Which of the following results would NOT be expected with our GFP virtual experiment? O The PGLO+ tube sample on LB with ampicillin and arabinose should glow under UV light O The PGLO- tube sample on LB with ampicillin should have a few colonies O The PGLO+ tube sample on LB with ampicillin and arabinose should have a few colonies O The PGLO- tube sample on LB should have hundreds of colonies
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- The table below shows the response of our ESKAPE safe relatives to 4 bacteria isolated from a master grid. We do not know the identity or any characteristics of the unknown bacteria. Each safe relative was spread onto a petri dish using aseptic technique. A grid pattern was taped to each plate and the unknown bacteria were patched into one of the squares. If there was no inhibition visible, including with a magnifying lens, the result was listed as -. If there was an inhibition zone between 1 and 10mm in diameter, the result is listed as +. If the inhibition zone was 10mm or greater, the result is listed as ++. In the lab, the MGC instructors plated all 6 of the ESKAPE pathogen safe relatives on LB agar plates. Then we patched Unknown Bacteria 5 from a Master plate onto the safe relative. The results are shown here: METRIC METRIC METRIC 1 B. subtilis S. epidermidis E. coli Complete the final column (Unknown Bacteria 5) of the table by selecting -, +, or ++ using the criteria in the…The table below shows the response of our ESKAPE safe relatives to 4 bacteria isolated from a master grid. We do not know the identity or any characteristics of the unknown bacteria. Each safe relative was spread onto a petri dish using aseptic technique. A grid pattern was taped to each plate and the unknown bacteria were patched into one of the squares. If there was no inhibition visible, including with a magnifying lens, the result was listed as -. If there was an inhibition zone between 1 and 10mm in diameter, the result is listed as +. If the inhibition zone was greater than 10mm, the result is listed as ++. Page 50 in your research guide states: "Some antibiotics are broad spectrum, meaning that they affect a wide range of bacteria. Other antibiotics have a narrow spectrum of activity. One anatomical feature that plays a significant role in the susceptibility of a microbe to a particular antibiotic is its cell wall composition (discussed in Section 8)". Research the cell wall…The table below shows the response of our ESKAPE safe relatives to 4 bacteria isolated from a master grid. We do not know the identity or any characteristics of the unknown bacteria. Each safe relative was spread onto a petri dish using aseptic technique. A grid pattern was taped to each plate and the unknown bacteria were patched into one of the squares. If there was no inhibition visible, including with a magnifying lens, the result was listed as -. If there was an inhibition zone between 1 and 10mm in diameter, the result is listed as +. If the inhibition zone was greater than 10mm, the result is listed as ++, Page 50 in your research guide states: "Some antibiotics are broad spectrum, meaning that they affect a wide range of bacteria. Other antibiotics have a narrow spectrum of activity. One anatomical feature that plays a significant role in the susceptibility of a microbe to a particular antibiotic is its cell wall composition (discussed in Section 8)". Research the cell wall…
- A phagehunter performs a spot titer using standard techniques (3 ul of each dilution spotted to lawn) of a lysate obtained from an optimum webbed plate experiment. The phagehunter counts 4 plaques on the 10-8 dilution spot. Which of the following scenarios is the best choice for the phagehunter to do next? a.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should try to make more optimum webbed plates. b.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should adopt a phage from direct isolation. c.) The phagehunter has achieved a high enough titer lysate to move forward with characterization experiments. d.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should go back to the pick-a-plaque experiment.A phagehunter performs a spot titer using standard techniques (3 ul of each dilution spotted to lawn) of a lysate obtained from an optimum webbed plate experiment. The phagehunter counts 8 plaques on the 10-7 dilution spot. Which of the following scenarios is the best choice for the phagehunter to do next? a.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should go back to the pick-a-plaque experiment. b.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should adopt a phage from direct isolation. c.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should try to make more optimum webbed plates. d.) The phagehunter has achieved a high enough titer lysate to move forward with characterization experiments.Hi, I am working on qPCR quantification for specific bacteria. However, when the CT value achieve greater than 30, it always shown high CT value differences in my duplication no matter I repeat the same set of experiment for those samples with low quality. My question is, what makes this happening? The other question is, how many CT value differences is acceptable for duplication?
- A high-throughput assay is being conducted in a 96 well plate to test compounds for anti-bacterial activity. Live bacterial cells are detected using “Live clear", a dye which is initially coloured blue, but turns clear in the presence of live bacterial cells. The control wells are shown below. Penicillin kills bacteria; glycerol does not affect them. Which of the control wells will appear blue at the end of the assay? "Live clear" glycerol penicillin - - + (A1 A2)(A3)A4) bacterial cells B1 (B2 (B3)B4 А4, В1, B2, ВЗ, B4 A2, A4, B2, B3, B4 А4, B2, ВЗ, В4 A1, АЗ, В1, В2, вЗ, В4 + + |The total number of cells in a culture is counted using the trypan blue exclusion assay and is found to be 6.8 x 106 cells/ml. Each well in a 6 well plate requires 2 x 105 cells. How should the solution be diluted so that 1ml can be added to each well?You want to subculture your cells from T25 flask to 96-well plates. You first collected your cells in a tube with 5ml of culture medium. Then carried out a trypan blue assay and counted your cells with a hematocytometer as shown in Figure below. Answer the questions according to the results: a. Calculate the concentration of the stock including the dead and living cells. Dont forget to show the units! b. Calculate the percentage (%) of the living cells in the stock. c. You want to seed 6000 living cells into each well of 96 well plates, then calculate the volume you should take from the stock for each well. 輯 1 mm I IRE
- Which of the following is a reason to run simultaneous tests on known positive and negative controls when identifying an unknown microbe using biochemical tests? O The controls maintain the differential qualities of the test medium The controls will ensure that the unknown microbe multiplies on the medium O The controls can show if there was contamination of the tubes with another species O The controls can show if there was a recent mutation in your unknown microbe MacBook Air DII F12 80 888 F7 F5 F4 F3 * & %2$ %23 6 7 8. 4 5Katelyn had been working for Dr. Johnson for a month, and while she had become quite good at measuring inhibitionzones, she didn’t know why she was doing all this work. She had gotten very curious after she began doing all themeasurements on a new set of antibiotics. # is experiment involved infecting mice with MRSA and tracking how theMRSA grew over time.Data were collected by counting the cells of MRSA taken from $ uid samples from the mice. # e cells were measuredby taking one gram of the $ uid and spreading it over plates, but now Katelyn counted the colonies that grew on theplate after 24 hours. Because there were so many, she actually measured the colonies as “log CFU/g.” A CFU is acolony forming unit, or essentially a cell that will divide into a colony that can be seen. Because there can be so many,Katelyn measured them on a logarithmic (log) scale. # e raw data in her lab notebook looked like the following:Table 1. E% ect of treatment on MRSA in mice after 24 hours of drug…You perform a Kirby-Bauer assay with two antibiotics. Antibiotic X has a zone of inhibition of 9 mm. Antibiotic Y has a zone of inhibition of 11 mm. Which antibiotic is better at killing this particular microorganism? Group of answer choices 1Antibiotic Y 2Antibiotic X and Y, which have identical antimicrobial activities 3Antibiotic X4 4It is impossible to tell from the information given