Biology Campbell 12 ed DNA Replication PDF

16 The Molecular Basis of Inheritance KEY CONCEPTS 16.1 16.2 DNA is the genetic material p. 315 16.3 A chromosome consists of a DNA molecule packed together with proteins p. 330 Many proteins work together in DNA replication and repair p. 320 Study Tip Draw DNA replication: Make up a single-...

16 The Molecular Basis of Inheritance KEY CONCEPTS 16.1 16.2 DNA is the genetic material p. 315 16.3 A chromosome consists of a DNA molecule packed together with proteins p. 330 Many proteins work together in DNA replication and repair p. 320 Study Tip Draw DNA replication: Make up a single-stranded DNA sequence that is 40 nucleotides in length. As you go through Concepts 16.1 and 16.2, use your sequence to draw a doublestranded DNA molecule, labeling the ends. Then diagram the process of replication of your DNA molecule. DNA Structure and Replication T C A G C G T A A G G T G A T G T A T A… Go to Mastering Biology For Students (in eText and Study Area) • Get Ready for Chapter 16 • Figure 16.15 Walkthrough: Addition of a Nucleotide to a DNA Strand • BioFlix® Animation: DNA Replication For Instructors to Assign (in Item Library) • Activity: The Hershey-Chase Experiment • Tutorial: Visualizing DNA Ready-to-Go Teaching Module (in Instructor Resources) • DNA Replication (Concept 16.2) Figure 16.1 The elegant double-helical structure of deoxyribonucleic acid (DNA) shook the scientific world when it was proposed in April 1953 by James Watson and Francis Crick. The DNA you inherited from your parents contains all your genes— your genetic information. How does DNA replication transmit genetic information? DNA replication allows genetic information to be inherited from a parent cell to daughter cells (by mitosis) and from generation to generation (starting with meiosis). Unduplicated chromosome (one DNA molecule and proteins) Each gene is a unit of hereditary information consisting of a specific DNA sequence. DNA segment from a chromosome DNA replication begins. Replication begins at multiple sites (origins), each forming a replication bubble with a fork at each end. New strands DNA replication and chromosome condensation is completed. Duplicated and condensed chromosome (two DNA molecules and proteins) 314 Two DNA molecules, which are distributed to daughter cells. 16.1 DNA is the genetic material Today, even schoolchildren have heard of DNA, and scientists routinely manipulate DNA in the laboratory. Early in the 20th century, however, identifying the molecules of inheritance posed a major challenge to biologists. The Search for the Genetic Material: Scientific Inquiry Once T. H. Morgan’s group showed that genes exist as parts of chromosomes (described in Concept 15.1), the two chemical components of chromosomes—DNA and protein—emerged as the leading candidates for the genetic material. Until the 1940s, the case for proteins seemed stronger: Biochemists had identified proteins as a class of macromolecules with great heterogeneity and specificity of function, essential requirements for the hereditary material. Moreover, little was known about nucleic acids, whose physical and chemical properties seemed far too uniform to account for the multitude of specific inherited traits exhibited by every organism. This view gradually changed as the role of DNA in heredity was worked out in studies of bacteria and the viruses that infect them, systems far simpler than fruit flies or humans. Let’s trace the search for the genetic material as a case study in scientific inquiry. Evidence That DNA Can Transform Bacteria In 1928, a British medical officer named Frederick Griffith was trying to develop a vaccine against pneumonia. He was studying Streptococcus pneumoniae, a bacterium that causes pneumonia in mammals. Griffith had two strains (varieties) of the bacterium, one pathogenic (disease-causing) and one nonpathogenic (harmless). He was surprised to find that when he killed the pathogenic bacteria with heat and then mixed the cell remains with living bacteria of the nonpathogenic strain, some of the living cells became pathogenic (Figure 16.2). Furthermore, this newly acquired trait of pathogenicity was inherited by all the descendants of the transformed bacteria. Apparently, some chemical component of the dead pathogenic cells caused this heritable change, although the identity of the substance was not known. Griffith called the phenomenon transformation, now defined as a change in genotype and phenotype due to the assimilation of external DNA by a cell. Later work by Oswald Avery, Maclyn McCarty, and Colin MacLeod identified the transforming substance as DNA. Scientists remained skeptical, however, since many still viewed proteins as better candidates for the genetic material. Also, many biologists were not convinced that bacterial genes would be similar in composition and function to those of more complex organisms. But the major reason for the continued doubt was that so little was known about DNA. ▼ Figure 16.2 Inquiry Can a genetic trait be transferred between different bacterial strains? Experiment Frederick Griffith studied two strains of the bacterium Streptococcus pneumoniae. The S (smooth) strain can cause pneumonia in mice; it is pathogenic because the cells have an outer capsule that protects them from an animal’s immune system. Cells of the R (rough) strain lack a capsule and are nonpathogenic. To test for the trait of pathogenicity, Griffith injected mice with the two strains: Living S cells (pathogenic control) Living R cells (nonpathogenic control) Heat-killed S cells Mixture of heat(nonpathogenic killed S cells and control) living R cells Results Mouse dies Mouse healthy Mouse healthy Mouse dies In blood sample, living S cells were found. They could reproduce, yielding more S cells. Conclusion The living R bacteria had been transformed into pathogenic S bacteria by an unknown, heritable substance from the dead S cells that enabled the R cells to make capsules. Data from F. Griffith, The significance of pneumococcal types, Journal of Hygiene 27:113–159 (1928). WHAT IF? How did this experiment rule out the possibility that the R cells simply used the dead S cells’ capsules to become pathogenic? Evidence That Viral DNA Can Program Cells Additional evidence that DNA was the genetic material came from studies of viruses that infect bacteria (Figure 16.3). These viruses are called bacteriophages (meaning “bacteria-eaters”), or phages for short. Viruses are much simpler than cells. A virus is little more than DNA (or sometimes RNA) enclosed by a protective coat, which is often simply protein. To produce more viruses, a virus must infect a Phage cell and take over the cell’s head metabolic machinery. DNA c Figure 16.3 A virus infecting a bacterial cell. A phage called T2 attaches to a host cell and injects its genetic material through the plasma membrane, while the head and tail parts remain on the outer bacterial surface (colorized TEM). CHAPTER 16 Tail sheath Tail fiber Genetic material Bacterial cell The Molecular Basis of Inheritance 100 nm CONCEPT 315 Phages have been widely used as tools by researchers in molecular genetics. In 1952, Alfred Hershey and Martha Chase performed experiments showing that DNA is the genetic material of a phage known as T2. This is one of many phages that infect Escherichia coli (E. coli), a bacterium that normally lives in the intestines of mammals and is a model organism for molecular biologists. At that time, biologists already knew that T2, like many other phages, was composed almost entirely of DNA and protein. They also knew that the T2 phage could quickly turn an E. coli cell into a T2-producing factory that released many copies of new phages when the cell ruptured. Somehow, T2 could reprogram its host cell to produce viruses. But which viral component—protein or DNA—was responsible? ▼ Figure 16.4 Hershey and Chase answered this question by devising an experiment showing that only one of the two components of T2 actually enters the E. coli cell during infection (Figure 16.4). In their experiment, they used a radioactive isotope of sulfur to tag protein in one batch of T2 and a radioactive isotope of phosphorus to tag DNA in a second batch. Because protein, but not DNA, contains sulfur, radioactive sulfur atoms were incorporated only into the protein of the phage. In a similar way, the atoms of radioactive phosphorus labeled only the DNA, not the protein, because nearly all the phage’s phosphorus is in its DNA. In the experiment, separate samples of nonradioactive E. coli cells were infected separately with the protein-labeled and DNA-labeled batches of T2. The researchers then tested the two samples Inquiry Is protein or DNA the genetic material of phage T2? Experiment Alfred Hershey and Martha Chase used radioactive sulfur and phosphorus to trace the fates of protein and DNA, respectively, of T2 phages that infected bacterial cells. They wanted to see which of these molecules entered the cells and could reprogram them to make more phages. 1 Mixed radioactively labeled phages with bacteria. The phages infected the bacterial cells. Phage 2 Agitated the mixture in a blender to free the phage parts outside the bacteria from the cells. Radioactive protein Empty protein shell Bacterial cell Batch 1: Phages were grown with radioactive sulfur (35S), which was incorporated into phage protein (pink). 3 Centrifuged the mixture 4 Measured the radioactivity in so that bacteria formed a the pellet and pellet at the bottom of the liquid. the test tube; free phages and phage parts, which are lighter, remained Radioactivity suspended in the liquid. (phage protein) found in liquid DNA Phage DNA Centrifuge Pellet (bacterial cells and contents) Radioactive DNA Batch 2: Phages were grown with radioactive phosphorus (32P), which was incorporated into phage DNA (blue). Centrifuge Pellet Results When proteins were labeled (batch 1), radioactivity remained outside the cells, but when DNA was labeled (batch 2), radioactivity was found inside the cells. Cells containing radioactive phage DNA released new phages with some radioactive phosphorus. Conclusion Phage DNA entered bacterial cells, but phage proteins did not. Hershey and Chase concluded that DNA, not protein, functions as the genetic material of phage T2. 316 UNIT THREE Genetics Radioactivity (phage DNA) found in pellet Data from A. D. Hershey and M. Chase, Independent functions of viral protein and nucleic acid in growth of bacteriophage, Journal of General Physiology 36:39–56 (1952). WHAT IF? How would the results have differed if proteins carried the genetic information? Mastering Biology Animation: The Hershey-Chase Experiment shortly after the onset of infection to see which type of radioactively labeled molecule—protein or DNA—had entered the bacterial cells and would therefore be capable of reprogramming them. Hershey and Chase found that the phage DNA entered the host cells, but the phage protein did not. Moreover, when these bacteria were returned to a culture medium and the infection ran its course, the E. coli released phages that contained some radioactive phosphorus. This result further showed that the DNA inside the cell played an ongoing role during the infection process. They concluded that the DNA injected by the phage must be the molecule carrying the genetic information that makes the cells produce new viral DNA and proteins. The Hershey-Chase experiment was a landmark study because it provided powerful evidence that nucleic acids, rather than proteins, are the hereditary material, at least for certain viruses. . Figure 16.5 The structure of a DNA strand. Each DNA nucleotide monomer consists of the sugar deoxyribose (blue) attached to both a nitrogenous base (A, T, G, or C), and a phosphate group (yellow). The phosphate group of one nucleotide is attached to the sugar of the next by a covalent bond, forming a “backbone” of alternating phosphates and sugars from which the bases project. A polynucleotide strand has directionality, from the 5′ end (with the phosphate group) to the 3′ end (with the ¬OH group of the sugar). 5′and 3′ refer to the numbers assigned to the carbons in the sugar ring (see magenta numbers). 5¿ end O O– P –O CH3 O 5¿ CH2 4¿ Additional Evidence That DNA Is the Genetic Material Further evidence that DNA is the genetic material came from the laboratory of biochemist Erwin Chargaff. DNA was known to be a polymer of nucleotides, each having three components: a nitrogenous (nitrogen-containing) base, a pentose sugar called deoxyribose, and a phosphate group (Figure 16.5). The base can be adenine (A), thymine (T), guanine (G), or cytosine (C). Chargaff analyzed the base composition of DNA from a number of different organisms. In 1950, he reported that the base composition of DNA varies from one species to another. For example, he found that 32.8% of sea urchin DNA nucleotides have the base A, whereas only 24.7% of those from the bacterium E. coli have an A. This evidence of molecular diversity among species, which most scientists had presumed to be absent from DNA, made DNA a more credible candidate for the genetic material. Chargaff also noticed a peculiar regularity in the ratios of nucleotide bases. In the DNA of each species he studied, the number of adenines approximately equaled the number of thymines, and the number of guanines approximately equaled the number of cytosines. In sea urchin DNA, for example, the four bases are present in these percentages: A = 32.8% and T = 32.1%; G = 17.7% and C = 17.3%. (The percentages are not exactly the same because of limitations in Chargaff’s techniques.) These two findings became known as Chargaff’s rules: (1) DNA base composition varies between species, and (2) for each species, the percentages of A and T bases are roughly equal, as are those of G and C bases. In the Scientific Skills Exercise, you can use Chargaff’s rules to predict percentages of nucleotide bases. The basis for these rules remained unexplained until the discovery of the double helix. Mastering Biology HHMI Animation: Chargaff’s Ratio Building a Structural Model of DNA Once most biologists were convinced that DNA was the genetic material, the challenge was to determine how the structure Nitrogenous bases Sugar-phosphate backbone 1¿ H 3¿ O P –O H H O 2¿ N N O H –O H O P O N O H H H 3¿ 2¿ OH H Cytosine (C) N O H 5¿ 4¿ DNA nucleotide H N H CH2 Phosphate group H H H H H H H CH2 Guanine (G) N H O O O N H O P –O N H H O H N O Thymine (T) O H O H N H H CH2 O H O 1¿ N H N N H H Sugar (deoxyribose) H Adenine (A) N N H Nitrogenous base 3¿ end Mastering Biology Animation: DNA and RNA Structure of DNA could account for its role in inheritance. By the early 1950s, the arrangement of covalent bonds in a nucleic acid polymer was well established (see Figure 16.5), and researchers focused on discovering the three-dimensional structure of DNA. Among the scientists working on the problem were Linus Pauling, at the California Institute of Technology, and Maurice Wilkins and Rosalind Franklin, at King’s College in London. First to come up with the complete answer, however, were two scientists who were relatively unknown at the time—the American James Watson and the Englishman Francis Crick. The collaborative work that solved the puzzle of DNA structure began soon after Watson went to Cambridge University, where Crick was studying protein structure with a technique called X-ray crystallography (see Figure 5.21). CHAPTER 16 The Molecular Basis of Inheritance 317 Scientific Skills Exercise Working with Data in a Table Given the Percentage Composition of One Nucleotide in a Genome, Can We Predict the Percentages of the Other Three Nucleotides? Even before the structure of DNA was elucidated, Erwin Chargaff and his co-workers noticed a pattern in the base composition of nucleotides from different species: The percentage of adenine (A) bases roughly equaled that of thymine (T) bases, and the percentage of cytosine (C) bases roughly equaled that of guanine (G) bases. Further, the percentage of each pair (A–T or C–G) varied from species to species. We now know that the 1:1 A/T and C/G ratios are due to complementary base pairing between A and T and between C and G in the DNA double helix, and differences between species are due to the unique sequences of bases along a DNA strand. In this exercise, you will apply Chargaff’s rules to predict the composition of bases in a genome. How the Experiments Were Done In Chargaff’s experiments, DNA was extracted from the given species, hydrolyzed to break apart the nucleotides, and then analyzed chemically. These studies provided approximate values for each type of nucleotide. (Today, whole-genome sequencing allows base composition analysis to be done more precisely directly from the sequence data.) Data from the Experiments Tables are useful for organizing sets of data representing a common set of values (here, percentages of A, G, C, and T) for a number of different samples (in this case, from different species). You can apply the patterns that you see in the known data to predict unknown values. In the table, complete base distribution data are given for sea urchin DNA and salmon DNA; you will use Chargaff’s rules to fill in the rest of the table with predicted values. While visiting the laboratory of Maurice Wilkins, Watson saw an X-ray diffraction image of DNA produced by Wilkins’s accomplished colleague Rosalind Franklin (Figure 16.6). Images produced by X-ray crystallography are not actually pictures of molecules. The spots in the image were produced by X-rays that were diffracted (deflected) as they passed through aligned fibers of purified DNA. Watson was familiar with the type of X-ray diffraction pattern of helical molecules, and seeing the . Figure 16.6 Rosalind Franklin and her X-ray diffraction photo of DNA. (a) Rosalind Franklin 318 UNIT THREE (b) Franklin’s X-ray diffraction photograph of DNA Genetics Base Percentage Source of DNA Adenine Guanine Cytosine Thymine Sea urchin 32.8 17.7 17.3 32.1 Salmon 29.7 20.8 20.4 29.1 Wheat 28.1 21.8 22.7 E. coli 24.7 26.0 Human 30.4 Ox 29.0 30.1 Average % Data from several papers by Chargaff: for example, E. Chargaff et al., Composition of the desoxypentose nucleic acids of four genera of sea-urchin, Journal of Biological Chemistry 195: 155–160 (1952). INTERPRET THE DATA ▶ Sea urchin 1. Explain how the sea urchin and salmon data demonstrate both of Chargaff’s rules. 2. Using Chargaff’s rules, fill in the table with your predictions of the missing percentages of bases, starting with the wheat genome and proceeding through E. coli, human, and ox. Show how you arrived at your answers. 3. If Chargaff’s rule—that the amount of A equals the amount of T and the amount of C equals the amount of G—is valid, then hypothetically we could extrapolate this to the combined DNA of all species on Earth (like one huge Earth genome). To see whether the data in the table support this hypothesis, calculate the average percentage for each base in your completed table by averaging the values in each column. Does Chargaff’s equivalence rule still hold true? Instructors: A version of this Scientific Skills Exercise can be assigned in Mastering Biology. photo that Wilkins showed him confirmed that DNA was helical in shape. The photo also added to earlier data obtained by Franklin and others suggesting the width of the helix and the spacing of the nitrogenous bases along it. The pattern in this photo implied that the helix was made up of two strands, contrary to a three-stranded model proposed by Linus Pauling a short time earlier. The presence of two strands accounts for the now-familiar term double helix. DNA is shown in some of its many different representations in Figure 16.7. Watson and Crick began building models of a double helix that would conform to the X-ray measurements and what was then known about the chemistry of DNA, including Chargaff’s rule of base equivalences. Having also read an unpublished annual report summarizing Franklin’s work, they knew she had concluded that the sugar-phosphate backbones were on the outside of the DNA molecule, contrary to their working model. Franklin’s arrangement was appealing because it put the negatively charged phosphate groups facing the aqueous surroundings, while the relatively hydrophobic nitrogenous bases were hidden in the interior. Watson constructed such a model, in which the two sugar-phosphate backbones are antiparallel—that is, their subunits run in opposite directions (see Figure 16.7). You can imagine the overall arrangement as a rope ladder with rigid rungs. The side ropes represent VISUALIZING DNA ▼ Figure 16.7 Here, the two DNA strands are shown untwisted so it’s easier to see the chemical details. Note that the strands are antiparallel—they are oriented in opposite directions, like the lanes of a divided street. DNA can be illustrated in many ways, but all diagrams represent the same basic structure. The level of detail shown depends on the process or the type of information being conveyed. 5¿ end Bases 0.34 nm apart Structural Images The space-filling model on the left shows the 3-D shape of the DNA double helix; the diagram on the right shows chemical details of DNA’s structure. In both images, phosphate groups are yellow, deoxyribose sugars are blue, and nitrogenous bases are shades of green and orange. O O 3¿ O O C G –O T C G C A 3¿ O G P O C –OH attached to 3¿ carbon O G CH2 O 3¿ A O– P O O O O– P O Van der Waals interactions between stacked base pairs help hold the molecule together. O P O CH2 O O O– P C O T O OH CH2 5¿ 3¿ end –O 1. Describe the bonds that hold together the nucleotides in one DNA strand. Then compare them with the bonds that hold the two DNA strands together. O O– P O 5¿ end 2. How do the two ends of a DNA strand differ in structure? When molecular detail is not necessary, DNA is portrayed in a range of simplified diagrams, depending on the focus of the figure. 5¿ A G C G 3¿ 5¿ The “ribbons” in these simplified double helix diagrams represent the sugar-phosphate backbones; these models emphasize the 3-D nature of DNA. 5¿ T A T A G C G C C G C G A T A T 3¿ T 5¿ 3¿ 3¿ 5¿ 3¿ 5¿ 3¿ 5¿ 3¿ 5¿ 3¿ 5¿ These flattened “ladder style” diagrams of DNA depict the sugar-phosphate backbones like the side rails of a ladder, with the base pairs as rungs. Light blue is used to indicate the more recently synthesized strand. 3. Compare the information conveyed in the three ladder diagrams. DNA Sequences O O Hydrogen bonds between bases hold strands together. O CH2 Sugar-phosphate backbone T A CH2 O O O Nitrogenous bases C O O Simplified Images G 3¿ Covalent sugar-phosphate bonds link nucleotides. CH2 Mastering Biology Animation: DNA Double Helix A OH A 2¿ CH2 One full turn every 10 base pairs (3.4 nm) DNA nucleotide Sugar-phosphate backbone T 1¿ O P –O Diameter = 2 nm T O Sugar 4¿ The DNA double helix is “right-handed.” Use your right hand as shown to follow the sugar-phosphate backbone up the helix and around to the back. (It won't work with your left hand.) 3¿ 5¿ CH2 –O 5¿ O– P –O 3¿ end Phosphate group attached to 5¿ carbon Nitrogenous base attached to 1¿ carbon Sometimes the double-stranded DNA molecule is shown simply as two straight lines. Genetic information is carried in DNA as a linear sequence of nucleotides that may be transcribed into mRNA and translated into a polypeptide. When focusing on the DNA sequence, each nucleotide can be represented simply as the letter of its base: A, T, C, or G. 3¿ - A C G T A A G C G G T T A A T - 5¿ 5¿ - T G C A T T C G C C A A T T A - 3¿ Instructors: Additional questions related to this Visualizing Figure can be assigned in Mastering Biology. CHAPTER 16 The Molecular Basis of Inheritance 319 the sugar-phosphate backbones, and the rungs represent pairs of nitrogenous bases. Now imagine twisting the ladder to form a helix. Franklin’s X-ray data indicated that the helix makes one full turn every 3.4 nm along its length. With the bases stacked just 0.34 nm apart, there are ten layers of base pairs, or rungs of the ladder, in each full turn of the helix. The nitrogenous bases of the double helix are paired in specific combinations: adenine (A) with thymine (T), and guanine (G) with cytosine (C). It was mainly by trial and error that Watson and Crick arrived at this key feature of DNA. At first, Watson imagined that the bases paired like with like—for example, A with A and C with C. But this model did not fit the X-ray data, which suggested that the double helix had a uniform diameter. Why is this requirement inconsistent with like-with-like pairing of bases? Adenine and guanine are purines, nitrogenous bases with two organic rings, while cytosine and thymine are nitrogenous bases called pyrimidines, which have a single ring. Pairing a purine with a pyrimidine is the only combination that results in a uniform diameter for the double helix (Figure 16.8). of guanine equals the amount of cytosine. (Modern DNAsequencing techniques have confirmed that the amounts are exactly equal.) Although the base-pairing rules dictate the combinations of nitrogenous bases that form the “rungs” of the double helix, they do not restrict the sequence of nucleotides along each DNA strand. The linear sequence of the four bases can be varied in countless ways, and each gene has a unique base sequence. In April 1953, Watson and Crick surprised the scientific world with a succinct, one-page paper that reported their molecular model for DNA: the double helix, which has since become an icon of molecular biology. Watson and Crick, along with Maurice Wilkins, were awarded the Nobel Prize in 1962 for this work. (Sadly, Rosalind Franklin had died at the age of 37 in 1958 and was thus ineligible for the prize.) The beauty of the double helix model was that the structure of DNA suggested the basic mechanism of its replication. Mastering Biology HHMI Video: Great Discoveries in Science: The Double Helix CONCEPT CHECK 16.1 . Figure 16.8 Possible base pairings in the DNA double helix. 1. Given a polynucleotide sequence such as GAATTC, explain what further information you would need in order to identify which is the 5′ end. (See Figure 16.5.) Purine + purine: too wide 2. VISUAL SKILLS While trying to develop a vaccine for S. pneumonia, Griffith was surprised to discover the phenomenon of bacterial transformation. Based on the results in the second and third panels of Figure 16.2, what result was he expecting in the fourth panel? Explain. Pyrimidine + pyrimidine: too narrow For suggested answers, see Appendix A. Purine + pyrimidine: width consistent with X-ray data CONCEPT Watson and Crick reasoned that there must be additional specificity of pairing dictated by the structure of the bases. Each base has chemical side groups that can form hydrogen bonds with its appropriate partner: Adenine can form two hydrogen bonds with thymine and only thymine; guanine forms three hydrogen bonds with cytosine and only cytosine. In shorthand, A pairs with T, and G pairs with C (Figure 16.9). . Figure 16.9 Base pairing in DNA. Nitrogenous base pairs are held together by hydrogen bonds. H H N N N Sugar N CH3 O H N O Sugar Thymine (T) H N N H N N O N N N N Adenine (A) N Sugar H N O H Sugar H Guanine (G) Cytosine (C) The Watson-Crick model took into account Chargaff’s ratios and ultimately explained them. Wherever one strand of a DNA molecule has an A, the partner strand has a T. Similarly, a G in one strand is always paired with a C in the complementary strand. Therefore, in the DNA of any organism, the amount of adenine equals the amount of thymine, and the amount 320 UNIT THREE Genetics 16.2 Many proteins work together in DNA replication and repair Hereditary information in DNA directs the development of your biochemical, anatomical, physiological, and, to some extent, behavioral traits. Your resemblance to your parents has its basis in the accurate replication of DNA prior to meiosis and therefore its transmission from your parents’ generation to yours. Replication prior to mitosis ensures faithful transmission of genetic information from a parent cell to two daughter cells. Of all nature’s molecules, nucleic acids are unique in their ability to dictate their own replication from monomers. The relationship between structure and function is evident in the double helix: The specific complementary pairing of nitrogenous bases in DNA has a functional significance. Watson and Crick ended their classic paper with this statement: “It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.”1 In this section, you will learn about the basic principle of DNA replication, the copying of DNA, as well as some important details of the process. 1 J. D. Watson and F. H. C. Crick, Molecular structure of nucleic acids: a structure for deoxyribose nucleic acids, Nature 171:737–738 (1953). . Figure 16.10 A model for DNA replication: the basic concept. In this simplified illustration, a short segment of DNA 5¿ 3¿ has been untwisted. Simple shapes symbolize the four kinds of bases. Dark blue represents DNA strands present in the parental molecule; 3¿ 5¿ light blue represents newly synthesized DNA. Mastering Biology Animation: DNA Replication: An Overview 3¿ 5¿ 3¿ 5¿ 3¿ A T A T A T A T C G C G C G C G T A T A T A T A A T A T A T A T G C G C G C G C 5¿ (a) The parental molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C. 3¿ 5¿ (b) First, the two DNA strands are separated. Each parental strand can now serve as a template for a new, complementary strand. The Basic Principle: Base Pairing to a Template Strand In a second paper, Watson and Crick stated their hypothesis for how DNA replicates: “Now our model for deoxyribonucleic acid is, in effect, a pair of templates, each of which is complementary to the other. We imagine that prior to duplication the hydrogen bonds are broken, and the two chains unwind and separate. Each chain then acts as a template for the formation on to itself of a new companion chain, so that eventually we shall have two pairs of chains, where we only had one before. Moreover, the sequence of the pairs of bases will have been duplicated exactly.”2 Figure 16.10 illustrates the basic idea. If you cover a DNA strand in Figure 16.10a, its linear sequence of nucleotides is revealed by applying the base-pairing rules to the uncovered strand. The two strands are complementary; each stores the information necessary to reconstruct the other. When a cell copies a DNA molecule, each strand serves as a template for ordering nucleotides into a new, complementary strand. Nucleotides line up along the template strand according to the base-pairing rules and are linked to form the new strands. One double-stranded DNA molecule becomes two, each an exact replica of the “parental” molecule. This model of DNA replication remained untested for several years following publication of the DNA structure. The necessary experiments were simple in concept but difficult to perform. Watson and Crick’s model predicts that when a double helix replicates, each of the two daughter molecules will have one old strand, from the parental molecule, and one newly made strand. This semiconservative model can be distinguished from a conservative model of replication, in which the two parental strands somehow come back together after the process (that is, the parental molecule is conserved). In yet a third model, called the dispersive model, all four strands of DNA following replication have a mixture of old and new DNA (Figure 16.11). 3¿ 5¿ 3¿ 5¿ (c) Nucleotides complementary to the parental (dark blue) strand are connected to form the sugar-phosphate backbones of the new “daughter” (light blue) strands. . Figure 16.11 DNA replication: three alternative models. Each short segment of double helix symbolizes the DNA within a cell. Beginning with a parent cell, we follow the DNA for two more generations of cells—two rounds of DNA replication. Parental DNA is dark blue; newly made DNA is light blue. Parent cell First replication Second replication (a) Conservative model. The two parental strands reassociate after acting as templates for new strands, thus restoring the parental double helix. (b) Semiconservative model. The two strands of the parental molecule separate, and each functions as a template for synthesis of a new, complementary strand. (c) Dispersive model. Each strand of both daughter molecules contains a mixture of old and newly synthesized DNA. 2 J. D. Watson and F. H. C. Crick, Genetical implications of the structure of deoxyribonucleic acid, Nature 171:964–967 (1953). CHAPTER 16 The Molecular Basis of Inheritance 321 Although mechanisms for conservative or dispersive DNA replication are not easy to devise, these models remained possibilities until they could be ruled out. After two years of preliminary work at the California Institute of Technology in the late 1950s, Matthew Meselson and Franklin Stahl devised a clever experiment that distinguished between the three models, described in Figure 16.12. Their results supported the semiconservative model of DNA replication, as predicted by Watson and Crick, and their experiment is widely recognized among biologists as a classic example of elegant design. The basic principle of DNA replication is conceptually simple. However, the actual process involves some complicated biochemical gymnastics, as we will now see. DNA Replication: A Closer Look The bacterium E. coli has a single chromosome of about 4.6 million nucleotide pairs. In a favorable environment, an E. coli cell can copy all of this DNA and divide to form two genetically identical daughter cells in considerably less than an hour. Each of your somatic cells has 46 DNA molecules in its nucleus, one long double-helical molecule per chromosome. In all, that represents about 6 billion nucleotide pairs, or over 1,000 times more DNA than is found in most bacterial cells. If we were to print the one-letter symbols for these bases (A, G, C, and T) the size of the type you are now reading, the 6 billion nucleotide pairs of information in a diploid human cell would fill about 1,400 biology textbooks. Yet it takes one of your cells just a few hours to copy all of this DNA during S phase of interphase. This replication of an enormous amount of genetic information is achieved with very few errors—only about one per 10 billion nucleotides. The copying of DNA is remarkable in its speed and accuracy. More than a dozen enzymes and other proteins participate in DNA replication. Much more is known about how this “replication machine” works in bacteria (such as E. coli) than in eukaryotes, and we will describe the basic steps of the process for E. coli, except where otherwise noted. What scientists have learned about eukaryotic DNA replication suggests, however, that most of the process is fundamentally similar for prokaryotes and eukaryotes. Getting Started The replication of chromosomal DNA begins at particular sites called origins of replication, short stretches of DNA that have a specific sequence of nucleotides. The E. coli chromosome, like many other bacterial chromosomes, is circular and has a single origin. Proteins that initiate DNA replication recognize this sequence and attach to the DNA, separating the two strands and opening up a replication “bubble” (Figure 16.13a). Replication of DNA then proceeds in both directions until the entire molecule is copied. In contrast to a bacterial chromosome, a eukaryotic chromosome may have hundreds or even a few thousand replication origins. Multiple replication bubbles form and eventually fuse, thus speeding up the copying of the very 322 UNIT THREE Genetics ▼ Figure 16.12 Inquiry Does DNA replication follow the conservative, semiconservative, or dispersive model? Experiment Matthew Meselson and Franklin Stahl cultured E. coli for several generations in a medium containing nucleotide precursors labeled with a heavy isotope of nitrogen, 15N. They then transferred the bacteria to a medium with only 14N, a lighter isotope. They took one sample after the first DNA replication and another after the second replication. They extracted DNA from the bacteria in the samples and then centrifuged each DNA sample to separate DNA of different densities. 1 Bacteria cultured in medium with 15N (heavy isotope) 2 Bacteria transferred to medium with 14N (lighter isotope) Results 3 DNA sample centrifuged after first replication 4 DNA sample centrifuged after second replication Less dense More dense Conclusion Meselson and Stahl compared their results to those predicted by each of the three models in Figure 16.11, as shown below. The first replication in the 14N medium produced a band of many molecules of hybrid (15N-14N) DNA. This result eliminated the conservative model. The second replication produced both light and hybrid DNA, a result that refuted the dispersive model and supported the semiconservative model. They therefore concluded that DNA replication is semiconservative. Predictions: First replication Second replication Conservative model Semiconservative model Dispersive model Data from M. Meselson and F. W. Stahl, The replication of DNA in Escherichia coli, Proceedings of the National Academy of Sciences USA 44:671–682 (1958). INQUIRY IN ACTION Read and analyze the original paper in Inquiry in Action: Interpreting Scientific Papers. Instructors: A related Experimental Inquiry Tutorial can be assigned in Mastering Biology. WHAT IF? If Meselson and Stahl had first grown the cells in 14 N-containing medium and then moved them into 15N-containing medium before taking samples, what would have been the result after each of the two replications? . Figure 16.13 Origins of replication in E. coli and eukaryotes. The red arrows indicate the movement of the replication forks and thus the overall directions of DNA replication within each bubble. (a) Origin of replication in an E. coli cell Origin of replication (b) Origins of replication in a eukaryotic cell Parental (template) strand Daughter (new) strand Origin of replication Double-stranded DNA molecule Parental (template) strand Doublestranded DNA molecule Daughter (new) strand Replication fork Replication bubble Replication fork Bubble Two daughter DNA molecules 0.5 om 0.25 om Two daughter DNA molecules The circular chromosome of E. coli and other bacteria has only one origin of replication. The parental strands separate there, forming a replication bubble with two forks (red arrows). Replication proceeds in both directions until the forks meet on the other side, resulting in two daughter DNA molecules. The TEM shows a bacterial chromosome with a replication bubble. In a linear chromosome of a eukaryote, replication bubbles form at many sites along the giant DNA molecule during S phase of interphase. The bubbles expand as replication proceeds in both directions (red arrows). Eventually, the bubbles fuse and synthesis of the daughter strands is complete. The TEM shows three replication bubbles along the DNA of a cultured Chinese hamster cell. DRAW IT In the TEM, add arrows in the forks of the third bubble. Mastering Biology BioFlix® Animation: The Replication Fork in E. coli long DNA molecules (Figure 16.13b). As in bacteria, eukaryotic DNA replication proceeds in both directions from each origin. At each end of a replication bubble is a replication fork, a Y-shaped region where the parental strands of DNA are being unwound. Several kinds of proteins participate in the unwinding (Figure 16.14). Helicases are enzymes that untwist the double helix at the replication forks, separating the two parental strands and making them available as template strands. After the parental strands separate, single-strand binding proteins bind to the unpaired DNA strands, keeping them from re-pairing. The untwisting of the double helix causes tighter twisting and strain ahead of the replication fork. Topoisomerase is an enzyme that helps relieve this strain by breaking, swiveling, and rejoining DNA strands. . Figure 16.14 Some of the proteins involved in the initiation of DNA replication. The same proteins function at both replication forks in a replication bubble. For simplicity, only the lefthand fork is shown, and the DNA bases are drawn much larger in relation to the proteins than they are in reality. Topoisomerase breaks, swivels, and rejoins the parental DNA ahead of the replication fork, relieving the strain caused by unwinding. Primase synthesizes RNA primers, using the parental DNA as a template. 3¿ 5¿ 5¿ Replication fork 3¿ Synthesizing a New DNA Strand The unwound sections of parental DNA strands are now available to serve as templates for the synthesis of new complementary DNA strands. However, the enzymes that 3¿ RNA primer Helicase unwinds and separates the parental DNA strands. CHAPTER 16 5¿ Single-strand binding proteins stabilize the unwound parental strands. The Molecular Basis of Inheritance 323 synthesize DNA cannot initiate the synthesis of a polynucleotide; they can only add DNA nucleotides to the end of an already existing chain that is base-paired with the template strand. An initial nucleotide chain that can be used as a pre-existing chain is produced during DNA synthesis; this is actually a short stretch of RNA, not DNA. The RNA chain is called a primer and is synthesized by the enzyme primase (see Figure 16.14). Primase starts a complementary RNA chain with a single RNA nucleotide and adds RNA nucleotides one at a time, using the parental DNA strand as a template. The completed primer, generally five to ten nucleotides long, is thus base-paired to the template strand. The new DNA strand will start from the 3′ end of the RNA primer. Enzymes called DNA polymerases catalyze the synthesis of new DNA by adding nucleotides to the 3′ end of a preexisting chain. In E. coli, there are several DNA polymerases, but two of them appear to play the major roles in DNA replication: DNA polymerase III and DNA polymerase I. The situation in eukaryotes is more complicated, with at least 11 different DNA polymerases discovered so far, although the general principles are the same. Most DNA polymerases require a primer and a DNA template strand, along which complementary DNA nucleotides are lined up, one by one. In E. coli, DNA polymerase III (abbreviated DNA pol III) adds a DNA nucleotide to the RNA primer and then continues adding DNA nucleotides, which are complementary to the parental DNA template strand, to the growing end of the new DNA strand. The rate of elongation is about 500 nucleotides per second in bacteria and 50 per second in human cells. Each nucleotide to be added to a growing DNA strand consists of a sugar attached to a base and to three phosphate groups. You have already encountered such a molecule—ATP (adenosine triphosphate; see Figure 8.9). The only difference between the ATP of energy metabolism and dATP, the adenine nucleotide used to make DNA, is the sugar component, which is deoxyribose in the building block of DNA but ribose in ATP. Like ATP, the nucleotides used for DNA synthesis are chemically reactive, partly because their triphosphate tails have an unstable cluster of negative charge. DNA polymerase catalyzes the addition of each monomer to the growing end of a DNA strand by a condensation reaction in which two phosphate groups are lost as a molecule of pyrophosphate ( P — P i). Subsequent hydrolysis of the pyrophosphate to two molecules of inorganic phosphate ( P i) is a coupled exergonic reaction that helps drive the polymerization reaction (Figure 16.15). Antiparallel Elongation As we have noted previously, the two ends of a DNA strand are different, giving each strand directionality, like a one-way street (see Figure 16.5). In addition, the two strands of DNA in a double helix are antiparallel, meaning that they are oriented in opposite directions to each other, like the two sides 324 UNIT THREE Genetics . Figure 16.15 Addition of a nucleotide to a DNA strand. DNA polymerase catalyzes addition of a nucleotide to the 3′ end of a growing DNA strand, with the release of two phosphates. New strand 5¿ Phosphate Sugar Template strand 3¿ T A T C G C G G C G C T A P T P 3¿ A Base OH 3¿ P 5¿ OH Incoming nucleotide DNA polymerase A OH P Pi 3¿ Pyrophosphate C C 5¿ 5¿ 2 Pi DRAW IT Circle the area where the new bond was made. Mastering Biology Figure Walkthrough of a divided street (see Figure 16.15). Therefore, the two new strands formed during DNA replication must also be antiparallel to their template strands. The antiparallel arrangement of the double helix, together with a constraint on the function of DNA polymerases, has an important effect on how replication occurs. Because of their structure, DNA polymerases can add nucleotides only to the free 3′ end of a primer or growing DNA strand, never to the 5′ end (see Figure 16.15). Thus, a new DNA strand can elongate only in the 5′ S 3′ direction. With this in mind, let’s examine one of the two replication forks in a bubble (Figure 16.16). Along one template strand, DNA polymerase III can synthesize a complementary strand continuously by elongating the new DNA in the mandatory 5′ S 3′ direction. DNA pol III remains in the replication fork on that template strand and continuously adds nucleotides to the new complementary strand as the fork progresses. The DNA strand made by this mechanism is called the leading strand. Only one primer is required for DNA pol III to synthesize the entire leading strand (see Figure 16.16). To elongate the other new strand of DNA in the mandatory 5′ S 3′ direction, DNA pol III must work along the other template strand in the direction away from the replication fork. The DNA strand elongating in this direction is called the lagging strand. In contrast to the leading strand, which elongates continuously, the lagging strand is synthesized discontinuously, as a series of segments. These segments of the lagging strand are called Okazaki fragments, after Reiji Okazaki, the Japanese scientist who discovered them. The fragments are about 1,000–2,000 nucleotides long in E. coli and 100–200 nucleotides long in eukaryotes. . Figure 16.16 Synthesis of the leading strand during DNA replication. This diagram focuses on the left replication fork shown in the overview box. DNA polymerase III (DNA pol III), shaped like a cupped hand, is shown closely associated with a protein called the “sliding clamp” that encircles the newly synthesized double helix like a doughnut. The sliding clamp moves DNA pol III along the DNA template strand. 5¿ Overview Leading strand Leading strand Origin of replication 2 5¿ Lagging strand 1 Primase joins RNA nucleotides into the first primer for the lagging strand. 3¿ Leading strand 5¿ Template strand Site where replication begins DNA pol III starts to synthesize the leading strand. 3¿ 5¿ RNA primer 3¿ 5¿ Helicase 3¿ 3¿ 2 The leading strand is 3¿ 5¿ elongated continuously in the 5¿: 3¿ direction as helicase opens up the fork further to the left. 2 DNA pol III adds DNA nucleotides to the primer, forming Okazaki fragment 1. 3¿ 5¿ 3¿ 5¿ 3¿ 5¿ 1 RNA primer for fragment 2 3¿ 5¿ 4 Fragment 2 is primed. Then DNA pol III adds DNA nucleotides, detaching when it reaches the fragment 1 primer. Okazaki fragment 2 2 Mastering Biology BioFlix® Animation: Synthesis of the Leading Strand 3¿ 5¿ 1 Figure 16.17 illustrates the steps in the synthesis of the lagging strand at one fork. Whereas only one primer is required on the leading strand, each Okazaki fragment on the lagging strand must be primed separately (steps 1 and 4 ). After DNA pol III forms an Okazaki fragment (steps 2 to 4 ), another DNA polymerase, DNA pol I, replaces the RNA nucleotides of the adjacent primer with DNA nucleotides one at a time (step 5 ). But DNA pol I cannot join the final nucleotide of this replacement DNA segment to the first DNA nucleotide of the adjacent Okazaki fragment. Another enzyme, DNA ligase, accomplishes this task, joining the sugar-phosphate backbones of all the Okazaki fragments into a continuous DNA strand (step 6 ). 3¿ 5¿ Okazaki fragment 1 5¿ 5¿ 5¿ Primer for leading strand 3 After reaching the next RNA primer to the right, DNA pol III detaches. 5¿ 3¿ 5¿ Site where replication begins 1 Sliding clamp 3¿ Parental DNA 3¿ RNA primer for fragment 1 3¿ 5¿ Single-strand binding proteins Leading strand Overall directions of replication Overall directions of replication 1 After an RNA primer is made, 5¿ 1 Template strand 3¿ 3¿ Lagging strand 3¿ Lagging strand Primer Origin of replication 5¿ 3¿ Overview Template strand . Figure 16.17 Synthesis of the lagging strand. 3¿ 5¿ 5 DNA pol I replaces the RNA with DNA, adding nucleotides to the 3¿ end of fragment 1 (and, later, of fragment 2). 2 1 3¿ 5¿ 6 DNA ligase forms a bond between the newest DNA and the DNA of fragment 1. 3¿ 5¿ 7 The lagging strand in this region is now complete. 2 1 Mastering Biology BioFlix® Animation: Synthesis of the Lagging Strand 3¿ 5¿ Overall direction of replication CHAPTER 16 The Molecular Basis of Inheritance 325 molecular brake, slowing progress of the replication fork and coordinating the placement of primers and the rates of replication on the leading and lagging strands. Second, the DNA replication complex may not move along the DNA; rather, the DNA may move through the complex during the replication process. In eukaryotic cells, multiple copies of the complex, perhaps grouped into “factories,” may be anchored to the nuclear matrix, a framework of fibers extending through the interior of the nucleus. Some experimental evidence supports a model in which two DNA polymerase molecules, one on each template strand, “reel in” the parental DNA and extrude newly made daughter DNA molecules. In this so-called trombone model, the lagging strand is also looped back through the complex (Figure 16.19). Whether the complex moves along the DNA or whether the DNA moves through the complex, either anchored or not, are still open, unresolved questions that are under active investigation. Synthesis of the leading strand and synthesis of the lagging strand occur concurrently and at the same rate. The lagging strand is so named because its synthesis is delayed slightly relative to synthesis of the leading strand; each new fragment of the lagging strand cannot be started until enough template has been exposed at the replication fork. Figure 16.18 and Table 16.1 summarize DNA replication. Please study them carefully before proceeding. The DNA Replication Complex It is traditional—and convenient—to represent DNA polymerase molecules as locomotives moving along a DNA railroad track, but such a model is inaccurate in two important ways. First, the various proteins that participate in DNA replication actually form a single large complex, a “DNA replication machine.” Many protein-protein interactions facilitate the efficiency of this complex. For example, by interacting with other proteins at the fork, primase apparently acts as a . Figure 16.18 A summary of bacterial DNA replication. The detailed diagram shows the left-hand replication fork of the replication bubble shown in the overview (upper right). Viewing each daughter strand in its entirety in the overview, you can see that half of it is made continuously as the leading strand, while the other half (on the other side of the origin) is synthesized in fragments Overview as the lagging strand. Origin of replication Leading strand Lagging strand 3 The leading strand is Leading strand synthesized continuously template in the 5¿ to 3¿ direction 2 Molecules of single- by DNA pol III. strand binding protein Leading strand stabilize the unwound Lagging strand template strands. Overall directions of replication 1 Helicase Leading strand unwinds the parental double helix. 3¿ 5¿ 5¿ 3¿ 3¿ DNA pol III Primer 5¿ Parental DNA Primase 3¿ 5 DNA pol III 5¿ 4 4 Primase begins synthesis of the RNA primer for the fifth Okazaki fragment. DRAW IT Draw a diagram showing the right-hand fork of the bubble in this figure. Number the Okazaki fragments, and label all 5′ and 3′ ends. 5 DNA pol III is completing synthesis of fragment 4. When it reaches the RNA primer on fragment 3, it will detach and begin adding DNA nucleotides to the 3¿ end of the fragment 5 primer in the replication fork. 3¿ Lagging strand DNA pol I 3 2 DNA ligase 5¿ Lagging strand template 6 DNA pol I removes the primer from the 5¿ end of fragment 2, replacing it with DNA nucleotides added one by one to the 3¿ end of fragment 3. After the last addition, the backbone is left with a free 3¿ end. UNIT THREE Genetics 3¿ 5¿ 7 DNA ligase joins the 3¿ end of fragment 2 to the 5¿ end of fragment 1. Mastering Biology Animation: DNA Replication: A Closer Look 326 1 Table 16.1 Bacterial DNA Replication Proteins and Their Functions Protein Function Helicase 3¿ 5¿ 3¿ 3¿ Topoisomerase 5¿ 3¿ 3¿ 5¿ Primase DNA pol III 3¿ 5¿ 3¿ 3¿ 5¿ 3¿ 5¿ DNA pol I 5¿ 3¿ DNA ligase Leading strand template 5¿ Single-strand binding protein 5¿ 5¿ Unwinds parental double helix at replication forks . Figure 16.19 The “trombone” model of the DNA replication complex. In this proposed model, two molecules of DNA polymerase III work together in a complex, one on each strand, with helicase and other proteins. The lagging strand template DNA loops through the complex, resembling the slide of a trombone. 3¿ 5¿ Binds to and stabilizes singlestranded DNA until it is used as a template DNA pol III Parental DNA Relieves overwinding strain ahead of replication forks by breaking, swiveling, and rejoining DNA strands 5¿ 3¿ Synthesizes an RNA primer at 5′ end of leading strand and at 5′ end of each Okazaki fragment of lagging strand Connecting protein Using parental DNA as a template, synthesizes new DNA strand by adding nucleotides to an RNA primer or a pre-existing DNA strand Lagging strand template Removes RNA nucleotides of primer from 5′ end and replaces them with DNA nucleotides added to 3′ end of adjacent fragment Joins Okazaki fragments of lagging strand; on leading strand, joins 3′ end of DNA that replaces primer to rest of leading strand DNA Proofreading and Repairing DNA We cannot attribute the accuracy of DNA replication solely to the specificity of base pairing. Initial pairing errors between incoming nucleotides and those in the template strand occur at a rate of one in 105 nucleotides. However, errors in the completed DNA molecule amount to only one in 1010 (10 billion) nucleotides, an error rate that is 100,000 times lower. This is because during DNA replication, many DNA polymerases proofread each nucleotide against its template as soon as it is covalently bonded to the growing strand. Upon finding an incorrectly paired nucleotide, the polymerase removes the nucleotide and then resumes synthesis. (This action is similar to fixing a texting error by deleting the wrong letter and then entering the correct one.) Mismatched nucleotides sometimes evade proofreading by a DNA polymerase. In mismatch repair, other enzymes remove and replace incorrectly paired nucleotides that have resulted from replication errors. Researchers highlighted the importance of such repair enzymes when they found that a hereditary defect in one of them is associated with a form of colon cancer. Apparently, this defect allows cancer-causing errors to accumulate in the DNA faster than normal. Incorrectly paired or altered nucleotides can also arise after replication. In fact, maintenance of the genetic information encoded in DNA requires frequent repair of 5¿ 3¿ Leading strand 3¿ 5¿ 5¿ 3¿ Helicase DNA pol III 3¿ 5¿ 3¿ 5¿ Lagging strand Overall direction of replication DRAW IT Draw a line tracing the lagging strand template in this figure. Mastering Biology BioFlix® Animation: DNA Replication various kinds of damage to existing DNA. DNA molecules are constantly subjected to potentially harmful chemical and physical agents, such as X-rays, as we’ll discuss in Concept 17.5. In addition, DNA bases may undergo spontaneous chemical changes under normal cellular conditions. However, these changes in DNA are usually corrected before they become permanent changes— mutations—perpetuated through successive replications. Each cell continuously monitors and repairs its genetic material. Because repair of damaged DNA is so important to the survival of an organism, it is no surprise that many different DNA repair enzymes have evolved. Almost 100 are known in E. coli, and about 170 have been identified so far in humans. Most cellular systems for repairing incorrectly paired nucleotides, whether they are due to DNA damage or to replication errors, use a mechanism that takes advantage of the base-paired structure of DNA. In many cases, a segment of the strand containing the damage is cut out (excised) by a DNA-cutting enzyme—a nuclease—and the resulting gap is then filled in with nucleotides, using the undamaged strand as a template. The enzymes involved in filling the gap are a DNA polymerase and DNA ligase. There are several such DNA repair systems; one is called nucleotide excision repair. CHAPTER 16 The Molecular Basis of Inheritance 327 . Figure 16.20 Nucleotide excision repair of DNA damage. 5¿ 3¿ 3¿ 5¿ Nuclease 5¿ 3¿ 3¿ 5¿ DNA polymerase 5¿ 3¿ 3¿ 5¿ DNA ligase 5¿ 3¿ 3¿ 5¿ 1 Teams of enzymes detect and repair damaged DNA, such as this thymine dimer (often caused by ultraviolet radiation), which distorts the DNA molecule. 2 A nuclease enzyme cuts the damaged DNA strand at two points, and the damaged section is removed. 3 Repair synthesis by a DNA polymerase fills in the missing nucleotides, using the undamaged strand as a template. Replicating the Ends of DNA Molecules 4 DNA ligase seals the free end of the new DNA to the old DNA, making the strand complete. An example is shown in Figure 16.20. An important function of the DNA repair enzymes in our skin cells is to repair genetic damage caused by the UV rays of sunlight: For example, adjacent thymine bases on a DNA strand can become covalently linked into thymine dimers, causing the DNA to buckle and interfere with DNA replication. The importance of repairing this kind of damage is underscored by a disorder called xeroderma pigmentosum (XP), which in most cases is caused by an inherited defect in a nucleotide excision repair enzyme. Individuals with XP are hypersensitive to sunlight; mutations in their skin cells caused by ultraviolet light are left uncorrected, often resulting in skin cancer. The effects are extreme: Without sun protection, children who have XP can develop skin cancer by age 10. Evolutionary Significance of Altered DNA Nucleotides EVOLUTION Faithful replication of the genome and repair of DNA damage are important for the functioning of the organism and for passing on a compl