Question

The growth of a bacterial colony can be modeled as a first-order process in which the probability of cell division is linear with respect to time such that dN/N = ζήτα dt, where dN is the number of cells that divide in the time interval dt, and ζήτα is a constant. Use the preceding expression to show that the number of cells in a colony is given by N=N₀eζήτα t, where N is the number of cell colonies and N₀ is the number of colonies present at t = 0

Answer 1

N is the number of cell colonies, N₀ is the number of colonies present at t = 0, and ζ is a constant representing the probability of cell division per unit time.

Consider the given expression for the growth of a bacterial colony:

dN/N = ζdt

This expression states that the rate of change of the number of cells (dN/dt) is proportional to the current number of cells (N) and a constant ζ.

The constant ζ represents the probability of cell division per unit time.

To find the number of cells as a function of time, we can separate the variables and integrate both sides of the equation:

∫ dN/N = ∫ ζdt

Integrating both sides gives:

ln(N) = ζt + C

where C is an integration constant. To solve for C, we use the initial condition that N = N₀ at t = 0:

ln(N₀) = 0 + C

Solving for C, we get:

C = ln(N₀)

Substituting this value of C back into the equation for ln(N), we get:

ln(N) = ζt + ln(N₀)

Exponentiating both sides, we get:

N = N₀e^ζt

Therefore, the number of cells in a colony is given by:

N = N₀e^ζt

where N is the number of cell colonies, N₀ is the number of colonies present at t = 0, and ζ is a constant representing the probability of cell division per unit time.