Question

In a milk at $37^\circ {\text{C}}$, Lactobacillus acidophilus has a generation time of about 75 minutes. Calculate the population relative to the initial value at 30, 60, 75, 90 and 150 minutes.

Answer 1

Lactobacillus acidophilus is a kind of bacteria which is present in our intestines. Bacterial colony growth takes place commonly by cell division and so it follows first order kinetics.
For decay first order kinetics, the relationship between the rate constant ‘k’ with concentration and time is
${\text{k}} = \dfrac{{2.303}}{{\text{t}}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}}$
Growth is the opposite of decay and so, for growth kinetics
${\text{k}} = - \dfrac{{2.303}}{{\text{t}}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}}$
Here, ${{\text{N}}_{\text{0}}}$ is the initial concentration and N is the concentration after time ‘t’ of the reactant.

Complete step by step answer:
Given, Lactobacillus acidophilus bacteria has a generation time of about 75 minutes, i.e., it equals 75 minutes. We need to find out the population of the bacteria relative to the initial value at 30, 60, 75, 90 and 150 minutes.
First we will calculate the value of the rate constant ‘k’ from the relation for ‘t’ equal to 75 minutes. If there was only one cell in the beginning, then after 75 minutes, there will be two cells. Thus, ${{\text{N}}_{\text{0}}}$ is equal to 1 and N is equal to 2.
So,
${\text{k}} = - \dfrac{{2.303}}{{75}}\log \dfrac{1}{2} \\\ \Rightarrow {\text{k}} = - 0.031\left( { - 0.30103} \right) \\\ \Rightarrow {\text{k}} = 0.0093{\min ^{ - 1}} \\\$
Now, we can use this value of the rate constant ‘k’ to find out the population N relative to the initial value ${{\text{N}}_{\text{0}}}$ at different times.
For time equal to 30 minutes, the equation becomes
$0.0093 = - \dfrac{{2.303}}{{30}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow 0.279 = - 2.303\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow \log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} = - 0.121 \\\ \Rightarrow \log \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 0.121 \\\ \Rightarrow \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 1.32 \\\$
Hence, the population of the bacteria relative to the initial value at 30 minutes is $1.32$.
Similarly, the population of the bacteria relative to the initial value at 60 minutes is:
$0.0093 = - \dfrac{{2.303}}{{60}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow 0.558 = - 2.303\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow \log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} = - 0.2423 \\\ \Rightarrow \log \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 0.2423 \\\ \Rightarrow \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 1.75 \\\$
The population of the bacteria relative to the initial value at 75 minutes is:
$0.0093 = - \dfrac{{2.303}}{{75}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow 0.6975 = - 2.303\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow \log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} = - 0.301 \\\ \Rightarrow \log \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 0.301 \\\ \Rightarrow \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 1.99 \\\$
The population of the bacteria relative to the initial value at 90 minutes is:
$0.0093 = - \dfrac{{2.303}}{{90}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow 0.837 = - 2.303\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow \log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} = - 0.3634 \\\ \Rightarrow \log \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 0.3634 \\\ \Rightarrow \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 2.30 \\\$
The population of the bacteria relative to the initial value at 150 minutes is:
$0.0093 = - \dfrac{{2.303}}{{150}}\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow 1.395 = - 2.303\log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} \\\ \Rightarrow \log \dfrac{{{{\text{N}}_0}}}{{\text{N}}} = - 0.606 \\\ \Rightarrow \log \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 0.606 \\\ \Rightarrow \dfrac{{\text{N}}}{{{{\text{N}}_0}}} = 4.04 \\\$

Note:
The doubling time of exponential growth is the converse of half – life for exponential decay equation. It is the time taken by the bacterial population to double in size or value. Thus, at doubling time ‘t’, N will be equal to twice of ${{\text{N}}_{\text{0}}}$. Hence,

$${\text{k}} = - \dfrac{{2.303}}{{\text{t}}}\log \dfrac{{{{\text{N}}_0}}}{{2{{\text{N}}_0}}} \\\ \Rightarrow {\text{t}} = - \dfrac{{2.303}}{{\text{k}}}\log \dfrac{1}{2} \\\ \Rightarrow {\text{t}} = \dfrac{{2.303}}{{\text{k}}}\log 2 \\\ \Rightarrow {\text{t}} = \dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}} \\\$$

Thus, the doubling time is independent of the initial population.