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- The colony forming units (CFU)/ml in sauerkraut brine are given for different types of bacteria except for leuconostoc (shaded column). To obtain the Leuconostoc counts for different days, 40 microlitre of sauerkraut brine was 10-fold diluted and the CFU was determined by plating different dilutions on agar plates. For
- Day 0 56 CFU were counted in 10-1 dilution,
- Day 1 81 CFU were counted in 10-3 dilution,
- Day 2 86 CFU were counted in 10-3 dilution
- Day 5 78 CFU were counted in 10-5 dilution
Calculate the CFU/ml of leuconostoc in the brine for the 4 different days. Show your calculation
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but why is the answers all x 1000 at the end? for example day 1 is 2.025×1000=2025CFU/ml.
should the answer not stay at 2.025?
by Bartleby Expertare you sure? because my calculations are the same numbers but i got
Day 0: 14 CFU/ml
Day 1: 2.025 CFU/ml
Day 2: 2.15 CFU/ml
Day 5: 0.0195 CFU/ml
by Bartleby Expert- Identify which antibiotic was used per set-up (see figure). Describe the result in the graphs provided to help you explain your answer. Listed below are some essential information.Antibiotic A: 0.5 kDa protein, targets peptidoglycanAntibiotic B: 20 kDa protein, targets peptidoglycanAntibiotic C: Cationic antimicrobial peptideAntibiotic D: Targets lipopolysaccharideStaphylococcus aureus: gram-positive bacteriumVibrio cholera: gram-negative bacteriumMethanosarcina: an archaean bacteriumCationic antimicrobial peptides (CAMPs): these positively charged antibiotics are attracted to the negatively charged cell wall and membrane. They are hydrophobic, and they insert into the membranes to create pores.Case Study: You are asked to inoculate lactose media (broth) with E. coli that was growing exponentially in glucose media (broth). You monitor the growth of these cells in the lactose broth using a viable plate count technique and plot the Growth Curve for E. coli growing c lactose. After a day of monitoring the cells, you note that the culture entered into lag phase growth initially and is now currently growing exponentially. In addition, you find that the cultures doubling time is 30 minutes. You know that the doubling time of E. coli in glucose media is 20 minutes. You did not change the oxygen content or temperature at Which you were growing the cells, E. coli is still aerobically respiring at 37°C. Question: If the lac operon is 'on' what does this imply about the cell? O The cell is dying and the lac operon is going to induce apoptosis. Apoptosis is truly cell death and the cells will enter into the death phase of the Growth Curve. O Lactose is being catabolized and glucose is no…Case Study: You are asked to inoculate lactose media (broth) with E. coli that was growing exponentially in glucose media (broth). You monitor the growth of these cells in the lactose broth using a viable plate count technique and plot the Growth Curve for E. coli growing on lactose. After a day of monitoring the cells, you note that the culture entered into lag phase growth initially and is now currently growing exponentially. In addition, you find that the cultures doubling time is 30 minutes. You know that the doubling time of E. coli in glucose media is 20 minutes. You did not change the oxygen content or temperature at which you were growing the cells, E. coli is still aerobically respiring at 37°C. Question: If you want to grow E. coli as quick as possible would you grow it on lactose or glucose? O Mixing the two carbon sources would provide the most optimal growth conditions. O Lactose because of the generation time being faster. O Lactose and an anaerobic environment. O Glucose…
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