Calculate the size of a population of bacteria that divide every 30 minutes will be in 10 hours if, at the start, it has 50 individuals and R = 2.
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- There are 10 E. coli cells in a test tube. After 4 generations in exponential phase, how many E. coli are there?In the experiment performed by Messelson and Stahl, bacterial cells grown in 15N-containing growth medium for many generations were then allowed to grow in 14N-containing medium. Suppose they would have conducted the experiment the other way around: that is, to first grow bacterial cells in 14N-containing medium for many generations and then switching them to a 15N-containing growth medium. After two generations of growth in 15N media the results of a CsCl density gradient centrifugation of the DNA would be expected to show O one band of intermediate density and one band of high-density (i.e. heavy band) one band on intermediate density O one band of intermediate-density and one band of low-density (i.e. light band) one band of low density (L.e. light band) and one band of high density (1.e. heavy band). one band of low density (I.e. light band), one band of intermediate density, and one band of high density (i.e. heavy band).In the experiment performed by Messelson and Stahl, bacterial cells grown in 15N-containing growth medium for many generations were then allowed to grow in 14N-containing medium. Suppose they would have conducted the experiment the other way around: that is, to first grow bacterial cells in 14N-containing medium for many generations and then switching them to a 15N-containing growth medium. After two generations of growth in 15N media the results of a CsCl density gradient centrifugation of the DNA would be expected to show one band of intermediate-density and one band of low-density (i.e. light band) one band of low density (i.e. light band), one band of intermediate density, and one band of high density (i.e. heavy band). one band of intermediate density and one band of high-density (i.e. heavy band) one band on intermediate density one band of low density (i.e. light band) and one band of high density (i.e. heavy band). O O
- Briefly explain, using the bacteria's response to environment, why the time =0 minutes is often not usable as the first point in log phase.Consider a batch culture of sphere-shaped bacteria in a growth medium or broth, in which the mean cell diameter of the bacteria is 2.0 μm (micron). Show all your calculations and assumptions in answering the questions below: What is the volume of one of these sphere-shaped cells, expressed in liters? If there are 3 x 10^12 cells in one liter of this culture broth, what percentage of the volume is occupied by bacteria and what percentage is occupied by the cell culture broth (water only)? If the cells contain 80% water, what is the dry cell concentration, expressed as “grams dry weight/Liter of broth”? You should assume that the density of the cells dry mass is approximately 1.0 gr/cm3Suppose that the population of the bacteria changes proportional to its current population. (Malthunian Theorem) Find the population of the bacteria at any time (t).
- What is the generation time (in minutes) of a bacterial population that increases from 10,000 cells to 10,000,000 cells in fours hours of growth? Show computation.Suppose the generation time of a bacterium is 90 minutes and the number of cells in a culture is 103 cells at the start of the log phase. How many bacteria will there be after 8 hours of exponential growth?Report the slope and the y-intercept below. Slope () = = y-intercept () = Can this standard curve be used for E.coli cells (why/why not?)
- Complete Table 1by doubling the number of bacteria (Clostridium sp.) and diatoms (Chaetoceros sp.) every 30 minutes and 60 minutes, respectively, and by writing the log10 of this number in the appropriate columns with the results , Draw a graph showing the growth (N) of the populations of both species from time = 0 (also written as t0) to time = 310 (t310) minutes. Draw both populations on the same graph. NB: It will be necessary to plot Chaetoceros sp. on a second y-axis. Label the axes and provide a suitable legends.https://studylib.net/doc/8245959/lab-7--got-milk%3F follow and open the link then answer the question number 4: Bacteria grown solely on lactose will grow, as will bacteria grow only on ONPG. However, cells grown solely on TMG will not grow, and will instead eventually die. Based on the structures provided below, why are cells capable of growing on ONPG and lactose, but not TMG alone?Using a schematic diagram, summarize the following steps in preparing competent cells for transformation: Inoculate a single colony of E. coli into 5 ml LB broth and incubate overnight at 37°C with moderate shaking (250 rpm). Add 200 μl of the culture into 50 ml LB broth and incubate overnight at 37°C with moderate shaking (250 rpm) to an OD600 = 1.3 to 1.5. Aliquot culture into five 15-ml pre-chilled, conical tubes. Leave tube on ice 5 to 10 min. Centrifuge cells 7 min at 1,600 × g (3,000 rpm), 4°C. Pour off supernatant and resuspend each pellet in 10 ml ice-cold CaCl2 solution (50 mM CaCl2), perform resuspension very gently, and keep on ice. Centrifuge cells 5 min at 1,100 × g (2,500 rpm), 4°C. Discard supernatant and resuspend each pellet in 10 ml ice-cold CaCl2 solution. Keep resuspended on ice for 30 min. Centrifuge cells 5 min at 1,100 × g, 4°C. Discard supernatant and resuspend each pellet in 10 ml ice-cold CaCl2 solution. Dispense cells (250 μl) into pre-chilled, sterile…