2013年《微生物学》实验双语测试题

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微生物学与免疫学题(中英)

微生物学与免疫学题(中英)

微生物学与免疫学题(中英)微生物学与免疫学复习题一、名词解释acquired Immune deficiency syndrome, AIDS(获得性免疫缺陷综合征,艾滋病) 由人类免疫缺陷病毒引起的传染病综合征,其特征为失去正常免疫应答,随后易感染机会性传染病和某些癌症。

acquired immunity(获得性免疫) 指特异性免疫类型,在与适当的抗原接触后或抗体从一个体转移到另一个体后产生。

active carrier(病期带菌者) 带有明显的临床病症和可能传播传染病给他人的个体。

active immunization(自动免疫接种) 通过自然接触病原体或接种疫苗诱发自动免疫。

active transport(主动运输) 溶液分子依电化学梯度跨膜的运输,这需要载体蛋白和能量的输入。

Active transport is the transport of solute molecules to higher concentrations, or against a concentration gradient, with the use of metabolic energy input.(2004A)adhesin(粘附素) 微生物表面的分子组分,参与对底物或细胞的粘附。

对特异性宿主组织的粘附通常是发病的预备阶段,而粘附素是重要的毒力因子。

Adherence is the process by which bacteria stick to the surfaces of host cells,Once bacteria have entered the body,adherence is major initial step in the infection process.The terms adherence,adhesion, and attachment are often used interchangeably(交替).adjuvant(佐剂) 加入抗原以提高其免疫原性的物质,常见的例子有明矾,灭活的百日咳菌和抗原的油乳剂,或者是单独使用(弗氏不完全佐剂),或者是与灭活的分枝杆菌一起使用(弗氏完全佐剂)。

微生物学检验练习题(附答案)

微生物学检验练习题(附答案)

微生物学检验练习题(附答案)一、单选题(共105题,每题1分,共105分)1.梭状芽胞杆菌与无芽胞厌氧菌相同的特性是( )A、在自然界的存在形式B、形态染色性C、专性厌氧D、抵抗力E、致病性正确答案:C2.荧光显微镜观察细菌的必备条件不包括( D)(难)A、具有紫外光或蓝紫光的荧光光源B、激发光滤片C、吸收光滤片D、荧光集光器E、普通光镜为基础正确答案:D3.关于铜绿假单胞菌的特性,错误的是( )A、是人体正常菌群的成员B、革兰阴性杆菌,产生水溶性色素C、致病与多种毒力因子有关D、对常用抗生素敏感E、其感染多见于皮肤黏膜受损部位正确答案:D4.临床微生物的最基本分类单位是( )A、科B、种C、属D、型E、亚种正确答案:B5.下列培养基中不属于选择培养基的是(B)(中)A、SS培养基B、庖肉培养基C、EMB培养基D、高盐培养基E、罗一琴培养基正确答案:B6.物表和空气的消毒,宜选用的消毒方法是(E)(中)A、巴氏消毒法B、间歇灭菌法C、高压蒸气灭菌法D、流通蒸气灭菌法E、紫外线照射法正确答案:E7.甲基红试验中培养基的pH 需降至多少、指示剂才指示阳性结果(D)(难)A、7.4B、6.5C、5D、4.5E、2.5正确答案:D8.革兰染色法脱色时所用乙醇浓度为A、95%B、90%C、20%D、40%E、70%正确答案:A9.关于培养基的叙述,错误的是(B)(中)A、按物理性状分为三类B、物理性状相同的培养基,其用途相似C、不同的培养基其成分不尽相同D、按用途分为六类E、成分柜同的培养基,其物理性状也相同正确答案:B10.碳源和氮源利用试验不包括( )A、硝酸盐还原试验B、丙二酸盐利用试验C、马尿酸盐水解试验D、枸橼酸盐利用试验E、醋酸盐利用试验正确答案:A11.下列哪种物质是原核细胞型微生物所特有的化学成分(C)(中)A、脂质B、无机盐C、肽聚糖D、蛋白质E、几丁质正确答案:C12.青霉素对细菌的作用机制是(B)(难)A、切断β-1,4 糖苷键B、干扰四肽侧链与五肽交联桥的连接C、影响细胞膜的合成D、干扰磷壁酸的活性E、干扰细菌DNA 的复制正确答案:B13.肺炎链球菌在血平板上生长时,菌落周围常会出现溶血环,其特征为( )A、褐色B、黄色C、砖红色D、暗红色E、草绿色正确答案:E14.真菌的繁殖结构就是:( )A、始体B、孢子C、原体D、包涵体E、芽孢正确答案:B15.细菌染色过程中固定的作用主要是(C)(中)A、使其易与染料结合B、使其失去抗原性C、杀死细菌D、利于保存E、破坏细菌结构正确答案:C16.甲基红试验所用培养基是(D)(难)A、肉汤培养基B、蛋白胨水C、碱性蛋白胨水D、葡萄糖蛋白胨水E、生理盐水正确答案:D17.关于脂多糖的叙述哪项是不正确的( )A、由磷脂和脂蛋白组成B、可称为细菌内毒素C、存在于革兰阴性菌的外膜D、由脂类和多糖组成E、具有多种生物学活性正确答案:A18.鹅口疮的病原菌就是:( )A、螺旋体B、立克次体C、细菌D、衣原体E、真菌正确答案:E19.尿液标本作L型菌培养时,培养基中加入20%蔗糖的原因是( )A、稀释尿液B、使尿中蛋白质变性C、中和尿液的酸度D、提供高营养E、保持高渗环境正确答案:E20.经螨传播的立克次体病是( )A、流行性斑疹伤寒B、恙虫病C、地方性斑疹伤寒D、落矶山斑点热E、Q热正确答案:B21.关于细菌产生的色素,错误的是( )A、有助于鉴别细菌B、可进行光合作用C、分两类:脂溶性和水溶性D、色素产生需一定的条件E、其功能尚不清楚正确答案:B22.和耶尔森菌属属于同科的细菌是( )A、链球菌B、棒状杆菌C、志贺菌D、奈瑟菌E、鲍特菌正确答案:C23.在一般中性环境中细菌带负电荷,易与以下何种染料结合(D)(中)A、中性染料B、姬姆萨染料C、酸性染料D、碱性染料E、以上均不对正确答案:D24.属于营养培养基的是(A)(中)A、巧克力平板B、SS平板C、麦康凯平板D、碱性蛋白胨水E、伊红一美蓝琼脂正确答案:A25.沙门菌属在普通平板上的菌落特征是(E)A、中等大小,粗糙型B、针尖状小菌落,半透明C、中等大小,粘液型D、针尖状小菌落,粘液型E、中等大小,无色半透明正确答案:E26.副溶血弧菌在TCBS平板上的菌落为( )A、红色B、黄色C、绿色D、蓝色E、黑色正确答案:C27.非发酵菌包括( )A、假单胞菌属,黄杆菌属,变形杆菌属B、假单胞菌属,埃希菌属,产碱杆菌属C、莫拉菌属,黄杆菌属,肠杆菌属D、假单胞菌属,不动杆菌属,产碱杆菌属E、假单胞菌属,气单胞菌属,黄杆菌属正确答案:D28.不同来源的同种细菌称为A、种B、菌株C、型D、异种E、亚种正确答案:B29.甲基红试验中培养基的pH需降至多少,指示剂才指示阳性结果(C)(难)A、2.5B、5.OC、4.5D、6.5E、7.4正确答案:C30.关于菌毛的叙述,下列哪项是错误的(C)(中)A、多见于革兰阴性菌B、有普通菌毛和性菌毛两种C、能直接用光学显微镜观察D、性菌毛与细菌的接合有关E、普通菌毛与性菌毛的致病力有关正确答案:C31.注射前在局部用碘酒和乙醇处理属于A、无菌操作B、灭菌C、消毒D、防腐E、无菌正确答案:C32.不动杆菌属细菌引起的院内感染,最常见的部位是( )A、泌尿道B、消化道C、呼吸道D、生殖道E、皮肤伤口正确答案:C33.在细菌检验中,制备血平板最常用血液采自(C)(中)A、豚鼠B、家兔C、绵羊D、大白鼠E、马正确答案:C34.细菌代谢活动活跃的原因是()。

微生物学检验测试题与答案

微生物学检验测试题与答案

微生物学检验测试题与答案一、选择题(每题2分,共40分)1. 下列哪种微生物不是原核生物?A. 革兰氏阳性菌B. 革兰氏阴性菌C. 真菌D. 支原体答案:C2. 下列哪种培养基属于选择性培养基?A. 营养琼脂B. 血平板C. 营养肉汤D. 胰蛋白胨琼脂答案:D3. 下列哪种微生物检验方法属于血清学检测?A. 直接涂片镜检B. 分离培养C. 聚合酶链反应D. 酶联免疫吸附试验答案:D4. 下列哪种微生物具有芽孢结构?A. 芽孢杆菌B. 梭菌C. 链球菌D. 肠道杆菌答案:A5. 下列哪种抗生素属于青霉素类?A. 阿莫西林B. 头孢菌素C. 红霉素D. 庆大霉素答案:A6. 下列哪种微生物检验方法可以用于检测细菌耐药性?A. 纸片扩散法B. 琼脂稀释法C. E测试D. 所有以上方法答案:D7. 下列哪种微生物可以引起医院感染?A. 金黄色葡萄球菌B. 大肠杆菌C. 肺炎克雷伯菌D. 所有以上微生物答案:D8. 下列哪种微生物检验方法属于分子生物学检测?A. 直接涂片镜检B. 分离培养C. 聚合酶链反应D. 酶联免疫吸附试验答案:C9. 下列哪种微生物可以引起食物中毒?A. 沙门氏菌B. 志贺氏菌C. 霍乱弧菌D. 所有以上微生物答案:D10. 下列哪种微生物可以引起败血症?A. 金黄色葡萄球菌B. 大肠杆菌C. 链球菌D. 所有以上微生物答案:D二、填空题(每题2分,共20分)1. 微生物的分类主要包括______、______、______三大类。

答案:原核生物、真核生物、非细胞型微生物2. 培养基按其物理状态可分为______、______、______三种类型。

答案:液体培养基、半固体培养基、固体培养基3. 微生物检验的四大基本技术包括______、______、______、______。

答案:形态学检查、生理生化试验、血清学检测、分子生物学检测4. 细菌耐药性检测方法有______、______、______等。

微生物测试题(附答案)

微生物测试题(附答案)

微生物测试题(附答案)一、单选题(共58题,每题1分,共58分)1.专性厌氧菌不能呼吸,只能发酵的原因是( )。

A、厌氧菌缺乏细胞色素和细胞色素氧化酶,不能氧化那些氧化还原电势高的物质B、厌氧菌缺乏过氧化氢酶,过氧化物酶和超氧化物歧化酶,不能清除有氧环境下产物的超氧离子和过氧化氢C、有氧条件下,细菌某些酶的-SH基被氧化为S-S基,从而失去活性D、以上均是正确答案:D2.基因库中大约( )的人类全长cDNA序列在酵母中有同源的基因。

A、5%B、15%C、50%D、75%正确答案:B3.真核微生物核糖体成分是( )。

A、蛋白质B、RNAC、40%蛋白质和60%RNAD、50%蛋白质和50%RNA正确答案:C4.下列物质中,最容易被毛霉分解的是 ( )。

A、蛋白质B、纤维素C、果胶D、淀粉正确答案:D5.菌丝过滤法适用于淘汰下列哪一种营养缺陷型菌株( )。

A、放线菌(有菌丝改)B、病毒C、酵母菌D、细菌正确答案:A6.真菌的营养方式为( )。

A、异养吸收型B、自给自足型C、化能自养型D、光能自养型正确答案:A7.能够高效进行自发转化的细菌是 ( )。

A、固氮菌B、链球菌C、奢血菌D、以上均是正确答案:D8.核苷酸在生物体内的作用,以下说法正确的是( )A、只能合成核酸和参与某些酶的组成B、只能参与某些酶的组成C、只能合成核酸D、主要是合成核酸和参与某些酶的组成正确答案:D9.真核微生物核糖体的功能是( )。

A、合成糖类物质B、进行氧化磷酸化反应C、合成蛋白质D、合成脂类物质正确答案:C10.微生物生态系统自身的特点不包括 ( )。

A、协调性B、适应性C、微环境D、稳定性正确答案:A11.关于氮源的描述,以下说法错误的是( )A、主要功能是提供合成原生质和细胞结构成分的氮素来源B、不提供能量(ATP中的含氮碱基C、来源最广的是分子态氮D、硝化细菌可以利用氮源提供能量正确答案:B12.关于微生物代谢调节的叙述,错误的是( )A、组成酶的合成只受遗传物质的控制,与生长环境中的营养物质无关B、微生物代谢调节主要有酶合成调节和酶活性调节C、在酶活性调节过程中,代谢产物与酶结合,能使酶分子结构和或许产生可逆变化D、酶合成调节是一种快速、精细的调节方式正确答案:A13.培养基中使用牛肉膏的作用是为微生物提供( )A、C源B、N源C、生长因素D、A,B,C都提供正确答案:D14.病毒、细菌和真菌个体的直径比约为( )。

微生物学检验习题库与参考答案

微生物学检验习题库与参考答案

微生物学检验习题库与参考答案一、单选题(共105题,每题1分,共105分)1.菌毛可用下列哪种显微镜观察A、荧光显微镜B、普通光学显微镜C、电子显微镜D、暗视野显微镜E、相差显微镜正确答案:C2.细菌的革兰染色性不同是由于( )A、细胞膜结构的不同B、细胞核结构的不同C、中介体的有无.D、胞质颗粒的有无或不同E、细胞壁结构的不同正确答案:E3.在一般中性环境中细菌带负电荷,易与以下何种染料结合( )A、中性染料B、姬姆萨染料C、酸性染料D、碱性染料E、以上均不对正确答案:D4.关于细菌产生的色素,错误的是( )A、有助于鉴别细菌B、可进行光合作用C、分两类:脂溶性和水溶性D、色素产生需一定的条件E、其功能尚不清楚F、大多由革兰阳性菌合成G、细菌生长繁殖过程中即可释放出来H、抗原性弱正确答案:C5.细菌的特殊结构不包括B、鞭毛C、荚膜D、芽胞E、中介体正确答案:E6.支原体与细菌L型的相同点是(C)A、都不能在无细胞的人工培养基上生长B、都是分支繁殖C、无细胞壁D、遗传物质都是RNAE、形态一致正确答案:C7.某患者因腹痛,腹泻入院,自叙有乏力,里急后重。

检查见其大便为水样便、无脓血,但体温正常,亦无脱水和呕吐,肝功能亦正常。

取粪便粘液部分镜检可见大量红细胞及白细胞,此患者应考虑是(C)A、肠炎型沙门菌感染B、轻型细菌性痢疾C、大肠埃希菌性肠炎D、阿米巴性痢疾E、霍乱正确答案:C8.青霉素对细菌的作用机制是(B)(难)A、切断β-1,4 糖苷键B、干扰四肽侧链与五肽交联桥的连接C、影响细胞膜的合成D、干扰磷壁酸的活性E、干扰细菌DNA 的复制正确答案:B9.触酶是(B)(易)A、氧化酶B、过氧化氢酶C、细胞色素氧化酶D、超氧化物歧化酶正确答案:B10.下列物质与细菌侵袭力无关是(D)(中)A、溶血素B、菌毛C、透明质酸酶D、芽胞E、荚膜正确答案:D11.实验室诊断皮肤癣菌感染时,皮屑或病发须经哪种溶液处理后再镜检:( )A、10%NaclB、70%乙醇C、10%KOHD、25%KOHE、15%NAOH正确答案:C12.关于细菌产生的色素,错误的是( )A、有助于鉴别细菌B、可进行光合作用C、分两类:脂溶性和水溶性D、色素产生需一定的条件E、其功能尚不清楚正确答案:B13.G+菌和G-菌细胞壁共同成分或结构是A、脂多糖B、磷壁酸C、外膜D、肽聚糖E、以上均不是正确答案:D14.细菌毒素中,毒性最强的是(D)(易)A、破伤风痉挛毒素B、金葡菌肠毒素C、白喉外毒素D、肉毒毒素E、霍乱肠毒素正确答案:D15.细菌合成细胞组分和获得能量的基质主要是( )A、糖原B、脂类C、淀粉D、蛋白质E、糖类正确答案:E16.绝大多数微生物对人类和动植物是A、可有可无的B、病原菌C、引起内源性感染D、有益的E、有害的正确答案:D17.从有正常菌群存在的部位所采取的标本应接种在哪种培养基中分离培养病原体( D)(易)A、基础培养基B、营养培养基C、增菌培养基D、选择鉴别培养基E、特殊培养基正确答案:D18.下列有关药敏试验意义的论述,错误的一项是A、药敏试验可预测抗菌治疗的效果B、药敏试验结果为“敏感”时,治疗可能有效C、药敏试验结果是提供药物选择的依据D、药敏试验为中介时,只要加大药物剂量完全可以达到治疗效果E、体外药敏结果不完全与体内治疗效果一致正确答案:D19.利用细菌生化反应鉴定细菌是根据(D)(难)A、细菌酶活性差异B、细菌毒素活性差异C、细菌酶含量的差异D、细菌分解代谢产物的差异E、细菌毒素种类的差异正确答案:D20.胆汁溶菌试验常用于鉴别A、A群链球菌与B群链球菌B、A群链球菌与D群链球菌C、肺炎链球菌和甲型溶血型链球菌D、肺炎链球菌和乙型溶血型链球菌E、表皮葡萄球菌与腐生葡萄球菌正确答案:C21.下列物质与细菌侵袭力无关是( )A、菌毛B、荚膜C、溶血素D、透明质酸酶E、芽胞正确答案:E22.细菌的革兰染色性不同是由于(C)(中)A、细胞核结构的不同B、细胞膜结构的不同C、细胞壁结构的不同D、中介体的有无.E、胞质颗粒的有无或不同正确答案:C23.气相色谱仪鉴定厌氧菌是检测(E)(难)A、细菌的抗原物质B、细菌的核酸C、细菌的酶活性D、细菌的化学组成E、细菌的代谢产物正确答案:E24.对厌氧条件要求不严格的细菌是( )A、双歧杆菌B、产黑色素类普雷沃菌C、艰难梭菌D、产气荚膜梭菌E、破伤风梭菌正确答案:D25.下列对外毒素的描述哪项是正确的(C)(中)A、对器官组织的毒性作用无选择性B、毒性作用弱C、大多数外毒素是在菌体内合成后分泌至菌体外D、多数耐热E、外毒素大都由革兰阴性菌产生正确答案:C26.下列哪种细菌不是革兰氏阴性菌( )A、伤寒沙门菌B、霍乱弧菌C、铜绿假单胞菌D、大肠埃希菌E、结核分枝杆菌正确答案:E27.血培养时采血量一般为血培养瓶中肉汤培养基量的A、1/4~1/3B、1/5~1/4C、1/7~1/6D、1/6~1/5E、1/3~1/ 2正确答案:B28.通常用于结核分枝杆菌分离培养的痰液,预处理时常用的消化液是( )A、4%NaOHB、3%NaOHC、3%KOHD、2%KOHE、0.5%NaOH正确答案:A29.关于液体培养接种,错误的是(C)(中)A、用手斜持液体培养基管B、根据接种物性质,可选用接种针或接种环C、在与液面交接处的管壁上轻轻研磨,并沾取少许培养基液体调和D、无菌操作关闭试管,直立培养E、有细菌生长时,培养管可呈现混浊、沉淀和菌膜等现象正确答案:C30.细菌哪种结构的功能类似真核细胞的线粒体(C)(难)A、核蛋白体B、胞质颗粒C、中介体D、质粒E、异染颗粒正确答案:C31.下列细菌特殊结构中,具有较强抵抗力的是A、荚膜B、鞭毛C、性菌毛D、芽胞E、菌毛正确答案:D32.下面有关细菌的命名说法错误的是A、细菌种的命名采用拉丁文双命名法B、由两个拉丁字组成C、前一字为属名D、中文译名的属名放在前面E、后一字为种名正确答案:D33.用萋内抗酸染色法,结果呈阳性的菌体是(E)(易)A、黄色B、紫色C、蓝色D、绿色E、红色正确答案:E34.能使纤维蛋白凝块溶解的是A、链激酶B、磷脂酶C、胶原酶D、透明质酸酶E、血浆凝固酶正确答案:A35.A群链球菌与其他群链球菌的主要鉴别试验是(B)(易)A、血浆凝固酶试验B、杆菌肽敏感试验C、马尿酸水解试验D、甘露醇发酵试验E、新生霉素敏感试验正确答案:B36.革兰染色所用染液的顺序是( )A、稀释复红一碘液一酒精一结晶紫B、结晶紫一碘液一酒精一稀释复红C、结晶紫一酒精一碘液一稀释复红D、稀释复红一酒精一结晶紫一碘液E、稀释复红一结晶紫一碘液一酒精正确答案:B37.G‐菌内毒素成分是A、肽聚糖B、磷壁酸C、脂质AD、外膜蛋白E、DAP正确答案:C38.有关外毒素的错误描述是(D)(难)A、主要由革兰阳性菌产生B、也可由革兰阴性菌产生C、具有特异的组织亲和性D、其成分为脂多糖E、可经甲醛脱毒成类毒素正确答案:D39.使用比浊法可粗略估计细菌的数量与下列何种物理性状有关A、光学性质B、带电现象C、半渗透性D、渗透压E、表面积正确答案:A40.经螨传播的立克次体病是( )A、恙虫病B、落矶山斑点热C、地方性斑疹伤寒D、Q热E、流行性斑疹伤寒正确答案:A41.自动化微生物药敏系统进行药敏测的实质是A、K-B纸片扩散法B、微量肉汤稀释法C、琼脂稀释法D、联合药敏试验E、E-test法正确答案:B42.下列哪种细菌不会引起食物中毒( )A、鼠伤寒沙门菌B、痢疾志贺菌C、丙型副伤寒杆菌D、产肠毒素的金葡菌E、蜡样芽胞杆菌正确答案:B43.革兰阳性菌和革兰阴性菌细胞壁的共同成分是(D)(易)A、磷壁酸B、脂多糖C、脂蛋白D、肽聚糖E、脂质正确答案:D44.观察螺旋体最好的方法是( )A、抗酸染色法B、Giemsa染色法C、直接镜检D、暗视野显微镜法E、革兰氏染色法正确答案:D45.D群链球菌的重要特征是(C)(易)A、对杆菌肽耐药B、对杆菌肽敏感C、胆汁七叶苷试验阳性D、水解尿酸E、CAMP 试验阳性正确答案:C46.金黄色葡萄球菌与表皮葡萄球菌的主要鉴别试验是(C)(易)A、VP试验B、吲哚试验C、凝固酶试验D、触酶试验E、O/F试验正确答案:C47.KIA琼脂一般不用于观察( )A、乳糖发酵B、硫化氢产生C、葡萄糖发酵D、色氨酸E、产气试验正确答案:D48.细菌染色多采用的染料是(C)(中)A、酸性染料B、中性染料C、碱性染料D、复合染料E、单纯染料正确答案:C49.不属于细菌菌落特征的是(D)(易)A、湿润B、扁平C、完全溶血D、致密E、透明正确答案:D50.构成细菌毒力的是(B)(中)A、基本结构B、侵袭力和毒素C、特殊结构D、分解代谢产物E、侵入机体的途径·正确答案:B51.测量细菌大小的单位是( )A、dmB、cmC、μmD、mmE、nm正确答案:C52.抗生素使用不当可导致(D)(中)A、十二指肠溃疡B、肠结核C、慢性胃炎D、伪膜性肠炎E、腹膜炎正确答案:D53.某村,春节后有3名散居儿童在l周内先后突然出现发热、头痛、喷射状呕吐、全身满布皮肤出血点、颈项强直明显。

微生物实验课试题库及标准答案

微生物实验课试题库及标准答案

微生物实验课试题库及标准答案试题类型试题数量备注名词解释29试题涵盖了《微生物学》各章节有关实验技术的内容,以及沈萍、范秀容等编《微生物学实验》第三版实验指导书的内容。

选择题40判断题59填空题70思考题13总数211微生物实验技术名词解释01.电子显微镜:电子显微镜是利用电子波波长短,分辨力高的特点以电子流代替光学显微镜的光束使物体放大成象的超显微镜检装置。

02.普通光学显微镜:用自然光或者灯光作光源镜检物体的显微镜。

03.合成培养基:由化学成分已知的营养物质配制而成的培养基。

04.人工培养基:人工配制的供微生物生长繁殖并积累代谢产物的一种营养基质。

05.天然培养基:由化学成分不完全清楚的天然物质如马铃薯,麸皮等配制而成的培养基。

06.半合成培养基:由化学成分已知的化学物质和化学成分不完全清楚的天然物质配制而成的培养基。

07.革兰氏染色法。

:革兰氏染色是细菌的一种鉴别染色法,细菌首先用结晶紫染色,再用碘液固定,然后用95%的酒精脱色,最后用蕃红复染。

凡是菌体初染的结晶紫被酒精脱去了紫色后,又被蕃红复染成红色的细菌称为革兰氏负反应细菌;凡是菌体初染的紫色不能被酒精脱色,也不能被蕃红复染成红色的细菌称为革兰氏正反应细菌。

08.简单染色:用单一染料使微生物细胞染上所用染料颜色的染色方法。

09.稀释平板计数法:将一定量的样品经十倍稀释后,用平板培养最后三个稀释度的样品稀释液。

待菌落长出后,计数出某一稀释度的菌落数后再乘以稀释倍数,即为样品中的含菌数。

10.显微直接计:利用血球计数板或细菌计数板在显微镜下测计出每小格的微生物细胞数量后,再换算出单位体积中微生物细胞总数的测数方法。

11.巴氏消毒:巴氏消毒是法国微生物学家巴斯德发明的一种消毒方法。

是在62-63℃的条件下,保温半小时杀死微生物的营养体(主要是病原菌)的方法。

12.间歇灭菌:利用100℃的温度杀死微生物的营养体.每次1小时连续三天,中间的空隙时间让未杀死的芽胞萌发成营养体,在下一次100℃的温度下被杀死。

微生物学英文题库完整

微生物学英文题库完整

IntroductionMicrobes are a diverse group of organisms that can be divided into the viruses, unicellular groups (Archaea, Eubacteria, protista, some fungi and some chlorophyta) and a small number of organisms with a simple multicellular structure (the large fungi and chlorophyta).Microbiology: the science (logos) of small (micro) life (bios). The study of living things so small that they cannot be seen with the naked eye.The three-domain system is a biological classification introduced by Carl Woese that divides cellular life forms into archaea, bacteria, and eukaryote domains. In particular, it emphasizes the separation of prokaryotes into two groups, originally called Eubacteria (now Bacteria) and Archaebacteria (now Archaea).Microorganisms were first visualized by Antoni van Leeuwenhoek (1632–1723), a Dutch cloth merchant and an expert lens grinder.The first proof of the involvement of bacteria in disease and the definitive proof of the germ theory of disease came from the German Robert Koch.Koch’s po stulates:1, The microorganism must be present in every instance of the disease and absent from healthy individuals.2, The microorganism must be capable of being isolated and grown in pure culture.3 ,When the microorganism is inoculated into a healthy host, the same disease condition must result.4, The same microorganism must be re-isolated from the experimentally infected host. Chapter oneFungi include yeasts, molds, and mushrooms.Viruses are noncellular organisms; they are intra-cellular parasites of animals, plants, or bacteria.1, The Prokaryotes can be divided into two kingdoms, Bacteria and Archaea. The classification was first proposed by Woese based on the differences in 16S rRNA sequence.2, Prokaryotic microorganism(原核微生物): It is an single-c elled organism which doesn’t have a nuclear membrane outside the cell nucleus and only has the bare DNA called as the nuclear area. 3, Many bacteria are shaped like long rods twisted into spirals or helices; they are called spirilla (螺旋菌)if rigid and spirochetes (螺旋体) when flexible.4, Capsules (荚膜) and slime layers (粘液层) usually are composed of polysaccharides (多糖), but they may be constructed of other materials.5, Gram staining procedure(革兰氏染色的步骤)In the first step of the Gram-staining procedure, the smear is stained with the basic dye crystal violet, the primary stain. It is followed by treatment with an iodine solution functioning as a mordant. That is, the iodine increases the interaction between the cell and the dye so that the cell is stained more strongly. The smear is next decolorized by washing with ethanol or acetone. This step generates the differential aspect of the Gram stain; gram-positive bacteria retain the crystal violet, whereas gram-negative bacteria lose their crystal violet and become colorless. Finally, the smear is counterstained with a simple, basic dye different in color from crystal violet. Safranin, the most common counterstain, colors gram-negative bacteria pink to red and leaves gram-positive bacteria dark purple.6, The Mechanism of Gram Staining (革兰氏染色的机制)Although several explanations have been given for the Gramstain reaction results, it seems likely that the difference between gram-positive and gram-negative bacteria is due to the physical nature of their cell walls. If the cell wall is removed from grampositive bacteria, they become gram negative. The peptidoglycan itself is not stained; instead it seems to act as a permeability barrier preventing loss of crystal violet. During the procedure the bacteria are first stained with crystal violet and next treated with iodine to promote dye retention. When gram-positive bacteria then are decolorized with ethanol, the alcohol is thought to shrink the pores of the thick peptidoglycan. Thus the dye-iodine complex is retained during the short decolorization step and the bacteria remain purple. In contrast, gram-negative peptidoglycan is very thin, not as highly cross-linked, and has larger pores. Alcohol treatment also may extract enough lipid from the gramnegative wall to increase its porosity further. For these reasons, alcohol more readily removes the purple crystal violet-iodine complex from gram-negative bacteria.简而言之,不同细菌的染色差异(G+或G-)是由于细胞壁化学成分的差异而引起了物理特性(脱色能力)的不同。

微生物学实验考试试题

微生物学实验考试试题

微生物学尝试测验试题一、单项选择题〔从以下各题四个备选答案中选出一个正确答案,并将其代号写在答题纸的相应位置。

答案选错或未选者,该题不得分。

〕1半固体培养基的琼脂参加量通常是。

A、1%B、0.5%C、0.1%D、1.5%2在使用油镜不雅察时,通常在玻片标本的镜检部位滴加__________。

A、乙醚B、酒精C、香柏油D、二甲苯3物镜外壳上标明的1.25暗示。

A、数值孔径B、浸油深度C、工作距离D、放大倍数4细菌平板制备好后,放入恒温箱培养,需要。

A、正放B、任意C、倒放D、斜放5不雅察细菌荚膜的形态时可采用__________。

A、插片法B、印片法C、水浸片法D、负染色法6含有酶或抗生素的溶液、血清等可采用的灭菌方法为__________。

A、干热灭菌法B、湿热灭菌法C、紫外线杀菌D、过滤除菌7显微直接计数法不克不及用来测定下面哪些微生物__________。

A、病毒B、霉菌孢子C、细菌D、酵母菌8从土样中别离土壤细菌选用的培养基是__________。

A、牛肉膏蛋白胨培养基B、马丁孟加拉红培养基C、无氮培养基D、根本培养基9改进Leifson法染色鞭毛成功的先决条件是__________。

A、玻片干净无油污B、涂黑素要薄C、印片时不克不及滑动D、需要稳按期的菌种10尝试室常用的马铃薯蔗糖培养基属于__________。

A、半合成培养基B、合成培养基C、天然培养基D、根本培养基11 革兰氏染色的关键操作步调是__________。

A、酒精脱色B、碘液媒染C、结晶紫染色D、复红复染12从土样中别离真菌选用的培养基是__________。

A、牛肉膏蛋白胨培养基B、马丁孟加拉红培养基B、无氮培养基D、根本培养基13以下方法中测定样品中微生物活菌数量的方法是__________。

A、平板菌落计数法B、比浊法C、测含氮量法D、称干重法14营养试验成果说明黑曲霉在培养基缺乏下面那种元素时仍发展良好。

A、锌B、磷C、钾D、碳15高氏一号培养基属于_________。

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《微生物学》实验双语测试题《Experiment for microbiology》test1.Multiple Choice (choose one answer)(每小题2分,共20分)1.The first microscopes were developed by:A.EhrlichB.MetchnikoffC.LeewenhoekD.Lister2.Light microscopy is dependent on the interaction of light with on object. The ability of light to pass through an object is referred to as:A.transported lightB.transmitted lightC.reflected lightD.refracted light3.The resolving power (R) of a microscope is dependent on the wavelength of light (;~) and the numerical aperture (NA) of the lens. The formula for R is:A.R = 0.5~. xNAB.R = 0.5;~/NAC.R = NA/0.5JrD.R = Square root of 0.5)./NA4.The gram stain uses ~ as a mordant to fix the primary stain:A.iodineB.alcoholC.acetoneD.safranin5.The acid-fast stain is useful in the identification of which of the following organisms:A.Staphylococcus aureusB.Mycoplasma mycoidesC.Mycobacteria tuberculosisD.Moraxella osloensis6.Which of the following types of microscopes utilizes electron beams to visualize objects?A.NomarskiB.TEMC.PCMD.Confocal7.Sterilization of material with an autoclave utilizes steam to kill microorganisms. The correct procedure for sterilization with an autoclave is:A.15 min at 121℃at 15 lb/in2B.15 min at 256℃at 15 lb/in2C.15 min at 121℃at 1 lb/in2D.15 rain at 121℃at 30 lb/in28.An antibiotic was added to a culture of bacteria to determine its effect. What method ofenumeration would you use to determine the efficacy of the antibiotic?A.direct countB.viable countC.turbidimetric countD.absorbance9.The transfer of DNA from one organism to another through the use of a viral vector is referred to as:A.electroporationB.conjugationC.transformationD.transduction10. In bacterial cultures, growth can be demonstrated by an increase in:A.massB.cell sizeC.cell numberD.cell length得分评卷人2.Fill in the Blank(每小题1分,共20分)1. The structure of the bright-field microscopeFig 1 Bight-field microscope1. Ocular lens.2. Tube.3. Nosepiece.4. Objective lens.5. Stage.6. Condenser lens.7. Iris diaphragm.8. Reflector.9. Arm. 10. Stage adjustment knob. 11. Condenser adjustment knob. 12. Fine-adjustment knob. 13.Coarse-adjustment knob. 14. Base.2. IMViC Series include 15.Indole production、16.The Methyl Red Test、17.The Voges-Proskauer Test、18.Citrate Utilization3. Pasteurization Inoculate Bacillus subtilis and Escherichia coli respectively to nutrient broth, place the tubes to a 63℃water bath for 30 minutes.得分评卷人3.Matching(每小题2分,共24分)Matching I:l. Primary stain for gram stain a. Negative stain2. Stains bacterial cell b. Carbohl fuchsin3. Used to fix stain c. Crystal violet4. Decolorize d. Malachite green5. Spore stain e. Safranin6. Acid-fast stain f. Positive stain7. Gram- bacteria take up this counterstain g. Alcohol8. Stains background h. Mordant1.c2.f.3.h4.g5.d.6. b7.e8.aMatching II:1. Media used to inhibit growth of unwanted organisms a. Enrichment2. Media where all components are not known b. Selective3. Media used to contrast organisms on same plate c. Differential4. Media used to enhance growth d. Complex1.b2.d3.c4.a得分评卷人4.Discussion(每小题12分,共36分)1. What’s the principle of Gram stain?A: Gram stain is a popular compound staining technique and differential staining processes are required by the use of crystal violet, Gram’s iodine, 95% ethyl alcohol, and safranin sequentially. It divides bacterial cells into two major groups named gram positive (G ) and gram negative (G-), which makes it an essential tool for classification and differentiation of microorganisms. According to Gram stain, Gram positive bacteria bind crystal violet, couldn’t be decolorized by 95% ethyl alcohol, and appear violet, while Gram negative bacteria are decolorized by 95% ethyl alcohol, bind safranin , and appear red.Generally, the principle of gram stain are as follows. ①The isoelectric point of gram positive bacteria is lower (pI=2~3) than that of gram negative bacteria (pI=4~5). So it contains much more negative charge and bind to crystal violet tightly. ②The cell wall of gram negative bacteria is composed of outer membrane containing much lipid composition, which is sensitive to ethyl alcohol. When decolorizing with 95% ethyl alcohol, the complex of crystal violet and iodine is extracted from cell wall. ③The cell wall of gram positive bacteria contain less lipid composition that is not sensitive to ethyl alcohol decolorizing. In addition, it contains multiplayer of high percentage of peptidoglycan and teichoic acid with large amount of negative charges. The complex of crystal violet and iodine binds to cell wall tightly and can not be extracted by 95% ethyl alcohol. So it appear purple.2.What’s the difference of the mycelia and spore between antinomycetes and molds?A: The primary cell structure of antinomuycetes is similar to bacteria. Antinomycetes havefilament which contain substrate filament and aerial filament and sporebearing filament. Sporebearing filament differentiate to spore. The shape of sporebearing filament are sphere, ellipse and rod. The morphology and color of aerial filament and spore are important for the classification and characterization of antinomycetes.Molds are eukaryotic cells and most of them are multicellular and filamented. The filament of molds is often wider than that of antinomycetes. Spores have different shapes and are divided into asexual or sexual ones.The morphology of mycelia and spore for antinomycetes and molds can be observed by microscope and cultivated on concavity slide or cover slip.After inoculated on agar culture media and cultured at approximate temperature for some time, microorganisms grow to form colonies from single cell which can been seen by the naked eye. The characteristics of colonies depends on their cell structure and growth behavior which is further used to the identification of microorganisms.3.How to judge if a pharmaceutical preparation is qualified for microbial contaminations or not? A: According to the pharmacopoeia, drugs for external or oral use need microbial limitation check except for aseptic inspection. That is the number of bacteria and molds and yeast, the pathogenic bacteria such as Escherichia coli, Salmonella species, Staphylococcus aureus, Pseudomonas aeruginosa, Bacillus tetani and living mite.The microbial standards of drugs are as shown in Table 44.1. In addition, no Salmonella species should be checked in preparations from any animal organs. The number of molds in oral antibiotics should be controlled to be less than 100 CFU/g and bacteria in antifungi orals less than 100 CFU/g too. On the other hand, mildewed and mite contaminated drugs are not qualified. The number of bacteria and molds can be determined by agar plate. Pathogenic bacteria can be confirmed by its morphology biochemical tests while living mite is observed by microscope.The number of bacteria and molds in the plates is reported in the following way. At first, it is suitable for bacteria counting when there are 30 to 300 colonies in plates while 30 to 100 for molds.(1) When the number of colonies at one dilution series is among 30 to 300 for bacteria or 30 to 100 for molds, report it.(2) When the number of colonies at two dilution series is among 30 to 300 for bacteria or 30 to 100 for molds, calculate the ratio as follows:Where NH and NL are colonies of high and low dilution series respectively, DH、DL are dilution times for two dilution series respectively. When R≤2, the average number of two dilution series is reported and the number of low dilution series is reported for R>2.(3) When the average number of colonies at three dilution series is among 30 to 300 for bacteria or 30 to 100 for molds, report it based on the two of low dilution series.(4) When the average number of colonies at all dilution series is not in the range of 30 to 300 for bacteria or 30 to 100 for molds, report colonies number which is most close to the range above.(5) When the average number of colonies at all dilution series is more than 300 for bacteria or 100 for molds, report colonies number based on the highest dilution series.(6) When the average number of colonies at all dilution series is less than 30, report coloniesnumber based on the lowest dilution series.(7) When the average number of colonies at 1:10 or 1:100 dilution series is not less than colonies of drug or 1:10, media dilution determination should be used.。

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