Question

a population of e. coli bacteria grows exponentially with time. you believe that the mean time between divisions is ( t_{d}=29 mathrm{~min} ), and that cell death occurs on average after ( t_{m}=200 mathrm{hr} ). the population starts with 1000 cells. complete parts (a) through (c). (a) use the equation ( n(t)=n_{0} e^{(b - m) t} ) to predict how many cells are present after 4 hours. the number of cells present after 4 hours is (do not round until the final answer. then round to the nearest integer as needed.)

Explanation:

Step1: Calculate the growth rate \(b\)

The mean - time between divisions \(t_{D}=29\) min. Convert it to hours: \(t_{D}=\frac{29}{60}\) hr.
The formula for the growth rate \(b\) in the exponential growth model \(N(t)=N_{0}e^{(b - m)t}\) (assuming \(m = 0\) for pure growth, since death is not considered in the growth - rate calculation part for now) is \(b=\frac{\ln2}{t_{D}}\).

$$b=\frac{\ln2}{\frac{29}{60}}=\frac{60\ln2}{29}$$

Step2: Substitute into the population formula

We know \(N_{0}=1000\), \(t = 4\) hr, and \(b=\frac{60\ln2}{29}\) (and assume \(m = 0\)).
The population formula is \(N(t)=N_{0}e^{bt}\).
Substitute the values: \(N(4)=1000e^{\frac{60\ln2}{29}\times4}\)
First, simplify \(\frac{60\ln2}{29}\times4=\frac{240\ln2}{29}\)
Since \(a\ln x=\ln(x^{a})\), then \(e^{\frac{240\ln2}{29}}=(e^{\ln2})^{\frac{240}{29}}\)
And \(e^{\ln2}=2\), so \(N(4)=1000\times2^{\frac{240}{29}}\)

$$2^{\frac{240}{29}}=2^{8+\frac{8}{29}}=2^{8}\times2^{\frac{8}{29}}$$

$$2^{8}=256$$

, and \(2^{\frac{8}{29}}=e^{\frac{8\ln2}{29}}\approx e^{\frac{8\times0.693}{29}}\approx e^{0.191}\approx1.21\)
\(N(4)=1000\times256\times1.21\)
Another way: \(N(4)=1000e^{\frac{240\ln2}{29}}\approx1000e^{\frac{240\times0.693}{29}}\approx1000e^{5.73}\)
Since \(e^{x}=\sum_{n = 0}^{\infty}\frac{x^{n}}{n!}\), and \(e^{5.73}\approx307.8\)
\(N(4)=1000\times307.8 = 307800\)

Answer:

\(308000\)