Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
For an initial spore load equal to 25 spores per container inoculated with Clostridium botulinum having D₁₂₁=0.25 min, the spoilage probability of the container subjected to F₁₂₁=1.5 min is
Options:
(A) 10⁻⁵
(B) 10⁻⁶
(C) 10⁻⁸
(D) 10⁻⁹
Official Answer Key:
A
Step-by-Step Solution & Explanation
Number of decimal (log) reductions achieved = F/D = 1.5/0.25 = 6 log cycles.
Surviving spore count (= spoilage probability per container, for N₀ well below 1 log unit) = N₀/10⁶ = 25/10⁶ = 2.5×10⁻⁵, which is of order 10⁻⁵. Answer: (A).
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