Q84. The decimal reduction time (D) for reducing 1012 spores of Clostridium
botulinum to 1 spore at 111°C will be_______ min (in integer). The D value
is 0.2 min at 121°C. The increase in temperature required to change D to
1/10th of its initial value (Z value) is 10°C.
Decimal Reduction Time (D-value) for Clostridium botulinum Spores at 121°C: Detailed Calculation Guide
The decimal reduction time, or D-value, for Clostridium botulinum spores at 121°C is 0.21 minutes, representing the time needed to reduce the spore population by 90% (1 log cycle). Given an initial spore count of 1012 at 121°C, the time to achieve a 12-log reduction to 1 spore follows the formula t=n×D , where (n = 12 ). With a standard z-value of 10°C for C. botulinum, temperature changes alter the D-value exponentially.
Key Concepts Explained
D-value quantifies microbial heat resistance, specifically the time for a 90% kill at a given temperature. For C. botulinum type A spores in low-acid foods, D121°C = 0.21 min, so a “botulinum cook” of 2.52-3 minutes ensures a 12D process for commercial sterility. The z-value (10°C) measures temperature sensitivity: log(D2 / D1) = (T1 – T2)/z.
Step-by-Step Solution
Initial population ( N0 = 1012) spores; target N = 1 spore, so (log(N0/N) = 12 . Time minutes at 121°C.
At 121°C, D = 0.21 min, so min (integer 3 min typically).
Temperature Change Calculation
To change D to 0.2 min (nearly same as 0.21), required ΔT is negligible (~0.14°C) since ΔT=z× log(D121/0.2) ≈ 10×log (0.21/0.2)C. Increase from initial D=0.2 min at 121°C to new D requires ~0°C change (integer).
| Parameter | Value at 121°C | 12D Time (min) | z-Value (°C) |
|---|---|---|---|
| D-value | 0.21 | 2.52 | 10 |
| To 1 spore from 1012 | N/A | 2.52 | N/A |
Introduction (SEO Optimized): In food microbiology for CSIR NET Life Sciences, understanding decimal reduction time D value Clostridium botulinum spores 121°C is crucial for thermal processing questions. This guide solves the exact problem: D at 121°C, time for 1012 to 1 spore, and temperature shift with z=10°C—in just 2.52 minutes!
Why D-Value Matters in Sterilization
C. botulinum spores cause botulism; D121°C=0.21 min ensures 12D reduction (probability <10-12 survival). Formula: t = log(N0/N) / (1/D) = nD.
- 12 logs from 1012 to 1: 12 × 0.21 = 2.52 min.
- z=10°C standard for extrapolation.
Detailed Problem Breakdown for CSIR NET
Q19 Blanks: D121°C = 0.21 min (integer often 0). Time for 1012 to 1 spore = 3 min (rounded). From D=0.2 min at 121°C, ΔT to new D= 0 °C (negligible).
Every Option/Aspect: If options given (e.g., MCQ), 0.21 min matches standard; alternatives like 0.2 approximate but 0.21 precise. Temp increase ~0°C as D nearly identical.
Practical Applications & Formulas
Use DT = Dref \times 10(Tref – T)/z for temps. E.g., D111°C ≈ 2.21 min.
This covers CSIR NET prep comprehensively.
