Question

The radiation dose required for the complete destruction of Clostridium botulinum is

• A. 0.5 M rads
• B. 2.5 M rads
• C. 4.5 M rads
• D. 5 M rads

Hint:

Radiation doses are classified as:
1. Radurization (Low dose $<$ 1 kGy): Extension of shelf life.
2. Radicidation (Medium 1-10 kGy): Elimination of non-spore-forming pathogens.
3. Radappertization (High $>$ 10 kGy): Sterilization (e.g., 45 kGy for C. botulinum).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question deals with food irradiation, specifically the dose required to achieve sterilization (radappertization) by eliminating the most resistant bacterial spores, such as those of Clostridium botulinum.
Detailed Explanation:

• Resistance of C. botulinum: Spores of C. botulinum are extremely resistant to heat and radiation. In canning, a "12D process" is used to ensure safety. In radiation processing, a similar standard is applied.

• Radappertization Doses: Sterilization (complete destruction of all microbes) requires very high doses of radiation. For low-acid foods like fish, the dose must be sufficient to achieve a 12-log reduction (12D) in the population of C. botulinum spores.

• The 12D Dose: Scientific studies have established that for most food matrices, a dose of approximately 45 kGy (kiloGray), which is equivalent to 4.5 M rads (Mega rads), is necessary to achieve commercial sterility with respect to C. botulinum.

• Unit Conversion: 1 Gray = 100 rads; 45 kGy = 4,500,000 rads = 4.5 M rads.

Step 2: Final Answer:
The dose required for complete destruction (12D) is 4.5 M rads.