Bacterial Growth Calculation: Generations to Reach Target Population
This problem involves calculating the number of generations for a bacterial population to grow exponentially from an initial size to a target size.
Understanding Exponential Growth
Bacteria reproduce through binary fission, where one cell divides into two. This results in exponential growth. The formula relating the final population ($N_t$), initial population ($N_0$), and the number of generations ($n$) is:
$ N_t = N_0 \times 2^n $
Solving for Number of Generations
We are given:
- Initial population, $N_0 = 100$
- Target population, $N_t = 10^5$
We need to find the number of generations, $n$. Substitute the values into the formula:
$ 10^5 = 100 \times 2^n $
Step 1: Isolate the Growth Factor
Divide both sides by the initial population ($N_0 = 100$):
$ \frac{10^5}{100} = 2^n $
$ 1000 = 2^n $
Step 2: Solve for 'n' using Logarithms
To find $n$, we can take the logarithm base 2 of both sides:
$ n = \log_2(1000) $
Step 3: Approximate the Result
We know that $2^{10} = 1024$. Since 1024 is very close to 1000, the number of generations ($n$) is approximately 10.
Therefore, after approximately 10 generations, the population will reach $10^5$.
Important Questions from Population growth curves
Population growth of a species can be modelled as $$ \frac{dN(t)}{dt} = rN(t)\left(1 - \frac{N(t)}{K}\right) $$ where $N(t)$ is the population size at time $t$; $r$ is the growth rate; and $K$ is the carrying capacity of the environment.
For $K = 9000$, $\frac{dN(t)}{dt}$ is maximized at $N =$ _____
(Answer in integer)
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- A population grows as per the equation $dn/dt = rn (1-n/1000)$ where $n$ is the population density, $r$ is the intrinsic growth rate and 1000 is the carrying capacity. The growth rate of the population is maximum at a population density of ________
