Chapter 5, Problem 1AQ

A medium was inoculated with 5 × 10⁶ cells/ml of Escherichia coli cells. Following a 1-h lag, the population grew exponentially for 5 h, after which the population was 5.4 × 10⁹ cells/ml. Calculate g and k for this growth experiment.

Summary Introduction

To calculate:

The generation time (g) and instantaneous growth rate constant (k) for the given growth experiment should be calculated.

Concept introduction:

The generation time is the specified time that requires to divide the cell into two. The bacteria reproduce by binary fission or cell division. The generation time depends on the environmental conditions such temperature, nutrients and pH and also it varies from one organism to the other organisms.

Explanation of Solution

The generation time (g) and instantaneous growth rate constant (k) for the given growth experiment is calculated below;

  $\begin{array}{l}\text{Doubling time = }\frac{\text{total duration time (t)}}{\text{number of generation (n)}}\mathrm{.......}(1)\\ \text{The bacterial growth equation, which is widely used to determine the doubling time is given below:}\\ \text{n = }\frac{\log N-\log N_0}{\log 2}\mathrm{.....}(2)\\ \text{Where,}\\ N_0=\text{ Number of cells initially}\\ \text{n = Number of generations}\\ \text{N = Final number of cells}\\ {\text{N}}_0=5.0\times {10}^6\\ N=5.4\times {10}^9\\ \log 2=0.301\\ {\text{N and N}}_0\text{ values should be inserted in equation (2):}\\ \text{Number of generation (n) = }\frac{(\log 5.4\times {10}^9)-(\log 5\times {10}^6)}{\log 2}\\ =\frac{(\log 5.4\times \log {10}^9)-(\log 5\times \log {10}^6)}{\log 2}\\ =\frac{(\log 5.4+9{\log }_{10})-(\log 5+6{\log }_{10})}{\log 2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=\frac{(0.73+9\times 1)-(0.69+6\times 1)}{0.301}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=\frac{(9.73)-(6.69)}{0.301}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=10.01\\ 2^n=\frac{\text{Final number of cells (N)}}{{\text{Initial number of cells (N}}_0)}\mathrm{........}(3)\\ \text{Initial and final number of cells should be inserted in equation (3)}\\ 2^n=\frac{5.4\times {10}^9}{5\times {10}^6}\\ =\frac{5.4\times {10}^9\times {10}^{-6}}{5}\\ =5.4\times {10}^3\times \frac{1}{5}\\ =5.4\times 0.2\times {10}^3\\ 2n=1080\\ n=\frac{\log 1080}{\log 2}\\ =\frac{3.033}{0.301}\\ =10.01\\ \text{Therefore, number of generations = 10}\text{.01}\\ \text{Hence, }\\ \text{n = 10}\\ \text{g = }\frac{t}{n}\\ =\frac{5}{10}=0.5h\\ k=\frac{0.301}{g}\\ =0.6\\ \text{Therefore n is 10, g is 0}\text{.5 h, and k is 0}\text{.6}\end{array}$

Conclusion

The generation time (g) and instantaneous growth rate constant (k) for the given growth experiment is calculated.