Chapter 8.3, Problem 58PPS

(a)

To determine

To Find: How long will it take four of these bacteria to multiply into $1024$ bacteria if the generation time of mycobacterium tuberculosis is $16$ hours.

Answer to Problem 58PPS

  $264hours$

Explanation of Solution

Given: The generation time for specific bacteria can be found by using $G=\frac{t}{3.3{\log }_{b}f}$ where $t$ the time period, $b$ number of bacteria at the beginning of the experiment, $f$ is the number of bacteria at end.

Given: $G=\frac{t}{3.3{\log }_{b}f}$

Substitute $G=16,b=4,f=1024$

We get

  $\begin{array}{l}G=\frac{t}{3.3{\log }_{b}f}\\ 16=\frac{t}{3.3{\log }_{4}1024}\\ t=16(3.3{\log }_{4}1024)\\ t=52.8{\log }_4{4}^5\\ t=264\times 1\\ t=264\end{array}$

Hence, it take $264hours$

(b)

To determine

To Find: How long will it take $20$ of these bacteria to multiply into $8000$ bacteria if the generation time for salmonella was $5$ hours.

Answer to Problem 58PPS

  $49.5hours$

Explanation of Solution

Given: The generation time for specific bacteria can be found by using $G=\frac{t}{3.3{\log }_{b}f}$ where $t$ the time period, $b$ number of bacteria at the beginning of the experiment, $f$ is the number of bacteria at end.

Given: $G=\frac{t}{3.3{\log }_{b}f}$

Substitute $G=5,b=20,f=8000$

We get

  $\begin{array}{l}G=\frac{t}{3.3{\log }_{b}f}\\ 5=\frac{t}{3.3{\log }_{20}8000}\\ t=5(3.3{\log }_{20}8000)\\ t=16.5{\log }_{20}{20}^3\\ t=49.5\times 1\\ t=49.5\end{array}$

Hence, it take $49.5hours$

(c)

To determine

To Find: E.coli are fast growing bacteria. If $6$ can grow to $1296$ in $4$ hours, what is the generation for it.

Answer to Problem 58PPS

  $0.3hours$

Explanation of Solution

Given: The generation time for specific bacteria can be found by using $G=\frac{t}{3.3{\log }_{b}f}$ where $t$ the time period, $b$ number of bacteria at the beginning of the experiment, $f$ is the number of bacteria at end.

Given: $G=\frac{t}{3.3{\log }_{b}f}$

Substitute $t=4.4,b=6,f=1296$

We get

  $\begin{array}{l}G=\frac{t}{3.3{\log }_{b}f}\\ G=\frac{4.4}{3.3{\log }_{6}1296}\\ G=\frac{4.4}{3.3\times 4\times {\log }_{6}6}\\ G=\frac{4.4}{14.52}\\ G=0.3hours\end{array}$

Hence, $0.3hours$