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Semiconservative replication

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Presentation on theme: "Semiconservative replication"— Presentation transcript:

1 Semiconservative replication
The mechanism of DNA replication is actually based on the biosynthesis of two novel double stranded molecules in which one strand is new and the other one is one of the two strands present in the mother molecule. So the two daughters cells will have two identical molecules, identical in the sequence of the nucleotides but also identical in the origin of the two strands.

2 Polymerization reaction
The reaction is catalyzed by a family of enzymes named DNA polymerases. It is very simple. This is the template strand (that of the mother cell) and this the growing (nascent) strand. This nascent strand grows of one nucleotide when a deoxynucleotide triphosphate forms a correct base pairing with the template strand. When the nucleotide is in this position, the enzyme catalyzes the attack of the hydroxyl group on this phosphate thus leading to the formation of a phosphodiester bond. And so on. A new nucleotide, in this case a Thymidine forms the base pairing with the template and the polymerase catalyzes the formation of a new bond.

3 Let’s figure out the catalysis: The enzyme is represented in green and this is the catalytic site. The substrates are the nascent polynucleotide and the entry deoxyribonucleotide which is paired with the complementary nucleotide present on the template strand. Upon the formation of the phosphodiesther bond the enzyme moves to the following position. This is an important point. The enzyme is always associated with the DNA and we can say walks on the DNA. This phenomenon is named processivity, that of course shortens the time needed to synthesize very long DNA molecules.

4 Let’ make a little effort
Let’ make a little effort. I already showed you the Watson and Crick basepairing. This is the ribose and this is the carbon 1’ involved in the aminoglycosisid bond with the nitrogen 9 of guanine. This is the n-glycosidic bond with the nitrogen 1 of cytosine. These two nucleotides form three hydrogen bonds, involving at one hand the oxygen of carbon 6 the hydrogen of nitrogen 1 and the hydrogen of carbon 2. (6, 1 , 2) and on the other hand, this hydrogen of the amino group of carbon 4, the nitrogen 3 and the oxygen of carbon 2 (4, 3, 2).

5 In the case of adenine-thymine base pair, there are two hydrogen bonds, and the number to remember are 6, 1 and 4, 3. The amino group at carbon 6 with the oxygen at carbon 4 and the nitrogen 1 with the hydrogen of nitrogen 3. Thus replication proceeds with the incorporation of of a thymidine when on the template strand there is an adenosine.

6 Basepairing of rare tautomeric forms of nucleotides
However, purines and pyrimidine could also exist in rare tautomeric forms. Look for example at cytidine. This group is in most cases an amino group, but in rare cases it can be an imino grup. In this condition cytidine can pair with adenine. Of course, this base pairing is sufficient to allow the incorporation of the cytidine, but it is highly unstable, because cytosine is converted in its tautomeric form amino.

7 Proofreading All DNA polymerases possess at least two enzyme activities. One is that polymerizing the nucleotides, another one is a sort of proof-reading process. The reason why for this activity is the fact that DNA polymerase in some cases incorporates a wrong nucleotide. When this accident happens it is completely unable to add a further nucleotide to the nascent strand. Thus the wrong nucleotide is removed by the enzyme by this second enzyme activity named exonuclease. This activity hydrolyzes the phosphodiester bond, so removing the wrong nucleotide. At this point the polymerization can restart.

8 As you can see, the to catalytic sites are close each other and the exonuclease functions only if the incorporated nucleotide is mispaired with the one present on the template strand. Editing reactions and repair activity result in a very efficient machinery with a unedited/unrepaired mistake rate of about 1 out of 10 to 10 incorporations. The fidelity of DNA polymerases is actually based on the fact that they are able to add a further nucleotide only if the last nucleotide of the nascent strand is perfectly paired with the template. This implies that DNA polymerases are unable to begin the biosynthesis.

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10 DNA polymerases are only able to add a nucleotide to a nascent growing polynucleotide

11 Primase To overcome this inability, all the cells use a trick. The biosynthesis of any DNA molecules is started by the biosynthesis of a short RNA molecule. The enzyme that catalyzes this event is named DNA primase, because it synthesizes a primer. It is a RNA polymerase, which haven’t the problem of fidelity. Once formed the short primer, the DNA polymerase can start to incorporate deoxynucleotides at the end of the short RNA fragment. This primer will be after removed by an exonuclease which works in the 5’ to 3’ direction.

12 In summary, this is a primer generated by the primase, it is elongated by the DNA polymerase. When it meets a downstream fragment, the exonuclease activity removes the primer and the two DNA fragments are covalently linked by a further enzyme named DNA ligase.

13 DNA ligase

14 Elicase Another protein necessary to replicate the DNA is the elicase. Actually this enzyme opens the two strands of the template molecule so that the single strands are exposed and available to function as templates. It is considered an enzyme because it catalyzes the breaking of the hydrogen bonds between the paired nucleotides.

15 Single strand binding proteins (SSBP)

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17 clamp (PCNA) Processivity

18 Replication should work on both strands at the same time
Any DNA that should be replicated has two template strands, which are antiparallel. This means that on one strand (the upper one in this case) the biosynthesis is continuous, while on the other strand (the lower one) it is discontinuous, in the sense that many fragments of DNA are synthesized and ligated each other. This fragments are named Okazaki fragments.

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23 DRAB Molecular Biology 1° week
Describe the main differences between the E. coli genome and the eukaryotic genomes Describe the structure of the nucleosome and the role of histone modifications Describe the role of the following proteins: elicase, topoisomerase, primase

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25 Replication origin

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27 Replication origins in eukaryotes

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