Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 14, Problem 28P
How would mutations that inactivate each of the following genes affect the determination of the lytic or lysogenic life cycle in mutated
a. cIc.croe. cII and cro
b. cIId.intf. N
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Chapter 14 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 14 - 12.1 Bacterial genomes frequently contain groups...Ch. 14 - Transcriptional regulation of operon gene...Ch. 14 - Why is it essential that bacterial cells be able...Ch. 14 - Identify similarities and differences between an...Ch. 14 - The transcription of -galactosidase and permease...Ch. 14 - 12.6 Is attenuation the product of an allosteric...Ch. 14 - The trpL region contains four repeated DNA...Ch. 14 - The CAP binding site in the lac promoter is the...Ch. 14 - What role does cAMP play in transcription of lac...Ch. 14 - How would a cap- mutation that produces an...
Ch. 14 - Explain the circumstances under which attenuation...Ch. 14 - Consider the transcription of genes of the...Ch. 14 - Describe the lytic and lysogenic life cycles of ...Ch. 14 - 12.14 Define antisense RNA, and describe how it...Ch. 14 - 12.15 Attenuation of trp operon transcription is...Ch. 14 - 12.16 In the lac operon, what are the likely...Ch. 14 - Identify which of the following lac operon haploid...Ch. 14 - Prob. 18PCh. 14 - 12.19 List possible genotypes for lac operon...Ch. 14 - Suppose each of the genotypes you listed in parts...Ch. 14 - 12.21 Four independent mutants (mutants A to D)...Ch. 14 - Suppose the lac operon partial diploid...Ch. 14 - Prob. 23PCh. 14 - 12.24 A repressible operon system, like the trp...Ch. 14 - 12.25 What is the likely effect of each of the...Ch. 14 - 12.26 Suppose that base substitution mutations...Ch. 14 - 12.27 Two different mutations affect. Mutant...Ch. 14 - How would mutations that inactivate each of the...Ch. 14 - The bacterial insertion sequence IS 10 uses...Ch. 14 - 12.34 Northern blot analysis is performed on...Ch. 14 - 12.37 The electrophoresis gel shown in part (a) is...Ch. 14 - Prob. 32PCh. 14 - The following hypothetical genotypes have genes A,...Ch. 14 - For an E. coli strain with the lac operongenotype...
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- The size of a DNA fragment that can be inserted into an unmodified λ vector is very limited. Large segment in the central region of the λ DNA molecule can be removed without affecting the ability of the phage to infect Escherichia coli cells. However, the deletion of this non- essential region does affect the life cycle of the bacteriophage. Explain this statement.arrow_forwardThe linear dsDNA genome of λ binds on the LamB receptor of E. Coli and conducts a normal lysogenic cycle. Exposure to stress will cause the excision of λ prophage from the E. Coli genome. The excised λ genome is then replicated, packaged, and released from the cell as mature λ phage particles and ready to infect other bacterial cells. Among λ phage particles,the transducing phage mediates a specific type of recombination. Understand this scenario and answer the following questions. 1. What are the basic requirements for the insertion of λ into the E. Coli genome? 2. What special features are found in the λ insertion site? 3. What type of recombination occurs with λ insertion in the E. Coli genome? 4. How you will differentiate λ transducing phage from normal λ phage? 5. What exclusive mechanism λ phage utilizes for recombination?arrow_forwardA researcher is studying the rII locus of phage T4. FourrII− strains are obtained: A, B, C, and D. In the first experiment, E. coli strain K(λ) is coinfected with two rII− strains simultaneously and the results are recorded. Infection with A and B phage = lysis occurs Infection with A and C phage = lysis occurs Infection with B and C phage = no lysis occurs Infection with B and D phage = no lysis occurs Infection with C and D phage = no lysis occurs In a second experiment, coinfections are performed first in E. coli strain B, then the progeny phage are used to infect E. coli strain K(λ). Progeny of A and B phage = plaques form Progeny of B and C phage = plaques form Progeny of C and D phage = plaques form Progeny of B and D phage = no plaques from Which conclusions are consistent with these data? Why? A) Strains A and B carry mutations in the same gene. B) Strains B and D both carry the same mutation. C) Strains B, C, and D carry mutations in the same gene. D)…arrow_forward
- In Hershey-Chase experiment, bacteriophages protein coats were tagged with radioactive isotope S-32. These phages were used to infect E. coli cells and the cells were further centrifuged to form pellets. Why was the radioactivity level of S-32 found greater outside the cells compared to the E. coli cell pellets? Explain briefly. If the experiment is repeated in the same manner but this time the phage protein coats are labelled with isotope X and the phage DNA with isotope Y, which isotope’s radioactivity will be found in greater amounts in the E. coli cell pellets after centrifugation? Explain briefly.arrow_forwardIn a genetics lab, Kim and Maria infected a samplefrom an E. coli culture with a particular virulent bacteriophage. They noticed that most of the cells werelysed, but a few survived. The survival rate in theirsample was about 1 × 10−4. Kim was sure the bacteriophage induced the resistance in the cells, whileMaria thought that resistant mutants probably alreadyexisted in the sample of cells they used. Earlier, for adifferent experiment, they had spread a dilute suspension of E. coli onto solid medium in a large petri dish,and, after seeing that about 105colonies were growingup, they had replica-plated that plate onto three otherplates. Kim and Maria decide to use these plates totest their theories. They pipette a suspension of thebacteriophage onto each of the three replica plates.What should they see if Kim is right? What shouldthey see if Maria is right?arrow_forwardAhmed had isolated bacteria from the sputum of a patient that seemed somewhat resistant to their antibiotic treatment, since multiple weeks of treatment had yielded no satisfactory improvement of their active disease. The researcher had then resuspended the patient's sputum in saline and used a mouse model to measure the efficiency of a phage selected for its capacity to target M. mycobacterium, as a single therapy, and as a combination therapy with antibiotics. Here is a schematic representation of the experiment: Group 1 Control (saline) Inoculate mice with patient sputum Group 2 wait for active disease to show Treated with phage Sputum samples Collect lung sample every 2 weeks Tuberculosis patient Group 3 Treated with Abx Split mice showing active disease in 4 Group 4 Treated with Abx and Count bacteria in sample groups phage Figure 4. Sputum from patients were isolated and resuspended in saline. Mice were given equal amounts of resuspended sputum intra-nasally and were left to…arrow_forward
- Ahmed had isolated bacteria from the sputum of a patient that seemed somewhat resistant to their antibiotic treatment, since multiple weeks of treatment had yielded no satisfactory improvement of their active disease. The researcher had then resuspended the patient's sputum in saline and used a mouse model to measure the efficiency of a phage selected for its capacity to target M. mycobacterium, as a single therapy, and as a combination therapy with antibiotics. Here is a schematic representation of the experiment: Group 1 Control (saline) Inoculate mice with patient sputum Group 2 wait for active disease to show Treated with phage Sputum samples Collect lung sample every 2 weeks Tuberculosis patient Group 3 Treated with Abx Split mice showing active disease in 4 Group 4 Treated with Abx and Count bacteria in sample groups phage Figure 4. Sputum from patients were isolated and resuspended in saline. Mice were given equal amounts of resuspended sputum intra-nasally and were left to…arrow_forwardAhmed had isolated bacteria from the sputum of a patient that seemed somewhat resistant to their antibiotic treatment, since multiple weeks of treatment had yielded no satisfactory improvement of their active disease. The researcher had then resuspended the patient's sputum in saline and used a mouse model to measure the efficiency of a phage selected for its capacity to target M. mycobacterium, as a single therapy, and as a combination therapy with antibiotics. Here is a schematic representation of the experiment: Group 1 Control (saline) Inoculate mice with patient sputum Group 2 wait for active Treated with phage disease to show Collect lung sample every 2 weeks Tuberculosis Sputum samples patient Group 3 Treated with Abx Group 4 Split mice showing active disease in 4 Treated with Abx and Count bacteria in sample groups phage Figure 4. Sputum from patients were isolated and resuspended in saline. Mice were given equal amounts of resuspended sputum intra-nasally and were left to…arrow_forwardBacteria exposed to viruses incorporate sections of the virus’s DNA into the CRISPR array sequences in their genome. This mechanism allows bacteria to fight off the viruses, like an immune response: the information in CRISPR spacers served as “coordinates” for recognizing and cutting up invading DNA sequences. Describe what might happen under the conditions described after a bacteriophage infects a bacterial cell and releases its DNA into the bacterial cell. Explain why: 1. The invading phage DNA is recognized by the Cas proteins but not inserted into the CRISPR array region of the bacterial genome: The bacteria will be unable to elicit an immune response and will succumb to the phase infection 2. The cas genes on the bacterial genome contains a missense mutation that increases its cleavage/cut activityThe bacteria will elicit an immune response that will successfully fight the phage infectionarrow_forward
- M13 is a filamentous phage that infects the bacterium Escherichia coli. Infection with M13 is not lethal. However, the infection causes turbid plaques in E. coli because infected bacteria grow slower than the surrounding uninfected bacteria. This phage has been engineered to act as a vector system. Explain how the amplification of gene of interest works in this phage with illustration.arrow_forwardDescribe in general terms how bacteriophage lambda regulates the switch between lytic and lysogenic cyclesarrow_forwardA researcher identified a mutation in PR of phage λ that causes itstranscription rate to be increased 10-fold. Do you think this mutationwould favor the lytic or lysogenic cycle? Explain your answer.arrow_forward
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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license