The maximum cell concentration (g $l^{-1}$) expected in a bioreactor with initial cell concentration of $1.75$ g $l^{-1}$ and an initial glucose concentration of $125$ g $l^{-1}$ is ($Y_{x/s} = 0.6$ g cell/g substrate) _________
To determine the maximum cell concentration ($X_{max}$) expected in the bioreactor, we use the concept of biomass yield from the substrate.
The amount of cells produced is directly proportional to the amount of substrate consumed. The yield coefficient ($Y_{x/s}$) relates cell mass produced to substrate consumed.
The formula connecting initial and final cell and substrate concentrations is:
$X - X_0 = Y_{x/s} \times (S_0 - S)$
Where:
Assuming the substrate is completely consumed ($S \approx 0$), the maximum cell concentration ($X_{max}$) can be estimated by rearranging the formula:
$X_{max} \approx X_0 + (Y_{x/s} \times S_0)$
Substitute the given values into the formula:
Calculation:
$X_{max} \approx 1.75 \text{ g } l^{-1} + (0.6 \text{ g cell/g substrate} \times 125 \text{ g } l^{-1})$
$X_{max} \approx 1.75 + 75$
$X_{max} \approx 76.75 \text{ g } l^{-1}$
The calculated maximum cell concentration is approximately $76.75$ g $l^{-1}$. This value falls within the expected range.
If the rate at which $E. coli$ divides is $0.5 \text{ h}^{-1}$, then its doubling time is _______________ h.
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)