Cell Division in E Coli
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Exercise 2 – Cell Division and Population Growth in E. Coli
Analysis:
1. Note the shape of the curve plotted from Table 5 on the semi-log paper. This curve is S-shaped and is called sigmoid (from Greek for S; Sigmoid means “like sigma”.) Why might this curve look like this? Hint: prior to inoculation, the bacteria in their stock culture were in a semi-dormant condition.
Growth is slower at the beginning as the E coli acclimates to the food and nutrients. Once they get started they double rapidly until the food supply begins to dwindle and growth slows down and number of bacteria stabilizes.
2. Using the information from Table 5 and your semi-log graph, practice finding the doubling time of your bacteria. Note that a rise from 1 x 108 to 2 x108 bacteria represents a doubling of bacterial numbers; this period is, therefore the doubling time.
At what time did the pop equal 1×108? 108 min
At what time did the pop equal 2×108? 124 min
Difference between the two:
16 min
Thus from these data, the doubling time was 16 minutes. Note that doubling times are normally calculated during exponential growth, when the doubling time is the shortest. Exponential growth can be identified on your semi-log graph as the middle part of the sigmoid curve (the straight line portion with the highest slope)
3. What was the doubling time of your bacterial culture (table 3, pg. 113). Show your computations. Why is it different from the model data when the experiment producing it was run the same way?