Genetics Exam 3

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Genetics Exam 3
2015-10-28 23:13:19
Genetics Exam
Genetics Exam 3
Genetics Exam 3
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  1. What are the 4 requirements for Genetic Material?
    • 1. Must encode large amounts of information
    • 2. Must be able to replicate faithfully
    • 3. Must encode the phenotype
    • 4. Must have variability
  2. Why was the discovery of the structure of DNA so important for understanding genetics?
  3. What was initially proposed to be the source of Genetic Material? Why?
    Protein, as nucleic acids were known to be very simple molecules it seemed unlikely that they could carry the required diversity that the 20 amino acids could allow for
  4. Who discovered that a DNA monomer (subunit) consisting of a phosphate, sugar, and nitrogenous base?
  5. Who proposed the Tetranucleotide Theory, and what did this theory state?
    • Levene
    • Tetranucleotide Theory implied that the structure of DNA is not variable enough to be the genetic material, as protein (with its 20 amino acids) could supposedly produce much more variation 
    • It suggested that a single Nucleic Acid subunit contained all 4 bases (adenine, guanine, cytosine, and thymine)
  6. Who discovered that the ratio of nitrogenous bases varies amongst species? What is a special property of these ratios? What were these properties collective known as?
    • Chargaff discovered that DNA from different organisms greatly varied in base composition, as well as within each species there was some regularity to the ratios of the bases (A%=T%, and G%=C%)
    • A/T or G/C should be roughly 1.0
    • (A+G)/(T+C) should equal 1.0 as well
    • Collectively this was known as Chargaff's Rule
  7. Whose experiments demonstrated that genetic material could be transferred between bacterial cells? What was this process called?
    • Griffith
    • This process was termed Genetic Transformation 
  8. Levene made which contribution of our understanding of DNA structure?

    A. Determined that the nucleus contains DNA.
    B. Determined that DNA consists of nucleotides.
    C. Determined that the nucleotide bases of DNA are present in regular ratios.
    D. Determined that DNA contains four nitrogenous bases.
    B. Determined that DNA consists of nucleotides.
    (this multiple choice question has been scrambled)
  9. Who demonstrated that the transforming substance was DNA and NOT protein?
    Avery, McLeod and McCarty
  10. How did Avery, McLeod and McCarty demonstrate that the transforming substance was DNA and NOT protein?
    • They used heat to kill/lyse bacteria and extract their material. This material was then treated with enzymes such as Trypsin or Chymotrypsin (both known to break down proteins) to see their effects. Trypsin and Chymotrypsin had no effects.
    • They also treated the samples with Ribonuclease, and enzyme that destroys RNA, but also had no effect.
    • Lastly they treated the sample with DNase, which destroys DNA, and these samples showed the effect 

    The effect was to look for transformed IIIS bacteria. If the bacteria could transform, it means the enzyme did not destroy the genetic material. If it did, the genetic material is destroyed and thus transformation can not occur.
  11. If Avery, MacLeod, and McCarty had found that samples of heat-killed bacteria treated with RNase and DNase transformed bacteria, but samples treated with protease did not, what conclusion would they have made?

    A. Protease carries out transformation.
    B. RNA and DNA are the genetic materials.
    C. Protein is the genetic material.
    D. RNase and DNase are necessary for transformation.
    C. Protein is the genetic material.
    (this multiple choice question has been scrambled)
  12. Describe the findings of the Hershey-Chase experiement
    Hershey & Chase labeled bacteriophages (viruses that infect bacteria) with radioactive isotopes to conclusively show that the genetic material transmitted to phage progeny was DNA and not protein
  13. How did Hershey-Chase differentiate (follow) between DNA and protein? How/why were they able to do this.
    • They used 32P to follow phage DNA during reproduction.
    • Protein contains sulfur but not phosphorus, so they were able to use 35S to follow the protein
  14. Could Hershey and Chase have used a radioactive isotope of carbon instead of 32P? Why or why not?
    No, carbon is found in both nucleic acids and protein. They would have been unable to distinguish between the two.
  15. What did Watson and Crick use to help solve the structure of DNA?

    a. X-ray diffraction.
    c. Models of DNA
    b. Laws of structural chemistry
    d. All the above
    d. All the above
  16. Franklin and Wilkins used X-ray diffraction techniques to study crystals of DNA. Their images suggested a ______ arrangement of the atoms with the nitrogenous bases extending _____. from backbone of molecule.
    Franklin and Wilkins used X-ray diffraction techniques to study crystals of DNA. Their images suggested a helical arrangement of the atoms with the nitrogenous bases extending 90o. from backbone of molecule.
  17. Detail what the Primary, Secondary, and Tertiary structures of DNA describe.
    • Primary: refers to its nucleotide structure and how the nucleotides are joined together
    • Secondary: refers to DNA's three dimensional configuration, and the helical structure 
    • Tertiary: refers to the complex packing arrangements of double-stranded DNA
  18. Detail the difference between a Ribose Sugar and a Deoxyribose Sugar, and what this difference causes
    • Ribose: RNA's sugar, has a -OH group attached to the 2'-carbon atom
    • Deoxyribose: has a hydrogen atom (-H) attached to the 2'-carbon atom

    • This difference makes Ribose more unstable (the additional Oxygen from -OH makes it more reactive), thus making DNA the more likely molecule to carry genetic information.
    • Enzymes catalyzing DNA replication and RNA transcription recognize the chemical difference at the 2’ carbon
  19. In a nucleotide, the nitrogenous base always forms a covalent bond with the ______. atom of the sugar
    In a nucleotide, the nitrogenous base always forms a covalent bond with the 1'-Carbon atom of the sugar
  20. What makes up a Nucleoside? What makes up a Nucelotide?
    • Nucleoside: a Sugar (ribose or deoxyribose) + a Nitrogeneous Base 
    • Nucelotide: a Nucleoside + a Phosphate 

  21. Describe the structure of a Nucleotide
    • Consists of a sugar, phosphate, and nitrogenous base
    • Sugar has 5 carbons, ribose and deoxyribose differ by what is attached to the 2'-carbon of the sugar
    • Bases are bound to the 1'-carbon of the sugar
    • Phosphate is bound to the 5'-carbon of the sugar

  22. What are the proper names of a DNA or RNA Nucleotide?
    • DNA nucleotide proper name: deoxyribonucleotide or deoxyribonucleoside 5’-monophosphate
    • RNA nucleotide proper name: ribonucleoside 5’-monophosphate
  23. How do the sugars of RNA and DNA differ?

    A. DNA’s sugar has a phosphorus atom; RNA’s sugar does not.
    B. RNA contains uracil; DNA contains thymine.
    C. RNA has a six-carbon sugar; DNA has a five-carbon sugar.
    D. The sugar of RNA has a hydroxyl group that is not found in the sugar of DNA.
    D. The sugar of RNA has a hydroxyl group that is not found in the sugar of DNA.
    (this multiple choice question has been scrambled)
  24. How are strands of Nucleotides joined together?
    • DNA is made up of many nucleotides connected by covalent bonds, which join the 5′-phosphate group of one nucleotide to the 3′-carbon atom of the next nucleotide
    • These bonds are called Phosphodiester Linkages
  25. Describe the structure of the backbone of DNA
    The backbone of the polynucleotide strand is composed of alternating sugars and phosphates; the bases project away from the long axis of the strand
  26. What does "Polarity" of DNA refer to?
    At one end of the strand, a free (meaning that it’s unattached on one side) phosphate group is attached to the 5′-carbon atom of the sugar in the nucleotide. This end of the strand is therefore referred to as the 5′ end. The other end of the strand, referred to as the 3′ end, has a free OH group attached to the 3′-carbon atom of the sugar.

  27. The antiparallel nature of DNA refers to

    A. its charged phosphate groups.
    B. the pairing of bases on one strand with bases on the other strand.
    C. the formation of hydrogen bonds between bases from opposite strands.
    D. the opposite direction of the two strands of nucleotides.
    D. the opposite direction of the two strands of nucleotides.
    (this multiple choice question has been scrambled)
  28. _____ hydrogen bonds form between A and T bases, while ____ hydrogen bonds form between C and G bases.
    Two hydrogen bonds form between A and T bases, while three hydrogen bonds form between C and G bases (this property means a greater CG% means higher DNA strength)
  29. Describe the structure and characteristics of B-DNA
    • It is the most common secondary structure of DNA
    • Exists when plenty of water surrounds the molecule and there is no unusual base sequence in the DNA
    • It is an Alpha-Helix, meaning it is a right-handed helix (clockwise)
    • Each complete turn consists of 10 nucleotides or roughly 3.4 nm 
    • The spiraling of the nucleotide strands creates major and minor grooves in the helix, features that are important for the binding of some proteins that regulate the expression of genetic information
  30. Describe the secondary structure A-DNA
    • Exists if less water is present.
    • A-DNA is an alpha (right-handed) helix
    • It is shorter and wider than B-DNA
    • Its bases are tilted away from the main axis of the molecule
    • There is little evidence that A-DNA exists under physiological conditions
  31. Describe the secondary structure Z-DNA
    • It is a radically different secondary structure arising from particular nucleotide sequences (e.g. stretches of alternating G & C)
    • Forms a left-handed helix
    • Z-DNA-specific antibodies bind to regions of the DNA that are being transcribed into RNA, suggesting that Z-DNA may play some role in gene expression.
  32. How does Z-DNA differ from B-DNA?
    Z-DNA has a left-handed helix; B-DNA has a right-handed helix. The sugar–phosphate backbone of Z-DNA zigzags back and forth, whereas the sugar–phosphate backbone of B-DNA forms a smooth continuous ribbon.
  33. Is RNA always single stranded? If no, give an example/description of a circumstance that shows double-stranded RNA.
    No, sequences within a single strand of nucleotides may be complementary to each other and can pair by forming hydrogen bonds, producing special double-stranded regions

    • One common type of secondary structure found in single strands of nucleotides is a Hairpin, which forms when sequences of nucleotides on the same strand are inverted complements
    • A hairpin consists of a region of paired bases (the stem) and sometimes includes intervening unpaired bases (the loop).

  34. The function of RNA depends on its _____.
    The function of RNA depends on its Secondary Structure.
  35. Hairpins are formed in DNA as a result of

    A. sequences on the same strand that are identical.
    B. sequences on the same strand that are inverted and complementary.
    C. sequences on the opposite strand that are complements.
    D. sequences on the opposite strand that are identical.
    B. sequences on the same strand that are inverted and complementary.
    (this multiple choice question has been scrambled)
  36. What is H-DNA?
    H-DNA is a triple stranded DNA molecule typically found in sequences containing only purines or only pyrimidines
  37. What is DNA Methylation and what a purpose of it for Bacteria and Eukaryotes?
    DNA Methylation: a process of primary DNA structure modification in which methyl groups (–CH3) are added (by specific enzymes) to certain positions on the nucleotide bases

    • Bacteria: DNA is frequently methylated to distinguish it from foreign, unmethylated DNA that may be introduced by viruses; bacteria use proteins called restriction enzymes to cut up any unmethylated viral DNA
    • Eukarya: methylation is often related to gene expression; Sequences that are methylated typically show low levels of transcription while sequences lacking methylation are actively being transcribed. Methylated genes are flagged for either increased or decreased levels of transcription
  38. In most animal cells, ~5% of all _____ are methylated.

    A. Thymines
    B. Cytosines
    C. Adenines
    D. Guanines
    B. Cytosines
    (this multiple choice question has been scrambled)
  39. For DNA transcription to occur, what must the chromatin structure do?
    For DNA to be transcribed, chromatin structure must relax to allow access for the transcriptional enzymes.
  40. If attraction between histones and DNA decreases, then chromatin packing decreases meaning ____
    it is now more relaxed, permitting transcription factors to bind to DNA.
  41. How can you decrease the attraction between histones and DNA?
    Make histones less positive
  42. What are the three methods one can change histone attraction to DNA?
    • Acetyltransferase: Enzymes that attach acetyl groups to lysine tails of histones, makes histones less positive, increasing transcription
    • DNA Methylation: DNA methylation can alter chromatin structure to inhibit binding of transcription factors, decreasing transcription
    • Add variant histones to nucleosome 
  43. What does Epigenetic Changes refer to?
    Stable alterations of chromatin structure that may be passed on to cells or individual organisms
  44. Chromatin remodeling can help determine _____. (in regards to epigenetics)
    Chromatin remodeling can help determine which genes are expressed where and when, and which genes are not
  45. True or False: In Epigenetic Changes DNA sequences do not change, but their expression can change, leading to changes in phenotype. This is an irreversible process

    DNA sequences do not change, but their expression can change, leading to changes in phenotype. This is a reversible process though.
  46. What is a Centromere?
    • The centromere is a constricted region of the chromosome to which spindle fibers attach and is essential for proper chromosome movement in mitosis and meiosis
    • It is mostly heterochromatin (thus no genes)
    • Defined NOT by the DNA sequence but by nucleosomes containing a variant histone called cenH3 that alters chromatin structure to promote formation of kinetochore
  47. What are Telomeres?
    Telomeres are the natural ends of a chromosome that serve as a cap that stabilizes chromosome end so it is not degraded and provide a means for replicating the end of a chromosome
  48. What are Telomeric Sequences?
    • Repeated units of a series of adenine or thymine nucleotides followed by several guanine nucleotides, taking the form 5′-(A or T)mGn-3′, where m is from 1 to 4 and n is 2 or more
    • In humans, 5’-TTAGGG-3’ repeated 100’s-1000’s X
    • G-rich strand extends beyond C-rich strand as a 3’ overhang
    • Proteins bind to the overhang to help prevent degradation and prevent ends from sticking together
    • In some cells, the G-rich 3’ overhang can loop back to pair with a short stretch of DNA (in a triplex) to form a T-loop to protect DNA end from degradation
  49. _______ prevents Telomeric ends from being repaired as a double stranded DNA break
    Shelterin prevents Telomeric ends from being repaired as a double stranded DNA break
  50. How does supercoiling arise? What is the difference between positive and negative supercoiling?
    Supercoiling arises from topoisomerases catalyzing the overwinding (positive supercoiling) or underwinding (negative supercoiling) of the DNA double helix.

    Supercoiling may occur: (1) when the DNA molecule does not have free ends, as in circular DNA molecules, or (2) when the ends of the DNA molecule are bound to proteins that prevent them from rotating about each other, as in linear eukaryotic chromosomes.
  51. What functions does supercoiling serve for the cell?
    Supercoiling compacts the DNA. Negative supercoiling helps to unwind the DNA duplex for replication and transcription.
  52. Describe the composition and structure of the nucleosome.
    The nucleosome core particle contains two molecules each of histones H2A, H2B, H3, and H4, which form a protein core with 145–147 bp of DNA wound around the core. Chromatosomes contain the nucleosome core with a molecule of histone H1.
  53. Describe in steps how the double helix of DNA, which is 2 nm in width, gives rise to a chromosome that is 700 nm in width.
    DNA is first packaged into nucleosomes; the nucleosomes are packed to form a 30-nm fiber. The 30-nm fiber forms a series of loops that pack to form a 250 nm fiber, which in turn coils to form a 700-nm chromatid.
  54. What are epigenetic changes and how are they brought about?
    Epigenetic changes are changes in gene expression that are passed on to cells or future generations, but do not involve alteration of the nucleotide sequence. Epigenetic changes are brought about by altering DNA structure, such as methylation of the DNA, or altering chromatin structure by modifying histones.
  55. Describe the function of the centromere. How are centromeres different from other regions of the chromosome?
    Centromeres are the points at which spindle fibers attach to the chromosome. They are necessary for proper segregation of the chromosomes in mitosis and meiosis. Most eukaryotic centromeres are characterized by heterochromatin consisting of highly repetitive DNA. Centromeres are thought to exist at specific locations on the chromosome because of epigenetic changes to chromatin structure at those locations. For examples, nucleosomes at centromeres often possess the variant histone CenH3. This special chromatin structure promotes the formation of the kinetochore, to which spindle fibers attach.
  56. Describe the function and molecular structure of a telomere.
    Telomeres are the ends of the linear chromosomes in eukaryotes. They cap and stabilize the ends of the chromosomes to prevent degradation by exonucleases or joining of the ends. Telomeres also enable replication of the ends of the chromosome by an enzyme called telomerase. Telomeric DNA sequences consist of repeats of a simple sequence, usually in the form of 5′Cn(A/T)m
  57. What is the difference between euchromatin and heterochromatin?
    • Euchromatin undergoes regular cycles of condensation during mitosis and decondensation during interphase
    • Heterochromatin remains highly condensed throughout the cell cycle, except transiently during replication.
    • Nearly all transcription takes place in euchromatic regions, with little or no transcription of within heterochromatin.
  58. Compare and contrast prokaryotic and eukaryotic chromosomes. How are they alike and how do they differ?
    Prokaryotic chromosomes are usually circular, whereas eukaryotic chromosomes are linear

    Prokaryotic chromosomes generally contain the entire genome, whereas each eukaryotic chromosome has only a portion of the genome; The eukaryotic genome is divided into multiple chromosomes.

    Prokaryotic chromosomes are generally smaller and have only a single origin of DNA replication. Eukaryotic chromosomes are often many times larger than prokaryotic chromosomes and contain multiple origins of DNA replication.

    Prokaryotic chromosomes are typically condensed into nucleoids, which have loops of DNA compacted into a dense body. Eukaryotic chromosomes contain DNA packaged into nucleosomes, which are further coiled and packaged into successively higher-order structures. The condensation state of eukaryotic chromosomes varies with the cell cycle.
  59. In a typical eukaryotic cell, would you expect to find more molecules of the H1 histone or more molecules of the H2A histone? Explain your reasoning
    Because each nucleosome contains two molecules of histone H2A and only one molecule of histone H1 is associated with each nucleosome, eukaryotic cells will have more H2A than H1.
  60. In a Eukaryotic cell, would you expect to find more molecules of H2A or more molecules of H3? Explain your reasoning.
    Because each nucleosome contains two molecules of H2A and two molecules of H3, eukaryotic cells should have equal amounts of these two histones.
  61. Suppose a chemist develops a new drug that neutralizes the positive charges on the tails of histone proteins. What would be the most likely effect of this new drug on chromatin structure? Would this drug have any effect on gene expression? Explain your answers.
    Such a drug would disrupt the ionic interactions between the histone tails and the phosphate backbone of DNA and thereby cause a loosening of the DNA from the nucleosome. The drug may mimic the effects of histone acetylation, which neutralizes the positively charged lysine residues. Changes in chromatin structure would result from the altered nucleosome-DNA packing and possible changes in interaction with other chromatin modifying enzymes and proteins. Changes in transcription would result because DNA may be more accessible to transcription factors.
  62. Describe Conservative Replication
    In conservative replication, the entire double-stranded DNA molecule serves as a template for a whole new molecule of DNA, and the original DNA molecule is fully conserved during replication
  63. Describe Dispersive Replication
    In dispersive replication, both nucleotide strands break down (disperse) into fragments, which serve as templates for the synthesis of new DNA fragments, and then somehow reassemble into two complete DNA molecules. In this model, each resulting DNA molecule is interspersed with fragments of old and new DNA; none of the original molecule is conserved
  64. Describe Semiconservative Replicaiton
    Semiconservative replication is intermediate between these two models; the two nucleotide strands unwind and each serves as a template for a new DNA molecule.
  65. How did Meselson and Stahl demonstrate that replication in E. coli takes place in a semiconservative manner?
    Meselson and Stahl grew E. coli cells in a medium containing the heavy isotope of nitrogen (15N) for several generations. The 15N was incorporated in the DNA of the E. coli cells. The E. coli cells were then switched to a medium containing the common form of nitrogen (14N) and allowed to proceed through a few cycles of cellular generations. Samples of the bacteria were removed at each cellular generation. Using equilibrium density gradient centrifugation, Meselson and Stahl were able to distinguish DNAs that contained only 15N from DNAs that contained only 14N or a mixture of 15N and 14N because DNAs containing the 15N isotope are “heavier.” The more 15N a DNA molecule contains, the further it will sediment during equilibrium density gradient centrifugation. DNA from cells grown in the 15N medium produced only a single band at the expected position during centrifugation. After one round of replication in the 14N medium, one band was present following centrifugation, but the band was located at a position intermediate to that of a DNA band containing only 15N and a DNA band containing only 14N. After two rounds of replication, two bands of DNA were present. One band was located at a position intermediate to that of a DNA band containing only 15N and a DNA band containing only 14N, while the other band was at a position expected for DNA containing only 14N. These results were consistent with the predictions of semiconservative replication and incompatible with the predictions of conservative and dispersive replication.
  66. Bacteria have a single origin of replication called the _____, which is the DNA sequence where replication starts .
    Bacteria have a single origin of replication called the ori, which is the DNA sequence where replication starts .
  67. What do Initiator Proteins do?
    • Binds to oriC and causes a short section of DNA to unwind. This unwinding allows helicase and other single-strand-binding proteins to attach to the polynucleotide strand
  68. What is Thera Replication?
    • In theta replication, double-stranded DNA begins to unwind at the replication origin, producing single-stranded nucleotide strands that then serve as templates on which new DNA can be synthesized.
    • The unwinding of the double helix generates a loop, termed a replication bubble.
    • Unwinding may be at one or both ends of the bubble, making it progressively larger.
    • DNA replication on both of the template strands is simultaneous and bidirectional with unwinding. The point of unwinding, where the two single nucleotide strands separate from the double-stranded DNA helix, is called a replication fork.
  69. What are the requirements for Replication?
    • 1. a template consisting of single-stranded DNA
    • 2. raw materials (substrates) to be assembled into a new nucleotide strand
    • 3. enzymes and other proteins that “read” the template and assemble the substrates into a DNA molecule.
  70. What is the function of DNA Polymerase in Replication?
    • The enzymes that synthesize DNA
    • Can add nucleotides only to the free 3′-OH end of the growing strand (not the 5′ end), and so new DNA strands always elongate in the same 5′-to- 3′ direction (5′→3′)
    • Catalyzes the formation of a phosphodiester bonds by joining the 5’-phosphate group of the incoming nucleotide to the 3’-OH group of the last nucleotide on the growing strand.
  71. What does it mean that DNA replication is Antiparallel?
    Because the two single-stranded DNA templates are antiparallel and strand elongation is always 5′→3′, if synthesis on one template proceeds from, say, right to left, then synthesis on the other template must proceed in the opposite direction, from left to right As DNA unwinds during replication, the antiparallel nature of the two DNA strands means that one template is exposed in the 5′→3′ direction and the other template is exposed in the 3′→5′ direction
  72. Differentiate between the Lagging and Leading Strands
    • Lagging: New strand synthesized continuously in the same direction as the movement of the replication fork (5′→3′ direction)
    • Leading: New strand synthesized in discontinuous bursts in the direction opposite to the movement of the replication fork (3′→5′)
  73. Describe the DNA Replication Enzyme: DNA Helicase
    • A DNA helicase breaks the hydrogen bonds that exist between the bases of the two nucleotide strands of a DNA molecule (at the fork)
    • Helicase cannot initiate the unwinding of double-stranded DNA; the initiator protein first separates DNA strands at the origin, providing a short stretch of single-stranded DNA to which a helicase binds.
    • Helicase binds to the lagging-strand template at each replication fork and moves in the 5′→3′ direction along this strand, thus also moving the replication fork
  74. Describe the DNA Replication Enzyme: Single-Stranded Binding Proteins
    • After DNA has been unwound by helicase, single-strand-binding proteins (SSBs) attach tightly to the exposed single-stranded DNA
    • These proteins protect the single-stranded nucleotide chains and prevent the formation of secondary structures such as hairpins that interfere with replication.
  75. Describe the DNA Replication Enzyme: DNA Gyrase
    • Another protein essential for the unwinding process is the enzyme DNA gyrase, a topoisomerase
    • In replication, DNA gyrase reduces the torsional strain (torque) that builds up ahead of the replication fork as a result of unwinding by making a double-stranded break in one segment of the DNA helix, passing another segment of the helix through the break, and then resealing the broken ends of the DNA. This action removes a twist in the DNA and reduces the supercoiling.
  76. Describe the DNA Replication Enzyme: DNA Polymerase
    • Adds new DNA nucleotide to the 3’-OH end of the growing daughter strand
    • There are many variations
  77. Describe the DNA Replication Enzyme: DNA Ligase
    Joins Okazaki fragments by sealing nicks in the sugar–phosphate backbone of newly synthesized DNA
  78. Describe the DNA Replication Enzyme: DNA Primase
    Synthesizes a short RNA primer to provide a 3′-OH group for the attachment of DNA nucleotides
  79. What are the three steps in DNA Replicaiton
    • 1) Initiation: Use of mainly Initiator proteins, Helicase, and SS Binding proteins. Eukaryotic replication is similar except for linear nature of the chromosome, multiple origins of replication, and differences in the DNA polymerases.
    • 2) Unwinding: Mainly use of Gyrase, Helicase, and SSBPs. Must take place to provide a Single-Stranded Template 
    • 3) Elongation: Mainly use of Polymerase and Primase. DNA polymerase III adds DNA nucleotides BUT can only add a new nucleotide to a free –OH on the 3’ end of a sugar. Primase synthesizes a short strand of RNA nucleotides, resulting in a primer available to provide the 3’-OH for DNA polymerase III
  80. What special activity does DNA Polymerase I exhibit?
    DNA Polymerase I has 5’→3’ exonuclease activity, which removes RNA nucleotides from preceding Okazaki fragment one at a time, and its 5’→3’ polymerization activity which replaces each RNA nucleotide with a new DNA nucleotide
  81. What issue occurs after an RNA Primer is replaced with DNA nucleotides?
    After the last nucleotide of the RNA primer has been replaced, a nick remains in the sugar–phosphate backbone of the strand. DNA ligase seals this nick with a phosphodiester bond between the 5’-P group of the initial nucleotide added by DNA polymerase III and the 3’-OH group of the final nucleotide added by DNA polymerase I.
  82. How does the error rate of DNA Replication remain so low?
    • 1) Any error in DNA polymerase III base-pairing is immediately corrected using its 3’ to 5’ exonuclease activity (backwards removal) and repair in a process called proofreading
    • 2) Mismatch repair enzymes correct any error that remains after DNA replication is completed. In bacteria, original DNA is methylated, so repair enzymes know which strand to fix.
  83. What is the issue that arrises in Chromosome End Replication for Eukaryotes?
    • RNA primers must be removed & replaced with DNA which is not a problem for a circular chromosome in bacteria as there will always be a 3’ –OH end. 
    • BUT, not at the end of a lagging strand in a linear chromosome of a Eukaryote. Once DNA polymerase I removes RNA primers, there is no 3’–OH to which it can add a DNA nucleotide Repeated rounds of replication produce progressively shorter DNA molecules. Eventually, gene sequences eroded
  84. How is The End Replication Problem remedied?
    • A telomere has a protruding end with a G-rich repeated sequence.
    • The RNA part of telomerase is complementary to the G-rich strand and pairs with it, providing a template for the synthesis of copies of the repeats.
    • Nucleotides are added to the 3’ end of the G-rich strand.
    • After several nucleotides have been added, the RNA template moves along the DNA.
    • More nucleotides are added, and the Telomere is removed
    • Synthesis takes place on the complementary strand, filling in the gap due to the removal of the RNA primer at the end.
  85. What is Telomerase?
    Telomerase is an enzyme that restores telomere length and is composed of protein and RNA components complementary to G-rich strand that serve as a template for DNA synthesis
  86. Describe how Telomeres impact Somatic or Germ cells
    • Somatic (body) Cells: have a genetically-programmed life span with limits on how many times they can divide before death (apoptosis)
    • Germ line cells (produce gametes): single cell organisms, and proliferative cell types (e.g. intestinal lining, bone marrow) can prevent this shortening:
  87. 90% of all cancer cells have active ______ which allows them divide indefinitely.
    -90% of all cancer cells have active telomerase which allows them divide indefinitely.
  88. What is Werner’s Syndrome?
    • Werner’s Syndrome is a rare genetic disorder leading to accelerated aging
    • Symptoms of old age appear in adolescence – graying, wrinkling, cataract, muscle atrophy
    • Some patients found with abnormally-short telomeres
    • Associated with a defective helicase required for DNA repair as well as telomere replication