The tac promoter, an artificial promoter made from portions of the frp and lacUV5 promoters, has been introduced into a plasmid. It is a hybrid of the lac and trp (tryptophan) promoters, containing the –35 region of one and the -10 region of the other. This promoter directs transcription initiation more efficiently than either the trp or lac promoters. Why?
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- The tac promoter, an artificial promoter made from portions of the trp and lacUV5 promoters, has been introduced into a plasmid. It is a hybrid of the lac and trp (tryptophan) promoters, containing the −35 region of one and the −10 region of the other. This promoter directs transcription initiation more efficiently than either the trp or lac promoters. Why?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 1.Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds. 2.Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. 3.Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. 4.Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination.. The tac promoter, an artificial promoter made from a chemically syn- thesized oligonucleotide, has been introduced into a plasmid. It is a hybrid of the lac and trp promoters, containing the – 35 region of one and the – 10 region of the other. This promoter directs transcription initiation more efficiently than either the trp or lac promoters. Why?
- An electrophoretic mobility shift assay can be used to study the binding of proteins to a segment of DNA. In the results shown here, an EMSA was used to examine the requirements for the binding of RNA polymerase |l (from eukaryotic cells) to the promoter of a protein-encoding gene. The assembly of general transcription factors and RNA polymerase Il at the core promoter is described in Week 4. In this experiment, the segment of DNA containing a promoter sequence was 1100 bp in length. The fragment was mixed with various combinations of proteins and then subjected to an EMSA. Lane 1: No proteins added Lane 2: TFIID Lane 3: TFIIB Lane 4: RNA polymerase IIl Lane 5: TFIID + TFIIB Lane 6: TFIID + RNA 1 2 3 4 5 6. 7 polymerase II Lane 7: TFIID + TFIIB + RNA polymerase Il 1100 bp Explain the results.Negative supercoiling of DNA favors the transcription of genes because it facilitates unwinding. However, not all promoter sites are stimulated by negative supercoiling. The promoter site for topoisomerase II itself is a noteworthy exception. Negative supercoiling decreases the rate of transcription of this gene. Propose a possible mechanism for this effect and suggest a reason why it may occur.Below is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'
- Bacterial DNA containing an operon encoding three enzymes is introduced into chromosomal DNA in yeast (a eukaryote) in such a way that it is properly flanked by a promoter and a transcriptional terminator. The bacterial DNA is transcribed and the RNA correctly processed, but only the protein nearest the promoter is produced. Can you suggest why?The dlagram below represents the tryptophan operon with the trp leader MRNA transcript enlarged to represent the AUG transiation start codon, two consecutive tryptophan amino acld codons (UGGUGG), and 4 regions (1, 2, 3, and 4) that base pair to form different hairpin-loop structures in the MRNA leader region. What would happen in this MRNA leader region when cells encounter very low levels of tryptophan in Its environment? Leader region trpE trpD trpC trpB trpA DNA 5' 3' Transcription trp leader sequence mRNA UGGUGG (tryptophan codons) AUG UUUUUU 1 3 4. The translating ribosome would stall at the two tryptophan codons, causing the formation of a hairpin-loop botween regions 3 and 4 to promote transcription of the trp operon. The translating ribosome would stall at the two tryptophan codons causing formation of hairpin-loop between regions 2 and 3, which functions as an anti-lerminator of transcription The translating ribosome would stall at the two tryptophan codons causing formation…In addition to Tc1, the C. elegans genome contains otherfamilies of DNA transposons such as Tc2, Tc3, Tc4, andTc5. Like Tc1, their transposition is repressed in thegerm line but not in somatic cells. Predict the behaviorof these elements in the mutant strains where Tc1 is nolonger repressed due to mutations in the RNAi pathway.Justify your answer.
- The following logo plot represents the preferred cis-regulatory sequences (i.e. transcription factor binding site) of bHLH transcription factor FOSL1. C 1 2 3 4 5 6 7 8 9 10 11 position Would you expect this sequence to be recognized by a monomer, a homodimer, or a heterodimer of the protein? Explain your answer. (short phrases are sufficient; please write your answer into the template below) A- В I A -l expect FOSL1 to bind as a: (monomer, homodimer, heterodimer; please choose) B - short explanation: information content (bit) !!A bacterial species has a hypothetical sigma promoter that has the following sequence: TTGGCA - 18 bases - TATAAT What change in the level of transcription would there be if the sequence was mutated to: TTCGCA -18 bases -TATAAT Group of answer choices 1.The mutation would inhibit the promoter thereby inhibiting transcription 2.No change the consensus TATAAT sequence in the same. 3.The mutation would move the promoter away from consensus and reduce the level of transcription 4.The mutation would bind the promoter to the consensus and produce normal levels of transcriptionTo characterize the promoter of the gadA operon you made a series of deletion mutants removing pieces of the promoter to see what would happen. The results are found below: gad promoter gada gadX gadz 450 +1 lacz activity transcription start site pH 2.0 pH 7.0 A gad promoter beta-galactosidase (lacZ) +++ 450 gad promoter beta galactosidase (lacZ) +++ +++ 300 +1 gad promoter beta galactosidase (lacZ) 150 D gad beta galactosidase (lacz) -450 150 E gad promoter beta-galactosidase (lacZ) -450 -300 Based on these results, what can you conclude about the gad promoter? O a. The promoter is only regulated by repression Ob. The promoter is regulated by a mix of activation and repression O c. The promoter is only regulated by activation O d. The promoter has multiple operators and multiple enhancers
