优化的紫外线-C处理使病原和腐烂相关的细菌失活,同时保留人乳中的生物活性蛋白质,维生素和脂质
Ningjian Liang1, Hussein Mohamed2, Rachel F Pung3
1Nutrition Program, College of Health, Oregon State University, Corvallis, Oregon 97331, United States.
Journal of agricultural and food chemistry
|May 16, 2024
概括
紫外线C (UV-C) 辐射有效地使捐赠人母乳中的细菌失活,同时保留了重要的蛋白质和营养素. 这种方法提供了一个比传统巴氏杀菌更安全的替代方案,提高了牛奶的质量.
科学领域:
- 食品科学 食品科学 食品科学
- 微生物学 微生物学
- 生物化学 生物化学
背景情况:
- 持有人巴氏杀菌 (HoP) 提高了供体母乳的微生物安全性,但降解了生物活性蛋白质.
- 紫外线-C (UV-C) 辐射显示出潜在的病原体无活化,同时保留牛奶成分,但最佳参数尚未确定.
研究的目的:
- 确定最小的UV-C剂量,以显著减少母乳中的细菌.
- 评估UV-C处理对生物活性蛋白,维生素E和脂质氧化的影响.
主要方法:
- 在6000和12,000 J/L的紫外线照射下对捐赠人母乳进行了照射.
- 评估了细菌日志的减少 (植物和子).
- 免疫球蛋白 (IgA,IgG,IgM),乳酸,甲素D,弹性酶和维生素E的量化水平;测量了脂酶和酶活性;并分析了脂质氧化产物.
主要成果:
- 6000 J/L和12,000 J/L的UV-C剂量分别实现了植物性细菌和细菌子的>5-log减少.
- 与HoP相比,两种UV-C剂量都更好地保留了免疫球蛋白,乳酸铁素,甲素D,弹性酶,以及胆盐刺激的脂酶和酶的活动.
- 紫外线治疗导致阿尔法-托科菲罗尔的最小减少,而马-托科菲罗尔没有变化,并且没有增加脂质氧化.
结论:
- 紫外线辐射是一种可行的方法,可以提高供体母乳的微生物安全性.
- 紫外线治疗可以比HoP更有效地保护关键的生物活性蛋白和维生素E.
- 优化紫外线照射为加工捐赠母乳提供了一个有希望的替代方案.
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