Consider the five E. coli merodiploid strains listed here. Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+ Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y- Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y- Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+ Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y- Which of these strains will express b-galactosidase constitutively?
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Consider the five E. coli merodiploid strains listed here.
Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+
Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y-
Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y-
Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+
Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y-
Which of these strains will express b-galactosidase constitutively? Select all correct answers.
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- Type S Streptococcus pneumoniae bacterium is lethal and will kill its host. If heat inactivated the S strain dies and becomes nonlethal. Type R Streptococcus pneumoniae is a nonvirulent strain of bacteria. What would occur if one were to inject both the R strain and heat-killed S strains into a host organism such as the mouse? The R strain would be transformed into the virulent S strain and kill the host. Neither the S nor the R strain would change. The R strain would be transformed into the virulent S strain and not affect the host. The S strain would be transformed into the nonvirulent R strain and not affect the host The S strain would be transformed into the nonvirulent R strain and kill the host.For each of the E. coli strains containing the lacoperon alleles listed, indicate whether the strain isinducible, constitutive, or unable to expressβ-galactosidase and permease.a. I+ o+ Z− Y+/ I+ ocZ+ Y+b. I+ o+ Z+ Y+/ I− ocZ+ Y−c. I+ o+ Z− Y+/ I− ocZ+ Y−d. I−P− o+ Z+ Y−/ I+ P+ ocZ− Y+e. Iso+ Z+ Y+/ I− o+ Z+ Y−Consider the five E. coli merodiploid strains listed here. Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+ Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y- Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y- Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+ Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y- Which of these strains will be inducible for expression of b-galactosidase? Select all correct answers. A.) Strain #2 B.) Strain #3 C.) Strain #1 D.) Strain #4 E.) None of these F. )Strain #5
- In five Hfr strains, each of which was used to build a time-of-transfer map, the genes entered the recipient cells as follows: Strain 1: S L A C T F Strain 2: N P F T C A Strain 3: T F P N U Y Strain 4: S H Y U N P Strain 5: U N P F T C Which of the following represents a correct gene map of these results? N P F T S L A C H U Y S L A C T F P N H Y U C T F P N U Y H S L A T C A L S P N U Y H F U N P C A L S F T H YFor the following sets of partial diploid bacteria, how do I fill out this table? I am not sure how to tell the difference between inducible and constitutive?After sequencing E.coli ROAR340 strain and using Silico Clermont Phylotyper tools show that is it belongs to group B2 that has chuA, TspE4 and yjaA genes. Is ROAR340 pathogenic?
- Streptococcus pneumoniae is a Gram-positive bacterium that colonizes the mucosal surface of the upper respiratory tract in humans. The presence of this bacterium in the nose and throat is widespread in the population, and in most people, colonization with Strep. pneumoniae is asymptomatic. The figure attached shows a comparison of in vitro growth curves of the wild-type strain of Strep. pneumoniae, as well as a Strep. pneumoniae mutant strain with a defect in one bacterial gene. The graph on the right shows the growth curve following addition of lysozyme during the logarithmic phase of bacterial growth. Which statement could account for the data in these graphs? Strain B is wild-type Strep. pneumoniae, and strain A is a mutant that cannot modify its peptidoglycan to be lysozyme-resistant. Strain B is wild-type Strep. pneumoniae, and strain A is a mutant that that expresses increased levels of LPS. Strain A is wild-type Strep. pneumoniae, and strain B is a mutant that cannot modify its…An E. coli colony grew on minimal medium supplemented with arginine and leucine. However, bacteria from this colony are unable to grow and form colonies on minimal medium supplemented with arginine and methionine. What is the genotype of the bacteria in this E. coli colony?DNA from a strain of Bacillus subtilis with genotype a+ b+ c+ d+ e+ is used to transform a strain with genotype a− b− c− d− e−. Pairs of genes are checked for cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformation a+ and b+ No b+ and d+ No a+ and c+ No b+ and e+ Yes a+ and d+ Yes c+ and d+ No a+ and e+ Yes c+ and e+ Yes b+ and c+ Yes d+ and e+ No On the basis of these results, what is the order of the genes on the bacterial chromosome?
- Order the following experimental steps to identify auxotrophic mutants in Saccharomyces Cerevisiae that cannot sythesize their own Leucine. an option can be selected more than once. step 1: Step 2: Step 3: Step 4: Step 5: Step 6:In the name E. coli 4X5B, which part is called the serotype, serovar, or strain? E coli 4X5BHow does Escherichia coli O157:H7 end up in groundbeef? To what class of pathogenic E. coli does this strainbelong? How does this class differ from other classes?