Module 6 – DNA and Genetics Outline1. Brief history a. Mendel - 1866 b. Watson and Crick – 1953 c. Human Genome Project - current 2. DNA Structure a. Nucleic Acids i. Nuclides 1. A-T 2. C-G ii. Bonds b. Carbohydrate Component c. Phosphate Component d. Primary structure e. Secondary Structure i. Single strand ii. Double strand f. ~3 billion base pairs g. Chromosome 3. DNA Functions a. Genetic Code i. Gene ii. Codon b. RNA/protein synthesis 4. DNA recombination and repair a. Replication b. Faulty repair c. Errors 5. Genome a. Stability b. Instability – tumorgenesis c. Epigenetics Reading: “Genomic Stability and Instability: A Working Paradigm” Cheng and Loeb, 1997. References: “Cells” Lewin et al., 2007. “Principles of Genetics” Snustad and Simmons, 2009. “Cell Biology” Pollard and Earnshaw, 2002. Introduction DNA is the blueprint for life. A blueprint contains all the information needed for construction of a building. One difference between a true blueprint and DNA is the recipe needed for making each component. DNA not only holds instructions for building a cell, but also contains the recipe for making the plasma membrane, for instance. As previously discussed, the cell contains all the processes required for life and is considered the smallest unit of life. In this module, the structure and function of DNA will be addressed in more detail. DNA structure begins with simple molecules and ends with a series of complex bonds. As for function, DNA carries the basic information of life, but transfer of that information undergoes the lengthy process of transcription and translation. Each of these processes will be discussed and their importance revealed. 1. History Although the study of genetics developed during the twentieth century, the roots of study began in the nineteenth century with Gregor Mendel. Mendel’s research, observation of pea plants in the monastery garden, was performed in relative obscurity. He studied the plants and followed which physical traits were carried from one generation to the next. Eventually, Mendel began to interbreed pea plants with differing characteristics to see which would be passed to the next generation. With further observation and interbreeding, Mendel began to propose that each gene, or hereditary trait, was composed of two parts, known now as alleles. Mendel’s first breeding of pea plants was a cross between a tall pea plant, growing two meters in height, and a short plant, growing only half a meter. The next generation of pea plants was tall, indicating two forms of alleles. As the second generation of pea plants were bred and grown, the was born and Mendel’s research technique was applied to many organisms. his paper was revisited.1). By linking the nucleotides together. Figure 3. Figure 6. the big question was “What is a gene?” In 1953. In 1866. are passed from one generation to the next via genes. Mendel had demonstrated that physical traits. as a science. in 1886.result was a mixture of tall and short plants. Now. Figure 6. a plethora of study began on inheritance in microorganisms.1 shows Mendel’s pea plant experiment. These results confirmed Mendel’s theory of hereditary factors existing in two forms. Sixteen years after Mendel’s death. These linkages were the product of chemical bonds between phosphate and sugar molecules located in the nucleotide itself. plants and animals. Mendel published his discoveries but the article was not much noticed and he went on to do other things. James Watson and Francis Crick tried to answer that question. a chain is formed and contains a particular sequence . As Mendel’s paper became better known. Watson and Crick had studied DNA and knew nucleotides were connected together. such as height.1 Mendel’s pea plant experiment (Snustad and Simmons. The study of genetics. The chemical bonds are needed to create double stranded DNA. FIGURE 6. The collection of DNA molecules that is characteristic to an organism is referred to as its genome. all the information necessary to analyze the organism’s genes should be present. efforts from both projects led to several lengthy articles about the human genome. the Human Genome Project began in 1990 and was a worldwide effort to sequence the approximately 3 billion nucleotide pairs in human DNA.2 Basic structure of DNA (a) displaying hydrogen bonds and (b) showing helical form (Snustad and Simmons. Watson and Crick proposed that DNA molecules consist of two chains of nucleotides and these chains were held together with weak chemical bonds. In addition to proposing double stranded DNA.unique to that chain.4). the idea of separate genes that encode traits was still being investigated. The Human Genome Project initially began as a collaboration of researchers in several different countries and was funded by each government. Watson and Crick discovered the two strands of DNA were wound around each other in a helical configuration.000 to 40. In the early 1900’s. Having this knowledge in hand.2 shows a representation of DNA structure. In 2001. the human gene number has been revised to a lower number of 20. both with and without helical arrangement. Figure 6. After Watson and Crick discovered the structure of DNA. By obtaining the sequence of bases. The articles indicated 2. However.7 billion nucleotide pairs had been sequenced and the human genome was estimated to have 30.000 to 25. geneticists were working at identifying what genes were made of. Upon further sequencing and completion of the Human Genome Project in 2003. geneticists began to work on ways to determine the sequence of bases in DNA molecules. Although the structure of DNA was determined. Figure 1.000 genes. or sequencing the DNA. These genes have been . Following this success. a privately funded project was initiated and soon developed alongside the publicly funded project. Genome sequencing began with bacteria and was first successful in sequencing the bacteriophage !"174. This sequence is what differentiates each chain.000 genes. catalogued by location. while thymine is found only in DNA and uracil replaces thymine in RNA. Thymine. The nucleobase is made from a combination of nitrogen and carbon atoms that form either five. which is known as base pairing. sixmembered rings. Adenine. The hydrogen bonds are not covalent and can be broken and rejoined with relative ease. Base pairing only occurs . thymine (T) and uracil (U). guanine (G). 2. The five major bases are adenine (A). it lacks a methyl group on its ring.or six-member rings. the base for aromatic rings. The nucleobases above can be classified into two types: pyrimidines. adenine and guanine are classified here. This interaction is called complementary base pairing. which are fused five. complementary base pairs are shown and hydrogen bonds are indicated as dashed lines. and purines. Nucleobases are involved in pairing throughout DNA and RNA polymers. The rings consist of two nitrogen and four carbon molecules. Bases are bonded together with hydrogen bonds. or deoxyribonucleic acid. The basic structure of each nuclide is shown in Figure 6. There are five nucleobases. In Figure 6. nitrogen and phosphorus. Nucleotides consist of carbon. structure. three basic elements are formed and combined to make a single nucleotide. From these molecules. cytosine and uracil are placed into the pyrimidine category. DNA Structure DNA.and six-membered compounds. however. An imidazole ring consists of five members. and potential function. hydrogen. In the case of purines. DNA and RNA each have one unique base. Figure 6.3 Nucleobase structure. efforts have shifted to the discovery of how genes influence characteristics of the human being. Although uracil is classified as a pyrimidine. guanine and cytosine are the common nucleobases. Pyrimidines are heterocyclic aromatic rings. is composed of a series of repeating units called nucleotides. a phosphorus unit and a nucleobase. cytosine (C). In the DNA double helix.4.3. two nitrogen and three carbon molecules. oxygen. Three of the nucleobases are found in both DNA and RNA. Now. The elements are a carbohydrate unit. A purine is most simply described as a pyrimidine ring fused with an imidazole ring. pyrimidines from one strand interact with purines from the other strand. 4 Complementary base pairs.5. Adenine will only pair with thymine (or uracil) and cytosine will only pair with guanine. However. It helps maintain the sequence of DNA throughout replication and allows reversible interactions between the bases. Figure 6. Each type of base pair forms with a different number of hydrogen bonds. However. the 2’ carbon contains a . The effects of base stacking are important in the secondary and tertiary structure of RNA. DNA replication depends on the separation of its complementary strands. As will be seen later. the carbohydrate is a pentose (five carbon) sugar. the assumption that DNA with high G-C content is more stable than that with low G-C content is misleading. The stability of DNA does not depend on inter-strand base interactions. exchange repulsion and electrostatic interactions. Base stacking interactions are due to dispersion attraction. Intra-strand base interactions are more stable in DNA with high G-C content due to base stacking interactions. As one can guess. In both DNA and RNA. G-C forms with three bonds. DNA contains 2-deoxyribose sugar and RNA contains ribose sugar. This specific interaction is critical for all the functions of DNA. Seen in Figure 6.as indicated in Figure 6. The ribonucleic acids contain hydroxyl groups connected to each carbon of the pentose ring. GC stacking tends to be more favorable with adjacent bases than CG stacking. but on intra-strand base interactions.4. three hydrogen bonds are more stable than two bonds. All nucleobases are chemically linked to a carbohydrate unit. the deoxyribose sugar contains only four hydroxyl groups. whereas A-T forms with only two. one or more phosphate groups are joined to the sugar through phosphodiester bonds. The phosphate groups are simply a single phosphorous atom surrounded by four oxygen atoms. As the phosphate binds to the pentose sugar. see Figure 6. This is one of the distinguishing characteristics of RNA. . DNA pairing works in the same fashion but uses deoxyribose sugar. The sugar is bonded to the nucleobase via ester bonds.6 shows the pairing of sugar and nucleobase molecules for RNA. DNA and RNA have a direction. the nucleotides form long polymers that are linked together through phosphodiester bonds between the deoxyribose sugar and phosphate groups. These asymmetric bonds form between the third and fifth carbon atoms of adjacent pentose sugars. a nucleotide is created. it is then referred to as a nucleoside.5 Pentose sugars of DNA and RNA (a) 2-deoxyribose (b) ribose Nucleosides contain one of the nucleobases and either a deoxyribose or ribose sugar. but not DNA.7. Due to the bonding nature of the phosphate groups. and are the same in both DNA and RNA. In DNA.hydroxyl group for RNA. Once the nucleobase becomes linked to a sugar. Nucleotides are the building blocks of DNA and RNA. Ester bonds are flexible and allow the DNA strands to move and bend. Figure 6. Figure 6. (a) (b) Figure 6.6 Nucleosides of RNA. After the nucleoside is formed. The bond is generally located between a nitrogen molecule of the nucleobase and the 1’ carbon of the sugar. arizona. Nucleobases are attached to the sugars and phosphates connect the nucleosides. (a) (b) Figure 6. The polymers align.blc.8 (a) is a picture of DNA showing the 5’ and 3’ ends.8 Single (a) and helical (b) DNA polymer with 5’ and 3’ ends (www. If a phosphate group is terminal.edu/Molecular_Graphics/DNA_Structure/DNA_Tutorial.Figure 6. Due to the asymmetry of linkage between each sugar. The direction is determined by the terminal end of the DNA strand.HTML). However. in opposite directions. as mentioned above. DNA and RNA have a direction. Shown in Figure 6. Phosphodiester bonds are strong covalent bonds that connect the 3’ carbon of one sugar to the 5’ carbon of the next. and. This rope-like structure is called a DNA polymer.7 Phosphate group The backbone of DNA is a series of alternating carbohydrate and phosphate groups. it is called the 3’ end. The sides of the ladder are the alternating carbohydrate and phosphate groups that make up the backbone of two DNA polymers. the end of the strand is said to be the 5’ end. The secondary structure of DNA is similar to a ladder. is held together via phosphodiester bonds. if a hydroxyl group from the sugar is located at the DNA terminus. with each other and the . either G-C or A-T. DNA winds around the surface of the histone octamer in a helical path. see Figure 6. two turns of DNA. The sugarphosphate backbone of each DNA polymer is on the outside of the helix. This octamer consists of a central tetramer flanked on each side by a heterodimer. Chromosomal DNA molecules are much longer than the diameter of the nucleus itself and must be highly compacted.34 nm from the next. on average.4 nm length.9 Spacing of nucleotides within a DNA helix (Snustad and Simmons.nucleobases begin to form bonds. To begin the process.9). while the bases are on the inside. Figure 6. Figure 6. the DNA molecules still undergo further packaging. Beyond the organization of the nucleotides into DNA polymers and then a double helix. The bases are added in a nearly perpendicular fashion to the long axis of the DNA polymer. The histone complex contains.10 give a representation of DNA wrapping itself around a histone octamer.8 (b). As the helix is formed. B-DNA is the conformation that DNA takes on under normal physiological conditions. Each base pair is stacked 0. a right-handed double helix is formed. approximately 10. fills a 3. makes the rungs of the ladder. DNA is coiled around a series of histones. The spacing of each nucleotide and turn within the double helix are shown in Figure 6. Each base pair. This is considered the regular structure of DNA and is referred to as B-DNA. As appropriate hydrogen bonds are formed between the bases. Figure 9. . each complete turn. such as those found within the nucleus. Histones are typically found as a set of eight proteins. which consists of approximately 150 base pairs. Essential features of a DNA double helix are two strands of DNA with the nucleobases bonded together.9.5 base pairs. The string is DNA between the nucleosomes and the beads are histone complexes that wrap the DNA. As can be seen. Nucleosome structure is often referred to as a string of beads.11 Electron micrograph (a) and illustration (b) of nucleosome substructure (Snustad and Simmons. Figure 6.to 104fold better to protein factors than nucleosomal DNA. formation of nucleosomes reduces the accessibility of DNA to transcription and protein regulatory factors. Both an electrograph and drawing of a nucleosome substructure are shown in Figure 6. Both strands of DNA must be free for the binding of proteins. the DNA and histone complex is referred to as a nucleosome.21). forming a nucleosome (nicerweb. . After DNA has successfully wrapped around the histone octamer.11.10 DNA organization around a histone complex.com). A comparison of DNA not bound to histones and DNA bound shows that unbound DNA binds 10.Figure 6. Figure 9. respective (Pollard and Earnshaw.11 Four models of nucleosome packing into the 30-nm filament. . a 30-nm filament has been shown to further condense the nucleosomes. This formation can be compared to a circular staircase. A final zigzag stacking is possible. similar to the double helix of DNA. This method of packaging is similar to the solenoid model. Figure 6. Cross-linking is thought to occur between nucleosomes on opposite sides of the long axis. Figure 13. The first is classic winding. Classical winding is similar to the DNA wrapping of histones. but is a two strand winding. In this fiber. a definitive answer is still in the future. or solenoid. However. can only be theorized. discrepancies have arisen about the method of nucleosome packing. The second is a cross-linked formation. shown in Figure 6.11. model. no obvious structure is seen.Chromatin structure beyond the nucleosome is poorly understood.6). Although there are theories about the construction of the 30-nm fiber. Panels E and F are electron micrographs of chromatin without and with histone H1. Structural studies of chromatin fibers are difficult due to the fragile nature of the fibers and higher levels of chromatin packing. There are currently four possible models. that deal with nucleosome packing. The disk shaped nucleosomes are thought to arrange themselves along the long-axis of the filament. However. The third method of stacking is a random stacking of nucleosomes. A single strand of nucleosomes are linked together in a spherical manner along the central axis of the fiber. As the name implies. the nucleosomes are linked in a zigzag pattern around a central axis. at the moment. Figure 6.12 Packaging of DNA from a double helix to a chromosome (biology200.As a summary. This condensation is similar to trying to fit 100 elephants into the back of a VW Beetle. Each codon specifies a single amino acid for protein synthesis. The total length of DNA in any nucleus is approximately 2 meters.edu) 3. DNA is a long polymer of nucleotides connected by phosphodiester bonds. The packaging begins with nucleosome formation and ends with the chromosome.13. The sequence of these nucleotides are translated and used for protein synthesis. The rules of the genetic code are defined as a set of three nucleotides to be translated at one time. DNA Functions The primary function of DNA is to carry the genetic instructions used in development and function of organisms. Four of those codons begin with the same two nucleotides and . is coded for by six varying codons. for example. not every amino acid is classified by only one DNA codon. Serine. Therefore. This sequence is called a codon.12 shows the organization of DNA. The DNA codon table is shown in Figure 6. These instructions are coded in the DNA by a set of by rules referred to as the genetic code. As mentioned above. As can be seen. These 2 meters must be condensed to fit within a 6 µm diameter nulceus. the genetic code is responsible for defining how the DNA sequences are interpreted.gsu. In essence. Figure 6. chromatin must be condensed about 104fold in length. it is not ambiguous. If only two bases were used per codon. Degeneracy is defined as the redundancy of the genetic code. When the nucleotide number is increased to three. serine would still be added to the protein. there are 64 possible codes (43=64). Each codon is specific for one particular amino acid. RNA has a similar codon table. Two nucleotides do not give enough variance for all 21 required codes (20 amino acids plus one stop). So. Figure 6. but uracil replaces thymine. these mutations would most likely be silent and not affect protein synthesis. Since DNA contains all genetic information required for life in an organism. This . Point mutations are a single base substitution that causes one nucleotide to be replaced with another. This repetition of coding is called degeneracy. only 16 amino acids would contain unique codes (42=16). Degeneracy occurs due to a need to code for 20 amino acids and a single stop codon. a point mutation in the third codon position would not alter the translation of the DNA sequence. This is the cause of genomic degeneracy. If the initial genetic code for serine was TCG and a point mutation at position three caused the codon to read TCA.vary in the third. Although the genetic code has redundancy. as seen with serine.13 DNA codon table. there must be a way to move that information out of the nucleus and into the cell for use. Taking serine as an example again. the codons can vary in any of the three nucleotide positions. A benefit to the redundancy in the genetic code is the fault-tolerance for point mutations. However. as seen in Serine. RNA is formed as a single strand. Transcription is defined as the process of creating a complementary RNA copy of a specific sequence of DNA. The first step is the unwinding of DNA. only about 80 nucleotides. hydrogen bonds formed between DNA and RNA nucleotides are broken. Once the strand of RNA nucleotides begins to form. that transfers a specific amino acid to a growing polypeptide. The hydrogen bonds are broken which allows for RNA synthesis. Figure 6.7). binds DNA to assist in unwinding and RNA synthesis. As discussed previously. completing step three. and occurs within five simple steps. Only mRNA. RNA synthesis occurs in a 5’ – 3’ direction. During step four. The main difference is only one strand of DNA is copied. Also. When the RNA is freed. tRNA is a small RNA chain. ribosomes are sites of protein synthesis. Similar to DNA replication. as mentioned previously. a single DNA strand is used to make an RNA strand. The sequence of mRNA determines the amino acid sequence of the protein to be produced. not two complementary strands. mRNA carries information copied from DNA to the ribosome.14 RNA synthesis from an unwound portion of DNA (Snustad and Simmons. transfer (tRNA). However. it undergoes further processing to protect the 5’ and 3’ ends and completes step five by exiting the nucleus through a nuclear pore. or translation begins. thymine is replaced by uracil. like DNA. They are messenger (mRNA).14. The final type of . ribosomal (rRNA). shown in Figure 6. RNA polymerase. Transcription ends with five biologically active RNAs. tRNA and rRNA will be discussed here. This is similar to DNA replication. When an mRNA reaches and binds to a ribosome. The second type of RNA is tRNA. not both as in DNA replication. Proteins attach to the DNA helix and “unzip” the strands to allow access to the nucleotides.process is called transcription. Figure 11. Step two pairs RNA nucleotides to the DNA nucleotides. protein synthesis. This frees the newly synthesized RNA from the DNA helix. a backbone of alternating ribose sugar and phosphate molecules are added. small nuclear (snRNA) and micro (miRNA) RNAs. The amino acids are linked to tRNA by peptide bonds and act as linkers between amino acids and the codons in mRNA during translation. as discussed above.15. Protein synthesis is the process that cells use to build proteins. rRNAs are structural and catalytic components that make ribosomes. 4. As in transcription. This initiates translation of the mRNA sequence. ribosome shown as gray circle surrounding the mRNA. mRNA. At this point. DNA Replication and Repair . termination occurs. Figure 6. initiation. shown in Figure 6.15 Translation of mRNA. Initiation involves the small subunit of the ribosome. Figure 6. Amino acids are added to the growing chain. The additions occur until the end of the mRNA is reached.15 shows translation of mRNA by a ribosome. and the fourth is synthesized elsewhere. Eukaryotic ribosomes contain four differing rRNA molecules. Elongation is just what it sounds like. elongation and termination. leaves the nucleus to enter the process of translation.RNA is rRNA. Activation attaches the correct amino acid to the correct tRNA. The ribosome then has the ability to bind mRNA and carry out protein synthesis. Translation is defined as the decoding of mRNA by the ribosome to produce a specific amino acid chain. discussed previously. translation proceeds in distinct phases. tRNA carries the amino acid to the mRNA and attaches to the correct codon. These are activation. the amino acid chain is transported to the internal space of the ER and folded into an active protein. binding to the 5’ end of the mRNA. rRNA combines with protein in the cytosol and forms the ribosome. tRNA does not recognize the stop codon sequence and the ribosome/mRNA complex disassembles. Implied from the name. Three of these are synthesized in the nucleolus. At this point. It begins in the nucleus with RNA transcription. In a sense. dispersive and semiconservative. The dispersive method produces two copies of DNA. DNA replication is the process by which all living organisms copy their DNA.15 Replication methods Now with the basic understanding of semiconservative replication. Figure 6. They are conservative. DNA within the chromosome is composed of a combination of both original or both new strands. That is. as the original helix is broken for replication. but neither strand is completely composed of template or new DNA. elongation and termination. it is the basis for inheritance between cell and organism generations. The DNA polymers become mixed and the end results are similar to crossover events that occur during meiosis. The semiconservative method. Conservative replication maintains the two original template DNA strands as a single helix while the newly synthesized strands form a second helix. DNA replication can be described in more detail.Several methods of DNA replication have been postulated. A comparison of the three replication methods is shown in Figure 6. the complimentary nucleotides that form the new DNA strand become attached via hydrogen bonds. DNA replication occurs in three steps which are initiation. one daughter cell receives the original template DNA while the other daughter receives the copied DNA. In this model. . conserves one strand of the original DNA in each replicated helix. The final two double strand DNA helices each consist of one original and one newly synthesized strand of DNA. In a similar fashion to RNA and protein synthesis.15. confirmed by the Meselson-Stahl experiment in 1958. Clearly. The new DNA strand is extended in the 3’ direction with the addition of complementary nucleotides. Small strands of RNA. These sites are called replication origins. When the second strand is complete. At this point. they are called a replication fork. Two sets of DNA replication machinery head out from the origin in opposite directions. that enables one strand of DNA to pass through another. However. 2000 hours greatly exceeds the time reserved to complete the S phase of the cell cycle. so they do not recognize 5’ phosphate groups. The double cut removes knots and entanglements that can form during replication. The role of the single strand binding proteins (SSBP) is just as their name implies. a replication bubble has formed. called primers. they do not attach directly to DNA templates and require an existing DNA strand paired with the template. the SSBP releases the DNA and hydrogen bonds are formed to hold the DNA helix together. DNA replication moves from initiation into elongation. The purpose of the directionality is the need to attach new nucleotides to the 3’ hydroxyl on the primer strand. Once the replication proteins and enzymes have gathered together and attached to DNA. or origins of bidirectional replication. to separate the DNA. DNA topoisomerases are responsible for nicking a single strand of DNA which allows the strands the ability to swivel around each other. Strand nicking removes the build-up of DNA twists during replication. Once the replication machinery has been established. During elongation. In addition to nicking a single strand. DNA synthesis extends new DNA polymers in a 5’ to 3’ direction. These proteins include DNA topoisomerases and single strand binding proteins. DNA at the origin contains specific sequences that allow replication proteins to attach to the DNA. DNA polymerases responsible for elongation do not have 3’ to 5’ synthesis activity. approximately 2000 hours would be needed replicate the entire genome. topoisomerases cut both backbones. They bind to single stranded DNA until the second strand is synthesized.Initiation of DNA replication occurs at specific sites within the DNA. The replication fork forms in the nucleus and . Elongation of DNA requires a special set of proteins referred to as replication machinery. Even with this speed. such as DNA helicase. DNA polymerases are the key players in elongation. A family of DNA helicases are responsible for breaking hydrogen bonds between nucleotides and unwinding the DNA helix. DNA polymerases are a family of enzymes that carry out DNA replication. a double strand cut. the SSBP prevents secondary structure formation within the DNA. new strands of DNA are synthesized at a rate of about 3000 nucleotide additions per minute. DNA polymerases are then able to synthesize the new strand of DNA. The initiator proteins recruit other proteins. are created and attached to the DNA template. If each chromosome contained only one replication origin. By attaching to single strand DNA. As the name implies. multiple origins are needed to allow complete replication within the time allotted for S phase. replication occurs in opposing directions. there is the problem on the second strand of the DNA helix. The RNA primer is placed at the beginning of each segment. Figure 45.17 are leading and lagging strands of synthesized DNA. As briefly mentioned. DNA polymerase “reads” the DNA in short. However. For a computer animated video of DNA replication. however. another primer is placed at the replication fork to enable synthesis of another segment of DNA. The leading strand is defined as the DNA template that is synthesized in a 5’ to 3’ direction. The replication fork moves along the chromosome in a 3’ to 5’ direction. the primer binds the template and synthesis continues. unlike the leading strand which needs only one primer. opposite to the activity of DNA polymerase. The lagging strand is characterized by growth in the opposite direction to the movement of the replication fork. Once the replication fork opens another section of DNA. Some of the discontinuities are caused by the replication fork itself.16 Replication fork (Pollard and Earnshaw. As each segment is synthesized. The movement is along the leading strand of DNA.16 is the basic structure of the replication fork. the replication fork moves in a 3’ to 5’ direction during replication. The replication machinery reads it in a 5’ to 3’ direction. only one lagging strand is shown. In a sense. It is responsible for breaking the hydrogen bonds that hold the two DNA strands together.html. separated segments. As the template DNA is being unwound in a 5’ to 3’ direction. Synthesis of the new strand is complementary to the movement of the fork and DNA polymerase activity. Seen in Figure 6. . please go to http://dnalc. the second strand of DNA becomes synthesized in short. Replication machinery takes up room and the RNA primer is not able to attach to the DNA. The short DNA fragments formed on the lagging strand are called Okazaki fragments. Figure 6.1). This strand is called the lagging strand. The Okazaki fragments are joined together by DNA ligase. DNA ligase is an enzyme that repairs single stranded discontinuities.only during DNA replication.org/resources/3d/04mechanism-of-replication-advanced. DNA polymerase is able to “read” the template strand and add nucleotides to the new strand continuously. Shown briefly in Figure 6. On the lagging strand. non-continuous segments. This can easily be translated to a break within the DNA strand.17 Continuous vs discontinuous replication (Snustad and Simmons.18 as base damage. including normal metabolic activities and environmental radiation. DNA damage is shown in Figure 6. DSB is the most hazardous to the cell due to an increased possibility of genomic rearrangements. however. DSB cleaves both DNA strands and results in two free ends of DNA. DNA can be damaged by any number of factors. This damage can consist of individual base damage or DNA structural damage. These lesions can cause structural damage to the DNA which can interfere with transcription of genes within the genome. one strand of the DNA helix is severed.Figure 6. DNA in human cells can be exposed to 1 million molecular lesions. . Figure 10. In any given day. the two DNA strands have not separated from each other. Two basic types of damage that will be discussed here are single-strand breaks (SSB) and doublestrand breaks (DSB). In the case of SSB.1). Instead. NHEJ is called “non-homologous” because it does not require a homologous template for repair. For repair of a single-strand break. However. among others. The lost DNA is then synthesized and the . These checkpoints are used to pause the cell cycle and allow repair of DNA. this is only seen with DNA overhangs that are not compatible. the repair proteins search out homologous DNA or the sister chromosome. Double-strand breaks severe both DNA strands at the same location resulting in two free ends of DNA. briefly. As previously discussed. The other method of repair is homologous recombination (HR). DNA can be damaged with double-strand breaks.(a) (b) Figure 6. BER removes the damaged base and replaces the missing nucleotide with the assistance of DNA polymerase. Once resection is complete. strand invasion occurs. The DNA polymerase activity is similar to that used during DNA synthesis. HR does not rely on short microhomologies. The base can be damaged by oxidation or hydrolysis. the cell has a choice of two pathways. NHEJ repair is mostly accurate. The primary repair method of a SSB uses the same proteins used by the base excision repair (BER) mechanism. BER. In addition to single-strand breaks. Strand invasion takes the 3’ end of the broken DNA and “invades” the homologous DNA. DNA ligase acts on the new base to seal the nick in the DNA strand. there are two DNA checkpoints within the cell cycle. Resection is the process by which DNA surrounding the damage is removed from the 5’ end of the break. The damaged DNA is resected. but can have some imprecision that leads to the loss of nucleotides. The DNA ends at the DSB typically have short single-stranded sequences that serve as microhomologies within the break. BER skips to the final step of ligation.18 (a) Single-strand and (b) double-strand damage Repair of DNA damage begins with the identification of damage. is the repair of a single damaged base. The broken ends of DNA are ligated together using the microhomologies found in the DNA overhangs for alignment. In order to repair damage. They are non-homologous end joining (NHEJ) and homologous recombination (HR). The broken phosphodiester bonds are reformed and the DNA strand break is repaired. Insertions can cause frameshift mutations within the DNA sequence. DNA polymerase pauses between Okazaki fragments. replication slippage can result in insertions. Frameshift mutations alter the normal reading frame of a gene and the amino acids encoded for by the gene. When DNA strands become misaligned. DNA polymerase moves at a speed comparable to the replication fork. An insertion is the addition of one or more nucleotides into the DNA sequence. One way in which insertions are created is multiple replications of the same DNA. The pause in replication can cause the polymerase to dissociate from the DNA and lead to replication slippage. Replication slippage is the most common cause of error.19 Figure 6. a deletion of part of the gene results in the development of Li-Fraumeni syndrome. . The first is deletion of DNA. repeated sequence. DNA polymerase is responsible for coping DNA. replication can skip over a section of DNA. if the number of nucleotides is not divisible by three. on the lagging strand.chromosomes separate when repair of the DSB has been completed. Replication slippage occurs in regions of DNA that have short.19 First steps of Homologous Recombination In addition to DNA damage. The first steps of HR are shown in Figure 6. During replication. replication has its own set of inherent errors. The dissociated polymerase leads to two types of error. In the cause of p53. The deletion can range from a single base to an entire piece of a chromosome. In addition to deletions. However. A genetic deletion is defined as a mutation in which part of the DNA sequence is missing. or a change in chromosome number. DNA duplication. the nick is repaired and replication occurs without further damaging the DNA. Instability in the chromosomal structure can lead to rearrangements and duplications. If a single-stranded nick were to be replicated. By pausing the cell cycle. more importantly. Epigenetics is the study of gene expression alterations caused by mechanisms other . As discussed above. is bombarded with up to a million. In the case of the G2 checkpoint. A common occurance of aneuploidy in humans involves chromosome 21. DNA is reviewed for mismatched nucleotides and unreplicated DNA. Without these vital checkpoints. a series of checkpoints are in the cell cycle as quality control mechanisms. Instability leads to deletion or insertion of DNA. but the number is either higher or lower than expected. In addition to repair of damage. Two separate genes cannot function properly when joined together and are common in cancer cells. In addition to aneuploidy. The chromosomes themselves have not been altered. chromosomal structure can be affected by instability. A separate type of DNA alteration does not affect the structure or chromosomal identity of the DNA itself. Mentioned previously. DNA breaks and lesions per day. mentioned above. again. among other changes. Aneuploidy is defined as an abnormal number of chromosomes in a cell. unequal separation of chromosomes can occur. Genome Maintaining genome integrity is just as important as maintaining genetic accuracy during replication. the cell is said to have aneuploidy. on average.5. The genome. When three copies of chromosome 21 appear in a developing fetus. One of the key characteristics of chromosome rearrangement is the fusion of two genes. Duplication of DNA is just that. The DNA sequence can be altered in a superficial manner that does not affect gene expression to the same extent of DNA deletion or gene fusion. not related to replication. When one daughter cell receives more than one copy of a chromosome. The break then causes instability within the genome. Genome instability is a process of chromosomal alterations that can lead to a wide variety of problems. Rearrangements most commonly occur between non-homologous chromosomes. the physical manifestation is Down Syndrome. Repair. the DNA nick would become a full-fledged double-strand break. paused and repairs are made. the two DNA checkpoints within the cell cycle are located in the G1 and G2 phases. ensures the damage is fixed and genomic integrity remains intact. The normal copy number for any chromosome in a human is two. as previously discussed. The cell cycle is. a second unneeded copy. breaks. The G1 checkpoint looks at the DNA to locate lesions and. can cause insertions that affect the structure of proteins. During cell separation. more specifically meiosis. or 106. integrity of the genome is compromised and mutations are more likely to occur. a brief explanation of epigenetics and possible outcome is shown. DNA methylation can activate or repress genes. The phenotype of a cell. suppressed genes can be activated or expressed genes can be silenced.21. These traits include characteristics such as morphology and behavior. The genome of an organism contains the information used for gene expression. These features cause changes that affect the phenotype of a cell without altering the genotype. When this occurs. the methyl group can be removed during chromatin remodeling. The Greek prefix epi.21 Some epigenetic mechanisms and its consequences. Chromatin remodeling involves moving the nucleosomes to . depends on the genes that are expressed. A cell phenotype is an observable trait that has some influence over the development of an organism. however. do not represent changes in the genetic information. DNA modification most commonly affects only one of the two alleles. . Epigenetics can be split into two general classes based on mechanism: Figure 6. However. DNA modification by covalent attachment of a protein or establishing a self-perpetuating protein state.of epigenetics refers to a feature that is “on top of” or “in addition to” genetics. Epigenetic events alter the expression of genes within the genome. As mentioned in Figure 6. In Figure 6. These epigenetic changes. The genotype of a cell is the genetic code.than the underlying DNA sequence. or organism. The addition of a methyl group causes the genes to become less accessible for transcription leading to suppression of the gene. The existence of epigenetic effects leads to the belief that proteins responsible for such modifications have some sort of self-templating or self-assembling capacity. Give a brief description of the beginning of the study of genetics 2. you should be able to answer the following 1.ensure proper winding of DNA. there is no reason for the protein complex to split and then reconstruct itself. Have a general understanding of epigenetics and its ramifications . self-perpetuation of an altered protein becomes difficult during replication. Explain the importance of genomic stability and why genomic instability can potentially lead to tumorgenesis 10. The second case would require complete assembly of a new protein complex. Explain the differences between nuclides in DNA and RNA 3. Describe the differences between genes and codons and how they are related to each other 6. Describe the principle behind DNA recombination 9. This might involve maintaining an alternative protein conformation throughout the life of the cell or modifying a specific protein. In the first case. Be able to label the general structure of DNA. The protein complex could divide equally between sister chromatin or remain completely on one chromatin. Discuss potential ramifications of faulty repair 8. # of base pairs in humans 5. including hydrogen bonds between pyrimidines and purines 4. However. Differentiate between the three methods of DNA repair 7. Before taking the quiz. The second class of epigenetics involves establishing a recurrent protein state.