The growth rate of Escherichia coli, a common bacterium found in the human intestine, is proportional to its size. Under ideal laboratory conditions, when this bacterium is grown in a nutrient broth medium, the number of cells in a culture doubles approximately every 30 min. (a) If the initial population is 100, determine the function Q(t) that expresses the growth of the number of cells of this bacterium as a function of time t (in minutes

The equation that expresses the growth of the number of cells of this bacterium as a function of time t (in minutes) is Q(t) = 100e^0.023t.

Let the number of bacteria cells present be Q

Given that;

Initial population of the bacteria is 100, this is is an exponential growth problem.

Q(0) = 100

Q(30) = 200

From;

Qt =Qoe^rt

Where r is the growth constant and t is the time taken.

200 = 100e^r(30)

200/100 = e^r(30)

2 = e^r(30)

ln 2 = lne^r(30)

0.693 = 30r

r = 0.693/30

r = 0.023

So we can write a general equation;

Q(t) = 100e^0.023t

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Step-by-step explanation:

The equation for the number of cells after t minutes is given by the following formula:

In which Q(0) is the initial population and r is the growth rate.

Initial population of 100

So

Doubles after 30 minutes.

So Q(30) = 200.

We use this to find r

So

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step-by-step explanation:

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