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 20 min.
(a) If the initial population is 1000, 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).
Q(t) =
(b) How long would it take for a colony of 1000 cells to increase to a population of 1 million? (Round your answer to the nearest whole number.)
min
(c) If the initial cell population were 10000, what is our model?
Q(t) =
(a) To determine the function Q(t) that expresses the growth of the number of cells of this bacterium as a function of time t, we know that the number of cells doubles every 20 minutes. This indicates exponential growth.
The general form of an exponential growth function is Q(t) = Q0 * e^(kt), where Q(t) is the population size at time t, Q0 is the initial population size, e is the base of the natural logarithm, and k is the growth rate constant.
In this case, the initial population is 1000 and the population doubles every 20 minutes. We can find the value of k using the formula:
Q(t) = Q0 * e^(kt)
When t = 20 minutes, Q(20) = 2 * Q0. Therefore:
2 * Q0 = Q0 * e^(20k)
Simplifying this equation, we can cancel out Q0:
2 = e^(20k)
Taking the natural logarithm (ln) of both sides:
ln(2) = ln(e^(20k))
Using the property ln(e^x) = x:
ln(2) = 20k
Now we can solve for k:
k = ln(2) / 20
Replacing k in the exponential growth formula, we get:
Q(t) = Q0 * e^[(ln(2) / 20) * t]
So, for this bacterium, Q(t) = 1000 * e^[(ln(2) / 20) * t].
(b) To determine how long it would take for a colony of 1000 cells to increase to a population of 1 million, we need to find the value of t when Q(t) = 1 million.
1000 * e^[(ln(2) / 20) * t] = 1 million
Dividing both sides of the equation by 1000, we get:
e^[(ln(2) / 20) * t] = 1000
Taking the natural logarithm (ln) of both sides:
ln(e^[(ln(2) / 20) * t]) = ln(1000)
Using the property ln(e^x) = x:
(ln(2) / 20) * t = ln(1000)
Solving for t:
t = (20 * ln(1000)) / ln(2)
Using a calculator, we find t ≈ 69.31 minutes.
Therefore, it would take approximately 69 minutes for a colony of 1000 cells to increase to a population of 1 million.
(c) If the initial cell population were 10000, the model would still be exponential growth, but with a different initial population size.
Using the same formula Q(t) = Q0 * e^[(ln(2) / 20) * t], we substitute Q0 = 10000:
Q(t) = 10000 * e^[(ln(2) / 20) * t]