Ch 6A Essays

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Ch 6A Essays
2015-02-15 14:13:23
Test Two
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  1. How are genes controlled?
    Many genes can be transcribed and RNA can direct the synthesis of identical proteins. But, they can be transcribed and translated with a different efficiency, allowing the cell to make vast quantities of some protiens and tiny quantities of others. A cell can change or regulate the expression of each of its genes according to the needs of the moment
  2. Differences between DNA and RNA
    • sugars and nucleotides
    • Not only does A pair with U, but occasionally G can pair with u. DNA is always a double-helix and RNA is single-stranded, which allows it to fold up into a particular shape
  3. Explain transcription. (Brief)
    • a.       It begins with the opening and unwinding of a small portion of the DNA double helix to act as a template. The nucleotide sequence is complementary to the RNA sequence that will be synthesized.
    • b.      When a good match is made, the incoming ribonucleotide is covalently linekd to the growing RNA chain in an enzymatically catalyzed reaction.
    • c.       The RNA chain produced by transcription—the transcript---is therefore elongated one nucleotide at a time.
  4. How does transcription differ from DNA replication?
    • a.       The RNA strand does not remain hydrogen bonded to the DNA template; it is displaced behind where the nucleotides are being added, allowing the helix to reform.
    • b.      Because they are copied from only a limited region of the DNA, RNA molecules are shorter than DNA molecules. 
  5. 1)      What are differences between DNA polymerase and RNA polymerase? 
    • a.       First, RNA polymerase catalyzes the linakge of ribonucleotdes
    • b.      Unlike DNA polymerases involved in DNA replication, RNA polymerases can start an RNA chain without a primer, which can exist because transcription doesn’t need to be as accurate since RNA doesn’t permanently store genetic information
    • c.       RNA polymerases are not as accurate as DNA polymerases, but they do have modest proofreading mechanisms
    • d.      Except for the Mg2+, DNA and RNA polymerases are not related 
  6. What is the proofreading mechanism of RNA polymerase?
    a.       If an incorrect ribonucleotide is added to the growing RNA chain, the polymerase can back up, and the active site of the enzyme can perform an excision reaction that resembles the reverse of the polymerization reaction using water instead of phosphate. 
  7. 1)      How do bacterial RNA polymerases know where to start and stop? 
    a.       The bacterial RNA polymerase core enzyme is a multisubunit complex that synthesizes RNA using DNA. The detachable subunit called sigma factor associates with the core enzyme and assists it in reading the signals int eh DNA that tell it where to begin transcribing. The sigma factor and core enzyme are the RNA polymerase holoenzymes, which weakly adhere to bacterial DNA when the two collide. A holoenzyme typically slides rapidly along the DNA molecule until dissociating. 
  8. Explain the action of the polymerase holoenzyme.
    • a.       When the polymerase holoenzyme slides into a region of the DNA double helix called a promoter, the polymerase binds tightly to this DNA; and, the polymerase holoenzyme, through its sigma factor, recognizes the promoter DNA sequence by making contacts with the portions of the bases that are exposed to the outside of the helix.
    • b.      After binding to the promoter DNA, it opens up the double helix to expose a short stretch of nucleotides. (This does not require energy). Instead, the polymerase and DNA undergo reversible structural changes that result in a state more energetically favorable than that of the initial binding.
    • c.       With the DNA unwound, one strand is a template for complementary base pairing. After the first ten or so are synthesized (inefficient period for polymerase), the core enzyme breaks its interactions with the promoter DNA, weakens its interactions with the sigma factor, and begins to move down the DNA, synthesizing RNA.
    • d.      Chain elongation continues until the enzyme encounters a terminator, where the polymerase halts and releases both the newly made RNA chain and the DNA template.
    • e.      After releasing, the polymerase reassociates with a free sigma factor to form a holoenzyme that can begin the process of transcription again
  9. What happens if an RNA polymerase dissociates prematurely?
    it must start over again
  10. How does termination signal the stop for the polymerase?
    a.       It usually has a string of A-T nucleotide pairs preceded by a twofold symmetric DNA sequence that folds into a hairpin structure. The formation of the hairpin helps to pull the RNA transcript from the active site. The DNA-RNA hybrid in the active site isn’t strong enough to hold the RNA in place, and it dissociates, causing release of the polymerase from the DNA. 
  11. 1)      What did researchers discover regarding the sequences of start and stop signals in prokaryotes? 
    a.       They are varied, yet they contain related sequences, such as a consensus sequence. The sequences differ in ways that determine their strength, allowing transcription to go at different rates. Transcription terminators also have a wide range of sequences, with the potential to form a simple hairpin RNA structure being the most important common feature
  12. 1)      What solves the problem of two different RNA molecules being able to be transcribed from any gene? 
    a.       A gene typically has only a single promoter because its sequence is asymmetric, allowing binding in only one orientation. Genome sequences reveal that the DNA strand used as the template for RNA synthesis varies from gene to gene depending on the location and orientation of the promoter. 
  13. Explain eukaryotic RNA polymerase.
    a.       They have RPI, RPII, and RPIII. The three are structurally similar to one another and share common subunits, but transcribe different types of genes. I and III transcribe the genes encoding tRNA, rRNA, and small RNAs. II transcribes most genes, including those that encode protiens. 
  14. 1)      Differentiate between RNA polymerase II and bacterial RNA polymerase. 
    • a.       Bacterial RNA polymerase requires only a single additional protein (sigma factor) for transcription initiation to occur in vitro, eukaryotic RNA polymerases require many additional proteins, collectively called the general transcription factors
    • b.      Eucaryoic transcription initationdeals with packing of DNA into nucleosomes and higher-order forms of chromatin structure, features absent from bacterial chromosomes
  15. 1)      Explain the assembly process of the transcription factors in eucaryotes. 
    • a.       It begins when TFIID binds to a short double-helical DNA sequence composed of T and As (TATA box); and, the subunit of TFIID that recognizes it is TBP. TFIID causes distortion of the DNA of the TATA box, which is recognized by the promoter, and it binds DNA sequences on both sides of the distortion together to allow for subsequent protein assembly units.
    • b.      Other factors then assemble with RNA polymerase II to form a complete transcription initiation complex.
    • c.       TFIIH is one; and, it consists of 9 subunits and it performs several steps needed to initiate transcription.
    • d.      After completeion of the complex, RNA polymerase II must gain access to the template strand. TFIIH contains a DNA helicase and hydrolyzes ATP to unwind the DNA.
    • e.      Then, RNA polymerase II remains at the promoter, making short lengths of RNA until it undergoes a series of conformational changes that allow it to move away from the promoter and enter the elongation phase of transcription. In this step, phosphate groups are added to the tail of the RNA polymerase (the CTD tail), allowing the polymerase to disengage from the cluster of general transcription factors.
    • f.        During this process, it undergoes a series of conformational changes that tighten its interaction with DNA, and it acquires new proteins that allow it to transcribe for long distances without dissociating.
    • g.       After the beginning of elongation, the general TFs are released to form an assembly elsewhere. 
  16. 1)      Aside from TFs, what else does RNA polymerase require? 
    • a.       First, gene regulatory proteins called transcriptional activators must bind sequences in DNA to attract RNA polymerase II to the start point.
    • b.      Second, eukaryotic transcription ignition requires the presence of a Mediator, which allows the activator proteins to communicate properly with the polymerase II and with the general transcription factors.
    • c.       Finally, transcription initiation in a eukaryotic cell requires the local recruitment of chromatin-modifying enzyme, like complexes and histone-modifying enzymes
    • d.      In regards to assembly, the RNA polymerase II must be released from the large complex of proteins, requiring the proteolysis of the activator protein. 
  17. 1)      Explain elongating and elongating RNA polymerases.
    a.       Elongating RNA polymerases are associated with a series of elongation factors, some of which can make transcription through nucleosomes easier without use of energy; they dislodge H2A-H2B dimers and replace them as the polymerase moves. They are also associated with chromatin-remodeling complexes, which move with the polymerase. 
  18. 1)      What may be the benefit of positive superhelical tension in DNA in front of the polymerase and negative helical tension behind it? 
    a.       The positive superhelical tension ahead makes the DNA more difficult to open, but the tension should facilitate the unwrapping of DNA in nucleosomes, as the release of DNA from the histone core helps to relax positive superhelical tension. 
  19. 1)      What is special about transcription elongation in eucaryotes? 
    • a.       Because it is tightly coupled to RNA processing, it involves covalent modification of the ends of RNA and removal of intron sequences that are discarded from the middle of the RNA transcript by RNA splicing.
    • b.      Capping of the 5’ end and polyadenylation of the 3’ end occur, which allow the cell to assess whether both ends of an mRNA molecule are present (and message is therefore intact) before it exports the RNA sequence from the nucleous and translates it into protein. 
  20. 1)      What is important of the CTD of RNA polymerase II? 
    a.       Phosphorylation of the C-terminal domain (tail of RNA polymerase II) helps dissociate RNA polymerase II from other protiens at the start of transcription, as well as allows a new set of proteins to associate with the RNA polymerase tail that function in transcription elongation and RNA processing, which hop on the tail to begin processing. CTD serves as a tether, holding a variety of proteins close by until needed. 
  21. 1)      Explain RNA capping. 
    a.       It occurs after about 25 nucleotides of RNA are synthesized. Three enzymes perform the reaction: one phosphatase removes a phosphate from the 5’ end of the nascent RNA, another (a guanyl transferase) adds a GMP in a reverse linkage (5’ to 5’) and a third (a methyl transferase) adds a methyl group to the guanosine. 
  22. Explain RNA splicing.
    • a.       Each splicing event removes one intron through transesterifications, which join two exons while removing the intron as a lariat (lasso).
    • b.      In the first step, a specific adenine nucleotide in the intron sequence attacks the 5’ splice site and cuts the sugar-phosphate backbone of the RNA at this pint.
    • c.       The cut 5’ end of the intron becomes covalently linked to the adenine nucleotide, thereby creating a loop in the RNA molecule.
    • d.      The released free 3’-OH end of the exon sequence then reacts with the start of the next exon sequence, joining the two exons together and releasing the intron sequence in the shape of a lariat.
    • The two exon sequences thereby become joined into a continuous coding sequence;  the released intron is degraded
  23. 1)      What is interesting about RNA splicing? 
    a.       It is performed by specialized RNA molecules, which recognize the nucleotide sequences that specify where splicing is to occur and also participate in splicing. They are short and there are five of them (U1, U2, U4, U5, and U6). They are snRNAs, each of which has at least seven protein subunits to forma snRNP, which form the core of the spliceosome. During the splicing reaction, recognition of the 5’ splice junction, the branch-point site, and the 3’ splice junction is performed largely through base-pairing between the snRNAs and the consensus RNA sequences in the pre-mRNA substrate. The spliceosome undergoes may shifts in which one set of base-pair interactions is broken and another is formed in its place. 
  24. 1)      What is ATP used for in the spliceosome?
    • a.       It is used for the assembly and rearrangements of the splieceosme. ATP hydrolysis can be used to break existing RNA-RNA interactions and form new ones. These ATP-requiring RNA-RNA rearrangements occur within the snRNPs and between snRNPs and pre-mRNA. And, these rearrangements crate active catalytic sites of the spliceosome, which ca prevent wayward splicing.
    • b.      ATP is also used to dissemble the snRNPs from the lariat and each other, causing the snRNAs to go to their original conformation. 
  25. 1)      Explain the catalytic site formed in the spliceosome.
    a.       It is formed from RNA; and, U2 and U6 snRNAs form the 3D RNA structure that juxtaposes the 5’ splice site of pre-mRNA with the branch point site and performs the first transesterification reaction. Similarly, the 5’ and 3’ splice junctions are brought together by U5 for second transesterication. At the completion of the splice, the spliceosome directs a set of proteins to bind to the mRNA near the position formerly occupied by the intron, called the exon junction complex. 
  26. 1)      What are the fidelity mechanisms built into the spliceosome supplemented by? 
    • a.       They are supplemented by two additional strategies that increase the accuracy of splicing.
    •                                                               i.      First, as transcription proceeds, the phosphorylated tail of RNA polymerase carries several components of the spliceosome, and these components are transferred directly from the polymerase to the RNA during synthesis, helping to keep track of introns and exons.
    •                                                             ii.      Another way is “exon definition” strategy, in which the splicing machinery initially seeks out the relatively homogenously sized exon sequences. As RNA synthesis proceeds, a group of additional components assemble (like SR—binds to RNA in exons, called splicing enhancers) on exon sequences and help to mark off eac 3’ and 5’ splice site starting at the 5’ end of the RNA. They recruit U1 to mark the downstream exon boundary and U2AF to specific the upstream one.
  27. 1)      What is different in splicing of higher eucaryotes as opposed to yeast? 
    a.       We have a second set of snRNPs that direct the splicing of the small fraction of our intron sequences. It recognizes a different set of RNA sequences at the 5’ and 3’ splice junctions and at the branch point; it is called the U12-type spliceosome because of the involvement of the U12 SnRNP. For the U12 spliceosome, its splicing is delayed, allowing co-regulation of several hundred gens, whose expression requires this spliceosome.  
  28. How is splicing flexible?
    a.       A mutation in a nucleotide sequence critical for splicing does not really prevent splicing of that intron. Instead, it creates a new pattern of splicing. Most commonly, an exon is simply skipped. In other cases, the mutation causes a cryptic splice junction to be efficiently used. Only the best pattern of splice junctions are used. Also, some alternative splicing are constitutive; and, the cell regulates the splicing patterns so that different forms of the protein are produced at different times in different tissues
  29. 1)      What are the two major classes of self-splicing intro sequences? 
    • a.       Group 1 intron sequences begin the splicing reaction by binding a G nucleotide to the intron sequence; this G is thereby activated to form the attacking group that will break the first of the phosphodiester bonds cleaved during splicing.
    • b.      In group II intron sequences, an especially reactive A residue in the intron sequence is the attacking group, and a lariat intermediate is genetated.
    • c.       For both, the nucleotide sequence of the intron is critical: the intron RNA folds into a specific 3D structure, which brings the 5’ and 3’ splice junctions together and provides precisely positioned reactive groups to perform the chemistry. 
  30. 1)      Explain RNA processing at the 3’ end of eukaryotic mRNA. 
    a.       The position of the 3’ end is specififed by a signal transcribed into RNA and recognized by RNA-binding proteins and RNA-processing enzymes, two protiens of which are called CstF (cleavage stimulation factor) and CPSF (cleavage and polyadenylation specificity factor). Both travel with the RNA polymerase tail and are transferred to the 3’ end processing sequence on an RNA molecule as it emerges from the RNA polymerase.
  31. What happens when CstF and CPSF bind?
    • a.       other proteins assemble with them to create the 3’ end of the mRNA.
    •                                                               i.      First, RNA is cleaved
    •                                                             ii.      Then, poly-A polymerase (PAP) adds, one at a time, approximately 200 A nucleotides to the 3’ end produced by the cleavage. The nucleotide precurrosr for these additions is ATP..
    • 1.       PAP does not require a template; and poly-A binding proteins assemble onto it to determine the final length of its tail. Some remain bounda as mRNA goes from the nucleos to the cytosol; and, they help to direct the synthesis of a protein on the ribosome.
    • b.      After the 3’ end of pre-mRNA is cleaved, the RNA polymerase II continues to transcribe, eventually releasing its grip on the template and terminating transcription. The newly made RNA that emerges from the polymerase lacks a 5’ cap and is degraded. 
  32. 1)      HOW DOES THE CELL distinguish between the rate mature mRNA and the debris from processing? 
    a.       As RNA is made, it loses certain proteins and acquires others. It can be distinguished by the proteins it lacks. Only when proteins present on an mRNA signify that processing was successfully completed is mRNA exported from the nucleus for translation. Other debris are degraded by the nuclear exosome, a large protein complex whose interior is rich in 3’-5’ RNA exonucleases. 
  33. In eukaryotes, what is tightly coupled?
    mRNA synthesis, processing,and transport
  34. Explain rRNAs/
    • a.       They are the most abundant RNAs in cells and are made by RNA polymerase I, which lacks a C-terminal tail. RNA components of the ribosome are final gene products; and so, millions of copies are needed, accounting for the multiple copies of rRNA genes hat code for rRNAs.
    • b.      In eucaryotes, there are four: 18S, 5.8S, 28S (these three are made by chemically modifying and cleaving a signle large precursor rRNA) and a fourth (5S RNA) is made by a separate cluster of genes by a different polymerase
  35. 1)      Explain rRNA modification. 
    • a.       rRNA undergoes extensive chemical modification—100 methylations of the 2’-OH positions on nucleotide sugars and 100 isomerizations of uridine nucleotides to pseudouridine. These may aid in folding and assembly of rRNA.
    •                                                               i.      These modifications are made at specific positions that are specified by 150 “guide” RNAs” which position themselves through base-pairing to the precursor rRNA and bring an RNA-modifying enzyme to the appropriate position. Other guide RNAs promote cleavage into mature rRNAs. All of these guide RNAs are members of the small nucleolar RNA family many of which are encoded int eh introns of other genes. 
  36. Explain ribosome assembly.
    a.       The 45S precursor rRNA is packaged in a large ribonucleoprotein particle containing many ribosomal proteins imported from the cytoplasm. While this particle remains at the nucleolus, selected pieces are added and others discarded as it is processed into immature large and small ribosomal subunits. The two ribosomal subunits are thought to attain their final functional form only as each is individually tnrasported through the nuclear pores into the cytoplasm. 
  37. 1)      What are other subnuclear structures?
    a.       The Cajal bodies and Gemini of Cajal bodies. They are paired and are the locations in which snoRNAs and snRNAs undergo covalent modifications and final assembly with proteins. Small Cajal RNAs select the sites of these modifications through base pairing. snRNPs may also be recycled here.