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Prescott’s Microbiology, 9th Edition Chapter 16 –Mechanisms of Genetic Variation GUIDELINES FOR ANSWERING THE MICRO INQUIRY QUESTIONS Figure 16.6 How would you screen for a tryptophan auxotroph? How would you select for a mutant that is resistant to the antibiotic ampicillin but sensitive to tetracycline (assume the parental stain is resistant to both antibiotics)? A bacterial isolate that can grow in the presence of tryptophan but not in the absence is a likely auxotroph. Minimal media which either contains all the amino acids except tryptophan or all with tryptophan is used in the screen, as shown in the figure with replica plating. Similarly, replica plating onto media with amp or tet can isolate a ampR tetS colony. Figure 16.8 How is mismatch repair similar to DNA polymerase proofreading? How is it different? When a pair of bases do not match for typical Watson-Crick base pairing, the double helical structure is altered, and there is a bulge in the structure. DNA polymerase has an exit hole for the newly synthesized double strand of DNA, and thus can catch these bulges, causing pausing, which in turn activate the exonuclease proofreading activity. Similarly, in mismatch repair, the MutS protein scans DNA for bulges, to find the mismatches. Differences are that mismatch repair works post-synthesis, and instead of one multi-subunit complex (DNA pol), is a set of proteins working together, including DNA polymerase itself. Figure 16.9 How is damaged DNA recognized by the UVRAB complex? Recognizes the hallmark thymidine dimers, where the T bases are unusually close to each other. Figure 16.13 What features are common to all types of transposable elements? All the components in an insertion sequence (DRs, IRs, and transposase gene) are common in all transposons. Figure 16.18 What does the term episome mean? Episomes are genetic elements such as plasmids that can replicate free in the cytoplasm or can be integrated into the bacterial chromosome. Figure 16.25 According to this model, what would happen if DNA that lacked homology to the S. pneumoniae chromosome were taken into the cell? If non homologous DNA was taken into the cell, it would not be able to align itself with a homologous region on the bacterial chromosome, and thus would not be integrated. The DNA could be degraded, but if not, could still only be passed to one daughter cell during cell division, and would thus be diluted out of the population over time. Some bacteria have non-homologous random integration as well. Figure 16.27 What is the term used to describe a lysogenic phage genome when it is integrated into the host genome? This is a prophage. For viruses that integrate their genomes into eukaryotes, such as HIV, this DNA is called proviral DNA. 1 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Prescott’s Microbiology, 9th Edition Figure 16.29 Compare the number of transducing particles that arise during generalized (figure 16.28) and specialized transduction. Why is there such a big difference? The viral assembly machinery in generalized transduction is design to incorporate the viral DNA genome into the virus particle, so the accidental incorporation of a bacterial chromosome segment is rare. Thus the number of transducing particles is low. In specialized transduction the accidental inclusion of host chromosome with the viral DNA during deintegration is also a rare event, but the number of particles is greatly amplified because of replication (step 4). Figure 16.30 Why can’t the gal and bio genes be transduced by the same transducing particle? Remember that the viral capsid can only contain a small portion of the bacterial genome in specialized transduction (~5-10% of bacterial DNA). The gal gene is located on one end of the integrated prophage and the bio gene is at the other. When excising, only a limited amount of DNA can fit into the capsid and the piece of DNA containing both the gal and bio genes would be too large. Figure 16.32 As a replicative transposon, what would happen if Tn3 “hopped” from this R1 plasmid into a different plasmid? This can and does happen. Now the other plasmid, which might have a very different host range, would carry Tn3, and also potentially allow phenotypic expression of ampicillin resistance. Tn3 can also jump in the host chromosome(s). 2 © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.