{"id":2527,"date":"2016-06-02T17:11:43","date_gmt":"2016-06-02T17:11:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/biologyxwaymakerxmaster\/?post_type=chapter&#038;p=2527"},"modified":"2017-04-18T22:28:33","modified_gmt":"2017-04-18T22:28:33","slug":"dna-base-pairs-and-replication","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-wmopen-biology1\/chapter\/dna-base-pairs-and-replication\/","title":{"raw":"DNA Base Pairs and Replication","rendered":"DNA Base Pairs and Replication"},"content":{"raw":"<h2>Explain the role of complementary base pairing in the precise replication process of DNA<\/h2>\r\nIn this outcome, we'll learn more about the precise structure of DNA and how it replicates.\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Understand the historical basis of our understanding of DNA<\/li>\r\n \t<li>Outline the basic steps in DNA replication<\/li>\r\n \t<li>Identify the major enzymes that play a role in DNA replication<\/li>\r\n \t<li>Identify the key proofreading processes in DNA replication<\/li>\r\n \t<li>Understand the basic role of telomeres in protecting DNA from replication errors<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The History of DNA<\/h2>\r\n[caption id=\"attachment_2594\" align=\"alignright\" width=\"300\"]<img class=\"wp-image-2594\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/06150802\/Figure_14_01_01.jpg\" alt=\"Portrait of Friedrich Miescher\" width=\"300\" height=\"413\" \/> Figure 1. Friedrich Miescher (1844\u20131895) discovered nucleic acids.[\/caption]\r\n\r\nModern understandings of DNA have evolved from the discovery of nucleic acid to the development of the double-helix model. In the 1860s, Friedrich Miescher (Figure 1), a physician by profession, was the first person to isolate phosphate-rich chemicals from white blood cells or leukocytes. He named these chemicals (which would eventually be known as RNA and DNA) nuclein because they were isolated from the nuclei of the cells.\r\n\r\nTo see Miescher conduct an experiment step-by-step, click through this <a href=\"https:\/\/www.dnalc.org\/view\/16342-Animation-15-DNA-and-proteins-are-key-molecules-of-the-cell-nucleus-.html\" target=\"_blank\">review of how he discovered the key role of DNA and proteins in the nucleus<\/a>.\r\n\r\nA half century later, British bacteriologist Frederick Griffith was perhaps the first person to show that hereditary information could be transferred from one cell to another \u201chorizontally,\u201d rather than by descent. In 1928, he reported the first demonstration of bacterial <strong>transformation<\/strong>, a process in which external DNA is taken up by a cell, thereby changing morphology and physiology. He was working with <em>Streptococcus pneumoniae,<\/em> the bacterium that causes pneumonia. Griffith worked with two strains, rough (R) and smooth (S). The R strain is non-pathogenic (does not cause disease) and is called rough because its outer surface is a cell wall and lacks a capsule; as a result, the cell surface appears uneven under the microscope. The S strain is pathogenic (disease-causing) and has a capsule outside its cell wall. As a result, it has a smooth appearance under the microscope. Griffith injected the live R strain into mice and they survived. In another experiment, when he injected mice with the heat-killed S strain, they also survived. In a third set of experiments, a mixture of live R strain and heat-killed S strain were injected into mice, and\u2014to his surprise\u2014the mice died. Upon isolating the live bacteria from the dead mouse, only the S strain of bacteria was recovered. When this isolated S strain was injected into fresh mice, the mice died. Griffith concluded that something had passed from the heat-killed S strain into the live R strain and transformed it into the pathogenic S strain, and he called this the transforming principle (Figure 2). These experiments are now famously known as Griffith's transformation experiments.\r\n\r\n[caption id=\"attachment_2595\" align=\"aligncenter\" width=\"800\"]<img class=\"size-full wp-image-2595\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/06150941\/Figure_14_01_02.jpg\" alt=\"On the left is a photo of a live mouse, representing a mouse injected with heat-killed, virulent S strain. On the right is a photo of a dead mouse, representing a mouse injected with heat-killed, virulent S strain and live, non-virulent R strain.\" width=\"800\" height=\"317\" \/> Figure 2. Two strains of <em>S. pneumoniae<\/em> were used in Griffith\u2019s transformation experiments. The R strain is non-pathogenic. The S strain is pathogenic and causes death. When Griffith injected a mouse with the heat-killed S strain and a live R strain, the mouse died. The S strain was recovered from the dead mouse. Thus, Griffith concluded that something had passed from the heat-killed S strain to the R strain, transforming the R strain into S strain in the process. (credit \"living mouse\": modification of work by NIH; credit \"dead mouse\": modification of work by Sarah Marriage)[\/caption]\r\n\r\nScientists Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) were interested in exploring this transforming principle further. They isolated the S strain from the dead mice and isolated the proteins and nucleic acids, namely RNA and DNA, as these were possible candidates for the molecule of heredity. They conducted a systematic elimination study. They used enzymes that specifically degraded each component and then used each mixture separately to transform the R strain. They found that when DNA was degraded, the resulting mixture was no longer able to transform the bacteria, whereas all of the other combinations were able to transform the bacteria. This led them to conclude that DNA was the transforming principle.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Forensic Scientists and DNA Analysis<\/h3>\r\nDNA evidence was used for the first time to solve an immigration case. The story started with a teenage boy returning to London from Ghana to be with his mother. Immigration authorities at the airport were suspicious of him, thinking that he was traveling on a forged passport. After much persuasion, he was allowed to go live with his mother, but the immigration authorities did not drop the case against him. All types of evidence, including photographs, were provided to the authorities, but deportation proceedings were started nevertheless. Around the same time, Dr. Alec Jeffreys of Leicester University in the United Kingdom had invented a technique known as DNA fingerprinting. The immigration authorities approached Dr. Jeffreys for help. He took DNA samples from the mother and three of her children, plus an unrelated mother, and compared the samples with the boy\u2019s DNA. Because the biological father was not in the picture, DNA from the three children was compared with the boy\u2019s DNA. He found a match in the boy\u2019s DNA for both the mother and his three siblings. He concluded that the boy was indeed the mother\u2019s son.\r\n\r\nForensic scientists analyze many items, including documents, handwriting, firearms, and biological samples. They analyze the DNA content of hair, semen, saliva, and blood, and compare it with a database of DNA profiles of known criminals. Analysis includes DNA isolation, sequencing, and sequence analysis; most forensic DNA analysis involves polymerase chain reaction (PCR) amplification of short tandem repeat (STR) loci and electrophoresis to determine the length of the PCR-amplified fragment. Only mitochondrial DNA is sequenced for forensics. Forensic scientists are expected to appear at court hearings to present their findings. They are usually employed in crime labs of city and state government agencies. Geneticists experimenting with DNA techniques also work for scientific and research organizations, pharmaceutical industries, and college and university labs. Students wishing to pursue a career as a forensic scientist should have at least a bachelor's degree in chemistry, biology, or physics, and preferably some experience working in a laboratory.\r\n\r\n<\/div>\r\nExperiments conducted by Martha Chase and Alfred Hershey in 1952 provided confirmatory evidence that DNA was the genetic material and not proteins. Chase and Hershey were studying a bacteriophage, which is a virus that infects bacteria. Viruses typically have a simple structure: a protein coat, called the capsid, and a nucleic acid core that contains the genetic material, either DNA or RNA. The bacteriophage infects the host bacterial cell by attaching to its surface, and then it injects its nucleic acids inside the cell. The phage DNA makes multiple copies of itself using the host machinery, and eventually the host cell bursts, releasing a large number of bacteriophages. Hershey and Chase labeled one batch of phage with radioactive sulfur, <sup>35<\/sup>S, to label the protein coat. Another batch of phage were labeled with radioactive phosphorus, <sup>32<\/sup>P. Because phosphorous is found in DNA, but not protein, the DNA and not the protein would be tagged with radioactive phosphorus.\r\n\r\nEach batch of phage was allowed to infect the cells separately. After infection, the phage bacterial suspension was put in a blender, which caused the phage coat to be detached from the host cell. The phage and bacterial suspension was spun down in a centrifuge. The heavier bacterial cells settled down and formed a pellet, whereas the lighter phage particles stayed in the supernatant. In the tube that contained phage labeled with <sup>35<\/sup>S, the supernatant contained the radioactively labeled phage, whereas no radioactivity was detected in the pellet. In the tube that contained the phage labeled with <sup>32<\/sup>P, the radioactivity was detected in the pellet that contained the heavier bacterial cells, and no radioactivity was detected in the supernatant. Hershey and Chase concluded that it was the phage DNA that was injected into the cell and carried information to produce more phage particles, thus providing evidence that DNA was the genetic material and not proteins (Figure 3).\r\n\r\n[caption id=\"attachment_2596\" align=\"aligncenter\" width=\"800\"]<img class=\"size-full wp-image-2596\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/06151133\/Figure_14_01_03.jpg\" alt=\"Illustration shows bacteria being infected by phage labeled with ^{35}S, which is incorporated into the protein coat, or ^{32}P, which is incorporated into the DNA. Infected bacteria were separated from phage by centrifugation and cultured. The bacteria that had been infected with phage containing ^{32}P-labeled DNA made radioactive phage. The bacteria that had been infected with ^{35}S-labeled phage produced unlabeled phage. The results support the hypothesis that DNA, and not protein, is the genetic material.\" width=\"800\" height=\"638\" \/> Figure 3. In Hershey and Chase's experiments, bacteria were infected with phage radiolabeled with either <sup>35<\/sup>S, which labels protein, or <sup>32<\/sup>P, which labels DNA. Only <sup>32<\/sup>P entered the bacterial cells, indicating that DNA is the genetic material.[\/caption]\r\n<p id=\"fs-id1430603\">Around this same time, Austrian biochemist Erwin Chargaff examined the content of DNA in different species and found that the amounts of adenine, thymine, guanine, and cytosine were not found in equal quantities, and that it varied from species to species, but not between individuals of the same species. He found that the amount of adenine equals the amount of thymine, and the amount of cytosine equals the amount of guanine, or A = T and G = C. This is also known as Chargaff\u2019s rules. This finding proved immensely useful when Watson and Crick were getting ready to propose their DNA double helix model.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Practice Question<\/h3>\r\nThe experiments by Hershey and Chase helped confirm that DNA was the hereditary material on the basis of the finding that:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>radioactive phage were found in the pellet<\/li>\r\n \t<li>radioactive cells were found in the supernatant<\/li>\r\n \t<li>radioactive sulfur was found inside the cell<\/li>\r\n \t<li>radioactive phosphorus was found in the cell<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"549343\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"549343\"]d.\u00a0radioactive phosphorus was found in the cell[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Basics of DNA Replication<\/h2>\r\n[caption id=\"attachment_4411\" align=\"alignright\" width=\"400\"]<img class=\" wp-image-4411\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1087\/2017\/02\/02000938\/Figure_14_03_01a.jpg\" alt=\"Illustration shows the conservative, semi-conservative, and dispersive models of DNA synthesis. In the conservative model, when DNA is replicated and both newly synthesized strands are paired together. In the semi-conservative model, each newly synthesized strand pairs with a parent strand. In the dispersive model, newly synthesized DNA is interspersed with parent DNA within both DNA strands.\" width=\"400\" height=\"404\" \/> Figure 4. The three suggested models of DNA replication. Grey indicates the original DNA strands, and blue indicates newly synthesized DNA.[\/caption]\r\n\r\nThe elucidation of the structure of the double helix provided a hint as to how DNA divides and makes copies of itself. This model suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. What was not clear was how the replication took place. There were three models suggested: conservative, semi-conservative, and dispersive (see Figure 4).\r\n\r\nIn conservative replication, the parental DNA remains together, and the newly formed daughter strands are together. The semi-conservative method suggests that each of the two parental DNA strands act as a template for new DNA to be synthesized; after replication, each double-stranded DNA includes one parental or \u201cold\u201d strand and one \u201cnew\u201d strand. In the dispersive model, both copies of DNA have double-stranded segments of parental DNA and newly synthesized DNA interspersed.\r\n\r\nMeselson and Stahl were interested in understanding how DNA replicates. They grew\u00a0<em>E. coli<\/em> for several generations in a medium containing a \u201cheavy\u201d isotope of nitrogen (<sup>15<\/sup>N) that gets incorporated into nitrogenous bases, and eventually into the DNA (Figure 5).\r\n\r\n[caption id=\"attachment_1430\" align=\"aligncenter\" width=\"800\"]<img class=\"size-full wp-image-1430\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184216\/Figure_14_03_02.jpg\" alt=\"Illustration shows an experiment in which E. coli was grown initially in media containing ^{15}N nucleotides. When the DNA was extracted and run in an ultracentrifuge, a band of DNA appeared low in the tube. The culture was next placed in ^{14}N medium. After one generation, all of the DNA appeared in the middle of the tube, indicating that the DNA was a mixture of half ^{14}N and half ^{15}N DNA. After two generations, half of the DNA appeared in the middle of the tube, and half appeared higher up, indicating that half the DNA contained 50% ^{15}N, and half contained ^{14}N only. In subsequent generations, more and more of the DNA appeared in the upper, ^{14}N band.\" width=\"800\" height=\"698\" \/> Figure 5. Meselson and Stahl experimented with E. coli grown first in heavy nitrogen (<sup>15<\/sup>N) then in <sup>14<\/sup>N. DNA grown in <sup>15<\/sup>N (red band) is heavier than DNA grown in <sup>14<\/sup>N (orange band), and sediments to a lower level in cesium chloride solution in an ultracentrifuge. When DNA grown in <sup>15<\/sup>N is switched to media containing <sup>14<\/sup>N, after one round of cell division the DNA sediments halfway between the <sup>15<\/sup>N and <sup>14<\/sup>N levels, indicating that it now contains fifty percent <sup>14<\/sup>N. In subsequent cell divisions, an increasing amount of DNA contains <sup>14<\/sup>N only. This data supports the semi-conservative replication model. (credit: modification of work by Mariana Ruiz Villareal)[\/caption]\r\n\r\nThe\u00a0<em>E. coli<\/em> culture was then shifted into medium containing <sup>14<\/sup>N and allowed to grow for one generation. The cells were harvested and the DNA was isolated. The DNA was centrifuged at high speeds in an ultracentrifuge. Some cells were allowed to grow for one more life cycle in <sup>14<\/sup>N and spun again. During the density gradient centrifugation, the DNA is loaded into a gradient (typically a salt such as cesium chloride or sucrose) and spun at high speeds of 50,000 to 60,000 rpm. Under these circumstances, the DNA will form a band according to its density in the gradient. DNA grown in <sup>15<\/sup>N will band at a higher density position than that grown in <sup>14<\/sup>N. Meselson and Stahl noted that after one generation of growth in <sup>14<\/sup>N after they had been shifted from <sup>15<\/sup>N, the single band observed was intermediate in position in between DNA of cells grown exclusively in <sup>15<\/sup>N and <sup>14<\/sup>N. This suggested either a semi-conservative or dispersive mode of replication. The DNA harvested from cells grown for two generations in <sup>14<\/sup>N formed two bands: one DNA band was at the intermediate position between <sup>15<\/sup>N and <sup>14<\/sup>N, and the other corresponded to the band of <sup>14<\/sup>N DNA. These results could only be explained if DNA replicates in a semi-conservative manner. Therefore, the other two modes were ruled out.\r\n\r\nDuring DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the parental or \u201cold\u201d strand. When two daughter DNA copies are formed, they have the same sequence and are divided equally into the two daughter cells.\r\n<div class=\"textbox shaded\"><a href=\"http:\/\/www.johnkyrk.com\/DNAreplication.html\" target=\"_blank\">Click through\u00a0this tutorial on DNA replication.<\/a><\/div>\r\n<h2>Major Enzymes<\/h2>\r\nThe process of <strong>DNA replication<\/strong> is catalyzed by a type of enzyme called <strong>DNA polymerase<\/strong> (<em class=\"italic\">poly<\/em> meaning many, <em class=\"italic\">mer<\/em> meaning pieces, and -<em class=\"italic\">ase<\/em> meaning enzyme; so an enzyme that attaches many pieces of DNA). Observe Figure 6: the double helix of the original DNA molecule separates (blue) and new strands are made to match the separated strands. The result will be two DNA molecules, each containing an old and a new strand. Therefore, DNA replication is called semiconservative. The term <em class=\"bold\">semiconservative<\/em> refers to the fact that half of the original molecule (one of the two strands in the double helix) is \u201cconserved\u201d in the new molecule. The original strand is referred to as the <em class=\"bold\">template strand<\/em> because it provides the information, or template, for the newly synthesized strand.\r\n\r\n[caption id=\"attachment_2570\" align=\"aligncenter\" width=\"602\"]<img class=\"wp-image-2570\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172248\/DNA_replication_split_horizontal.svg_-1024x508.png\" alt=\"Stylized DNA replication fork with nucleotides matched, 5'-&gt;3' synthesis shown, no enzymes in diagram. \" width=\"602\" height=\"299\" \/> Figure 6. By Madprime(wikipedia) (<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:DNA_replication_split_horizontal.svg?uselang=en\" target=\"_blank\">DNA replication split horizontal<\/a>) CC BY-SA 2.0[\/caption]\r\n\r\n<strong>DNA replication<\/strong> relies on the double-stranded nature of the molecule. One double stranded DNA molecule, when replicated, will become two double-stranded molecules, each containing one original strand and one newly synthesized strand. You remember that the two strands of DNA run antiparallel: one from the 5\u2032 to the 3\u2032, and the other from the 3\u2032 to the 5\u2032. The synthesis of the new DNA strand can only happen in one direction: from the 5\u2032 to the 3\u2032 end. In other words, the new bases are always added to the 3\u2032 end of the newly synthesized DNA strand. So if the new nucleotide is always added to the 3\u2032 end of an existing nucleotide, where does the <em class=\"bold\">first<\/em> nucleotide come from? In fact, <strong>DNA polymerase<\/strong> needs an \u201canchor\u201d to start adding nucleotides: a short sequence of DNA or RNA that is complementary to the template strand will work to provide a free 3\u2032 end. This sequence is called a <em>primer\u00a0<\/em>(Figure 7).\r\n\r\n[caption id=\"attachment_2571\" align=\"aligncenter\" width=\"600\"]<img class=\"wp-image-2571 \" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172346\/RNA.jpg\" alt=\"Diagram of a primer moving along the template strand of DNA.\" width=\"600\" height=\"273\" \/> Figure 7. Primer and Template[\/caption]\r\n\r\nHow does <strong>DNA polymerase<\/strong> know in what order to add nucleotides? Specific base pairing in DNA is the key to copying the DNA: if you know the sequence of one strand, you can use base pairing rules to build the other strand. Bases form pairs (base pairs) in a very specific way. Figure 8\u00a0shows how <em>A<\/em> (adenine) pairs with <em>T<\/em> (thymine) and <em>G<\/em> (guanine) pairs with <em>C<\/em> (cytosine). It is important to remember that this binding is specific: <em>T<\/em> pairs with <em>A<\/em>, but not with <em>C<\/em>. The molecular recognition occurs because of the ability of bases to form specific hydrogen bonds: atoms align just right to make hydrogen bonds possible. Also note that a larger base (purine, <em>A<\/em> or <em>G<\/em>) always pairs with a smaller base (pyrimidine, <em>C<\/em> or <em>T<\/em>).\r\n\r\n[caption id=\"attachment_2572\" align=\"aligncenter\" width=\"528\"]<img class=\"wp-image-2572 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172431\/Molecular-DNA.png\" alt=\"Diagram showing the hydrogen bonds between nucleotides. Adenine is bound to thymine, and cytosine is bound to guanine. \" width=\"528\" height=\"365\" \/> Figure 8. DNA chemical structure. Modification of <a href=\"https:\/\/en.wikipedia.org\/wiki\/File:DNA_chemical_structure.svg\" target=\"_blank\">DNA chemical structure<\/a>\u00a0by Madeleine Price Ball; CC-BY-SA-2.0[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Practice Questions<\/h3>\r\nTrue\/False: DNA replication requires an enzyme.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"527189\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"527189\"]True.\u00a0Most biological reactions rely on the enzyme to speed up the reaction. In the case of DNA\u00a0replication, this enzyme is DNA polymerase.\r\n\r\n[\/hidden-answer]\r\n\r\nWhat are the building blocks on DNA?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Deoxyribonucleotides<\/li>\r\n \t<li>Fatty acids<\/li>\r\n \t<li>Ribonucleotides<\/li>\r\n \t<li>Amino acids<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"93495\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"93495\"]Answer a. DNA is a double helix made up of two long chains of deoxyribonucleotides.\r\n\r\n[\/hidden-answer]\r\n\r\nTrue\/False: DNA replication requires energy.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"62103\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"62103\"]True.\u00a0Making large molecules from small subunits (anabolism) requires energy. What supplies the\u00a0energy? The building blocks themselves serve as a source of energy. As they get incorporated into\u00a0the DNA polymer, two phosphate groups are broken off to release energy, some of which is used for\u00a0making the polymer. Deoxyribonucleotides differ from nucleotides like ATP only by one missing oxygen atom.\r\n\r\n[\/hidden-answer]\r\n\r\nWe have the building blocks, a source of energy, and a catalyst. What's missing? We need instruction about the order of nucleotides in the new polymer. Which molecule provides these instructions?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Protein<\/li>\r\n \t<li>DNA<\/li>\r\n \t<li>Carbohydrate<\/li>\r\n \t<li>Lipid<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"494506\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"494506\"]Answer b.\u00a0We refer to this DNA as a template. The original information stored in the order of bases will\u00a0direct the synthesis of the new DNA via base pairing.\r\n\r\n[\/hidden-answer]\r\n\r\nThere is one more thing required by the DNA polymerase. It cannot just start making a DNA copy of the template strand; it needs a short piece of DNA or RNA with a free hydroxyl group in the right place to attach the nucleotides to. (Remember that synthesis always occurs in one direction\u2014new building blocks are added to the 3\u2032 end.) This component starts the process by giving DNA polymerase something to bind to. What might you call this short piece of nucleic acid?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>A solvent<\/li>\r\n \t<li>A primer<\/li>\r\n \t<li>A converter<\/li>\r\n \t<li>A sealant<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"529681\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"529681\"]Answer b. A primer is used to start this process by giving DNA polymerase something to bind the new nucleotide to.[\/hidden-answer]\r\n\r\n<\/div>\r\nNow that you understand the basics of <strong>DNA replication<\/strong>, we can add a bit of complexity. The two strands of\u00a0DNA have to be temporarily separated from each other; this job is done by a special enzyme, <em>helicase<\/em>, that helps unwind and separate the DNA helices (Figure 9). Another issue is that the <strong>DNA polymerase<\/strong> only works in one direction along the strand (5\u2032 to 3\u2032), but the double-stranded DNA has two strands oriented in opposite directions. This problem is solved by synthesizing the two strands slightly differently: one new strand grows continuously, the other in bits and pieces. Short fragments of RNA are used as primers for the <strong>DNA polymerase<\/strong>.\r\n\r\n[caption id=\"attachment_2573\" align=\"aligncenter\" width=\"1024\"]<img class=\"wp-image-2573 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172842\/DNA_replication_en.svg_-1024x498.png\" alt=\"Diagram of both the leading and lagging strands the helicase splits the two strands and a DNA polymerase travels over both strands to create complementary strands.\" width=\"1024\" height=\"498\" \/> Figure 9. By Mariana Ruiz (<a href=\"https:\/\/en.wikipedia.org\/wiki\/File:DNA_replication_en.svg\" target=\"_blank\">DNA replication<\/a>) Public Domain[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Practice Questions<\/h3>\r\nWhich of these separates the two complementary strands of DNA?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>DNA polymerase<\/li>\r\n \t<li>helicase<\/li>\r\n \t<li>RNA primer<\/li>\r\n \t<li>single-strand binding protein<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"197431\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"197431\"]Answer b. Helicase breaks the hydrogen bonds holding together the two strands of DNA.\r\n\r\n[\/hidden-answer]\r\n\r\nWhich of these attaches complementary bases to the template strand?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>DNA polymerase<\/li>\r\n \t<li>helicase<\/li>\r\n \t<li>RNA primer<\/li>\r\n \t<li>single-strand binding protein<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"381621\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"381621\"]Answer a. DNA polymerase builds the new strand of DNA.\r\n\r\n[\/hidden-answer]\r\n\r\nWhich of these is later replaced with DNA bases?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>DNA polymerase<\/li>\r\n \t<li>helicase<\/li>\r\n \t<li>RNA primer<\/li>\r\n \t<li>single-strand binding protein<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"2143\"]Show Answers[\/reveal-answer]\r\n[hidden-answer a=\"2143\"]Answer c. the RNA primer is replaced with DNA nucleotides.[\/hidden-answer]\r\n\r\n<\/div>\r\nTable 1 summarizes the roles of different enzymes in DNA replication:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table 1. Important Enzymes in DNA Replication<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Enzyme<\/th>\r\n<th>Function<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>Topoisomerase<\/td>\r\n<td>Relaxes the super-coiled DNA<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>DNA helicase<\/td>\r\n<td>Unwinds the double helix at the replication fork<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Primase<\/td>\r\n<td>Provides the starting point for DNA polymerase to begin synthesis of the new strand<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>DNA polymerase<\/td>\r\n<td>Synthesizes the new DNA strand; also proofreads and corrects some errors<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>DNA ligase<\/td>\r\n<td>Re-joins the two DNA strands into a double helix and joins Okazaki fragments of the lagging strand<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>\"Proofreading\" DNA<\/h2>\r\nDNA replication is a highly accurate process, but mistakes can occasionally occur, such as a DNA polymerase inserting a wrong base. Uncorrected mistakes may sometimes lead to serious consequences, such as cancer. Repair mechanisms correct the mistakes. In rare cases, mistakes are not corrected, leading to mutations; in other cases, repair enzymes are themselves mutated or defective.\r\n\r\nMost of the mistakes during DNA replication are promptly corrected by DNA polymerase by proofreading the base that has been just added (Figure 10). In <strong>proofreading<\/strong>, the DNA pol reads the newly added base before adding the next one, so a correction can be made. The polymerase checks whether the newly added base has paired correctly with the base in the template strand. If it is the right base, the next nucleotide is added. If an incorrect base has been added, the enzyme makes a cut at the phosphodiester bond and releases the wrong nucleotide. This is performed by the exonuclease action of DNA pol III. Once the incorrect nucleotide has been removed, a new one will be added again.\r\n\r\n[caption id=\"attachment_1436\" align=\"aligncenter\" width=\"544\"]<img class=\"size-full wp-image-1436\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184707\/Figure_14_06_01.jpg\" alt=\"Illustration shows DNA polymerase replicating a strand of DNA. The enzyme has accidentally inserted G opposite A, resulting in a bulge. The enzyme backs up to fix the error.\" width=\"544\" height=\"222\" \/> Figure 10. Proofreading by DNA polymerase corrects errors during replication.[\/caption]\r\n\r\nSome errors are not corrected during replication, but are instead corrected after replication is completed; this type of repair is known as\u00a0<strong>mismatch repair <\/strong>(Figure 11). The enzymes recognize the incorrectly added nucleotide and excise it; this is then replaced by the correct base. If this remains uncorrected, it may lead to more permanent damage. How do mismatch repair enzymes recognize which of the two bases is the incorrect one? In <em>E. coli<\/em>, after replication, the nitrogenous base adenine acquires a methyl group; the parental DNA strand will have methyl groups, whereas the newly synthesized strand lacks them. Thus, DNA polymerase is able to remove the wrongly incorporated bases from the newly synthesized, non-methylated strand. In eukaryotes, the mechanism is not very well understood, but it is believed to involve recognition of unsealed nicks in the new strand, as well as a short-term continuing association of some of the replication proteins with the new daughter strand after replication has completed.\r\n\r\n[caption id=\"attachment_4553\" align=\"aligncenter\" width=\"1237\"]<img class=\"size-full wp-image-4553\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1087\/2017\/03\/23213100\/Figure_14_06_02ab.jpg\" alt=\"The top illustration shows a replicated DNA strand with G-T base mismatch. The bottom illustration shows the repaired DNA, which has the correct G-C base pairing.\" width=\"1237\" height=\"275\" \/> Figure 11. In mismatch repair, the incorrectly added base is detected after replication. The mismatch repair proteins detect this base and remove it from the newly synthesized strand by nuclease action. The gap is now filled with the correctly paired base.[\/caption]\r\n\r\nIn another type of repair mechanism,\u00a0<strong>nucleotide excision repair<\/strong>, enzymes replace incorrect bases by making a cut on both the 3' and 5' ends of the incorrect base (Figure 12).\r\n\r\n[caption id=\"attachment_4554\" align=\"aligncenter\" width=\"605\"]<img class=\"size-full wp-image-4554\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1087\/2017\/03\/23213317\/Figure_14_06_03ab.jpg\" alt=\"Illustration shows a DNA strand in which a thymine dimer has formed. Excision repair enzyme cut out the section of DNA that contains the dimer so it can be replaced with normal base pairs.\" width=\"605\" height=\"143\" \/> Figure 12. Nucleotide excision repairs thymine dimers. When exposed to UV, thymines lying adjacent to each other can form thymine dimers. In normal cells, they are excised and replaced.[\/caption]\r\n\r\nThe segment of DNA is removed and replaced with the correctly paired nucleotides by the action of DNA pol. Once the bases are filled in, the remaining gap is sealed with a phosphodiester linkage catalyzed by DNA ligase. This repair mechanism is often employed when UV exposure causes the formation of pyrimidine dimers.\r\n<h2>Telomeres<\/h2>\r\nAs you've learned, the enzyme DNA pol can add nucleotides only in the 5\u2032 to 3\u2032 direction. In the leading strand, synthesis continues until the end of the chromosome is reached. On the lagging strand, DNA is synthesized in short stretches, each of which is initiated by a separate primer. When the replication fork reaches the end of the linear chromosome, there is no place for a primer to be made for the DNA fragment to be copied at the end of the chromosome. These ends thus remain unpaired, and over time these ends may get progressively shorter as cells continue to divide.\r\n\r\nThe ends of the linear chromosomes are known as\u00a0<strong>telomeres<\/strong>, which have repetitive sequences that code for no particular gene. In a way, these telomeres protect the genes from getting deleted as cells continue to divide. In humans, a six base pair sequence, TTAGGG, is repeated 100 to 1000 times. The discovery of the enzyme telomerase (Figure 13) helped in the understanding of how chromosome ends are maintained. The <strong>telomerase<\/strong> enzyme contains a catalytic part and a built-in RNA template. It attaches to the end of the chromosome, and complementary bases to the RNA template are added on the 3\u2032 end of the DNA strand. Once the 3\u2032 end of the lagging strand template is sufficiently elongated, DNA polymerase can add the nucleotides complementary to the ends of the chromosomes. Thus, the ends of the chromosomes are replicated.\r\n\r\n[caption id=\"attachment_1432\" align=\"aligncenter\" width=\"544\"]<img class=\"size-full wp-image-1432\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184502\/Figure_14_05_01.jpg\" alt=\"Telomerase has an associated RNA that complements the 5 prime overhang at the end of the chromosome. The RNA template is used to synthesize the complementary strand. Telomerase then shifts, and the process is repeated. Next, primase and DNA polymerase synthesize the rest of the complementary strand.\" width=\"544\" height=\"694\" \/> Figure 13. The ends of linear chromosomes are maintained by the action of the telomerase enzyme.[\/caption]\r\n\r\n[caption id=\"attachment_1434\" align=\"alignright\" width=\"300\"]<img class=\"wp-image-1434\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184543\/Figure_14_05_02.jpg\" alt=\"Photo of Elizabeth Blackburn.\" width=\"300\" height=\"223\" \/> Figure 14. Elizabeth Blackburn, 2009 Nobel Laureate, is the scientist who discovered how telomerase works. (credit: US Embassy Sweden)[\/caption]\r\n\r\nTelomerase is typically active in germ cells and adult stem cells. It is not active in adult somatic cells. For her discovery of telomerase and its action, Elizabeth Blackburn (Figure 14) received the Nobel Prize for Medicine and Physiology in 2009.\r\n<h3>Telomerase and Aging<\/h3>\r\nCells that undergo cell division continue to have their telomeres shortened because most somatic cells do not make telomerase. This essentially means that telomere shortening is associated with aging. With the advent of modern medicine, preventative health care, and healthier lifestyles, the human life span has increased, and there is an increasing demand for people to look younger and have a better quality of life as they grow older.\r\n\r\nIn 2010, scientists found that telomerase can reverse some age-related conditions in mice. This may have potential in regenerative medicine.[footnote]Jaskelioff et al., \u201cTelomerase reactivation reverses tissue degeneration in aged telomerase-deficient mice,\u201d <em>Nature<\/em> 469 (2011): 102\u20137.[\/footnote]\u00a0Telomerase-deficient mice were used in these studies; these mice have tissue atrophy, stem cell depletion, organ system failure, and impaired tissue injury responses. Telomerase reactivation in these mice caused extension of telomeres, reduced DNA damage, reversed neurodegeneration, and improved the function of the testes, spleen, and intestines. Thus, telomere reactivation may have potential for treating age-related diseases in humans.\r\n\r\nCancer is characterized by uncontrolled cell division of abnormal cells. The cells accumulate mutations, proliferate uncontrollably, and can migrate to different parts of the body through a process called metastasis. Scientists have observed that cancerous cells have considerably shortened telomeres and that telomerase is active in these cells. Interestingly, only after the telomeres were shortened in the cancer cells did the telomerase become active. If the action of telomerase in these cells can be inhibited by drugs during cancer therapy, then the cancerous cells could potentially be stopped from further division.\r\n<div class=\"textbox learning-objectives\">\r\n<h3>In Summary: DNA Base Pairs and Replication<\/h3>\r\nDNA was first isolated from white blood cells by Friedrich Miescher, who called it nuclein because it was isolated from nuclei. Frederick Griffith's experiments with strains of <em data-effect=\"italics\">Streptococcus pneumoniae<\/em> provided the first hint that DNA may be the transforming principle. Avery, MacLeod, and McCarty proved that DNA is required for the transformation of bacteria. Later experiments by Hershey and Chase using bacteriophage T2 proved that DNA is the genetic material. Chargaff found that the ratio of A = T and C = G, and that the percentage content of A, T, G, and C is different for different species.\r\n\r\nThe model for DNA replication suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. In conservative replication, the parental DNA is conserved, and the daughter DNA is newly synthesized. The semi-conservative method suggests that each of the two parental DNA strands acts as template for new DNA to be synthesized; after replication, each double-stranded DNA includes one parental or \u201cold\u201d strand and one \u201cnew\u201d strand. The dispersive mode suggested that the two copies of the DNA would have segments of parental DNA and newly synthesized DNA. Experimental evidence showed DNA replication is semi-conservative.\r\n\r\nReplication in eukaryotes starts at multiple origins of replication. A primer is required to initiate synthesis, which is then extended by DNA polymerase as it adds nucleotides one by one to the growing chain. The leading strand is synthesized continuously, whereas the lagging strand is synthesized in short stretches called Okazaki fragments. The RNA primers are replaced with DNA nucleotides; the DNA remains one continuous strand by linking the DNA fragments with DNA ligase. Below is a summary table of the major enzymes addressed in this reading, listed in rough order of activity during replication.\r\n\r\nThe ends of the chromosomes pose a problem during DNA replication as polymerase is unable to extend them without a primer. Telomerase, an enzyme with a built-in RNA template, extends the ends by copying the RNA template and extending one end of the chromosome. DNA polymerase can then extend the DNA using the primer. In this way, the ends of the chromosomes are protected. This is important as evidence indicates telomere length may play a role in regulating cell division and the process of aging.\r\n\r\n<\/div>\r\n<h2><strong>Check Your Understanding<\/strong><\/h2>\r\nAnswer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does\u00a0<strong>not<\/strong>\u00a0count toward your grade in the class, and you can retake it an unlimited number of times.\r\n\r\nUse this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section.\r\n\r\nhttps:\/\/assessments.lumenlearning.com\/assessments\/3358","rendered":"<h2>Explain the role of complementary base pairing in the precise replication process of DNA<\/h2>\n<p>In this outcome, we&#8217;ll learn more about the precise structure of DNA and how it replicates.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Understand the historical basis of our understanding of DNA<\/li>\n<li>Outline the basic steps in DNA replication<\/li>\n<li>Identify the major enzymes that play a role in DNA replication<\/li>\n<li>Identify the key proofreading processes in DNA replication<\/li>\n<li>Understand the basic role of telomeres in protecting DNA from replication errors<\/li>\n<\/ul>\n<\/div>\n<h2>The History of DNA<\/h2>\n<div id=\"attachment_2594\" style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2594\" class=\"wp-image-2594\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/06150802\/Figure_14_01_01.jpg\" alt=\"Portrait of Friedrich Miescher\" width=\"300\" height=\"413\" \/><\/p>\n<p id=\"caption-attachment-2594\" class=\"wp-caption-text\">Figure 1. Friedrich Miescher (1844\u20131895) discovered nucleic acids.<\/p>\n<\/div>\n<p>Modern understandings of DNA have evolved from the discovery of nucleic acid to the development of the double-helix model. In the 1860s, Friedrich Miescher (Figure 1), a physician by profession, was the first person to isolate phosphate-rich chemicals from white blood cells or leukocytes. He named these chemicals (which would eventually be known as RNA and DNA) nuclein because they were isolated from the nuclei of the cells.<\/p>\n<p>To see Miescher conduct an experiment step-by-step, click through this <a href=\"https:\/\/www.dnalc.org\/view\/16342-Animation-15-DNA-and-proteins-are-key-molecules-of-the-cell-nucleus-.html\" target=\"_blank\">review of how he discovered the key role of DNA and proteins in the nucleus<\/a>.<\/p>\n<p>A half century later, British bacteriologist Frederick Griffith was perhaps the first person to show that hereditary information could be transferred from one cell to another \u201chorizontally,\u201d rather than by descent. In 1928, he reported the first demonstration of bacterial <strong>transformation<\/strong>, a process in which external DNA is taken up by a cell, thereby changing morphology and physiology. He was working with <em>Streptococcus pneumoniae,<\/em> the bacterium that causes pneumonia. Griffith worked with two strains, rough (R) and smooth (S). The R strain is non-pathogenic (does not cause disease) and is called rough because its outer surface is a cell wall and lacks a capsule; as a result, the cell surface appears uneven under the microscope. The S strain is pathogenic (disease-causing) and has a capsule outside its cell wall. As a result, it has a smooth appearance under the microscope. Griffith injected the live R strain into mice and they survived. In another experiment, when he injected mice with the heat-killed S strain, they also survived. In a third set of experiments, a mixture of live R strain and heat-killed S strain were injected into mice, and\u2014to his surprise\u2014the mice died. Upon isolating the live bacteria from the dead mouse, only the S strain of bacteria was recovered. When this isolated S strain was injected into fresh mice, the mice died. Griffith concluded that something had passed from the heat-killed S strain into the live R strain and transformed it into the pathogenic S strain, and he called this the transforming principle (Figure 2). These experiments are now famously known as Griffith&#8217;s transformation experiments.<\/p>\n<div id=\"attachment_2595\" style=\"width: 810px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2595\" class=\"size-full wp-image-2595\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/06150941\/Figure_14_01_02.jpg\" alt=\"On the left is a photo of a live mouse, representing a mouse injected with heat-killed, virulent S strain. On the right is a photo of a dead mouse, representing a mouse injected with heat-killed, virulent S strain and live, non-virulent R strain.\" width=\"800\" height=\"317\" \/><\/p>\n<p id=\"caption-attachment-2595\" class=\"wp-caption-text\">Figure 2. Two strains of <em>S. pneumoniae<\/em> were used in Griffith\u2019s transformation experiments. The R strain is non-pathogenic. The S strain is pathogenic and causes death. When Griffith injected a mouse with the heat-killed S strain and a live R strain, the mouse died. The S strain was recovered from the dead mouse. Thus, Griffith concluded that something had passed from the heat-killed S strain to the R strain, transforming the R strain into S strain in the process. (credit &#8220;living mouse&#8221;: modification of work by NIH; credit &#8220;dead mouse&#8221;: modification of work by Sarah Marriage)<\/p>\n<\/div>\n<p>Scientists Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) were interested in exploring this transforming principle further. They isolated the S strain from the dead mice and isolated the proteins and nucleic acids, namely RNA and DNA, as these were possible candidates for the molecule of heredity. They conducted a systematic elimination study. They used enzymes that specifically degraded each component and then used each mixture separately to transform the R strain. They found that when DNA was degraded, the resulting mixture was no longer able to transform the bacteria, whereas all of the other combinations were able to transform the bacteria. This led them to conclude that DNA was the transforming principle.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Forensic Scientists and DNA Analysis<\/h3>\n<p>DNA evidence was used for the first time to solve an immigration case. The story started with a teenage boy returning to London from Ghana to be with his mother. Immigration authorities at the airport were suspicious of him, thinking that he was traveling on a forged passport. After much persuasion, he was allowed to go live with his mother, but the immigration authorities did not drop the case against him. All types of evidence, including photographs, were provided to the authorities, but deportation proceedings were started nevertheless. Around the same time, Dr. Alec Jeffreys of Leicester University in the United Kingdom had invented a technique known as DNA fingerprinting. The immigration authorities approached Dr. Jeffreys for help. He took DNA samples from the mother and three of her children, plus an unrelated mother, and compared the samples with the boy\u2019s DNA. Because the biological father was not in the picture, DNA from the three children was compared with the boy\u2019s DNA. He found a match in the boy\u2019s DNA for both the mother and his three siblings. He concluded that the boy was indeed the mother\u2019s son.<\/p>\n<p>Forensic scientists analyze many items, including documents, handwriting, firearms, and biological samples. They analyze the DNA content of hair, semen, saliva, and blood, and compare it with a database of DNA profiles of known criminals. Analysis includes DNA isolation, sequencing, and sequence analysis; most forensic DNA analysis involves polymerase chain reaction (PCR) amplification of short tandem repeat (STR) loci and electrophoresis to determine the length of the PCR-amplified fragment. Only mitochondrial DNA is sequenced for forensics. Forensic scientists are expected to appear at court hearings to present their findings. They are usually employed in crime labs of city and state government agencies. Geneticists experimenting with DNA techniques also work for scientific and research organizations, pharmaceutical industries, and college and university labs. Students wishing to pursue a career as a forensic scientist should have at least a bachelor&#8217;s degree in chemistry, biology, or physics, and preferably some experience working in a laboratory.<\/p>\n<\/div>\n<p>Experiments conducted by Martha Chase and Alfred Hershey in 1952 provided confirmatory evidence that DNA was the genetic material and not proteins. Chase and Hershey were studying a bacteriophage, which is a virus that infects bacteria. Viruses typically have a simple structure: a protein coat, called the capsid, and a nucleic acid core that contains the genetic material, either DNA or RNA. The bacteriophage infects the host bacterial cell by attaching to its surface, and then it injects its nucleic acids inside the cell. The phage DNA makes multiple copies of itself using the host machinery, and eventually the host cell bursts, releasing a large number of bacteriophages. Hershey and Chase labeled one batch of phage with radioactive sulfur, <sup>35<\/sup>S, to label the protein coat. Another batch of phage were labeled with radioactive phosphorus, <sup>32<\/sup>P. Because phosphorous is found in DNA, but not protein, the DNA and not the protein would be tagged with radioactive phosphorus.<\/p>\n<p>Each batch of phage was allowed to infect the cells separately. After infection, the phage bacterial suspension was put in a blender, which caused the phage coat to be detached from the host cell. The phage and bacterial suspension was spun down in a centrifuge. The heavier bacterial cells settled down and formed a pellet, whereas the lighter phage particles stayed in the supernatant. In the tube that contained phage labeled with <sup>35<\/sup>S, the supernatant contained the radioactively labeled phage, whereas no radioactivity was detected in the pellet. In the tube that contained the phage labeled with <sup>32<\/sup>P, the radioactivity was detected in the pellet that contained the heavier bacterial cells, and no radioactivity was detected in the supernatant. Hershey and Chase concluded that it was the phage DNA that was injected into the cell and carried information to produce more phage particles, thus providing evidence that DNA was the genetic material and not proteins (Figure 3).<\/p>\n<div id=\"attachment_2596\" style=\"width: 810px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2596\" class=\"size-full wp-image-2596\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/06151133\/Figure_14_01_03.jpg\" alt=\"Illustration shows bacteria being infected by phage labeled with ^{35}S, which is incorporated into the protein coat, or ^{32}P, which is incorporated into the DNA. Infected bacteria were separated from phage by centrifugation and cultured. The bacteria that had been infected with phage containing ^{32}P-labeled DNA made radioactive phage. The bacteria that had been infected with ^{35}S-labeled phage produced unlabeled phage. The results support the hypothesis that DNA, and not protein, is the genetic material.\" width=\"800\" height=\"638\" \/><\/p>\n<p id=\"caption-attachment-2596\" class=\"wp-caption-text\">Figure 3. In Hershey and Chase&#8217;s experiments, bacteria were infected with phage radiolabeled with either <sup>35<\/sup>S, which labels protein, or <sup>32<\/sup>P, which labels DNA. Only <sup>32<\/sup>P entered the bacterial cells, indicating that DNA is the genetic material.<\/p>\n<\/div>\n<p id=\"fs-id1430603\">Around this same time, Austrian biochemist Erwin Chargaff examined the content of DNA in different species and found that the amounts of adenine, thymine, guanine, and cytosine were not found in equal quantities, and that it varied from species to species, but not between individuals of the same species. He found that the amount of adenine equals the amount of thymine, and the amount of cytosine equals the amount of guanine, or A = T and G = C. This is also known as Chargaff\u2019s rules. This finding proved immensely useful when Watson and Crick were getting ready to propose their DNA double helix model.<\/p>\n<div class=\"textbox exercises\">\n<h3>Practice Question<\/h3>\n<p>The experiments by Hershey and Chase helped confirm that DNA was the hereditary material on the basis of the finding that:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>radioactive phage were found in the pellet<\/li>\n<li>radioactive cells were found in the supernatant<\/li>\n<li>radioactive sulfur was found inside the cell<\/li>\n<li>radioactive phosphorus was found in the cell<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q549343\">Show Answer<\/span><\/p>\n<div id=\"q549343\" class=\"hidden-answer\" style=\"display: none\">d.\u00a0radioactive phosphorus was found in the cell<\/div>\n<\/div>\n<\/div>\n<h2>Basics of DNA Replication<\/h2>\n<div id=\"attachment_4411\" style=\"width: 410px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4411\" class=\"wp-image-4411\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1087\/2017\/02\/02000938\/Figure_14_03_01a.jpg\" alt=\"Illustration shows the conservative, semi-conservative, and dispersive models of DNA synthesis. In the conservative model, when DNA is replicated and both newly synthesized strands are paired together. In the semi-conservative model, each newly synthesized strand pairs with a parent strand. In the dispersive model, newly synthesized DNA is interspersed with parent DNA within both DNA strands.\" width=\"400\" height=\"404\" \/><\/p>\n<p id=\"caption-attachment-4411\" class=\"wp-caption-text\">Figure 4. The three suggested models of DNA replication. Grey indicates the original DNA strands, and blue indicates newly synthesized DNA.<\/p>\n<\/div>\n<p>The elucidation of the structure of the double helix provided a hint as to how DNA divides and makes copies of itself. This model suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. What was not clear was how the replication took place. There were three models suggested: conservative, semi-conservative, and dispersive (see Figure 4).<\/p>\n<p>In conservative replication, the parental DNA remains together, and the newly formed daughter strands are together. The semi-conservative method suggests that each of the two parental DNA strands act as a template for new DNA to be synthesized; after replication, each double-stranded DNA includes one parental or \u201cold\u201d strand and one \u201cnew\u201d strand. In the dispersive model, both copies of DNA have double-stranded segments of parental DNA and newly synthesized DNA interspersed.<\/p>\n<p>Meselson and Stahl were interested in understanding how DNA replicates. They grew\u00a0<em>E. coli<\/em> for several generations in a medium containing a \u201cheavy\u201d isotope of nitrogen (<sup>15<\/sup>N) that gets incorporated into nitrogenous bases, and eventually into the DNA (Figure 5).<\/p>\n<div id=\"attachment_1430\" style=\"width: 810px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1430\" class=\"size-full wp-image-1430\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184216\/Figure_14_03_02.jpg\" alt=\"Illustration shows an experiment in which E. coli was grown initially in media containing ^{15}N nucleotides. When the DNA was extracted and run in an ultracentrifuge, a band of DNA appeared low in the tube. The culture was next placed in ^{14}N medium. After one generation, all of the DNA appeared in the middle of the tube, indicating that the DNA was a mixture of half ^{14}N and half ^{15}N DNA. After two generations, half of the DNA appeared in the middle of the tube, and half appeared higher up, indicating that half the DNA contained 50% ^{15}N, and half contained ^{14}N only. In subsequent generations, more and more of the DNA appeared in the upper, ^{14}N band.\" width=\"800\" height=\"698\" \/><\/p>\n<p id=\"caption-attachment-1430\" class=\"wp-caption-text\">Figure 5. Meselson and Stahl experimented with E. coli grown first in heavy nitrogen (<sup>15<\/sup>N) then in <sup>14<\/sup>N. DNA grown in <sup>15<\/sup>N (red band) is heavier than DNA grown in <sup>14<\/sup>N (orange band), and sediments to a lower level in cesium chloride solution in an ultracentrifuge. When DNA grown in <sup>15<\/sup>N is switched to media containing <sup>14<\/sup>N, after one round of cell division the DNA sediments halfway between the <sup>15<\/sup>N and <sup>14<\/sup>N levels, indicating that it now contains fifty percent <sup>14<\/sup>N. In subsequent cell divisions, an increasing amount of DNA contains <sup>14<\/sup>N only. This data supports the semi-conservative replication model. (credit: modification of work by Mariana Ruiz Villareal)<\/p>\n<\/div>\n<p>The\u00a0<em>E. coli<\/em> culture was then shifted into medium containing <sup>14<\/sup>N and allowed to grow for one generation. The cells were harvested and the DNA was isolated. The DNA was centrifuged at high speeds in an ultracentrifuge. Some cells were allowed to grow for one more life cycle in <sup>14<\/sup>N and spun again. During the density gradient centrifugation, the DNA is loaded into a gradient (typically a salt such as cesium chloride or sucrose) and spun at high speeds of 50,000 to 60,000 rpm. Under these circumstances, the DNA will form a band according to its density in the gradient. DNA grown in <sup>15<\/sup>N will band at a higher density position than that grown in <sup>14<\/sup>N. Meselson and Stahl noted that after one generation of growth in <sup>14<\/sup>N after they had been shifted from <sup>15<\/sup>N, the single band observed was intermediate in position in between DNA of cells grown exclusively in <sup>15<\/sup>N and <sup>14<\/sup>N. This suggested either a semi-conservative or dispersive mode of replication. The DNA harvested from cells grown for two generations in <sup>14<\/sup>N formed two bands: one DNA band was at the intermediate position between <sup>15<\/sup>N and <sup>14<\/sup>N, and the other corresponded to the band of <sup>14<\/sup>N DNA. These results could only be explained if DNA replicates in a semi-conservative manner. Therefore, the other two modes were ruled out.<\/p>\n<p>During DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the parental or \u201cold\u201d strand. When two daughter DNA copies are formed, they have the same sequence and are divided equally into the two daughter cells.<\/p>\n<div class=\"textbox shaded\"><a href=\"http:\/\/www.johnkyrk.com\/DNAreplication.html\" target=\"_blank\">Click through\u00a0this tutorial on DNA replication.<\/a><\/div>\n<h2>Major Enzymes<\/h2>\n<p>The process of <strong>DNA replication<\/strong> is catalyzed by a type of enzyme called <strong>DNA polymerase<\/strong> (<em class=\"italic\">poly<\/em> meaning many, <em class=\"italic\">mer<\/em> meaning pieces, and &#8211;<em class=\"italic\">ase<\/em> meaning enzyme; so an enzyme that attaches many pieces of DNA). Observe Figure 6: the double helix of the original DNA molecule separates (blue) and new strands are made to match the separated strands. The result will be two DNA molecules, each containing an old and a new strand. Therefore, DNA replication is called semiconservative. The term <em class=\"bold\">semiconservative<\/em> refers to the fact that half of the original molecule (one of the two strands in the double helix) is \u201cconserved\u201d in the new molecule. The original strand is referred to as the <em class=\"bold\">template strand<\/em> because it provides the information, or template, for the newly synthesized strand.<\/p>\n<div id=\"attachment_2570\" style=\"width: 612px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2570\" class=\"wp-image-2570\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172248\/DNA_replication_split_horizontal.svg_-1024x508.png\" alt=\"Stylized DNA replication fork with nucleotides matched, 5'-&gt;3' synthesis shown, no enzymes in diagram.\" width=\"602\" height=\"299\" \/><\/p>\n<p id=\"caption-attachment-2570\" class=\"wp-caption-text\">Figure 6. By Madprime(wikipedia) (<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:DNA_replication_split_horizontal.svg?uselang=en\" target=\"_blank\">DNA replication split horizontal<\/a>) CC BY-SA 2.0<\/p>\n<\/div>\n<p><strong>DNA replication<\/strong> relies on the double-stranded nature of the molecule. One double stranded DNA molecule, when replicated, will become two double-stranded molecules, each containing one original strand and one newly synthesized strand. You remember that the two strands of DNA run antiparallel: one from the 5\u2032 to the 3\u2032, and the other from the 3\u2032 to the 5\u2032. The synthesis of the new DNA strand can only happen in one direction: from the 5\u2032 to the 3\u2032 end. In other words, the new bases are always added to the 3\u2032 end of the newly synthesized DNA strand. So if the new nucleotide is always added to the 3\u2032 end of an existing nucleotide, where does the <em class=\"bold\">first<\/em> nucleotide come from? In fact, <strong>DNA polymerase<\/strong> needs an \u201canchor\u201d to start adding nucleotides: a short sequence of DNA or RNA that is complementary to the template strand will work to provide a free 3\u2032 end. This sequence is called a <em>primer\u00a0<\/em>(Figure 7).<\/p>\n<div id=\"attachment_2571\" style=\"width: 610px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2571\" class=\"wp-image-2571\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172346\/RNA.jpg\" alt=\"Diagram of a primer moving along the template strand of DNA.\" width=\"600\" height=\"273\" \/><\/p>\n<p id=\"caption-attachment-2571\" class=\"wp-caption-text\">Figure 7. Primer and Template<\/p>\n<\/div>\n<p>How does <strong>DNA polymerase<\/strong> know in what order to add nucleotides? Specific base pairing in DNA is the key to copying the DNA: if you know the sequence of one strand, you can use base pairing rules to build the other strand. Bases form pairs (base pairs) in a very specific way. Figure 8\u00a0shows how <em>A<\/em> (adenine) pairs with <em>T<\/em> (thymine) and <em>G<\/em> (guanine) pairs with <em>C<\/em> (cytosine). It is important to remember that this binding is specific: <em>T<\/em> pairs with <em>A<\/em>, but not with <em>C<\/em>. The molecular recognition occurs because of the ability of bases to form specific hydrogen bonds: atoms align just right to make hydrogen bonds possible. Also note that a larger base (purine, <em>A<\/em> or <em>G<\/em>) always pairs with a smaller base (pyrimidine, <em>C<\/em> or <em>T<\/em>).<\/p>\n<div id=\"attachment_2572\" style=\"width: 538px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2572\" class=\"wp-image-2572 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172431\/Molecular-DNA.png\" alt=\"Diagram showing the hydrogen bonds between nucleotides. Adenine is bound to thymine, and cytosine is bound to guanine.\" width=\"528\" height=\"365\" \/><\/p>\n<p id=\"caption-attachment-2572\" class=\"wp-caption-text\">Figure 8. DNA chemical structure. Modification of <a href=\"https:\/\/en.wikipedia.org\/wiki\/File:DNA_chemical_structure.svg\" target=\"_blank\">DNA chemical structure<\/a>\u00a0by Madeleine Price Ball; CC-BY-SA-2.0<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice Questions<\/h3>\n<p>True\/False: DNA replication requires an enzyme.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q527189\">Show Answer<\/span><\/p>\n<div id=\"q527189\" class=\"hidden-answer\" style=\"display: none\">True.\u00a0Most biological reactions rely on the enzyme to speed up the reaction. In the case of DNA\u00a0replication, this enzyme is DNA polymerase.<\/p>\n<\/div>\n<\/div>\n<p>What are the building blocks on DNA?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Deoxyribonucleotides<\/li>\n<li>Fatty acids<\/li>\n<li>Ribonucleotides<\/li>\n<li>Amino acids<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q93495\">Show Answer<\/span><\/p>\n<div id=\"q93495\" class=\"hidden-answer\" style=\"display: none\">Answer a. DNA is a double helix made up of two long chains of deoxyribonucleotides.<\/p>\n<\/div>\n<\/div>\n<p>True\/False: DNA replication requires energy.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q62103\">Show Answer<\/span><\/p>\n<div id=\"q62103\" class=\"hidden-answer\" style=\"display: none\">True.\u00a0Making large molecules from small subunits (anabolism) requires energy. What supplies the\u00a0energy? The building blocks themselves serve as a source of energy. As they get incorporated into\u00a0the DNA polymer, two phosphate groups are broken off to release energy, some of which is used for\u00a0making the polymer. Deoxyribonucleotides differ from nucleotides like ATP only by one missing oxygen atom.<\/p>\n<\/div>\n<\/div>\n<p>We have the building blocks, a source of energy, and a catalyst. What&#8217;s missing? We need instruction about the order of nucleotides in the new polymer. Which molecule provides these instructions?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Protein<\/li>\n<li>DNA<\/li>\n<li>Carbohydrate<\/li>\n<li>Lipid<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q494506\">Show Answer<\/span><\/p>\n<div id=\"q494506\" class=\"hidden-answer\" style=\"display: none\">Answer b.\u00a0We refer to this DNA as a template. The original information stored in the order of bases will\u00a0direct the synthesis of the new DNA via base pairing.<\/p>\n<\/div>\n<\/div>\n<p>There is one more thing required by the DNA polymerase. It cannot just start making a DNA copy of the template strand; it needs a short piece of DNA or RNA with a free hydroxyl group in the right place to attach the nucleotides to. (Remember that synthesis always occurs in one direction\u2014new building blocks are added to the 3\u2032 end.) This component starts the process by giving DNA polymerase something to bind to. What might you call this short piece of nucleic acid?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>A solvent<\/li>\n<li>A primer<\/li>\n<li>A converter<\/li>\n<li>A sealant<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q529681\">Show Answer<\/span><\/p>\n<div id=\"q529681\" class=\"hidden-answer\" style=\"display: none\">Answer b. A primer is used to start this process by giving DNA polymerase something to bind the new nucleotide to.<\/div>\n<\/div>\n<\/div>\n<p>Now that you understand the basics of <strong>DNA replication<\/strong>, we can add a bit of complexity. The two strands of\u00a0DNA have to be temporarily separated from each other; this job is done by a special enzyme, <em>helicase<\/em>, that helps unwind and separate the DNA helices (Figure 9). Another issue is that the <strong>DNA polymerase<\/strong> only works in one direction along the strand (5\u2032 to 3\u2032), but the double-stranded DNA has two strands oriented in opposite directions. This problem is solved by synthesizing the two strands slightly differently: one new strand grows continuously, the other in bits and pieces. Short fragments of RNA are used as primers for the <strong>DNA polymerase<\/strong>.<\/p>\n<div id=\"attachment_2573\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2573\" class=\"wp-image-2573 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/06\/02172842\/DNA_replication_en.svg_-1024x498.png\" alt=\"Diagram of both the leading and lagging strands the helicase splits the two strands and a DNA polymerase travels over both strands to create complementary strands.\" width=\"1024\" height=\"498\" \/><\/p>\n<p id=\"caption-attachment-2573\" class=\"wp-caption-text\">Figure 9. By Mariana Ruiz (<a href=\"https:\/\/en.wikipedia.org\/wiki\/File:DNA_replication_en.svg\" target=\"_blank\">DNA replication<\/a>) Public Domain<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice Questions<\/h3>\n<p>Which of these separates the two complementary strands of DNA?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>DNA polymerase<\/li>\n<li>helicase<\/li>\n<li>RNA primer<\/li>\n<li>single-strand binding protein<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q197431\">Show Answer<\/span><\/p>\n<div id=\"q197431\" class=\"hidden-answer\" style=\"display: none\">Answer b. Helicase breaks the hydrogen bonds holding together the two strands of DNA.<\/p>\n<\/div>\n<\/div>\n<p>Which of these attaches complementary bases to the template strand?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>DNA polymerase<\/li>\n<li>helicase<\/li>\n<li>RNA primer<\/li>\n<li>single-strand binding protein<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q381621\">Show Answer<\/span><\/p>\n<div id=\"q381621\" class=\"hidden-answer\" style=\"display: none\">Answer a. DNA polymerase builds the new strand of DNA.<\/p>\n<\/div>\n<\/div>\n<p>Which of these is later replaced with DNA bases?<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>DNA polymerase<\/li>\n<li>helicase<\/li>\n<li>RNA primer<\/li>\n<li>single-strand binding protein<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2143\">Show Answers<\/span><\/p>\n<div id=\"q2143\" class=\"hidden-answer\" style=\"display: none\">Answer c. the RNA primer is replaced with DNA nucleotides.<\/div>\n<\/div>\n<\/div>\n<p>Table 1 summarizes the roles of different enzymes in DNA replication:<\/p>\n<table>\n<thead>\n<tr>\n<th colspan=\"2\">Table 1. Important Enzymes in DNA Replication<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Enzyme<\/th>\n<th>Function<\/th>\n<\/tr>\n<tr>\n<td>Topoisomerase<\/td>\n<td>Relaxes the super-coiled DNA<\/td>\n<\/tr>\n<tr>\n<td>DNA helicase<\/td>\n<td>Unwinds the double helix at the replication fork<\/td>\n<\/tr>\n<tr>\n<td>Primase<\/td>\n<td>Provides the starting point for DNA polymerase to begin synthesis of the new strand<\/td>\n<\/tr>\n<tr>\n<td>DNA polymerase<\/td>\n<td>Synthesizes the new DNA strand; also proofreads and corrects some errors<\/td>\n<\/tr>\n<tr>\n<td>DNA ligase<\/td>\n<td>Re-joins the two DNA strands into a double helix and joins Okazaki fragments of the lagging strand<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>&#8220;Proofreading&#8221; DNA<\/h2>\n<p>DNA replication is a highly accurate process, but mistakes can occasionally occur, such as a DNA polymerase inserting a wrong base. Uncorrected mistakes may sometimes lead to serious consequences, such as cancer. Repair mechanisms correct the mistakes. In rare cases, mistakes are not corrected, leading to mutations; in other cases, repair enzymes are themselves mutated or defective.<\/p>\n<p>Most of the mistakes during DNA replication are promptly corrected by DNA polymerase by proofreading the base that has been just added (Figure 10). In <strong>proofreading<\/strong>, the DNA pol reads the newly added base before adding the next one, so a correction can be made. The polymerase checks whether the newly added base has paired correctly with the base in the template strand. If it is the right base, the next nucleotide is added. If an incorrect base has been added, the enzyme makes a cut at the phosphodiester bond and releases the wrong nucleotide. This is performed by the exonuclease action of DNA pol III. Once the incorrect nucleotide has been removed, a new one will be added again.<\/p>\n<div id=\"attachment_1436\" style=\"width: 554px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1436\" class=\"size-full wp-image-1436\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184707\/Figure_14_06_01.jpg\" alt=\"Illustration shows DNA polymerase replicating a strand of DNA. The enzyme has accidentally inserted G opposite A, resulting in a bulge. The enzyme backs up to fix the error.\" width=\"544\" height=\"222\" \/><\/p>\n<p id=\"caption-attachment-1436\" class=\"wp-caption-text\">Figure 10. Proofreading by DNA polymerase corrects errors during replication.<\/p>\n<\/div>\n<p>Some errors are not corrected during replication, but are instead corrected after replication is completed; this type of repair is known as\u00a0<strong>mismatch repair <\/strong>(Figure 11). The enzymes recognize the incorrectly added nucleotide and excise it; this is then replaced by the correct base. If this remains uncorrected, it may lead to more permanent damage. How do mismatch repair enzymes recognize which of the two bases is the incorrect one? In <em>E. coli<\/em>, after replication, the nitrogenous base adenine acquires a methyl group; the parental DNA strand will have methyl groups, whereas the newly synthesized strand lacks them. Thus, DNA polymerase is able to remove the wrongly incorporated bases from the newly synthesized, non-methylated strand. In eukaryotes, the mechanism is not very well understood, but it is believed to involve recognition of unsealed nicks in the new strand, as well as a short-term continuing association of some of the replication proteins with the new daughter strand after replication has completed.<\/p>\n<div id=\"attachment_4553\" style=\"width: 1247px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4553\" class=\"size-full wp-image-4553\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1087\/2017\/03\/23213100\/Figure_14_06_02ab.jpg\" alt=\"The top illustration shows a replicated DNA strand with G-T base mismatch. The bottom illustration shows the repaired DNA, which has the correct G-C base pairing.\" width=\"1237\" height=\"275\" \/><\/p>\n<p id=\"caption-attachment-4553\" class=\"wp-caption-text\">Figure 11. In mismatch repair, the incorrectly added base is detected after replication. The mismatch repair proteins detect this base and remove it from the newly synthesized strand by nuclease action. The gap is now filled with the correctly paired base.<\/p>\n<\/div>\n<p>In another type of repair mechanism,\u00a0<strong>nucleotide excision repair<\/strong>, enzymes replace incorrect bases by making a cut on both the 3&#8242; and 5&#8242; ends of the incorrect base (Figure 12).<\/p>\n<div id=\"attachment_4554\" style=\"width: 615px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4554\" class=\"size-full wp-image-4554\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1087\/2017\/03\/23213317\/Figure_14_06_03ab.jpg\" alt=\"Illustration shows a DNA strand in which a thymine dimer has formed. Excision repair enzyme cut out the section of DNA that contains the dimer so it can be replaced with normal base pairs.\" width=\"605\" height=\"143\" \/><\/p>\n<p id=\"caption-attachment-4554\" class=\"wp-caption-text\">Figure 12. Nucleotide excision repairs thymine dimers. When exposed to UV, thymines lying adjacent to each other can form thymine dimers. In normal cells, they are excised and replaced.<\/p>\n<\/div>\n<p>The segment of DNA is removed and replaced with the correctly paired nucleotides by the action of DNA pol. Once the bases are filled in, the remaining gap is sealed with a phosphodiester linkage catalyzed by DNA ligase. This repair mechanism is often employed when UV exposure causes the formation of pyrimidine dimers.<\/p>\n<h2>Telomeres<\/h2>\n<p>As you&#8217;ve learned, the enzyme DNA pol can add nucleotides only in the 5\u2032 to 3\u2032 direction. In the leading strand, synthesis continues until the end of the chromosome is reached. On the lagging strand, DNA is synthesized in short stretches, each of which is initiated by a separate primer. When the replication fork reaches the end of the linear chromosome, there is no place for a primer to be made for the DNA fragment to be copied at the end of the chromosome. These ends thus remain unpaired, and over time these ends may get progressively shorter as cells continue to divide.<\/p>\n<p>The ends of the linear chromosomes are known as\u00a0<strong>telomeres<\/strong>, which have repetitive sequences that code for no particular gene. In a way, these telomeres protect the genes from getting deleted as cells continue to divide. In humans, a six base pair sequence, TTAGGG, is repeated 100 to 1000 times. The discovery of the enzyme telomerase (Figure 13) helped in the understanding of how chromosome ends are maintained. The <strong>telomerase<\/strong> enzyme contains a catalytic part and a built-in RNA template. It attaches to the end of the chromosome, and complementary bases to the RNA template are added on the 3\u2032 end of the DNA strand. Once the 3\u2032 end of the lagging strand template is sufficiently elongated, DNA polymerase can add the nucleotides complementary to the ends of the chromosomes. Thus, the ends of the chromosomes are replicated.<\/p>\n<div id=\"attachment_1432\" style=\"width: 554px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1432\" class=\"size-full wp-image-1432\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184502\/Figure_14_05_01.jpg\" alt=\"Telomerase has an associated RNA that complements the 5 prime overhang at the end of the chromosome. The RNA template is used to synthesize the complementary strand. Telomerase then shifts, and the process is repeated. Next, primase and DNA polymerase synthesize the rest of the complementary strand.\" width=\"544\" height=\"694\" \/><\/p>\n<p id=\"caption-attachment-1432\" class=\"wp-caption-text\">Figure 13. The ends of linear chromosomes are maintained by the action of the telomerase enzyme.<\/p>\n<\/div>\n<div id=\"attachment_1434\" style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1434\" class=\"wp-image-1434\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/110\/2016\/05\/02184543\/Figure_14_05_02.jpg\" alt=\"Photo of Elizabeth Blackburn.\" width=\"300\" height=\"223\" \/><\/p>\n<p id=\"caption-attachment-1434\" class=\"wp-caption-text\">Figure 14. Elizabeth Blackburn, 2009 Nobel Laureate, is the scientist who discovered how telomerase works. (credit: US Embassy Sweden)<\/p>\n<\/div>\n<p>Telomerase is typically active in germ cells and adult stem cells. It is not active in adult somatic cells. For her discovery of telomerase and its action, Elizabeth Blackburn (Figure 14) received the Nobel Prize for Medicine and Physiology in 2009.<\/p>\n<h3>Telomerase and Aging<\/h3>\n<p>Cells that undergo cell division continue to have their telomeres shortened because most somatic cells do not make telomerase. This essentially means that telomere shortening is associated with aging. With the advent of modern medicine, preventative health care, and healthier lifestyles, the human life span has increased, and there is an increasing demand for people to look younger and have a better quality of life as they grow older.<\/p>\n<p>In 2010, scientists found that telomerase can reverse some age-related conditions in mice. This may have potential in regenerative medicine.<a class=\"footnote\" title=\"Jaskelioff et al., \u201cTelomerase reactivation reverses tissue degeneration in aged telomerase-deficient mice,\u201d Nature 469 (2011): 102\u20137.\" id=\"return-footnote-2527-1\" href=\"#footnote-2527-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>\u00a0Telomerase-deficient mice were used in these studies; these mice have tissue atrophy, stem cell depletion, organ system failure, and impaired tissue injury responses. Telomerase reactivation in these mice caused extension of telomeres, reduced DNA damage, reversed neurodegeneration, and improved the function of the testes, spleen, and intestines. Thus, telomere reactivation may have potential for treating age-related diseases in humans.<\/p>\n<p>Cancer is characterized by uncontrolled cell division of abnormal cells. The cells accumulate mutations, proliferate uncontrollably, and can migrate to different parts of the body through a process called metastasis. Scientists have observed that cancerous cells have considerably shortened telomeres and that telomerase is active in these cells. Interestingly, only after the telomeres were shortened in the cancer cells did the telomerase become active. If the action of telomerase in these cells can be inhibited by drugs during cancer therapy, then the cancerous cells could potentially be stopped from further division.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>In Summary: DNA Base Pairs and Replication<\/h3>\n<p>DNA was first isolated from white blood cells by Friedrich Miescher, who called it nuclein because it was isolated from nuclei. Frederick Griffith&#8217;s experiments with strains of <em data-effect=\"italics\">Streptococcus pneumoniae<\/em> provided the first hint that DNA may be the transforming principle. Avery, MacLeod, and McCarty proved that DNA is required for the transformation of bacteria. Later experiments by Hershey and Chase using bacteriophage T2 proved that DNA is the genetic material. Chargaff found that the ratio of A = T and C = G, and that the percentage content of A, T, G, and C is different for different species.<\/p>\n<p>The model for DNA replication suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. In conservative replication, the parental DNA is conserved, and the daughter DNA is newly synthesized. The semi-conservative method suggests that each of the two parental DNA strands acts as template for new DNA to be synthesized; after replication, each double-stranded DNA includes one parental or \u201cold\u201d strand and one \u201cnew\u201d strand. The dispersive mode suggested that the two copies of the DNA would have segments of parental DNA and newly synthesized DNA. Experimental evidence showed DNA replication is semi-conservative.<\/p>\n<p>Replication in eukaryotes starts at multiple origins of replication. A primer is required to initiate synthesis, which is then extended by DNA polymerase as it adds nucleotides one by one to the growing chain. The leading strand is synthesized continuously, whereas the lagging strand is synthesized in short stretches called Okazaki fragments. The RNA primers are replaced with DNA nucleotides; the DNA remains one continuous strand by linking the DNA fragments with DNA ligase. Below is a summary table of the major enzymes addressed in this reading, listed in rough order of activity during replication.<\/p>\n<p>The ends of the chromosomes pose a problem during DNA replication as polymerase is unable to extend them without a primer. Telomerase, an enzyme with a built-in RNA template, extends the ends by copying the RNA template and extending one end of the chromosome. DNA polymerase can then extend the DNA using the primer. In this way, the ends of the chromosomes are protected. This is important as evidence indicates telomere length may play a role in regulating cell division and the process of aging.<\/p>\n<\/div>\n<h2><strong>Check Your Understanding<\/strong><\/h2>\n<p>Answer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does\u00a0<strong>not<\/strong>\u00a0count toward your grade in the class, and you can retake it an unlimited number of times.<\/p>\n<p>Use this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section.<\/p>\n<p>\t<iframe id=\"lumen_assessment_3358\" class=\"resizable\" src=\"https:\/\/assessments.lumenlearning.com\/assessments\/load?assessment_id=3358&#38;embed=1&#38;external_user_id=&#38;external_context_id=&#38;iframe_resize_id=lumen_assessment_3358\" frameborder=\"0\" style=\"border:none;width:100%;height:100%;min-height:400px;\"><br \/>\n\t<\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2527\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Introduction to DNA Base Pairs and Replication. <strong>Authored by<\/strong>: Shelli Carter and Lumen Learning. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Biology. <strong>Provided by<\/strong>: OpenStax CNX. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@10.8\">http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@10.8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@10.8<\/li><li>DNA Replication. <strong>Provided by<\/strong>: Open Learning Initiative. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/oli.cmu.edu\/jcourse\/workbook\/activity\/page?context=434a5f7e80020ca6000225da6a4220c9\">https:\/\/oli.cmu.edu\/jcourse\/workbook\/activity\/page?context=434a5f7e80020ca6000225da6a4220c9<\/a>. <strong>Project<\/strong>: Introduction to Biology (Open + Free). <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-2527-1\">Jaskelioff et al., \u201cTelomerase reactivation reverses tissue degeneration in aged telomerase-deficient mice,\u201d <em>Nature<\/em> 469 (2011): 102\u20137. <a href=\"#return-footnote-2527-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":17,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Introduction to DNA Base Pairs and Replication\",\"author\":\"Shelli Carter and Lumen Learning\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Biology\",\"author\":\"\",\"organization\":\"OpenStax CNX\",\"url\":\"http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@10.8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@10.8\"},{\"type\":\"cc\",\"description\":\"DNA Replication\",\"author\":\"\",\"organization\":\"Open Learning 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