采用液态发酵工艺,以细菌数为检测指标,从3种乳酸菌中筛选出能利用罗非鱼内脏发酵的菌种及其营养配方,并通过单因素实验和正交实验对其发酵条件进行了优化。结果显示:Lactobacillus fermentum菌株发酵罗非鱼内脏能力更强。优化后的发酵条件为:发酵时间8h,接种量5%,30℃静置培养;营养配方为:0.2%K2HPO4.3H2O,0.5%CH3COONa,葡萄糖含量为2.0%,3.3%罗非鱼内脏,在此条件下,发酵液的细菌数达到109CFU/mL以上。

The Tilapia visceras,which is one type of wastes produced during the processing,cannot be handled properly or use efficiently.In this study,the process of liquid fermentation was used and total plate count was taken as the detection index to screen one strain that can utilize Tilapia viscera to grow well from three Lactobacillus strains and its corresponding nutrition recipe.And then,the fermentation conditions were optimized with single-factor and orthogonal tests.The results show that Lactobacillus fermentum is the best strain.The optimal condition of fermentation is as the followings:8h standing fermentation at 30℃,5% inoculum size;and the optimal nutrition recipe contains 0.2%K2PO4·3H2O,0.5%CH3COONa,2.0%glucose and 3.3% visceras.The total plate count can reach over 109 CFU/mL at the optimal condition.