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- Which of the following apply to the development of urogenital tract Candida albicans infections? Disruption of population dynamics allowing the RB form of Chlamydia to cause tissue damage Loss of Lactobacillus populations leading to an increase in the local pH Loss of Gardnerella vaginalis populations leading to a lose of Clue cells Loss of the anaerobic environment that promotes the outgrowth of Trichomonas vaginalis None of the answers apply1. Precise words:Find the nonspecific terms in the following sentences. Replace the nonspecific choices with more preciseterms or phrases (It is not necessary to change the sentence structure).(i) All OVE mutants showed enhanced iP concentrations.(ii) Plants were kept in the cold overnight.(iii) To provide proof of concept for our hypothesis, we studied a virus in its host cell.(iv) The present paper reports on continuing experiments that were performed to clarify thissurprising effect.(v) The first transition state is a little lower in energy than the second transition state. 2. Simple words:Improve the word choice in the following examples by replacing the underlined terms or phrases withsimpler word choices (do not change the sentence structure).(i) These data substantiate our hypothesis.(ii) The difference in our results compared to those of Reuter et al. (1995) can be accounted forby the fact that different conditions were used.(iii) For the purpose of discussing cell migration we…Results from a Kirby Bauer antibiotic assay on a Gram-negative bacterial culture are described as follows: A) the bacterium is resistant to penicillin, an antibiotic that targets synthesis of the peptidoglycan cell wall and B) the bacterium is resistant to tetracycline, an antibiotic that targets the small subunit of the ribosome, inhibiting protein synthesis. Which of the results represents intrinsic resistant and which represents acquired resistants?
- In 1944, Avery, Macleod, and McCarty provided strong evidence that DNA is the hereditary material in Streptococcus pneumoniae by Group of answer choices showing that avirulent cells could become virulent by the process of transduction none of these is true. showing that virulent cells could become avirulent if the DNA was destroyed after transformation showing that avirulent cells could not gain the ability to become virulent cells if conjugation was interrupted. showing that avirulent cells could not gain the ability to become virulent if DNA was destroyed after transformation.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.3 2. The table below was obtained by P1 phage transduction followed by media selection of an E. coli strain lacking the three genes. Determine the map order of the genes given the information provided. Demonstrate CLEARLY which are closer or farther from each other (no numbers needed). 1 2 MAP: Experiment Selected marker Ala+ Gly+ Ser+ Unselected markers (select- ed for by subsequent plating) 27% Ser+, 2% Gly+ 3% Ser+, 1% Ala+ 35% Ala+, 4% Gly+
- Shown below are the complementation test results involving 4 independently isolated lethal mutants in a bacteriophage. Complementation was assayed by simultaneouly infecting bacteria with two phage strains, each with a different mutation, neither of which could alone lyse the cells. In the table below, a "+" indicates the strains complemented each other and therefore lysed open the bacteria. A "0" indicates no complementation and therefore no cell lysis occurred. Test pair Results 1___2___3___4 1,2 + 1 0 + + 0 1,3 + 2 0 + + 1,4 0 3 0 + 2,3 + 4 0 2,4 + 3,4 + How many genes are there? a. 3 b.1 c. 2 d. 4Shown below are the complementation test results involving 4 independently isolated lethal mutants in a bacteriophage. Complementation was assayed by simultaneouly infecting bacteria with two phage strains, each with a different mutation, neither of which could alone lyse the cells. In the table below, a "+" indicates the strains complemented each other and therefore lysed open the bacteria. A "0" indicates no complementation and therefore no cell lysis occurred. Test pair Results 1___2___3___4 1,2 + 1 0 + + 0 1,3 + 2 0 + + 1,4 0 3 0 + 2,3 + 4 0 2,4 + 3,4 + Which mutants are in the same gene? . a. 2, & 3 b.1, 2, 3 & 4 c.1 & 4 d.1, 2 & 4Three pairs of bacterial cells with the given genotypes undergo conjugation. Place match the genotype of each cell after conjugation to its initial genotype. F+ x F Hfr F- F' F- Answer Bank F F F+ Hfr What is the role of the F-factor in conjugation? It contains genes necessary for replication of the donor's F plasmid. can occur. It allows auxotrophic bacterial cells to survive on minimal medium so that conjugation It contains genes that force recombination between the donor and recipient chromosomes. It contains genes necessary for the formation of the pilus. O It degrades the chromosome of the recipient cell after conjugation.
- The strain of λ phage t is cI857. That tells you that the cI DNA segment is disabled by a specific mutation. What is the exact genetic change in cI857? What specific property of the cI gene product does this mutation change, and how does this help titering for a plaque assay?Bacteria 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 infection20. P2 and P4 are bacteriophages of E. coli. They have the following properties: (1) When P2 phage infects a bacterium, the bacterium eventually bursts, yielding about 100 P2 progeny. (2) When P4 infects a bacterium, the bacterium survives because P4 is a defective phage. (3) When P2 phage and P4 phage coinfect the same bacterium, lysis of the bacterium yields 100 P4 progeny and no P2 progeny, because the presence of P4 inhibits the growth of P2. Suppose 3 x 108 P2 and 2 x 108 P4 are added to 108 bacterial cells. How many bacterial cells will produce P2 progeny? A. 1.28 x 107 В. 6.75 х 105 С. 5.00 х 106 D. 2.56 x 107 E. 3.23 x 104