Bio Final Nov 26 DNA Replication Tool Kit Begins at Ori C origin of replication o o Bacteria have one Ori C because they have a singular circular chromosome When DNA is unzipped breaking hydrogen bonds we have two replication forks that proceed in opposite directions and meet on the opposite side of the chromosome then the chromosome is then replicated Eukaryotes have 46 chromosomes and each chromosome has many Ori C s is the enzyme that unzips the DNA molecule by breaking the hydrogen bonds that are Enzymes 1 Helicase holding the bases together 2 Gyrase the enzyme that relieves supercoils up stream of the helicase a a Tab on blue jean zippers Telephone cord 3 DNA Polymerase I the idea that is can create a polymer of nucleotides Functions SLOWLY a Very large complex enzyme b c Has three activities functioning as i 5 1 3 1 Polymerase Ability DNA Molecule is structurally polarized and the two polynucleotides are antiparallel strands are oriented in opposite directions Adds Nucleotides to the 31 end of other nucleotides Allows the addition of new nucleotides to the 31 hydroxyl end of another already made nucleotide Template Strand Daughter Strand DNA polymerase 1 can add a brand new nucleotide to the 31 end of the the strand of DNA being copied the copied DNA strand one that is already present Comes in the form of a nucleotide triphosphate a dTTP dATP dCTP dGTP deoxythymine triphosphate etc Build up polynucleotide by adding one nucleotide at a time using DNA polymerase 1 When nucleotide is added to the 3 prime end of the nucleotide that is already there 2 phosphates break off and one bonds to the hydroxyl group typical dehydration synthesis ii 5 1 3 1 Exonuclease Ability Like Paccman Removes on nucleotide at a time from the 5 prime phosphate end Removes the nucleotide with a hydrolytic reaction where you add H2O and break two monomers apart Used to remove primers iii 3 1 5 1 Exonuclease Ability Can begin at the 3 prime hydroxyl and bite off one nucleotide at a time Functions as a proof reader Ex DNA polymerase 1 tried to bond a C with an A instead of a T violation of base pairing rules if this happens the enzyme uses 31 51 exonuclease ability to remove the last nucleotide added to the 3 prime end it clips it off and the right one is put in its place 4 DNA Polymerase III FAST a Two Activities i 5 1 3 1 Polymerase Ability Allows the addition of new nucleotides to the 31 hydroxyl end of another already made nucleotide Same as DNA Polymerase I ii 3 1 5 1 Exonuclease Ability Proof reader Same as DNA Polymerase I b DOES NOT HAVE 51 31 Exonuclease Ability c DNA Polymerase adds a nucleotide to the 3 Prime Hydroxyl end of another nucleotide that is already there 5 Primase getting things started a Puts in a short RNA Primer b If the DNA is AT the RNA is UA and the RNA primer is 5 10 nucleotides long and it primes the pump c When the primer is put in DNA Polymerase I or III can start adding nucleotides one at a time using the 5 3 polymerase ability 6 DNA Ligase replication through dehydration synthesis the enzyme that can bond one nucleotide to the nucleotide next to it during DNA a Two nucleotides next to each other release water form covalent bond between a phosphate and hydroxyl group Replication Process o o o o o o o o o o Two nucleotides are anti parallel 51 31 and 31 51 Helicase enters the DNA at the replication fork breaking the bonds between the A s T s C s and G s An RNA Primer has to be antiparallel to template strand has U s on it bc RNA DNA Polymerase one adds DNA to 31 end of the primer New polynucleotide being elongated into the replication fork as the helicase unzips the molecule Leading Strand molecule the DNA polymerase III can continue to elongate the daughter of the leading strand as long as it adds to the 31 end Lagging Strands is elongated continuously as long as the helicase is unzipping the DNA are elongated discontinuously Slower because of the primers have to be antiparallel to the template strand if one end of the nucleotide is the 51 end then the primer has to be 31 The primer will be elongated one at a time at the 31 Each segment is called and Oaksaki Fragment 100 300 nucleotides in length DNA Polymerase I is going to come in and use its 51 31 exonuclease ability to remove the primers It goes in and attacks the 51 end of the primer it will digest primer then use its polymerase ability to elongate the 31 end to connect to the 51 of the following fragments One nucleotide is bonded to the next by DNA ligase Primers are added until the helicase is finished unzipping and runs into the other helicase o o MAKE FLASHCARDS AND BE ABLE TO TELL WHAT ALL THE ENZYMES DO ALSO TAKE A DNA MOLECULE LIKE HE DID AND DO IT BY PUTTING IN THE ORIENTATION ANTIPARALLEL AND SHOW HOW YOU D ELONGATE THE STRANDS LAGGING AND LEADING At the end of the DNA molecule DNA polymerase I using its 51 31 exonuclease ability will come in and digest out a nucleotide There is no way fill in the gap once the primer at the end of the strand has been removed we would need another primer on the outside but we cant build without a template molecule to attach it to So what happens is every time a chromosome is copied on the lagging strand there is gap left which cant be filled in because there is no 31 end to attach the nucleotides to Eukaryotes had linear chromosomes and at each end of the chromosome every time a chromosome is replicated it will get shorter by the length of a primer because we cant fill in the gap They get shorter and shorter until eventually they start to eat away at important genes in your chromosomes This problem is solved by the addition of telomeres Telomeres end of a chromosome and is this sequence in length which is repeated between 100 300 times is a sequence of DNA TTAAGGG beginning at the 5 prime end It is found at the As the chromosome gets shorter with each replication they get shorter and so do the telomere they form a buffer so the important genes are not removed help http www 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