Monod Model Growth Rate Calculation
The question asks for the net specific growth rate ($\mu_{net}$) of a microbial culture using the Monod model. The formula for the Monod model is:
$ \mu = \mu_{max} \frac{S}{K_s + S} $
Where:
- $ \mu $ is the specific growth rate
- $ \mu_{max} $ is the maximum specific growth rate
- $ S $ is the limiting substrate concentration
- $ K_s $ is the Monod constant
The net specific growth rate accounts for the endogenous decay rate ($k_d$):
$ \mu_{net} = \mu - k_d $
Parameter Unit Conversion
First, ensure all units are consistent, preferably in days ($day$) and grams ($g$).
- Maximum specific growth rate ($ \mu_{max} $): $ 0.1 \text{ h}^{-1} = 0.1 \times 24 \text{ day}^{-1} = 2.4 \text{ day}^{-1} $
- Monod constant ($ K_s $): $ 1 \text{ mg L}^{-1} = 0.001 \text{ g L}^{-1} $
- Limiting substrate concentration ($ S $): $ 23 \text{ g L}^{-1} $
- Endogenous decay rate ($ k_d $): $ 0.1 \text{ day}^{-1} $
Specific Growth Rate Calculation
Calculate the specific growth rate ($ \mu $) using the Monod equation with the converted values:
$ \mu = (2.4 \text{ day}^{-1}) \frac{23 \text{ g L}^{-1}}{0.001 \text{ g L}^{-1} + 23 \text{ g L}^{-1}} $
$ \mu = 2.4 \times \frac{23}{23.001} \text{ day}^{-1} $
$ \mu \approx 2.3999 \text{ day}^{-1} $
Net Growth Rate Calculation
Now, calculate the net specific growth rate ($ \mu_{net} $) by subtracting the endogenous decay rate ($ k_d $):
$ \mu_{net} = \mu - k_d $
$ \mu_{net} \approx 2.3999 \text{ day}^{-1} - 0.1 \text{ day}^{-1} $
$ \mu_{net} \approx 2.2999 \text{ day}^{-1} $
Final Answer Rounding
Rounding the net specific growth rate to one decimal place:
$ \mu_{net} \approx 2.3 \text{ day}^{-1} $
This value falls within the specified range of 2.2 to 2.4.
Important Questions from Kinetics of Cell Growth Substrate Utilization and Product Formation
If the rate at which $E. coli$ divides is $0.5 \text{ h}^{-1}$, then its doubling time is _______________ h.
- Which of the following factors can affect the growth of a microbial culture in a batch cultivation process?
Let $y(t)$ be a bacterial population whose growth is given by
$ \frac{dy}{dt} = \lambda(y + 2) $
where $ \lambda $ is the growth rate constant. If $y(0) = 1$ and $y(1) = 4$, then the value of $ \lambda $ is
If the doubling time of a bacterial population is 3 hours, then its average specific growth rate during this period is _________ $h^{-1}$.
(Round off to two decimal places)
- A microorganism is grown in a batch culture using glucose as a carbon source. The apparent growth yield is $0.5 \frac{\text{g biomass}}{\text{g substrate}}$. The initial concentrations of biomass and substrate are $2 \text{ g L}^{-1}$ and $200 \text{ g L}^{-1}$, respectively. Assuming that there is no endogenous metabolism, the maximum biomass concentration that can be achieved is ________ $\text{g L}^{-1}$.
