Question

Bacteria usually reproduce by a process called binary fission. In this type of reproduction, one bacterium divides to form two bacteria. Under ideal conditions, some bacteria reproduce every 15 minutes. Find the constant k for this type of bacteria under ideal conditions. Assume t is in minutes.

Answer 1

Answer:

K= 0.067 per minute

Explanation:

This was obtained from the equation k= (Log Nt - Log No) / (0.301 x t)

where Nt = 2, No = 1 and t = 15

therefore, k = ( Log 2 - Log 1) / (0,301 x 15)

then k = 0.301 / 0.301 x 15

k = 1 / 15

Finally, K = 0.067m^{-1}

Answer 2

Answer:

$k = 0.462$

Explanation:

Given-

A bacteria reproduces after every fifteen minutes.

Thus, after every fifteen minutes, the bacterial population will get double.

Now as we know that

$N_t = N_o e^{kt}$

Here,

$N_t$ represents the population after time t

$N_o$  represents the initial population

Substituting the given values in above equation we get -

$\frac{N_t}{N_o} =2$

$2 = e^{15k]$

$15k = ln 2$

On solving we get -

$k = \frac{ln 2}{15}$

$k = 0.462$