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Replication Requirements: 1) Double helix must open up and stay open. 2) Replication needs to occur on both strands, 3’5’ and 5’3’. Kornberg (1959) Discovered the first enzyme that catalyzed the reaction for adding a nucleotide to another nucleotide, DNA polymerase I. Features of DNA polymerase I: 1. Could remove nucleotides in a 5’3’ direction and/or in a 3’5’ direction from the end of a DNA strand (DNA exonuclease) 2. Could add bases in only one direction, 5’3’. For this to occur an open 3’ end needed to be present. 26 3. Could remove RNA (RNA nuclease). Problems: only adds bases in one direction. needed a nucleotide with an open 3’ end present to start adding bases. DNA polymerase I was too slow (did not add bases fast enough). Other polymerases were soon identified. DNA polymerase III proved to be the principal enzyme for replication. 27 DNA polymerases Pol I mol. wt. 109,000 subunit Pol II Pol III 120,000 140,000 yes no yes yes exonuclease activity 3’5’ 5’3’ yes yes turnover activity nucleotides per minute 600 30 9000 function repair ? synthesis 28 Other enzymes involved in replication: Topoisomerase - allows DNA to move from a supercoiled state to a relaxed state. Enzyme also called DNA gyrase. Helicase - unwinds the DNA double helix. Destabilizing enzymes (SSBPs) DNA single strand binding proteins to keep the strands separated. Initiator protein - Binds to the DNA at the origin of replication. Primase - Makes a RNA primer. Different from RNA polymerase. DNA Polymerase I - Removes RNA primer and repairs any errors (proofreads). Ligase - Joins ends of DNA fragments. Also need - Mg2+ ions and all four dNTPs 29 Eukaryotic replication (mammalian) DNA Polymerases location function nucleus replication nucleus repair nucleus replication nucleus repair mitochondria mitochondrial replication The replication process is similar to system in prokaryotes except for the 3’ end of the linear DNA. 30 Need an additional enzyme, Telomerase The problem is that once the primer is removed from the end of the 3’strand of DNA there is no way to replicate that section because there is not an open 3’ end for the DNA polymerase to start attaching nucleotides. The enzyme can synthesize TTGGGG repeats to the end of a strand by using an internal RNA sequence as a template. The internal RNA sequence has a region that is complementary to the region where the RNA primer had been. 31 The telomerase uses its RNA sequence as a template to extend the DNA. A primer can now be attached to the extended section of DNA and serve as an open 3’ end to allow completion of replication of the linear DNA. DNA polymerase fills in the gap in the DNA, the end is trimmed, and ligase connects the nick in the DNA. 32 Origins of Replication Site where replication begins on a circular or linear piece of DNA. Site is rich in A-T base pairs to facilitate separation of the DNA strands. With circular DNA (prokaryotes), there is usually a single origin of replication. With linear DNA (eukaryotes ), there are multiple origins of replication along the DNA. 33 Transcription - The formation of an RNA molecule upon a DNA template by complementary base pairing, mediated by RNA polymerase. Basic components for transcription: 1) RNA polymerase 2) DNA template 3) Nucleotides: ATP, UTP, GTP, CTP Prokaryotic RNA polymerase 2 units 1) core enzyme 2) sigma factor 34 Core enzyme 5 polypeptide subunits chain ’ chain 2 chains chain The core enzyme can bind to a DNA template without the sigma () factor but it binds at random. Function of the sigma factor appears to be the recognition of the promoter regions usually found at the beginning of a gene. 35 Promoters Usually have a similar sequence and contain a high percentage of A and T. example: 5’-TATAAT - 3’ In prokaryotes the sequences are similar for all genes. In eukaryotes the sequence depends on the gene product: protein, tRNA, or rRNA. 36 Steps in Transcription 1) Binding of the sigma factor to the core enzyme. 2) Binding of the RNA polymerase to the DNA template. 3) Release of the sigma factor. 4) RNA synthesis. 5) Termination of RNA synthesis and release of the RNA and RNA polymerase from the DNA template. 37 1. Binding of Sigma factor to the core enzyme Once the sigma factor is attached the RNA polymerase can bind to the promoter region of the gene. 2. Binding of the RNA polymerase to a gene’s promoter region a) RNA polymerase first binds to the consensus sequence - 35 base pairs from the initiation site for transcription. example: 5’-TTGACA -3’ b) tighter binding then occurs at the Pribnow box - 10 base pairs from the initiation site for transcription. This is the site where the DNA opens. Example: 5’-TATAAT-3’ 38 3. Initiation of RNA synthesis The first region transcribed is called the leader region. This region can be 18 to 175 bases long and is used as a recognition site during translation. DNA promoter leader gene 4. Elongation or gene transcription Only one of the DNA strands is transcribed (called the sense strand). RNA is synthesized 5’ 3’ so the DNA sense strand is read 3’ 5’. Pairing of the bases: DNA RNA A U T A G C C G 5. Termination of transcription 39 DNA sequence near the end of the gene codes for a transcript that has two-fold symmetry (a complementary sequence) that allows RNA to form a region of base pairing producing a hairpin loop. This followed by a series of uracils. AAU U G C A CG GC C G region of C G complementarity CG GC AUUUUUUUUUUU The hairpin loop and series of uracils is found with rho-independent termination (intrinsic termination). Two types of termination: 40 rho-independent termination rho-dependent termination Rho-independent termination (intrinsic termination) After synthesis of the region of two fold symmetry the RNA polymerase then transcribes a series of adenines (A’s) on the DNA sense strand. The association of the RNA polymerase in this region is weak. When the hairpin loop forms in the newly transcribed RNA it folds back interfering with the RNA polymerase causing it to release. 41 Rho-dependent termination A separate protein factor rho () interacts with the RNA transcript and the RNA polymerase to cause the RNA polymerase to release from the DNA. There are regions in the RNA sequence that signal rho to bind. There are also nucleotide sequences slightly downstream from these regions called potential terminators because they cause the RNA polymerase to pause letting the rho factor to catch up and terminate transcription. They are called ‘potential ‘terminators because if the rho factor is blocked from binding then the RNA polymerase will continue transcribing after pausing. Rho factor can be blocked from binding by ribosomes 42 Ribosomes can be present to block attachment of the rho factor in prokaryotes because transcription and translation can occur at the same time for a given gene. Why? No nuclear membrane separating the site of transcription from the site of translation in prokaryotes. 43 Transcription in Eukaryotes 3 RNA polymerases have been identified Polymerase RNA product I ribosomal RNA II messenger RNA III transfer RNA small ribosomal RNA There appears to be no ‘sigma factor’ in eukaryotic polymerases. Instead the polymerase specificity is determined by the promoter sequence of the gene. RNA polymerase I promoter found in the 3’ flanking region of the DNA. sequence 5’-TATTG-3’ or 5’-TTTTGG-3’ not highly conserved across species. 44 RNA polymerase II promoter found in the 3’ flanking region of the DNA. promoters: a) TATA box (Goldberg-Hogress box) -30 bps 5’-TATAAA-3’ b) CAAT element -80 bps 5’-GGCCAATCT-3’ c) GC element 1 to 2 copies location varies 5’-GGGCGG conserved across species RNA polymerase III - promoter located within the gene 45 Eukaryotic transcription does involve additional proteins that facilitate or modulate initiation of transcription. There are two types of proteins, transcription factors and activators. Transcription factors - These are proteins that facilitate initiation of transcription by binding to the TATA region forming a binding site for RNA polymerase II. Activators - Are proteins that bind to enhancer sites upstream or downstream from the gene increasing the efficiency of initiation of transcription. 46 Termination of Transcription in Eukaryotes Termination of transcription in eukaryotes is not understood as well as termination in prokaryotes. It appears to vary for the different RNA polymerases and cleavage of the newly produced RNA strand may occur without halting transcription or inducing the release of the RNA polymerase. 47 How can transcription be stopped? 1) Bind to the DNA a) prevent separation of the DNA strands b) block the binding site of RNA polymerase 2) Bind to RNA polymerase a) change the shape of the RNA polymerase so it can not function properly. a) bind or remove the sigma factor Transcriptional control occurs: naturally: binding of proteins to DNA to block transcription is a common form of gene regulation chemically: antibiotics are available that can bind to DNA or RNA polymerase 48 Genetic Code How is the genetic information carried on the DNA? Sequence of bases act as a code: DNA bases RNA bases protein amino acids How many nucleotides equal one amino acid? Have 4 different nucleotides and 20 different amino acids. 49 Want the fewest number of bases per amino acid for the greatest efficiency. 2 bases per amino acid (42) = 16 combinations 3 bases per amino acid (43) = 64 combinations 4 bases per amino acid (44) = 256 combinations So 3 bases per amino acid gives enough combinations for 20 amino acids. A 3 base sequence codes for one amino acid and is called a codon. The 3 base codons for the 20 amino acids is considered to be universal in that in general the codons code for the same amino acid regardless of the species. 50