Pak J Biol Sci. 2021 Jan;24(11):1138-1143. doi: 10.3923/pjbs.2021.1138.1143.
<b>Background and Objective:</b> The composition of the waste consists mostly of plant biomass. Cellulose is the largest component of plant biomass and cellulolytic bacteria are needed to degrade it. This study aimed to determine enzyme activity possessed by bacterial isolates from Biological Education and Research Forest floor Andalas University. <b>Materials and Methods:</b> The isolation stage was carried out with NA (Nutrient agar) medium, Screening with CMC (Carboxymethyl Cellulose) medium with congo red dye and enzyme activity testing was carried out using the Nelson-Somogyi method. <b>Results:</b> We found 16 bacterial isolates obtained from Biological Education and Research Forest Floor Andalas University, 10 of them were positive for cellulolytic bacteria with the highest cellulolytic index value of 2.59 on FFB 2 isolates. <b>Conclusion:</b> The bacterial isolate with the best enzyme activity value was FFB 2 isolate 0.166 U mL<sup>1</sup> for 72 hrs.
背景与目的:废物的组成主要来自植物生物质。纤维素是植物生物质的最大组成部分,需要纤维素分解菌来降解它。本研究旨在确定来自安达拉斯大学生物教育与研究林地面的细菌分离物所具有的酶活性。
材料与方法:采用 NA(营养琼脂)培养基进行分离阶段,用刚果红染料的 CMC(羧甲基纤维素)培养基进行筛选,并用 Nelson-Somogyi 法进行酶活性测试。
结果:我们从安达拉斯大学生物教育与研究林地面发现了 16 个细菌分离物,其中 10 个为纤维素分解菌阳性,在 FFB2 分离物上的纤维素分解指数值最高为 2.59。
结论:酶活性值最佳的细菌分离物为 FFB2 分离物,在 72 小时内的酶活性值为 0.166 U mL1。