BMB 462 Lecture 29 Outline of Last Lecture I. Differentiating between the types of RNAsII. Varying RNA functionsIII. Structure of a typical transcription unitIV. Mechanism of transcription and RNA synthesisV. Role of topoisomerases in transcriptionOutline of Current Lecture I. Comparing transcription and replicationII. Transcription in bacteria: the components involvedIII. The initiation phaseIV. Sigma factor recognition of promotersV. Amino Acid interactions with DNAVI. Sigma factor structure/mechanismVII. DNase I footprintingVIII. Initiation and elongationIX. TerminationCurrent LectureConcepts to remembers from previous courses/lectures:-I. Comparing transcription and replicationa. Similaritiesi. Both require topoisomerases.1. You're required to pull apart the double stranded DNA; this creates positive supercoiling ahead of the transcription bubble or replication fork. In bacteria, DNA gyrase relieves this by introducing negative supercoils. Topoisomerases also take care of rewinding behind the transcription bubble; in replication, two helices are made so it is not necessary to reintroduce coiling.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ii. Both are built in a 5' to 3' direction, with the 3' hydroxyl group of the new nucleotide attacking the 5' phosphate on the previous nucleotide.iii. The new strand is synthesized anti-parallel to the template strand.iv. 2 high-energy bonds are needed to add the nucleotides to the growing strand; the bond between the alpha and beta phosphates is broken and releases pyrophosphate. Pyrophosphatase degrades that into 2 inorganic phosphates to make the reaction irreversible.b. Differencesi. DNA polymerases are not able to synthesize new strands de novo, and require a primer. RNA polymerases are able to synthesize de novo and thus do not use a primer. This results in RNA having triphosphate at the 5'end, which becomes important later on with processing eukaryotic mRNA.ii. RNA transcription occurs in discrete regions, the genes, while DNA replication replicates entire chromosomes. Replication also uses both strands as a template.iii. DNA polymerase can add 1000 nucleotides per second.iv. This error rate is comparable to DNA polymerases that lack proofreading capabilities.II. Transcription in bacteria: the components involveda. E. coli RNA polymerase was discovered in 1960, shortly after Watson and Crick had proposed the structure of DNA (1953).b. There is a holoenzyme and a core unit involved.i. The difference between the holoenzyme and the core is that the holoenzyme has a sigma. It's not always the same sigma as bacteria have multiple different types. The holoenzyme is involved in initiation; the sigma subunit in particular is associated with recognizing the primer. This allows the polymerase to recognize and start initiation at the +1 region.c. The core is devoid of a sigma factor; once initiation has started, the holoenzyme loses the sigma and is now considered the core enzyme and continues elongationof the RNA.i. B and B' subunits of the core make up the active siteii. The 2 alpha subunits are involved in assembly of the core. They also interact with transcription factors.d. In complex molecules that have multiple subunits, the structures and mechanisms are often discovered by studying those enzymes in organisms that survive extreme environments because the proteins can still function at high temperaturesi. i.e. Thermus aquaticus RNA polymerase (Taq polymerase, which is used for PCR because it can withstand the high denaturation temperatures)III. The initiation phasea. Initiation of transcription begins with separating the 2 strands to create the transcription bubble and binding to the promoter region.i. Promoter recognition is done by the sigma factor; there are 7 sigma factors in E. coli; sigma 70 is the major sigma factor used during transcription. It directs most of the transcription for proteins and for all tRNA and rRNA (The house keeping genes; the basic things that you need for the life of the cell; those used for the basic functions and metabolism)b. Very few promoters have an UP element, which is the first area recognized in these RNA and encodes the highly transcribed rRNA genes.c. All promoters need a -35 and a -10 region (each are hexanucleotide sequences; they have 6 base pairs each). Depending on how close they are to the consensus sequence (the favourite sequence for the sigma factor to recognize) and the spacing between the -10 and the -35 determines the strength of the promoter; it determines the ability of the sigma to carry the holoenzyme to the promoter.d. The consensus sequence, found in the coding strand, is an AT rich in the promoter (the exact sequence can be seen in Fig. 26-5). The -35 consensus sequence is TTGACA and the -10 sequence is TATAAT. The closer a promoter is to that sequence, the stronger it will be and the more likely it will be bound by sigma 70.e. Protein encoding genes do not contain an UP element.f. Some promoters are so weak they need activator proteins to help call the holoenzyme.IV. Sigma factor recognition of promotersa. One way that cells regulate genes differentially is by using different sigma factors;different sigma factors bring the holoenzyme to different subsets of promoters in the chromosome.i. i.e. sigma 32 directs transcriptions of over 30 genes; these genes code for proteins that help the cell survive heat shock after exposure to high temperature (thus it is called the heat shock sigma)1. This sigma has a different -35 and -10 consensus sequence that it recognizesii. Sigma 32 only becomes abundant and is upregulated when the cell is exposed to heat shockb. How do sigma factors recognize specific promoters? That is the work of sequence-specific interactions and transcription factors that promote or repress transcription by RNA polymerase.c. Base pairs in the DNA present different patterns of hydrogen bond donors and acceptors in the major groove of DNA, which is primarily where recognition takesplace.i. The major groove is defined by the points of attachment to the sugar-phosphate backbone (the longer distance between points of attachment constitute the major groove). The H-bond donors (D) and acceptors (A) in this groove allow specific interactions between base pairs in DNA and amino acids in protein.1. Nitrogens and Oxygens become the H-bond acceptors and the amino
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