What would happen to regulation from a promoter under negative control if the region where the regulatory protein binds was deleted? What If the promoter was under positive control? Promoters from Escherichia coli under positive control are not close matches to the promoter consensus sequence for E., coll. Why?
To explain:
What would happen to regulation process from a promoter under negative control if the region where the regulatory protein binds was deleted. What if the promoter was under positive control.
Concept introduction:
The regulation of transcription is effective in negative control, because the cell needs the signal for the presence of enough biosynthetic products for their growth. The transcription is also controlled by both negatively or positively because of the presence of lactose and absence of glucose in the culture medium.
Explanation of Solution
The operonal gene products are produced in high amount when the regulatory protein-binding region of a promoter are deleted under negative control. Because of the repressor protein cannot bind to the promoter region. The operonal gene products are not produced when the regulatory protein-binding region of a promoter are deleted. This is because of the activator which cannot attach itself to the promoter region.
To explain:
Promoters from Escherichia coli under positive control are not close matches to the promoter consensus sequence for E. coli. Why.
Concept introduction:
The regulation of transcription is effective in negative control, because the cell needs the signal for the presence of enough biosynthetic products for their growth. The transcription is also controlled by both negatively or positively because of the presence of lactose and absence of glucose in the culture medium.
Explanation of Solution
The DNA consequences matches closely with the promoter region that leads to the binding of the RNA polymerase to the promoter region and stimulates the transcription without the presence of other factors. Thus, a quite different transcription control is seen in the promoters under positive control.
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Chapter 6 Solutions
Brock Biology of Microorganisms (15th Edition)
- 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?arrow_forwardAnswer these questions concerning promoters. a) What role do promoters play in transcription? b) What is the common structure of bacterial promoter with respect to consensus sequences? c) Eukaryotic promoters are more variable than bacterial promoters. Why? d) What is the meaning of the term alternative promoter? How does the use of alternative promoters affect transcription?arrow_forwardThe diagram below represents a hypothetical operon in the bacterium E. coli. The operon consists of two structural genes (A and B), which code for the enzymes “Aase" and "Base", respectively, and also includes P (promoter) and O (operator) regions as shown. A В When a certain compound (X) is added to the growth medium of E. coli, the separate enzymes "Aase" and "Base" are both synthesized at a 50-fold higher rate than in the absence of X. (X has a molecular weight of about 200.) Which of the following statements is true of the operon described above? The region of the A gene that codes for the carboxyl-terminal amino acid of “Aase" is near the left end of the A gene. The P region contains nucleotide sequences to which the RNA polymerase holoenzyme (including the o subunit) binds specifically but which the core enzyme does not recognize. The addition of X to the growth medium causes a repressor protein to bind tightly to the O region. The mRNA copied from this operon will be covalently…arrow_forward
- Tryptophan (Trp) attenuation is an extra mechanism of regulation of trp operon when trp level is too high. Attenuation causes premature transcription termination. a) Propose a solution to overcome a mutated trp codon in the Trp attenuator sequence (domain 1)arrow_forward. a. How many ribosomes are required (at a minimum)for the translation of trpE and trpC from a singletranscript of the trp operon?b. How would you expect deletion of the two tryptophan codons in the RNA leader to affect theexpression of the trpE and trpC genes?arrow_forwardA 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 transcriptionarrow_forward
- You made four mutants for a promoter sequence in DNA and studied them for transcription. The results of the amount of gene expression or transcription (based on beta-Gal activity shown on Y-axis) for these DNAs (X-axis) are shown. The sequence of the wild-type and mutant DNAs, and consensus sequence from many promoters are shown here for your convenience. From this experiment you can conclude that: Nucleotide substitution can identify important bases of the binding sites or promoter in DNA (e.g., -10 and -35 promoter sequences of lac operon). True or false: Spacer (a) -10 region -35 region TTGACA Consensus sequence TATAAT Wild-type Lac promoter GGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGAATT Mutant 1 GGCTTTACACTTTATG-TTCCGGCTCGTATGTTGTGTGGAATT Mutant 2 GGCTTTACACTTTATGCTTCCGGCTCGTATAATGTGTGGAATT Mutant 3 GGCTTTACACTTTATG-TTCCGGCTCGTATAATGTGTGGAATT Mutant 4 GGCTTGACACTTTATG-TTCCGGCTCGTATAATGTGTGGAATT (b) 700 600- 500- 400- 300- 200- 100. 0 ● True O False B-Galactosidase activity Wild-type…arrow_forwardThe diagram below represents the tryptophan operon with the trp leader mRNA transcript enlarged to represent the AUG translation start codon, two consecutive tryptophan amino acid codons (UGGUGG), and 4 regions (1, 2, 3, and 4) that base pair to form different hairpin-loop structures in the mRNA leader region. Suppose a mutant bacteria has region 3 of the trp operon attenuator region mutated so that it cannot base pair normally. Would the bacteria grow in the absence of the amino acid tryptophan? (hint: in order for bacteria to grow in absence of tryptophan it should be able to synthesize its own tryptophan) Leader region trpE trpD trpC trpB trpA DNA 5' 3' Transcription trp leader sequence MRNA AUG UGGUGG UUUUUU 1 2 3 (tryptophan codons) There is insufficient information to answer the question. Yes No O Oarrow_forwardThe diagram below represents the tryptophan operon with the trp leader mRNA transcript enlarged to represent the AUG translation start codon, two consecutive tryptophan amino acid codons (UGGUGG), and 4 regions (1, 2, 3, and 4) that base pair to form different hairpin-loop structures in the MRNA leader region. Suppose a mutant bacteria has region 4 of the trp operon attenuator region mutated so that it cannot base pair normally. Would the bacteria grow in the absence of the amino acid tryptophan? (hint: in order for bacteria to grow in absence of tryptophan it should be able to synthesize its own tryptophan) Lead&r region trpE trpD trpC trpB trpA DNA 5' 3' Transcription trp leader sequence MRNA UGGUGG 1 (tryptophan codons) AUG UUUUUU No There is insutficient information to answer the question. O Yesarrow_forward
- The streptolysin S toxin made by S. pyogenes is encoded by a 9-gene operon, sagABCDEFGHI. Thinking about what a 3-line diagram would look like for this operon, answer the following questions. Write numeric answers only. For example, if your answer is 6 promoters, write only 6. 1) How many promoters control the expression of these genes? 2) How many locations does RNA Polymerase bind to get full expression of these genes? 3) How many ribosome binding sites are needed for full protein expression? 4) How many start codons will be needed for full protein expression? 5) How many mRNA strands will be produced with full operon expression? 6) How many proteins will be produced with full protein expression? 1arrow_forwardA 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 1.The mutation would move the promoter away from consensus and reduce the level of transcription 2.No change the consensus TATAAT sequence in the same. 3.The mutation would bind the promoter to the consensus and produce normal levels of transcription 4.The mutation would inhibit the promoter thereby inhibiting transcriptionarrow_forwardThe 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?arrow_forward
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