Q.58 Decimal reduction time of bacterial spores is 23 min at 121 °C and the death kinetics follow first
order. One liter of medium containing 10⁵ spores per mL was sterilized for 10 min at 121 °C in a batch
sterilizer. The number of spores in the medium after sterilization (assuming destruction of spores in
heating and cooling period is negligible) will be __________ × 10⁷.

 

The decimal reduction time (D-value) of 23 minutes at 121°C indicates that the bacterial spore population reduces by 90% (one log cycle) every 23 minutes under first-order death kinetics. For 1 liter of medium with an initial 10⁵ spores/mL (total 10⁸ spores), sterilization at 121°C for 10 minutes yields approximately 3.68 × 10⁷ surviving spores, matching the fill-in-the-blank as 3.68 × 10⁷ or rounded to 4 in some contexts, but precisely 3.68.

Problem Breakdown

First-order kinetics follow log(N/N₀) = -t/D, where N is final spores, N₀ is initial (10⁸ total), t=10 min, D=23 min.

This gives log(N/10⁸) = -10/23 ≈ -0.4348, so N/10⁸ ≈ 10^(-0.4348) ≈ 0.3678.

Thus, N ≈ 0.3678 × 10⁸ = 3.678 × 10⁷ spores per liter.

Step-by-Step Calculation

• Initial load: 10⁵ spores/mL × 1000 mL = 10⁸ spores.
• Log reduction: t/D = 10/23 ≈ 0.4348 (less than 1, so ~63% survivors expected).
• Survivors: N = N₀ × 10^(-t/D) = 10⁸ × 10^(-0.4348) = 3.678 × 10⁷.
• Negligible heating/cooling loss per assumption confirms direct holding time use.

Decimal reduction time of bacterial spores at 121°C is key for batch sterilization in biotech fermentation. This guide solves the exact problem: 1 liter medium with 10⁵ spores per mL sterilized 10 min at 121°C (D=23 min, first-order kinetics), finding survivors as __ ×10⁷. Ideal for GATE BT, biochemical engineering students optimizing medium sterility.

Core Concept: D-Value in Sterilization

Decimal reduction time (D-value) at 121°C measures time for 90% spore kill under first-order kinetics: log(N/N₀) = -t/D.

For bacterial spores (e.g., Bacillus), D₁₂₁=23 min means slow death, requiring t > 23 min for full log reduction.

Here, short 10-min hold gives partial kill (0.43 logs).

Detailed Solution Walkthrough

• Initial spores: 10⁵/mL × 1000 = 10⁸ total.
• Log survivors ratio: -10/23 = -0.4348.
• N = 10⁸ × antilog(-0.4348) = 10⁸ × 0.3678 = 3.678 × 10⁷.
• Answer: 3.68 (or 3.7 rounded) ×10⁷, fitting “__________ ×10⁷” blank.

Common Errors Explained

• Mistake 1: Using total volume wrong (e.g., forget ×1000 mL/L) → underestimates N₀.
• Mistake 2: Wrong kinetics (zero-order) → linear decay, not exponential.
• Mistake 3: Include heating/cooling → but negligible per question.
• Overkill assumption: 10 min << 23D (for 10⁵→1 needs ~115 min).

Biotech Applications

In fermenters, this ensures <10^{-3 contamination probability via Del factor (t/D × viability).}

Scale-up uses F₀ = ∫10^(T-121)/10 dt for variable heat.

Practice for exams: Verify with Python: import math; N=1e8*math.pow(10,-10/23) → 3.68e7.