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)
- 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_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_forwardA number of mutations affect the expression of the lac operon in E. coli. Consider each genotype below and complete the table using “+” to indicate that the gene is expressed, and “−” to indicate that gene is not expressed.arrow_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 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_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_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_forwardMolecular biology, please explain in detailarrow_forwardWhat would happen if the operator sequence of the trp operon contained a mutation that prevented the repressor protein from binding to the operator? (Explain what would happen in both the presence and absence of tryptophan)arrow_forward
- A number of mutations affect the expression of the lac operon in E. coli. The genotypes of several E. coli strains are shown below. ("+" indicates a wild-type gene with normal function and "-" indicates a loss-of-function allele.) Please predict which of the following strains would have the highest beta-galactosidase enzyme activity, when grown in the lactose medium. O CAP+ r* p* o* z O CAP* I P* o* z* O CAP* r* P O* z* O CAP I P* O z*arrow_forward. Listed in parts a through g are some mutations that were found in the 5′ UTR of the trp operon of E. coli. What will the most likely effect of each of these mutations be on the transcription of the trp structural genes? Q. A mutation that prevents the binding of the ribosome to the 5′ end of the mRNA 5′ UTRarrow_forwardHere is the model for TfR mRNA regulation (see Fig.), for your convenience. Orange protein (IRP or IREBP), when bound to RNA, protects mRNA from degradation. When Fe is bound to the protein, it cannot bind the mRNA. If you made a radioactive DNA probe specific to TfR mRNA and performed a Northern blot analysis (to detect mRNA) from cells treated with high iron (Fe) concentration what would you expect to see? (a) Low iron RNase A (b) High iron O High levels of RNase O Low levels of TfR mRNA Fe RNase O The result would be the same as untreated cells Degradation O High levels of IRP (or IREBP - iron responsive element binding protein) O High levels of TfR mRNAarrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning