人教版2018-2019学年高中化学同步必修二课时跟踪检测：(十九) 油脂和蛋白质 含答案

课时跟踪检测（十九）油脂和蛋白质

1．下列各物质中，不能发生水解反应的是()

A．葡萄糖B．纤维素

C．油脂D．酶

解析：选A A项，葡萄糖是单糖不能水解，正确；B项，纤维素是多糖，在一定条件下水解最终生成单糖，错误；C项，油脂在酸、碱等催化作用下水解生成甘油和高级脂肪酸或其盐，错误；D项，酶在一定条件下水解最终生成氨基酸，错误。

2．下列物质不属于高分子化合物的是()

A．纤维素B．油脂

C．淀粉D．蛋白质

解析：选B纤维素、淀粉、蛋白质属于天然高分子化合物，油脂不属于高分子化合物。

3．鉴别某种白色织物是否是蚕丝制品，可选用的方法是()

A．滴加盐酸B．滴加浓硫酸

C．滴加氢氧化钠溶液D．滴加浓硝酸

解析：选D蚕丝属于蛋白质，遇到浓硝酸并加热会发生颜色反应而变为黄色，所以可以用滴加浓硝酸的方法鉴别。

4．下列说法正确的是()

A．糖类、油脂、蛋白质都能发生水解反应

B．油脂有油和脂肪之分，都属于酯类

C．糖类、油脂、蛋白质都是由C、H、O三种元素组成的

D．糖类、油脂、蛋白质都是高分子化合物

解析：选B A项，糖类中的单糖不能水解，错误；B项，油脂有油和脂肪之分，都是高级脂肪酸的甘油酯，属于酯类，正确；C项，糖类、油脂都是由C、H、O三种元素组成的，而蛋白质中还含有N、S等元素，错误；D项，油脂和糖类中的单糖、双糖不是高分子化合物，错误。

5．下列有关糖类、油脂和蛋白质的叙述正确的是()

A．油脂的水解反应又叫做皂化反应

B．一定条件下，糖类都能发生水解反应

C．蛋白质发生水解反应后，直接生成葡萄糖

D ．通过灼烧时产生的气味可以鉴别蚕丝和棉纱

解析：选D A 项，油脂在碱性条件下的水解反应又叫做皂化反应，错误；B 项，单糖如葡萄糖，不能发生水解反应，错误；C 项，蛋白质发生水解反应，最终生成氨基酸，错误；D 项，蚕丝的主要成分是蛋白质，灼烧时有烧焦羽毛的气味，棉纱的主要成分是纤维素，灼烧时不具有烧焦羽毛的气味，可以通过灼烧时产生的气味鉴别二者，正确。

6．糖类、脂肪、蛋白质是维持人体生命活动的三大营养物质。以下叙述正确的是( )

A ．植物油不能使溴的四氯化碳溶液褪色

B ．淀粉水解的最终产物是葡萄糖

C ．葡萄糖能发生氧化反应和水解反应

D ．蚕丝、羊毛、棉花的成分都是蛋白质

解析：选B A 项，植物油中含有不饱和键，可以使溴的四氯化碳溶液褪色，错误；B 项，淀粉水解的最终产物是葡萄糖，正确；C 项，葡萄糖不能发生水解反应，可以发生氧化反应，错误；D 项，棉花的成分是纤维素，蚕丝、羊毛的成分是蛋白质，错误。

7．下列说法不正确的是( )

A ．蛋白质水解的最终产物是氨基酸

B ．米饭在嘴中越咀嚼越甜的原因是淀粉水解生成甜味物质

C ．油脂、乙醇是人体必需的营养物质

D ．水果因含有酯类物质而具有香味

解析：选C 蛋白质水解最终得到氨基酸，A 正确；米饭在嘴中咀嚼时淀粉水解产生有甜味的麦芽糖，这时嘴中有甜味，B 正确；水果中由于含有一些天然的酯类物质，因而有香味，D 正确；乙醇不是人体所必需的营养物质，C 错误。

8．下列营养物质在人体内发生的变化及其对人的生命活动所起的作用叙述不正确的是

( )

A ．淀粉――→水解葡萄糖――→氧化

CO 2和H 2O(释放能量维持生命活动)

B ．纤维素――→水解葡萄糖――→氧化CO 2和H 2O(释放能量维持生命活动)

C ．油脂――→水解甘油和高级脂肪酸――→氧化CO 2和H 2O(释放能量维持生命活动)

D ．蛋白质――→水解氨基酸――→合成人体所需的蛋白质(人体生长发育)

解析：选B 人体中没有水解纤维素的酶，纤维素在人体内不消化。

9．现有4种物质：A.乙烯，B.葡萄糖，C.油脂，D.蛋白质，试按要求填写有关物质的

序号：

(1)只由碳、氢、氧3种元素组成的物质有________；

(2)能发生银镜反应的物质有________；

(3)一定条件下能与H2O反应的物质有________；

(4)一定条件下既能发生水解反应，又能与H2发生加成反应的物质有________。

解析：(1)乙烯只含碳、氢两种元素；葡萄糖和油脂只含碳、氢、氧3种元素；蛋白质除含碳、氢、氧元素外，还有氮、硫、磷等元素。

(2)葡萄糖能发生银镜反应，其余3种物质均不能。

(3)乙烯能与H2O发生加成反应，油脂和蛋白质均能发生水解反应。

(4)乙烯和葡萄糖不能水解，植物油既能水解又能与H2加成，蛋白质能水解但不能与H2加成。

答案：(1)B、C(2)B(3)A、C、D(4)C

10．糖类、油脂和蛋白质都是人类基本营养物质。请回答下列问题：

(1)在试管中加入0.5 g淀粉和4 mL 20%的H2SO4溶液，加热3～4 min，然后用碱液中和试管中的H2SO4溶液。

①淀粉完全水解生成的有机物分子式为________。

②若要检验淀粉已经发生了水解，可取少量上述溶液加入__________(填试剂的名称)，加热后再根据实验现象判断。

(2)油脂在人体内通过水解生成________和甘油，再氧化分解，为人体提供能量。

(3)糖尿病人宜多吃蔬菜和豆类食品。蔬菜中富含纤维素，豆类食品中富含蛋白质。下列说法错误的是_________________________________________________________。

A．蛋白质都属于天然高分子化合物，且都不溶于水

B．用灼烧闻气味的方法可以区分合成纤维和羊毛

C．人体内不含纤维素水解酶，人不能消化纤维素，因此蔬菜中的纤维素对人没有用处D．用天然彩棉制成贴身衣物可减少染料对人体的副作用

解析：(1)淀粉水解的最终产物为葡萄糖，分子式为C6H12O6，可用银氨溶液或新制的氢氧化铜悬浊液鉴别葡萄糖的存在。

(2)油脂水解生成高级脂肪酸和甘油。

(3)有些蛋白质(如鸡蛋白)可以溶于水，A错误；纤维素可促进胃肠的蠕动，有助于消化，C错误。

答案：(1)①C6H12O6

②银氨溶液(或新制的氢氧化铜悬浊液)

(2)高级脂肪酸(3)A、C

1．下列说法不正确的是()

A．油脂会造成人体肥胖，所以不能食用

B．脂肪酸在体内可以被氧化从而供给人体热量

C．摄入人体的脂肪大部分成为脂肪组织存在于人体内

D．必需氨基酸在体内有促进发育、维持健康和参与胆固醇代谢的生理功能

解析：选A油脂是人体重要的营养物质，应该适量食用。

2．下列说法正确的是()

A．糖类化合物都具有相同的官能团

B．酯类物质是形成水果香味的主要成分

C．油脂的皂化反应生成脂肪酸和丙醇

D．蛋白质的水解产物都含有羧基和羟基

解析：选B糖类，如葡萄糖中除含有羟基外，还含有醛基官能团，A项错误；皂化反应是指油脂在碱性条件下的水解反应，故生成的应是脂肪酸盐和丙三醇，C项错误；蛋白质水解的最终产物是氨基酸，均含有羧基和氨基，D项错误。

3．下列有关蛋白质的说法正确的是()

①蛋白质是重要的营养物质，它有助于食物的消化和排泄

②蛋白质在淀粉酶作用下，可水解成葡萄糖

③在家庭中可采用灼烧法定性检查奶粉中是否含有蛋白质，蛋白质燃烧可产生特殊的气味

④蛋白质水解的最终产物都是氨基酸

A．①②B．②③

C．③④D．①④

解析：选C蛋白质是营养物质，提供人体所需的能量，在人体内可在蛋白酶的作用下水解生成氨基酸，蛋白质遇浓硝酸时可以显黄色，在灼烧时可以产生烧焦羽毛的气味，③、④正确。

4．下列说法正确的是()

A．向鸡蛋清的溶液中加入浓的硫酸钠溶液或福尔马林，蛋白质的性质发生改变并凝聚B．将牛油和烧碱溶液混合加热，充分反应后加入热的饱和食盐水，上层析出甘油

C．氨基酸为高分子化合物，种类较多，分子中都含有—COOH和—NH2

D．淀粉、纤维素、麦芽糖在一定条件下可和水作用转化为葡萄糖

解析：选D鸡蛋清的溶液中加入浓的Na2SO4，会使蛋白质发生盐析而非变性，A错；甘油易溶于水，上层析出的是高级脂肪酸钠，是肥皂的主要成分，B错；氨基酸是小分子化合物，蛋白质是高分子化合物，C错。

5．下列说法正确的是()

A．油脂、糖类和蛋白质均为高分子化合物

B．植物秸秆的主要成分是纤维素，纤维素在催化剂作用下经水解可得葡萄糖，葡萄糖在酒化酶的作用下能转化为酒精

C．往含硫酸的淀粉水解液中加入银氨溶液，水浴加热后无银镜产生，说明淀粉未水解D．向鸡蛋清的溶液中加入甲醛溶液，可观察到蛋白质发生凝聚，再加入蒸馏水，振荡后蛋白质又发生溶解

解析：选B A项，油脂以及单糖、二糖等不属于高分子化合物，错误；C项，银镜反应需在碱性条件下进行，错误；D项，蛋白质在甲醛溶液中发生变性而不是盐析，因变性发生的凝聚，再加入蒸馏水，不能使其溶解，错误。

6．化学与生活密切相关。下列有关说法错误的是()

A．用灼烧的方法可以区分蚕丝和人造纤维

B．食用油反复加热会产生稠环芳烃等有害物质

C．加热能杀死流感病毒是因为蛋白质受热变性

D．医用消毒酒精中乙醇的浓度为95%

解析：选D A项蚕丝的主要成分是蛋白质，灼烧时蚕丝有烧焦羽毛的气味，人造纤维则没有。B项食用油反复加热会发生复杂的反应，产生稠环芳烃等有害物质。C项加热能使流感病毒体内的蛋白质发生变性，从而杀死流感病毒。D项医用消毒酒精中乙醇的浓度为75%。

7．下列说法不正确的是()

A．硬脂酸甘油酯属于高级脂肪酸甘油酯，是高分子化合物

B．含淀粉或纤维素的物质可以制造酒精

C．向鸡蛋清的溶液中加入硫酸铜溶液，鸡蛋清凝聚，蛋白质变性

D．不同种类的氨基酸能以不同的数目和顺序彼此结合，形成更复杂的多肽化合物

解析：选A硬脂酸甘油酯属于高级脂肪酸甘油酯，不是高分子化合物，故A错误。淀粉或纤维素经水解及发酵，可得到酒精，B正确。硫酸铜为重金属盐，重金属离子能使蛋白质变性，C正确。氨基酸分子间可通过氨基与羧基结合形成肽键，从而形成肽链，故不同种类的氨基酸能以不同的数目和顺序彼此结合，形成更复杂的多肽化合物，D正确。

8．下列与有机物结构、性质相关的叙述错误的是()

A．乙酸分子中含有羧基，可与NaHCO3溶液反应生成CO2

B．蛋白质和油脂都属于高分子化合物，一定条件下都能水解

C．甲烷和氯气反应生成一氯甲烷与苯和硝酸反应生成硝基苯的反应类型相同

D．苯不能使溴的四氯化碳溶液褪色，说明苯分子中没有与乙烯分子中类似的碳碳双键解析：选B乙酸含有羧基，酸性比碳酸强，可与NaHCO3溶液反应生成CO2，A项正确；油脂是高级脂肪酸的甘油酯，不属于高分子化合物，B项错误；甲烷和氯气的反应与苯和硝酸的反应均为取代反应，C项正确；乙烯分子中含碳碳双键可以与溴发生加成反应从而使溴的四氯化碳溶液褪色，苯不能使溴的四氯化碳溶液褪色，说明苯分子中不含碳碳双键，D项正确。

9．Ⅰ.按下表左边的实验操作，可观察到表中右边的实验现象。请从“实验现象”栏中选择正确选项的字母代号填入“答案”栏中。

食，不爱吃水果、蔬菜，结果营养缺乏、发育不良，这主要是由于摄取________(填“脂肪”、“维生素”或“蛋白质”)不足引起的。

(2)①糖类、油脂、蛋白质都是人体必需的营养物质。油脂被摄入人体后，在酶的作用下水解为高级脂肪酸和__________(写名称)。

②氨基酸是组成蛋白质的基本结构单元，其分子中一定含有的官能团是氨基(—NH2)和________(写名称)。人体中共有二十多种氨基酸，其中人体自身________(填“能”或“不能”)合成的氨基酸称为人体必需氨基酸。

③淀粉在淀粉酶的作用下最终水解为__________(写化学式)，部分该产物在体内被氧化

放出能量，供人体活动需要。

解析：Ⅰ.(1)向鸡蛋清中滴加少许浓硝酸，微热，显黄色；(2)乙烯能使酸性KMnO4溶液褪色；(3)CCl4萃取碘水中的碘，四氯化碳不溶于水，液体分层，碘溶于四氯化碳中显紫红色；(4)土豆片的成分中含淀粉，遇到碘水中的碘显蓝色。

Ⅱ.(1)水果和蔬菜中含有丰富的维生素，所以摄入维生素少引起营养不良；(2)油脂水解成高级脂肪酸和甘油；氨基酸含有氨基和羧基；人体自身不能合成的氨基酸为人体必需氨基酸，必须从食物中摄取。淀粉水解最终产物为葡萄糖，化学式为C6H12O6。

答案：Ⅰ.(1)B(2)A(3)D(4)C

Ⅱ.(1)维生素

(2)①甘油②羧基不能③C6H12O6

10．有一种有机物的结构简式为

试回答下列问题：

(1)该化合物是________(填字母)。

A．烯烃B．油脂

C．蛋白质D．糖类

(2)该化合物的密度________。

A．比水大B．比水小C．与水相同

(3)常温下该化合物呈________。

A．液态B．固态C．气态

(4)下列物质中，能与该物质反应的有________。

A．NaOH(溶液)B．溴水

C．乙醇D．乙酸

E．H2

解析：该有机物是高级脂肪酸的甘油酯，属于油脂；其密度比水的小；因油脂的烃基中含碳碳双键，故该油脂常温下呈液态。该油脂分子结构中含有碳碳双键，因此它能在NaOH 的作用下发生水解反应，能与H2或Br2发生加成反应。

答案：(1)B(2)B(3)A(4)A、B、E

新人教版必修一 Unit5 Period 1课时跟踪检测

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高中英语学习材料 madeofjingetieji 课时跟踪检测（一） Warming Up & Reading — Language Points Ⅰ.单句语法填空 1．It was in prison that he set down the story. 2．After going through the woods, you’ll find a hospital. 3．The house fell down before she was able to save her little daughter. 4．She joined an English club in order to improve her English. 5．The couple told us it was the fourth time that they had_visited (visit) the West Lake. 6．Before eating, please add some salt to the dish. 7．As far as I’m concerned (concern), it is not a good idea to go camping tonight. 8.While swimming (swim) in the river, we saw a fish jump out of the river. 9．It is no good staying up too late every day. Ⅱ.完成句子 1.He was_very_upset_about_the_fact (为此事很难过) that they couldn’t go to the beach. 2．The teacher made every effort to calm_his_student_down (使他的学生冷静下来) after the exam. 3．I got up early in_order/so_as_not_to_be_late_for_school (为了上学不迟到)． 4．The police went_through_our_luggage (检查了我们的行李) before we got on the plane. 5．A_series_of_wet_days (一连串的雨天) drove me crazy. 6．The time I spend on reading every day adds_up_to_two_hours (合计两个小时)．7．I won’t go there unless_(I_am)_invited (除非受到邀请)． 8．He treated me so badly that I couldn’t_stand_it_any_longer (不能再忍受了)． Ⅲ.阅读理解 A Dear Mr Black， I used to have a really good group of friends. Now they’re all getting into smoking and drinking. I want to find a new group of friends, but I’m shy. How can I know who are the types of people I should make friends with, and who will accept me? Yours， Mike

高中语文语文版必修四课时跟踪检测(十二)+师说+Word版含答案

课时跟踪检测（十二）师说 (时间：40分钟满分：60分) 一、文言基础专练(28分) 1．下列各句中加点的词语，解释不正确的一项是(3分)( ) A．吾从而师．之师：以……为师 B．吾未见其明．也明：明智 C．君子不齿．齿：并列 D．圣人无常．师常：经常 解析：选D D项，常：固定的。 2．下列加点词语的含义与现在的用法分析正确的一组是(3分)( ) ①古之学者 ．．受业解惑也③今之众人 ．．，其下圣人也亦远矣④小．．．必有师②师者，所以传道 学．而大遗，吾未见其明也⑤弟子不必 ．． ．．不如师，师不必贤于弟子⑥年十七，好古文A．全不相同 B．②③⑤和现在的用法相同 C．全都相同 D．①③⑥和现在的用法相同 解析：选A ①学者，古义：求学之人；今义：在学术上有一定成就的人。②传道，古义：传播道理，文中指传播儒家思想；今义：通常指传播宗教思想。③众人，古义：普通人；今义：大家。④小学，古义：在文中指对小的方面学习；今义：对儿童、少年实施初等教育的学校。⑤不必，古义：不一定；今义：用不着，不需要。⑥古文，古义：先秦两汉的文字；今义：相对于白话文的文言文。 3．从句式特征看，与“师者，所以传道受业解惑也”一句相同的一项是(3分)( ) A．道之所存，师之所存也B．句读之不知，惑之不解 C．不拘于时，学于余D．圣人无常师 解析：选A A项与例句同为判断句。B项是宾语前置；C项是被动句；D项是一般句式。 4．下列句子中，“师”字的用法不同于其他三项的一项是(3分)( ) A．于其身也，则耻师．焉 B．或师．焉，或不焉 C．爱其子，择师．而教之 D．师．道之不复，可知矣 解析：选C C项为名词，译为“老师”，其他三项为动词，意为“从师”。 5．下列加点词语含义相同的一组是(3分)( )