Question

A bacterial Colony grows most commonly by cell division. the change in the population due to cell division in an actively growing colony is dn= lemda g N dt. the population of bacterial colony at time t is [No=N(t = o)]
​

Answer 1

Answer:

Explanation:

A bacterial Colony grows most commonly by cell division. the change in the population due to cell division in an actively growing colony is dn= lemda g N dt. the population of bacterial colony at time t is [No=N(t = o)]

​

Growth of bacterial cultures is defined as an increase in the number of bacteria in a population rather than in the size of individual cells. The growth of a bacterial population occurs in a geometric or exponential manner: with each division cycle (generation), one cell gives rise to 2 cells, then 4 cells, then 8 cells, then 16, then 32, and so forth. The time required for the formation of a generation, the generation time (G), can be calculated from the following formula:

                                           $G=\frac{t}{n} \\=\frac{t}{\ 3.3log{b}{B} }$

 for example:

   The bacterial growth follows the rate law,  $\frac{dN}{dt} =kN$

where k is a constant and 'N' is the number of bacteria cell at any time. If the population of bacteria (number of cell) is doubled in 5 min, find the time in which the population will be eight times of the initial one?

 Bacterial growth rate is given as: $\frac{dN}{dt} =kN$   which is a first order reaction  

               $\int\limits^ {}\frac{dN}{N} \, =k\int\limits {} \, dt$

 Integrating both sides, we get    

                $In=\frac{N}{N_{0} } =kt$

 After $t=5 min, N=2N_{0}$

∴     $k(5)=ln 2 ..........(1)$

After  $time =t min, N=8N_{0}$

∴  $k(8)=ln 8 ..........(2)$

Dividing (2) and (1), we get   $\frac{t}{5} =\frac{ln8}{ln2} \\where ln8=3ln2$

                                      $\frac{t}{5} =\frac{3ln2}{ln2}$

                                      $t=15 min$

The project code is #SPJ3