Q.34 If the rate at which E. coli divides is
0.5 h⁻¹, then its doubling time is ______ h.

Calculate 0.5 h⁻¹ Rate to 1.386 Hours

The doubling time for E. coli at a division rate of 0.5 h⁻¹ is 1.386 hours, derived from the standard formula:

t_d = ln(2) / k

Here, k is the growth rate constant. This reflects exponential bacterial growth during the log phase, common in microbiology studies. Understanding this calculation aids precise modeling of microbial populations in research or lab settings.

Core Formula Explained

Doubling time t_d measures the time for a bacterial population to double under exponential growth:

t_d = ln(2) / k

With ln(2) ≈ 0.693 and k = 0.5 h⁻¹:

t_d = 0.693 / 0.5 ≈ 1.386 hours

This continuous growth model applies to E. coli in ideal lab conditions, such as nutrient-rich media.

Step-by-Step Calculation

Start with the growth equation:

N_t = N₀ e^{k t}

Doubling occurs when N_t = 2 N₀, so set t = t_d:

2 = e^{k t_d}  ⇒  k t_d = ln(2)  ⇒  t_d = ln(2)/k

Plugging in values:

t_d = 0.693147 / 0.5 ≈ 1.386 hours

Common Options and Explanations

Only option C matches the precise formula validated across microbiology resources.
Errors in other options stem from mixing discrete (t_d = 1/k) and continuous growth models.