In exponential growth, the population doubling time (td) is related to the specific growth rate (μ) by which expression?

Introduction:Microbial kinetics frequently uses the specific growth rate μ (per hour or per minute) to describe exponential growth. The doubling time td translates this rate into a practical measure: the time required for biomass or cell count to double. Correctly relating td and μ is essential for scheduling harvests and designing dilution rates in continuous cultures.

Given Data / Assumptions:

• Growth follows N(t) = N0 * e^(μt) during the exponential phase.
• Doubling time td is the time when N(t) = 2 * N0.
• Natural logarithms are used for μ in the exponential model.

Concept / Approach:Set N(td) = 2 * N0 in the exponential model and solve for td in terms of μ. Because the base of the natural exponential is e, the constant that appears is ln 2 (approximately 0.693). The result connects the intuitive time-to-double with the rate constant of growth.

Step-by-Step Solution:

Verification / Alternative check:If μ = 0.693 h^-1, then td = ln 2 / 0.693 ≈ 1 h, which is consistent with doubling once per hour. Dimensional analysis shows td has units of time when μ has inverse time units.

Why Other Options Are Wrong:

• log2/μ and μ/log2: Ambiguous base-10 notation and wrong placement of μ; the model uses natural logs.
• μ/ln2: Inverts the relationship; that would be a rate, not a time.

Common Pitfalls:Mixing log bases (log10 vs ln) leads to incorrect constants. Also, applying td = ln 2 / μ outside exponential phase or under substrate limitation yields misleading estimates; always verify phase conditions.

Final Answer:ln2/μ