CH08 - RNA: Transcription and Processing BASIC PROBLEMS 1....

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Unformatted text preview: RNA: Transcription and Processing BASIC PROBLEMS 1. Because RNA can hybridize to both strands, the RNA must be transcribed from both strands. This does not mean, however, that both strands are used as a template within each gene. The expectation is that only one strand is used within a gene but that different genes are transcribed in different directions along the DNA. The most direct test would be to purify a specific RNA coding for a specific protein and then hybridize it to the 9. genome. Only one strand should hybridize to the purified RNA. In prokaryotes, translation is beginning at the 5’ end while the 3’ end is still being transcribed. In eukaryotes, processing (capping, splicing) is occurring at the 5’ end while the 3’ end is still being transcribed. There are many examples of proteins that act on nucleic acids, but some mentioned in this chapter are RNA polymerase, GTFs (general transcription factors), 0' (sigma factor), rho, TBP (TATA binding protein), and snRNPs (a combination of proteins and snRNAs). Sigma factor, as part of the RNA polyermase holoenzyme, recognizes and binds to the —35 and —10 regions of bacterial promoters. It positions the holoenzyme to correctly initiate transcription at the start site. In eukaryotes, TBP (TATA binding protein) and other GTFs (general transcription factors) have an analagous function. The CTD (carboxy tail domain) of the B subunit of RNA polymerase 11 contains binding sites for enzymes and other proteins that are required for RNA processing and is located near the site where nascent RNA emerges. If mutations in this subunit prevent the correct binding and/or localization of the proteins necessary for capping, then this modification will not occur even though all the required enzymes are normal. For a given gene, only one strand of DNA is transcribed. This strand (called the template) will be complementary to the RNA and also to the other strand (called the nontemplate or coding strand). Consequently, the nucleotide sequence of the 176 Chapter Eight 10. RNA must be the same as that of the nontemplate strand of the DNA (except that the Ts are instead Us). Ultimately, it is the nucleotide sequence that gives the RNA its function. Transcription of both strands would give two complementary RNAs that would code for completely different polypeptides. Also, double-stranded RNA initiates cellular processes that lead to its degradation. 3., b., c., and d. Transcription / initiation site a 3, / Transcription / initiation site mRNA Plasmid DNA Yes. Both replication and transcription is performed by large, multi—subunit molecular machines (the replisome and RNA polymerase II, respectively) and both require helicase activity at the fork of the bubble. However, transcription proceeds in only one direction and only one DNA strand is copied. a. False. Sigma factor is required in prokaryotes, not eukaryotes. b. True. Processing begins at the 5’ end, while the 3’ end is still being synthesized. c. False. Processing occurs in the nucleus and only mature RNA is transported out to the cytoplasm. d. False. Hairpin loops or rho factor (in conjunction with the rut site) is used to terminate transcription in prokaryotes. In eukaryotes the conserved sequences AAUAAA or AUUAAA, near the 3’ end of the transcript is recognized by an enzyme that cuts off the end of the RNA approximately 20 bases downstream. e. True. Multiple RNA polymerases may transcribe the same template simultaneously. a. The original sequence represents the —35 and —1() consensus sequences (with the correct number of intervening spaces) of a bacterial promoter. Sigma factor, as part of the RNA polymerase holoenzyme, recognizes and binds to these sequences. b. The mutated (transposed) sequences would not be a binding site for sigma factor. The two regions are not in the correct orientation to each other and therefore would not be recognized as a promoter. Chapter Eight 177 11. a. The promoters of eukaryotes and prokaryotes do not have the same conserved sequences. In yeast, the promoter would have the required TATA box located about —30 whereas bacteria would have conserved sequences at —35 and —10 that would interact with sigma factor as part of the RNA polymerase holoenzyme. b. There are two possible reasons that the mRNA is longer than expected. First, many eukaryotic genes contain introns, and bacteria would not have the splicing machinery necessary for their removal. Second, termination of transcription is not the same in bacteria and yeast; the sequences necessary for correct terminaton in E. 6011' would not be expected in the yeast gene. 12. promoter \ DNA 3: 5, /| ImRNA /'(‘)a3 transcription Start / \ translation stop I termination site 5, UT translation start 3’ UT 13. promoter / exon intron ‘ 3’ transcription start . . . termination Site (polyadenylation signal) m70_p_p-;7:I:::Ej:IAAAAAAAAAAAAAA i translation stop/ f 5’ UT translation start RNA processing 3’UT 14. a. Yes. The exons encode the protein, so null mutations would be expected to map within exons. b. Possibly. There are sequences near the boundaries of and within introns that are necessary for correct splicing. If these are altered by mutation, correct 178 Chapter Eight splicing will be disrupted. Although transcribed, it is likely that translation will not occur. c. Yes. If the promoter is deleted or altered such that GTFs cannot bind, transcription will be disrupted. (I. Yes. There are sequences near the boundaries of and within introns that are necessary for correct splicing. 15. Self-splicing introns are capable of excising themselves from a primary transcript without the need of additional enzymes or energy source. They are one of many examples of RNA molecules that are catalytic, and for this property, they are also known as ribozymes. With this additional function, RNA is the only known biological molecule to encode genetic information and catalyze biological reactions. In simplest terms, it is possible that life began with an RNA molecule, or group of molecules, that evolved the ability to self— replicate. 16. Antibiotics need to selectively target bacterial structures and functions that are essential for life but unique or sufficiently different from the equivalent structure and functions of their animal hosts. Bacterial RNA polymerase fits these criteria as its function is obviously essential, yet its structure is sufficiently different from the several eukaryotic RNA polymerases. These differences make it possible to develop drugs that specifically bind bacterial RNA polymerase but have little or no affinity for eukaryotic RNA polymerases. CHALLENGING PROBLEMS 17. a. The data cannot indicate whether one or both strands are used for transcription. You do not know how much of the DNA is transcribed nor which regions of DNA are transcribed. Only when the purine/pyrimidine ratio is not unity can you deduce that only one strand is used as template. b. If the RNA is double-stranded, the percentage of purines (A + G) would equal the percentage of pyrimidines (U + C) and the (A +G)/(U + C) ratio would be 1.0. This is clearly not the case for E. coli, which has a ratio of 0.80. The ratio for B. subtilis is 1.02. This is consistent with the RNA being double—stranded but does not rule out single-stranded if there are an equal number of purines and pyrimidines in the strand. 18. Chapter Eight 179 promoter / ZCI::_—__———:___9_5» / termination site 01 aden lation si nal transcription Stmlp y/ y g ) 19. 20. 21. - \\\\\\\\\\\\\\\\\\\\\‘ I \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ _ exon intron exon intron exon b. Alternative splicing of the primary transcript would result in mRNAs that were only partially identical. In this case, the two transcripts share 224 nucleotides in common. As this is the exact length of the second exon, one possible solution to this problem is that this exon is shared by the two alternatively-spliced mRNAs. The second transcript also contains 2.3 kb of sequence not found in the first. Perhaps what was considered the first intron is actually also part of the second transcript as that would result in the 2524 nucleotides stated as this transcript’s length. Of course, other combinations of alternative splicing would also fit the data. Double-stranded RNA, composed of a sense strand and a complementary antisense strand, can be used in C. elegans (and likely all organisms) to selectively prevent the synthesis of the encoded gene product (a discovery awarded the 2006 Nobel Prize in Medicine). This process, called gene silencing, blocks the synthesis of the encoded protein from the endogenous gene and is thus equivalent to “knocking out” the gene. To test whether a specific mRNA encodes an essential embryonic protein, eggs or very early embryos should be injected with the double-stranded RNA produced from your mRNA, thus activating the RNAi pathway. The effects of knocking out the specific gene product can then be followed by observing what happens in these, versus control, embryos. If the encoded protein is essential, embryonic development should be perturbed when your gene is silenced. Transgene silencing is a common phenomenon in plants. Silencing may occur at the transcriptional or post-transcriptional level. Since you cannot control where transgenes insert, some may insert into transcriptionally inactive parts of the genome. Post—transcriptional silencing may be the result of activation of the RNAi pathway due to the misexpression of both strands of your transgene. See Figure 8—22 in the companion text as one example of how this can happen. In these cases, the RNAi pathway will be activated and the EPSPS gene product will be silenced. RNAi has the potential to selectively prevent protein production from any targeted gene. Oncogenes are mutant versions of “normal” genes (called proto- oncogenes) and their altered gene products are partly or wholly responsible for causing cancer. In theory, it may be possible to design appropriate siRNA 180 Chapter Eight molecules (small interfering RNAs) that specifically silence the mutant oncogene product but do not silence the closely related proto-oncogene product. The latter is necessary to prevent serious side effects as the products of proto- oncogenes are essential for normal cellular function. ...
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CH08 - RNA: Transcription and Processing BASIC PROBLEMS 1....

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