PH9389 | 伊红美蓝琼脂培养基EMB Eosin-Methylene Blue Agar

用途：用途：用于分离革兰氏阴性肠道菌特别是大肠菌群和粪大肠菌群

成分（g/L）

蛋白胨 10.0

乳糖 10.0

磷酸氢二钾 2.0

琼脂 15.0

伊红 0.4

美蓝 0.065

PH值7.1±0.2 25℃

检验原理：蛋白胨作为基础营养物质提供细菌生长所需的氮源以及其他营养元素；乳糖作为可发酵的碳源；伊红和美蓝同时作为抑制剂和指示剂，在酸性条件下产生沉淀，形成具黑色中心有金属光泽或无光泽的菌落。伊红为酸性染料，美蓝为碱性染料，细菌分解乳糖产酸时，细菌带正电荷，可以与带负电的酸性染料结合即染上了伊红的颜色；如果细菌因产碱性物质较多，带负电荷时与带正电荷的碱性染料美蓝结合而使得菌落呈蓝色。细菌产酸量较大时，染料可析出结晶形成金属光泽。磷酸氢二钾维持缓冲体系；琼脂作为凝固剂。  

用法：称取本品37.4g，加热溶解1000ml蒸馏水中，分装三角瓶，121℃高压灭菌15分钟，备用。

Purpose: It is used for the separation of Gram-negative intestinal bacteria, especially Escherichia coli and fecal coliform bacteria.

Component (g/L)

Peptone 10.0

Lactose 10.0

Potassium dihydrogen phosphate 2.0

Agar 15.0

Eosin 0.4

Methylene blue 0.065

pH 7.1 ± 0.2 at 25℃

Principle of test: Peptone serves as the basic nutrient providing the nitrogen source and other nutrients required for bacterial growth; lactose acts as a fermentable carbon source; eosin and methylene blue function simultaneously as inhibitors and indicators, forming precipitates under acidic conditions and creating colonies with a black center and metallic or non-metallic luster. Eosin is an acidic dye, and methylene blue is a basic dye. When bacteria ferment lactose to produce acid, they carry a positive charge and can combine with the negatively charged acidic dye eosin, thus taking on its color. If bacteria produce more alkaline substances and carry a negative charge, they will combine with the positively charged basic dye methylene blue, resulting in blue colonies. When bacteria produce a large amount of acid, the dye can crystallize and form a metallic luster. Dipotassium hydrogen phosphate maintains the buffer system; agar serves as a solidifying agent.

Usage: Weigh 37.4g of this product, dissolve it in 1000ml of distilled water by heating, transfer it into triangular flasks, sterilize at 121℃ under high pressure for 15 minutes, and keep it ready for use.