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- The locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…The locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…The locations of the TATA box in two species of yeast, Saccharomyces pombe and Saccharomyces cerevisiae, differ dramatically. The TATA box of S. pombe is about 30 nucleotides upstream of the transcription start site, similar to the location in most other eukaryotic cells. However, the TATA box of S. cerevisiae is 40 to 120 nucleotides upstream of the start site. To better understand what sets the start site in these organisms, researchers at Stanford University conducted a series of experiments to determine which components of the transcription apparatus of these two species could be interchanged (Y. Li et al. 1994. Science 263:805–807). In these experiments, different general transcription factors and RNA polymerases were switched in S. pombe and S. cerevisiae, and the effects of each switch on the level of RNA synthesis and on the starting point of transcription were observed. The results from one set of experiments are shown in the table below. Components cTFIIB, cTFIIE, cTFIIF,…
- Mutations in bacterial promoters may increase or decrease therate of gene transcription. Promoter mutations that increasetranscription are termed up-promoter mutations, and those thatdecrease transcription are termed down-promoter mutations.The sequence of the −10 site of the promoter for the lac operonis TATGTT (see Figure 14.5). Would you expect each of thefollowing mutations to be an up-promoter or down-promotermutation?A. TATGTT to TATATTB. TATGTT to TTTGTTC. TATGTT to TATGATThe IMD2 promoter contains three upstream transcription start sites (TSS) that are utilized under high GTP conditions and a single downstream TSS (-106) that is normally only utilized under low GTP conditions. In a wild type cell, expression of IMD2 mRNA only occurs if transcription initiates from the -106 TSS. In 300 words or less, describe: 1.) The normal function of Ssl2, and 2.) why a mutation in Ssl2, that increases its catalytic rate, would allow expression of the IMD2 ORF under high GTP conditions. (Conditions under which the IMD2 ORF is NOT expressed in the wild type.)What would you expect krox20 expression to look like in embryos treated with an RA inhibitor? Why? Strictly no plagiarism.
- Mutant # Deletion Region %Transcription (lungs) %Transcription(kidneys) WT None 100% 100% 1 1-200 130% 130% 2 250-400 100% 100% 3 500-800 50% 50% 4 950-1100 0% 100% 1) For the following gene, you notice the following results when analyzing several muntants. What type of sequence (be specific) has been mutated in mutant 1? What type of sequence (be specific) has been mutated in mutant 3? What would we call this type of mutation? What type of sequence (be specific) has been mutated in mutant 4?The DNA sequence of the promoter region of E. coli xyzA gene is shown below. Transcription start site is the A (in bold) at position 43. 5 10 15 20 25 30 35 40 45 50 GAGCT GTTGA CAATT AATCA TCGAA CTAGT TAACT AGTAC GCAAG TTCAC Mutations were introduced in the sequence to identify residues important for gene expression. Indicate the effect of the following mutations on xyzA expression (increase, decrease, no effect, cannot be predicted). Provide reasoning for each answer. A. G3A (G at position 3 was changed to A) G9A Deletion of TCA at position 18-20 C22A T31A, A32T double mutant T35G G45C C48A B. What are the promoter sequences of the gene?Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.
- Comparing the -10 regions of two E. coli promoters which have identical -35 regions revealed the sequence TATAAT for the first and GATACT for the second one. Why does the first promoter cause a higher transcription rate than the second one? a. The transcription rate from the first promoter will be higher, because RNA polymerase will bind TATAAT with a higher affinity than GATACT. b. It will be higher, because formation of the open promoter complex is more easily achieved with TATAAT than with GATACT. c. It will be higher, because TATAAT of the -10 region is transcribed into UAUAAU, which forms fewer hydrogen bonds with the template strand than GAUACU. d. a and b, but not c e. a, b, and cAn electrophoretic mobility shift assay can be used to study the bindingof proteins to a segment of DNA. In the results shown here, anelectrophoretic mobility shift assay was used to examine the requirementsfor the binding of RNA polymerase II (from eukaryotic cells)to the promoter of a protein-encoding gene. The assembly of generaltranscription factors and RNA polymerase II at the core promoteris described in Chapter 14 (Figure 14.14). In this experiment, thesegment of DNA containing a promoter sequence was 1100 bpin length. The fragment was mixed with various combinations ofproteinsand then subjected to an electrophoretic mobility shift assay.Explain which proteins (TFIID, TFIIB, or RNA polymerase II) areable to bind to this DNA fragment by themselves. Which transcriptionfactors (i.e., TFIID or TFIIB) are needed for the binding ofRNA polymerase II?True/False: In general, weak promoters that support active transcription have -10 and -35 boxes that are close to the consensus sequences and a spacer between the two boxes that is 17 +/- 1 bp.
