The specific substrate consumption rate quantifies how efficiently a microbial culture consumes a substrate relative to its own biomass. In simpler terms, it measures the amount of substrate consumed per unit of biomass per unit of time.
Let's break down the components:
Therefore, the unit for specific substrate consumption rate is derived as:
$ \frac{\text{Mass of Substrate}}{\text{Mass of Biomass} \cdot \text{Time}} $Substituting the common units:
$ \frac{g}{g \cdot h} $Based on the derivation:
Thus, the correct unit for specific substrate consumption rate in a growing culture is $\frac{g}{g \cdot h}$.
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)