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Berkeley MCELLBI 110 - MIDTERM #2 answer key

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Your name: Student ID #: 1 MCB110, Fall 2007 MIDTERM #2 answer key Question 1 1. Answer: The full-length mRNA transcribed from the trp operon is polycistronic, consisting of multiple open reading frames for the expression of not only the leader peptide but also the trpA-E gene products. Each open reading frame has its own start (AUG) and stop codons as well as ribosom-binding site (S-D sequence) and recruits ribosomes independently for its own translation. Thus, the ribosome stuck on the leader mRNA will have no consequence on the translation of the downstream genes as they are translated by different ribosomes. 2. Answer: The two mechanisms of (1) capping enzyme binding to initiated RNA Pol II specifically through its unique, Ser5-phosphorylated CTD, which is absent in Pol I and III, and (2) activation of capping enzyme activity by binding to the phosphorylated CTD result in specific capping of the minor fraction of Pol II transcripts. 3. Answer: The described computational approach depends on comparative analysis of conserved DNA sequences derived from multiple species across a major spectrum of evolution. If only sequences from highly similar species are available for analysis, it is difficult to filter out noises and pick up specific, meaningful signals. 4. Answer: Accessibility to AUG on mRNA controls the efficiency of translation. Because chromatin-remodeling complexes can only work on chromatin DNA but not RNA molecules, they play no role in translational control. 5. Answer: Stimulation of paused Pol II permits a rapid response. These genes are always paused in a state of suspended transcription and therefore, when an emergency arises, require no time to remodel and acetylate chromatin over the promoter and assemble a transcription pre-initiation complex. Many genes that are controlled in this fashion encode proteins that must act rapidly to respond to sudden environmental changes, such as heat shock. 6. Answer: The mutations are likely to be in the exonic splicing enhancers. They prevent the formation of the cross-exon recognition complex. As a result, the affected exon is "skipped" during splicing and is not included in the final processed mRNA. Question 2 (32 points). 1. Answer: First, you will have to introduce a nucleotide substitution into the DNA segment that constitutes the half site of the G/C-rich sequence with two-fold rotational symmetry. This mutation will disrupt the precise base-pairing between the two segments of RNA transcribed from the two half sites. To preserve the G/C-rich nature of the sequence, a G will be changed to C or a C to G. When tested in transcription, this mutant is expected to be inactive or significantly less active than the wild-type in enabling polymerase to terminate transcription. (8 points).Your name: Student ID #: 2 To further confirm that it is the base-pairing of the RNA stem, but not an unrelated function of the terminator that is destroyed by your point mutation, you need to introduce a second, so-called compensatory mutation into the second half site of the palindromic G/C-rich sequence to restore base-pairing and integrity of the RNA stem. This new mutant is expected to display wild-type activity in carrying out transcription termination. (8 points). 2. Answer: You can use the chromatin immunoprecipitation (ChIP) assay to confirm these observations. The key is to use multiple (at least two) sets of PCR primers to amplify different regions along the chromatin template, with one set to specifically exam the protein occupancy at the promoter region and the other at an interior or 3' region of the gene. Please refer to the ChIP figure in your reader and the situation here is nearly identical to the different behaviors observed of TBP and Pol II in that experiment. If Brd4 stays put at the promoter region and does not travel together with P-TEFb and RNA Pol II during elongation, you should expect to see the purified DNA obtained through anti-Brd4 immunoprecipitation to get amplified by only the promoter-specific PCR primer set but not any other sets corresponding to the interior or 3’ end of the gene. In contrast, DNA obtained through anti-P-TEFb immunoprecipitation should become amplified by all primer sets and show a similar distribution pattern across the entire transcription unit. Question 3 (34 points). Part 1. Answer: Mutant 1 makes a CAP protein that can bind to a cognitive promoter element even when cAMP levels are low. Also, β-gal levels are always high in this strain. Thus, mutant 1 appears to have a mutation in the CAP gene, which leads to the production of a CAP protein that binds to DNA and activates transcription constitutively (all the time), regardless of the cAMP levels. Mutant 2 always has low levels of cAMP, and transcription of the lac genes is low all the time. Thus, mutant 2 probably has a mutation in the adenylyl cyclase gene, which results in a loss of adenylyl cyclase activity. This would result in a failure to activate the CAP protein by cAMP when glucose levels become low.Your name: Student ID #: 3 Part 2 E. coli partial diploids When glucose levels in medium Cellular cAMP levels… …and CAP promoter binding activity … and β-galactosidase activity (IPTG always in media) Mutant 1 + WT Genomic DNA Low High High Low Yes Yes High High Mutant 2 + WT Genomic DNA Low High High Low Yes No High Low Answer: Mutant 1 makes a CAP protein that binds to DNA and activates transcription all the time, regardless of cAMP levels. Thus this mutation would be dominant over the wild type CAP gene function and the introduction of the wild type CAP gene cannot rescue the defect caused by the mutation. Mutant 2 has a mutation in the adenylyl cyclase gene, which results in a loss of the enzymetic activity. Thus, introduction of a normal copy of the adenylyl cyclase gene into mutant 2 would again allow the production of cAMP when glucose levels become limiting, correcting the defects caused by the mutation in the chromosomal adenylyl cyclase gene. Question 4 (19 points). (a) Answer: σ32


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Berkeley MCELLBI 110 - MIDTERM #2 answer key

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