1 E coli has a doubling time of 20 minutes under optimal gro

1. E. coli has a doubling time of 20 minutes under optimal growth conditions. Compute the biomass of E. coli (in pounds) that would accumulate when a single cell of E. coli grows exponentially for 2 days (48 hours) in these conditions, assuming that nutrients are not limiting. Show your work. (The weight of a single E. coli cell is 10-12 gram/cell).

2. You are doing a viable cell count on a bacterial culture and you perform ten-fold serial dilutions. You count 130 colonies on a plate that was spread with 0.05 ml of a 10-6 dilution of the original culture. How many viable cells per milliliter are there in the undiluted bacterial culture?

Solution

1. Assuming that the nutrients have not depleted , doubling will continue at a constant rate . The factor to be considered here is not the nutrient limitation but accumulation of toxic metabolites in the nutrient medium irrespective of no nutrient limitation which limit the growth of the bacteria. E.coli is seen to be grown well up to 24 hours of the experiment , in the same media, even there, the cells begin to die due to wastes and toxic metabolites, resulting in cell death. Further, if all assumptions are to be made , then we can calculate the number of cells based on the number of generations that it would have in 48 hours, i.e. 144 generations. Assuming that there is only one and only a single cell initially we can calculate the number of cells as 2ⁿ where n=144, and multiply the number obtained by the weight per cell given.

2) 0.05 ml = 130 cells
0.1 ml = 130*2 = 260 cells
Since dilution has been 10^-6 , to calculate the number of viable cells in undiluted bacterial culture will be 260 * 10⁶, i.e. 2.6*10⁷ number of cells in undiluted bacterial culture.