DNA Structure, Replication, and Repair (Lippincott's Illustrated Reviews Biochemistry 5th Edition) PDF
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This chapter from *Lippincott's Illustrated Reviews Biochemistry 5th Edition* details the structure, replication, and repair of DNA molecule. It begins by introducing the central dogma of molecular biology, highlighting the flow of information from DNA to RNA to protein.
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168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 395 UNIT VI: Storage and Expression of...
168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 395 UNIT VI: Storage and Expression of Genetic Information DNA Structure, Replication, and Repair 29 I. OVERVIEW Nucleic acids are required for the storage and expression of genetic information. There are two chemically distinct types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA, see Chapter 30). DNA, the repository of genetic information, is present not only in chromosomes in the nucleus of eukaryotic organisms, but also in Replication mitochondria and the chloroplasts of plants. Prokaryotic cells, which lack nuclei, have a single chromosome, but may also contain nonchromosomal DNA in the form of plasmids. The genetic informa- tion found in DNA is copied and transmitted to daughter cells through DNA DNA replication. The DNA contained in a fertilized egg encodes the information that directs the development of an organism. This devel- opment may involve the production of billions of cells. Each cell is specialized, expressing only those functions that are required for it to Transcription perform its role in maintaining the organism. Therefore, DNA must be able to not only replicate precisely each time a cell divides, but also to have the information that it contains be selectively expressed. Transcription (RNA synthesis) is the first stage in the expression of RNA genetic information (see Chapter 30). Next, the code contained in the nucleotide sequence of messenger RNA molecules is translated Translation (protein synthesis, see Chapter 31), thus completing gene expres- sion. The regulation of gene expression is discussed in Chapter 32. The flow of information from DNA to RNA to protein is ter med the “central dogma” of PROTEIN molecular biology (Figure 29.1), and is descrip- tive of all organisms, with the exception of some viruses that have RNA as the repository of their Figure 29.1 genetic information. The “central dogma” of molecular biology. 395 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 396 396 29. DNA Structure, Replication, and Repair 5'-End A 5'-End H3C O N H Thymine B O – H O p 5' N O P O CH2 O 5' 3' T – O 4' 1' H H p 3' 2' N N N Adenine 5' 3' A H H p N N O 5' 5' 3' C O P O CH2 O – p O 4' 1' H H 3' 2' N 5' 3' G H Cytosine N OH H N O O 5' 3'-End O P O CH2 O O O – 4' 1' H N N H Guanine C 3' 2' H N N N 5'-End 3'-End H pTpApCpG 3' 5' Phosphodiester bond O 5' O P O CH2 O – O 4' 1' 3' 2' D 5'-End 3'-End OH 3'-End TACG Figure 29.2 A. DNA chain with the nucleotide sequence shown written in the 5'→ 3' direction. A 3' →5'-phosphodiester bond is shown highlighted in the blue box, and the deoxyribose-phosphate backbone is shaded in yellow. B. The DNA chain written in a more stylized form, emphasizing the ribose–phosphate backbone. C. A simpler representation of the nucleotide sequence. D. The simplest (and most common) representation, with the abbreviations for the bases written in the conventional 5'→3' direction. II. STRUCTURE OF DNA DNA is a polymer of deoxyribonucleoside monophosphates covalently linked by 3'→5'–phosphodiester bonds. With the exception of a few viruses that contain single-stranded (ss) DNA, DNA exists as a double- stranded (ds) molecule, in which the two strands wind around each other, forming a double helix. In eukaryotic cells, DNA is found associ- ated with various types of proteins (known collectively as nucleoprotein) present in the nucleus, whereas in prokaryotes, the protein–DNA com- plex is present in a nonmembrane-bound region known as the nucleoid. A. 3'→5'-Phosphodiester bonds Phosphodiester bonds join the 3'-hydroxyl group of the deoxypentose of one nucleotide to the 5'-hydroxyl group of the deoxypentose of an adjacent nucleotide through a phosphate group (Figure 29.2). The resulting long, unbranched chain has polarity, with both a 5'-end (the end with the free phosphate) and a 3'-end (the end with the free hydroxyl) that are not attached to other nucleotides. The bases located along the resulting deoxyribose–phosphate backbone are, by convention, always written in sequence from the 5'-end of the chain to the 3'-end. For example, the sequence of bases in the DNA shown 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 397 II. Structure of DNA 397 in Figure 29.2 is read “thymine, adenine, cytosine, guanine” (5'-TACG-3'). Phosphodiester linkages between nucleotides (in DNA Axis of symmetry or RNA) can be cleaved hydrolytically by chemicals, or hydrolyzed Deoxyribose– 3' End enzymatically by a family of nucleases: deoxyribonucleases for DNA phosphate 5' End and ribonucleases for RNA. [Note: Only RNA is cleaved by alkali.] backbone B. Double helix A=T In the double helix, the two chains are coiled around a common axis C=G called the axis of symmetry. The chains are paired in an antiparallel C=G manner, that is, the 5'-end of one strand is paired with the 3'-end of the other strand (Figure 29.3). In the DNA helix, the hydrophilic T=A deoxyribose–phosphate backbone of each chain is on the outside of Base pairs are Minor perpendicular to the the molecule, whereas the hydrophobic bases are stacked inside. groove axis of symmetry The overall structure resembles a twisted ladder. The spatial rela- G=C tionship between the two strands in the helix creates a major (wide) groove and a minor (narrow) groove. These grooves provide access A=T for the binding of regulatory proteins to their specific recognition T=A sequences along the DNA chain. Certain anticancer drugs, such as dactinomycin (actinomycin D), exert their cytotoxic effect by interca- C=G lating into the narrow groove of the DNA double helix, thus interfer- Major ing with DNA and RNA synthesis.1 groove C=G 1. Base pairing: The bases of one strand of DNA are paired with the bases of the second strand, so that an adenine is always paired A=T with a thymine and a cytosine is always paired with a guanine. G=C [Note: The base pairs are perpendicular to the axis of the helix (see Figure 29.3).] Therefore, one polynucleotide chain of the DNA T=A double helix is always the complement of the other. Given the sequence of bases on one chain, the sequence of bases on the complementary chain can be determined (Figure 29.4). [Note: The 5' End specific base pairing in DNA leads to the Chargaff Rule: In any 3' End sample of dsDNA, the amount of adenine equals the amount of thymine, the amount of guanine equals the amount of cytosine, and the total amount of purines equals the total amount of pyrimidines.] Figure 29.3 The base pairs are held together by hydrogen bonds: two between DNA double helix, illustrating some A and T and three between G and C (Figure 29.5). These hydrogen of its major structural features. bonds, plus the hydrophobic interactions between the stacked bases, stabilize the structure of the double helix. Deoxyribose– Base pairs Deoxyribose– phosphate phosphate backbone backbone 2. Separation of the two DNA strands in the double helix: The two 5' P 5' A T 3' 3' strands of the double helix separate when hydrogen bonds between the paired bases are disrupted. Disruption can occur in 3' 5' P the laboratory if the pH of the DNA solution is altered so that the P G C 3' nucleotide bases ionize, or if the solution is heated. [Note: 5' 5' Phosphodiester bonds are not broken by such treatment.] When 3' P DNA is heated, the temperature at which one half of the helical P T A 3' 5' structure is lost is defined as the melting temperature (Tm). The 5' 3' loss of helical structure in DNA, called denaturation, can be moni- Hydrogen P P bonds tored by measuring its absorbance at 260 nm. [Note: ssDNA has a 3' 5' higher relative absorbance at this wavelength than does dsDNA.] Because there are three hydrogen bonds between G and C but Figure 29.4 Two complementary DNA 1See INFO Chapter 39 in Lippincott’s Illustrated Reviews: Pharmacology sequences. LINK for a discussion of the anticancer drug, actinomycin D. 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 398 398 29. DNA Structure, Replication, and Repair only two between A and T, DNA that contains high concentrations H H of A and T denatures at a lower temperature than G- and C-rich H DNA (Figure 29.6). Under appropriate conditions, complementary H C O H N H N C DNA strands can reform the double helix by the process called C C C C renaturation (or reannealing). H C N H N C N N C C N 3. Structural forms of the double helix: There are three major struc- O H tural forms of DNA: the B form, described by Watson and Crick in Thymine Adenine 1953, the A form, and the Z form. The B form is a right-handed Hydrogen bonds helix with ten residues per 360° turn of the helix, and with the planes of the bases per pendicular to the helical axis. Chromosomal DNA is thought to consist primarily of B-DNA H (Figure 29.7 illustrates a space-filling model of B-DNA). The A H N H O N H form is produced by moderately dehydrating the B form. It is also C C C C C a right-handed helix, but there are 11 base pairs per turn, and the H C N H N C N planes of the base pairs are tilted 20° away from the perpendicu- N C C N lar to the helical axis. The conformation found in DNA–RNA O H N hybrids or RNA–RNA double-stranded regions is probably very H close to the A form. Z-DNA is a left-handed helix that contains Cytosine Guanine about 12 base pairs per turn (see Figure 29.7). [Note: The deoxyribose–phosphate backbone “zigzags,” hence, the name Figure 29.5 “Z”-DNA.] Stretches of Z-DNA can occur naturally in regions of Hydrogen bonds between DNA that have a sequence of alternating purines and pyrimidines, complementary bases. for example, poly GC. Transitions between the B and Z helical forms of DNA may play a role in regulating gene expression. C. Linear and circular DNA molecules At temperatures above the Tm, DNA is present as a single strand. Each chromosome in the nucleus of a eukaryote contains one long, linear molecule of dsDNA, which is bound to a complex mixture of proteins (histone and non-histone, see p. 409) to form chromatin. Eukaryotes have closed, circular DNA molecules in their mitochon- dria, as do plant chloroplasts. A prokaryotic organism typically con- Relative absorbance at 260 nm 1. High AT- content DNA tains a single, double-stranded, supercoiled, circular chromosome. Each prokaryotic chromosome is associated with non-histone pro- High GC- teins that can condense the DNA to form a nucleoid. In addition, content DNA most species of bacteria also contain small, circular, extrachromo- 1.24 somal DNA molecules called plasmids. Plasmid DNA carries genetic information, and undergoes replication that may or may not be syn- chronized to chromosomal division.2 Tm Tm 1.00 62 74 86 98 Temperature (°C) Plasmids may carry genes that convey antibiotic resistance to the host bacterium, and may facilitate the transfer of genetic information from Figure 29.6 one bacterium to another. Melting temperatures (Tm) of DNA molecules with different nucleotide compositions. (At a wavelength of 260 nm, single-stranded DNA has [Note: The use of plasmids as vectors in recombinant DNA technol- a higher relative absorbance than ogy is described in Chapter 33.] does double-stranded DNA.) 2See INFO Chapter 7 in Lippincott’s Illustrated Reviews: Microbiology for a LINK discussion of plasmids. 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 399 III. Steps in Prokaryotic DNA Synthesis 399 III. STEPS IN PROKARYOTIC DNA SYNTHESIS Minor groove When the two strands of the DNA double helix are separated, each can serve as a template for the replication of a new complementary strand. This produces two daughter molecules, each of which contains two DNA Major groove strands with an antiparallel orientation (see Figure 29.3). This process is called semiconservative replication because, although the parental duplex is separated into two halves (and, therefore, is not “conserved” as an entity), each of the individual parental strands remains intact in one of the two new duplexes (Figure 29.8). The enzymes involved in the DNA replication process are template-directed polymerases that can synthe- size the complementary sequence of each strand with extraordinary fidelity. The reactions described in this section were first known from studies of the bacterium Escherichia coli (E. coli), and the description given below refers to the process in prokaryotes. DNA synthesis in B-form DNA Z-form DNA higher organisms is less well understood, but involves the same types of mechanisms. In either case, initiation of DNA replication commits the cell Figure 29.7 to continue the process until the entire genome has been replicated. Structures of B-DNA and Z-DNA. A. Separation of the two complementary DNA strands In order for the two strands of the parental double helical DNA to be replicated, they must first separate (or “melt”) over a small region, because the polymerases use only ssDNA as a template. In T G A C prokaryotic organisms, DNA replication begins at a single, unique nucleotide sequence—a site called the origin of replication (Figure C 29.9A). [Note: This is referred to as a consensus sequence, G Parental Newly A T because the order of nucleotides is essentially the same at each synthesized double helix site.] This site includes a short sequence composed almost exclu- strands T A sively of AT base pairs that facilitate melting. In eukaryotes, replica- G C tion begins at multiple sites along the DNA helix (Figure 29.9B). C G Having multiple origins of replication provides a mechanism for T rapidly replicating the great length of the eukaryotic DNA molecules. G A G C C B. Formation of the replication fork A T A T As the two strands unwind and separate, they form a “V” where active synthesis occurs. This region is called the replication fork. It C G C G moves along the DNA molecule as synthesis occurs. Replication of A T A T dsDNA is bidirectional—that is, the replication forks move in oppo- site directions from the origin, generating a replication bubble (see T A T A Figure 29.9). G C G C C C G C G 1. Proteins required for DNA strand separation: Initiation of DNA replication requires the recognition of the origin of replication by a A T A T G C G CC group of proteins that form the prepriming complex. These pro- teins are responsible for maintaining the separation of the parental strands, and for unwinding the double helix ahead of the advancing replication fork. These proteins include the following: One parental strand is conserved a. DnaA protein: DnaA protein binds to specific nucleotide in each of the two new double helices sequences at the origin of replication, causing short, tandemly arranged (one after the other) AT-rich regions in the origin to Figure 29.8 melt. Melting is ATP-dependent, and results in strand separa- Semiconservative replication of DNA. tion with the formation of localized regions of ssDNA. 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 400 400 29. DNA Structure, Replication, and Repair A Origin of replication B Multiple origins of replication Local opening of double helix Replication bubble Replication Replication fork fork Bidirectional replication continues + + Figure 29.9 Replication of DNA: origins and replication forks. A. Small prokaryotic circular DNA. B. Very long eukaryotic DNA. b. DNA helicases: These enzymes bind to ssDNA near the repli- cation fork, and then move into the neighboring double- 3' DNA helicase stranded region, forcing the strands apar t—in effect, unwinds the e helix. unwinding the double helix. Helicases require energy provided by ATP (Figure 29.10). [Note: DnaB is the principal helicase of replication in E. coli. Its binding to DNA requires DnaC.] ATP ADP + Pi c. Single-stranded DNA-binding (SSB) proteins: These proteins bind to the ssDNA generated by helicases (see Figure 29.10). They bind cooperatively—that is, the binding of one molecule Direction of movement of SSB protein makes it easier for additional molecules of SSB of replication protein to bind tightly to the DNA strand. The SSB proteins are 5' fork not enzymes, but rather serve to shift the equilibrium between Single-stranded DNA- dsDNA and ssDNA in the direction of the single-stranded binding proteins keep the two strands of DNA forms. These proteins not only keep the two strands of DNA separate. separated in the area of the replication origin, thus providing the single-stranded template required by polymerases, but also protect the DNA from nucleases that degrade ssDNA. Figure 29.10 Proteins responsible for maintaining 2. Solving the problem of supercoils: As the two strands of the dou- the separation of the parental strands and unwinding the double ble helix are separated, a problem is encountered, namely, the helix ahead of the advancing appearance of positive supercoils (also called supertwists) in the replication fork. region of DNA ahead of the replication fork (Figure 29.11). The 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 401 III. Steps in Prokaryotic DNA Synthesis 401 accumulating positive supercoils interfere with further unwinding DNA double helix of the double helix. [Note: Supercoiling can be demonstrated by tightly grasping one end of a helical telephone cord while twisting the other end. If the cord is twisted in the direction of tightening the coils, the cord will wrap around itself in space to form positive Positive supercoiling supercoils. If the cord is twisted in the direction of loosening the coils, the cord will wrap around itself in the opposite direction to form negative supercoils.] To solve this problem, there is a group of enzymes called DNA topoisomerases, which are responsible for removing supercoils in the helix. a. Type I DNA topoisomerases: These enzymes reversibly cut one strand of the double helix. They have both nuclease (strand-cutting) and ligase (strand-resealing) activities. They Figure 29.11 do not require ATP, but rather appear to store the energy from Positive supercoiling resulting the phosphodiester bond they cleave, reusing the energy to from DNA strand separation. reseal the strand (Figure 29.12). Each time a transient “nick” is created in one DNA strand, the intact DNA strand is passed through the break before it is resealed, thus relieving (“relax- ing”) accumulated supercoils. Type I topoisomerases relax negative supercoils (that is, those that contain fewer turns of Nick the helix than relaxed DNA) in E. coli, and both negative and positive supercoils (that is, those that contain fewer or more turns of the helix than relaxed DNA) in eukaryotic cells. Nick sealed b. Type II DNA topoisomerases: These enzymes bind tightly to the DNA double helix and make transient breaks in both strands. The enzyme then causes a second stretch of the DNA double helix to pass through the break and, finally, reseals the Figure 29.12 break (Figure 29.13). As a result, both negative and positive Action of Type I DNA topoisomerases. supercoils can be relieved by this ATP-requiring process. Type II DNA topoisomerases are also required in both prokaryotes and eukaryotes for the separation of interlocked molecules of DNA following chromosomal replication. DNA gyrase, a Type II Relaxed circle topoisomerase found in bacteria and plants, has the unusual property of being able to introduce negative supercoils into relaxed circular DNA using energy from the hydrolysis of ATP. This facilitates the future replication of DNA because the nega- The left half of the circle tive supercoils neutralize the positive supercoils introduced 1 folds over the right half. during opening of the double helix. It also aids in the transient strand separation required during transcription (see p. 417). The back half 2 of the helix is cleaved. Anticancer agents, such as etoposide,3 target human topoisomerase II. Bacterial DNA gyrase is a unique target of a group of antimicrobial agents called quinolones, for example, cipro- floxacin.4 Negatively The front half of supercoiled DNA 3 the helix passes through the break, 3See which is resealed. Chapter 39 in Lippincott’s Illustrated Reviews: Pharmacology for a INFO LINK discussion of etoposide as an anticancer agent. 4See Chapter 33 in Lippincott’s Illustrated Reviews: Pharmacology for a Figure 29.13 discussion of the quinolones. Action of Type II DNA topoisomerase. 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 402 402 29. DNA Structure, Replication, and Repair Lagging strand Leading strand Origin of replication 3' 5' 3' 5' 3' 5' 5' 3' 5' 3' 5' 3' 5' 3' 5' 3' 5' 3' 5' 3 5' 3' 5' 3' Replication forks Leading strand Lagging strand Figure 29.14 Discontinuous synthesis of DNA. C. Direction of DNA replication The DNA polymerases responsible for copying the DNA templates are only able to “read” the parental nucleotide sequences in the 3'→5' direction, and they synthesize the new DNA strands only in P the 5'→3' (antiparallel) direction. Therefore, beginning with one G RNA primer CH2 O parental double helix, the two newly synthesized stretches of nucleotide chains must grow in opposite directions—one in the 5'→3' direction toward the replication fork and one in the 5'→3' P OH direction away from the replication fork (Figure 29.14). This feat is CH2 O A accomplished by a slightly different mechanism on each strand. 1. Leading strand: The strand that is being copied in the direction of OH OH the advancing replication fork is called the leading strand and is P P P synthesized continuously. Free 3'-OH end of ribose CH2 C 2. Lagging strand: The strand that is being copied in the direction O away from the replication fork is synthesized discontinuously, with DNA polymerase small fragments of DNA being copied near the replication fork. P P These short stretches of discontinuous DNA, termed Okazaki OH H fragments, are eventually joined (ligated) to become a single, con- dCTP tinuous strand. The new strand of DNA produced by this mecha- P P Entering deoxy- nism is termed the lagging strand. ribonucleotide P D. RNA primer CH2 O G DNA polymerases cannot initiate synthesis of a complementary strand of DNA on a totally single-stranded template. Rather, they require an RNA primer—that is, a short, double-stranded region P OH consisting of RNA base-paired to the DNA template, with a free CH2 O A hydroxyl group on the 3'-end of the RNA strand (Figure 29.15). This hydroxyl group serves as the first acceptor of a deoxynucleotide by action of DNA polymerase. [Note: Recall that glycogen synthase Newly formed P OH also requires a primer (see p. 126).] phospho- diester C bond CH2 O 1. Primase: A specific RNA polymerase, called primase (DnaG), Deoxribose synthesizes the short stretches of RNA (approximately ten nucleotides long) that are complementary and antiparallel to the OH H DNA template. In the resulting hybrid duplex, the U in RNA pairs with A in DNA. As shown in Figure 29.16, these short RNA Figure 29.15 sequences are constantly being synthesized at the replication fork Use of an RNA primer to initiate on the lagging strand, but only one RNA sequence at the origin of DNA synthesis. replication is required on the leading strand. The substrates for 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 403 III. Steps In Prokaryotic DNA synthesis 403 RNA primer 5' 3' Newly synthesized strand Leading strand template DNA polymerase Single-stranded DNA- binding proteins (SSB) DNA polymerase III recognizes the RNA DNA helicase primer and begins to synthesize DNA Parental DNA helix 5' 3' Lagging strand 3' template Primase 5' RNA primer Figure 29.16 Elongation of the leading and lagging strands. this process are 5'-ribonucleoside triphosphates, and pyrophos- phate is released as each ribonucleoside monophosphate is added through formation of a 3'→5' phosphodiester bond. [Note: The RNA primer is later removed as described on p. 405.] 2. Primosome: The addition of primase converts the prepriming complex of proteins required for DNA strand separation (see p. 399) to a primosome. The primosome makes the RNA primer required for leading strand synthesis, and initiates Okazaki frag- ment formation in lagging strand synthesis. As with DNA synthe- sis, the direction of synthesis of the primer is 5'→3'. E. Chain elongation Prokaryotic (and eukaryotic) DNA polymerases elongate a new DNA strand by adding deoxyribonucleotides, one at a time, to the 3'- end of the growing chain (see Figure 29.16). The sequence of nucleotides that are added is dictated by the base sequence of the template strand with which the incoming nucleotides are paired. 1. DNA polymerase III: DNA chain elongation is catalyzed by DNA polymerase III. Using the 3'-hydroxyl group of the RNA primer as the acceptor of the first deoxyribonucleotide, DNA polymerase III begins to add nucleotides along the single-stranded template that specifies the sequence of bases in the newly synthesized chain. DNA polymerase III is a highly “processive” enzyme—that is, it remains bound to the template strand as it moves along, and does not diffuse away and then rebind before adding each new nucleotide. The processivity of DNA polymerase III is the result of 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 404 404 29. DNA Structure, Replication, and Repair A POLYMERASE FUNCTION B PROOFREADING FUNCTION An incoming nucleoside triphosphate If DNA polymerase mispairs a nucleotide is correctly matched to its complementary with the template, it uses its base on the DNA template and is added as 3'→5' exonuclease activity to excise DNA the monophosphate to the growing DNA chain. the mismatched nucleotide. A P P P P G P C C C A P P G P P A T A T A T Enzyme P Enzyme P P P P P G C G C P advances A T G C P retreats P P C P P P P P P G C G G C C G 5'→3' DNA P 3'→5' Exo- P P C P A P A T P polymerase activity G P A T P nuclease activity T P P 3' 5' 3' 5' A T 3' 5' 3' 5' DNA Newly template synthesized strand Figure 29.17 3'→5'-Exonuclease activity enables DNA polymerase III to “proofread” the newly synthesized DNA strand. its β subunit forming a ring that encircles and moves along the template strand of the DNA, thus serving as a sliding DNA clamp. The new strand grows in the 5'→3' direction, antiparallel to the parental strand (see Figure 29.16). The nucleotide substrates are 5'-deoxyribonucleoside triphosphates. Pyrophosphate (PP i) is released when each new deoxynucleoside monophosphate is added to the growing chain (see Figure 29.15). Hydrolysis of PPi to 2Pi means that a total of two high-energy bonds are used to drive the addition of each deoxynucleotide. The production of PPi with subsequent hydrolysis to 2Pi, as seen in the synthesis of DNA and RNA, is a common theme in biochemistry. Removal of the PPi product drives a reaction in the forward direction, making it essentially irreversible. All four deoxyribonucleoside triphosphates (dATP, dTTP, dCTP, and dGTP) must be present for DNA elongation to occur. If one of the four is in short supply, DNA synthesis stops when that nucleotide is depleted. 2. Proofreading of newly synthesized DNA: It is highly important for the survival of an organism that the nucleotide sequence of DNA be replicated with as few errors as possible. Misreading of the tem- plate sequence could result in deleterious, perhaps lethal, muta- tions. To ensure replication fidelity, DNA polymerase III has, in addition to its 5'→3' polymerase activity, a “proofreading” activity 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 405 III. Steps in Prokaryotic DNA Synthesis 405 (3'→5' exonuclease, Figure 29.17). As each nucleotide is added to the chain, DNA polymerase III checks to make certain the added nucleotide is, in fact, correctly matched to its complementary base on the template. If it is not, the 3'→5' exonuclease activity corrects the mistake. [Note: The enzyme requires an improperly base- paired 3'-hydroxy terminus and, therefore, does not degrade cor- rectly paired nucleotide sequences.] For example, if the template base is cytosine and the enzyme mistakenly inserts an adenine instead of a guanine into the new chain, the 3'→5' exonuclease activity hydrolytically removes the misplaced nucleotide. The 5'→3' polymerase activity then replaces it with the correct nucleotide containing guanine (see Figure 29.17). [Note: The proofreading exonuclease activity requires movement in the 3'→5' direction, not 5'→3' like the polymerase activity. This is because the excision must be done in the reverse direction from that of synthesis.] F. Excision of RNA primers and their replacement by DNA DNA polymerase III continues to synthesize DNA on the lagging strand until it is blocked by proximity to an RNA primer. When this Exonucleases cut from the end of the chain, occurs, the RNA is excised and the gap filled by DNA polymerase I. releasing single nucleotides. 1. 5'→3' Exonuclease activity: In addition to having the 5'→3' poly- merase activity that synthesizes DNA, and the 3'→5' exonuclease activity that proofreads the newly synthesized DNA chain like DNA polymerase III, DNA polymerase I also has a 5'→3' exonu- 5' 3' clease activity that is able to hydrolytically remove the RNA A C G A T C A Double- stranded primer. [Note: These activities are exonucleases because they DNA T G C T A G T 5' remove one nucleotide at a time from the end of the DNA chain, 3' rather than cleaving the chain internally as do the endonucleases (Figure 29.18).] First, DNA polymerase I locates the space (“nick”) between the 3'-end of the DNA newly synthesized by DNA poly- Endonucleases cleave within merase III and the 5'-end of the adjacent RNA primer. Next, DNA the chain to produce single- polymerase I hydrolytically removes the RNA nucleotides “ahead” stranded nicks. of itself, moving in the 5'→3' direction (5'→3' exonuclease activ- ity). As it removes the RNA, DNA polymerase I replaces it with deoxyribonucleotides, synthesizing DNA in the 5'→3' direction Figure 29.18 (5'→3' polymerase activity). As it synthesizes the DNA, it also Endonuclease versus exonuclease “proofreads” the new chain using 3'→5' exonuclease activity. This activity. removal/synthesis/proofreading continues, one nucleotide at a time, until the RNA primer is totally degraded and the gap is filled with DNA (Figure 29.19). 2. Differences between 5'→3' and 3'→5' exonucleases: The 5'→3' exonuclease activity of DNA polymerase I differs from the 3'→5' exonuclease used by both DNA polymerase I and III in two impor- tant ways. First, 5'→3' exonuclease can remove one nucleotide at a time from a region of DNA that is properly base-paired. The nucleotides it removes can be either ribonucleotides or deoxyri- bonucleotides. Second, 5' → 3' exonuclease can also remove groups of altered nucleotides in the 5'→3' direction, removing from one to ten nucleotides at a time. This ability is important in the repair of some types of damaged DNA. 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 406 406 29. DNA Structure, Replication, and Repair 3' 5' Leading RNA primer is elongated by 1 DNA polymerase III until another DNA polymerase III stretch of RNA is encountered. DNA helicase RNA primer is excised 2 by DNA polymerase I, one ribonucleotide at a time. 5' Gap is filled by 3 DNA polymerase I. 3' RNA primer DNA made by DNA polymerase III 3' DNA made by DNA polymerase I 5' Figure 29.19 Removal of RNA primer and filling of the resulting “gaps” by DNA polymerase I. G. DNA ligase 3' 5' The final phosphodiester linkage between the 5'-phosphate group on the DNA chain synthesized by DNA polymerase III and the 3'- A T T G C A hydroxyl group on the chain made by DNA polymerase I is catalyzed T A A C G T by DNA ligase (Figure 29.20). The joining of these two stretches of OH DNA requires energy, which in most organisms is provided by the 5' 3' cleavage of ATP to AMP + PPi. PO32– ATP IV. EUKARYOTIC DNA REPLICATION DNA ligase AMP + PPi The process of eukaryotic DNA replication closely follows that of prokaryotic DNA synthesis. Some differences, such as the multiple ori- gins of replication in eukaryotic cells versus single origins of replication 3' in prokaryotes, have already been noted. Eukaryotic single-stranded 5' DNA-binding proteins and ATP-dependent DNA helicases have been A T T G C A identified, whose functions are analogous to those of the prokaryotic enzymes previously discussed. In contrast, RNA primers are removed T A A C G T O O by RNase H and FEN1 rather than by a DNA polymerase. 5' P 3' O– O A. The eukaryotic cell cycle The events surrounding eukaryotic DNA replication and cell division Figure 29.20 (mitosis) are coordinated to produce the cell cycle (Figure 29.21). Formation of a phosphodiester bond The period preceding replication is called the G1 phase (Gap1). by DNA ligase. [Note: AMP is linked DNA replication occurs during the S (synthesis) phase. Following to ligase then the 5'phosphate and released.] 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 407 IV. Eukaryotic DNA Replication 407 DNA synthesis, there is another period (G2 phase, or Gap2) before Cells mitosis (M). Cells that have stopped dividing, such as mature neu- that rons, are said to have gone out of the cell cycle into the G0 phase. reactivate Mitosis G0 Cells can leave the G0 phase and reenter the early G1 phase to resume division. [Note: The cell cycle is controlled at a series of M “checkpoints” that prevent entry into the next phase of the cycle until G2 the preceding phase has been completed. Two key classes of pro- Gap 2 teins that control the progress of a cell through the cell cycle are the cyclins and cyclin-dependent kinases (Cdk).] G1 Gap 1 B. Eukaryotic DNA polymerases DNA synthesis G0 At least five key eukaryotic DNA polymerases have been identified Cells and categorized on the basis of molecular weight, cellular location, S that have ceased sensitivity to inhibitors, and the templates or substrates on which division they act. They are designated by Greek letters rather than by Roman numerals (Figure 29.22). Figure 29.21 1. Pol α: Pol α is a multisubunit enzyme. One subunit has primase The eukaryotic cell cycle. activity, which initiates strand synthesis on the leading strand and at the beginning of each Okazaki fragment on the lagging strand. The primase subunit synthesizes a short RNA primer that is extended by the pol α 5'→3' polymerase activity, generating a short piece of DNA. POLYM- PROOF- ERASE FUNCTION READING* 2. Pol ε and pol δ: Pol ε is thought to be recruited to complete DNA Pol α Contains primase – synthesis on the leading strand whereas pol δ elongates the (alpha) Initiates DNA Okazaki fragments of the lagging strand, each using 3' → 5' synthesis exonuclease activity to proofread the newly synthesized DNA. Pol β – [Note: DNA polymerase ε associates with the protein, proliferating (beta) Repair cell nuclear antigen (PCNA), which serves as a sliding DNA Pol γ Replicates + clamp in much the same way the β subunit of DNA polymerase III (gamma) mitochondrial DNA does in E. coli, thus ensuring high processivity.]. + Pol δ Thought to elongate (delta) Okazaki fragments 3. Pol β and pol γ: Pol β is involved in "gap filling" in DNA repair (see of the lagging strand below). Pol γ replicates mitochondrial DNA. Pol ε + (epsilon) Thought to elongate the leading strand C. Telomeres Telomeres are complexes of noncoding DNA plus proteins located at Figure 29.22 the ends of linear chromosomes. They maintain the structural integrity Activities of eukaryotic DNA of the chromosome, preventing attack by nucleases, and allow repair polymerases (pols). *3'→5' exo- systems to distinguish a true end from a break in dsDNA. In humans, nuclease activity. telomeric DNA consists of several thousand tandem repeats of a noncoding hexameric sequence, AG3T2, base-paired to a comple- mentary region of Cs and As. The GT-rich strand is longer than its CA complement, leaving ssDNA a few hundred nucleotides in length at the 3'-end. The single-stranded region is thought to fold back on itself, forming a loop structure that is stabilized by protein. 1. Telomere shortening: Eukaryotic cells face a special problem in replicating the ends of their linear DNA molecules. Following removal of the RNA primer from the extreme 5'-end of the lagging strand, there is no way to fill in the remaining gap with DNA. Consequently, in most normal human somatic cells, telomeres shorten with each successive cell division. Once telomeres are 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 408 408 29. DNA Structure, Replication, and Repair shortened beyond some critical length, the cell is no longer able 5' 3' to divide and is said to be senescent. In germ cells and other 3' 5' stem cells, as well as in cancer cells, telomeres do not shorten Eukaryotic DNA Telomere and the cells do not senesce. This is a result of the presence of a Telomere ribonucleoprotein, telomerase, which maintains telomeric length in these cells. Telomere repeats 2. Telomerase: This complex contains a protein that acts as a AGGGTT AGGGTT AGGGTT 3' reverse transcriptase, and a short piece of RNA that acts as a TCCCAA 5' template. The CA-rich RNA template base-pairs with the GT-rich, single-stranded 3'-end of telomeric DNA (Figure 29.23). The Newly synthesized strand with terminal RNA primer removed reverse transcriptase uses the RNA template to synthesize DNA in the usual 5'→3' direction, extending the already longer 3'-end. New Telomerase then translocates to the newly synthesized end, and Telomerase telomere 1 extends the repeat the process is repeated. Once the GT-rich strand has been 3'–end of lengthened, primase can use it as a template to synthesize an the DNA. RNA primer. The RNA primer is extended by DNA polymerase, Telomerase and the primer is removed. AGGGTT AGGGTT AGGGTT AGGGTT 3' Telomeres may be viewed as mitotic clocks in TCCCAA 5' 3' UCCCAA 5' that their length in most cells is inversely related to the number of times the cells have divided. The study of telomeres is providing insight into RNA template the biology of aging and cancer. RNA primer is is part of the 2 synthesized ribonucleoprotein by primase. telomerase D. Reverse transcriptases AGGGTT AGGGTT AGGGTT AGGGTT 3' Reverse transcriptases, as seen with telomerase, are RNA-directed DNA polymerases. A reverse transcriptase is involved in the replica- UCCCAA TCCCAA 5' 5' tion of retroviruses, such as human immunodeficiency virus (HIV).5 The 3'–end These viruses carry their genome in the form of ssRNA molecules. 3 of the RNA Following infection of a host cell, the viral enzyme, reverse transcrip- serves as an tase, uses the viral RNA as a template for the 5'→3' synthesis of acceptor for Primase DNA viral DNA, which then becomes integrated into host chromosomes. polymerase. Reverse transcriptase activity is also seen with transposons, DNA elements that can move about the genome (see p. 461). In eukary- AGGGTT AGGGTT AGGGTT AGGGTT 3' otes, such elements are transcribed to RNA, the RNA is used as a TCCCAA 5' TCCCAA UCCCAA 5' template for DNA synthesis by a reverse transcriptase encoded by the transposon, and the DNA is randomly inserted into the genome. [Note: Transposons that involve an RNA intermediate are called DNA retrotransposons or retroposons.] polymerase RNA primer 4 is removed. E. Inhibition of DNA synthesis by nucleoside analogs DNA chain growth can be blocked by the incorporation of certain nucleoside analogs that have been modified in the sugar portion of AGGGTT AGGGTT AGGGTT AGGGTT 3' the nucleoside (Figure 29.24).6 For example, removal of the hydroxyl TCCCAA TCCCAA TCCCAA 5' Figure 29.23 5See Chapter 38 in Lippincott’s Illustrated Reviews: Microbiology Mechanism of action of telomerase. INFO for a discussion of retroviruses. LINK 6See Chapter 38 in Lippincott’s Illustrated Reviews: Pharmacology for a discussion of nucloside analogs. 168397_P395-416.qxd7.0:29 DNA Structure 5-30-04 2010.4.5 8:20 AM Page 409 V. Organization of Eukaryotic DNA 409 group from the 3'-carbon of the deoxyribose ring as in 2',3'-dideoxyi- O O nosine (ddI, also known as didanosine), or conversion of the HN CH3 HN CH3 deoxyribose to another sugar such as arabinose, prevents further chain elongation. By blocking DNA replication, these compounds O N O N O O slow the division of rapidly growing cells and viruses. Cytosine ara- HO 5' HO 5' binoside (cytarabine, or araC) has been used in anticancer chemotherapy, whereas adenine arabinoside (vidarabine, or araA) is N3 OH an antiviral agent. Chemically modifying the sugar moiety, as seen in zidovudine (AZT, ZDV), also terminates DNA chain elongation. AZT Thymidine (zidovudine) (naturally occurring [Note: These drugs are generally supplied as nucleosides, which nucleoside) are then converted to the active nucleotides by cellular “salvage” enzymes (see p. 296).] O NH2 N N HN N V. ORGANIZATION OF EUKARYOTIC DNA N N N N O O A typical human cell contains 46 chromosomes, whose total DNA is HO HO approximately 1m long! It is difficult to imagine how such a large amount of genetic material can be effectively packaged into a volume OH the size of a cell nucleus so that it can be efficiently replicated and its 2'-3'-Dideoxyinosine, Deoxyadenosine genetic information expressed. To do so requires the interaction of DNA (ddI, didanosine) (naturally occurring with a large number of proteins, each of which performs a specific func- nucleoside) tion in the ordered packaging of these long molecules of DNA. Eukaryotic DNA is associated with tightly bound basic proteins, called Figure 29.24 histones. These serve to order the DNA into fundamental structural Examples of nucleoside analogs units, called nucleosomes, that resemble beads on a string. that lack a 3'-hydroxyl group. [Note: Nucleosomes are further arranged into increasingly more complex ddI is converted to its active form structures that organize and condense the long DNA molecules into (ddATP).] chromosomes that can be segregated during cell division. [Note: The complex of DNA and protein found inside the nuclei of eukaryotic cells Nucleosome core is called chromatin.] (H2A, H2B, H3, H4) 2 A. Histones and the formation of nucleosomes DNA There are five classes of histones, designated H1, H2A, H2B, H3, and H4. These small proteins are positively charged at physiologic pH as a result of their high content of lysine and arginine. Because of their positive charge, they form ionic bonds with negatively charged DNA. Histones, along with positively charged ions such as Mg2+, help Linker DNA neutralize the negatively charged DNA phosphate groups. 1. Nucleosomes: Two molecules each of H2A, H2B, H3, and H4 form the structural core of the individual nucleosome “beads.” Around this core, a segment of the DNA double helix is wound nearly twice, forming a negatively supertwisted helix (Figure 29.25). [Note: The N-terminal ends of these histones can be acetylated,