酵素菌中乳酸菌的分离鉴定

Division and Identification on Lactobacillus of Enzymatic Microorganism

目的 对分离自酵素菌的 2株乳杆菌 (Lacotbacillus)进行分类鉴定。方法 观察实验菌细胞的形态大小 ,染色特征。葡萄糖氧化试验、氧化酶试验、触酶试验、生长因子试验、硝酸盐还原试验、葡萄糖产气试验。阿拉伯糖、葡萄糖、纤维二糖、乳糖、麦芽糖、甘露醇、松三糖、密二糖、棉子糖、核糖、木糖、鼠李糖、甘油等糖醇类发酵产酸试验。厌氧生长试验。菌株的 15℃生长试验、4 5℃生长试验。乳酸产气试验。营养要求、底物利用、代谢途径及代谢产物检测等 ,进行表型性状的观察分析。结果 2株乳酸菌 (Lacotbacillus)被鉴定为植物乳杆菌 (Lactobacilluaplantarum)。结论 植物乳杆菌 (Lactobacilluaplantarum)是酵素菌的主要活性菌种。

Objective To identify two strains of Lactobacillua from Enzymatic Microorganism.Method The following assays were done,including observation of the size of bacteria,characteristic of stain.Oxidize experiment on glucose.Produce experiment of galactic acid.Reaction on Catalase.Reaction of Voges-Proskauer VP,reaction of Indol.Reaction of Methyl red.ferment experiment on compounds Carbon source: glucose,galactoes,L-arabinose,D-xylose,D- mannitol,sucrose,cellobiose,maltose,trehalose,melezitose,lactose,raffinose,inositol,rhamnose,sorbitol,salicin,glycerol,and erythritol.Experiment of growth factor.Experiment of requirement of definite gasseous:obligate aerobe,microaerophilic bacterium,facultative anaerobe and obligate anaerobe.Growth experiment in 5% and 7%NaCl.Growth experiment on 15℃ and 45℃.Nutritional need,ways of metabolism,determination metabolism.Result Two isolated Lactobacillua were proved to be Lactobacillua plantarum.Conclusion Lactobacillua plantarum is the main type of active bacteria of Enzymatic Microorganism.