A bacterial culture doubles every 3 hours. If the initial cell count is 5×10^6, how many cells will be present after 12 hours?

A bacterial culture doubles every 3 hours. If the initial cell count is $5 \times 1 0^{6}$ , how many cells will be present after 12 hours?

medium 2 marks Class 12 · Biology · Organisms and Populations

Answer

8 \times 1 0^{7}

cells

The correct answer is $8 \times 1 0^{7}$ cells — Thus, after 12 hours the culture contains $8 \times 1 0^{7}$ cells.

Determine the number of doubling periods in 12 hours: 12 ÷ 3 = 4 doublings.
Use the formula N = N_0 $\times$ $2^{n}$ , where N_0 is the initial count and n is the number of doublings.
Substitute the values: N = $5 \times 1 0^{6} \times 2^{4}$ .
Calculate $2^{4}$ = 16.
Multiply: $5 \times 1 0^{6}$ × 16 = $80 \times 1 0^{6} = 8 \times 1 0^{7}$ .
Thus, after 12 hours the culture contains $8 \times 1 0^{7}$ cells.

Common mistake: Treating the increase as linear (adding the initial count each period) instead of exponential.

Concept tested: Exponential bacterial growth