Question

1. There has been a wastewater spill on a beach. The Escherichia coli density of the wastewater is 7.8 x 1010 cells/100 mL. The spilled wastewater was diluted by 1000 times when mixed with seawater. The specific growth rate constant for Escherichia coli is 0.231 h-1 . Determine the density (cells/100 mL) of Escherichia coli in receiving seawater 24 and 72 hours after the spill. Note that exponential growth (X = X0e μt ) applies.

Answer 1

Cell denarity in wastewater 2 7 18 8 1010 cells/100ml 20.78 x 1010 cells / ml & Coll density in iseawater z 0.78 x 1010 cells/ml 1000 20.00.078 * 1010 cells / me 1. Specife grouath rate constant = 0.2314"! No. of cells after 24 hours => N = Noeht , A. 60 618 8 10:0 x 60 2 201.439 cells/me - All density after 24 howes a 201.439 Cello / one 20143.9 Cells /100ml . After 72 howes - 1 N z No echt 20.00078 * 1010 se 2 13170485.95 Cells/ml z 1.3 x 107 cells/me 21.3x10 cells 100 ml