8. The Structure and Replication of DNA - kartikmicrocu · The Structure and Replication of DNA 8. The Structure and Replication of DNA ... Figure 8-1. This same basic ... (Figure

The Structure and Replication of DNA

8. The Structure and Replication of DNA

Key Concepts

Bacterial cells that express one phenotype can be transformed into cells that express a different phenotype; the transforming agent is DNA.

Experiments with labeled T2 phage have established that DNA is the hereditary material.

James Watson and Francis Crick showed that the structure of DNA is a double helix. Each helix is a chain of nucleotides linked by phosphodiester bonds. The helices are held together by specific hydrogen bonds between pairs of bases.

DNA structure ensures the fidelity of replication because the complementary base of each base is specified by hydrogen bonding.

The replication of DNA is semiconservative in that each daughter duplex contains one parental and one newly synthesized strand.

Many of the enzymes taking part in DNA synthesis have been characterized.

Introduction

The elucidation of the structure of DNA in 1953 by James Watson and Francis Crick was one of the most exciting discoveries in the history of genetics. It paved the way for an understanding of gene action and heredity in molecular terms. Before we see how the solution of DNA structure was achieved, let's review what was known about genes and DNA at the time that Watson and Crick began their historic collaboration:

1. Genes—the hereditary “factors” described by Mendel—were known to be associated with specific character traits, but their physical nature was not understood. 2. The one-gene–one-enzyme theory (described more fully in Chapter 9) postulated that genes control the structure of proteins. 3. Genes were known to be carried on chromosomes. 4. The chromosomes were found to consist of DNA and protein. 5. Research by Frederick Griffith and subsequently by Oswald Avery and his co-workers pointed to DNA as the genetic material. These experiments, described here, showed that bacterial cells that express one phenotype can be transformed into cells that express a different phenotype and that the transforming agent is DNA. DNA: The genetic material

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The physical nature of the gene fascinated scientists for many years. A series of experiments beginning in the 1920s finally revealed that DNA was the genetic material.

Discovery of transformation

A puzzling observation was made by Frederick Griffith in the course of experiments on the bacterium Streptococcus pneumoniae in 1928. This bacterium, which causes pneumonia in humans, is normally lethal in mice. However, different strains of this bacterial species have evolved that differ in virulence (in the ability to cause disease or death). In his experiments, Griffith used two strains that are distinguishable by the appearance of their colonies when grown in laboratory cultures. In one strain, a normal virulent type, the cells are enclosed in a polysaccharide capsule, giving colonies a smooth appearance; hence, this strain is labeled S. In Griffith's other strain, a mutant nonvirulent type that grows in mice but is not lethal, the polysaccharide coat is absent, giving colonies a rough appearance; this strain is called R.

Griffith killed some virulent cells by boiling them and injected the heat-killed cells into mice. The mice survived, showing that the carcasses of the cells do not cause death. However, mice injected with a mixture of heat-killed virulent cells and live nonvirulent cells did die. Furthermore, live cells could be recovered from the dead mice; these cells gave smooth colonies and were virulent on subsequent injection. Somehow, the cell debris of the boiled S cells had converted the live R cells into live S cells. The process is called transformation. Griffith's experiment is summarized in Figure 8-1.

This same basic technique was then used to determine the nature of the transforming principle—the agent in the cell debris that is specifically responsible for transformation. In 1944, Oswald Avery, C. M. MacLeod, and M. McCarty separated the classes of molecules found in the debris of the dead S cells and tested them for transforming ability, one at a time. These tests showed that the polysaccharides themselves do not transform the rough cells. Therefore, the polysaccharide coat, although undoubtedly concerned with the pathogenic reaction, is only the phenotypic expression of virulence. In screening the different groups, Avery and his colleagues found that only one class of molecules, DNA, induced the transformation of R cells (Figure 8-2). They deduced that DNA is the agent that determines the polysaccharide character and hence the pathogenic character (see pages 219–220 for a description of the mechanism of transformation). Furthermore, it seemed that providing R cells with S DNA was tantamount to providing these cells with S genes.

MESSAGE

The demonstration that DNA is the transforming principle was the first demonstration that genes are composed of DNA.

Hershey-Chase experiment


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The experiments conducted by Avery and his colleagues were definitive, but many scientists were very reluctant to accept DNA (rather than proteins) as the genetic material. The clincher was provided in 1952 by Alfred Hershey and Martha Chase with the use of the phage (virus) T2. They reasoned that phage infection must entail the introduction (injection) into the bacterium of the specific information that dictates viral reproduction. The phage is relatively simple in molecular constitution. Most of its structure is protein, with DNA contained inside the protein sheath of its “head.”

Phosphorus is not found in proteins but is an integral part of DNA; conversely, sulfur is present in proteins but never in DNA. Hershey and Chase incorporated the radioisotope of phosphorus (32P) into phage DNA and that of sulfur (35S) into the proteins of a separate phage culture. They then used each phage culture independently to infect E. coli with many virus particles per cell. After sufficient time for injection to take place, they sheared the empty phage carcasses (called ghosts) off the bacterial cells by agitation in a kitchen blender. They used centrifugation to separate the bacterial cells from the phage ghosts and then measured the radioactivity in the two fractions. When the 32P-labeled phages were used, most of the radioactivity ended up inside the bacterial cells, indicating that the phage DNA entered the cells. 32P can also be recovered from phage progeny. When the 35S-labeled phages were used, most of the radioactive material ended up in the phage ghosts, indicating that the phage protein never entered the bacterial cell (Figure 8-3). The conclusion is inescapable: DNA is the hereditary material; the phage proteins are mere structural packaging that is discarded after delivering the viral DNA to the bacterial cell.

Why such reluctance to accept this conclusion? DNA was thought to be a rather simple chemical. How could all the information about an organism's features be stored in such a simple molecule? How could such information be passed on from one generation to the next? Clearly, the genetic material must have both the ability to encode specific information and the capacity to duplicate that information precisely. What kind of structure could allow such complex functions in so simple a molecule?

Figure 8-2. Demonstration that DNA is the transforming agent. DNA is the only agent that produces smooth (S) colonies when added to live rough (R) cells.




Figure 8-3. The Hershey-Chase experiment, which demonstrated that the genetic material of phage is DNA, not protein. The experiment uses two sets of T2 bacteriophages. In one set, the protein coat is labeled with radioactive sulfur (35S), not found in DNA. In the other set, the DNA is labeled with radioactive phosphorus (32P), not found in protein. Only the 32P is injected into the E. coli, indicating that DNA is the agent necessary for the production of new phages.

Structure of DNA

Although the DNA structure was not known, the basic building blocks of DNA had been known for many years. The basic elements of DNA had been isolated and determined by partly breaking up purified DNA. These studies showed that DNA is composed of only four basic molecules called nucleotides, which are identical except that each contains a different nitrogen base. Each nucleotide contains phosphate, sugar (of the deoxyribose type), and one of the four bases (Figure 8-4). When the phosphate group is not present, the base and the deoxyribose form a nucleoside rather than a nucleotide. The four bases are adenine, guanine, cytosine, and thymine. The full chemical names of the nucleotides are deoxyadenosine 5′-monophosphate (deoxyadenylate, or dAMP), deoxyguanosine 5′-monophosphate (deoxyguanylate, or dGMP), deoxycytidine 5′-monophosphate (deoxycytidylate, or dCMP), and deoxythymidine 5′-monophosphate (deoxythymidylate, or dTMP). However, it is more convenient just to refer to each nucleotide by the abbreviation of its base (A, G, C, and T, respectively). Two of the bases, adenine and guanine, are similar in structure and are called purines. The other two bases, cytosine and thymine, also are similar and are called pyrimidines.






Figure 8-4. Chemical structure of the four nucleotides (two with purine bases and two with pyrimidine bases) that are the fundamental building blocks of DNA. The sugar is called deoxyribose because it is a variation of a common sugar, ribose, that has one more oxygen atom.

After the central role of DNA in heredity became clear, many scientists set out to determine the exact structure of DNA. How can a molecule with such a limited range of different components possibly store the vast range of information about all the protein primary structures of the living organism? The first to succeed in putting the building blocks together and finding a reasonable DNA structure—Watson and Crick in 1953—worked from two kinds of clues. First, Rosalind Franklin and Maurice Wilkins had amassed X-ray diffraction data on DNA structure. In such experiments, X rays are fired at DNA fibers, and the scatter of the rays from the fiber is observed by catching them on photographic film, where the X rays produce spots. The angle of scatter represented by each spot on the film gives information about the position of an atom or certain groups of atoms in the DNA molecule. This procedure is not simple to carry out (or to explain), and the interpretation of the spot patterns is very difficult. The available data suggested that DNA is long and skinny and that it has two similar parts that are parallel to each other and run along the length of the molecule. The X-ray data showed the molecule to be helical (spiral-like). Other regularities were



present in the spot patterns, but no one had yet thought of a three-dimensional structure that could account for just those spot patterns.

The second set of clues available to Watson and Crick came from work done several years earlier by Erwin Chargaff. Studying a large selection of DNAs from different organisms (Table 8-1), Chargaff established certain empirical rules about the amounts of each component of DNA:

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A + T

Organism Tissue Adenine Thymine Guanine Cytosine G + C

Escherichia

coli (K12) — 26.0 23.9 24.9 25.2 1.00 Diplococcus

pneumoniae — 29.8 31.6 20.5 18.0 1.59 Mycobacterium

tuberculosis — 15.1 14.6 34.9 35.4 0.42 Yeast — 31.3 32.9 18.7 17.1 1.79 Paracentrotus

lividus (sea urchin) Sperm 32.8 32.1 17.7 18.4 1.85

Herring Sperm 27.8 27.5 22.2 22.6 1.23 Rat Bone marrow 28.6 28.4 21.4 21.5 1.33 Human Thymus 30.9 29.4 19.9 19.8 1.52 Human Liver 30.3 30.3 19.5 19.9 1.53 Human Sperm 30.7 31.2 19.3 18.8 1.62

* . Defined as moles of nitrogenous constituents per 100 g-atoms phosphate in hydrolysate. Source: E. Chargaff and J. Davidson, eds., The Nucleic Acids. Academic Press, 1955. 1. The total amount of pyrimidine nucleotides (T + C) always equals the total amount of purine nucleotides (A + G). 2. The amount of T always equals the amount of A, and the amount of C always equals the amount of G. But the amount of A + T is not necessarily equal to the amount of G + C, as can be seen in the last column of Table 8-1. This ratio varies among different organisms. Double helix

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The structure that Watson and Crick derived from these clues is a double helix, which looks rather like two interlocked bedsprings. Each bedspring (helix) is a chain of nucleotides held together by phosphodiester bonds, in which a phosphate group forms a bridge between −OH groups on two adjacent sugar residues. The two “bedsprings” (helices) are held together by hydrogen bonds, in which two electronegative atoms “share” a proton, between the bases. Hydrogen bonds form between hydrogen atoms with a small positive charge and acceptor atoms with a small negative charge. For example,

Each hydrogen atom in the NH2 group is slightly positive (δ +) because the nitrogen atom tends to attract the electrons of the N–H bond, thereby leaving the hydrogen atom slightly short of electrons. The oxygen atom has six unbonded electrons in its outer shell, making it slightly negative (δ −). A hydrogen bond forms between one H and the O. Hydrogen bonds are quite weak (only about 3 percent of the strength of a covalent chemical bond), but this weakness (as we shall see) plays an important role in the function of the DNA molecule in heredity. One further important chemical fact: the hydrogen bond is much stronger if the participating atoms are “pointing at each other” in the ideal orientations.

The hydrogen bonds are formed by pairs of bases and are indicated by dotted lines in Figure 8-5, which shows a part of this paired structure with the helices uncoiled. Each base pair consists of one purine base and one pyrimidine based, paired according to the following rule: G pairs with C, and A pairs with T. In Figure 8-6, a simplified picture of the coiling, each of the base pairs is represented by a “stick” between the “ribbons,” or so-called sugar-phosphate backbones of the chains. In Figure 8-5, note that the two backbones run in opposite directions; they are thus said to be antiparallel, and (for reasons apparent in the figure) one is called the 5′ → 3′ strand and the other the 3′ → 5′ strand.






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Figure 8-5. The DNA double helix, unrolled to show the sugar-phosphate backbones (blue) and base-pair rungs (red). The backbones run in opposite directions; the 5′ and 3′ ends are named for the orientation of the 5′ and 3′ carbon atoms of the sugar rings. Each base pair has one purine base, adenine (A) or guanine (G), and one pyrimidine base, thymine (T) or cytosine (C), connected by hydrogen bonds (dotted lines). (From R. E. Dickerson, “The DNA Helix and How It Is Read.” Copyright © 1983 by



Scientific American, Inc. All rights reserved.)

Figure 8-6. A simplified model showing the helical structure of DNA. The sticks represent base pairs, and the ribbons represent the sugar-phosphate backbones of the two antiparallel chains. The various measurements are given in angstroms (1 Å = 0.1 nm).

The double helix accounted nicely for the X-ray data and tied in very nicely with Chargaff's data. Studying models that they made of the structure, Watson and Crick realized that the observed radius of the double helix (known from the X-ray data) would be explained if a purine base always pairs (by hydrogen bonding) with a pyrimidine base (Figure 8-7). Such pairing would account for the (A + G) = (T + C) regularity observed by Chargaff, but it would predict four possible pairings: T···A,




T···G, C···A, and C···G. Chargaff's data, however, indicate that T pairs only with A and C pairs only with G. Watson and Crick showed that only these two pairings have the necessary complementary “lock and key” shapes to permit efficient hydrogen bonding (Figure 8-8).

Figure 8-7. The pairing of purines with pyrimidines accounts exactly for the diameter of the DNA double helix determined from X-ray data. (From R. E. Dickerson, “The DNA Helix and How It Is Read.” Copyright © 1983 by Scientific American, Inc. All rights reserved.)




Figure 8-8. The lock-and-key hydrogen bonding between A and T and between G and C. (From G. S. Stent, Molecular Biology of Bacterial Viruses. Copyright © 1963 by W. H. Freeman and Company.)

Figure 8-9. (a) A space-filling model of the DNA double helix. (b) An unwound representation of a short stretch of nucleotide pairs, showing how A–T and G–C pairing produces the Chargaff ratios. This model is of one of several forms of DNA, termed the B form. (Part a from C. Yanofsky, “Gene Structure and Protein Structure.” Copyright © 1967 by Scientific American, Inc. All rights reserved. Part b after A. Kornberg, “The Synthesis of DNA.” Copyright © 1968 by Scientific American, Inc. All rights reserved.)



Note that the G–C pair has three hydrogen bonds, whereas the A–T pair has only two. We would predict that DNA containing many G–C pairs would be more stable than DNA containing many A–T pairs. In fact, this prediction is confirmed. DNA structure neatly explains Chargaff's data (Figure 8-9), and that structure is consistent with the X-ray data.

Three-dimensional view of the double helix

In three dimensions, the bases form rather flat structures, and these flat bases partly stack on top of one another in the twisted structure of the double helix. This stacking of bases adds tremendously to the stability of the molecule by excluding water molecules from the spaces between the base pairs. (This phenomenon is very much like the stabilizing force that you can feel when you squeeze two plates of glass together underwater and then try to separate them.) Subsequently, it was realized that there were two forms of DNA in the fiber analyzed by diffraction. The A form is less hydrated than the B form and is more compact. It is believed that the B form of DNA is the form found most frequently in living cells.

The stacking of the base pairs in the double helix results in two grooves in the sugar-phosphate backbones. These grooves are termed the major and minor grooves and can be readily seen in the space-filling (three-dimensional) model in Figure 8-9a.

Implications of DNA structure

Elucidation of the structure of DNA caused a lot of excitement in genetics (and in all areas of biology) for two basic reasons. First, the structure suggests an obvious way in which the molecule can be duplicated, or replicated, inasmuch as each base can specify its complementary base by hydrogen bonding. This essential property of a genetic molecule had been a mystery until this time. Second, the structure suggests that perhaps the sequence of nucleotide pairs in DNA dictates the sequence of amino acids in the protein organized by that gene. In other words, some sort of genetic code may write information in DNA as a sequence of nucleotide pairs and then translate it into a different language of amino acid sequences in protein.

This basic information about DNA is now familiar to almost anyone who has read a biology textbook in elementary or high school, or even magazines and newspapers. But try to put yourself back into the scene in 1953 and imagine the excitement. Until then, the evidence that the uninteresting DNA was the genetic molecule had been disappointing and discouraging. But the Watson-Crick structure of DNA suddenly opened up the possibility of explaining two of the biggest “secrets” of life. James Watson told the story of this discovery (from his own point of view, strongly questioned by other participants) in a fascinating book called The Double Helix, which reveals the intricate interplay of personality clashes, clever insights, hard work, and simple luck in such important scientific advances.

Alternative structures






In addition to the A and B forms of DNA, a new form was found in crystals of synthetically prepared DNA that contain alternating G's and C's on the same strand. This Z DNA form has a zigzaglike backbone and generates a left-handed helix, whereas both A and B DNA form right-handed helices.

Replication of DNA

The faithful transmission of hereditary information depends on accurate replication of the genetic material. This section examines the mechanism of DNA replication.

Semiconservative replication

Figure 8-10 diagrams the possible basic mechanism for DNA replication proposed by Watson and Crick. The sugar-phosphate backbones are represented by lines, and the sequence of base pairs is random. Let's imagine that the double helix is like a zipper that unzips, starting at one end (at the bottom in Figure 8-10). We can see that, if this zipper analogy is valid, the unwinding of the two strands will expose single bases on each strand. Because the pairing requirements imposed by the DNA structure are strict, each exposed base will pair only with its complementary base. Because of this base complementarity, each of the two single strands will act as a template, or mold, and will begin to reform a double helix identical with the one from which it was unzipped. The newly added nucleotides are assumed to come from a pool of free nucleotides that must be present in the cell.

If this model is correct, then each daughter molecule should contain one parental nucleotide chain and one newly synthesized nucleotide chain. This prediction has been tested in both prokaryotes and eukaryotes. A little thought shows that there are at least three different ways in which a parental DNA molecule might be related to the daughter molecules. These hypothetical modes are called semicon-servative (the Watson-Crick model), conservative, and dispersive (Figure 8-11). In semiconservative replication, each daughter duplex contains one parental and one newly synthesized strand. However, in conservative replication, one daughter duplex consists of two newly synthesized strands, and the parent duplex is conserved. Dispersive replication results in daughter duplexes that consist of strands containing only segments of parental DNA and newly synthesized DNA.

Meselson-Stahl experiment

In 1958, Matthew Meselson and Franklin Stahl set out to distinguish among these possibilities. They grew E. coli cells in a medium containing the heavy isotope of nitrogen (15N) rather than the normal light (14N) form. This isotope was inserted into the nitrogen bases, which then were incorporated into newly synthesized DNA strands. After many cell divisions in 15N, the DNA of the cells were well labeled with the heavy isotope. The cells were then removed from the 15N medium and put into a 14N medium; after one and two cell divisions, samples were taken. DNA was extracted










from the cells in each of these samples and put into a solution of cesium chloride (CsCl) in an ultracentrifuge.

If cesium chloride is spun in a centrifuge at tremendously high speeds (50,000 rpm) for many hours, the cesium and chloride ions tend to be pushed by centrifugal force toward the bottom of the tube. Ultimately, a gradient of Cs+ and Cl− ions is established in the tube, with the highest ion concentration at the bottom. Molecules of DNA in the solution also are pushed toward the bottom by centrifugal force. But, as they travel down the tube, they encounter the increasing salt concentration, which tends to push them back up owing to the buoyancy of DNA (its tendency to float). Thus, the DNA finally “settles” at some point in the tube where the centrifugal forces just balance the buoyancy of the molecules in the cesium chloride gradient. The buoyancy of DNA depends on its density (which in turn depends on the ratio of G–C to A–T base pairs). The presence of the heavier isotope of nitrogen changes the buoyant density of DNA. The DNA extracted from cells grown for several generations on 15N medium can be readily distinguished from the DNA of cells grown on 14N medium by the equilibrium position reached in a cesium chloride gradient. Such samples are commonly called heavy and light DNA, respectively.

Meselson and Stahl found that, one generation after the heavy cells were moved to 14N medium, the DNA formed a single band of an intermediate density between the densities of the heavy and light controls. After two generations in 14N medium, the DNA formed two bands: one at the intermediate position, the other at the light position (Figure 8-12). This result would be expected from the semiconservative mode of replication; in fact, the result is compatible with only this mode if the experiment begins with chromosomes composed of individual double helices (Figure 8-13).

Autoradiography

The Meselson-Stahl experiment on E. coli was essentially duplicated in 1958 by Herbert Taylor on the chromosomes of bean root-tip cells, by using a cytological technique. Taylor put root cells into a solution containing tritiated thymidine ([3H]thymidine)—the thymine nucleotide labeled with a radioactive hydrogen isotope called tritium. He allowed the cells to undergo mitosis in this solution so that the [3H]thymidine could be incorporated into DNA. He then washed the tips and transferred them to a solution containing nonradioactive thymidine. Addition of colchicine to such a preparation inhibits the spindle apparatus so that chromosomes in metaphase fail to separate and sister chromatids remain “tied together” by the centromere.

The cellular location of 3H can be determined by autoradiography. As 3H decays, it emits a beta particle (an energetic electron). If a layer of photographic emulsion is spread over a cell that contains 3H, a chemical reaction takes place wherever a beta particle strikes the emulsion. The emulsion can then be developed like a photographic





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print so that the emission track of the beta particle appears as a black spot or grain. The cell can also be stained, making the structure of the cell visible, to identify the location of the radioactivity. In effect, autoradiography is a process in which radioactive cell structures “take their own pictures.”

Figure 8-14 shows the results observed when colchicine is added during the division in [3H]thymidine or during the subsequent mitotic division. It is possible to interpret these results by representing each chromatid as a single DNA molecule that replicates semiconservatively (Figure 8-15).

Harlequin chromosomes

With the use of a more modern staining technique, it is now possible to visualize the semiconservative replication of chromosomes at mitosis without the aid of autoradiography. In this procedure, the chromosomes are allowed to go through two rounds of replication in bromodeoxyuridine (BUdR). The BUdR labeling pattern, shown in Figure 8-16a, is the reciprocal of that of Figure 8-15, because the BUdR is used in both replications, rather than being replaced by normal thymidine for the second replication as in the autoradiographic process. The chromosomes are then stained with fluorescent dye and Giemsa stain; this process distinguishes hybrid chromatids with one BUdR-containing strand and one original strand (dark stain) from those in which both strands contain BUdR (light stain), and generates so-called harlequin chromosomes (Figure 8-16b). (Note, in passing, that harlequin chromosomes are particularly favorable for the detection of sister-chromatid exchange at mitosis; two examples are seen in Figure 8-16b). Using similar techniques. Taylor showed that chromosome replication at meiosis also is semiconservative.

Chromosome structure

Figures 8-14 and 8-15 bring up one of the remaining great unsolved questions of genetics: Is a eukaryotic chromosome basically a single DNA molecule surrounded by a protein matrix? Two things strongly suggest that this is, in fact, the case. First, if there were many DNA molecules in the chromosome (whether they were side by side, end to end, or randomly oriented), it would be almost impossible for the chromosome to replicate semiconservatively (with all the label going into one chromatid, as in Taylor's results). Studies on isolated chromosomes and long DNA molecules are consist-ent with the suggestion that each chromatid is a single molecule of DNA. The second fact supporting a single-molecule hypothesis is that DNA and genes behave as though they are attached end to end in a single string, or thread, that we call a linkage group. All genetic linkage data (Chapter 5) tell us that we need nothing more than a single linear array of genes per chromosome to explain the genetic facts. It has been convincingly demonstrated that a chromosome or a chromatid in fact contains just one DNA molecule, as we saw in Chapter 3.

Replication fork














A prediction of the Watson-Crick model of DNA replication is that a replication zipper, or fork, will be found in the DNA molecule during replication. In 1963, John Cairns tested this prediction by allowing replicating DNA in bacterial cells to incorporate tritiated thymidine. Theoretically, each newly synthesized daughter molecule should then contain one radioactive (“hot”) strand and another nonradioactive (“cold”) strand. After varying intervals and varying numbers of replication cycles in a “hot” medium, Cairns extracted the DNA from the cells, put it on a slide, and autoradiographed it for examination under the light microscope. After one replication cycle in [3H]thymidine, rings of dots appeared in the autoradiograph. Cairns interpreted these rings as shown in Figure 8-17. It is also apparent from Figure 8-17 that the bacterial chromosome is circular—a fact that also emerged from genetic analysis described earlier (Chapter 7).

In the second replication cycle, the forks predicted by the model were indeed seen. Furthermore, the density of grains in the three segments was such that the interpretation shown in Figure 8-18 could be made. Cairns saw all sizes of these moon-shaped, autoradiographic patterns, corresponding to the progressive movement of the replication zipper, or fork, around the ring. Structures of the sort shown in Figure 8-18 are called theta ( θ) structures.

Rolling-circle replication

The replication of some circular molecules, such as plasmids and certain viruses, proceeds by the mechanism depicted in Figure 8-19. Here, a nuclease cut provides a free 3′-OH end onto which nucleotides are added. As can be seen from Figure 8-19, as the synthesis proceeds, the other end of the template strand is displaced from the double-stranded circle and then copied. We can envision this displacement as the strand rolling off the circle. Because there is no termination point, synthesis often continues beyond a single circle unit, producing concatamers (a series of linked chains) of several circle lengths, which are then processed by recombination to yield normal-length circles.













Figure 8-10. The model of DNA replication proposed by Watson and Crick is based on the hydrogen-bonded specificity of the base pairs. Complementary strands are shown in different colors. This drawing is a simplified version of the current picture of replication but represents the basic concept suggested by the Watson-Crick structure. The fact that new strands can grow only in the 5′-to-3′ direction adds complexities to the detailed mechanism of replication.

Figure 8-11. Three alternative patterns for DNA replication. The Watson-Crick model would produce the first (semiconservative) pattern. Light blue lines represent the newly synthesized

strands. Figure 8-12. Centrifugation of DNA in a cesium chloride (CsCl) gradient. Cultures grown for many generations in 15N and 14N media provide control positions for heavy and light DNA bands, respectively. When the cells grown in 15N are transferred to a 14N medium, the first generation produces an intermediate DNA band and the second generation produces two bands: one



intermediate and one light. Figure 8-13. Only the semiconservative model of DNA replication predicts results like those shown in Figure 8-12: a single intermediate band in the first generation and one intermediate and one light band in the second generation. (See Figure 8-11 for explanation of colors.) Recall that, after growth in 15N, the cells are switched to 14N.





Figure 8-14. Diagrammatic representation of the autoradiography of chromosomes from cells grown for one cell division in the presence of the radioactive hydrogen isotope 3H (tritium) and then grown in a nonradioactive medium for a second mitotic division. Each dot represents the track of a particle of radioactivity.

Figure 8-15. An explanation of Figure 8-14 at the DNA level. Light blue lines represent radioactive strands. In the second replication (which takes place in nontritiated solution), both the 3H strand and the nontritiated strand incorporate nonradioactive nucleotides, yielding one hybrid and one nontritiated chromatid.

Figure 8-16. (a) A diagrammatic representation of the production of harlequin chromosomes. In




this procedure, the chromosomes go through two rounds of replication in the presence of bromodeoxyuridine (BUdR), which replaces thymidine in the newly synthesized DNA. The chromosomes are then stained with a fluorescent dye and Giemsa stain, producing the appearance shown. (The light blue lines represent the BUdR-substituted strands.) (b) Photograph of harlequin chromosomes in a Chinese hamster ovary (CHO) cell. The chromatids with two strands containing BUdR are light in this photograph, whereas those with one BUdR strand and one original strand are dark. A chromosome at the top (see arrows) has two sister-chromatid exchanges. (Photograph courtesy of Sheldon Wolff and Judy Bodycote.)

Figure 8-17. Left: Autoradiograph of a bacterial chromosome after one replication in tritiated thymidine. According to the semiconservative model of replication, one of the two strands should be radioactive. Right: Interpretation of the autoradiograph. The light blue line represents the

tritiated strand.

Figure 8-18. Left: Autoradiograph of a bacterial chromosome in the second round of replication in tritiated thymidine. In this theta (θ) structure, the newly replicated double helix that crosses the circle could consist of two radioactive strands (if the parental strand was the radioactive one). Right: The double thickness of the radioactive tracing on the autoradiogram appears to confirm the interpretation shown here. The light blue helices represent the “hot” strands.



Figure 8-19. Rolling-circle replication. Newly synthesized DNA is light blue. The displaced strand is replicated discontinuously, as described in the text. (After D. L. Hartl and E. W. Jones, Genetics: Principles and Analysis, 4th ed. Jones and Bartlett, 1998.) Mechanism of DNA replication

Watson and Crick first reasoned that complementary base pairing provides the basis of fidelity in DNA replication; that is, that each base in the template strand dictates the complementary base in the new strand. However, we now know that the process of DNA replication is very complex and requires the participation of many different components. Let's examine each of these components and see how they fit together to produce our current picture of DNA synthesis in E. coli, the best-studied cellular replication system. In the preceding section, we introduced the concept of the replication fork. Figure 8-20 gives a detailed schematic view of fork movement during DNA replication; we can refer to this illustration as we consider each component of the process.

DNA polymerases

In the late 1950s, Arthur Kornberg successfully identified and purified the first DNA polymerase, an enzyme that catalyzes the replication reaction.

This reaction works only with the triphosphate forms of the nucleotides (such as deoxyadenosine triphosphate, or dATP). The total amount of DNA at the end of the




reaction can be as much as 20 times the amount of original input DNA, so most of the DNA present at the end must be progeny DNA. Figure 8-21 depicts the chain-elongation reaction, or polymerization reaction, catalyzed by DNA polymerases. We now know that there are three DNA polymerases in E. coli. The first enzyme that Kornberg purified is called DNA polymerase I or pol I. This enzyme has three activities, which appear to be located in different parts of the molecule:

1. a polymerase activity, which catalyzes chain growth in the 5′ → 3′ direction; 2. a 3′ → 5′ exonuclease activity, which removes mismatched bases; and 3. A 5′ → 3′ exonuclease activity, which degrades double-stranded DNA.

Subsequently, two additional polymerases, pol II and pol III, were identified in E. coli. Pol II may repair damaged DNA, although no particular role has been assigned to this enzyme. Pol III, together with pol I, has a role in the replication of E. coli DNA (Figure 8-20). The complete complex, or holoenzyme, of pol III contains at least 20 different polypeptide subunits, although the catalytic “core” consists of only three subunits, alpha (α), epsilon (ε), and theta (θ). The pol III complex will complete the replication of single-stranded DNA if there is at least a short segment of duplex already present. The short oligonucleotide that creates the duplex is termed a primer.

Prokaryotic origins of replication

E. coli replication begins from a fixed origin but then proceeds bidirectionally (with moving forks at both ends of the replicating piece), as shown in Figure 8-22, ending at a site called the terminus. The unique origin is termed oriC and is located at 83 minutes on the genetic map. It is 245 bp long and has several components, as illustrated in Figure 8-23. First, there is a side-by-side, or tandem, set of 13-bp sequences, which are nearly identical. There is also a set of binding sites for a protein, the DnaA protein. An initial step in DNA synthesis is the unwinding of the DNA at the origin in response to binding of the DnaA protein. The consequences of bidirectional replication can be seen in Figure 8-24, which gives a larger view of DNA replication.

Eukaryotic origins of replication

Bacteria such as E. coli usually require a 40-minute replication-division cycle, but, in eukaryotes, the cycle can vary from 1.4 hours in yeast to 24 hours in cultured animal cells and may last from 100 to 200 hours in some cells. Eukaryotes have to solve the problem of coordinating the replication of more than one chromosome, as well as replicating the complex structure of the chromosome itself (see Chapter 3 for a description of chromosome structure).










In eukaryotes, replication proceeds from multiple points of origin. This process can be demonstrated by a procedure in which a eukaryotic cell is briefly exposed to [3H]thymidine, in a step called a pulse exposure, and is then provided an excess of “cold” (unlabeled) thymidine, in a step called the chase; the DNA is then extracted, and autoradiographs are made. Figure 8-25 shows the results of such a procedure, with what appear to be distinct, simultaneously replicating regions along the DNA molecule. Replication appears to begin at several different sites on these eukaryotic chromosomes. Similarly, a pulse-and-chase study of DNA replication in polytene (giant) chromosomes of Drosophila by autoradiography reveals many replication regions within single chromosome arms (Figure 8-26). As yet there is no firm proof that these regions are indeed different starting points on a single DNA molecule. However, experiments in yeast indicate the existence of approximately 400 replication origins distributed among the 17 yeast chromosomes, and in humans there are estimated to be more than 10,000 growing forks.

Priming DNA synthesis

DNA polymerases can extend a chain but cannot start a chain. Therefore, as already mentioned, DNA synthesis must first be initiated with a primer, or short oligonucleotide, that generates a segment of duplex DNA. The primer in DNA replication can be seen in Figure 8-27 (see also Figure 8-20). RNA primers are synthesized either by RNA polymerase or by an enzyme termed primase. Primase synthesizes a short (approximately 30 bp long) stretch of RNA complementary to a specific region of the chromosome. The RNA chain is then extended with DNA by DNA polymerase. E. coli primase forms a complex with the template DNA, and additional proteins, such as DnaB, DnaT, Pri A, Pri B, and Pri C. The entire complex is termed a primosome (see Figure 8-20).

Leading strand and lagging strand

DNA polymerases synthesize new chains only in the 5′ → 3′ direction and therefore, because of the antiparallel nature of the DNA molecule, move in a 3′ → 5′ direction on the template strand. The consequence of this polarity is that while one new strand, the leading strand, is synthesized continuously, the other, the lagging strand, must be synthesized in short, discontinuous segments, as can be seen in Figure 8-28 (see also Figure 8-20). The addition of nucleotides along the template for the lagging strand must proceed toward the template's 5′ end (because replication always moves along the template in a 3′ → 5′ direction so that the new strand can grow 5′ → 3′). Thus, the new strand must grow in a direction opposite that of the movement of the replication fork. As fork movement exposes a new section of lagging-strand template, a new lagging-strand fragment is begun and proceeds away from the fork until it is stopped by the preceding fragment. In E. coli, pol III carries out most of the DNA synthesis on both strands, and pol I fills in the gaps left in the lagging strand, which are then sealed by the enzyme DNA ligase. DNA ligases join broken pieces of DNA by catalyzing the formation of a phosphodiester bond between the 5′ phosphate end of a













hydrogen-bonded nucleotide and an adjacent 3′ OH group, as shown in Figure 8-29. It is the only enzyme that can seal DNA chains. Figure 8-30 shows the lagging-strand synthesis and gap repair in detail. The primers for the discontinuous synthesis on the lagging strand are synthesized by primase (step a). The primers are extended by DNA polymerase (step b) to yield DNA fragments that were first detected by Reiji Okazaki and are termed Okazaki fragments. The 5′ → 3′ exonuclease activity of pol I removes the primers (step c) and fills in the gaps with DNA, which are sealed by DNA ligase (step d). One proposed mechanism that allows the same dimeric holoenzyme molecule to participate in both leading- and lagging-strand synthesis is shown in Figure 8-31. Here, the looping of the template for the lagging strand allows a single pol III dimer to generate both daughter strands. After approximately 1000 base pairs, pol III will release the segment of lagging-strand duplex and allow a new loop to be formed.

Replication at chromosome tips

The ends of chromosomes present a special problem for the replication process. Figure 8-32 shows the problem: for the leading strand, the polynucleotide addition during replication can always extend to the end because it is automatically primed from behind. However, at the tip, the lagging strand reaches a point where its system of RNA priming cannot work, and an unpolymerized section remains and a shortened chromosome would be the result. To solve this problem, the tips of chromosomes, called telomeres, have adjacent repeats of simple DNA sequences. For example, in the ciliate Tetrahymena, there is repetition of the sequence TTGGGG; in humans, it is TTAGGG. These repeats do not code for an RNA or a protein product but nevertheless serve a definite function in replication. An enzyme called telomerase adds these simple repeat units to the chromosome ends. The telomerase protein is a member of a class of enzymes called reverse transcriptases, which are used in specialized situations to synthesize DNA from RNA. The telomerase carries a small RNA molecule, part of which acts as a template for the polymerization of the telomeric repeat unit that is added to the 3′ end. For example, in Tetrahymena, the RNA is 3′-AACCCC-5′, which acts as the template for the repeat unit, which is 5′-TTGGGG-3′ (Figure 8-33). The additional DNA is then able to act as template for synthesis on the lagging strand. This process counteracts the tendency to shorten at replication. Figure 8-34 demonstrates the positions of the telomeric DNA through in situ hybridization. An age-dependent decline in telomere length has been found in several somatic tissues in humans. In addition, human fibroblasts in culture show progressive telo-mere shortening up to their eventual death. Such observations have led to the telomere theory of aging, and the validity of this theory is now being tested.

Helicases and topoisomerases

Helicases are enzymes that disrupt the hydrogen bonds that hold the two DNA strands together in a double helix. Hydrolysis of ATP drives the reaction. Among E. coli helicases are the DnaB protein and the Rep protein. The Rep protein may help to










unwind the double helix ahead of the polymerase (refer to Figure 8-20). The unwound DNA is stabilized by the single-stranded binding (SSB) protein, which binds to the single-stranded DNA and retards reformation of the duplex.

The action of helicases during DNA replication generates twists in the circular DNA that need to be removed to allow replication to continue. Circular DNA can be twisted and coiled, much like the extra coils that can be introduced into a rubber band. This supercoiling can be created or relaxed by enzymes termed topoisomerases, an example of which is DNA gyrase (Figure 8-35). Topoisomerases can also induce (catenate) or remove (decatenate) knots, or links in a chain. There are two basic types of isomerases. Type I enzymes induce a single-stranded break into the DNA duplex. Type II enzymes cause a break in both strands. In E. coli, topo I and topo III are examples of type I enzymes, whereas gyrase is an example of a type II enzyme.

Untwisting of the DNA strands to open the replication fork causes extra twisting at other regions, and the supercoiling releases the strain of the extra twisting (Figure 8-36). During replication, gyrase is needed to remove positive supercoils ahead of the replication fork.

Exonuclease editing

Both DNA polymerase I and DNA polymerase III also possess 3′ → 5′ exonuclease activity, which serves a “proofreading” and “editing” function by searching for mismatched bases that were inserted erroneously during polymerization and excising them. The proofreading activity of pol III is the ε subunit, which must be bound to α for full proofreading activity (Figure 8-37). Strains lacking a functional ε have a higher mutation rate (see Chapter 16). Figure 8-38 shows the excision of a cytosine residue that has erroneously been paired with an adenine. As can be seen, hydrolysis takes place at the 5′ end of the mismatched base; removal of the incorrect base leaves a 3′-OH group on the preceding base, which is then free to continue the growing strand by accepting the correct nucleotide triphosphate (thymidine, in this case).

Note that this exonuclease activity takes place at the 3′ end of the growing strand (and is therefore 3′ → 5′). The coordination of exonuclease activity with strand growth helps to explain why replication is in the 5′ → 3′ direction. As we saw earlier, new bases are added when the 3′ OH on the terminal deoxyribose of the growing strand attacks the high-energy phosphate of the nucleotide triphosphate that is being added (see Figure 8-21). Chain growth is thus 5′ → 3′. It is conceivable that replication could be in the 3′ → 5′ direction (in Figure 8-21, the 5′ triphosphate at the bottom would be the last base on the chain, and the 3′ OH that attacks it would be on the free nucleotide triphosphate about to be added to the strand). However, if replication were in this direction, there would be exonuclease excisions at the 5′ end of the strand. When a mismatched base was removed, a 5′ OH would be left at the end of the growing strand. The 3′ OH of an incoming nucleotide triphosphate would thus be facing this 5′ OH instead of the high-energy 5′ triphosphate necessary for bond












formation. No bond would form and strand growth would stop. Therefore, replication is not in the 3′ → 5′ direction.

Eukaryotic DNA polymerases

There are at least five DNA polymerases, α, β, γ, δ, and ε, in higher eukaryotes. Polymerases α and δ in the nucleus have roles similar to pol I in E. coli. Polymerase β has a role in DNA repair and gap filling. The γ polymerase is found in mitochondria and appears to take part in replication of mitochondrial DNA.

Experimental applications of base-sequence complementarity

In 1960, Paul Doty and Julius Marmur observed that, when DNA is heated to 100°C, all the hydrogen bonds between the complementary strands are destroyed, and the DNA becomes single stranded (Figure 8-39). If the solution is cooled slowly, some double-stranded DNA is formed. This reannealing process occurs when two single strands happen to collide in such a way that the complementing base sequences can align and reconstitute the original double helix. The annealing of complementary strands is very specific, as shown in Figure 8-39, and forms the basis of many important techniques in molecular biology, such as the identification of specific DNA segments by hybridization and the isolation of specific DNA fragments that are used in cloning, as explained in Chapter 12. Figure 8-40a shows the basic profile of a typical DNA strand as the temperature increases. At a temperature characteristic for each DNA segment, the DNA starts to denature. The melting temperature, T m, is defined as the temperature at which half the molecules are denatured into single strands. The melting temperature depends on the proportion of G:C base pairs, because they are held together by three hydrogen bonds, whereas A:T base pairs are held together by two hydrogen bonds. The higher the G:C content, the higher the melting temperature, as shown in Figure 8-40b.

Doty and Marmur's finding that in solutions the separated single strands of a double helix will find each other because of complementary base pairing led to several experimental applications that have had an enormous effect on research in molecular genetics.

1. Analysis of genome structure. If total genomic DNA is melted and allowed to reanneal, several distinct stages are observed in the annealing process. First, there is a stage of very rapid annealing. This stage represents highly repetitive DNA because, for this type of DNA, there are many copies per genome. These copies can find each other faster than can unique genes that are present in only one copy per haploid genome. Later annealing fractions contain progressively less repetitive DNA, and unique sequences anneal last. Hence this property allowed scientists to isolate and characterize the various repetitive categories. This characterization in turn allowed the overall characterization of genomes from organisms of most taxonomic groups, providing another approach to evolutionary comparison at the genetic level.









2. Gene isolation. Many of the techniques for isolating genes (gene cloning; see Chapter 12) are based on DNA hybridization by base complementarity. The most common type uses a radioactive denatured DNA fragment as a probe to find a clone of some specific gene of interest in a mixture of clones representing the whole genome. 3. Southern and Northern hybridization. We learned in Chapter 1 that a denatured labeled probe can be used to identify specific genomic fragments in a mixture separated on an electrophoretic gel (the technique of Southern hybridization); in a parallel technique, specific RNA transcripts can be detected on electrophoretic gels (Northern hybridization). Because of the incisiveness of these techniques, they now form part of the everyday technology used by geneticists everywhere. 4. Chromosome mapping. Hybridization by probes has allowed the identification of DNA markers at specific chromosomal locations. [One type of DNA marker is the restriction fragment length polymorphism (RFLP); see Chapter 13.] DNA markers have provided many thousands of new loci to saturate the chromosomal map. Furthermore, such markers have provided linked markers for the diagnosis of disease alleles in humans. In a related technique, labeled probes can be added to partly denatured DNA still in chromosomes, revealing the chromosomal position of the DNA homologous to the probe (in situ hybridization; see Chapter 14).

Hence we see that the structure of DNA provides not only two key properties for biological function (replication and information storage), but also key techniques for the genetic dissection of organisms and their cells.

MESSAGE

The specificity of base complementarity forms the foundation for the continuity of life through replication and the foundation for information transfer from DNA into protein—the main determinant of biological form. This same specificity is used by geneticists as a tool to investigate gene and genome structure and function.







Figure 8-20. DNA replication fork.

Leading and Lagging Strand Synthesis.



Figure 8-21. Chain-elongation reaction catalyzed by DNA polymerase.

Figure 8-22. Chain-elongation reaction catalyzed by DNA polymerase. (From L. Stryer, Biochemistry, 4th ed. Copyright © 1995 by Lubert Stryer.)



Figure 8-23. OriC, the origin of replication in E. coli, has a length of 245 bp. It contains a tandem array of three nearly identical 13-nucleotide sequences and four binding sites for DNA protein.



Figure 8-24. Bidirectional replication of a circular DNA molecule.



Figure 8-25. A replication pattern in DNA revealed by autoradiography. A cell is briefly exposed to [3H]thymidine (pulse) and then provided with an excess of nonradioactive (“cold”) thymidine (chase). DNA is spread on a slide and autoradiographed. In the interpretation shown here, there are several initiation points for replication within one double helix of DNA.

Figure 8-26. Replication pattern in a Drosophila polytene chromosome revealed by autoradiography. Several points of replication are seen within a single chromosome, as indicated



by the arrows.

Figure 8-27. Initiation of DNA synthesis by an RNA primer.



Figure 8-28. DNA synthesis proceeds by continuous synthesis on the leading strand and discontinuous synthesis on the lagging strand.



Figure 8-29. The reaction catalyzed by DNA ligase (Enz) joins the 3′-OH end of one fragment to the 5′ phosphate of the adjacent fragment (From H. Lodish, D. Baltimore, A. Berk, S. L. Zipursky, P. Matsudaira, and J. Darnell, Molecular Cell Biology, 3d ed. Copyright © 1995 by Scientific

American Books, Inc.)

Figure 8-30. The overall structure of a growing fork (top) and steps in the synthesis of the lagging strand. (From H. Lodish, D. Baltimore, A. Berk, S. L. Zipursky, P. Matsudaira, and J. Darnell, Molecular Cell Biology, 3d ed. Copyright © 1995 by Scientific American Books, Inc.)



Figure 8-32. The replication problem at chromosome ends. There is no way of priming the last section of the lagging strand, and a shortened chromosome would result.

Figure 8-33. Telomerase carries a short RNA molecule that acts as a template for the addition of the complementary DNA sequence at the 3′ end of the double helix. In the ciliate Tetrahymena, the



DNA sequence added is TTGGGG.

Figure 8-35. DNA-gyrase-catalyzed supercoiling. Replicating DNA generates “positive” supercoils, depicted at the bottom of the diagram, as a result of rapid rotation of the DNA at the replication fork. DNA gyrase can nick and close phosphodiester bonds, relieving the supercoiling, as shown here (relaxed DNA). Gyrase can also generate supercoils twisted in the opposite direction, termed negative supercoils; this arrangement facilitates the unwinding of the helix. (After L. Stryer, Biochemistry, 4th ed. Copyright © 1995 by Lubert Stryer.)

Figure 8-36. Swivel function of topoisomerase during replication. Extra-twisted (positively supercoiled) regions accumulate ahead of the fork as the parental strands separate for replication. A topoisomerase is required to remove these regions, acting as a swivel to allow extensive replication. (From A. Kornberg and T. A. Baker, DNA Replication, 2d ed. Copyright © 1992 by W.



H. Freeman and Company.) Figure 8-37. Proofreading by the pol III α–ε complex. (From A. Kornberg and T. A Baker, DNA Replication, 2d ed. Copyright © 1992 by W. H. Freeman and Company.)

Figure 8-38. The 3′ → 5′ exonuclease action of DNA polymerase III.

Figure 8-39. The denaturation and renaturation of double-stranded DNA molecules.



Figure 8-40. (a) The absorption of ultraviolet light of 260-nm wavelength by solutions of single-stranded and double-stranded DNA. As regions of double-stranded DNA unpair, the absorption of light by those regions increases almost twofold. The temperature at which half the bases in a double-stranded DNA sample have denatured is denoted T m. (b) DNA melting curves. The absorbance relative to that at 25°C is plotted against temperature. (The wavelength of the incident light was 260 nm.) The T m is 69°C for E. coli DNA (50 percent GC pairs) and 76°C for Pseudomonas aeruginosa DNA (68 percent GC pairs). (Part b from L. Stryer, Biochemistry, 4th ed. Copyright © 1995 by Lubert Stryer.) Summary

Experimental work on the molecular nature of hereditary material has demonstrated conclusively that DNA (not protein, RNA, or some other substance) is indeed the genetic material. Using data supplied by others, Watson and Crick created a double-helical model with two DNA strands, wound around each other, running in antiparallel fashion. Specificity of binding the two strands together is based on the fit of adenine (A) to thymine (T) and guanine (G) to cytosine (C). The former pair is held by two hydrogen bonds; the latter, by three.

The Watson-Crick model shows how DNA can be replicated in an orderly fashion—a prime requirement for genetic material. Replication is accomplished semiconservatively in both prokaryotes and eukaryotes. One double helix is replicated into two identical helices, each with identical linear orders of nucleotides; each of the two new double helices is composed of one old and one newly polymerized strand of DNA.

Replication is achieved with the aid of several enzymes, including DNA polymerase, gyrase, and helicase. Replication starts at special regions of the DNA called origins of



replication and proceeds down the DNA in both directions. Because DNA polymerase acts only in a 5′ → 3′ direction, one of the newly synthesized strands at each replication fork must be synthesized in short segments and then joined by the enzyme ligase. DNA polymerization cannot begin without a short primer, which is also synthesized with special enzymes.

Concept Map

Draw a concept map interrelating as many of the following terms as possible. Note that the terms are listed in no particular order.

Chapter Integration Problem

Mitosis and meiosis were presented in Chapter 3. Considering what we have covered in this chapter concerning DNA replication, draw a graph showing DNA content against time in a cell that undergoes mitosis and then meiosis. Assume a diploid cell.See answer

Solution

See question

Solved Problems

1. If the GC content of a DNA molecule is 56 percent, what are the percentages of the four bases (A, T, G, and C) in this molecule? See answer

Solution

If the GC content is 56 percent, then, because G = C, the content of G is 28 percent and the content of C is 28 percent. The content of AT is 100 − 56 = 44 percent. Because A = T, the content of A is 22 percent and the content of T is 22 percent. See question



http://www.ncbi.nlm.nih.gov/books/bv.fcgi?call=bv.View..ShowSection&rid=iga.section.d1e34626





2. Describe the expected pattern of bands in a CsCl gradient for conservative replication in the Meselson-Stahl experiment. Draw a diagram. See answer

Solution

Refer to Figure 8-13 for an additional explanation. In conservative replication, if bacteria are grown in the presence of 15N and then shifted to 14N, one DNA molecule will be all 15N after the first generation and the other molecule will be all 14N, resulting in one heavy band and one light band in the gradient. After the second generation, the 15N DNA will yield one molecule with all 15N and one molecule with all 14N, whereas the 14N DNA will yield only 14N DNA. Thus, only all 14N or all 15N DNA is generated, again, yielding a light band and a heavy band:

See question

Problems

1. Describe the types of chemical bonds and forces of the DNA double helix.

See answer The DNA double helix has two types of bonds, covalent and hydrogen. Covalent bonds exist within each linear strand and strongly bond bases, sugars, and phosphate groups (both within each component and between components). Hydrogen bonds exist between the two strands and form between a base, from one strand and a base from the second strand in complementary pairing. These hydrogen bonds are individually weak but collectively quite strong.

2. Explain what is meant by the terms conservative and semiconservative replication.







3. What is meant by a primer, and why are primers necessary for DNA replication?

See answer A primer is a short segment of RNA that is synthesized by primase using the DNA as a template during DNA replication. After the primer has been synthesized, DNA polymerase then adds DNA to the 3′ end of the RNA. Primers are required because the major DNA polymerase catalyzing DNA replication is unable to initiate DNA synthesis and, rather, requires a 3′ end. The RNA is subsequently removed and replaced with DNA so that no gaps exist in the final product.

4. What are helicases and topoisomerases?

5. Why is DNA synthesis continuous on one strand and discontinuous on the opposite strand?

See answer Because the DNA polymerase is capable of adding new nucleotides only at the 3′ end of a DNA strand and because the two strands are antiparallel, at least two molecules of DNA polymerase must take part in the replication of any specific region of DNA. When a region becomes single stranded, the two strands have an opposite orientation. Imagine a single-stranded region that runs from left to right. At the left end, the 3′ end of one strand points to the right, and synthesis can initiate and continue toward the right end of that region. The other strand has a 5′ end pointing toward the right, and synthesis cannot initiate and continue toward the right end of the single-stranded region at the 5′ end. Instead, synthesis must initiate somewhere to the right of the left end of the single-stranded region and move toward the left end of the region. As the first strand continues synthesis (continuous synthesis), the single-stranded region extends toward the right. This now leaves the second strand unreplicated in this new region of single-strandedness, and there must be a second initiation of DNA synthesis moving from the current right end of the single-stranded region toward the first initiation point on that strand. This results in discontinuous synthesis along that strand.

6. If thymine makes up 15 percent of the bases in a specific DNA molecule, what percentage of the bases is cytosine?

7. If the GC content of a DNA molecule is 48 percent, what are the percentages of the four bases (A, T, G, and C) in this molecule? See answer The frequency of both A and T is (52%) = 26%.

8. E. coli chromosomes in which every nitrogen atom is labeled (that is, every nitrogen atom is the heavy isotope 15N instead of the normal isotope 14N) are allowed to replicate in an environment in which all the nitrogen is 14N. Using a






solid line to represent a heavy polynucleotide chain and a dashed line for a light chain, sketch the following: a. The heavy parental chromosome and the products of the first replication after transfer to a 14N medium, assuming that the chromosome is one DNA double helix and that replication is semiconservative. b. Repeat part a, but assume that replication is conservative. c. Repeat part a, but assume that the chromosome is in fact two side-by-side double helices, each of which replicates semiconservatively. d. Repeat part c, but assume that each side-by-side double helix replicates conservatively and that the overall chromosome replication is semiconservative. e. Repeat part d, but assume that the overall chromosome replication is conservative. f. If the daughter chromosomes from the first division in 14N are spun in a cesium chloride (CsCl) density gradient and a single band is obtained, which of possibilities in parts a through e can be ruled out? Reconsider the Meselson-Stahl experiment: what does it prove?

9. R. Okazaki found that the immediate products of DNA replication in E. coli include single-stranded DNA fragments approximately 1000 nucleotides in length after the newly synthesized DNA is extracted and denatured (melted). When he allowed DNA replication to proceed for a longer period of time, he found a lower frequency of these short fragments and long single-stranded DNA chains after extraction and denaturation. Explain how this result might be related to the fact that all known DNA polymerases synthesize DNA only in a 5′ → 3′ direction.

See answer The results suggest that the DNA is replicated in short segments that are subsequently joined by enzymatic action (DNA ligase). Because DNA replication is bidirectional, because there are multiple points along the DNA where replication is initiated, and because DNA polymerases work only in a 5′ → 3′ direction, one strand of the DNA is always in the wrong orientation for the enzyme. This requires synthesis in fragments.

10. When plant and animal cells are given pulses of [3H]thymidine at different times during the cell cycle, heterochromatic regions on the chromosomes are invariably shown to be “late replicating.” Can you suggest what, if any, biological significance this observation might have?

11. On the planet of Rama, the DNA is of six nucleotide types: A, B, C, D, E, and F. A and B are called marzines, C and D are orsines, and E and F are pirines. The following rules are valid in all Raman DNAs:




a. Prepare a model for the structure of Raman DNA. b. On Rama, mitosis produces three daughter cells. Bearing this fact in mind, propose a replication pattern for your DNA model. c. Consider the process of meiosis on Rama. What comments or conclusions can you suggest?

12. If you extract the DNA of the coliphage øX174, you will find that its composition is 25 percent A, 33 percent T, 24 percent G, and 18 percent C. Does this composition make sense in regard to Chargaff's rules? How would you interpret this result? How might such a phage replicate its DNA?

See answer Chargaff's rules are that A = T and G = C. Because this composition is not observed, the most likely interpretation is that the DNA is single stranded. The phage would first have to synthesize a complementary strand before it could begin to make multiple copies of itself.

13. The temperature at which a DNA sample denatures can be used to estimate the proportion of its nucleotide pairs that are G–C. What would the basis for this determination be, and what would a high denaturation temperature for a DNA sample indicate?

See answer Remember that there are two hydrogen bonds between A and T, whereas there are three hydrogen bonds between G and C. Denaturation necessitates the breaking of these bonds, which requires energy. The more bonds that need to be broken, the more energy that must be supplied. Thus the temperature at which a given DNA molecule denatures is a function of its base composition. The higher the temperature of denaturation, the higher the percentage of G–C pairs.

14. Suppose that you extract DNA from a small virus, denature it, and allow it to reanneal with DNA taken from other strains that carry either a deletion, an inversion, or a duplication. What would you expect to see on inspection with an electron microscope?

15. DNA extracted from a mammal is heat denatured and then slowly cooled to allow reannealing. The following graph shows the results obtained. These are two “shoulders” in the curve. The first shoulder indicates the presence of a very





rapidly annealing part of the DNA—so rapid, in fact, that the annealing occurs before strand interactions take place.

a. What could this part of the DNA be? b. The second shoulder is a rapidly reannealing part as well. What does this evidence suggest?

16. Design tests to determine the physical relation between highly repetitive and unique DNA sequences in chromosomes. (Hint: It is possible to vary the size of DNA molecules by the amount of shearing to which they are subjected.)

17. Viruses are known to cause cancer in mice. You have a pure preparation of virus DNA, a pure preparation of DNA from the chromosomes of mouse cancer cells, and pure DNA from the chromosomes of normal mouse cells. Viral DNA will specifically anneal with cancer-cell DNA, but not with normal-cell DNA. Explore the possible genetic significance of this observation, its significance at the molecular level, and its medical significance.

18. Ruth Kavenaugh and Bruno Zimm devised a technique to measure the maximal length of the longest DNA molecules in solution. They studied DNA samples from the three Drosophila karyotypes shown at the right. They found the longest molecules in karyotypes a and b to be of similar length and about twice the length of the longest molecule in karyotype c. Interpret these results.

See answer The data suggest that each chromosome is composed of one continuous molecule of DNA and that translocations can alter their size. In part c, it appears that part of the longest chromosome has been translocated to the shortest chromosome.

19. In the harlequin chromosome technique, you allow three rounds of replication in bromodeoxyuridine and then stain the chromosomes. What result do you expect?





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