Question

In milk at $37^\circ C$ 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

Hint:We have to calculate the relative population by doubling the time after every 75 minutes. Therefore, at any time (t), its population is given by,
$P = {P_0} \times {2^{\left( {t/75} \right)}}$
Here, ${P_0}$ is the initial population.
$P$ is the population at any time t.

Complete step by step answer:
Lactobacillus acidophilus has a generation time of 75 minutes, it doubles itself after every 75minutes. At any time (t), the population of Lactobacillus acidophilus is given by,
$P = {P_0} \times {2^{\left( {t/75} \right)}}$ $\left( 1 \right)$
We can say that population relative is the ratio of population at any time t to the initial population.
We can arrange the equation (1) to get the relative population.
$\dfrac{P}{{{P_0}}} = {2^{\left( {t/75} \right)}}$
Let us now calculate the population relative to initial value at a given time.
Let us substitute the value of t as 30 minutes. At 30 minutes, the population relative is calculated as,
$\dfrac{P}{{{P_0}}} = {2^{\left( {t/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = {2^{\left( {30/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = 1.319$
The relative population of the bacteria at 30 minutes is $1.319$.
Let us substitute the value of t as 60 minutes. At 60 minutes, the population relative is calculated as,
$\dfrac{P}{{{P_0}}} = {2^{\left( {t/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = {2^{\left( {60/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = 1.741$
The relative population of the bacteria at 60 minutes is $1.741$.
Let us substitute the value of t as 75 minutes. At 75 minutes, the population relative is calculated as,
$\dfrac{P}{{{P_0}}} = {2^{\left( {t/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = {2^{\left( {75/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = 2$
The relative population of the bacteria at 75 minutes is $2$.
Let us substitute the value of t as 90 minutes. At 90 minutes, the population relative is calculated as,
$\dfrac{P}{{{P_0}}} = {2^{\left( {t/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = {2^{\left( {90/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = 2.297$
The relative population of the bacteria at 90 minutes is $2.297$.
Let us substitute the value of t as 150 minutes. At 150 minutes, the population relative is calculated as,
$\dfrac{P}{{{P_0}}} = {2^{\left( {t/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = {2^{\left( {150/75} \right)}}$
$ \Rightarrow $$\dfrac{P}{{{P_0}}} = 4$
The relative population of the bacteria at 150 minutes is $4$.

Note: We can also calculate the population relative to the initial value by an alternate method. The alternate method is,
For growth kinetics,
$k = - \dfrac{{2.303}}{t}{\log _{10}}\left( {\dfrac{a}{{a + x}}} \right)$
$k = - \dfrac{{2.303}}{t}{\log _{10}}\left( {\dfrac{{{N_0}}}{N}} \right)$
The generation time is 75 minutes. The value of a is taken as 1 and the value of $a + x = 2$
Let us substitute the value of a and value of $a + x$ in the equation,
$k = - \dfrac{{2.303}}{t}{\log _{10}}\left( {\dfrac{1}{{1 + 1}}} \right) = 0.00924{\min ^{ - 1}}$
After thirty minutes, we can calculate the relative population to the initial value as,
$k = - \dfrac{{2.303}}{t}{\log _{10}}\left( {\dfrac{{{N_0}}}{N}} \right)$
Substituting the values we get,
$ \Rightarrow $$0.00924{\min ^{ - 1}} = - \dfrac{{2.303}}{{30}}{\log _{10}}\left( {\dfrac{{{N_0}}}{N}} \right)$
$ \Rightarrow $$\dfrac{N}{{{N_0}}} = 1.319$
The relative population to the initial value after 30 minutes is $1.319$.
We can also the relative population after 60, 75, 90 and 150 minutes by this method.