To describe the growth by cell division mathematically, we need to identify the correct relationship between the number of cells at a certain time [tex]\( N \)[/tex] and the initial number of cells [tex]\( N_0 \)[/tex], considering the number of generations [tex]\( n \)[/tex].
In cell division, particularly in the context of binary fission (common in bacteria), each cell divides into two new cells. This implies that with each generation, the number of cells doubles. Therefore, if you start with [tex]\( N_0 \)[/tex] cells, after one generation, you will have [tex]\( 2 \times N_0 \)[/tex] cells, after two generations, you will have [tex]\( 2^2 \times N_0 \)[/tex] cells, and so on.
In general, after [tex]\( n \)[/tex] generations, the number of cells [tex]\( N \)[/tex] can be expressed as:
[tex]\[ N = N_0 \times 2^n \][/tex]
Now, let's analyze the given options:
a) [tex]\( N_0 = N_0 \times n^2 \)[/tex]
- This implies that the number of cells after [tex]\( n \)[/tex] generations is proportional to the square of [tex]\( n \)[/tex], which is incorrect for binary division growth.
b) [tex]\( N_0 = N_t^* n^2 \)[/tex]
- This does not make sense in the context of our problem, as it incorrectly relates [tex]\( N_0 \)[/tex] and [tex]\( N_t \)[/tex] with [tex]\( n^2 \)[/tex].
c) [tex]\( N_t = 2^{n} N_0 \)[/tex]
- This correctly reflects the nature of binary fission, where the number of cells doubles with each generation.
d) [tex]\( N_t = N_0 \times 2n \)[/tex]
- This implies a linear relationship between [tex]\( N_t \)[/tex] and [tex]\( n \)[/tex], which is incorrect for binary division growth.
Therefore, the correct equation describing cell growth by binary division is:
[tex]\[ N_t = 2^{n} N_0 \][/tex]
Option c) is the correct answer.
