5 x 104 cells of Bacillus cereus were inoculated into fresh sterile nutrient medium and incubated at 30oC. After 2 hour lag phase, cells grew exponentially with a generation time of 30 minutes. How many cells would be present after 8 hours incubation? The growth of the bacterium in log phase can be expressed by the following equations: N1 = N0 2n & g = (T1 – T0) / n Where n = number of generations, N0 = initial number of cells at a particular time (T0), N1 = final number of cells at a particular time (T1), and g = generation time Group of answer choices 1.28 x 107 cells/ml 4.10 x 107 cells/ml 2.05 x 108 cells/ml 3.28 x 109 cells/ml 5.24 x 1010 cells/ml
Question
5 x 104 cells of Bacillus cereus were inoculated into fresh sterile nutrient medium and incubated at 30oC. After 2 hour lag phase, cells grew exponentially with a generation time of 30 minutes. How many cells would be present after 8 hours incubation?
The growth of the bacterium in log phase can be expressed by the following equations: N1 = N0 2n & g = (T1 – T0) / n
Where n = number of generations, N0 = initial number of cells at a particular time (T0), N1 = final number of cells at a particular time (T1), and g = generation time
Group of answer choices
1.28 x 107 cells/ml
4.10 x 107 cells/ml
2.05 x 108 cells/ml
3.28 x 109 cells/ml
5.24 x 1010 cells/ml
Solution
To solve this problem, we first need to determine the number of generations that occur during the exponential growth phase.
Given that the lag phase is 2 hours, the exponential growth phase is 8 hours - 2 hours = 6 hours.
Since the generation time is 30 minutes (or 0.5 hours), the number of generations (n) during the exponential growth phase is 6 hours / 0.5 hours/generation = 12 generations.
We can then use the equation N1 = N0 * 2^n to calculate the final number of cells (N1).
Here, N0 is the initial number of cells, which is 5 x 104 cells.
So, N1 = 5 x 104 cells * 2^12 = 5 x 104 cells * 4096 = 2.048 x 10^9 cells.
Therefore, the closest answer is 2.05 x 10^9 cells/ml.
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