Question

Given below are two statements:
Statement (I): The UHT sterilization treatment of milk is designed to achieve a 9-log reduction of the endospores of Geobacillus stearothermophilus
Statement (II): The UHT sterilization treatment designed for milk given above also results is 12-log reduction of Clostridium botulinum

• A. Both Statement (I) and Statement (II) are true
• B. Both Statement (I) and Statement (II) are false
• C. Statement (I) is true but Statement (II) is false
• D. Statement (I) is false but Statement (II) is true

Hint:

Remember that Geobacillus stearothermophilus has a much higher decimal reduction time (\(D\)-value) than Clostridium botulinum at sterilization temperatures. Designing a process to eliminate G. stearothermophilus automatically ensures a highly safe margin against botulism.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Ultra-High Temperature (UHT) processing of milk involves heating milk to 135–140 \(^{\circ}\text{C}\) for 2 to 5 seconds to achieve commercial sterility.
The intensity of the thermal process is designed around the inactivation of highly heat-resistant bacterial endospores.
Detailed Explanation:
Let us evaluate both statements individually:
- Statement I: Geobacillus stearothermophilus is a thermophilic, spore-forming bacterium. Its endospores are among the most heat-resistant biological structures known. Because of this high resistance, it is used as the standard biological indicator/reference organism for validating UHT sterilization processes. To ensure commercial sterility under practical conditions, the thermal treatment must be designed to achieve at least a 9-log reduction (\(B \geq 9\)) of G. stearothermophilus endospores. Thus, Statement I is true.
- Statement II: Clostridium botulinum is an anaerobic, spore-forming pathogen of extreme concern in low-acid foods. Its spores are significantly less heat-resistant than those of G. stearothermophilus. A thermal process designed to achieve a 9-log reduction in G. stearothermophilus spores is highly intense. Consequently, this treatment will achieve a reduction of Clostridium botulinum spores that far exceeds the standard 12-log safety criterion (the "12-D" or "botulinum cook" concept). Thus, Statement II is true.
Since both Statement I and Statement II are correct, the correct option is (A).
Final Answer:
Both Statement (I) and Statement (II) are true. Hence, the correct option is (A).