This solution explains how to calculate the cell yield coefficient ($Y_{x/s}$) for Saccharomyces cerevisiae in an exponential batch culture using the provided growth parameters.
The rate at which biomass increases per unit volume ($dX/dt$) during exponential growth is calculated using the specific growth rate ($\mu$) and the current cell density ($X$):
$ \frac{dX}{dt} = \mu X $
Substitute the given values:
$ \frac{dX}{dt} = (0.4 \, h^{-1}) \times (20 \, gl^{-1}) = 8 \, gl^{-1}h^{-1} $
The cell yield coefficient ($Y_{x/s}$) quantifies the efficiency of converting substrate into biomass. It is defined as the ratio of biomass produced to the substrate consumed:
$ Y_{x/s} = \frac{\text{Biomass Produced}}{\text{Substrate Consumed}} $
In terms of volumetric rates during active growth, this becomes:
$ Y_{x/s} = \frac{dX/dt}{-dS/dt} $
Where $-dS/dt$ is the rate of substrate consumption per unit volume.
The given substrate uptake rate (v) directly represents the rate of substrate consumption per unit volume ($-dS/dt$).
$ Y_{x/s} = \frac{8 \, gl^{-1}h^{-1}}{16 \, gl^{-1}h^{-1}} $
$ Y_{x/s} = 0.5 $
Therefore, the cell yield coefficient ($Y_{x/s}$) is 0.5.
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)