Module 6 Lecture radiobiology

8/12/2019 Module 6 Lecture radiobiology

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Module 6 DNA and Genetics

Outline

1. Brief history

a. Mendel - 1866b. Watson and Crick 1953c. Human Genome Project - current

2. DNA Structurea. Nucleic Acids

i. Nuclides1. A-T2. C-G

ii. Bondsb. Carbohydrate Componentc. Phosphate Componentd. Primary structuree. Secondary Structure

i. Single strandii. Double strand

f. ~3 billion base pairsg. Chromosome

3. DNA Functionsa. Genetic Code

i. Geneii. Codonb. RNA/protein synthesis

4. DNA recombination and repaira. Replicationb. Faulty repairc. Errors

5. Genomea. Stabilityb. Instability tumorgenesis

c. Epigenetics

Reading:

Genomic Stability and Instability: A Working Paradigm Cheng and Loeb, 1997.


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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 forconstruction of a building. One difference between a true blueprint and DNA is therecipe needed for making each component. DNA not only holds instructions for buildinga cell, but also contains the recipe for making the plasma membrane, for instance. Aspreviously discussed, the cell contains all the processes required for life and isconsidered the smallest unit of life.

In this module, the structure and function of DNA will be addressed in moredetail. DNA structure begins with simple molecules and ends with a series of complexbonds. As for function, DNA carries the basic information of life, but transfer of that

information undergoes the lengthy process of transcription and translation. Each ofthese processes will be discussed and their importance revealed.

1. History

Although the study of genetics developed during the twentieth century, the rootsof study began in the nineteenth century with Gregor Mendel. Mendels 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 differingcharacteristics to see which would be passed to the next generation. With furtherobservation and interbreeding, Mendel began to propose that each gene, or hereditarytrait, was composed of two parts, known now as alleles. Mendels first breeding of peaplants was a cross between a tall pea plant, growing two meters in height, and a shortplant, growing only half a meter. The next generation of pea plants was tall, indicatingtwo forms of alleles. As the second generation of pea plants were bred and grown, the


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result was a mixture of tall and short plants. Figure 6.1 shows Mendels pea plantexperiment. These results confirmed Mendels theory of hereditary factors existing intwo forms. In 1866, Mendel published his discoveries but the article was not muchnoticed and he went on to do other things. Sixteen years after Mendels death, in 1886,his paper was revisited. The study of genetics, as a science, was born and Mendelsresearch technique was applied to many organisms.

Figure 6.1 Mendels pea plant experiment (Snustad and Simmons, Figure 3.1).

As Mendels paper became better known, a plethora of study began oninheritance in microorganisms, plants and animals. Mendel had demonstrated thatphysical traits, such as height, are passed from one generation to the next via genes.Now, the big question was What is a gene? In 1953, James Watson and Francis Cricktried to answer that question. Watson and Crick had studied DNA and knewnucleotides were connected together. These linkages were the product of chemicalbonds between phosphate and sugar molecules located in the nucleotide itself. Bylinking the nucleotides together, a chain is formed and contains a particular sequence


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unique to that chain. This sequence is what differentiates each chain. Having thisknowledge in hand, Watson and Crick proposed that DNA molecules consist of twochains of nucleotides and these chains were held together with weak chemical bonds.The chemical bonds are needed to create double stranded DNA. In addition toproposing double stranded DNA, Watson and Crick discovered the two strands of DNAwere wound around each other in a helical configuration. Figure 6.2 shows arepresentation of DNA structure, both with and without helical arrangement. Althoughthe structure of DNA was determined, the idea of separate genes that encode traits wasstill being investigated.

FIGURE 6.2 Basic structure of DNA (a) displaying hydrogen bonds and (b) showinghelical form (Snustad and Simmons, Figure 1.4).

In the early 1900s, geneticists were working at identifying what genes weremade of. After Watson and Crick discovered the structure of DNA, geneticists began towork on ways to determine the sequence of bases in DNA molecules. By obtaining thesequence of bases, or sequencing the DNA, all the information necessary to analyzethe organisms genes should be present. The collection of DNA molecules that ischaracteristic to an organism is referred to as its genome. Genome sequencing beganwith bacteria and was first successful in sequencing the bacteriophage !" 174.Following this success, the Human Genome Project began in 1990 and was aworldwide effort to sequence the approximately 3 billion nucleotide pairs in human DNA.The Human Genome Project initially began as a collaboration of researchers in severaldifferent countries and was funded by each government. However, a privately fundedproject was initiated and soon developed alongside the publicly funded project. In 2001,efforts from both projects led to several lengthy articles about the human genome. Thearticles indicated 2.7 billion nucleotide pairs had been sequenced and the humangenome was estimated to have 30,000 to 40,000 genes. Upon further sequencing andcompletion of the Human Genome Project in 2003, the human gene number has beenrevised to a lower number of 20,000 to 25,000 genes. These genes have been


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catalogued by location, structure, and potential function. Now, efforts have shifted tothe discovery of how genes influence characteristics of the human being.

2. DNA StructureDNA, or deoxyribonucleic acid, is composed of a series of repeating units called

nucleotides. Nucleotides consist of carbon, oxygen, hydrogen, nitrogen andphosphorus. From these molecules, three basic elements are formed and combined tomake a single nucleotide. The elements are a carbohydrate unit, a phosphorus unit anda nucleobase. The nucleobase is made from a combination of nitrogen and carbonatoms that form either five- or six-member rings. Nucleobases are involved in pairingthroughout DNA and RNA polymers, which is known as base pairing. There are fivenucleobases. The five major bases are adenine (A), guanine (G), cytosine (C), thymine

(T) and uracil (U). Three of the nucleobases are found in both DNA and RNA; however,DNA and RNA each have one unique base. The basic structure of each nuclide isshown in Figure 6.3. Adenine, guanine and cytosine are the common nucleobases,while thymine is found only in DNA and uracil replaces thymine in RNA.

Figure 6.3 Nucleobase structure.

The nucleobases above can be classified into two types: pyrimidines, six-membered rings, and purines, which are fused five- and six-membered compounds.Pyrimidines are heterocyclic aromatic rings. The rings consist of two nitrogen and fourcarbon molecules, the base for aromatic rings. Thymine, cytosine and uracil are placedinto the pyrimidine category. Although uracil is classified as a pyrimidine, it lacks amethyl group on its ring. In the case of purines, adenine and guanine are classifiedhere. A purine is most simply described as a pyrimidine ring fused with an imidazole

ring. An imidazole ring consists of five members, two nitrogen and three carbonmolecules.

In the DNA double helix, pyrimidines from one strand interact with purines fromthe other strand. This interaction is called complementary base pairing. In Figure 6.4,complementary base pairs are shown and hydrogen bonds are indicated as dashedlines. Bases are bonded together with hydrogen bonds. The hydrogen bonds are notcovalent and can be broken and rejoined with relative ease. Base pairing only occurs


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as indicated in Figure 6.4. Adenine will only pair with thymine (or uracil) and cytosinewill only pair with guanine. This specific interaction is critical for all the functions ofDNA. It helps maintain the sequence of DNA throughout replication and allowsreversible interactions between the bases. As will be seen later, DNA replicationdepends on the separation of its complementary strands.

Figure 6.4 Complementary base pairs.

Each type of base pair forms with a different number of hydrogen bonds. G-Cforms with three bonds, whereas A-T forms with only two. As one can guess, threehydrogen bonds are more stable than two bonds. However, the assumption that DNAwith high G-C content is more stable than that with low G-C content is misleading. Thestability of DNA does not depend on inter-strand base interactions, but on intra-strandbase interactions. Intra-strand base interactions are more stable in DNA with high G-Ccontent due to base stacking interactions. Base stacking interactions are due todispersion attraction, exchange repulsion and electrostatic interactions. GC stackingtends to be more favorable with adjacent bases than CG stacking. The effects of basestacking are important in the secondary and tertiary structure of RNA.

All nucleobases are chemically linked to a carbohydrate unit. In both DNA andRNA, the carbohydrate is a pentose (five carbon) sugar. DNA contains 2-deoxyribosesugar and RNA contains ribose sugar. The ribonucleic acids contain hydroxyl groupsconnected to each carbon of the pentose ring. However, the deoxyribose sugarcontains only four hydroxyl groups. Seen in Figure 6.5, the 2 carbon contains a


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hydroxyl group for RNA, but not DNA. This is one of the distinguishing characteristics ofRNA. Once the nucleobase becomes linked to a sugar, it is then referred to as anucleoside.

(a) (b)

Figure 6.5 Pentose sugars of DNA and RNA (a) 2-deoxyribose (b) ribose

Nucleosides contain one of the nucleobases and either a deoxyribose or ribosesugar. The sugar is bonded to the nucleobase via ester bonds. Ester bonds are flexibleand allow the DNA strands to move and bend. The bond is generally located between anitrogen molecule of the nucleobase and the 1 carbon of the sugar. Figure 6.6 showsthe pairing of sugar and nucleobase molecules for RNA. DNA pairing works in thesame fashion but uses deoxyribose sugar.

Figure 6.6 Nucleosides of RNA.

After the nucleoside is formed, one or more phosphate groups are joined to thesugar through phosphodiester bonds. These asymmetric bonds form between the thirdand fifth carbon atoms of adjacent pentose sugars. Due to the bonding nature of thephosphate groups, DNA and RNA have a direction. The phosphate groups are simply asingle phosphorous atom surrounded by four oxygen atoms, see Figure 6.7, and are thesame in both DNA and RNA. As the phosphate binds to the pentose sugar, anucleotide is created. Nucleotides are the building blocks of DNA and RNA. In DNA,the nucleotides form long polymers that are linked together through phosphodiesterbonds between the deoxyribose sugar and phosphate groups.


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Figure 6.7 Phosphate group

The backbone of DNA is a series of alternating carbohydrate and phosphategroups. Nucleobases are attached to the sugars and phosphates connect thenucleosides. This rope-like structure is called a DNA polymer, and, as mentioned

above, is held together via phosphodiester bonds. Phosphodiester bonds are strongcovalent bonds that connect the 3 carbon of one sugar to the 5 carbon of the next.Due to the asymmetry of linkage between each sugar, DNA and RNA have a direction.The direction is determined by the terminal end of the DNA strand. If a phosphategroup is terminal, the end of the strand is said to be the 5 end. However, if a hydroxylgroup from the sugar is located at the DNA terminus, it is called the 3 end. Shown inFigure 6.8 (a) is a picture of DNA showing the 5 and 3 ends.

(a) (b)

Figure 6.8 Single (a) and helical (b) DNA polymer with 5 and 3 ends(www.blc.arizona.edu/Molecular_Graphics/ DNA _Structure/ DNA _Tutorial.HTML ).

The secondary structure of DNA is similar to a ladder. The sides of the ladderare the alternating carbohydrate and phosphate groups that make up the backbone oftwo DNA polymers. The polymers align, in opposite directions, with each other and the


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nucleobases begin to form bonds. As appropriate hydrogen bonds are formed betweenthe bases, a right-handed double helix is formed, see Figure 6.8 (b). Each base pair,either G-C or A-T, makes the rungs of the ladder. Essential features of a DNA doublehelix are two strands of DNA with the nucleobases bonded together. The sugar-phosphate backbone of each DNA polymer is on the outside of the helix, while thebases are on the inside. Each base pair is stacked 0.34 nm from the next. The basesare added in a nearly perpendicular fashion to the long axis of the DNA polymer. As thehelix is formed, each complete turn, approximately 10.5 base pairs, fills a 3.4 nm length.The spacing of each nucleotide and turn within the double helix are shown in Figure 6.9.This is considered the regular structure of DNA and is referred to as B-DNA. B-DNA isthe conformation that DNA takes on under normal physiological conditions, such asthose found within the nucleus.

Figure 6.9 Spacing of nucleotides within a DNA helix (Snustad andSimmons, Figure 9.9).

Beyond the organization of the nucleotides into DNA polymers and then a doublehelix, the DNA molecules still undergo further packaging. Chromosomal DNAmolecules are much longer than the diameter of the nucleus itself and must be highlycompacted. To begin the process, DNA is coiled around a series of histones. Histonesare typically found as a set of eight proteins. This octamer consists of a central tetramer

flanked on each side by a heterodimer. DNA winds around the surface of the histoneoctamer in a helical path. The histone complex contains, on average, two turns of DNA,which consists of approximately 150 base pairs. Figure 6.10 give a representation ofDNA wrapping itself around a histone octamer.


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Figure 6.10 DNA organization around a histone complex, forming a nucleosome(nicerweb.com).

After DNA has successfully wrapped around the histone octamer, the DNA andhistone complex is referred to as a nucleosome. Nucleosome structure is often referredto as a string of beads. The string is DNA between the nucleosomes and the beadsare histone complexes that wrap the DNA. Both an electrograph and drawing of anucleosome substructure are shown in Figure 6.11. As can be seen, formation ofnucleosomes reduces the accessibility of DNA to transcription and protein regulatoryfactors. Both strands of DNA must be free for the binding of proteins. A comparison ofDNA not bound to histones and DNA bound shows that unbound DNA binds 10- to 10 4-fold better to protein factors than nucleosomal DNA.

Figure 6.11 Electron micrograph (a) and illustration (b) of nucleosomesubstructure (Snustad and Simmons, Figure 9.21).


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Chromatin structure beyond the nucleosome is poorly understood. However, a30-nm filament has been shown to further condense the nucleosomes. The diskshaped nucleosomes are thought to arrange themselves along the long-axis of thefilament. However, discrepancies have arisen about the method of nucleosomepacking. There are currently four possible models, shown in Figure 6.11, that deal withnucleosome packing. The first is classic winding, or solenoid, model. Classical windingis similar to the DNA wrapping of histones. A single strand of nucleosomes are linkedtogether in a spherical manner along the central axis of the fiber. This formation can becompared to a circular staircase. The second is a cross-linked formation. Cross-linkingis thought to occur between nucleosomes on opposite sides of the long axis. Thismethod of packaging is similar to the solenoid model, but is a two strand winding,similar to the double helix of DNA. The third method of stacking is a random stacking ofnucleosomes. In this fiber, no obvious structure is seen. A final zigzag stacking ispossible. As the name implies, the nucleosomes are linked in a zigzag pattern around

a central axis. Although there are theories about the construction of the 30-nm fiber, adefinitive answer is still in the future. Structural studies of chromatin fibers are difficultdue to the fragile nature of the fibers and higher levels of chromatin packing, at themoment, can only be theorized.

Figure 6.11 Four models of nucleosome packing into the 30-nm filament. Panels E andF are electron micrographs of chromatin without and with histone H1, respective(Pollard and Earnshaw, Figure 13.6).


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As a summary, Figure 6.12 shows the organization of DNA. The packagingbegins with nucleosome formation and ends with the chromosome. The total length ofDNA in any nucleus is approximately 2 meters. These 2 meters must be condensed tofit within a 6 m diameter nulceus. Therefore, chromatin must be condensed about 10 4-fold in length. This condensation is similar to trying to fit 100 elephants into the back ofa VW Beetle.

Figure 6.12 Packaging of DNA from a double helix to a chromosome

(biology200.gsu.edu)

3. DNA Functions

The primary function of DNA is to carry the genetic instructions used indevelopment and function of organisms. These instructions are coded in the DNA by aset of by rules referred to as the genetic code. In essence, the genetic code isresponsible for defining how the DNA sequences are interpreted. As mentioned above,DNA is a long polymer of nucleotides connected by phosphodiester bonds. Thesequence of these nucleotides are translated and used for protein synthesis. The rulesof the genetic code are defined as a set of three nucleotides to be translated at onetime. This sequence is called a codon. Each codon specifies a single amino acid forprotein synthesis. The DNA codon table is shown in Figure 6.13. As can be seen, notevery amino acid is classified by only one DNA codon. Serine, for example, is coded forby six varying codons. Four of those codons begin with the same two nucleotides and


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vary in the third. This repetition of coding is called degeneracy. Degeneracy is definedas the redundancy of the genetic code. Although the genetic code has redundancy, it isnot ambiguous. Each codon is specific for one particular amino acid. However, asseen with serine, the codons can vary in any of the three nucleotide positions. RNA hasa similar codon table, but uracil replaces thymine.

Figure 6.13 DNA codon table.

Degeneracy occurs due to a need to code for 20 amino acids and a single stopcodon. If only two bases were used per codon, only 16 amino acids would containunique codes (4 2=16). Two nucleotides do not give enough variance for all 21 requiredcodes (20 amino acids plus one stop). When the nucleotide number is increased tothree, there are 64 possible codes (4 3=64). This is the cause of genomic degeneracy,as seen in Serine. A benefit to the redundancy in the genetic code is the fault-tolerancefor point mutations. Point mutations are a single base substitution that causes onenucleotide to be replaced with another. Taking serine as an example again, a pointmutation in the third codon position would not alter the translation of the DNA sequence.If the initial genetic code for serine was TCG and a point mutation at position threecaused the codon to read TCA, serine would still be added to the protein. So, thesemutations would most likely be silent and not affect protein synthesis.

Since DNA contains all genetic information required for life in an organism, theremust be a way to move that information out of the nucleus and into the cell for use. This


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process is called transcription. Transcription is defined as the process of creating acomplementary RNA copy of a specific sequence of DNA. Similar to DNA replication, asingle DNA strand is used to make an RNA strand. The main difference is only onestrand of DNA is copied, not both as in DNA replication. Also, RNA is formed as asingle strand, not two complementary strands. RNA synthesis occurs in a 5 3direction, like DNA, and occurs within five simple steps. The first step is the unwindingof DNA. Proteins attach to the DNA helix and unzip the strands to allow access to thenucleotides. The hydrogen bonds are broken which allows for RNA synthesis. Steptwo pairs RNA nucleotides to the DNA nucleotides. This is similar to DNA replication.However, thymine is replaced by uracil, as mentioned previously. RNA polymerase,shown in Figure 6.14, binds DNA to assist in unwinding and RNA synthesis. Once thestrand of RNA nucleotides begins to form, a backbone of alternating ribose sugar andphosphate molecules are added, completing step three. During step four, hydrogenbonds formed between DNA and RNA nucleotides are broken. This frees the newly

synthesized RNA from the DNA helix. When the RNA is freed, it undergoes furtherprocessing to protect the 5 and 3 ends and completes step five by exiting the nucleusthrough a nuclear pore.

Figure 6.14 RNA synthesis from an unwound portion of DNA (Snustad andSimmons, Figure 11.7).

Transcription ends with five biologically active RNAs. They are messenger(mRNA), transfer (tRNA), ribosomal (rRNA), small nuclear (snRNA) and micro (miRNA)RNAs. Only mRNA, tRNA and rRNA will be discussed here. mRNA carries information

copied from DNA to the ribosome. As discussed previously, ribosomes are sites ofprotein synthesis. When an mRNA reaches and binds to a ribosome, protein synthesis,or translation begins. The sequence of mRNA determines the amino acid sequence ofthe protein to be produced. The second type of RNA is tRNA. tRNA is a small RNAchain, only about 80 nucleotides, that transfers a specific amino acid to a growingpolypeptide. The amino acids are linked to tRNA by peptide bonds and act as linkersbetween amino acids and the codons in mRNA during translation. The final type of


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RNA is rRNA. Implied from the name, rRNAs are structural and catalytic componentsthat make ribosomes. Eukaryotic ribosomes contain four differing rRNA molecules.Three of these are synthesized in the nucleolus, discussed previously, and the fourth issynthesized elsewhere. rRNA combines with protein in the cytosol and forms theribosome. The ribosome then has the ability to bind mRNA and carry out proteinsynthesis.

Protein synthesis is the process that cells use to build proteins. It begins in thenucleus with RNA transcription. mRNA, as discussed above, leaves the nucleus toenter the process of translation. Translation is defined as the decoding of mRNA by theribosome to produce a specific amino acid chain. As in transcription, translationproceeds in distinct phases. These are activation, initiation, elongation and termination.

Activation attaches the correct amino acid to the correct tRNA. tRNA carries the aminoacid to the mRNA and attaches to the correct codon. Initiation involves the smallsubunit of the ribosome, shown in Figure 6.15, binding to the 5 end of the mRNA. Thisinitiates translation of the mRNA sequence. Elongation is just what it sounds like.

Amino acids are added to the growing chain. The additions occur until the end of themRNA is reached. At this point, termination occurs. tRNA does not recognize the stopcodon sequence and the ribosome/mRNA complex disassembles. At this point, theamino acid chain is transported to the internal space of the ER and folded into an activeprotein. Figure 6.15 shows translation of mRNA by a ribosome.

Figure 6.15 Translation of mRNA, ribosome shown as gray circle surrounding themRNA.

4. DNA Replication and Repair


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Several methods of DNA replication have been postulated. They areconservative, dispersive and semiconservative. Conservative replication maintains thetwo original template DNA strands as a single helix while the newly synthesized strandsform a second helix. In this model, one daughter cell receives the original templateDNA while the other daughter receives the copied DNA. The dispersive methodproduces two copies of DNA, but neither strand is completely composed of template ornew DNA. DNA within the chromosome is composed of a combination of both originalor both new strands. The DNA polymers become mixed and the end results are similarto crossover events that occur during meiosis. The semiconservative method,confirmed by the Meselson-Stahl experiment in 1958, conserves one strand of theoriginal DNA in each replicated helix. That is, as the original helix is broken forreplication, the complimentary nucleotides that form the new DNA strand becomeattached via hydrogen bonds. The final two double strand DNA helices each consist ofone original and one newly synthesized strand of DNA. A comparison of the three

replication methods is shown in Figure 6.15.

Figure 6.15 Replication methods

Now with the basic understanding of semiconservative replication, DNAreplication can be described in more detail. DNA replication is the process by which allliving organisms copy their DNA. In a sense, it is the basis for inheritance between celland organism generations. In a similar fashion to RNA and protein synthesis, DNAreplication occurs in three steps which are initiation, elongation and termination.


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Initiation of DNA replication occurs at specific sites within the DNA. These sitesare called replication origins, or origins of bidirectional replication. As the name implies,replication occurs in opposing directions. Two sets of DNA replication machinery headout from the origin in opposite directions. Once the replication machinery has beenestablished, new strands of DNA are synthesized at a rate of about 3000 nucleotideadditions per minute. Even with this speed, multiple origins are needed to allowcomplete replication within the time allotted for S phase. If each chromosome containedonly one replication origin, approximately 2000 hours would be needed replicate theentire genome. Clearly, 2000 hours greatly exceeds the time reserved to complete theS phase of the cell cycle. DNA at the origin contains specific sequences that allowreplication proteins to attach to the DNA. The initiator proteins recruit other proteins,such as DNA helicase, to separate the DNA. A family of DNA helicases are responsiblefor breaking hydrogen bonds between nucleotides and unwinding the DNA helix. At thispoint, a replication bubble has formed. DNA replication moves from initiation into

elongation.During elongation, DNA synthesis extends new DNA polymers in a 5 to 3

direction. The purpose of the directionality is the need to attach new nucleotides to the3 hydroxyl on the primer strand. DNA polymerases responsible for elongation do nothave 3 to 5 synthesis activity, so they do not recognize 5 phosphate groups. DNApolymerases are the key players in elongation. DNA polymerases are a family ofenzymes that carry out DNA replication. However, they do not attach directly to DNAtemplates and require an existing DNA strand paired with the template. Small strandsof RNA, called primers, are created and attached to the DNA template. DNA

polymerases are then able to synthesize the new strand of DNA. The new DNA strandis extended in the 3 direction with the addition of complementary nucleotides.

Elongation of DNA requires a special set of proteins referred to as replicationmachinery. These proteins include DNA topoisomerases and single strand bindingproteins. DNA topoisomerases are responsible for nicking a single strand of DNA whichallows the strands the ability to swivel around each other. Strand nicking removes thebuild-up of DNA twists during replication. In addition to nicking a single strand,topoisomerases cut both backbones, a double strand cut, that enables one strand ofDNA to pass through another. The double cut removes knots and entanglements that

can form during replication. The role of the single strand binding proteins (SSBP) is justas their name implies. They bind to single stranded DNA until the second strand issynthesized. By attaching to single strand DNA, the SSBP prevents secondarystructure formation within the DNA. When the second strand is complete, the SSBPreleases the DNA and hydrogen bonds are formed to hold the DNA helix together.Once the replication proteins and enzymes have gathered together and attached toDNA, they are called a replication fork. The replication fork forms in the nucleus and


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only during DNA replication. It is responsible for breaking the hydrogen bonds that holdthe two DNA strands together. The replication fork moves along the chromosome in a3 to 5 direction. Seen in Figure 6.16 is the basic structure of the replication fork.

Figure 6.16 Replication fork (Pollard and Earnshaw, Figure 45.1).

As briefly mentioned, the replication fork moves in a 3 to 5 direction duringreplication. The movement is along the leading strand of DNA. The leading strand isdefined as the DNA template that is synthesized in a 5 to 3 direction. Synthesis of thenew strand is complementary to the movement of the fork and DNA polymerase activity.In a sense, DNA polymerase is able to read the template strand and add nucleotidesto the new strand continuously. However, there is the problem on the second strand ofthe DNA helix. The replication machinery reads it in a 5 to 3 direction, opposite to theactivity of DNA polymerase.

As the template DNA is being unwound in a 5 to 3 direction, the second strandof DNA becomes synthesized in short, non-continuous segments. This strand is calledthe lagging strand. The lagging strand is characterized by growth in the oppositedirection to the movement of the replication fork. On the lagging strand, DNApolymerase reads the DNA in short, separated segments. The RNA primer is placedat the beginning of each segment, unlike the leading strand which needs only oneprimer. As each segment is synthesized, another primer is placed at the replication forkto enable synthesis of another segment of DNA. Some of the discontinuities are causedby the replication fork itself. Replication machinery takes up room and the RNA primeris not able to attach to the DNA. Once the replication fork opens another section ofDNA, the primer binds the template and synthesis continues. The short DNA fragmentsformed on the lagging strand are called Okazaki fragments. The Okazaki fragments are

joined together by DNA ligase. DNA ligase is an enzyme that repairs single strandeddiscontinuities. Shown briefly in Figure 6.17 are leading and lagging strands ofsynthesized DNA, however, only one lagging strand is shown. For a computeranimated video of DNA replication, please go to http://dnalc.org/resources/3d/04-mechanism-of-replication-advanced.html .


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Figure 6.17 Continuous vs discontinuous replication (Snustad andSimmons, Figure 10.1).

DNA can be damaged by any number of factors, including normal metabolicactivities and environmental radiation. This damage can consist of individual basedamage or DNA structural damage. In any given day, DNA in human cells can beexposed to 1 million molecular lesions. These lesions can cause structural damage tothe DNA which can interfere with transcription of genes within the genome. Two basictypes of damage that will be discussed here are single-strand breaks (SSB) and double-strand breaks (DSB). In the case of SSB, one strand of the DNA helix is severed;however, the two DNA strands have not separated from each other. DSB cleaves bothDNA strands and results in two free ends of DNA. DSB is the most hazardous to thecell due to an increased possibility of genomic rearrangements. DNA damage is shownin Figure 6.18 as base damage. This can easily be translated to a break within the DNAstrand.


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(a) (b)

Figure 6.18 (a) Single-strand and (b) double-strand damage

Repair of DNA damage begins with the identification of damage. As previouslydiscussed, there are two DNA checkpoints within the cell cycle. These checkpoints are

used to pause the cell cycle and allow repair of DNA. The primary repair method of aSSB uses the same proteins used by the base excision repair (BER) mechanism. BER,briefly, is the repair of a single damaged base. The base can be damaged by oxidationor hydrolysis, among others. BER removes the damaged base and replaces themissing nucleotide with the assistance of DNA polymerase. The DNA polymeraseactivity is similar to that used during DNA synthesis. DNA ligase acts on the new baseto seal the nick in the DNA strand. For repair of a single-strand break, BER skips to thefinal step of ligation. The broken phosphodiester bonds are reformed and the DNAstrand break is repaired.

In addition to single-strand breaks, DNA can be damaged with double-strandbreaks. Double-strand breaks severe both DNA strands at the same location resultingin two free ends of DNA. The DNA ends at the DSB typically have short single-strandedsequences that serve as microhomologies within the break. In order to repair damage,the cell has a choice of two pathways. They are non-homologous end joining (NHEJ)and homologous recombination (HR). NHEJ is called non-homologous because itdoes not require a homologous template for repair. The broken ends of DNA are ligatedtogether using the microhomologies found in the DNA overhangs for alignment. NHEJrepair is mostly accurate, but can have some imprecision that leads to the loss ofnucleotides. However, this is only seen with DNA overhangs that are not compatible.The other method of repair is homologous recombination (HR). HR does not rely onshort microhomologies. Instead, the repair proteins search out homologous DNA or thesister chromosome. The damaged DNA is resected. Resection is the process by whichDNA surrounding the damage is removed from the 5 end of the break. Once resectionis complete, strand invasion occurs. Strand invasion takes the 3 end of the brokenDNA and invades the homologous DNA. The lost DNA is then synthesized and the


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chromosomes separate when repair of the DSB has been completed. The first steps ofHR are shown in Figure 6.19

Figure 6.19 First steps of Homologous Recombination

In addition to DNA damage, replication has its own set of inherent errors.Replication slippage is the most common cause of error. During replication, DNApolymerase is responsible for coping DNA. DNA polymerase moves at a speedcomparable to the replication fork. However, on the lagging strand, DNA polymerasepauses between Okazaki fragments. The pause in replication can cause the

polymerase to dissociate from the DNA and lead to replication slippage. Replicationslippage occurs in regions of DNA that have short, repeated sequence. The dissociatedpolymerase leads to two types of error. The first is deletion of DNA. A genetic deletionis defined as a mutation in which part of the DNA sequence is missing. The deletioncan range from a single base to an entire piece of a chromosome. When DNA strandsbecome misaligned, replication can skip over a section of DNA. In the cause of p53 , adeletion of part of the gene results in the development of Li-Fraumeni syndrome. Inaddition to deletions, replication slippage can result in insertions. An insertion is theaddition of one or more nucleotides into the DNA sequence. One way in whichinsertions are created is multiple replications of the same DNA. Insertions can cause

frameshift mutations within the DNA sequence, if the number of nucleotides is notdivisible by three. Frameshift mutations alter the normal reading frame of a gene andthe amino acids encoded for by the gene.


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5. Genome

Maintaining genome integrity is just as important as maintaining genetic accuracyduring replication. The genome, on average, is bombarded with up to a million, or 10 6,DNA breaks and lesions per day. Repair, as previously discussed, ensures the damageis fixed and genomic integrity remains intact. In addition to repair of damage, a series ofcheckpoints are in the cell cycle as quality control mechanisms. Mentioned previously,the two DNA checkpoints within the cell cycle are located in the G1 and G2 phases.The G1 checkpoint looks at the DNA to locate lesions and, more importantly, breaks. Ifa single-stranded nick were to be replicated, the DNA nick would become a full-fledgeddouble-strand break. The break then causes instability within the genome. By pausingthe cell cycle, the nick is repaired and replication occurs without further damaging theDNA. In the case of the G2 checkpoint, DNA is reviewed for mismatched nucleotidesand unreplicated DNA. The cell cycle is, again, paused and repairs are made. Withoutthese vital checkpoints, integrity of the genome is compromised and mutations are morelikely to occur.

Genome instability is a process of chromosomal alterations that can lead to awide variety of problems. Instability leads to deletion or insertion of DNA, mentionedabove, or a change in chromosome number, among other changes. During cellseparation, more specifically meiosis, unequal separation of chromosomes can occur.When one daughter cell receives more than one copy of a chromosome, the cell is saidto have aneuploidy. Aneuploidy is defined as an abnormal number of chromosomes ina cell. The chromosomes themselves have not been altered, but the number is eitherhigher or lower than expected. A common occurance of aneuploidy in humans involveschromosome 21. The normal copy number for any chromosome in a human is two.When three copies of chromosome 21 appear in a developing fetus, the physicalmanifestation is Down Syndrome. In addition to aneuploidy, chromosomal structurecan be affected by instability.

Instability in the chromosomal structure can lead to rearrangements andduplications. Rearrangements most commonly occur between non-homologouschromosomes. One of the key characteristics of chromosome rearrangement is thefusion of two genes. Two separate genes cannot function properly when joinedtogether and are common in cancer cells. Duplication of DNA is just that, a secondunneeded copy. As discussed above, DNA duplication, not related to replication, cancause insertions that affect the structure of proteins.

A separate type of DNA alteration does not affect the structure or chromosomalidentity of the DNA itself. The DNA sequence can be altered in a superficial mannerthat does not affect gene expression to the same extent of DNA deletion or gene fusion.Epigenetics is the study of gene expression alterations caused by mechanisms other


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than the underlying DNA sequence. The Greek prefix epi - of epigenetics refers to afeature that is on top of or in addition to genetics. These features cause changesthat affect the phenotype of a cell without altering the genotype. A cell phenotype is anobservable trait that has some influence over the development of an organism. Thesetraits include characteristics such as morphology and behavior. The phenotype of acell, or organism, depends on the genes that are expressed. The genotype of a cell isthe genetic code. The genome of an organism contains the information used for geneexpression. Epigenetic events alter the expression of genes within the genome. Whenthis occurs, suppressed genes can be activated or expressed genes can be silenced.

These epigenetic changes, however, do not represent changes in the geneticinformation. In Figure 6.21, a brief explanation of epigenetics and possible outcome isshown. Epigenetics can be split into two general classes based on mechanism:

Figure 6.21 Some epigenetic mechanisms and its consequences.DNA modification by covalent attachment of a protein or establishing a self-perpetuatingprotein state. DNA modification most commonly affects only one of the two alleles. Asmentioned in Figure 6. , DNA methylation can activate or repress genes. The additionof a methyl group causes the genes to become less accessible for transcription leadingto suppression of the gene. However, the methyl group can be removed duringchromatin remodeling. Chromatin remodeling involves moving the nucleosomes to


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ensure proper winding of DNA. The second class of epigenetics involves establishing arecurrent protein state. This might involve maintaining an alternative proteinconformation throughout the life of the cell or modifying a specific protein. However,self-perpetuation of an altered protein becomes difficult during replication. The proteincomplex could divide equally between sister chromatin or remain completely on onechromatin. In the first case, there is no reason for the protein complex to split and thenreconstruct itself. The second case would require complete assembly of a new proteincomplex. The existence of epigenetic effects leads to the belief that proteinsresponsible for such modifications have some sort of self-templating or self-assemblingcapacity.

Before taking the quiz, you should be able to answer the following1. Give a brief description of the beginning of the study of genetics2. Explain the differences between nuclides in DNA and RNA3. Be able to label the general structure of DNA, including hydrogen bonds between

pyrimidines and purines4. # of base pairs in humans5. Describe the differences between genes and codons and how they are related to

each other6. Differentiate between the three methods of DNA repair

7. Discuss potential ramifications of faulty repair8. Describe the principle behind DNA recombination9. Explain the importance of genomic stability and why genomic instability can

potentially lead to tumorgenesis10. Have a general understanding of epigenetics and its ramifications

Module 6 Lecture radiobiology

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