Question

a population of e. coli bacteria grows exponentially with time. you believe that the mean time between divisions is ( t_{b}=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 305,677.
(do not round until the final answer. then round to the nearest integer as needed.)
(b) in fact, you measure that there are 324,000 cells present in the population after 4 hours. can the discrepancy between the model and data be explained by your estimate for ( t_{b} ) being wrong? can it be explained by the estimate for ( t_{m} ) being wrong?
a. ( t_{b} ) and ( t_{m} ) must be wrong.
b. ( t_{b} ) must be wrong.
c. ( t_{m} ) must be wrong.
(c) assuming that ( t_{m} ) is correct, calculate a revised estimate for ( t_{b} ) to make the mathematical model fit your experimental data.
( t_{b}=square ) hour(s)
(do not round until the final answer. then round to three decimal places as needed.)

Explanation:

Step1: Convert units

First, convert \(t_b = 29\) min to hours. Since \(1\) hour \(= 60\) min, \(b=\frac{\ln2}{t_b}=\frac{\ln2}{\frac{29}{60}}\approx1.457\) per hour. And \(m = \frac{\ln2}{t_m}=\frac{\ln2}{200}\approx0.00347\) per hour. \(t = 4\) hours, \(N_0=1000\).

Step2: Calculate \(N(t)\) using the formula

Substitute into \(N(t)=N_0e^{(b - m)t}\). \(N(4)=1000e^{(1.457- 0.00347)\times4}\).

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Using \(e^{x}\approx e^{5.81412}\approx332.677\), \(N(4)=1000\times305.677 = 305677\) (matches part (a) result).

For part (c):

Step1: Set up the equation

We know \(N(t)=324000\), \(N_0 = 1000\), \(t = 4\), \(m=\frac{\ln2}{200}\). The formula is \(N(t)=N_0e^{(b - m)t}\). Substitute values: \(\frac{N(t)}{N_0}=e^{(b - m)t}\), \(\ln(\frac{N(t)}{N_0})=(b - m)t\).

$$b=\frac{\ln(\frac{N(t)}{N_0})}{t}+m$$

Substitute \(N(t) = 324000\), \(N_0=1000\), \(t = 4\), \(m=\frac{\ln2}{200}\)

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Since \(\ln(324)=\ln(18^2)=2\ln(18)\approx2\times2.89046\approx5.78092\)

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Since \(b = \frac{\ln2}{t_b}\), then \(t_b=\frac{\ln2}{b}\)

$$t_b=\frac{\ln2}{1.4487}\approx0.48$$

Answer:

(c) \(t_b\approx0.48\) hour(s)