If the rate at which $E. coli$ divides is $0.5 \text{ h}^{-1}$, then its doubling time is _______________ h.
The doubling time ($t_d$) is the duration required for a microbial population, such as E. coli, to double its number. This time is directly related to the organism's division rate ($k$), often referred to as the growth rate constant.
The mathematical relationship connecting the division rate ($k$) and the doubling time ($t_d$) is expressed as:
$ t_d = \frac{\ln(2)}{k} $
Where $\ln(2)$ is the natural logarithm of 2.
The question provides the division rate for E. coli:
$ k = 0.5 \text{ h}^{-1} $
To find the doubling time, substitute the given division rate into the formula:
$ t_d = \frac{\ln(2)}{0.5 \text{ h}^{-1}} $
Using the approximate value $\ln(2) \approx 0.693$:
$ t_d \approx \frac{0.693}{0.5 \text{ h}^{-1}} $
Performing the division yields:
$ t_d \approx 1.386 \text{ h} $
Based on the provided division rate of $0.5 \text{ h}^{-1}$, the calculated doubling time for E. coli is approximately 1.386 hours.
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)