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)
The question asks for the average specific growth rate ($\mu_{avg}$) of a bacterial population given its doubling time ($t_d$). The doubling time is the time it takes for the population to double in size.
The relationship between the average specific growth rate and doubling time is defined by the formula:
$ \mu_{avg} = \frac{\ln(2)}{t_d} $
Where:
Given the doubling time $t_d = 3$ hours:
$ \mu_{avg} = \frac{\ln(2)}{3 \text{ h}} $
$ \mu_{avg} \approx \frac{0.693147}{3 \text{ h}} \approx 0.231049 \text{ h}^{-1} $
$ \mu_{avg} \approx 0.23 \text{ h}^{-1} $
The calculated average specific growth rate is approximately 0.23 $h^{-1}$. This value falls within the range of 0.2 to 0.25.
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