bond at an even-numbered carbon atom (e.g., the cis-Δ12 double bond of linoleate? Ans: In the fifth round of β-oxidation, cis-∆2-enoyl CoA is formed. Dehydration by the classic hydratase yields D-3-hydroxyacyl CoA, the appropriate isomer. Thus, hydratases of opposite stereospecificities serve to epimerize of the 3-hydroxyl group of the acyl CoA intermediate. 12. Covalent catastrophe(共价键灾难). What is a potential disadvantage of having many catalytic sites together on one very long polypeptide chain? Ans: The probability of synthesizing an error-free polypeptide chain decreases as the length of the chain increases. A single mistake can make the entire polypeptide ineffective. In contrast, a defective subunit can be spurned in forming a noncovalent multienzyme complex; the good subunits are not wasted.

Chapter 24 Amino Acid degradation and the Urea Cycle
1. Keto counterparts(酮型对应物). Name the α-keto acid that is formed by transamination of each of the following amino acids:
(a) Alanine

(b) Aspartate

(c) Glutamate

(d) Leucine

(e) Phenylalanine

(f) Tyrosine

Ans: (a) Pyruvate; (b) oxaloacetate; (c) α-ketoglutarate; (d) α-ketoisocaproate; (e) phenylpyruvate; and (f) hydroxy-phenylpyruvate.
2. A versatile building block(一个多用途的构建分子). (a) Write a balanced equation for the conversion of aspartate into glucose by way of oxaloacetate. Which coenzymes participate in this transformation?
(b) Write a balanced equation for the conversion of aspartate into oxaloacetate by way of fumarate.
Ans: (a) Aspartate +α-ketoglutarate + GTP +ATP +2 H2O + NADH +H+→glutamate +CO2 +ADP +GDP +NAD++2Pi. (b) Aspartate +CO2 +NH4+ +3ATP +NAD++4 H2O→oxaloacetate + urea + 2ADP +4Pi +AMP+ NADH +H+.
3. Migratory route(迁移路线). Consider the mechanism of conversion of L—methylmalonyl CoA into succinyl CoA by L-methylmalonyl CoA mutase.
(a) Design an experiment to distinguish between the migration of he COO- group and that of the –CO-S-CoA group in this reaction.
(b) What is the significance of the finding that no tritium is incorporated into succinyl CoA when the mutase reaction is carried out in tritiated water?
Ans: (a) Label the methyl carbon atom of L-methylmalonyl CoA with 14C. Determine the location of 14C in succinyl CoA. The group transferred is the one bonded to the labeled carbon atom. (b) 55

The proton that is abstracted from the methyl group of L-methylmalonyl CoA is directly transferred to the adjacent carbon atom.
4. Effective electron sinks ( 有 效 电 子 库 ) . Pyridoxal phosphate stabilizes carbanionic intermediates by serving as an electron sink. Which other prosthetic group catalyzes reactions in this way?
Ans: Thiamine pyrophosphate.
5. Helping hand(帮手). Propose a role for the positively charged guanidinium nitrogen in the cleavage of argininosuccinate into arginine and fumarate.
Ans: It acts as an electron sink.
6. Telltale tag(能说明问题的标签). Methylmalonyl mutase is incubated with deuterated methylmalonyl CoA. Coenzyme B12 extracted from mutase in this reaction mixture is found to contain deuterium in its 5′-methylene group. Account for the transfer of label from substrate to coenzyme.

Ans: Deuterium is abstracted by the radical form of the coenzyme. The methyl group rotates before hydrogen is returned to the product radical.
7. Divergent pathways (分支途径). Heterolytic cleavage of a C-H bond can yield two types of products. What are they?
Ans: A carbanion or a carbonium ion.
8. Completing the cycle (完成循环). Four ~P are consumed in synthesizing urea according to the stoichiometry. In this reaction, aspartate is converted into fumarate. Suppose that fumarate is converted back to aspartate. What is the resulting stoichiometry of urea synthesis? How many ~P are spent?

Ans:
CO2+NH4++3ATP+NAD++aspartate+3H2O→urea+2ADP+Pi+AMP+PPi+NADH+H++oxaloacetat e. Four ~P are spent.
9. A precise diagnosis(一个准确诊断). The urine of an infant gives a positive reaction with 2, 4-dinitrophenylhydrazine. Mass spectrometry then showed abnormally hogh blood levels of pyruvate, α-ketoglutarate, and the α-keto acids of valine, isoleucine, and leucine. Identify a likely molecular defect and propose a definitive test of your diagnosis. Ans: The mass spectrometric analysis strongly suggests that three enzymes—Pyruvate dehydrogenase,

α-ketoglutarate

dehydrogenase,

and

the

branched-chain

α-keto

dehydrogenase—are deficient. Most likely, the common E3 component of these enzymes is missing or defective. This proposal could be tested by purifying these three enzymes and assaying their capacity to catalyze the regeneration of lipoamide.

56

10. Therapeutic design(治疗设计). How would you treat an infant who is deficient in argininosuccinate synthetase? Which molecules would carry nitrogen out of the body? Ans: Benzoate, phenylacetate, and arginine would be given to supply a protein-restricted diet. Nitrogen would emerge in hippurate, phenylacetylglutamine, and citruline. 11. Sweet hazard(甜食的危害). Why should phenylketonurics avoid using aspatame, an artificial sweetener? (Hint: Aspartame is L-aspartryl-L-phenylalanine methyl ester.) Ans: Aspartame, a dipeptide ester (aspartyl-phenylalanine methyl ester), is hydrolyzed to Laspartate and L-phenylalanine. High levels of phenylalanine are harmful in phenylketonurics. 12. Déjà υu(记忆错误). N-acetylglutamate is required as a cofactor in the synthesis of carbamoyl phosphate. How is it synthesized from glutamate?

Ans: N-acetylglutamate is synthesized from acetyl CoA and glutamate. Once again, acetyl CoA serves as an activated acetyl donor. This reaction is catalyzed by N-acetylglutamate synthase.

Chapter 25 Photosynthesis
1. Electron transfer(电子传递). Calculate the ∆E0′ and ∆G0′ for the reduction of NADP+ by ferredoxin. Use data given in Table 21-1 on p.532.
Ans: ΔE0′ =+0.11V and ΔG0′ =-5.1 kcal/mol.
2. Weed killer(除草剂). Dichlorophenyldimethylurea (DCMU), an herbicide, interferes with photophosphorylation and O2 evolution. However, it does not block the Hill reaction. Propose a site for the inhibitory action of DCMU.

Ans: DCMU inhibits electron transfer in the link between photosystem II and I. O2 evolution can occur in the presence of DCMU if an artificial electron acceptor such as ferricyanide can accept electrons from Q.

3. Infrared harvest(近红外产额). Consider the relation between the energy of a photon and its wavelength.
(a) Some bacteria are able to harvest 1000-nm light. What is the energy (in kcal) of a mole (also called an einstein) of 1000nm photons?
(b) What is the maximum increase in redox potential that can be induced by a 1000nm photon? (c) What is the minimum number of 1000-nm photons needed to form ATP from ADP and Pi? Assume a ∆G of 12 kcal/mol for the phosphorylation reaction. Ans: (a) 28.7kcal/einstein (or 120kJ/einstein). (b) 1.24V. (c) One 1000-nm photon has the free energy content of 2.39ATP. A minimum of 0.42 photons is needed to drive the synthesis of an ATP. 4. Subito fortissimo(突然增强). A chloroplast suspension is exposed to a train of 1-μs light pulses. 57

The dark interval between light pulses is 100 ms. A burst of O2 evolution occurs after every fourth light pulse. Account for this periodicity.
Ans: The formation of O2 is a four-electron process. Each brief light pulse provides one electron for reduction, which is accumulated by the manganese center. 5. Variation on a theme(主题的变异). Sedoheptulose 1,7-bisphosphate is an intermediate in the Calvin cycle but not in the pentose phosphate pathway. What is the enzymatic basis of this difference?

Ans: Aldolase participates in the Calvin cycle, whereas transaldolase participates in the pentose phosphate pathway.
6. Total eclipse(日全食). An illuminated suspension of Chlorella is actively carrying out photosynthesis. Suppose that the light is suddenly switched off. How would the levels of 3-phosphoglycerate and ribulose 1, 5-bisphosphate change during the next minute? Ans: The concentration of 3-phosphoglycerate would increase, whereas that of ribulose 1, 5-bisphosphate would decrease.

7. CO2 deprivation(CO2耗尽). An illuminated suspension of Chlorella is actively carrying out photosynthesis in the presence of 1% CO2. The concentration of CO2 is abruptly reduced to 0.003%. What effect would this have on the levels of 3-phosphoglycerate and ribulose 1, 5-bisphosphate during the next minute?

Ans: The concentration of 3-phosphoglycerate would decrease, whereas that of ribulose 1, 5-bisphosphate would increase.
8. Self-preservation(自我保护). Blue-algae bacteria deprived of a source of nitrogen digest their least essential proteins. Which component of phycobilisomes is likely to be degraded first under starvation conditions?

Ans: Phycoerythrin, the most peripheral protein in the phycobilisome. 9. A potent analog(强力类似物). 2-carboxyarabinitol 1, 5-bisphosphate (CABP) has been useful in studies of rubisco.
(a) Write the structural formula of CABP..
(b) Which catalytic intermediate does it resemble?
(c) Predict the effect of CABP on rubisco.
Ans: (a)

58

(b) CABP resembles the addition compound formed in the reaction of CO2 and ribulose 1, 5-bisphosphate. (c) As predicted, CABP is a potent inhibitor of rubisco. 10. Salvage operation(补救措施). Write a balanced equation for the transmination of glyoxylate to yield glycine.

Ans: Aspartate +glyoxylate →oxaloacetate + glycine.

Chapter 26 Biosynthesis of Membrane Lipids and Steroids
1. Making fats(合成脂肪). Write a balanced equation for the synthesis of a triacylglycerol, starting from glycerol and fatty acids.
Ans: Glycerol +4ATP+3 fatty acids +4H2O→ triacylglycerol +ADP +3AMP+7Pi +4H+. 2. Making phospholipid(合成磷脂). Write a balanced equation for the sythesis of phosphatidyl serine by the denovo pathway, starting from serine, glycerol, and fatty acids. Ans:Glycerol+3ATP+2fatty acids+2H2O+CTP+serine→ phosphatidyl serine +CMP+ADP+2AMP +6Pi +3H+.

3. Activated donors(活化供体). What is the activated reactants in each of these biosyntheses? (a) Phosphatidyl serine from serine.
(b) Phosphatidyl ethanolamine from ethanolamine.
(c) Ceramide from sphingosine.
(d) Sphingomyelin from ceramide.
(e) Cerebroside from ceramide.
(f) Ganglioside GM1 from ganglioside GM2.
(g) Farnesyl pyrophosphate from geranyl pyrophosphate.
Ans: (a) GDP-diacylglycerol; (b) CDP-ethanolamine; (c) acyl CoA; (d) CDP-choline; (e) UDP-glucose or UDP-galactose; (f) UDP-galactose; and (g0 geranyl pyrophosphate.

59

4. Telltale labels(说明问题的标记). What is the distribution of isotopic labeling in cholesterol synthesized from each of these precursors?
(a) Mevalonate labeled with 14C in its carboxyl carbon atom. (b) Malonyl CoA labeled with 14C in its carboxyl carbon atom. Ans: (a and b) None, because the label is lost as CO2.
5. Familial hypercholesterolemia(家族性高胆固醇血症). Several classes of LDL receptor mutations have been identified as causes of this disease. Suppose that you were given cells from patients and antibody specific for the LDL receptor, and had access to an electron microscope. Which categories of mutations are likely to be found?

Ans: (a) No receptor is synthesized. (b) Receptors are synthesized but do not reach the plasma membrane because they lack signals for intracellular transport or do not fold properly. (c) Receptors reach the cell surface, but they fail to bind LDL normally because of defect in the LDL-binding domain. (d) Receptors reach the cell surface and bind LDL, but they fail to cluster in coated pits because of a defect in their carboxyl-terminal region. 6. RNA editing(RNA 编辑). A shortened version (apo B-48) of apolipoprotein B is formed by intestine, whereas the full-length protein (apo B-100) is synthesized by liver. A Gln codon (CAA) is changed into a stop codon. Propose a simple mechanism for this change. Ans: Deamination of cystidine to uridine changes CAA (Gln) into UAA (stop). 7. Inspiration for drug design(药物设计的启示). Some actions of androgens are mediated by dihydrotestosterone, which is formed by reduction of testosterone. This finishing touch is catalyzed by an NADPH-dependent 5α-reductase.

Chromosomal XY males with a genetic deficiency of this reductase are born with a male internal urogenital tract but predominantly female external genitalia. These people are usually reared as girls. At puberty, they masculinize because the testosterone level rises. The testes of these reductase-deficient men are normal, whereas their prostate glands remain small. How might this information be used to treat benign prostatic hypertrophy, a common consequence of the normal aging process in men? A majority of men over age 55 have some degree of prostatic enlargement, which often leads to urinary obstruction.

60

Ans: Benign prostatic hypertrophy can be treated by inhibiting the 5α-reductase. Finasteride, the 4-aza steroid analog of dihydrotestosterone, competitively inhibits the reductase but does not act on androgen receptors. Patients taking finasteride have a markedly lower plasma level of testosterone. The prostate becomes smaller, whereas testosterone-dependent processes such as fertility, libido, and muscle strength appear to be unaffected.

8. Drug idiosyncrasies(药物特异反应性). Debrisoquine, a β-andrenergic blocking agent, has been used to treat hypertension. The optimal dose varies greatly (20-400 mg daily) in a population of patients. The urine of most patients taking the drug contains a high level of 5-hydroxydebrisoquine. However, those most sensitive to this drug (about 8% of the group studied) excrete debrisoquine and very little of the 4-hydroxy derivative. Propose a molecular basis for this drug idiosyncrasy. Why shoul caution be excercised in giving other drugs to patients who are very sensitive to debrisoquine>

Ans: Patients who are most sensitive to debrisoquine have a deficiency of a liver P450 enzyme encoded by a member of the CYP2 subfamily. This characteristic is inherited as an autosomal recessive trait. The capacity to degrade other drugs may be impaired in people who hydroxylate debrisoquine at a slow rate because a single P450 enzyme usually handles a broad range of substrates.

9. Congenital adrenal hyperplasia ( 先 天 性 肾 上 腺 肿 大 ) . A genetic deficiency of the 21-hydroxylase causes virilization and enlargement of the adrenals. Which other enzymatic defects in the synthesis of corticoids would lead to similar symptoms? Ans: Deficiencies of 11β-hydroxylase, 17α-hydroxylase, 3β-dehydrogenase, and desmolase also lead to adrenal hyperplasia.

10. Removal of odorants(嗅觉味物质的除去). Many odorant molecules are highly hydrophobis and concentrate within the olfactory epithilium. They would give a persistent signal independent of their concentration in the environment if they were not rapidly modified. Propose a mechanism 61

for converting hydrophobic odorants into water-soluble derivatives that can be rapidly eliminated. Ans: Many hydrophobic odorants are deactivated by hydroxylation. O2 is activated by a cytochrome P450 monooxygenase. NADPH serves as the reductant. One oxygen atom of O2 goes into the odorant substrate, whereas the other is reduced to water.

Chapter 27 Biosynthesis of Amino Acids and Heme
1. From sugar to amino acid(从糖到氨基酸). Write a balance equation for the synthesis of alanine from glucose.
Ans: Glucose +2ADP+2Pi +2NAD++2glutamate→2 alanine +2 α-ketoglutarate + 2ATP +2NADH +2H2O +2H+.
2.

From air to blood(从空气到血液). What are the intermediates in the flow of nitrogen from

N2 to heme?
Ans: N2→NH4+→ glutamate→serine→glycine→δ-aminolevulinate→porphobilinogen→heme 3. One-carbon transfer(一碳单位转移). Which derivative of folate is a reactant in the conversion of
(a) Glycine into serine.
(b) homocysteine into methionine.
Ans: (a) N5; N10-Methylenetetrahydrofolate; (b) N5-methyltetrahydrofolate. 4. Telltale tag(说明问题的标签). In the reaction catalyzed by glutamine synthetase, an oxygen atom is transferred from the side chain of glutamate to orthophosphate, as shown by 18O-labeling studies. Account for this finding.

Ans: γ-Glutamyl phosphate is a likely reaction intermediate. 5. Therapeutic glycine(治疗性的甘氨酸). Isovaleric acidemia is an inherited disorder of leucine metabolism caused by a deficiency of isovaleryl CoA dehydrogenase. Many infants having this disease die in the first month of life. The administration of large amounts of glycine sometimes leads to marked clinical improvement. Propose a mechanism for this therapeutic action of glycine. Ans: The administration of glycine leads to the formation of isovalerylglycine. This water-soluble conjugate, in contrast with isovaleric acid, is excreted very rapidly by the kidneys. 6. Deprived algae(剥夺必需品的藻类). Blue-green algae (cyanobacteria) form heterocysts when deprived of ammonia and nitrate. They lack nuclei and are attached to adjacent vegetative cells. Heterocysts have photosystem I activity but are entirely devoid of photosystem II activity. What is their role?

Ans: They carry out nitrogen fixation. The absence of photosystem II provides an environment in which O2 is not produced. Recall that the nitrogenase is very rapidly inactivated by O2. 62

7. Cysteine and cystine(半胱氨酸与胱氨酸). Most cytosolic proteins lack disulfide bonds, whereas extracellular proteins usually contain them. Why?
Ans: The cytosol is a reducing environment, whereas the extra-cellular milieu is an oxidizing environment.
8. To and fro(来来往往). The synthesis of δ-aminolevulinate occurs in the mitochondrial matrix, whereas the formation of porphobilinogen takes place in the cytosol. Propose a reason for the mitochondrial location of the first step in heme synthesis.

Ans: Succinyl CoA is formed in the mitochondrial matrix.
9. A dangerous trap(危险的陷阱). In pernicious anemia, much of the body′s tetrahydrofolate is sequestered in the form of N5-methyltetrahydofolate. Why?
Ans: The reduction of N5, N10-methylenetetrahydrofolate to N5-methyltetrahydrofolate is essentially irreversible. Tetrahydrofolate is normally recovered in the conversion of homocysteine to methionine, a methylcobalamin-dependent reaction. In pernicious anemia, vitamin B12 deficiency blocks this methyl transfer reaction. Hence, the folate pool is trapped in the form of N5-methyltetrahydrofolate. Most one-carbon transfers are thereby blocked. 10. Messengers in the wings(舞台两侧的信使). Signal molecules such as epinephrine, nitric oxide, and ethylene are often generated in a small number of steps from a major biomolecule. Each year, new molecules that participate in signal transduction processes are being discovered. Propose two candidates that appeared in this chapter. What kind of receptor might be used to detect your putative messengers?

Ans: CO, which is formed in the breakdown of heme, and HCN, which formed in the synthesis of ethylene, are two plausible candidates. Both bind to heme, CO to the ferrous form and CN- to the ferric form. A heme protein could serve as a receptor for either. It will be interesting to learn whether CO and HCN have messenger roles.

Chapter 28 Biosynthesis of Nucleotides
1. Activated ribose phosphate(活化磷酸核糖). Write a balanced equation for the synthesis of PRPP from glucose via the oxidative branch of the pentose phosphate pathway. Ans: Glucose +2ATP+2NADP++H2O→PRPP +CO2+ADP+AMP+2NADPH+3H+. P

2. Making a pyrimidine(合成嘧啶). Write a balanced equation for the synthesis of orotate from glutamine, CO2, and aspartate.
Ans: Glutamine + aspartate +CO2 +2ATP+NAD+→orotate +2ADP+2Pi+glutamate+NADH +H+. 3. Identifying the donor (找出供体). What is the activated reactant in the biosynthesis of each of 63

these compounds?
(a) Phosphoribosylamine

(b) Carbamoylaspartate

(d) Nicotinate ribonucleotide

(e) Phosphoribosylanthranilate

(c) Oritidylate (from orotate)

Ans: (a,c,d, and e) PRPP; (b) carbamoyl phosphate.
4. Inhibiting purine biosynthesis(抑制嘌呤生物合成). Amidotransferases are inhibited by the antibiotic azaserine (O-diazoacetyl-L-Serine), which is an alalog of glutamine .

Which intermediates in purine biosynthesis would accumulate in cells treated with azaserine? Ans: PRPP and formylglycinamide ribonucleotide.
5. The price of methylation(甲基化的代价). Write a balanced equation for the synthesis of dTMP from dUMP that is coupled to the conversion of serine into glycine. Ans: dUMP +serine +NADPH +H+→ dTMP+NADP++glycine.

6. Sulfa action(磺胺类药物作用机理). Bacterial growth is inhibited by sulfanilamide and related sulfa drugs, and there is a concomitant ribonucleotide of 5—aminoimidazole-4-carboxamide ribonucleotide. This inhibition is reversed by the addition of p-aminobenzoate.

Propose a mechanism for the inhibitory effect of sulfanilamide. Ans: There is a deficiency of N10-formyltetrahydrofolate. Sulfanilamide inhibits the synthesis of folate by acting as an analog of p-aminobenzoate, one of the precursors of folate. 6. A generous donor(慷慨的供体). What are the major biosynthetic reactions that utilize PRPP? Ans: PRPP is the activated intermediate in the synthesis of (a)phosphoribosylamine in the de novo pathway of purine formation, (b) purine necleotides from free bases by the salvage pathway, (c) orotidylate in the formation of pyrimidines, (d) nicotinate ribonucleotide, (e) phosphoribosyl ATP in the pathway leading to histidine, and (f) phosphoribosylanthranilate in the pathway leading to tryptophan.

7. Catalytic duet(催化双重性). Carbamoyl phosphate synthetase from E. coli consists of a small subunit (40kd) and a large one (130kd). The isolated small subunit has glutaminase activity, whereas the isolated large subunit can synthesize carbamoyl phosphate from NH3 but not from glutamine. Furthermore, the isolated small subunit is an ATPase in the presence of bicarbonate. Propose a plausible reaction mechanism for the intact enzyme on the basis of the activities of its 64

separated subunits. How might the holoenzyme be constructed to favor the interplay of subunits? Ans: The small subunit releases ammonia from glutamine. The nascent ammonia then reacts with an activated form of CO2 that is formed by the large subunit. The bicarbonate-dependent ATPase activity implies that this activated species is carbonyl phosphate.. Reactions of this carbonic-phosphoric mixed anhydride with NH3 yields carbamate., which would then react with ATP to give carbamoyl phosphate. The holoenzyme probably contains a channel for the diffusion of nascent NH3 from the small subunit to the large one. Recall that tryptophan synthetase contains a channel for the diffusion of indole from the α to the β subunit. An intriguing possibility is that the catalytic activities of the subunits of carbamoyl phosphate synthetase are coordinated: NH3 production by the small subunit is stimualted when carbonyl phosphate is formed by the large subunit.

8. Déjà υu(记忆错误). In purine biosynthesis, aspartate becomes linked to the carboxylate group of an intermediate. In the next reaction, the caron skeleton of the aspartate unit leaves in the form of fumarate. These reactions are reminiscent of a similar pair of reactions in the degradation of amino acids. Which ones?

Ans: Analogous reactions occur in the urea cycle—from citruline to argininosuccinate, and then to arginine.
9. Pernicious anemia(恶性贫血). Purine biosyhtesis is impaired in vitamin B12 deficiency. Why?
Ans: In vitamin B12 deficiency, methyltetrahydrofolate cannot donate its methyl group to homocysteine to regenerate methionine. Because the synthesis of methyltetrahydrofolate is irreversible, the cell′s tetrahydrofolate will ultimately be converted into this form. No formyl or methylene tetrahydrofolate will be left for nucleotide synthesis. Pernicious anemia illustrates the intimate connection between amino acid and nucleotide metabolism. 10. A devastating deficiency(灾难性的缺陷). Adenosine deaminase (ADA) catalyzes the irreversible deamination of adenosine and 2′-deoxyadenosine to inosine and 2′-deoxyinosine, respectively. Inherited defects of ADA lead to abnormalities in purine necleoside metabolism that are selectively toxic to lymphocytes. The result is a severe combined immunodeficiency disease (SCID).

(a) The intracelluar level of S-adenosylhomocysteine is elevated in ADA-deficient patients. Why? (b) Predict the major consequence of an elevated level of S-adenosylhomocysteine. Ans: (a) S-Adenosylhomocysteine hydrolase activity is markedly diminished in ADA-deficient patients because of reversible inhibition by adenosine and suicide inactivation by 2′-deoxyadenosine.

(b)

The

activated

methyl

cycl

is

blocked

in

these

patients.

S-adenosylhomocysteine is a potent inhibitor of methyl transfer reactions involving 65

S-adenosylmethionine.
12. HAT medium(次黄嘌呤-氨基喋呤或氨甲喋呤-胸腺嘧啶培养基). Mutant cells unable to synthesize nucleotides by salvage pathways are very useful tools in molecular and cell biology. Suppose that cell A lacks thymidine kinase, the enzyme catalyzing the phosphorylation of thymidine to thymidylate, and that cell B lacks hypoxanthine-guanine phoshoribosyl transferase. (a) Cell A and cell B do not proliferate in HAT medium containing hypoxanthine, aminopterin or amethopterin (methotrexate), and thymine. However, cell C formed by the fusion of A and B gows in the medium. Why?

(b) Suppose that you wanted to introduce foreign genes into cell A. Devise a simple means of distinguishing between cells that have taken up foreign DNA and those that have not. Ans: (a) Cell A cannot grow in a HAT medium because it cannot synthesize dTMP either from thymidine or dUMP. Cell B cannot grow in this medium because it cannot synthesize purines by either the de novo pathway or the salvage pathway. Cell C can gorw in a HAT medium because it contains active thymidine kinase from cell B (enabling it to phosphorylate thymidine to dTMP) and hypoxanthine-guanine phosphoribosyl transferase from cell A (enabling it to synthesize purines from hypoxanthine by the salvage pathway). (b) transform cell A with a plasmid containing foreign genes of interest and a functional thymidine kinase gene. The only cells that will grow in a HAT medium are those that have acquired a thymidylate kinase gene; nearly all of these transformed cells will also contain the other genes on the plasmid. 13. Adjunct therapy(辅助治疗). Allopurinol is often given to patients with acute leukemia who are being treated with anticancer drugs. Why is allopurinol used? Ans: These patients have a high level of urate because of the breakdown of nucleic acids. Allopurinol prevents the formation of kidney stones and blocks other deleterious consequences of hyperuricemia by preventing the formation of urate.

13. A hobbled enzyme. Both side-chain oxygen atoms of aspartate 27 at the active site of dihydrofolate reductase form hydrogen bonds with the pteridine ring of folates in the following figure. The importance of this interaction has been assessed by studying two mutants at this position, Asn 27 and Ser 27. The dissociation constant of methotrexate was 0.07nM for the wild type, 1.9nM for the Asn 27 mutant, and 210nM for the Ser 27 mutant, at 25℃. Calculate the free energy of binding of methotrexate by these three proteins. What is the decrease in binding energy resulting from each mutation?

66

Ans: The free energies of binding are -13.8 (wild type), -11.9 (Asn 27), and -9.1 (Ser 27) kcal/mol. The loss in binding energy is 1.9 and 4.7 kcal/mol.
15. Hyperuricemia(高尿酸血症). Many patients with glucose 6-phosphatase deficiency have high serum levels of urate. Hyperricemia can also be induced in normal people by the ingestion of alcohol or by strenuous exercise. Propose a common mechanism that accounts for these findings. Ans: The cytosolic level of ATP in liver falls and that of AMP rises above normal in all three conditions. The excess AMP is degraded to urate.

Chapter 29 Integration of Metabolism
1. Distinctive organs(可区别的器官). What are the key enzymatic differences between liver, muscle, and brain that account for their differing utilization of metabolic fuels? Ans: Liver contains glucose 6-phosphatase, whereas muscle and brain do not. Hence, muscle and brain, in contrast with liver, do not release glucose. Another key enzymatic difference is that 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 brain.

2. Missing enzymes(缺失的酶). Predict the major consequences of each of the following enzymatic deficiencies:
(a) Hexokinase in adipose tissue.
(b) Glucose 6-phosphatase in liver.
(c) Carnitine acyltransferase I in skeletal muscle.
(d) –Glucokinase in liver..
(e) Thiolase in brain.
(f) Kinase in liver that synthesizes fructose 2,6-bisphosphate. Ans: (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 liver following gluconeogenesis. This disorder (called von Gierke′s disease) is characterized by an abnormally high content of glycogen in the liver and a low blood glucose level. (c) A deficiency of carnitine 67

acyltransferase I impairs the oxidation of long-chain fatty 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 thilase 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. Contrasting milieux(对比鲜明的环境). Cerebrospinal fluid has a low content of albumin and other proteins compared with plasma.
(a) What effect does this have on the concentration of fatty acids in the extracellular medium of the brain?
(b) Propose a plausible reason for the selection by brain of glucose rather than fatty acids as the prime fuel.
(c) How does the fuel preference of muscle complement that of the brain? Ans: (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 with fatty acids, which must be carried by tranport proteins such as albumin. Micelles of fatty acids would disrupt membrane structure. (c) Fatty acids , not glucose, are the major fuel of resting muscle.

4. Metabolic energy and power(代谢能与动力). The rate of energy expenditure of a typical 70-kg person at rest is about 70 watts (W), like a light bulb. (a) Express this rate in kilojoules per second and in kilocalories per second. (b) How many electrons flow through the electron transport chain of mitochondria per second under these conditions?

(c) Estimate the corresponding rate of ATP production.
(d) The total ATP content of the body is about 50 grams. Estimate how often an ATP molecule turns over in a person at rest.
Ans: (a) A watt is equal to 1joule (J) per second (0.239 calorie per second). Hence, 70 W is equivalent to 0.07kJ/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 1.14V). Hence, the current is 61.4A, which corresponds to 3.86×1020 electrons per second (1A =1 coulomb/s =6.28×1018 charges /s). (c) Three ATP are formed per NADH oxidized (two electrons). Hence, one ATP is formed per 0.67 electrons transferred. A flow of 3.86×1020 electrons per second therefore leads to the generation of 5.8×1020 ATP per second or 0.96 mmol per second. (d) The molecular weight of ATP is 507. The total body content of ATP of 68

50g is equal to 0.099 mol. Hence, ATP turns over about once per 100 seconds when the body is at rest.
5. Respiratory quotient (RQ)(呼吸熵). This classic metabolic index is defined as the volume of CO2 released divided by the volume of O2 consumed.
(a) Calculate the RQ values for the complete oxidation of glucose and of tripalmitoylglycerol. (b) What do RQ measurements reveal about the contributions of different energy sources during intense exercises? (Assume that protein degradation is negligible.) Ans: (a) The stoichimetry of complete oxidation of glucose is C6H12O6 +6O2→6CO2 +6H2O

And that of tripalmitoyl glucerol is
C51H98O6 +72.5 O2→51 CO2 +49 H2O
Hence, the RQ values are 1.0 and 0.703, respectively.
(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. Camel′s hump(驼峰). Compare the H2O yield from the complete oxidation of 1g glucose with that of 1g tripalmitoylglycerol. Relate these values to the evolutionary selection of the contents in a camel′s hump.

Ans: 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 1.24mmol. The reaction stoichiometries indicate that 6 mol of H2O are produced per mol of glucose oxidized, and 49 mol of H2O per mol of tripalmitoylglycerol oxidized. Hence, the water yields per gram of fuel are 33.3 mmol(0.6g) for glucose, and 60.8mmol (1.09g) 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. A hump consisting mainly of glycogen would be an intolerable burden—far more than the straw that broke the camel′s back!

7. The wages of sin(罪的工价). How long does one have to jog to offset the calories obtained from eating 10 macadamia nuts(澳洲坚果)? (Assume an incremental power consumption of 400 W).
Ans: A typical macadamia nut has a mass of about 2g. Because it consists of mainly fats(~9kcal/g), a nut has a value of about 19kcal. The ingestion of 10 nuts results in an intake of about 180 kcal. As was discussed in problem 4, a power comsumption of 1W corresponds to 0.239 cal per second, and so 400-W running requires 95.6 cal/s, or 0.0956 kcal/s. Hence, one would have to run 1882s, 69

or about 31 min, to spend the calories provided by 10 nuts. 8. Sweet hazard(甜食的危害). Ingesting large amounts of glucose before a marathon might seem to be a good way of increasing the fuel stores. However, experienced runners do not ingest glucose before a race. What is the biochemical reason for their avoidance of this potential fuel? ( Hint: consider the effect of glucose ingestion on the level of insulin.) Ans: A high blood glucose level would trigger the secretion of insulin, which would stimulate the synthesis of glycogen and triacylglycerols. A high insulin level would impede the mobilization of fuel reserves during the marathon.

Chapter 30 DNA Structure, Replication and Repair
1. Activated intermediates(活化中间产物). DNA polymerase I, DNA ligase and topoisomerase I catalyze the formation of phosphodiester bonds. What is the activated intermediates in the linkage reaction catalyzed by each of these enzymes? What is the leaving group? Ans: DNA polymerase I uses deoxyribonucleoside triphosphates; pyrophosphate is the leaving group. DNA ligase uses a DNA-adenylate (AMP joined to 5′-phosphate) as a reaction partner; AMP is the leaving group. Topoisomerase Ⅰ uses a DNA-tyrosyl intermediate (5′-phosphate linked to the phenolic OH); the tyrosine residue of the enzyme is the leaving group. 2. Fuel for a new ligase(新连接酶的燃料). Whether the joining of two DNA chains by known DNA ligases is driven by NAD+ or ATP depends on the species. Suppose that a new DNA ligase requiring a different energy donor is found. Propose a plausible substitute for NAD+ or ATP in this reaction.

Ans: FAD, CoA, and NADP+ are plausible alternatives.
3. AMP-induced relaxation(AMP 诱导的松弛). DNA ligase from E. coli relaxes supercoiled circular DNA in the presence of AMP but not in its absence. What is the mechanism of this reaction, and why is it dependent on AMP?

Ans: DNA ligase relaxes supercoiled DNA by catalyzing the cleavage of a phosphodiester bond in a DNA strand. The attacking group is AMP, which becomes attached to the 5′-phosphoryl group at the site of scission. AMP is required because this reaction is the reverse of the final step in the joining of pieces of DNA.

4. Life in a hot-tub(浴缸中的生命). An archeabacterium (Sulfolobus acidocaldarius) found in acidic hot springs contains a topoisomerase that catalyzes the ATP-driven introduction of positive supercoils into DNA. How might the enzyme be advantageous to this unusual bacterium? Ans: Positive supercoiling resists the unwinding of DNA. The melting temperature of DNA increases in going from negatively supercoiled to relaxed to positively supercoiled DNA. Positive 70

supercoiling is probably an adaptation to high temperature. 5. A cooperative transition(协同性转变). The transition from B-DNA to Z-DNA occurs over a small change in the superhelix density, which shows that the transition is highly cooperative. (a) Consider a DNA molecule at the midpoint of this transition. Are B- and Z- DNA regions frequently intermingled or are there long stretches of each? (b) What does this finding reveal about the energetics of forming a junction between the two kinds of helices?

(c) Would you expect the transition from B- to A-DNA to be more or less cooperative than the one from B- to Z- DNA? Why?
Ans: (a) Long stretches of each occur because the transition is highly cooperative. (b) B-Z junctions are energetically highly unfavorable. (c) A-B transitions are less cooperative than B-Z transitions because the helix stays right-handed at an A-B junction but not at a B-Z junction. 6. A revealing analog(说明问题的类似物). AppNHp, the β,γ-imino analog of ATP, is hydrolyzed very slowly by most ATPases. The addition of AppNHp to DNA gyrase and circular DNA leads to the negative supercoiling of a single molecule of DNA per gyrase. DNA remains bound to gyrase in the presence of this analog. What does this finding reveal about the catalytic mechanism? Ans: ATP hydrolysis is required to release DNA gyrase after it has acted on its DNA substrate. Negative supercoiling requires only the binding of ATP, not its hydrolysis. 7. Mutagenic trail(突变剂线索). Suppose that the single-stranded RNA from tobacco mosaic virus was treated with a chemical mutagen, that mutants were obtained having serine or leucine instead of proline at a specific position, and that further treatment of these mutants with the same mutagen yielded phenylalanine at this position.

Pro→Ser/Leu→Phe
(a) What are the plausible codon assignments for these four amino acids? (b) Was the mutagen 5-bromouracil, nitrous acid, or an acridine dye? Ans: (a) Pro (CCC), Ser (UCC), Leu (CUC), and Phe (UUC). Alternatively, the last base of each of these codons could be U. (b) These C→U mutations are produced by nitrous acid. 8. Revealing tracks(说明问题的轨道). Suppose that replication is initiated in a medium containing moderately radioactive tritiated thymine. After a few minutes of incubation, the bacteria are transferred to a medium containing highly radioactive tritiated thymidine. Sketch the autoradiographic pattern that would be seen for (a) unidirectional replication, and (b)bidirectional replication, from a single origin.

Ans: If replication were unidirectional, tracks with a low grain density at one end and high grain density at the other end would be seen. On the other hand, if replication were bidirectional, the 71

middle of a track would have a low density, as shown below. For E.coli, the grain tracks are denser on both ends than in the middle, indicating that replication is bidirectional.

9. Induced spectrum(诱导光谱). DNA photolyases convert the energy of light in the near UV or visible region (300 t0 500 nm) into chemical energy to break the cyclobutane ring of pyrimidine dimmers. In the absence of substrate, these photoreactivating enzymes do not absorb light of wavelengths longer than 300nm. Why is the substrate-induced absorption band advantageous? Ans: Potentially deleterious side reactions are avoided. The enzyme itself might be damaged by light if it could be activated by light in the absence of bound DNA harboring a pyrimidine dimer. The DNA-induced absorption band is reminiscent of the glucose-induced activation of the phosphotransferase activity of hexokinase.

10. Molecular motors in replication(复制过程中的分子马达). (a) How fast does template DNA spin (expressed in revolutions per second) at an E.coli replication fork?
(b) What is the velocity of movement (in um/s) of DNA polymerase III holoenzyme relative to the template?
(c) Compare these rates with those of flagella rotation in E.coli and sarcomere shortening in vertebrate skeletal muscle.
Ans: (a) 96.2 revolutions per second (1000 nucleotides /s divided by 10.4 nucleotides / turn for B-DNA gives 96.2 rps). (b) 0.34 μm/s(1000 nucleotides/s corresponds to 3400Ǻ/s because the axial distance between nucleotides in B-DNA is 3.4 Ǻ). (c) The rate of rotation of DNA at a replication fork (96.2rps) is nearly the same as that of a bacterial flagellum (100rps). Skeletal muscle sarcomere shortening is about 15-fold as rapid as polymerase movement at a replication fork (5 μm/s. compared with 0.34μm/s). Smooth muscle sarcomere shortening (0.4μm/s) occurs at about the same speed as replication fork movement.

Chapter 31 Gene Rearrangements
1. Unwinding DNA(DNA 解旋). A variety of ATP-driven proteins unwind duplex DNA in E. coli. Name three of them and cite their function.
Ans: Helicase participates in DNA replication. The recB component of the recBCD complex generates single-stranded DNA in general recombination. The recA protein binds to single

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–stranded DNA and catalyzes its invasion of duplex DNA and a switch in base pairing. 2. Linking, writhing, and twisting(连接、圈曲和扭曲). One of the strands of a 1040-bp circular duplex DNA is nicked. Rec A protein is then added in an amount such that an average of 100 protein molecules are bound per DNA duplex. DNA ligase and NAD+ are added to close the nicked circle. RecA protein is then removed. (a) Compare the linking number of the resulting DNA molecule with that of the corresponding relaxed duplex. (b) Compare the hydrodynamic properties of this DNA molecule with that of its relaxed counterpart. Ans: (a) The recA-DNA complex contains 18.6 base pairs per turn, compared with 10.4 for B-DNA. Hence, the linking number (Lk) is 56, compared with 100for the relaxed circle formed in the absence of recA protein. (b) Recall that about 70% of the change in Lk in an isolated circular DNA molecule is expressed in Wr (writhing) and 30% in Tw (twisting). Hence, after removal of recA, Wr is 30.8 less than that of the relaxed circle, which means that the duplex is negatively (right-handed) supercoiled. The degree of supercoiling σ is -0.44, compared with -0.06 for most naturally occurring DNA. Thus, closure in the presence of recA generates a highly supercoiled molecule, which is much more compact than its relaxed counterpart and will sediment much more rapidly.

3. Autorepression(自我阻遏). The lexA protein represses its own synthesis. How does this control feature enable the SOS response to be reversed following repair of DNA? Ans: Proteolysis of the lexA protein at the onset of the SOS response relieves the repression of synthesis of lexA mRNA. DNA repair then leads to a decrease in the amount of the ssDNA-recA complex, the coprotease. Hence, the level of intact lexA increases, which in turn terminates the SOS response.

4. A revealing analog(说明问题的类似物). ATPγS is a hydrolysis-resistant analog of ATP in which a nonesterified oxygen atom of the terminal phosphoryl group is replaced by a sulfur atom. Predict the consequences of using ATPγS instead of ATP in the reactions catalyzed by (a) recA protein and (b) the rec BCD complex.

Ans: (a) In the presence of ATPγS, recA protein binds to DNA but its dissociation is blocked because this analog is not readily hydrolyzed. Strand exchange does not take place. (b) Rec B protein binds to DNA in the presence of this analog buit its release is blocked. Consequently, single-stranded DNA is not formed.

5. Prelude to recombination(重组前奏). Why is recombination favored by negative supercoiling? Ans: Negative supercoiling promotes recombination by inducing strand separation (negative Wr can be changed into negative Tw). Furthermore, negatively supercoiled DNA occupies a much smaller volume than does relaxed DNA. Hence, the effective concentration of sites undergoing recombination is increased by negative supercoiling.

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6. Gene knockout(基因敲除). Devise an experimental strategy for permanently inactivating a specific mammalian gene.
Ans: Homologous recombination between a genomic DNA sequence and a similar cloned DNA sequence introduced into a cell makes it possible to transfer any modification of the cloned gene into the genome of a cell. For example, a stop codon could be placed early in a sequence encoding a protein of interest.

7. Heterodimer from homodimer(从同二聚体到异二聚体). The polymerase domain of the p55 subunit of reverse transcriptase has a rather different conformation from that of the p66 subunit. Furthermore, only one chain of the nascent p66-p66 homodimer is cleaved to generate the functional p55-p66 heterodimer. Initially identical chains usually do not have such different fates. Why does HIV-1 craft the reverse transcriptase heterodimer from two initially identical chains rather than two different ones?

Ans: Viruses make the most of a small genome. The encoding of a different second subunit would require about 2 additional kb of RNA. Replication would be slower and packaging of the nucleic acid more difficult.

Chapter 32 RNA Synthesis and Splicing
1.

Complements(互补结构). The sequence of part of an mRNA is :

5′-AUGGGGAACAGCAAGAGUGGGGCCCUGUCCAAGGAG-3′
What is the sequence of the DNA coding strand? Of the DNA template strand? Ans:

The

sequence

of

the

coding

(+,

sense)

strand

is

5′-ATGGGGAACAGCAAGAGTGGGGCCCTGTCCAAGGAG-3′. And the sequence of template (-, antisense) strand is 3′-TACCCCTTGTCGTTCTCACCCCGGGACAGGTTCCTC-5′. 2. Potent inhibitor(强力抑制剂). Heparin inhibits RNA polymerase by binding to its β′ subunit. Why does heparin bind so effectively to β′?

Ans: Heparin, a glycosaminoglycan, is highly anionic. Its negative charges, like the phosphodiester bridges of DNA templates, bind to lysine and argining residues of β′. 3. A loose cannon (松散的加农炮). Sigma protein by itself does not bind to promoter sites. Predict the effect of a mutation enabling sigma to bind to the -10 and -35 regions in the absence of other subunits of RNA polymerase.

Ans: This mutant sigma would competitively inhibit the binding of holoenzyme and prevent the specific initiation of RNA chains at promoter sites.
4. Stuck sigma(粘着的 sigma 因子). What would be the likely effect of a mutation that would prevent σ from dissociating from the RNA polymerase core?
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Ans: The core enzyme without sigma binds more tightly to the DNA template than does the holoenzyme. The retention of sigma after chain initiation would make the mutant RNA polymerase less processive. Hence, RNA synthesis would be much slower than normal. 5. Transcription time(转录时间). What is the minimum length of time required for the synthesis by E.coli polymerase of an mRNA encoding a 100-kd protein?

Ans: A 100-kd protein contains about 910 residues, which are encoded by 2730 nucleotides. At a maximal transcription rate of 50 nucleotides per second, the protein would be synthesized in 54.6 seconds.

6. Between bubbles(转录泡之间). How far apart are transcription bubbles on E. coli genes that are being transcribed at a maximum rate?
Ans: Initiation at strong promoters occurs every two seconds. In this interval, 100 nucleotides are transcribed. Hence, centers of transcription bubbles are 34nm (340Ǻ) apart. 7. A revealing bubble(说明问题的转录泡). Consider the synthetic RNA-DNA bubble shown in the following figure. Let us refer to the coding DNA strand, the template strand, and the RNA strand as strands 1, 2, and 3, respectively.

(a) Suppose that strand 3 is labeled with

32

P at its 5′ end and that polyacrylamide gel

electrophoresis is carried out under nondenaturing conditions. Predict the autoradiographic pattern for (i) strand 3 alone, (ii) strands 1 and 3, (iii) strands 2 and 3, (iv) strands 1, 2, and 3, and (v) strands 1, 2, and 3, and the core RNA polymerase.

(b) What is the likely effect of rifampicin on RNA synthesis in this system? (c) Heparin blocks elongation of the RNA primer if it is added to core RNA polymerase prior to the onset of transcription but not if added after transcription starts. Account for this difference. (d) Suppose that synthesis is carried out in the presence of ATP, CTP, and UTP. Compare the length of the longest product obtained with that expected when all four ribonucleoside triphosphates are present.

Ans: (a) The lowest band on the gel will be that of (i), whereas the highest will be that of (v). Band (ii) will be at the same positin as (i) because the RNA is not complementary to the nontemplate strand, whereas band(iii) will be higher because a complex is formed between RNA and the template strand. Band (iv) will be higher than the others because strand 1 is complexed to 2, and strand 2 to 3. Band (v) is the highest because core polymerase associates with the three

75

strands. (b) None, because rifampicin acts prior to the formation of the open complex. (c) RNA polymerase is processive. Once the template is bound, heparin cannot enter the DNA-binding site. (d) The longest RNA product formed is a 36-mer rather than the full 72-mer. Because GTP is absent, synthesis stops when the first C downstream of the bubble is encountered in the template strand.

8. An extra piece(额外的一段). What is the amino acid sequence of the extra segment of protein synthesized in the thalassemic patients with a mutation leading to aberrant splicing? The reading frame after the splice site begins with TCT.

Ans: Ser-Ile-Phe-His-Pro-Stop.
9. A long-tailed messenger(长尾信使). Another thalassemic patient had a mutation leading to the production of an mRNA for the β chain of hemoglobin that was 900 nucleotides longer than the normal one. The poly (A) tail of this mutant mRNA was located a few nucleotides after the only AAUAAA sequence in the additional sequence. Propose a mutation that would lead to the production of this altered mRNA.

Ans: A mutation that disrupted the normal AAUAA recognition sequence for the endonuclease could account for this finding. In fact, a change from U to C in this sequence caused this defect in a thalassemic patient. Cleavage occurred at the AAUAAA 900 nucleotides downstream from this mutant AACAAA site.

10. The right metal(适当的金属). In the proposed transition state for catalysis by a ribozyme, Mg2+ stabilizes the negative charge on the 3′ oxygen of the leaving group. If this oxygen is replaced by sulfur, Mg2+ is ineffective in catalyzing cleavage. However, cleavage readily occurs if Mn2+ is used in place of Mg2+. Why?

Ans: Mg2+ coordinates strongly to oxygen but not sulfur, whereas Mn2+ coordinates strongly to sulfur but not to oxygen.
11. RNA editing(RNA 编辑). Many U′s are inserted into some mitochondrial mRNAs in trypanosomes. The uridines come from the poly (U) tail of a donor strand. Nucleoside triphosphates do not participate in this reaction. Propose a reaction mechanism that accounts for these findings.(hint: relate RNA editing to RNA splicing)

Ans: One possibility is that the 3′ of the poly(U) donor strand cleaves the phosphodiester bond on the 5′side of the insertion site. The newly formed 3′ terminus of the acceptor strand then cleaves 76

the poly (U) strand on the 5′ side of the nucleotide that initiated the attack. In other words, A U could be added by two trans-esterification reactions. This postulated mechanism is akin to the one in RNA splicing.

Chapter 33 Protein Synthesis
1. Synthetase mechanism(合成酶作用机理). The formation of Ile-tRNA proceeds through an enzyme-bound Ile-AMP intermediate. Predict whether 32P-labeled ATP is formed from 32PPi when each of the following sets of components is incubated with the specific activating enzyme: (a) ATP and 32PPi.

(b) tRNA, ATP, and 32PPi.
(c) Isoleucine, ATP, and 32PPi.
Ans: (a) No; (b) no; and (c) yes.
2. Light and heavy ribozyme(轻重核糖体).

Ribosomes were isolated from bacteria grown in a

“heavy′ medium (13C and 15N) and from bacteria grown in a “light” medium (12C and 14N). These 70S ribosomes were added to an in vitro system actively engaged in protein synthesis. An aliquot removed several hours later was analyzed by density-gradient centrifugation. How many bands of 70S ribosomes would you expect to see in the density gradient? Ans: Four bands: light, heavy, a hybrid of light 30S and heavy 50S, and a hybrid of heavy 30S and light 50S.

3. The price of protein synthesis (蛋白质合成的代价). How many high-energy phosphate bonds are consumed in the synthesis of a 200-residue protein? Starting from amino acids? Ans: About 799 high-energy phosphate bonds are consumed—400 to activate the 200 amino acids, 1 for initiation, and 398 to form 199 peptide bonds.

4. Contrasting modes of elongation(对比鲜明的延长模式). There are two basic mechanisms for the elongation of biomolecules.

In type I, the activating group (labeled X) is released from the growing chain. In type 2, the activating group is released from the incoming unit as it is added to the growing chain. Indicate whether each of the following biosyntheses occurs by means of type 1 or a type 2 mechanism. (a) Glycogen synthesis

(b) Fatty acid synthesis.
(c)C5→C10→C15 in cholesterol synthesis
(d) DNA synthesis
77

(e) RNA synthesis
(f) Protein synthesis
Ans: TypeI: b, c, and f; type2: a, d, and e.
5. suppressing frameshifts(抑制移码突变). The insertion of a base in a coding sequence leads to a shift in reading frame, which in most cases produces a nonfunctional protein. Propose a mutation in tRNA that would suppress frameshifting.

Ans: 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. Tagging a ribosomal site(给核糖体位点贴标签). Design an affinity-labeling reagent for one of the tRNA binding sites in E. coli ribosomes. How would you synthesize such a reagent? Ans: One approach is to synthesize a tRNA that is acylated with a reactive amino acid analog. For example, bromoacetyl-phenylalanyl-tRNA is an affinity-labeling reagent for the P sites of E.coli ribosomes.

7. Viral mutation(病毒突变). An mRNA transcript of a T7 phage gene contains the base sequence:
5′-AACUGCACGAgGUAACACAAGAUGGCU-3′
Predict the effect of a mutation that changes the g marked to A. Ans: The sequence GAGGU is complementary to a sequence of five bases at the 3′ end of 16SrRNA 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.

8. Molecular attack(分子进攻). What is the nucleophile in the reaction catalyzed by peptidyl transferase?
Ans: The nitrogen atom of the deprotonated α-amino group of aminoacyl –tRNA is the nucleophile in peptide-bond formation.
9. Two synthetic modes(两种合成方式). Compare and contrast protein synthesis by ribosomes with that by the solid-phase method.
Ans: 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 dicycohexylcarbodiimide.

10. Enhancing fidelity(增强忠实性). Compare the accuracy of (a) DNA replication, (b) RNA 78

synthesis, and (c) protein synthesis. Which mechanisms are used to assure the fidelity of each of these processes?
Ans: 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 deponds 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 synthesis, the mischarging of some tRNAs is corrected by the hydrolytic action of the aminoacyl-tRNA 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. 11. Triggered GTP hydrolysis(引发 GTP 水解). Ribosomes markedly accelerate the hydrolysis of GTP bound to the complex of EF-Tu and aminoacyl-tRNA. What is the biological significance of this enhancement of GTPase activity by ribosomes?

Ans: 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.

12. Déjà vu(记忆错误). Which protein in G protein cascades plays a role akin to that of elongation factor Ts?
Ans: EF-Ts catalyze 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. 13. Blocking translation(阻断翻译). Devise an experimental strategy for

switching off the

expression of a specific mRNA without changing the gen encoding the protein or the gene′s control elements.
Ans: The translation of an mRNA molecule can be blocked by antisense RNA, an RNA molecule with the comlementary sequence. The antisense-sense RNA duplex cannot serve as atemplate for translation; single-sranded mRNA is required. Furthermore, the antisense-sense 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.

Chapter 34 Protein Targeting
1. Targeting signals(分选信号). What are the distinctive features of each of the following targeting sequences?
(a) Eukaryotic signal directing a nascent protein to the ER. 79

(b) Prokaryotic signal directing a nascent protein to the plasma membrane. (c ) Signal directing a protein in the Golgi to lysosomes.
(d) Signal directing a membrane protein in the Golgi to the plasma membrane. (e) Signal directing a protein in the cytosol to mitochondria. (f) Signal targeting a cytosolic protein for rapid destruction. Ans: (a) Cleavable amino-terminal signal sequences are usually 13 to 36 residues long and contain a highly hydrophobic central region 10 to 15 residues long. The amino-terminal part has at least one basic residue. The cleavage site is preceded by small neutral residues. (b) prokaryotic signal sequences are similar to eukaryotic ones. In addition, a stop-transfer sequence is needed to keep the protein in the plasma membrane. (c) Mannose 6-phosphate residues direct proteins to lysosomes. (d) integral membrane proteins initially in the ER go to the plasma membrane unless they carry instructions to the contrary. No specific signal is needed. (e) An amino-terminal sequence containing positively charged residues, serine, and threonine, in addition to hydrophobic residues. (f) An amino-terminal arginine, histidine, isoleucine, leucine, lysine, phenylalanine, or tryptophan.

2. Insouciance(漫不经心). A mutant LDL receptor is found to be uniformly distributed in the plasma membrane rather than being concentrated in coated-pit regions. The mutant binds LDL normally but is not internalized. Which region of the receptor is likely to be altered? Ans: The cytosolic portion of the receptor, which enables it to interact with coated pits, is likely to be altered.

3. A different route(不同的路径). Cells derived from patients with I-cell disease take up lysosomal enzymes purified from normal cells. The added enzymes appear in the lysosomes of these cells.
(a) What is the likely pathway for the transport of these enzymes from extracellular medium to the lysosomes of I-cells?
(b) This experiment led to the hypothesis that lysosomal enzymes are normally secreted and taken up by this route. Which sugar phosphate would you add to the extracellular medium to test this hypothesis?

Ans: (a) Endocytosis mediated by a receptor specific for mannose 6-phosphate residues. Endocytic vesicles containing the added lysosomal enzymes then fuse with lysosomes. (b) The addition of mannose 6-phosphate to the extracellular medium should prevent lysosomal enzymes from reaching their destination if the normal pathway involves secretion outside the cell and import back into the cell. However, mannose 6-phosphate does not inhibit normal lysosomal targeting, showing that lysosomal enzymes reach their destination without leaving the cell. 4. Predicting the locations of new proteins ( 新 蛋 白 质 的 位 置 预 测 ) . Membrane-bound 80

immunoglobulin and its secreted counterpart differ in the carboxyl-terminal regions of their heavy chains. These variations on a theme arise from alternative splicing of mRNA. (a) Suppose that the transmembrane sequence of the antibody is placed by recombinant DNA methods at the amino terminus of a cytosolic protein. Where is this chimeric protein likely to be located?

(b) Suppose that this transmembrane sequence is placed at the carboxy terminus of chymotrypsinogen. Where is this chimeric protein likely to be located? Ans: (a) The chimeric protein will probably be found in the cytosol. The transmembrane sequence of a membrane –bound immunoglobulin functions as a stop-transfer sequence and not as a signal sequence. (b) The chimeric protein will probably be found in the plasma membrane because chymotrypsinogen is synthesiszed with a signal sequence.

5. Contrasting amino termini(对比鲜明的氨基末端). The amino-terminal residue of secreted enkaryotic proteins is usually different from that of the cytosolic proteins. A majority of secretory proteins have leucine, phenylalanine, aspartate, lysine, or arginine as their amino terminal residue. In contrast, these residues are rarely found in cytosolic proteins. What is a potential benefit of having different amino-terminal residues for these two classes of proteins? Ans: Secretory proteins that are erroneously targeted to the cytosol will be rapidly degraded because their amino terminii mark them for a rapid destruction. Rapid elimination of mistargeted secretory proteins is important because some of them (e.g. trypsinogen) could wreak havoc in the cytosol.

6. Defective peroxisomes(有缺陷的过氧化物酶体). A newborn with multiple developmental abnormalities and severe neurological damage was found to have peroxisomes that are unable to oxidize long-chain fatty acids, carry out steps in the synthesis of plasmalogens, and form bile acids. Catalase was found in the cytosol rather than in the peroxisomes of this infant′s cells. Propose a molecular defect that would account for these findings. Ans: This genetic disorder, known as Zellweger syndrome, is caused by defective import of peroxisomal matrix proteins bearing SKL sequences. The mutation is probably in the fene for the SKL receptor or the associated translocase.

7. Redirected enzymes(重新分选的酶). Dihydrofolate reductase (DHFR), a cytosolic enzyme, can be redirected into mitochondria by attaching a matrix-targeting sequence to its amino terminus. DHFR can be placed in the mitochondrial matrix by adding to its amino terminus the 27-residue presequence of an alcohol dehydrogenase isozyme that normally resides in the matrix. The import of this chimeric DHFR into the matrix is blocked by methotrexate. However, methotrexate does not block the transport of anthentic mitochondrial proteins, nor does it interfere with the binding 81

of the chimeric protein to receptors on the outer mitochondrial membrane. How does methotrexate interfere with the import of this presequence-DHFR chimeric protein? Ans: Methotrexate binds very tightly to DHFR and prevents it from unfolding. Recall that proteins must be partly or completely unfolded during their translocation across the inner mitochondrial membrane as well as other membranes.

8. Cancer chemotherapy(癌症化疗). Mutant forms of ras that are persistently in the GTP state accelerate the progression of many tumors. Propose a new class of anticancer agents based on altered targeting of ras.

Ans: Mutant ras proteins are oncogenic only if they are membrane-anchored. Farnesylation of ras is essential for membrane anchoring. Hence, specific inhibitors of farnesyl transferase are potentially valuable anticancer agents.

Chapter 35

Control of Gene Expression in Prokaryotes

1. Missing genes(缺失的基因). What is likely to be the effect of each of these mutations? (a) deletion of the lac repressor gene.
(b) Deletion of the trp repressor gene.
(c) Deletion of the araC protein gene.
(d) Deletion of the N gene of λ.
Ans: (a) The lac repressor is missing in an i- mutant. Hence, this mutant is constitutive for the proteins of the lac operon. (b) This mutant is constitutive for the proteins of the trp operon because the trp repressor is missing. (c) The arabinose operon is not expressed in this mutant because araC protein is needed to activate transcription. (d) This mutant is lytic but not lysogenic, because it cannot synthesize the λ repressor. (e) This mutant is lysogenic but not lytic because it cannot synthesize the N protein, a positive control factor in transcription. 2. Turned off(关闭). Superrepressed mutants (i s) of the lac operon behave as noninducible mutants. The is gene is dominant over i+ in partial diploids. What is the molecular nature of this mutation likely to be?

Ans: One possibility is that an is mutant produces an altered lac repressor that has almost no affinity for inducer but normal affinity for the operator. Such a lac repressor would bind to the operator and block transcription even in the presence of inducer. 3. Uncontrolled synthesis(不受控制的合成). A mutant of E.coli synthesizes large amounts of β-galactosidase whether or not inducer is present. A partial diploid formed from this mutant and Fi+o+z- also synthesizes large amounts of β-galactosidase whether or not inducer is present. What kind of mutation might give these results?

Ans: This mutant has an altered lac operator that fails to bind the repressor. Such a mutator is 82

called Oc (operator constitutive).
4. A limited palette(有限的调色板). A mutant unable to grow on galactose, lactose, arabinose, and several other carbon sources is isolated from a wild-type culture. The cyclic AMP level in this mutant is normal. What kind of mutation might give these results? Ans: The cyclic AMP binding protein (CAP) is probably defective or absent in this mutant. 5. A grant of immunity(获得免役). An E.coli cell bearing a λ prophage is immune to lytic infection by λ. Why?

Ans: An E. coli cell bearing a λ prophage contains λ repressor molecules, which also block the transcription of the immediate-early genes of invading λ DNA. 6. Translational efficacy(翻译效率). Transcription of cI can be initiated at pRE, the promoter for repressor establishment, or at pRM, the promoter for repressor maintenance. The pRM transcript begins at its 5′ end with an AUG codon for the λ repressor, whereas in the pRE transcript this initiation codon is preceded by a sequence that is complementary to the 3′ end of 16S rRNA. Initiation at the pRE site requires phage-encoded proteins that are not expressed in the lysogenic state.

(a) Which transcript will be translated more efficiently?
(b) What is the likely physiologic significance of this difference? Ans: (a) Translation of the pRE transcript is from 5 to 10 times as rapid as that of the pRM transcript because it contains the full protein synthesis initiation signal. (b) The more effective translation of the pRE transcript provides a burst of λ repressor molecules needed to establish the lysogenic state.

7. The price of tenacity(执着的代价). Predict the consequence of a mutation that increases by a factor of 100 the binding affinity of lac repressor for lac operator without changing the binding affinity for nonspecific sites on DNA.

Ans: The rate constant for association of wild-type lac repressor with the operator site is near the diffusion-controlled limit. Hence, a 100-fold increase in binding affinity implies that the mutant dissociates from the operator site about 100 fold more slowly than does the wild type. The lag between addition of inducer and the initiation of transcription of the lac operon will be much longer in the mutant than in the wild type.

8. A kinetic difference(动力学差异). The binding of most repressors to operator sites is much faster at low ionic strength than at high. In contrast, the affinity of the operator for the repressor is not markedly affected by ionic strength. Why?

Ans: A repressor finds its target site by first binding to a nonspecific site anywhere in the DNA molecule and then diffusing along the DNA to reach the specific site. The repressor binds less 83

tightly to nonspecific DNA at low ionic strength than at high ionic strength because of increased electrostatic repulsion. In contrast, binding to the operator is nearly independent of ionic strength because the interaction is mediated by hydrogen-bond and van der Waals interactions with the bases of the operator site.

9. Shatterproof mutant(防崩解的突变体). Crystals of normal trp aporepressor shatter on addition of tryptophan. However, crystals of a mutant aporepressor are unaffected by the addition of tryptophan. The unit cell dimensions of this crystal are virtually identical with those of a crytal of normal repressor containing bound tryptophan. Predict the physiologic activity of this mutant repressor protein.

Ans: The mutant protein acts as a repressor even when tryptophan is not bound. Expression of the trp operon is permanently repressed in this mutant bacterium. 10. Terminating a terminator(终止子的终止). The amount of termination factor RF2 is regulated by an unusual mechanism. Either a truncated polypeptide ending at residue 25 or a complete protein ending at residue 340 is produced from the mRNA for RF2. The amino acid sequence in the vicinity of residue 25 and the corresponding base sequence are -Gly-Tyr-Leu-Asp-Tyr-Asp5′-GGGUAUCUUUGACUACGAC-3′

Propose a control mechanism based on these data. What is the outcome of this regulatory circuit?
Ans: When RF2 is present, UGA is read as a stop codon, leading to the formation of a truncated 25-residue polypeptide.

.
When the level of RF2 is very low, termination at UGA does not occur. Instead, a shift of the reading frame occurs, resulting in continued protein synthesis with Asp26 as the next residue.

.
11. On the importance of being integral(完整的重要性). Lambda repressor binds cooperatively to operator sites that are separated by 5 or 6 turns of double helix. In contrast, binding is noncooperative when the operator sites are 4.5, 5.5, or 6.5 turns apart. Propose a molecular basis for this difference.

Ans: Operator sites separated by five or six turns of double helix are oriented so that they can bind to both sites of the dimeric receptor. Only one operator can bind when the other is 4.5, 5.5 or 6.5 84

turns away because the second site faces the opposite direction (180 degrees out of phase) 12. A pregnant pause(意味深长的停顿). In transcribing the trp leader gene, RNA polymerase pauses at a discrete site immediately after it makes segments 1 and 2. Why? Ans: Attenuation requires that transcription and translation be tightly coupled. Pausing is a means of synchronizing these two processes. Transcription is halted until a ribosome initiates translation and comes near the paused polymerase. The outcome then depends on whether or not the leader is completely translated.

13. helix swap(螺旋交换). Five amino acid residues on the outside of the C-terminal helix of the helix-turn-helix motif of the phage 434 repressor were replaced with those occupying the same position in the P22 repressor encoded by a phage that infects Salmonella. The DNA-binding specificity of this chimeric protein corresponded to that of the P22 repressor rather than the 434 repressor. What does this helix redesign experiment reveal about the recognition process? Ans: The second helix of the HTH motif of both repressors is the major determinant of their specificity. Hence, it is often called the recognition helix.

Chapter 36 Eukaryotic Chromosomes and Gene Expression
1. Eons apart(相距久远). Compare prokaryotic and eukaryotic gene expression in regard to: (a) Degree of coupling of transcription and translation.
(b) Number of gene products on a primary transcript.
(c) Number of proteins arising from the translation of a primary transcript. (d) Density of coding sequences in DNA.
(e) Organization of genes into operons.
Ans: (a)In prokaryotes, translation begins while transcription is still in progress. In eukaryotes, theses processes are separated in space and time. (b) Eukaryotic transcripts are monogenic, while prokaryotic transcripts are typically polygenic. (c) In prokaryotes, several proteins can be specified by a single primary transcript, one for each gene encoding the mRNA. In eukaryotes, a primary transcript encoded by a single gene can give rise to multiple proteins through alternative splicing. Cleavage of a poly-protein can also yield multiple proteins in eukaryotes. (d) Most of prokaryotic DNA encodes proteins and functional RNAs. In contrast, most of the eukaryotic DNA does not encode functional macromolecules. (e) Most prokaryotic genes are clustered in operons. Eukaryotes do not have operons.

2. YACs(酵母人工染色体). Synthetic yeast chromosomes can be constructed by the joining of three kinds of DNA elements. What are they?
Ans: Centromeres, telomeres, and ars (autonomous replicating sequences serving as origins of replication) are needed to fashion a synthetic yeast chromosome. 3. Tolerance(耐受力). The error rate of the mitochondrial DNA polymerase is much higher than 85

that of the nuclear DNA polymerase. Why can lower fidelity be tolerated in the mitochondrial enzymes?
Ans: The mitochondrial genome is much smaller than the nuclear genome. Furthermore, a cell contains a large number of mitochondria. A substantial portion of them, say 5%, could be nonfuctional without injuring the cell.

4. Rapid evolution(快速进化). In the course of evolution, genes encoding mitochondrial proteins probably moved from the mitochondrial to the nuclear genome. Human mitochondrial DNA, for example, is fivefold smaller and simpler than yeast mitochondrial DNA. Different sets of mitochondrial proteins are encoded by mitochondrial DNA in the two species. Hence, there seems to be no structural reason why a mitochondrial protein must be synthesized within the organelle rather than be imported. Do you expect the human mitochondrial genome to be even shorter 10million years from now? Which aspect of mitochondrial gene expression would impede the transfer of a mitochondrial gene to the nuclear genome?

Ans: Mitochondrial DNA is evolving at a very rapid rate. It is possible but not likely that a human mitochondrial gene will be transferred to the nuclear genome in the next 10 million years. The existence of different genetic codes in the mitochondrion and the cytosol is a formidable barrier to gene transfer. In particular, UGA specifies tryptophan in mitochondrial proteins but is a stop signal in cytosolic protein synthesis.

5. The sense of the supercoil(超螺旋的意义). Design an experiment to determine whether DNA coiled around histones forms a left-handed or a right-handed superhelix. Ans: One experimental approach is to add histones to linear duplex DNA and form closed circles using DNA ligase. The histones are then removed from the supercoiled circles. Topoisomerase I is added to an aliquot to form the relaxed counterpart. The melting temperature of the supercoiled DNA is compared with that of the relaxed circle. A left-handed superhelix has a lower melting temperature than the relaxed circle, whereas a right-handed superhelix has a higher melting temperature. Recall that left-handed supercoiled DNA (negatively supercoiled; Wr is negative) is poised to be unwound.

6. Dual allegiance(双重忠实). Transcription factor IIIA (TFIIIA) binds to 5S RNA as well as to the internal control region of the gene encoding it. Why? How might this binding of the gene product be used to regulate expression of the 5S genes?

Ans: 5S RNA has the same base sequence (except for U in place of T) as the coding strand of its gene, the one that strongly binds TFIIIA. Hence, 5S RNA competes with its gene for the binding of TFIIIA. When 5S RNA is abundant, little TFIIIA is bound to the internal control region, and so transcription is slowed. The binding of TFIIIA to 5S RNA as well as the internal control region of the gene is a basis of a feedback loop that regulates the amount of 5S RNA. 86

7. Tenacity(执着). TFIIIA stays on a 5S ribosomal RNA gene while it is being transcribed. How might TFIIIA remain bound in the face of an advancing RNA polymerase? Ans: TFIIIA has nine zinc fingers. Some of its zinc fingers could stay bound to the 5S RNA gene and the others could be released to allow RNA polymerase to pass. 8. Seven but not six(七而非六). Recall that yeast cells respond to amino acid starvation by synthesizing GCN4, a leucine zipper protein. The insertion of six amino acid residues at the amino terminal junction of the zipper sequence destroys the capacity of GCN4 to activate transcription of amino acid biosynthetic genes. In contrast, seven residues can be inserted with retention of activity. Account for for this difference. Predict the effects of inserting four and five residues. Ans: The leucine zipper symmetrically orients the two basic DNA-binding regions at the half sites of the palindromic DAN target sequence. The insertion of six residues changes the angular relationship of the pair of basic sites. In contrast, the correct angle is preserved when seven residues are inserted because this number corresponds to a heptad repeat. The insertion of four or five residues also destroys activity.

9. Replacing the leucine zipper(置换亮氨酸拉链). Design an experiment to test the hypothesis that the

sole purpose of the leucine zipper is to bring two basic DNA-binding regions together in

an appropriate spatial relationship for interaction with the target DNA sequence. Ans: Covalently cross-link a pair of basic DNA-binding regions by forming a flexible disulfide at their C-terminal end. The resulting protein, devoid of a coiled coil, serves as a transcriptional activator.

10. Crafting a repressor ( 制 作 阻 遏 蛋 白 ) . Describe a simple method for converting a transcriptional activator into a transcriptional repressor.
Ans: Retain the DNA-binding domain but delete the activation domain to block the formation of a functional transcription complex.
11. Never nuclear(不会进入细胞核).

A mutant estrogen receptor binds hormone but is unable

to move from the cytosol to the nucleus. Propose a structural basis for this defect. Ans: The mutant receptor lacks a nuclear localization signal, which is required for passage through nuclear pores.

12. Identifying target genes. Suppose that you have cloned a new cDNA that encodes a protein resembling members of the nuclear receptor superfamily. The ligand that activates this new receptor is unknown. How would you identify the target of this receptor and the battery of genes that it acitivates?

87

Ans: One approach is to synthesize a truncated cDNA that encodes the DNA-binding domain but not the hormone binding domain of the new nuclear receptor. The DNA-binding domain produced by expressing this cDNA in E. coli should then bind with high affinity to target DNA sequences. A complementary approach is to carry out a domain swap experiment. The target cell could be transfected with a gene for a hybrid receptor containing the hormone-binding domain of the new receptor. The addition of glucocorticoid to the transfected cell would then activate genes normally stimulated by the unknown ligand.

88

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