Q.42 The population of a bacterial culture increases from one thousand to one billion in five hours. The doubling time of the culture (correct to 1 decimal place) is ______ min.

The doubling time for the bacterial culture is 15.1 minutes.

Problem Breakdown

Bacterial populations grow exponentially during the log phase via binary fission, following Nt=N0×2t/dNt=N0×2t/d, where NtNt is final population (1 billion or 109109), N0N0 is initial (1000 or 103103), $t$ is time (5 hours = 300 minutes), and $d$ is doubling time. Rearrange to solve for d=t/log⁡2(Nt/N0)d=t/log2(Nt/N0). Here, Nt/N0=106Nt/N0=106, so log⁡2(106)≈19.93log2(106)≈19.93 doublings occur in 300 minutes, yielding $d=300/19.93=15.1$ min (to 1 decimal).

Step-by-Step Calculation

Convert units: 5 hours = 300 min. Compute growth ratio: 109/103=106109/103=106. Find doublings: log⁡2(106)=6×log⁡2(10)≈6×3.3219=19.9316log2(106)=6×log2(10)≈6×3.3219=19.9316. Divide: $300/19.9316=15.04$ min, rounded to 15.1 min.

Bacterial cultures exhibit rapid exponential growth, doubling from 1000 to 1 billion cells in just 5 hours, making bacterial doubling time calculation essential for CSIR NET life sciences exams. This problem tests understanding of binary fission kinetics, where population follows Nt=N0×2nNt=N0×2n and $n=t/d$. For initial N0=103N0=103, final Nt=109Nt=109, and $t=300$ min, the ratio 106106 requires ~19.93 doublings, so $d=15.1$ min.

Why No Options?

This fill-in-the-blank style (common in CSIR NET) has no multiple choices, avoiding distractors like 15.0 min (ignores precise log) or 20.0 min (E. coli lab average). Common errors include using hours (75.2 min, too high) or base-10 log (wrong base).

CSIR NET Relevance

Master bacterial doubling time calculation for units on growth curves, generation time, and biotech applications like fermentation. Practice verifies: 19.93 × 15.1 min ≈ 300 min, confirming 1 billion cells.