高考生物一轮复习 必修部分 第二章第三节课时跟踪检测

高考生物一轮复习必修部分第二章第三节课时跟踪检测

(时间：30分钟，满分：100分)

一、选择题(每小题5分，共50分)

1．细胞正常分化的结果是()

A．增加细胞蛋白质的种类

B．赋予不同种类细胞特异的结构和功能

C．使细胞代谢增强

D．使细胞的全能性更易表现

解析：细胞分化是基因的选择性表达，使细胞具有特定的结构和功能，细胞的种类增多；分化程度越高的细胞全能性越不容易表现。

答案：B

2．细胞分化的原因是()

A．基因发生重组

B．细胞选择性保留部分基因

C．细胞结构的改变

D．基因的选择性表达

解析：细胞分化是基因选择性表达的结果，细胞分化过程并不改变细胞的基因组成，但可导致细胞的形态、结构和功能发生改变。

答案：D

3．(2011·唐山模拟)动物和人体内具有分裂和分化能力的细胞称为干细

胞。对右图的相关叙述中，正确的是()

A．细胞①②③中所含细胞器的种类和数量都相同

B．细胞①②③的细胞核中基因的种类和数量有差异

C．a、b、c过程中都有DNA的复制和转录过程

D．细胞①在体外培养成细胞系，遗传物质将发生改变

解析：由图中信息可知，a是细胞增殖，b、c是细胞分化过程，干细胞分化成的不同细胞内细胞器含量有差异。干细胞经增殖或分化后，细胞内基因的种类或数量不变。b、c过程为细胞的分化，不进行增殖，无DNA的复制过程。细胞在体外培养成细胞系，遗传物质发生改变。

答案：D

4．下列对于细胞分化不同水平的分析，不正确的是()

A．从细胞水平分析，细胞分化是由于细胞形态、结构和功能改变的结果

B．从细胞亚显微结构水平分析，细胞分化是由于细胞器的数目、形态、结构、功能及细胞质基质成分和功能改变的结果

C．从蛋白质分子的角度分析，细胞分化是由于蛋白质种类、数量、功能改变的结果，这是细胞分化的根本原因

D．从核酸分子的角度分析，细胞分化是由于基因选择性表达的结果

解析：在个体发育中，由一个或一种细胞增殖产生的后代，在形态、结构和生理功能上发生稳定性差异的过程，叫做细胞分化。从核酸分子的角度分析，细胞分化是由于基因选择性表达的结果，这是细胞分化的根本原因。

答案：C

5．(2011·南宁调研)许多动物具有再生能力，例如，壁虎被天敌追赶紧迫时，能断尾而逃，一段时间后，又重新长出一条尾巴；海参遇到敌害时，把内脏全部抛出，以转移敌害的注意

力，而趁机逃之夭夭，约50天后，可再生出一副新的内脏。下列关于再生的叙述，错误的是()

A．再生是细胞分裂和分化的结果

B．再生是基因选择性表达的结果

C．再生是自然选择的结果

D．再生体现了细胞的全能性

解析：再生即组织器官的再生成，是损伤部位细胞分裂和分化的结果；细胞分化的实质是基因在特定时间和空间选择性表达的结果；两种生物都是为了躲避敌害而舍弃了部分器官，所以再生是在天敌环境下，长期选择而产生的一种适应环境的表现，是自然选择的结果；细胞的全能性是指细胞具有发育成完整个体的潜能，题干所述现象都只是形成部分器官，而不是产生一个完整新个体，不能体现全能性。

答案：D

6．以下不能说明细胞全能性的实验是()

A．将诱变处理的番茄种子培育成植株

B．将紫罗兰的叶肉细胞培育成植株

C．将番茄与马铃薯细胞融合并培育成植株

D．将转基因的棉花叶肉细胞培育成植株

解析：细胞全能性应为“离体细胞→完整植株”，种子形成植株应为正常发育过程，并未体现细胞全能性。

答案：A

7．(2011·乌鲁木齐模拟)珍珠是河蚌产生的一种有机物，它既是一种有机宝石，又可作为护肤品的主要原料，因为它在抑制细胞脂褐素的增加上有重要作用，这表明珍珠在保健上可用于()

A．抑制细胞癌变B．延缓细胞衰老

C．促进细胞分裂D．诱导细胞分化

解析：根据题干信息可知，珍珠可以抑制细胞色素的增加，而色素的增加是细胞衰老的一种特征。

答案：B

8．美国研究人员的研究成果表明，在化疗的同时让实验鼠服用糖尿病治疗药物二甲双胍，可抑制实验鼠体内的乳腺癌细胞扩散。下列有关描述正确的是()

A．正常鼠长期接触患癌鼠，其细胞癌变几率增加

B．癌症是由致癌因子引发的，患病几率与实验鼠年龄无关

C．一只切除胸腺的鼠与一只正常鼠患癌症的几率相同

D．二甲双胍不可能使细胞表面的糖蛋白减少

解析：细胞癌变不会传染，正常鼠长期接触患癌鼠，细胞癌变几率不会增加。年龄大的鼠免疫力降低，接触致癌因子的机会也多，细胞癌变几率增大。切除胸腺的鼠免疫力降低，细胞癌变几率也增大。癌细胞膜表面糖蛋白减少，使癌细胞易转移，二甲双胍可抑制实验鼠体内的乳腺癌细胞扩散，可能是增加了癌细胞膜表面的糖蛋白。

答案：D

9．通过特定方法，科学家将小鼠和人已分化的体细胞成功地转变成了类胚胎干细胞。有关分化的体细胞和类胚胎干细胞的描述，正确的是()

A．类胚胎干细胞能够分化成多种细胞

B．分化的体细胞丢失了某些基因

C．二者功能有差异，但形态没有差异

D．二者基因组相同，且表达的基因相同

解析：类胚胎干细胞具有分裂、分化的能力，能够分化成多种细胞；细胞分化的实质为基因的选择性表达，基因并未丢失；细胞分化使细胞的形态、结构、生理功能出现差异。

答案：A

10．关于细胞癌变、衰老的叙述，正确的是()

A．癌细胞的细胞膜上糖蛋白的含量比正常细胞多

B．细胞的衰老不等同于机体的衰老

C．衰老的细胞内，酶的活性降低，细胞核体积缩小

D．细胞癌变是细胞高度分化的结果

解析：癌细胞膜上糖蛋白减少；衰老的细胞内，酶的活性降低，细胞核体积增大；癌变是细胞畸形分化的结果；细胞的衰老不等于个体的衰老。

答案：B

二、非选择题(共50分)

11．(12分)下图是某细胞增殖和分化的概念图，据图回答下列有关问题：

(1)图A表示的是________过程，图B表示的是________过程。图C处于有丝分裂的

________期。

(2)图B中b、c、d、e具有相同的________。由许多相同的b共同组成的结构叫

________。

(3)若a为植物细胞，而d能在体外条件下培养成一个完整植物体，说明d具有_____性。

(4)图C中细胞有________条染色体，①和⑤两条染色体是经过分裂间期_____形成的。

(5)高度分化的动物细胞分化的潜能变小，它的全能性受到限制，但它的细胞核仍具有

全能性，请用性成熟的小白鼠作实验材料，提出一个能证明上述说法的实验设计思

路：____________________________________________________________________

________________________________________________________________________

解析：(1)由图可知，图A表示的是细胞增殖过程，图B表示的是细胞分化过程，图C中两组染色体分别移向细胞的两极，因此处于有丝分裂的后期。(2)在细胞分化过程中遗传物质不变，因此b、c、d、e具有相同的遗传信息；由许多相同的细胞构成的结构叫组织。(3)根据题意，植物组织培养利用了细胞的全能性。(4)图C中①和⑤两条染色体是由一条染色体上的两条染色单体分开形成的，因此是分裂间期复制的结果。(5)欲证明高度分化的动物细胞全能性受到限制，但其细胞核仍具有全能性，可设置单独培养体细胞及进行核移植的两组实验，依据能否培育发育成小白鼠得出结论。

答案：(1)细胞增殖细胞分化后(2)遗传信息组织

(3)全能(4)8复制(5)先把完整的小白鼠体细胞单独培养，证明高度分化的细胞的全能性受到限制；再把体细胞的细胞核导入去核的小白鼠卵细胞中，看重组细胞是否能最终发育成小白鼠

12．(15分)下图表示干细胞的两个发育途径。据图回答下列问题：

(1)由A细胞形成的B细胞仍然保持着其特有的________________能力。

(2)若C细胞是胰腺细胞，则结构⑦所起的作用是______________________________

________________________________________________________________________。

(3)A细胞分化成C细胞的根本原因是_________________________________________

________________________________________________________________________。

(4)若C细胞是正在衰老的细胞，该细胞中结构⑧发生的变化是________。

(5)由A细胞到形成多个卵细胞的过程，必须经过细胞的________分裂。

解析：结构⑦是高尔基体，对蛋白质可以进行加工、转运、分类和包装。细胞分化的根本原因是细胞中基因的选择性表达。A细胞是干细胞，经有丝分裂、分化产生原始的生殖细胞，然后经过减数分裂产生卵细胞。

答案：(1)分裂和分化

(2)对来自内质网的蛋白质进行加工、转运、分类和包装(答出其中两个即可)

(3)基因的选择性表达

(4)细胞核体积增大、染色质固缩、染色加深(只答出其中两项即可)(5)有丝分裂和减

数

13．(23分)[实验1]分别在相同的条件下，体外培养胎儿、中年人和老年人的肺成纤维细胞，结果见下表：

[实验2]分别将老鼠、鸡、人和龟的体细胞在体外培养，结果见下图：

[实验3]将老年男性体细胞和年轻女性体细胞混合培养，当老年男性细胞停止分裂时，年轻女性细胞依然分裂旺盛。

[实验4]将年轻人体细胞去核后与老年人完整细胞融合，另将老年人体细胞去核后与年轻人的完整细胞融合，分别在体外培养，结果前者不分裂，后者分裂旺盛。

分析以上实验，回答下列问题。

(1)每个实验可以得出怎样的结论？

①实验1：______________________________________________________________

________________________________________________________________________。

②实验2：_______________________________________________________________

________________________________________________________________________。

③实验3：________________________________________________________________

________________________________________________________________________。

④实验4：________________________________________________________________

________________________________________________________________________。(2)影响细胞衰老的内在因素有哪些？(至少写出两项)

________________________________________________________________________。(3)实验3的设计是否严密？________。应怎样改进？

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________。

解析：通过阅读材料，可以掌握本题主要探究影响细胞衰老的因素，站在不同角度，用不同的方法通过实验加以验证。实验1主要探究年龄与细胞增殖能力的关系。实验2主要探究物种寿命和最高分裂次数之间的关系。实验3主要探究年龄、性别与细胞衰老的关系。实验4主要探究细胞核在细胞衰老中的作用，但实验3的设计缺乏对照性。

答案：(1)①细胞分裂能力随着生物体的年龄增大而减弱

②细胞分裂次数与具体物种的特异性有关，一般来讲，寿命越长的物种体细胞分裂的最高次数越高

③年轻女性体细胞比老年男性体细胞分裂旺盛

④细胞核是决定细胞衰老的重要因素

(2)影响细胞衰老的内在因素有生物的物种、年龄、性别、细胞核的变化等

(3)实验设计不严密应增加2个对照组，具体如下：

第一组：同年龄的男性细胞与女性细胞一起混合培养，然后观察实验结果，以探究性别对细胞分裂能力有什么影响。

第二组：同性别的男性或女性细胞分不同年龄一起混合培养，探究在同一条件下年龄对细胞分裂能力的影响。

[教师备选题]

14．如图左为某植物根示意图，右为左图中的P、Q、R的放大示意图。回答：哪个细胞表面积最大？哪个细胞最老？哪个细胞生长速率最快？哪对细胞是刚刚由一个细胞分裂而成的？以上依次是()

A．7；1；7；4和7 B．2或3；1；2或3；5和6

C．7；1；7；2和3 D．1；1；2或3；5和6

解析：根毛区(成熟区)细胞由于有很多突起，细胞表面积最大；分生区细胞可以分裂增生，并分化成不同的细胞，Q区的细胞发育成熟为P区细胞；细胞2、3均为成熟的细胞，不再

分裂，但具有快速伸长生长的能力；细胞5、6是由一个细胞分裂而来的两个细胞；细胞7正处于有丝分裂的分裂期。

答案：D

15．癌症是威胁人类健康的一类重要疾病。人们对它进行了大量的研究，以下是有关癌细胞结构及生理的描述，你认为正确的是()

①每个癌细胞都有数目相同的染色体组

②新陈代谢速度减慢，细胞萎缩

③每个癌细胞的DNA含量不一定相等

④都具有两个中心体，且分别位于细胞的两极

⑤在某些病毒的诱导下能产生原癌基因

⑥细胞内的色素逐渐积累

⑦失去控制，无限增殖

A．①②③⑦B．①②④

C．②⑤⑥⑦D．③⑦

解析：不同的癌细胞可能处在不同的细胞周期，因此其染色体数目不同，DNA含量也不一定相同；癌细胞的代谢速度快，细胞没有萎缩；细胞分裂的前、中、后期细胞中有两个中心体，末期、间期有一个中心体；原癌基因在细胞中本来就存在，但是不表达，在病毒的诱导下会被激活；色素增加是衰老细胞的特点，癌细胞的特点之一是能无限增殖，不受机体的控制。

答案：D

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Unit 2 Robots Learning about Language Using Language Ⅰ.单句语法填空 1．It was an absurd arrangement, bound to_lead (lead) to catastrophe. 2．She is junior to me but she looks mature. 3．The young man has a talent for management, so you can rely on him to do the work perfectly. 4．I don't know when they got divorced (divorce)． 5．Try to set aside some money each month for future use. Ⅱ.完成句子 1．如果这样的话，你的观点一定会和他的起冲突。 If so, your points of view are_bound_to conflict with his. 2．小时候，她写了很多优秀的诗歌，显示出了文学天赋。 She wrote many excellent poems at an early age, showing_a_talent_for literature. 3．为了身体健康，我们应该留出一些时间锻炼身体。 For the benefit of our health, we should_set_aside_some_time to take exercise. 4．童年的一切一下子涌入我的大脑，仿佛它们刚刚发生似的。 Everything in my childhood crowed upon my mind as_if_they_had_just_happened. 5．汤姆变成了法官，然而他依然是个渔夫。 Tom became a judge while_he_remained_a_fisherman. Ⅲ.完形填空 A library is a place to find out about almost anything. In it there are mostly __1__， but there are also pictures, papers, magazines, maps, records, special science and art shows, films, plays, and contests may also be __2__ in a library. __3__， a library only had books, and these were difficult to get. There were __4__ books that no one was __5__ to take them out of the library. After the printing machine was invented, books could be made __6__ and libraries got more books. To borrow a book to be __7__ home today, a person just needs a library card. The person goes to the main desk to __8__ a library helper check the book out. The helper prints the card with the date by which the book __9__ be returned to the library. If the book is returned __10__， the person must usually __11__ money for having broken the __12__.

课时跟踪检测(二十三) 电场能的性质

课时跟踪检测（二十三）电场能的性质 [A级——基础小题练熟练快] 1．(2019·湖北八校联考)下列说法正确的是() A．带电粒子仅在电场力的作用下一定做匀变速运动 B．带电粒子仅在电场力的作用下运动时，动能一定增加 C．电场力做正功，带电粒子的电势一定降低 D．电场力做正功，带电粒子的电势能一定减少 解析：选D只有电场是匀强电场时，带电粒子仅在电场力的作用下做匀变速运动，A 错误；如果电场力做负功，则动能减小，B错误；电场力做正功，电势能一定减小，负电荷从低电势向高电势运动，故C错误，D正确。 2.(2020·山东济南模拟)如图所示，等量异种点电荷P、Q连线中点 处有一电子，在外力F作用下处于静止状态。现让电荷Q沿连线向右 移动一小段距离，此过程中电子一直处于静止状态。下列说法正确的是() A．外力F逐渐减小，电子的电势能逐渐增大 B．外力F逐渐增大，电子的电势能逐渐增大 C．外力F逐渐增大，电子的电势能逐渐减小 D．外力F逐渐减小，电子的电势能逐渐减小 解析：选D由题意可知，外力F向右，则电场力向左，可知P带正电，Q带负电；当电荷Q沿连线向右移动一小段距离时，电子所在的位置场强减小，电势升高，则电子受的电场力减小，外力F逐渐减小，电子的电势能降低，故选项D正确，A、B、C错误。 3.(2019·浙江东阳中学模拟)如图所示，MN是由一个正点电荷Q产生的 电场中的一条电场线，一个带正电的粒子＋q飞入电场后，在电场力的作用 下沿一条曲线运动，先后通过a、b两点，不计粒子的重力，则() A．粒子在a点的加速度小于在b点的加速度 B．a点电势φa小于b点电势φb C．粒子在a点的动能E k a小于在b点的动能E k b D．粒子在a点的电势能E p a小于在b点的电势能E p b 解析：选C由题图可知粒子受力应向左方，因粒子带正电，故电场线的方向应向左，故正点电荷Q应在N一侧，故a处的场强大于b处的场强，故粒子在a处的电场力大于b