Q.84 During the exponential growth, it took 6 hours for the population of bacterial cells to increase from 2.5 × 106 to 5 × 108. The generation time of the bacterium, rounded off to the nearest integer, is ________ minutes.

Q.84 During the exponential growth, it took 6 hours for the population of bacterial cells
to increase from 2.5 × 106 to 5 × 108. The generation time of the bacterium,
rounded off to the nearest integer, is ________ minutes.

The generation time of the bacterium is 47 minutes. This result comes from calculating the number of doublings during exponential growth and dividing the total time by that value. The problem aligns with standard bacterial growth kinetics formulas used in CSIR NET Life Sciences exams.

Calculation Steps

Bacterial populations during exponential growth follow Nt = N0 × 2n, where Nt is the final population, N0 is the initial population, and n is the number of generations.

Here, N0 = 2.5 × 106, Nt = 5 × 108, and t = 6 hours (360 minutes). First, solve for n:

n = log2(Nt/N0) = log2(5×108/2.5×106) = log2(200) ≈ 7.64 generations.

Generation time g = t/n = 360/7.64 ≈ 47.1 minutes, which rounds to 47 minutes.

Formula Verification

The standard generation time formula is g = (t × log 2) / log(Nt/N0), equivalent to g = t / log2(Nt/N0). Substituting values yields the same 47 minutes. This matches examples like E. coli (20-30 minutes generation time under optimal conditions) but reflects the specific growth rate here.

Bacterial Generation Time Calculation Guide

Bacterial generation time calculation is crucial for understanding microbial exponential growth in competitive exams like CSIR NET Life Sciences. This guide solves the exact problem: during exponential growth, a bacterial population increases from 2.5 × 106 to 5 × 108 cells in 6 hours, requiring the generation time rounded to the nearest integer in minutes.

Why Generation Time Matters

Generation time, or doubling time, measures how quickly bacteria divide by binary fission during log phase. Faster times (e.g., 20 minutes for E. coli) indicate optimal conditions, vital for biotechnology and infection control studies.

Step-by-Step Calculation

  1. Identify initial (N0 = 2.5 × 106) and final (Nt = 5 × 108) populations, plus time t = 6 hours = 360 minutes.
  2. Compute fold increase: Nt/N0 = 200.
  3. Find generations n = log2(200) ≈ 7.64 using n = ln(Nt/N0)/ln 2.
  4. Divide: g = 360/7.64 ≈ 47.1 minutes → 47 minutes.

No options are provided (fill-in-the-blank style), but common distractors include errors like using base-10 log (yielding ~16 minutes) or forgetting to convert hours to minutes.

Applications in CSIR NET and Biology

This mirrors GATE/CSIR NET questions testing logarithmic growth kinetics. Master it for topics like microbial growth curves, enzyme kinetics, and biotech.

Parameter Summary Table

Parameter Value Role in Formula
Initial Population 2.5 × 106 N0
Final Population 5 × 108 Nt
Time 6 hours (360 min) t
Generations (n) ~7.64 log2(Nt/N0)
Generation Time 47 minutes t/n

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