Answer

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**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.

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