Consider a continuous culture provided with a sterile feed containing $10 \text{ mM}$ glucose. The steady state cell density and substrate concentration at three different dilution rates are given in the table below. The maximum specific growth rate $\mu_m$$(\mathrm{in\ h^{-1}})$, will be _____. Dilution rate $(\mathrm{h^{-1}})$ Cell density $(\mathrm{g\ L^{-1}})$ Substrate concentration (mM 0.05 0.248 0.067 0.5 0.208 1.667 5 0 10
Calculating Maximum Specific Growth Rate ($\mu_m$)
In a chemostat operating at steady state, the dilution rate ($D$) equals the specific growth rate ($\mu$). The Monod equation describes the relationship between $\mu$ and substrate concentration ($S$):
$ \mu = \frac{\mu_m S}{K_s + S} $
At steady state ($D = \mu$), this becomes:
$ D = \frac{\mu_m S}{K_s + S} $
To find $\mu_m$, we linearize the Monod equation using the Lineweaver-Burk transformation:
$ \frac{1}{\mu} = \frac{K_s}{\mu_m} \frac{1}{S} + \frac{1}{\mu_m} $
Substituting $D$ for $\mu$:
$ \frac{1}{D} = \frac{K_s}{\mu_m} \frac{1}{S} + \frac{1}{\mu_m} $
This equation is in the form $y = mx + c$, where $y = 1/D$, $x = 1/S$, the slope $m = K_s / \mu_m$, and the intercept $c = 1 / \mu_m$.
Data Transformation for Analysis
We use the first two data points from the table. The third data point (Dilution rate = 50 h⁻¹) is disregarded as it represents an unrealistically high growth rate for typical microbial cultures and would imply substrate concentration equals influent concentration ($S = S_{in}$), which is inconsistent with growth.
| Dilution Rate, $D$ (h⁻¹) | Substrate Concentration, $S$ (mM) | $1/D$ (h) | $1/S$ (mM⁻¹) |
|---|---|---|---|
| 0.05 | 0.067 | 20.000 | 14.925 |
| 0.50 | 1.667 | 2.000 | 0.600 |
Calculation of $\mu_m$
Using the transformed data points $(1/S, 1/D)$: (14.925, 20.000) and (0.600, 2.000).
Calculate the slope ($m$):
$ m = \frac{2.000 - 20.000}{0.600 - 14.925} = \frac{-18.000}{-14.325} \approx 1.2565 $
Calculate the y-intercept ($c$) using the point (0.600, 2.000):
$ c = 2.000 - (m \times 0.600) = 2.000 - (1.2565 \times 0.600) \approx 2.000 - 0.7539 = 1.2461 $
The maximum specific growth rate ($\mu_m$) is the reciprocal of the y-intercept:
$ \mu_m = \frac{1}{c} = \frac{1}{1.2461} \approx 0.8025 \text{ h}^{-1} $
The calculated value of $\mu_m \approx 0.8025 \text{ h}^{-1}$ falls within the specified range of 0.795 to 0.805.
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 ________ $\text{g L}^{-1}$.
