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PROBLEM 1: How could you determine whether the
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an accomplice to its own destruction. However, the phenomenon makes much more sense if we realize that the RecA protein has several other functions critically important to the bacterial cell (see chapter 12). The phage has evolved the ability to take advantage of the existence of the RecA protein by evolving a repressor sensitive to it. Evolutionary biologists view this as coevolution, two interacting organisms evolving to take advantage of or minimize properties of the other. The bacterium, however, might be at a disadvantage. Since RecA has many functions involving interactions with other proteins, it may be highly limited in how it can change.This is one plausible explanation as to why RecA liberates phage .
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PROBLEM 3: What are the differences in action of the promoters pRE and pRM Answer: The promoters pRE and pRM are both promoters of the repressor operon of phage . Transcription from these promoters allows production of the cI repressor protein, the repressor that favors lysogeny. Initially, the promoter pRE is activated. For it to be a transcription site, it must be activated by the product of the cII gene, which lies in the right operon. This promoter, pRE, produces a messenger RNA with a long leader that is translated very ef ciently. Once the repressor binds at the operators of the left and right operons, the cII gene is no longer transcribed, and therefore pRE is no longer a site for transcription. However, the repressor gene can still be transcribed from the pRM promoter, which does not need the product
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genes for the breakdown of the sugar arabinose are under inducible control in E. coli Answer: Inducible means that the genes to break down the substrate arabinose, a ve-carbon sugar, in this case are not active in the absence of the inducer (again, arabinose). Therefore, in the absence of arabinose in the cells environment, the arabinose utilization enzymes should not be active within the bacterial cells, but after arabinose is added to the medium, the enzymes should be present. We thus need to assay the contents of the cells before and after arabinose is added to the medium, performing the assay after the cells are broken open and the DNA destroyed so as not to confound the experiment. Using a standard biochemical analysis for arabinose, we should nd that the bacterial cell is incapable of metabolizing arabinose before induction but capable of metabolizing it afterward. If the cells were capable of metabolizing arabinose in both cases, we would say that arabinose utilization is constitutive. If the cells were incapable of utilizing arabinose in both cases, we would conclude that the bacterium is incapable of using the sugar arabinose as an energy source. (In fact, arabinose utilization is inducible.)
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PROBLEM 2: Why would the RecA protein of E. coli cleave the repressor Answer: Since the cleaving of the repressor is a signal to begin the lytic phase of the life cycle of the phage, it seems odd that the lysogenized bacterial cell would be
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III. Molecular Genetics
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14. Gene Expression: Control in Prokaryotes and Phages
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The McGraw Hill Companies, 2001
Fourteen
Gene Expression: Control in Prokaryotes and Phages
of the cII gene. This promoter produces a transcript with no leader and thus is translated very inef ciently. At that point, however, only a very small quantity of repressor is needed to maintain lysogeny.Thus, the two promoters are the sites for the initiation of the repressor operon under different circumstances: one early in the infection stage and one after lysogeny is under way.
PROBLEM 4: What are the phenotypes of the following partial diploids for the lac operon in E. coli in the presence and absence of lactose
a. (F ) i o p z /i o p z (chromosome) b. (F ) i oc p z /i o p z (chromosome)
Answer: Consider one DNA molecule at a time. If one DNA molecule can never make the enzyme, it can be ignored. In (a), the plasmid DNA (F ) will never make enzyme (it is z ), and the chromosomal DNA will never make repressor (it is i ). The functional repressor in the plasmid (i ) will bind to both DNAs, and hence the chromosomal operon will not be transcribed in the absence of lactose and will be induced to transcribe in the presence of lactose. In (b), the plasmid DNA (F ) will always be transcribing (operator constitutive) because the repressor can never bind the operator (oc); hence, the operon will be transcribed all the time. The chromosomal DNA can never make RNA (it is p ).
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