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  1. 1. Tommy was a full-term baby but weighed only4.5 pounds (2 kg) at birth. At about 9 months of age,an unusual and persistent rash appeared on his face,and he frequently caught colds and infections. Theillnesses caused no serious problems; so his parentswere not concerned. Throughout childhood, Tommy remained small;by age 18, he was only 4 feet 6 inches (137 cm) inheight. Tommy’s first major health problem aroseshortly after he turned 22—he was diagnosed withintestinal cancer. The tumor was surgically removedbut additional, unrelated tumors appearedspontaneously over the next 10 years. Their appearance startled Tommy’s doctors;the chance of multiple, independent cancers arisingin the same person is generally remote.
  2. 2. The propensity of Tommy’s cells to becomecancerous hinted at a high mutation rate in his genes.Indeed, when pathologists studied Tommy’schromosomes, they observed a wide range ofabnormalities. Tommy had inherited BLOOMSYNDROME. Bloom syndrome is a rare autosomal recessivecondition characterized by short stature, a facialrash induced by sun exposure, a small narrow head,and a predisposition to cancers of all types. The disorder is extremely rare; only severalhundred cases have been reported worldwide. Cellsfrom persons with Bloom syndrome exhibit excessivemutations in all genes, and numerous gaps and breaksoccur in chromosomes that lead to extensive geneticexchange in cell division. Rates of DNA synthesis areretarded.
  3. 3. The characteristics of Bloom syndrome suggestthat its underlying cause is a defect in DNA replication.In 1995, researchers at the New York Blood Centertraced Bloom syndrome to a gene on chromosome 15 thatencodes an enzyme called DNA helicase. A variety ofhelicase enzymes are responsible for unwinding double-stranded DNA during replication and repair. The cells of a person with Bloom syndrome carrytwo mutated copies of the gene and possess little or noactivity for a particular helicase. Normal DNAreplication is disrupted, leading to chromosome breaksand numerous mutations. The genetic damage resultingfrom faulty DNA replication leads to tumors. It is not yet clear whether the basic defect inDNA synthesis is associated with replication or DNArepair or both.
  4. 4. To understand Tommy’s case, we need to answer the following questions: What models of DNA replication exist among life forms? Where is the origin of replication in the DNA strand? What is the direction of replication at this site? How does the chain grow in length? How does the chain terminate? What is the enzymology behind DNA replication? Are there other protein factors that must be present? What is the role of DNA replication in the expression of disease?
  5. 5. MODELS OF DNA REPLICATION These models may differ with respect to theinitiation and progression of replication, but all produce new DNA molecules by semi- conservative replication.
  6. 6. http://highered.mcgraw-THETA REPLICATION: E. coli /dl/free/0072437316/120073/micro03.swf::Bidirectional% 20Replication%20of%20DNA
  7. 7. ROLLING CIRCLE:Viruses and F factor of E. Coli
  9. 9. FUNDAMENTAL RULES OF DNA REPLICATION 1. Replication is semi-conservative Meselson and Stahl convincingly demonstrated that each E. coli DNA strand serves as a template for the synthesis of a new DNA molecule.
  10. 10. 2. Replication begins at an origin- the replication fork
  11. 11. 3. DNA replication is bi-directional, and proceeds in a 5’-3’ direction DNA synthesis takes place simultaneously butin opposite directions on the 2 template strands.
  12. 12. 4. DNA Replication is Semi-discontinuous Replication fork lagging strand Replication fork Leading strand
  13. 13. The polarity of DNA synthesis creates an asymmetrybetween the leading strand and the laggingstrand at the replication fork
  14. 14. Requirements of Replication Although the process of replication includesmany components, they can be combined into three major groups:1. a template consisting of single-stranded DNA,2. raw materials (substrates) to be assembled into a new nucleotide strand, and3. enzymes and other proteins that “read” the template and assemble the substrates into a DNA molecule.
  15. 15. New DNA is synthesized from deoxyribonucleotide triphosphates(dNTPs). Since the 5’ end does not get added to and the 3’ endrepeatedly does, the DNA strand is said to grow in a 5’- 3’ manner.
  16. 16. DNAPolymerase
  17. 17. DNA Polymerases in E. coli DNA Polymerase III
  18. 18. TopoisomeraseProtein complexes of the replication fork
  19. 19. DNA helicase unwinds the DNA duplex ahead of DNA polymerase creating single stranded DNAthat can be used as a template
  20. 20. ssDNA binding proteins bind to the sugar phosphate backboneleaving the bases exposed for DNA polymerase. The binding of SSB to newly formed ssDNA prevents reassociation of the single strands and “iron out” the unwound DNA.
  21. 21. Since DNA polymeraserequires a template and a free 3’ OH group to add nucleotides on to, RNA primers are required to initiate DNApolymerization. Primase, an enzyme which is part of a large complex of proteins called the primosome, synthesizes a small stretch of RNA (the primer) of 3-10 nucleotide in length, which will act as a starting site for the DNA polymerase.
  22. 22. DNA polymerase falls off the DNA easily. A “sliding clamp” is required tokeep DNA polymerase on and allow duplication of long stretches of DNA
  23. 23. A “clamp loader:”complex is required to get theclamp onto the DNA
  24. 24. Ahead of the replication fork the DNA becomes supercoiled The supercoiling needsto be relieved or tension would build up (like coiling as spring) and block fork progression.
  25. 25. Supercoiling is relieved by the action of Topoisomerases.1. Type I topoisomerases:  Make nicks in one DNA strands  Can relieve supercoiling2. Type II topoisomersases or DNA gyrase  Make nicks in both DNA strands (double strand break)  Can relieve supercoiling and untangle linked DNA helices Both types of enzyme form covalent intermediates with the DNA
  26. 26. Type I Topoisomerase
  27. 27. Type II Topoisomerase
  28. 28. Topoisomerases as drug targets1. Dividing cells require greater topoisomerase activity due to increased DNA synthesis2. Topoisomerase inhibitors which act by stablilizing the DNA-topoisomerase complex are used as chemotherapeutic agents:  camptothecin -Topo I inhibitor doxorubicin -- Topo II inhibitor  Some antibiotics are inhibitors of the bacterial-specific topoisomerase DNA gyrase: e.g. ciprofloxacin
  29. 29. DNALIGASE
  30. 30. , with less than one error perbillion nucleotides. This accuracy results from the processes of nucleotide selection, proofreading, and mismatch repair.
  31. 31. DNA mismatch repair corrects errors made during DNA replication.(A) If uncorrected, the mismatch will lead to apermanent mutation in one of the two DNA molecules produced by the next round of DNA replication. (B) If the mismatch is “repaired” using the newly synthesized DNA strand as the template, both DNA molecules produced by the next round of DNA replication will contain a mutation. (C) If the mismatch is corrected using the original template (old) strand as thetemplate, the possibility of a mutation is eliminated. The scheme shown in (C) is used by cells to repair mismatches.
  32. 32. Chemical modifications of nucleotides, if left unrepaired, produce mutations. (A) Deamination of cytosine, if uncorrected, results in the substitution of one base for another when the DNA is replicated. Deamination of cytosine produces uracil. Uracil differs from cytosine in its base-pairing properties and preferentially base-pairs withadenine. The DNA replication machinery therefore inserts an adenine when it encounters a U on the template strand. (B) Depurination, if uncorrected, can lead to the loss of a nucleotide pair. When the replication machinery encounters a missing purine on thetemplate strand, it can skip to the next complete nucleotide, thus producing a nucleotidedeletion in the newly synthesized strand. In other cases, the replication machinery places an incorrect nucleotide across from the missing base, again resulting in a mutation.
  33. 33. 1. Helicase enzyme unwinds DNA. This reaction needs ATP. At each replicating fork, the exposed single-stranded DNA is protected by single-strand binding proteins (ssb). Primase enzyme binds, preparing to make RNA primers.2. Primase enzyme makes RNA primer molecules. Each primer hybridizes (base pairs) with DNA, at the origin of replication. The 3 OH end will attach new deoxy nucleotides (dNTPs). The primers will each start a leading strand,3. DNA polymerase III attaches new dNTPs to the 3 OH end of the growing chain of the leading strand, which elongates toward the replicating fork, 5 to 3. (For each origin, there are TWO leading strands) For each NTP, a pyrophosphate (PP) is released, providing the necessary energy.4. More primers hybridize to the opposite strand of DNA. Pol III starts elongating 5 to 3 but it keeps running into the back of an RNA primer. This is the lagging strand. There are TWO lagging strands.5. DNA polymerase I (Pol I) starts at “nicks” in the growing strands. It edits the strand by removing bases ahead of it (5 end), including RNA and mismatched bases, while elongating the strand "behind" 5 to 3. It replaces all RNA nucleotides with dNTPs.6. Ligase seals the phosphate bonds at all “nicks” in the DNA.7. Editing endonucleases excise mismatched nucleotides, replacing with the proper match. How do they know which is old DNA vs. new DNA? The old DNA contains methyl groups on some of its cytosine bases.8. Gyrase restores negative superturns in DNA. ATP is needed.9. Methylases add methyl groups to the new DNA, at the same positions as the original strands. Now the two daughter helices are indistinguishable from each other, and from the original helix.
  34. 34. THE EUKARYOTIC REPLICON Time for DNA replication is limited in the S phase ofeukaryotes (6-8 hrs in mammals. Such RFs move only about 1/10th of the prokaryotic forks, andchromosomes can be in excess of 108 bp. Completion of replication at the allotted time requires multiple RFs called replicons. The Origin Recognition Complex (ORC) is a complex of 6 ATPases which is the functional equivalent of DnaA.
  35. 35.
  37. 37. How does a linearchromosome close replication at its two ends?
  38. 38. As DNA synthesis requires a RNA primer that will eventually be digested away, standard DNA replication would result in linear chromosomes that would shrink with every round of replication. This is resolved in bacteria by the circular genome which does not have an end. Inlinear chromosomes, the telomere solves the DNA end replication problem.
  39. 39.  Telomeres have highly repeated DNA sequences 5- TTAGGG-3. Human chromosomes have between 100 and 1500 copies of this sequence. Telomerase, a special DNA polymerase, can add additional copies of the 5-TTAGGG-3 to the end of a chromosome. The telomerase enzyme is actually a complex containing protein and RNA (a "ribozyme").
  40. 40.  The RNA portion has a 5-CCCTAA-3 region that acts as a template for adding the DNA repeat to the chromosome ends. The telomerase enzyme is found mostly in the germ cells of multicellular organisms. In somatic cells, the absence of telomerase results in shorter chromosomal ends with each division and may be the limiting factor in an organisms life span.
  42. 42. Errors of DNA Replication and Disease Origins or replication are strictly controlled so that they “fire” only once per cell cycle Errors lead to over-replication of specific chromosomal regions = gene amplification  This is commonly seen in cancer cells and can be an important prognostic indicator.  It can also contribute to acquired drug resistance, e.g. Methotrexate induces amplification of the dihydrofolate reductase locus.
  43. 43.  The rate of misincorporation of bases by DNA polymerase is extremely low, however repeated sequences can cause problems. In particular, trinucleotide repeats cause difficulties which can lead to expansion of these sequences. Depending where the repeat is located, expansion of the sequence can have severe effects on the expression of a gene or the function of a protein. Looping out of repeats before replication.
  44. 44. Several inherited diseases are associated withexpansion of trinucleotide repeat sequences. Very different disorders, but they share the characteristic of becoming more severe in succeeding generations due to progressive expansion of the repeats
  45. 45.