Under optimal conditions for growth, an E. coli cell will divide around every 20
ID: 2003105 • Letter: U
Question
Under optimal conditions for growth, an E. coli cell will divide around every 20. minutes. For the following part, assume an E. coli cell to be a cylinder 2.00 ,Mu m long and 1.00 Mu m in diameter. If no cells died, how long would it take a single E. coli cell, under optimal conditions in a 10.-L culture flask, to reach its maximum cell density of 1.00 x 10^10 cells/mL (a "saturated" culture) Assuming that optimum conditions could be maintained, how long would it take for the total volume of just the cells in culture to 1.00 km^3 (starting with a single E. coli cell)Explanation / Answer
volume of flask = 10 * 0.001 m63 = 0.01 m^3
volume of 1 E-colibacteria = pi*(0.5*10^-6)^2 * 2 *10^-6 = 1.57*10^-18 m^3
a.)
timer required to reach the maximum cell density = 1(2^n -1 ) = 10^14
n = 46.506
therefore total time = 20 * 46.506 = 930.13 min = 15.50 hour
b.) total volume of flask = 1 Km^3 = 10^9 m^3 = 10^12 L = 10^15 ml
so total no of cell in 1Km^3 = 10^15 ml * 10^10 = 10^25 cells
so time taken to fill this volume = 1(2^n -1 ) = 10^25
n = 83.048
total time = 1660.964 min = 27.6827 hour
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