Second trdition

Peter J.Russell

DNA neplication

PRINCIPAL POINTS

. DNA replication in prokaryotes and eukaryotesoccurs by a semiconservative mechanism in whichthe two strands of a DNA double helix are separatedand a new complementary strand of DNA is synthe-sized on each of the two parental template strands.This mechanism ensures that genetic informationwill be copied faithfully at each cell division.

. The enz)'mes called DNA polymerases catalyze thesprthesis of DNA. Using deo4,ribonucleoside5'-triphosphate (dNTP) precursors, all DNA poly-merases make new sftands in the 5,-to-3' direction.

. DNA poll'rnerases cannot initiate the synthesis ofa new DNA strand. All new DNA strands need a shortprimer of RNA, the synthesis of which is catalyzed,bythe enzyme DNA primase.

. DNA replication in E. coli requires two DNA polymer_ases and several other enzymes and proteins. In bothprokaryotes and eukaryotes, the s1'nthesis of DNA is

continuous on one template strand and discontinuouson the other template strand, a process called semidis_continuous replication.

o In eukaryotes, DNA replication occurs in the S phaseof the cell cycle and is biochemically and molecularlysimilar to replication in prokaryotes. To enable longchromosomes ro replicare efficientl¡ DNA replicatiánis initiated at many sites (origins) along the chromo_somes and proceeds bidirectionally

. Special enzymes-telomerases-replicate the ends ofchromosomes in eukaryotes. A telomerase is a com_plex of proteins and RNA. The RNA acts as a templatefor the synthesis of the complementary telomererepeat of the chromosome. In mammals, telomeraseactivity is limited to immortal cells (such as rumorcells). The absence of telomerase activity in nontumorcells results in a progressive shortening of chromo_some ends as the cell divides, thereby limiting thenumber of cell divisions before the cell dies.

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#il.É,it ¡F.Pdpc*n. fia$ rAf,.lslTs,,,.,Ki$¡iity.;ia:reüti á:$Één*;$$¡r,.sü-iiiiüt the senetic

ff information encoded in the nucleotides can be trans-

4, tted.fion eüch c *lto,ált of,itipm0é-!y,'Jarnes : ,.,"*',W¡t nánd Fiancis Crick recognized that the,comole,:

rhentary retatio:nship letwee n D NA stra n ds: probab lywould be the basis for DNA replication. However; even

'after scientists confirmed this fact five years afterWatson and Crick developed their rnodel, many ques-

tions about the mechanics of DNA replicationremained. ln this chapter, you will learn about thesteps and enzymes involved in the replication ofprokaryotic and eukaryotic DNA molecules. Then, in theiActivity, you will have a chance to investigate thespecifics of DNA replication in E. coli.

Your goal in the chapter is to learn about the mechanismsof DNA replication and chromosome duplication inprokaryotes and eukaryotes and about some of theenzymes and other proteins needed for replication. Someof these enzymes are also involved in the repair of dam-age to DNA, a topic we discuss in Chapter 7.

Semiconservative DNA Replication

When Watson and Crick proposed their double-helixmodel for DNA in 1953, they realized that DNA

Figure 3.1Three models for DNA replication. (a) Semiconservative model (the correct model).(b) Conservative model. (c) Dispersive model. The parental strands are shown in taupe,and the newly synthesized strands are shown in red.

replication would be straightforward if their model wascorrect. That is, if the DNA molecule were unwoundand the two strands separated, each strand wouldbe a template for the synthesis of a new, complemen-tary strand of DNA that would remain bound to theparental strand. This model for DNA replication isknown as the semiconservative model, because eachprogeny molecule retains one of the parental strands(Figure 3.la).

At the time, two'other models for DNA replicationwere proposed: the conservative model (Figure 3.lb)and the dispersive model (Figure 3.tc). In the conser-vative model, the two parental strands of DNA remaintogether or pair again after replication and, as a whole,serve as a template for the synthesis of new progenyDNA double helices. Thus, one of the two progeny DNAmolecules actually is the parental double-stranded DNAmolecule, and the other consists of new material. In thedispersive model, the parental double helix is cleavedinto double-stranded DNA segments that act as tem-plates for the synthesis of new double-stranded DNAsegments. Somehow, the segments reassemble into com-plete DNA double helices, with parental and progenyDNA segments interspersed. Thus, although the twoprogeny DNAs are identical with respect to their base-pair sequence, double-stranded parental DNA hasbecome dispersed throughout both progeny molecules.It is hard to imagine how the DNA sequences ofchromosomes could be kept the same with such a

a) The semiconservative model

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Firstgeneration

Secondgeneration

mechanism. We include thiscompleteness.

model only for historical

The Meselson-Stahl ExperimentIn 1958, Matthew Meselson and Frank Stahl obtainedexperimental evidence that the semiconservative repli-cation model was correct. Meselson and Stahl grew

E. coli in a medium in whichthe only nitrogen source wasl5NHoCl (ammonium chloride)(Figure 3.2). ln rhis com-pound, the normal isotope ofnitrogen, 1aN, is replaced with

15N, the heavy isotope. (Note: Density is weight dividedby volume, so 15N, with one extra neutron in itsnucleus, is %+ denser than ]aN.) As a result, all the bac-teria's nitrogen-containing compounds, including itsDNA, contained 15N instead of laN. 15N DNA can beseparated from laN DNA by using equilibrium densitygradient centrifugation (described in Box 3.1). Briefly,in this technique, through high-speed centrifugation, a

solution of cesium chloride (CsCl) forms a density gra-dient, with the lightest marerial ar rhe rop of the tubeand the densest material at the bottom. If DNA is pre-sent in the solution, it forms a band at a position whereits buoyant density is the same as that of the surround-ing cesium chloride.

Next, the l5N-labeled bacteria were transferred to a

medium containing nitrogen in the normal laN form, andüe bacteria were allowed to reproduce for several gener-

ia re ¡ eae d¡ 4-* ?e É$€*t{$6tt{* *,.!9}FN

The

Meselson-StahlExperiment

In equilibrium density gradient centrifugation, a concen-trated solution of cesium chloride (CsCI) is centrifuged athigh speed to produce a linear concentration gradient ofthe CsCl. The actual densities of CsCl at the extremes ofthe gradient are related to the CsCl concentration that iscentriluged. /

For example, to examine DNfof density L7O {cñ Qtypical density for DNA), one makes a gradient which spansthat density-for example, from 1.60 to 1.80g/cm3. If DNA

Box Figure 3"1

Schematic diagrarn forseparating DNAs of differentbuoyant densities by equilib-rium centrifugation in acesium chloride densitygradient. The separation of1aN-DNA and 15N-DNA isiliustrated.

is mixed with the CsCl and rhe mixture is centrifuged, theDNA comes to equilibrium at the point in the gradientwhere its buoyant density equals the density of the sur-rounding CsCl. (See rhe accompanying figure.) The DNA issaid to have banded in the gradient. If DNAs that havedifferent densities are present, as is the case with ItN DNAand laN DNA, they band (come to equilibrium) in differentpositions. The DNA is detected in the gradient by its ultra-violet absorption.

14N-DNA

15N-DNA

'@cooo).co(s(t)

ocCentrifugation for 50-60 hat 100,000 X g results ingeneration of gradient of CsCland banding of DNA

S emiconsery atiy e DN A Replication Id

ations. During this time, samples of E. coli were taken andthe DNA was extracted andanalyzedin CsCl density gradi-ents. (See Figure 3.2.) After one replication cycle (one gen,eration) in the laN medium, all of the DNA had a densitythat was exactly intermediate between that of 15N DNA andthat of 14N DNA. After two replication cycles, half the DNAwas of the intermediate density and half was of the densityof DNA containing entirely taN. These observations, pre-sented in Figure 3.2, and those obtained from subsequentreplication cycles were exactly what the semiconservativemodel predicted.

If the conservative model for DNA replication werecorrect, after one replication cycle there would havebeen two bands of DNA: one in the heavy-density posi-tion of the gradient containing parental DNA moleculeswith both srrands labeled with 15N, the other in thelight-density position conraining progeny DNA mole-cules with both strands labeled with laN. (See Figure3.fb.) The heary parental DNA band would have beenseen at each subsequent replication cycle, in the amountfound at the start of the experiment. All new DNA mol-ecules would than have had both srrands labeled withlaN. Therefore, the amounr of DNA in the light-densityposition would have increased with each replicationcycle. In the conservative model of DNA replication,then, the most significant prediction was that at no timewould any DNA of íntermediate density be found. The factthat intermediate-density DNA was found ruled out theconservative model.

If the dispersive model for DNA replication werecorrect, then all DNA present in the laN medium after

DNA in 6M CsCl

f- Ghapter 3 DNA Reptication

Flgure 3.2

The Meselson-Stahl experiment. The demonstration of semiconservative replication

inE. coli. Cells were grown in a l5N-containing medium for several replication cycles

and then were transferred to a laN-containing medium. At various times over several

replication cycles, samples were taken; the DNA was extracted and, analyzed by CsCl

eq-uilibrium density gradient centrifugation. Shown in the figure are a schematic inter-

pietation of the DNA composition after various replication cycles, photographs of the

DNA bands, and densitometric scans of the bands.

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cycle would be dispersed throughout the progeny DNAdouble helices produced. Thus, the 15N-r5N DNA seg-

ments dispersed among ne- 14N-r4N DNA after onereplication cycle would then be distributed among twiceas many DNA molecules after two replication cycles. As

Photographsof DNA bands

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one replication cycle would have been of intermediatedensity (see Figure 3.1c), and this was indeed seen inthe Meselson-Stahl experiment. The dispersive modelpredicted that, after a second replication cycle in the

same medium, DNA segments from the first replication

Replicationcycle 3

.:

I

I

a result, the DNA molecules would be found in oneband located halfway between the intermediate-densityand light-density band positions. With subsequentreplication cycles, there would continue to be one band,and it would become lighter in density with eachreplication cycle. The results of the Meselson-Stahlexperiment did not bear out this prediction, so thedispersive model was ruled out.

Semiconservative DNA Replication in EukaryotesDNA replication in eukaryotes also occurs by a semicon-servative mechanism. One way in which the process canbe visualized is by staining replicating chromosomes incells growing in tissue culture with 5-bromodeoxyuridine(BUdR). BUdR is abase analog, with a strucrure similar tothat of the normal base th)'mine in DNA. During replica-tion, wherever a T is called for in the new strand, BUdRcan be incorporated instead. Counterintuitively, mitoticchromosomes stained with a fluorescent dye and Giemsastain become lighter as the amount of BUdR incorporatedincreqses.

To examine DNA replication, Chinese hamsterovary (CHO) cells in culture were treared wirh BUdR.After two cycles of replication, mitotic chromosomes

Figure 3.3Visualization of semiconservative DNA replication in eukary-otes. Shown are harlequin chromosomes in Chinese hamsterovary cells that have been allowed to go through two roundsof DNA replication in the presence of the base analog5-bromodeoxyuridine, followed by staining. Arrows indicatethe sites where crossing-over has occurred.

DNA Polymerases, the DNA Replicating Enzymes ú

were stained, with the result shown in Figure 3.3. Sisterchromatids are visible, one darkly stained and onelightly stained. The dark-light staining partern bringsto mind the costumes of harlequins, so the chromo-somes are called harlequin chromosomes. The interpreta-tion of the pattern is as follows: The darker stainingchromatids have less BUdR in their DNA than thelighter staining chromatids. That is, if semiconservativereplication occurs, then, after one replication cycle,two progeny DNAs will be produced, each with oneT strand and one BUdR strand. Following that, a sec-ond replication cycle in the presence of BUdR willproduce one sister chromatid consisting of DNA withone T-containing strand and one BUdR-containingstrand (darkly staining) and another sister chromatidconsisting of DNA with two BUdR-containing strands(lightly staining). This is the staining pattern that wasobserved, showing that semiconservative DNA replica-tion also is the correct model in eukaryotes.

KEYNOTE

DNA replication in E coliand other prokaryotes andin eukaryotes occurs by a semiconservative mecha-nism in which the strands of a DNA double helixseparate and a new complementary strand of DNA issynthesized on each of the two parental templaiestránds. Semiconservative replication results in twodouble-stranded DNA moleculei, eadh of which hasone strand from the parent molecule and one newlysynthesized strand.

DNA Polymerases, the DNAReplicating Enrymes

first to identify the enz)rynes necessary for DNA replica-tion. Their work focused on bacteria, because it wasassumed that bacterial replication machinery would beless complex than that of eukaryotes. Kornberg sharedthe 1959 Nobel Prize in Physiology or Medicine for his"discovery of the mechanisms in the biological slnthesisof deoxyribonucleic acid."

DNA Polymerase IKornberg's approach was to identify all the ingredientsneeded to synthesize E. coli DNA in vitro. The first suc-cessful DNA synthesis was accomplished in.a reactionmixture coritaining DNA fragments-a mixture of fourdeoxyribonucleoside 5'-triphosphate precursors (dATB

b Ghapter 3 DNA RePlication

dcTE dTTB and dCTR collectively abbreviated dNTP, for

deoxyribonucleoside triphosphate)-and an E' coli lysate

(cells of the bacteria, broken open to release their con-

tents). To measure the minute quantities of DNA

expected to be synthesized in the reaction, Kornberg used

radioactively labeled dNTPs.

Kornberg analyzed the lysate and isolated an enzyme

that was capable of DNA s)'nthesis. This enz)'rne was

originally .ull"d th" Kornberg enzyme, but is now called

DNA polymerase I (or DNA Pol I; by definition,

"rry-ét that cafalyze DNA synthesis are called DNA

polymerases).With DNA Pol I isolated, more detailed information

could be obtained about DNA synthesis in vitro'Researchers found that four components were needed' Ifany one of the following four components was omitted,

DNA synthesis would not occur:

1. All four dNTPs. (If any one dNTP is missing, no syn-

thesis occurs.) These molecules are the precursors

for the nucleotide (phosphate-sugar-base) buildingblocks of DNA described in Chapter 10 (p. 258)'

2. A fragment of DNA to act as a template.

3. DNA Pol l.4. Magnesium ions (Mg2*), needed for optimal DNA

polymerase activity.Subsequent experiments showed that the fragments of

DNA acted as a template for the synthesis of the new

DNA; that is, the new DNA made in vitro was a faithful

base-pair-for-base-pair copy of the original DNA'

Roles of DNA PolYmerases

AII DNA polymerases from prokaryotes and eukaryotes

catalyze the polymerization of nucleotide precursors

XrX$ffi$m*fr#g3 (dNTPs) into a DNA chain

é DNA Biosvnthe- (Figure 3'4a)' The same reac-

(€ ;fr:;"J]'ffi tion in shorthand notation is

DNA strand rs :l;J;"* ,l'f"'in#b rhe

Made features:

1. At the growing end of the DNA chain, DNA poly-

merase catalyzes the formation of a phosphodiester

bond between the 3'-OH group of the deoxyribose

on the last nucleotide and the 5'-phosphate of the

dNTP precursor. The energy for the formation of the

phosphodiester bond comes from the release of two

of three phosphates from the dNTP The important

concept here is that the lengtheningDNA chain acts as

aprimer in the reaction-a preexisting poll'nucleotide

chain to which a new nucleotide can be added at the

free 3'-OH.2, At each step in lengthening the new DNA chain,

DNA polymerase finds the correct precursor dNTP

that cán form a complementary base pair with the

nucleotide on the template strand of DNA.

Nucleotides are added rapidly-for example, 850

per second in E. coli and 60-90 per second inhuman tissue culture cells. The process does notoccur with 100 percent accuracy, but the error fre-

quency is very low.3. The direction of synthesis of the new DNA chain is

only from 5' to 3' , because of the properties of DNA

polymerase.One of the best understood systems of DNA replica-

tion is that of E. coli. For several years after the discov-

ery of DNA polymerase l, scientists believed that that

enzyme was the only DNA replication enz"rrre inE. colí.

However, genetic studies disproved that hypothesis'

One way to study the action of an enzyme in vivo is toinduce a mutation in the gene that codes for the

eÍzyme. ln this way, the phenotypic consequences ofthe mutation can be compared with the wild-type phe-

notype. The first DNA Pol I mutant, polAl, was isolated

in 1969 by Peter Delucia andJohn Cairns. This mutant

shows less than I percent of normal polymerizing activ-

ity and near-normal 5'->3' exonuclease activity. DNA

polymerase was expected to be essential to cell function'so a mutation in the gene for that enzyme was expected

to be lethal or at least crippling. Unexpectedly, however,

E. coli cells carrying tt'e polAT mutation grew and

divided normally. Nonetheless, polAl mutants have a

high mutation rate when they are exposed to ultraviolet(UV) light and chemical mutagens, a property whichwas interpreted to mean that DNA polymerase I has an

important function in repairing damaged (chemically

changed) DNA.To study the consequences of mutations in genes

coding for essential proteins and enzymes, geneticists

find it easiest to work with temperature-sensitivemutants-mutants that function normally until the

temperature is raised past some threshold level, at whichtime some temperature-sensitive defect is manifested. AtE. colí's normal growth temperature of 37"C, tempera-

ture-sensitiv e polAexl mutant strains produce DNA Pol Iwith normal activity. In vitro at 42"C, however, the tem-

perature-sensitive DNA Pol I has near-normal polymer-

izing activity, but is defective in 5'-+3' exonuclease

activity (the progressive removal of nucleotides from a

free 5' end toward the 3' end). A 42"C temperature-

sensitive polAexl mutants die (are lethal) showing that

5''-->3' exonuclease activity of DNA Pol I is essential to

DNA replication. Taken together, the results of studies of

- the polAl and polAexl DNA Pol I mutants indicated that

there must be other DNA-polymetizíng enzymes in the

cell. With improvements in preparing cell extracts and inenzyme assay techniques, two new E' coli DNA poly-

merases were identified. Malcolm Gefter, Rolf Knippers,

and C. C. Richardson, all working independently, dis-

covered DNA Pol Il in 1970, and Tom Kornberg and

!

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DN A P oly mer ases, the DN A Replicating Enzy mes fil

Figure 3.4DNA chain elongation catalyzed. by DNA polymerase. (a) Mechanism at molecularlevel. (b) The same mechanism, using a shorthand method to represent DNA.

a) Mechanism of DNA elongation

New strand Template strand

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[18 chapter 3 DNA Reptication

Gefter, working together, discovered DNA Pol III in1971. Since that time, two orher E. coli DNA poly-merases-DNA Pol IV and DNA Pol V-have been dis-covered. DNA Pol Il (poln gene), DNA Pol IV (dinBgene), and DNA Pol Y (umuDC gene) are polymerasesthat function in DNA repair, while DNA Pol I and DNAPol III are the polymerases necessary for replication.

DNA polyrnerase I is encoded by the polA gene andconsists of one polypeptide. The core DNA polymerase IIIcontains the catalytic functions of the enzyme and con-sists of three pollpeptides: a (alpha, encoded by t}i'e dnaEgene), e (epsilon, encoded by the dnaQ gene), and 0(theta, encoded by the holE gene). The complete DNAPol III errzyrr.e, called the DNA Pol III holoenzyme, con-tains an additional six different pollpeptldes.

Both DNA Pol I and DNA Pol III replicate DNA inthe 5'-+3'direction in the cell. Both enzymes also have3'-'>5'exonuclease activity: They can remove nucleo-tides from the 3' end of a DNA chain. This enzymeactivity is part of an error correction mechanism. If anincorrect base is inserted by DNA polymerase (an eventthat occurs at a frequency o[ about 10-6 for both DNApolymerase I and DNA polymerase III), in many cases

the error is recognized immediately. Then, the 3'--+5'exonuclease activity excises the erroneous nucleotidefrom the new strand. This process resembles that of thebackspace or delete key on a computer keyboard. Afterexcision, the DNA polymerase resumes motion in the for-ward direction and inserts the correct character. Thus, inDNA replication, 3'-+5' exonuclease activity is a

proofreading mechanism that helps keep the frequency

of errors very low. With proofreading, the frequency ofreplication errors by DNA polymerase I or III is reducedto less than 10-e. The 3'-+5' exonuclease activity ofDNA Pol I is a domain of the single pollpeptide onzyme,while, for DNA Pol III, ir is rhe function of the e subunit,stimulated by the 0 subunir.

Uniquely of the two, DNA Pol I has 3'--+5' exonu-clease activity and can remove either DNA or RNAnucleotides from the 5' end of a nucleic acid strand. Thisactivity is important in DNA replication and is examinedlater in the chapter.

KEYNOTE

The enzymes that catalyze the synthesis of DNA are

called DNA polymerases. All known DNA polymerasessynthesize DNA in the 5'*3' direction. Polymerasesmay also have other activities, such as proofreading, orremoving nucleotides from a strand in the 5'+3'direction.

Molecular Model of DNA ReplicationTable 3.1 presents the functions of some of the E. coliDNA replication genes and key DNA sequences involvedin replication. A number of the genes were identified bymutational analysis. In this section, we discuss a molecu-lar model of DNA replication involving these genes andsequences.

@ Functions of Some of the Genes and DNA Sequences Involved in DNA Reptication in E coli

Gene Product or Function Gene

{tt$:

DNA polymerase IDNA pollmerase IIIInitiator protein; binds to oriC

IHF protein (DNA binding protein); binds to oriC

FIS protein (DNA binding protein); binds to oriC

Helicase and activator of primase

Complexes with dnaB protein and delivers i¡ to DNA

Primase; makes RNA primer for extension by DNA poll,rnerase IIISingle-stranded binding (SSB) proteins; bind to unwound single-srranded

arms of replication forks

DNA ligase; seals single-stranded gaps

Gyrase (t1pe II topoisomerase); replication swivel to avoid tanglingof DNA as replication fork advances

Origin of chromosomal repiication

Terminus of chromosomal repl ication

TBP (ter binding protein); stalls replication forks

polA

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lnitiation of ReplicationThe initiation of replication is directed by a DNAsequence called the replicator. The replicator usuallyincludes the origin of replication, the specific regionwhere the DNA double helix denatures into singlestrands and within which replication commences. Thelocally denatured segment of DNA is called a replicationbubble. The segments of untwisted single strands onwhich the new strands are made (in accordance withcomplementary base-pairing rules) are called thetemplate strands.

When DNA untwists to expose the two single-stranded template strands for DNA replication, a Y-shapedstructure called a replication fork is formed. A replica-tion fork moves in the direction of untwisting of the DNA.When DNA untwists in the middle of a DNA molecule, as

in a circular chromosome, there are two replicationforks-like two Ys joined together at their tops. In many(but not all) cases, each replication fork is active, so DNAreplication proceeds bidirectionally.

An outline of the initiation of replication in E. coli isshown in Figure 3.5. The E. coli replicator is oriC, whichspans 245 bp and contains a cluster of three copies ofa 13-bp AT-rich sequence and four copies of a 9-bpsequence. An AT-rich region is relatively easy to denatureto single strands and is characteristic of replicators in allorganisms. For the initiation of replication, an initiatorprotein or proteins bind to the replicator and stimulateüe local denaturing at the AT-rich region. The E. coli ini-tiator protein is DnaA (dnsA gene), which binds to the9-bp regions in multiple copies, leading to the denaturingof the region with the 13-bp sequences. DNA helicases(¡he dnaB gene) are recruited and loaded onto the DNA.The helicases begin untwisting the DNA in both direc-Lions from the origin of replication. The energy for theuntwisting comes from the hydrolysis of AfP-a reactionthat causes a change in the shape of the helicase, enablingthe enz)'rne to move along a single strand of DNA. Byrepeated ATP hydrolysis, the helicase can move along thesingle strand and untwist any double-stranded DNA itencounters.

Next, each DNA helicase recruits the enzyme DNAprimase (a product of the dnaG gene), forming a com-plex called the primosome. DNA primase is importantin DNA replication because no known DNA poly-merases can initiate the synthesis of a DNA strand; theycan only add nucleotides to a preexisting strand. TheDNA primase on each template strand is activated by itsassociated DNA helicase and synthesizes a short RNAprimer (about 5-10 nucleotides) to which newnucleotides can be added by DNA polymerase. TheRNA primer is removed later and replaced with DNA,we will return to discuss this event further. At thispoint, the bidirectional replication of DNA has justbegun.

Figure 3.5Initiation ofreplication inE. coli. The DnaA initiator proteinbinds to oriC (the replicator) and stimulates denaturation of theDNA. DNA helicases are recruited and begin to untwist the DNAto form two head-to-head replication forks.

¡4| DnaA

13-bp g-bprepeats repeats

Molecular Model of DNAReplication [[

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5',

The ínitiation of DNA synthesis first ínvolves tl.le denat-uration of double-stranded DNA at an origin of replica-tion, catalyzed by DNA helicase. Next, DNA primase

binds to the helicase and the denatured DNA'and syn-thesizes a short RNA primer. The RNA primer is extendedby DNA polymerase as Rew DNA is made. Later, the RNAprimer is removed.

onnneticase-- 61

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We must be clear about the difference between a

template and a primer with respect to DNA replication.A template strand is the one on which thé new strand issynthesized according to complementary base-pairingrules. A primer is a short segment of nucleotides boundto the template strand. The primer acts as a substrate forDNA polymerase, which extends the primer as a newDNA strand, the sequence of which is complementary tothe template strand.

EYNO

DNA primase e@ -j

Ghapter 3 DNA RePlication

Semidiscontinuous DNA ReplicationThe foregoing discussion of the initiation of replication

consideréd the production of two replication forks when

DNA denatures at an origin' The replication events are

being made continuously, whereas the lagging strand can

be made only in pieces, or discontinuously, DNA replica-

tion as a whole occurs in a semidiscontinuous manner'

The fragments of lagging strand made in the process

just described are called Okazaki fragments' after their

discoverers, Reiji and Tüneko Okazaki and colleagues'

Experimentally, the Okazakis added a radioactive DNA

p.".rrrro. (3U-thymidlne) to cultures of E. coli for

b.5 percent of a generation time. Next, they added a large

amount of nonradioactive thymidine to prevent the

incorporation of any more of the radioactive precursor

into the DNA. At various times, they extracted the DNA

and determined the size of the newly labeled molecules'

At intervals soon after the labeling period, most of the

radioactive 3H-thymidine was present in low-molecular-

weight DNA about 100 to 1,000 nucleotides long' As

timá increased, a greater and greater proportion of the

labeled molecules was found in DNA of a high molecu-

lar weight. These results indicated that DNA replication

normally involves the s1'nthesis of short DNA seg-

ments-the now-named Okazaki fragments-that are

subsequently linked together.ln Figure 3.6c, the process repeats itself: Helicase

untwists the DNA, DNA is continuously synthesized on

the leading-strand template, and DNA is discontinu-ously synthesized on the lagging-strand template, withnew RNA primers being synthesized every f ,000-2'000nucleotides along the template. Eventuall¡ the uncon-

nected Okazaki fragments on the lagging-strand tem-

plate are joined into a continuous DNA strand' Joiningihem requires the activities of DNA polymerase I and

DNA ligase. Consider two adjacent Okazaki fragments'

The 3' end of the newer fragment is adjacent, but notjoined, to the primer at the 5' end of the previously

made fragment. DNA polymerase lll leaves the DNA,

and DNA polymerase I continues the 5'--->3' synthesis

of the fragment, simultaneously removing the primer

section of the older fragment by its 5'->3' exonuclease

activity (Figure 3.6d). When DNA polymerase I has

replaced all the RNA primer nucleotides with DNA

nucleotides, a single-stranded nick is left between the

two fragments. The fragments are joined by DNA ligase

to produce a longer DNA strand (Figure 3.6e)' The

cat;lydc reaction of DNA ligase is diagrammed inFigure 3.7. The entire sequence of events is repeated

until all the DNA is rePlicated'In sum, DNA replication in E' coli is a complicated

process, and what we have shown is a simplified versiort

óf lt. t.t fact, the key replication proteins are closely

associated, to form a replisome. Figure 3.8 shows the lag-

ging-strand DNA, folded so that its DNA polymerase III iscomplexed with the DNA poll'rnerase III on the leading

,t uttd. These are two copies of the core errzqe described

earlier (see p. 50), held together by the six other

pollpeptides to form the DNA Pol III holoenzyme' Only

#f3*&3r*?€x*ffi

MolecularModel of DNA

Replication

identical with each rePlicationfork, so we will focus on just

one fork in the discussion thatfollows.

As we just discussed (see

Figure 3.5), helicase untwists

rs

the DNA to produce single-stranded template strands'

Single-strand DNA-binding (SSB) proteins bind to the

sinfle-stranded DNA, stabilizing it (Figure 3'6) and pre-

veriting it from reannealing. lnE. coli, the SSB protein (ssb

gene) is a tetramer of four identical subunis that binds to

á 32-nucleotide segment of DNA. More than 200 of the

proteins bind to each replication fork.The RNA primer made by DNA primase in Figure 3'5

is at the 5' ená of the new DNA strand being s;mthesized

on the template strand at the bottom of Figure 3'6' The

DNA primase at the fork s1'nthesizes another RNA primer,

this one on the top template DNA strand (Figure 3'6a)'

These RNA primers are lengthened by DNA pol1'rnerase

IlI, which synthesizes new DNA complementary to the

template stiands while simultaneously displacing the

¡ouna SSB proteins (Figure 3.6a). Recall that DNA poly-

merases can make DNA only in the 5 '-> 3' direction; how-

ever, the two DNA strands are of opposite polarity To

maintain the 5'-+3' polarity of DNA s1'nthesis on each

template, and to maintain one overall direction of replica-

tionfork movement, DNA is made in opposite directions

on the two template strands. (See Figure 3'6a') The new

strand being made in the 5'->3' direction (in the same

direction urih" *ou"*ent of the replication fork) is called

the leading strand, and the new strand being made in the

direction opposite that of the movement of the replication

fork is calléá the tagging strand. The leading strand needs

a single RNA primer for is slnthesis, whereas the lagging

strand needs a series of primers, as we will see'

As the replication fork moves, helicase untwists more

DNA (Figure 3.6b). DNA gyrase (a form of topoiso-

merase) ,Jlu*", the tension produced in the DNA ahead of

the replication fork. This tension could be considerable,

because the replication fork rotates at about 3,000 rpm'

On the leading-strand template (the bottom strand in

Figure 3.6), the leading strand is synthesized continu-

oully toward the replication fork. Because DNA synthesis

can proceed only in the 5'--+3''direction, however, lag-

ging-strand s1'nthesis has gone as far as it can' For DNA

ieplication to continue on the lagging-strand template

(th" top strand in Figure 3.6), a new initiation of DNA

synthesis must occur. An RNA primer is made by the

óNA primase at the replication fork. (See Figure 3'6b')

DNA poli.'rnerase III then adds DNA to the RNA primer to

make another DNA fragment. Since the leading strand is

Molecular Model of DNA Replication ][

Dlodel for the events o""o*&romosome. (a) Initiation. (b) Further urrt*istirrg und erongation of the new DNAs¿nds. (c) Furthe¡ untwisting and continued DN,isynthesis. (d) Removar of theprimer by DNA polymerase I' (e) Joining of adjacent bNA frugrrr".,,s by the action o{DNA ligase. Green = RNA; red = new DNA.

r) lnit¡at¡on; RNAprimer made byDNA pr¡mase startsreplicat¡on of laggíngstrand (synthesis oflst Okazaki fragment)

b) Further untwist¡ngand elongation of newDNA strands;2ndOkazaki fragmentelongated

Polymerase IIILagg¡ngstrand

Fork movement

-

Leadingstrand

1st Okazakifragment

DNA helicase

Polymerase tIIsynthesized by DNA polymerase III

,SSB (single-strand DNA binding proteins)

/r.Lrzana primer for 2nd Okazaki fragment made by DNA primase5

1st Okazakifragment

RNA primer made by primase

Polymerase III dissociates

¡ff Discontinuoüs synthesis on this strand

2nd Okazakifragment elongation

primer for 3rd Okazaki fragmenl

Continued untwisting and fork movement

Polymerase III dissociates

c) Process continues;2nd Okazaki fragmentfinished,3rd beingsynthesized; DNAprimase beginning4th fragment

d) Primer removed byDNA polymerase I

e) Joining of ad¡acentDNA fragments byDNA ligase

RNA primer being replaced withDNA by polymerase I..

Nick sealed by DNA ligase

for simplicitybrings the 3'near the site

4th Okazakifragment

sth Okazakifragment

rl

the core enzymes are shown in the figure,The folding of the lagging-strand t"*"plut"end of each completed Okazaki fragment

prime,rs intermittently Similarly, because the lagging_strand polymerase is complexed with the other replicátlJnproteins at the fork, that polyrnerase can be reused con_

Itl$ly at the same replication fork, slmthesizing a srringol Okazaki fragmenb as it moves with the rest of the

where thestays near

nextthe

Okazaki fragment will start. The primasereplication fork, s1'nthes2ing new RNA

F chaPter 3 DNA RePlication

Figure 3.7

¡,"tiot of ONe hgase in sealing the nick between adjacent

DNA fragments (e.g., Okazaki fragments) to fonn a longet,

covalently continuous chain. The DNA ligase catalyzes the

formation of a phosphodiester bond between the 3'-OH and the

5'-phosphate groups on either side of a nick, sealing the nick'

3'ffiis',ffib'tffi

Replication of Circular DNA and theSupercoiling ProblemIn E. coli, the parental DNA strands remain in a circular

form throughout the replication cycle. This is true ofmany, but not all, circular DNA molecules. During repli-

cation, these circular DNA molecules exhibit a theta-like(0) shape, because of a replicating bubble's initiation at

the replication origin. (See Figure 3.9.)

As the two DNA strands in a circular chromosome

untwist during replication, positive supercoils form else-

where in the molecule. For the replication fork to move,

then, the chromosome ahead of the fork must rotate. Given

a rate of movement of the replication fork of 500 nucleotides

per second, at l0 base pairs per tum, üe helix ahead of the

fork must rotate at 50 revolutions per second, or 3,000 rpm-

The supercoiling problem is solved by the action oftopoisomerases (see Chapter 2, pp. 29 -30)-enz)'rnes that

introduce negative supercoils into DNA or that conveftnegatively supercoiled DNA into relaxed DNA. Topoiso-

merases play an important role in replication by prevent-

ing excessively supercoiled DNA from forming and

thereby allowing both parental strands to remain intact

during the replication cycle as the replication forkmigrates. That is, the unreplicated part of the theta struc-

ture ahead of the replication fork repeatedly has negative

supercoils introduced into it by the action of DNA g)'rase

(a type Il topoisomerase), relieving the positive supercoil-

ing that occurs as DNA is untwisted during replication.

Rolling Citcle ReplicationA rolling circle model of replication (Figure 3.10)

applies to the replication of several viral DNAs, such as

bacteriophages @X174 (see Figure 2.16) and l. (see

Figure 2.I7). The starting point is a circular, double-stranded DNA molecule. In the case of @XI74, this isproduced when a complementary strand is synthesized

with the use of the circular, single-stranded genomic

DNA as a template. ln the case of 1,, the circular mole-

cule is produced when the short, complementarysingle-stranded ends of the linear double-strandedDNA genome (see Figure 3.11, p. 56) pair together after

infecting the bacterial cell.The first step in rolling circle replication is the

generation of a specific cut (nick) in one of the twostrands at the origin of replication. The 5' end of the cut

strand is then displaced from the circular molecule, cre-

ating a replication fork and leaving a single-stranded

Figure 3.8Model for the replisome, the complex ofkey replication proteins, u¡ith the DNAat the replication fork. The DNA poly-

merase III on the lagging-strand template

(top of figure) isjust finishing the synthe-

sis of an Okazaki fragment.

ParentalDNA

s', yffimgmffis'3'-s' rffiffiffiftl s'

tNick sealed

replication machine. That is, the complex of replication

proteins that forms at the replication fork moves mostly as

a unit along the DNA and enables new DNA to be

s;mthesized efficiently on both the leading-strand and lag-

ging-strand templates. Finally, with bidirectional replica-

tion, we can üsualize two replisomes, each the mirrorimage of the other, on replication forks that are moving in

opposite directions (Figure 3.9).

# l¡i¡t¡n¡ tome of the spicific ilements an¿i"

"$ processes needed for DNA replication in the

Q iRctivity lJnroveling DNA Repticotion on yourh co-Rolvt.

I or.rn tisase

Single-strand nick

ai

DNA

SSBprotein

Template

TemplateDNA

RNAprimer

DNA primase

DNA helicase

DNA polymerase III

Okazaki fragment

s',

3'.

Direction of fork movement

5'3'

Leadingstrand

I

¡l

DNA polymerase III

Moleanlar Múel of DNA Replication

FÉEe 3.9 Figure 3.10lhilircctional rephcation of ci¡cular DNA nolecules.

stretch of DNA that serves as a template for the additionof deoxy'ribonucleotides to the free 3' end by DNA polym-erase III, using the intact circular DNA as a template.This new DNA s1'nthesis occurs continuously as the 5'cut end is displaced from the circular molecule; the

intact circular DNA acts as the leading-strand template.The 5' end of the cut DNA strand is rolled out as a free

*tongue" ofincreasing lengü as replication proceeds. This

single-stranded DNA tongue becomes covered with SSB

proteins. New DNA is slmthesized on the displaced DNA in

The replication process of double-stranded circular DNA mol-ecules througlr the rolling circle rnechanism. The active forcethat unwinds the 5' tail is the movement of the replisome pro-pelled by its helicase components.

Q ttict< is made in the+ strand of theparental duplexlO = orioin)'3',

s',

Q The s'end isdisplaced andcovered by SSBs

e Polymerizationat the 3'endadds newdeoxyribonucleotides

Attachment ofreplisome andformation ofOkazaki fragments

I

Old Okazakifragment

Newly initiatedOkazaki fragment

the 5'-+3' direction, meaning from the circle out towardthe end of the displaced DNA. With further displacement,new DNA is slmthesized again, beginning at the circleand moving outward along the displaced DNA strand.Thus, synthesis on this strand is discontinuous; that is, thedisplaced strand is the lagging strand template. (See

@Replisome

I

F Ghapter 3 DNA Replication

Figure 3.6.) As the single-stranded DNA tongue rolls out,DNA q'nthesis continues on the circular DNA template.

Since the parental DNA circle can continue to roll, itis possible to generate a linear double-stranded DNAmolecule that is longer than the circumference of the cir-cle. For example, in the later stages of phage l" DNA repli-cation, linear tongues that ate many times thecircumference of the original circle are produced byrolling circle replication. These molecules are cut intoindiüdual linear l" chromosomes, and those unit-lengthmolecules are then packaged into phage heads.

Let us consider the rolling circle mechanism ofDNA replication in the context of the phage ), life cycle.

Figure 3.11I chrornosome structure varies at stages of l¡ic infection of E. coli. (a) Parts ofthe l" chromosome, showing the nucleotide sequence of the two single-stranded, comple-

mentary ("sticky") ends and the chromosome circularizing after infection by pairing ofthe ends, with the single-stranded nicks filled in to produce a covalently closed circle. (b)

Generation of the "sticky" ends of the ), DNA during the lytic cycle. During replication ofthe L chromosome, a giant concatameric DNA molecule is produced; it contains tandem

repeats of the 1, genome. The diagram shows the joining of two adjacent ], chromosomes

and the extent of the cos sequence. The cos sequence is recognized by the ter gene prod-uct, an endonuclease that makes two cuts at the sites shown by the arrows. These cuts

produce a complete l, chromosome from the concatamer.

a) Linear l, chromosome (-48,000 base pairs) forms circular l, chromosome

\ ,,nn,"-rtranded comptementary " o" /lnfection of hostcell results incircularization ofchromosome

Nicks are sealedby DNA ligase

b) Production of progeny, linear l, chromosomes from concatamers (multiple copies linked end to end at complementary ends)

cos sequence cos sequence

Part ofconcatamericmolecule

tet enzyme

l, chromosome with single-stranded complementaryends produced by cleaving cos sequencesat staggered sites ({) with fer enzyme

Recall that phage ), is a temperate phage, meaning thatwhen it infects E. coli, it has a choice of entering thelytic pathway, which results in cell lysis, or following thelysogenic pathway to a quiescent state that does notresult in phage reproduction. (The life cycle of phage l,is described in Chapter 18, pp. 492-497 and is dia-grammed in Figure 18.12, p. 493.) Regardless of whichpathway ), follows when it infects a cell, the first stepafter infection is the conversion of the linear moleculeinto a circular molecule by pairing of the complemen-tary "sticky" ends and then sealing of the single-stranded nicks by DNA ligase (Figure 3.1la). The pairedends are called the cos sequence. In the lysogenic cycle,

cos sequence

r*

Cleavage point

I chromosome cut out of concatameric molecule

the circular DNA finds a particular site in the E. colichromosome and is integrated into the main chromo-some by a crossing-over event.

In the lytic cycle, the rolling circle replicationmechanism produces a vety long molecule with head-to-tail copies of the l. chromosome. A DNA moleculelike this, made up of repeated monomers, is calleda concatqmer. From this concatameric molecule,unit-length progeny phage l" chromosomes are gener-

ated as follows: The phage l, chromosome has a

gene called ter (for terminus-generating activity,Figure 3.1lb), the product of which is a DNA endonu-clease (an enzyme that digests a nucleic acid chain bycutting somewhere along its length rather than at thetermini). The endonuclease recognizes the cos

sequence. Once ter is aligned on the DNA at the cos

site, the endonuclease makes a staggered cut such thatlinear l, chromosomes with the correct complementary("sticky"), I2-base-long, single-stranded ends are pro-duced. The chromosomes are then packaged in theassembled phage heads, and progeny l, phages are

assembled and released from the cell when it lyses.

KEVNOTE

DNA Replicat¡on in Eukaryotes

The biochemistry and molecular biology of DNAreplication are similar in prokaryotes and eukaryotes.However, an added complication in eukaryotes is thatDNA is distributed among many chromosomes ratherthan just one. In this section, we summarize some of theimportant aspects of DNA replication in eukaryotes.

RepliconsEach eukaryotic chromosome consists of one linear DNAdouble helix. For example, there are about 3 billion base

pairs of DNA in the haploid human genome (24 chromo-somes), and the average chromosome is roughly 108 base

pairs long. Replication fork movement is much slower ineukaryotes than in E. coli, so, if there were only one ori-gin of replication per chromosome, replicating each chro-

mosome would take many days.

During DNA replication, new DNA is made in the5'--+3' direction, so chain growth is continuous on

one strand and discontinuous (i.e., in segments thatare later joined) on the other strand. This semidis-continuous model is applicable to many otherprokaryotic replication systems, each of which differsin the number and properties of the enzymes andproteins needed.

DNA Replication in Euharyotes

Actual measuremenls show that the chromosomes ineukaryotes replicate much faster than would be the case

with only one origin of replication per chromosome. Thediploid set of chromosomes in Drosophila embryos, forexample, replicates in 3 minutes. This is six times faster thanthe replication of the E. coli chromosome, even though thereis about I00 times more DNA inDrosophila than in E. coli.

Eukaryotic chromosomes duplicate rapidly becauseDNA replication initiates at many origins of replicationthroughout the genome. At each origin of replication, theDNA denatures (as in E. coli), and the replication pro-ceeds bidirectionally. Eventually, each replication forkruns into an adjacent replication fork, initiated at anadjacent origin of replication. In eukaryotes, the stretchof DNA from the origin of replication to the two terminiof replication (where adjacent replication forks fuse) oneach side of the origin is called a replicon or replicationunit (Figure 3.I2). In general, the replicon size is muchsmaller, and the rate of fork movement much slower, ineukaryotic organisms than in bacteria. For example, theE. coli genome consists of one replicon, of size 4.6 Mb(million base pairs, the entire genome size), with a repli-cation fork movement rate of 2.2 Mb per hour. By con-trast, eukaryotic replicons are relatively small (an average

of 30 kb in adult frogs and 160 kb in yeast), with slowerrates of replication fork movement (18 kb per hour inDrosophila and -4 kb per hour in the mustard Crepis).

Figure 3.12Replicating DNA of Drosophilamelanogaster. (a) Electronmicrograph showing replication units (replicons). (b) An interpre-tation of the electron micrograph shown in (a).

a)

b)

F chaPter 3 DNA RePlication

DNA replication does not occur simultaneously in allthe replicons in an organism's genome. Instead, there is

a cell-specific timing of initiation of replication at the var-ious origins. Figure 3.13 shows one segment of one chro-

mosome in which three replicons always begin replicatingat distinct times. When the replication forks fuse at the

margins of adjacent replicons, the chromosome has repli-cated into two sister chromatids. In general, replication ofa segment of chromosomal DNA occurs after the s1'nchro-

nous activation of a cluster of origins.

lnitiation of ReplicationReplicators are less well defined in eukaryotes than inprokaryotes. ln the yeast Saccharomyces cerevisiae, specific

chromosomal sequences have been identified that, whenthey are included as part of an extrachromosomal, circularDNA molecule, confer upon that molecule the ability to

replicate autonomously within the yeast cell. These

approximately 100-bp sequences are yeast replicators and

are called autonomously replicating sequences (ARSs).

Three sequence elements are typically found at yeast repli-cators in the order A, Bl, and 82. Replicators of more com-

plex, multicellular organisms are poorly understood.The initiator protein in eukaryotes is the multisub-

unit origin recognition complex (ORC). At yeast repli-cators, the ORC binds to A and Bl, and recruits otherreplication proteins, among which is the protein needed

for DNA unwinding in 82. The origin of replication isbetween Bl and 82.

DNA replication takes place in a specific stage of the cell

cycle. The cell cycle consists of four stages (see Fig.ure I2.4,p. 302): G1 during which the cell prepares for DNA replica-

tion, S during which DNA replication occurs, G2 duringwhich the cell prepares for cell division, and M, the division

of the cell by mitosis. A key issue involves the control of

Origins of rePrlication units

!ti,

I i i Template DNA (btue)

New DNA (red)

replication initiation at the replicators. No origin of replica-

tion must be used more than once in the cell cycle. This isaccomplished by a fairly complicated series of events that wewill only outline. In essence, the initiation of replicationinvolves two tempora\ separate steps. The first step is

replicator selection, in which particular proteins assemble oneach replicator to form prereplicative complexes (pre-RC).

This selection step occurs in the G1 phase of the cell cycle

and begins with recognition of the replicator by the ORC.

Once bound, ORC recruits the other proteins. In contrast to

the situation in bacteria, the binding of the initiator to the

replicator does not lead immediately to unwinding of the

DNA. Rather, the pre-RCs are activated to initiate replicationat the origins only after passage of the cell from G1 to the S

phase. The limiting of replication initiation to the S phase is

controlled by cy clin-dqendent hinases (Cdhs), key enzymes

that carefu\ regulate the progression of a cell through the

cell cycle. Cdks are needed to activate pre-RCs to initiatereplication, but Cdk activity inhibits the formation of newpre-RCs. No active Cdks are present in G1, allowing the pre-

RCs to form. Active Cdks are present in the rest of the cellcycle, so when the cell enters S, the pre-RCs are activated

and replication starts. Once a replicator has "fired," a newpre-RC cannot form on it until the next G1, when active

Cdks are again absent.

Eukaryotic Replication EnzymesAs we saw earlier, many of the enzyrnes and proteinsinvolved in prokaryotic DNA replication have been identi-fied. Less is known about the enzyrnes and proteinsinvolved in eukaryotic DNA replication. However, it is clear

that the steps described for DNA synthesis in prokaryotesalso occur in DNA slrlthesis in eukaryotes-namely, denat-

uration of the DNA double helix and the semiconservative,semidiscontinuous replication of DNA.

Figure 3.13Temporal ordering of DNA replicationinitiation eyents in replication unitsof eukaryotic chromosornes.

it

Template DNA (blue)

DNA (red)

Eukaryotic cells have 15 or more DNA polymerases.Replication of the nuclear DNA requires 3 of these: Pol o(alpha)/primase, Pol 6 (delta), and Pol e (epsilon). Pol cr

(alpha)/primase initiates new strands in replication byprimase making about 10 nucleotides of an RNA primer,which is extended by about 30 nucleotides of DNA byPol c. The RNA,/DNA primers are extended by Pol 6 andPol e. One of these enzymes synthesizes the leading-strand DNA, and the other synthesizes the lagging-st¡and DNA, but which slrlthesizes which is not clear.

Other eukaryotic DNA polymerases replicate mito-chondrial or chloroplast DNA or are involved in specificDNA repair processes.

Replicating the Ends of ChromosomesBecause DNA polymerases can synthesize new DNA onlyby extending a primer, there are special problems inreplicating the ends-the telomeres-of eukaryoticchromosomes (Figure 3.1+). A parental chromosome(Figure 3.I4a) is replicated, resulting in two new DNAmolecules, each of which has an RNA primer at the 5'end of the newly synthesized strand in the telomereregion (Figure 3.f4b). The RNA primers are removed,leaving a single-stranded stretch of DNA-a gap-at the5' end of the new strand. The gap cannot be filled in byDNA polymerase, because that enzyme cannot initiatenew DNA synthesis. lf nothing were done about thesegaps, the chromosomes would get shorter and shorterwith each replication cycle.

There is a special mechanism, however, for replicatingüe ends of chromosomes. Most eukaryotic chromosomeshave tandemly repeated, species-specific, simple sequences

at their telomeres. (See Chapter 2, pp. 35-36.) ElizabethBlackburn and Carol W Greider have shown that an enztrrne

called telomerase maintains chromosome lengths by addingtelomere repeats to the chromosome ends. This mechanismdoes not involve the regular replication machinery

Figure 3.15 is a simplified diagram of the mechanismthat has been deduced for the protozoanTetrallrymena.Therepeated sequence in Tetrahymena is 5'-TTGGGG-3',reading toward the end of the DNA on the top strand in thefigure. Telomerase acts at the stage shown in Figure 3.14c-that is, where a chromosome end has been produced with a

gap at the 5' end of the new DNA (Figure 3.15a). Telom-erase is

^n errzqe made up of both protein and RNA. The

RNA component includes a base sequence that is comple-mentary to the telomere repeat unit of the organism inwhich it is found. Therefore, the telomerase binds specifi-cally to the overhanging telomere repeat at the end of thechromosome (Figure 3.I5b). Next, the telomerase catalyzesüe slnthesis of three nucleotides of new DNA-TTG-using the telomerase RNA as a template (Figure 3.15c). Thetelomerase then slides toward the end of the chromosome,so that its AAC at the 3' end of the RNA template nowpairs with the newly synthesized TTG on the DNA

I

I

c) RNA primers removed, ¡eaving gaps at telomeres

(Figure 3.I5d). Telomerase then makes the rest of theTTGGG telomere repeat (Figure 3.15e). The process recurs,to add more telomere repeats. In this way, the chromosomeis lengthened by the addition of a number of telomererepeats. Then, by primer sy'nthesis and DNA slrrthesis cat-alyzed by DNA polymerase in the conventional way, the for-mer gap is filled in, and the new chromosomal DNA islengthened (Figure 3.150. After removal of the RNAprimer, a new 5' gap is left (Figure 3.15g), bur any nershortening of the chromosome has been averted.

Introducing into cells mutant telomerase RNA geneswith certain of their template bases changed showed thattelomerase RNA is used as'a template to synthesize newchromosomal telomere repeats. The new repeats madeby the mutated telomerase had sequences that were com-plementary to the altered RNA, rather than the normal

and *"Jo*ol

s',

3',

5'

3'

DNA Replication in Euharyot"r Iil

Figure 3.14The problem of replicating completely a linear chromosomein eukaryotes. (a) Schematic diagram of a parent double-stranded DNA molecule representing the full length of a chromo-some. (b) After semiconservative replication, new DNA segmentshydrogen bonded to the template srrands have RNA primers attheir 5' ends. (c) The RNA primers are removed, DNA poly-merase fills the resulting gaps, and DNA ligase joins the adjacentfragments. However, at the two telomeres, there are still gaps atthe 5' ends of the new DNA. The gaps result from RNA primerremoval, because no new DNA synthesis could fill them in.

a) Parent chromosome with multiple originsof replication

5'

3's',

5'

b) After repl¡cation

s',

3',

RNA primer

RNA primer

F GhaPter 3 DNA RePlication

ü*

Figure 3.15 .

"fTetrahymenatelomeres. The process is described in the text' (a) The

stafiing point is the chromosome end with 5' gap left after primer

removal. (b) Binding of telomerase to the overhanging telomere repeat

at the end of the chromosome' (c) Slmthesis of three-nucleotide DNA

segment at chromosome end, using the RNA template of telomerase'

(d) The telomerase moves so that the RNA template can bind to the

newly synthesized TTG in a different way. (e) Télomerase c^:u.lyzes the

qmüesis of a new telomere repeat, using the RNA template' The

pro."r. ,".rrtr, to add more telomere repeats' (f) After telomerase has

i"ft, ,r"* DNA is made on the template, stafiing with an RNA primer'

(g) After the primer is removed, the result is a longer chromosome

than at the start, with a new 5' gaP.

sequence. The synthesis of DNA from an RNA template

is called reverse ttanscription' so telomerase is an exam-

ple of a reverse transcriptase enzrlme' (The telomerase

i"rr"rr" transcriptase is abbreviated TERT')

Telomere length, while not identical from chromo-

some end to chromosome end, is nonetheless regulated to

al average length for the organism and cell tlpe' In wild-

type yeast, for example, the simple telomere sequences

(rc,-r, a repeating sequence of one T followed by one to

three -es)

o.inpy arr average of about 300 bp' Mutants that

affect telomet" iettgth have been identified' For example, ifttre TLC| gene (which encodes the telomerase RNA) is

deleted or the EST1 (ever shorter telomeres) gene is

mutated, telomeres shorten continuously until the cells

die. This phenotlpe provides evidence that telomerase

activity is necessary for long-term cell viability' The prod-

uct of the EST1 gene, the protein Estlp, is either a compo-

nent of the telomere RNA-protein complex or a separate

factor that is essential to telomerase function' Mutations of

ttre TELT and, TEI2 genes cause cells to maintain their

telomeres at a new shorter-than-wild-type length, making

it clear that telomere length is regulated genetically

Current evidence suggests many levels of regulation

of telomere activity and telomere length' For example,

attention is being given to the observation that telom-

erase activity in mammals is limited to immortal cells

(such as tumor cells). The absence of telomerase activity

in other cells results in progressive shortening of chro-

mosome ends during successive divisions, because of the

failure to replicate those ends, and also results in a lim-

ited number of cell divisions before the cell dies'

l

b)

c)

s','

3',

d)

e)

Telomerase

Ligation New DNA catalyzedby DNA polymerase

RNA primer,subsequentlyremoved

Special enzymes-telomerases-are used to replicate the

ends of chromosomes in eukaryotes. A telomerase is a

complex of proteins and RNA. The RNA acts as a tem-

plate for synthesizing the complementary telomere -

repeat of the chromosome, so telomerase is a type of

reverse transcri ptase enzyme.

Assembling Newly Replicated DNA into Nucleosomes

Eukaryotic DNA is complexed with histones in nucleo-

,o-"r, which are the basic units of chromosomes' (See

Chapter Z, p. 32.) Recall that there are eight histones in

the nucleosome, two each of H2A, H2B, H3' and H4'

Gap left by primer removal

Therefore, when the DNA is replicated, the histone com-plement must be doubled so that all nucleosomes are

duplicated. Doubling involves two processes: the q'nthe-sis of new histone proteins and the assembly of newnucleosomes.

Most histone synthesis is coordinated with DNAreplication. The transcription of the genes for the five his-tones is initiated near the end of the G1 phase, just before

S. Tianslation of the histone mRNAs occurs throughoutS, producing the histones to be assembled into nucleo-somes as the chromosomes are duplicated.

DNA Replication in Euharyot"r @

Electron microscopy studies have shown thatnewly replicated DNA is assembled into nucleosomesalmost immediately. Nonetheless, for replication toproceed, nucleosomes must disassemble during theshort time when a replication fork passes. New nucleo-somes are assembled as follows (Figure 3.16): Eachparental histone core of a nucleosome separates into anti'3-H4 tetramer (two copies each of H3 and H4) andtwo copies of an H2A-H2B dimer. The H3-H4 tetrameris transferred directly to one of the two replicated DNAdouble helices past the fork, whereupon it begins

Figure 3.16Assembly of new nucleosomes at a replication fork. New nucleosomes are assem-

bled first with the use of either a parental or a new H3-H4 tetramer and then by com-

pleting the stnrcture with a pair of H2A-H2B dimers.

otd histones: ?=nzn f Hza I Hs ü H¿

New histones: H2A ffi Hza # Hs ii# n¿

direction ofDNA rePlication Parental

---------------- nucleosomeH2A-H2B dimer

DNAreplicationmachinery

H2A-H2B dimer

'$rro-r2B dimer

F chapter 3 DNA Replication

nucleosome assembly. The H2A-H2B dimers are

released, joining the pool of newly synthesizedH2A-HZB's that have assembled. A pool of new }i'3-}i'4tetramers is also present, and one of these tetramers ini-tiates nucleosome assembly on the other DNA doublehelix past the fork. The rest of the new nucleosomes isassembled from H2A-H2B dimers, which may beparental or new. Thus, a new nucleosome will haveeither a parental or new H3-H4 tetramer and a pair ofH2A-H2B dimers that may be parental-parental,parental-new, or new-new. The assembly process is notspontaneous in the cell, however, requiring proteinsknown as histone chaperones to direct it. Self-assemblyof nucleosomes has been shown in vitro, but the condi-tions are nonphysiological.

Summary

In this chapter, we discussed DNA replication. Manyaspects of DNA replication are similar in prokaryotes andeukaryotes: Both tlpes of organism employ a semiconser-vative and a semidiscontinuous mechanism, synthesis ofnew DNA in the 5'--+3' direction, and the use of RNAprimers to initiate DNA chains. The enzymes that cat-alyze DNA synthesis are the DNA polymerases. In E. coli,there are three DNA polymerases, two of which are

known to be involved in DNA replication along withseveral other enz)rynes and proteins. The DNA poly-merases have 3'--+ 5' exonuclease activity, which permitsproofreading to take place during DNA spthesis if anincorrect nucleotide is inserted opposite the templatestrand. In eukaryotes, two DNA polymerases are involvedin nuclear DNA replication. Neither has associated proof-reading activity; that function presumably is the propertyof a separate protein.

In prokaryotes, DNA replication begins at specificchromosomal sites. Such sites are known also for yeast. Aprokaryotic chromosome has one initiation site for DNAreplication, whereas a eukaryotic chromosome has manyinitiation sites dividing the chromosome into replicationunits, or replicons. It is not clear whether the initiationsites that are used are the same sequences from cellgeneration to cell generation. The existence of repliconsmeans that DNA replication of the entire set of chromo-somes in a eukaryotic organism can proceed quickly,in some cases faster than with the single E. coli chromo-some, despite the presence of orders of magnitudemore DNA.

Since eukaryotic chromosomes are linear, there is a

special problem of maintaining the lengths of chromo-somes, because the removal of RNA primers results in a

shorter new DNA strand. This problem is overcome byspecial enzymes called telomerases that maintain thelength of chromosomes. Telomerases are a combination

of proteins and RNA. The RNA component acts as atemplate to guide the synthesis of new telomere repeatunits at the chromosome ends.

DNA replication in eukaryotes occurs in the S phase ofthe cell division cycle. Eukaryotic chromosomes are com-plexes of DNA with histones and nonhistone chromosomeproteins, so not only must DNA be replicated, but thechromosome structure must be duplicated. In particulaqthe nucleosome organization of chromosomes must beduplicated as the replication forks migrate. Nucleosomesare disassembled to allow the replication fork to pass, andthen new nucleosomes are assembled soon after a replica-tion fork passes. Nucleosome assembly is an orderlyprocess directed with the aid of histone chaperones.

Analytical Approaches to SolvingGeneties Problems

Q3.1a. Meselson and Stahl used l5N-labeled DNA ro prove

that DNA replicates semiconservatively. The method ofanalysis was cesium chloride equilibrium densitygradient centrifugation, in which bacterial DNAlabeled in both strands with 15N (the heavy isotope ofnitrogen) bands to a different position in the gradientthan DNA labeled in both strands with IaN (the normalisotope of nitrogen). Starting with a mixture ofr5N-containing and laN-containing DNAs, then, twobands result after CsCl density gradient centrifugation.

When double.stranded DNA is heated to 100oC,

the two strands separate because the hydrogen bondsbetween the strands break-a process called denatura-tion. When the solution is cooled slowly, any twocomplementary single strands will find each other andre-form the double helix-a process called renatura-tion or reannealing. lf the mixture of l5N-containing

and laN-containing DNAs is first heated to 100"C andthen cooled slowly before centrifuging, the result isdifferent. In this case, two bands are seen in exactlythe same positions as before, and a new third band isseen at a position haltway between the other two.From its position relative to the other two bands, thenew band is interpreted to be intermediate in densitybetween the other two bands. Explain the existence ofthe three bands in the gradient.

b. DNA from E. coli containing lsN in both strands ismixed with DNA from another bacterial species,Bacillus subtilís, containing laN in both strands. Twobands are seen after CsCl density gradient centrifu-gation. If the two DNAs are mixed, heated to 100'C,slowly cooled, and then centrifuged, two bands againresult. The bands are in the same positions as in theunheated DNA experiment. Explain these results.

*

43.1a. When DNA is heated to 100'C, it is denatured to sin-

gle strands. If denatured DNA is allowed to coolslowly, complementary st¡ands renature to producedouble-stranded DNA again. Thus, when mixed,denatured r5N-r5N DNA and 14N-r4N DNA fromthe same species is cooled slowly, the single strandspair randomly during renaturation so that15N-15N, 14N-I4N, and lsN-IaN double-strandedDNA are produced. The latter type of DNA has a den-sity intermediate between those of the two othertypes, accounting for the third band. Theoretically, ifall DNA strands pair randomly, there should be a

r:2:L distribution of 15N-15N, 15N-r4N, and1'1N-14N DNAs, and this ratio should be reflected inthe relative intensities of the bands.

b. DNAs from different bacterial species have differentsequences. In other words, DNA from one species

typically is not complementary to DNA from anotherspecies. Therefore, only two bands are seen because

only the two E. coli DNA strands can renature to formr5N-15N DNA and only the two B. subtilis DNAstrands can renature to form 14N-l4N DNA. No15N-14N hybrid DNA can form, so in this case thereis no third band of intermediate density.

Q3,2 What would be the effect on chromosome replicationtn E. coli strains carrying deletions of the following genes?

a, dnaEb. polAc. dnaG

Questions andProblems

dnaG encodes DNA primase, the enz).rne that syn-thesizes the RNA primer on the DNA template. With-out the synthesis of the short RNA primer, DNApolymerase III cannot initiate DNA synthesis, so

chromosome replication will not take place.

lig encodes DNA ligase, the en4¡me that catalyzes theligation of Okazaki fragments. In a strain carrying a

deletion of lig DNA would be s1'nthesized, but stableprogeny chromosomes would not result, because theOkazaki fragments could not be ligated together, so

the lagging strand slmthesized discontinuously on thelagging-strand template would be in fragments.

ssb encodes the single-strand binding proteins thatbind to and stabilize the single-stranded DNAregions produced as the DNA is unwound at thereplication fork. In the absence of single-strand bind-ing proteins, DNA replication would be impeded orabsent, because the replication bubble could not bekept open.oriC is the origin-of-replication region in E. coli,-that is, the location at which chromosome replica-tion initiates. Without the origin, the initiatorprotein cannot bind and no replication bubble canform, so chromosome replication cannot take place.

Questions and Problems

3.1 Describe the Meselson-Stahl experiment, and explainhow it showed that DNA replication is semiconservative.

*3.2 In the Meselson-Stahl experiment, I5N-labeled cellswere shifted to a t4N medium at what we can designate as

generatión 0.

a, For the semiconservative model of replication, whatproportion of lsN-15N, 15N-14N, and 1aN-raN

would you expect to find after one, two, three, four,six, and eight replication cycles?

b. Answer (a) in terms of the conservative model ofDNA replication.

3,3 A spaceship lands on Earth and with it a sample ofexüaterrestrial bacteria. You are assigned the usk of deter-mining the mechanism of DNA replication in this organism.

You grow the bacteria in an unlabeled medium forseveral generations and then grow it in the presence of15N for exactly one generation. You exüact the DNA andsubject it to CsCl centrifugation. The banding patternyou find is as follows:

Control

1s¡_15¡ 1a¡-14¡

Experimentalsample

d.

d. lige. ssb

l. oriC

A3.2 When genes are deleted, the function encoded bythose genes is lost. All the genes listed in the question are

involved in DNA replication inE. coli, and their functionsare briefly described in Table 3.1 and discussed further inüe text.a. dnaE encodes a subunit of DNA polymerase III, the

principal DNA polymerase in E. coli that is responsi-

ble for elongating DNA chains. A deletion of thednaE gene undoubtedly would lead to a nonfunc-tional DNA poll.rnerase III. In the absence of DNApol;'rnerase III activity, DNA strands could not be

synthesized from RNA primers; therefore, new DNAstrands could not be s1'nthesized, and there would be

no chromosome replication.b. polA encodes DNA poll'rnerase I, which is used

in DNA synthesis to extend DNA chains made byDNA polymerase III while simultaneously excisingthe RNA primer by 5'-to-3' exonuclease activity. As

discussed in the text, in mutant strains lacking the

originally studied DNA polymerase-DNA poly-merase l-chromosome replication still occurred.Thus, chromosomes would replicate normally in an

E. coli strain carrying a deletion oÍ polA.

F Ghapter 3 DNA Replication

rt

lr

l._

It appears to you that this is evidence that lNe replicates

in the semiconservative manner, but you are wrong'

Why? What other experiment could you perform (using

the same sample and technique of CsCl centrifugation)

that would further distinguish between semiconservative

and dispersive modes of replication?

*3.4 The elegant Meselson-Stahl experiment was

among the first experiments to contribute to what

is now a highly detailed understanding of DNA replica-

tion. Consider this experiment again in light of

current molecular models by answering the following

questions:a. Does the fact that DNA replication is semiconserva-

tive mean that it must be semidiscontinuous?

b. Does the fact that DNA replication is semidiscontin-

uous ensure that it is also semiconservative?

c. Do any properties of known DNA polymerases

ensure that DNA is synthesized semiconservatively?

*3.5 List the components necessary to make DNA invitro, using the enzyme system isolated by Kornberg'

*3.6 How do we know that the Kornberg enz;'rne is not

the main enzqeinvolved in DNA synthesis for chromo-

some duplication in the growth of E. coli?

3.7 Kornberg isolated DNA polymerase I from E' coli'

DNA pollnnerase I has an essential function in DNA

replication. What are the functions of the enzyme in DNA

replication?

3.8 Suppose you have a DNA molecule with the base

,"qrr"tri" TATCA, going from the 5' to the 3' end of one

of lhe polynucleotide chains. The building blocks of the

DNA are drawn as in the following figure:

*+",. *+", *+"- *+",Use this shorthand system to diagram the completed dou-

ble-stranded DNA molecule, as proposed by Watson and

Crick.

*3.9 Base analogs are compounds that resemble the

natural bases found in DNA and RNA, but are not nor-

mally found in those macromolecules. Base analogs can

replace their normal counterparts in DNA during inviiro DNA synthesis. Researchers studied four base

analogs for their effects on in vitro DNA synthesis

using E. coli DNA polymerase. The results were as fol-

lows, with the amounts of DNA synthesized expressed

l

l

G

cG

as percentagesbases only:

of the DNA synthesized from normal

Normal Bases Substitutedby the Analog

AnalosA

B

cD

0

0

0

0

0

54

0

97

Which bases are analogs of adenine? of thymine? of cyto-

sine? of guanine?

3.10 Concerning DNA rePlication:

a. Describe (draw) models of continuous, semidiscon-

tinuous, and discontinuous DNA replication.

b. What was the contribution of Reiji and Tuneko Okazaki

and colleagues with regard to these replication models?

3.11 The following events, steps, or reactions occur dur-

íngE. colí DNA replication. For each entry in column A,

select its match(es) from column B. Each entry iu A may

have more than one match, and each entry in B can be

used more than once.

0

0

I00

0

25

0

0

0

Ir,,

i'I

Column Aa. Unwinds the

double helix

b. Prevents reassociationof complementary bases

c. Is an RNA polymerase

d. ls a DNA polymerase

e. Is the 'tepair" enzyme

f. Is the major elongationenzyme

g. Is a 5'--->3'polymerase

h. Is a 3'-+5'polymerase

i. Has 5'--+3'exonuclease function

j. Has 3'--+5'exonuclease function

k. Bonds free 3'-OH endof a polynucleotide to afree 5'-monophosphateend of polynucleotide

l. Bonds 3'-OH end ofa polynucleotide to a free

5' nucleotide triPhosPhatem. Separates daughter

molecules and causes

supercoiling

Column BA. Pol)¡merase IB. Polymerase IIIC. HelicaseD. Primase

E. Ligase

E SSB prótein

G. Gyrase

H. None of these

3.12 How long would it take E. coli to replicate its entiregenome (+.2 x 106 bp), assuming a replication rate of1,000 nucleotides per second at each fork with no pauses?

*3.13 A diploid organism has 4.5 x 108 bp in its DNA.The DNA is replicated in 3 minutes. Assuming that allreplication forks move at a rate of 104 bp per minute,how many replicons (replication units) are present in theorganism's genome?

x3.14 Describe the molecular action of the enzyme DNAligase. What properties would you expect an E. coli cell tohave if it had a temperature-sensitive mutation in the

gene for DNA ligase?

x3.15 Chromosome replication in E. colí commences

from a constant point, called the origin of replication. Itis known that DNA replication is bidirectional. Devise a

biochemical experiment to prove that the E. coli chromo-some replicates bidirectionally. (Hint: Assume that the

amount of gene product is directly proportional to thenumber of genes.)

3.16 Reiji Okazaki concluded that both DNA strandscould not replicate continuously. What evidence led himto this conclusion?

*3.17 A space probe returns fromJupiter and brings withit a new microorganism for study. It has double-strandedDNA as its genetic material. However, studies of replica-úon of the alien DNA reveal that, although the process is

semiconservative, DNA synthesis is continuous on boththe leading-strand and the lagging-strand templates.What conclusions can you draw from this result?

3.18 Phage such as )u and T4 are packaged from con-catamers.a, What are concatamers, and what type of DNA repli-

cation is responsible for producing concatamers?

b. In what ways does this type of DNA replication dif-fer from that used by E. coli?

x3.19 Although l, is replicated into a concatamer, linearunit-length molecules are packaged into phage heads.

a. What enzymatic activity is required to produce linearunit-length molecules, how does it produce mole-cules that contain a single complete l, genome, andwhat gene encodes the en4..rne involved?

b. What types of ends are produced when this er'zymeacts on DNA, and how are these ends important inthe l. life cycle?

*3.20 M13 is an E. coli bacteriophage whose capsid holds

a closed circular DNA molecule with 2,22L T,I,296 C,

1,315 G, and 1,575 A nucleotides. M13 lacks a gene forDNA polymerase and so must use bacterial DNA polym-

Questions and Problems

erases for replication. Unlike )v or T4, this phage does notform concatamers during replication and packaging.a. Suppose the M13 chromosome were replicated in a

manner similar to the way the E. coli chromosome isreplicated, using semidiscontinuous replication froma double-stranded circular DNA template. Howwould the semidiscontinuous DNA replication mech-anism discussed in the text need to be modified?

b. Suppose the M13 chromosome were replicated in a

manner similar to the way the l" chromosome is repli-cated, using rolling circle replication. How would therolling circle replication mechanism discussed in thetext need to be modified?

*3.21 Compare and contrast eukaryotic and prokaryoticDNA poll'rnerases.

3.22 What mechanism do eukaryotic cells employ tokeep their chromosomes from replicating more than onceper cell cycle?

*3.23 Autoradiography is a technique that allowsradioactive areas of chromosomes to be observed underthe microscope. The slide is covered with a photographicemulsion, which is exposed by radioactive decay Inregions of exposure, the emulsion forms silver grains onbeing developed. The tiny silver grains can be seen on topof the (much larger) chromosomes. Devise a method tofind out which regions in the human karyotype replicateduring the last 30 minutes of the S phase. (Assume a cellcycle in which the cell spends l0 hours in G1, t hours inS, 4 hours in G2, and I hour in M.)

3,24 ln typical human fibroblasts in culture, the G1

period of the cell cycle lasts about 10 hours, S lasts aboutt hours, G2 takes 4 hours, and M takes I hour. Supposeyou added radioactive (3H) thymidine to the medium,left it there for 5 minutes, and then washed it out andreplaced it with an ordinary medium.a. What percentage of cells would you expect to

become labeled by incorporating the 3H-th).midine

into their DNA?b. How long would you have to wait after removing the

3H medium before you would see labeled metaphasechromosomes?

c. Would one or both chromatids be labeled?d. How long would you have to wait if you wanted to

see metaphase chromosomes containing 3H in theregions of the chromosomes that replicated at thebeginning of the S period?

3.25 Suppose you performed the experiment in Question3.24,but left the radioactive medium on the cells for 16

hours instead of 5 minutes. How would your answerschange?

Ghapter 3 DNA Replication

*3.26 In Figure 3.3, semiconservative DNA replication is

visualized in eukaryotic cells with the harlequin chromo-some-staining technique.a. Explain what the harlequin chromosome-staining

technique is and how it provides evidence for semi-

conservative DNA replication in eukaryotes.b. Propose a hypothesis to explain why, in Figure 3.3,

some chromatids appear to contain segments of bothDNA containing T and DNA containing BUdR, whileothers appeff to consist entirely of DNA with T orDNA with BUdR.

3.27 When the eukaryotic chromosome duplicates, the

nucleosome structures must duplicate. Discuss how the

synthesis of histones is related to the cell cycle, anddiscuss new nucleosomes are assembled at replicationforks.

*3.28 A mutant Tetralrymena has an altered repeated

sequence in its telomeric DNA. What change in thetelomerase enzqe would produce this phenotype?

3.29 What is the evidence that telomere length is regu-lated in cells, and what are the consequences of the mis-regulation of telomere length?