大学《现代分子生物学(第3版)考试重点

1、证明DNA是遗传物质的两个关键性实验是什么？

Avery肺炎链球菌转化实验(R\S\小鼠-证明基因就是DNA 分子)、Hershey和Chase的噬菌体DNA侵染细菌实验

2、什么是DNA重组技术？

DNA重组技术又称基因工程，目的是将不同的DNA片段（如某个基因或基因的一部分）按照人们的设计定向连接起来，在特定的受体细胞中与载体同时复制并得到表达，产生影响受体细胞的新的遗传性状。

DNA重组技术是核酸化学、蛋白质化学、酶工程及微生物学、遗传学、细胞学长期深入研究的结晶，而限制性内切酶、DNA连接酶及其他工具酶的发现与应用则是这一技术得以建立的关键。

3、比较原核生物和真核生物基因组DNA的特点。

原核生物基因组结构特点

（1）基因组小

（2）结构简练

（3）存在含多顺反子的转录单元

（4）有重叠基因（Sanger 发现）

真核生物基因组结构特点

（1）基因组结构庞大

（2）结构较复杂，含大量重复序列

（3）转录单元为单顺反子

（4）功能DNA序列大多被非功能DNA所隔开（外显子和内含子），即基因有不连续性（5）非编码区多

（6）存在C值矛盾

4、原核、真核生物DNA聚合酶的特性。

原核生物中的DNA聚合酶(大肠杆菌）

聚合酶Ⅰ：主要是对DNA损伤的修复；以及在DNA复制时切除RNA引物并填补其留下的空隙。

聚合酶Ⅱ：修复紫外光引起的DNA损伤

聚合酶Ⅲ：DNA 复制的主要聚合酶，还具有3’-5’外切酶的校对功能，提高DNA复制的保真性

真核生物中的DNA聚合酶

αβγδε

定位细胞核细胞核线粒体细胞核细胞核

3‘-5’外切酶活性- - + + +

功能引物合成修复作用线粒体DNA的复制核DNA的复制RNA引

物去掉后把缺口补全5、细胞通过哪几种修复系统对DNA损伤进行修复？简述DNA错配修复的过

程。

错配修复的过程：一：根据母链甲基化原则找出错配碱基①发现碱基错配②在水解A TP的作用下，MutS,MutL与碱基错配位点的DNA双链结合③MutS—MutL在DNA双链上移动，发现甲基化DNA后由MutH切开非甲基化的子链

二：碱基错配修复过程：当错配碱基位于切口3‘下游端时，在MutL—MutS、解链酶Ⅱ、DNA外切酶Ⅵ或RecJ核酸酶的作用下，从错配碱基3‘下游端开始切除单链DNA直到原切口，并在Pol Ⅲ和SSB的作用下合成新的子链片段。若错配碱基位于切口的5’上游端，则在DNA外切酶Ⅰ或Ⅹ的作用下，从错配碱基5‘上游端开始切除单链DNA 直到原切口，再合成新的子链片段（详见课本p53）

6、基因：启动子：增强子：全酶：核心酶：核酶：三元复合物：SD序列：

7、比较原核生物和真核生物mRNA的特点。

原核生物mRNA的特征：（1）半衰期短（2）多以多顺反子的形式存在

（3）5’端无“帽子”结构，3’端没有或只有较短的poly（A ）结构。

（4）原核生物常以AUG（有时GUG，甚至UUG）作为起始密码子

真核生物mRNA的特征：（1）5’端存在“帽子”结构

（2）多数mRNA 3’端具有poly（A ）尾巴（组蛋白除外）

（3）以单顺反子存在（4）而真核生物几乎永远以AUG作为起始密码子。

8、真核生物的原始转录产物需要经过哪些加工才能成为成熟的mRNA？

答：（1）、在5’端加帽，5’端的一个核苷酸总是7-甲基鸟核苷三磷酸（m7Gppp）。

（2）、3’端加尾，多聚腺苷酸尾巴。准确切割，加poly（A）

（3）、RNA的剪接，参与RNA剪接的物质：snRNA、snRNP

（4）、RNA的编辑，编辑（editing）是指转录后的RNA在编码区发生碱基的突变、加入或丢失等现象。

（5.）、RNA的再编码，mRNA有时可以改变原来的编码信息，以不同的方式进行翻译（6.）、RNA的化学修饰，人细胞内rRNA分子上就存在106种甲基化和95种假尿嘧啶产物。

9、原核启动子和真核启动子的构成

原核生物启动子结构：

TA TA区(-10区）:酶的紧密结合位点（富含AT碱基，利于双链打开）

TTGACA区（-35区）：提供了RNA聚合酶全酶识别的信号

●真核生物启动子

(1)核心启动子:定义：指保证RNA聚合酶Ⅱ转录正常起始所必需的、最少的DNA序列，包括转录起始位点及转录起始位点上游TATA区(TATA 常在-25bp左右，相当于原核的-10序列T85A97T93A85A63A83A50)

`作用：选择正确的转录起始位点，保证精确起始

(2)上游启动子元件:包括CAAT盒（CCAA T）和GC盒（GGGCGG）等(CAAT：-70 - -80bp GGGCGG：-80 - -110bp)

` 作用：控制转录起始频率。

10、遗传密码有哪些特性？理解掌握其摆动性

连续性、简并性、通用性与特殊性、摆动性

转运氨基酸的tRNA上的反密码子需要通过碱基互补与mRNA上的遗传密码子反向配对结合，在密码子与反密码子的配对中，前两对严格遵守碱基配对原则，第三对碱基有一定的自由度，可以“摆动”，这种现象称为密码子的摆动性。（在密码子中，是第三对碱基，在反密码子中，是第一对碱基）

11、tRNA中起作用的重要两个臂是什么臂？

tRNA有两个关键部位：

●3’端CCA：接受氨基酸，形成氨酰-tRNA。

●与mRNA结合部位—反密码子部位

12、.肽链延伸由许多循环组成，每加一个氨基酸就是一个循环，每个循环包括哪些步骤？

肽链延伸由许多循环组成，每加一个氨基酸就是一个循环，每个循环包括：AA-tRNA与核

糖体结合、肽键的生成和移位。

(1)AA-tRNA与核糖体A位点的结合：起始复合物形成以后，第二个AA-tRNA在延伸因子EF-Tu及GTP的作用下，生成AA-tRNA。EF-Tu。GTP复合物，然后结合到核糖体的A 位上。这时GTP被水解释放，通过延伸因子EF-Ts再生GTP，形成EF-Tu。GTP复合物，进入新一轮循环

（2）肽键形成：在核糖体。mRNA.。AA-tRNA复合物中，AA-tRNA占据A位，fMet-tRNA*fMet 占据p位。在肽基转移酶的催化下，A位上的AA-tRNA转移到P位，fMet-tRNA*fMet上的氨基酸生成肽键。起始tRNA在完成使命后离开核糖体p位点，A位点准备接受新的AA-tRNA，开始下一轮合成反应

（3）移位：核糖体通过EF-G介导的GTP水解提供的能量向mRNA模板的3‘端移动一个密码子，使二肽酰—tRNA2完全进入P位，准备开始新一轮的肽链延伸。

13、真核生物和原核生物在翻译的起始过程有哪些区别？

14、同工tRNA：分子伴侣：信号肽：核定位序列：

15、比较PCR扩增和细胞内DNA复制的异同。

16、设计一个典型的PCR 扩增程序和PCR 反应体系。

典型的程序

17、 在基因操作实践中检测核酸相对分子量的最常用方法是什么？其原理

是什么？

答，核酸凝胶电泳技术：核酸分子中糖-磷酸骨架中的磷酸基团，呈负离子化状态，核酸分子在一定的电场强度的电场中，他们会像正电极方向移动。

电泳中使用无反应活性的稳定的支持介质，电泳迁移率与分子大小成反比。

因此，可在同一凝胶电泳中、一定电场强度下分离出不同分子量大小或相同分子量但构型有差异的核酸分子，从而可测定出核酸的相对分子质量。

18、 如何克隆一个新基因（cDNA 的中间片段）？

race 技术：利用PCR 技术，在已知CDNA 部分序列的情况下特异性克隆5’和3’端缺

失序列的方法。 现在已知CDNA 3’端的部分序列，通过以下实验可获得改基因的全长：

① 利用已知的CDNA 3’端的序列设计出两个特异性引物GSP1和GSP2。

② 获取高质量的总RNA

③在反转录酶的作用下，用GSP1起始cDNA 第一条链的合成

④用RNase 混合物物降解模板mRNA 并纯化已合成的cDNA 第一条链

⑤由末端转移酶在已合成的cDNA 链的cDNA 链3‘端连续加入dCTP, 形成oligo dC 尾

巴

⑥以连有oligo dC 的锚定引物和特异引物GSP2进行nest PCR 扩增，以其得到目的基因

5‘端片段并检测

⑦将经过PCR 纯化的产物克隆到载体DNA 中，进行序列分析

⑧将已知的CDNA 3’端的序列和测序得到的5’端序列通过DNAstar 软件拼接即可得

到改基因的全长。

19、SNP技术

SNP即单核苷酸多态性，指基因组DNA序列中由于单个核苷酸（A，T，C和G）的突变而引起的多态性。是指利用数据库中已有的SNP进行特定人群的序列和发生频率的研究，主要包括基因芯片技术、Taqman技术、分子信标技术和焦磷酸测序法等。

20、基因敲除技术的基本原理。

基因敲除（gene knock-out）又称基因打靶，通过外源DNA与染色体DNA之间的同源重组，进行精确的定点修饰和基因改造，具有专一性强、染色体DNA可与目的片段共同稳定遗传等特点

基因敲除分为完全基因敲除和条件型基因敲除（又称不完全基因敲除）两种。

完全基因敲除是指通过同源重组法完全消除细胞或者动物个体中的靶基因活性，条件型基因敲除是指通过定位重组系统实现特定时间和空间的基因敲除

21、RNAi技术的基本原理。

RNAi技术利用双链小RNA高效、特异性降解细胞内同源mRNA从而阻断靶基因表达，使细胞出现靶基因缺失的表型。原理就是，短片段的双链RNA在体内能在酶（Dicer）及相关复合物（RISC）的作用下，变成单链分子，并与目标基因mRNA互补，在Dicer酶作用下，是mRNA发生剪切，转录受抑制或翻译受到抑制，从而在转录水平或转录后水平干扰基因表达。

22、操纵子：弱化子: 葡萄糖效应（代谢物阻遏效应、降解物抑制作用）：安慰性诱导物：

23、乳糖操纵子的调控模型。

主要内容：

①Z、Y、A基因的产物由同一条多顺反子的mRNA分子所编码

②这个mRNA分子的启动子紧接着操纵区（O区），而位于阻遏基因I与操纵基因O之间的启动子区（P），不能单独起动合成β-半乳糖苷酶和透过酶的生理过程。

③操纵基因是DNA上的一小段序列（仅为26bp），是阻遏物的结合位点。

④当阻遏物与操纵基因结合时，lac mRNA的转录起始受到抑制。

⑤诱导物通过与阻遏物结合，改变它的三维构象，使之不能与操纵基因结合，从而激发lac mRNA的合成。当有诱导物存在时，操纵基因区没有被阻遏物占据，所以启动子能够顺利起始mRNA的合成。

24、色氨酸操纵子的负控阻遏系统和弱化调控机制。

负控阻遏系统：

当培养基中色氨酸含量较高时：色氨酸与游离的辅阻遏蛋白相结合，并使之与操纵区DNA紧密结合

当培养基上色氨酸供应不足时：辅阻遏物失去色氨酸并从操纵区上解离，色氨酸操纵子去阻遏

这两个图大家稍微的

把核糖体、MRNA画出来，标出1、2、3、4区就可以了，不用画的那么复杂，但是答题的时候一定要画的厄

弱化调控机制：如上图，当培养基中色氨酸的浓度很低时，负载有色氨酸的tRNA*Trp也就少，这样翻译通过两个相邻的色氨酸的密码子的速度就会很慢，当4区被转录完成时，核糖体才进行到1区，这时的前导结构是2—3配对，不形成3—4配对的终止结构，所以转录继续进行，直到将色氨酸操纵子中的结构基因全部转录。

当培养基中色氨酸浓度高时，核糖体可顺利通过两个相邻的色氨酸密码子，在4区被转录之前，核糖体就达到2区，这样使2-3不能配对，3—4可以自由配对形成茎-环状终止子结构，转录停止，色氨酸操纵子中的结构基因被关闭而不再合成色氨酸

25、为什么半乳糖操纵子需要双启动子？（这题自己可以稍微的缩减一点）

半乳糖不仅作为唯一的碳源供细胞生长，而且是大肠杆菌细胞壁合成的前体。在没有外源半乳糖的情况下，细胞通过半乳糖差向异构酶的作用与UDP—葡萄糖合成UDPgal。因为合成细胞壁的过程中对异构酶的需要量很小，本底水平的永久型合成就能够满足生理需要。实际上，gal mRNA的永久型合成水平已高于lac操纵子所合成的lac mRNA的水平，显然这部分mRNA是在没有半乳糖的情况下合成的。

假设只有一个启动子S1，这个启动子的活性依赖于CA M P—CRP，在培养基有葡萄糖存在时可合成异构酶；假如只有一个启动子S2，即使在有葡萄糖存在的情况下，半乳糖也能使操纵子处于充分诱导的状态，这无疑是一种浪费。所以，无论考虑到必要性还是经济性，需要一个不依赖与CA MP—CRP的启动子S2，进行本底水平的永久型合成，也需要一个依赖于CA MP—CRP的启动子S1，进行高水平合成的调节。也就是说，只有在S2活性受到CA MP—CRP抑制时，调控作用才是有效的。

26、顺式作用元件：反式作用因子：基因家簇：断裂基因：

27、图示简要说明真核生物启动子的结构。

28、说明DNA甲基化对基因转录活性的影响机理。为什么说甲基化密度与启动子强度之间

的平衡决定了该启动子是否具有转录活性（可用图示说明）？

DNA甲基化导致某些区域DNA构象变化，从而影响了蛋白质与DNA的相互作用，

抑制了转录因子与启动区DNA的结合效率。甲基化达到一定程度时会发生从常规的B—DNA向Z—DNA 结构收缩，螺旋加深，使许多蛋白质因子赖以结合的元件缩入大沟而不利于基因转录的起始。甲基化的引入不利于模板与RNA聚合酶的结合，降低了其体外转录活性。

甲基化密度与启动子强度之间的平衡决定了该启动子是否具有转录活性。P294

箭头表示转录方向，箭头粗细程度表示转录强度，小黑圆点表示DNA甲基化，白点表示DNA未甲基化，椭圆表示MecpI松散结合，长方形表示紧密结合

29、举例说明蛋白质磷酸化如何影响基因表达。

蛋白质磷酸化与细胞分裂调控：细胞通过p53及p21蛋白控制CDK活性，调控细胞分裂的进程。如果p21蛋白过量，大量细胞周期蛋白E—CDK2复合物与p21蛋白相结合，使CDK2丧失磷酸化pRb蛋白的功能。没有被磷酸化的pRb蛋白与转录因子E2F结合并使后者不能激活一系列与DNA合成有关的酶，导致细胞不能由G1期进入S期，细胞分裂受阻。如果细胞中p53基因活性降低，p21蛋白含量急剧下降，细胞周期蛋白E—CDK2复合物就能有效地将pRb蛋白磷酸化。此时，pRb蛋白不能与E2F相结合，使后者发挥转录调控因子的作用，激活许多DNA合成有关的基因表达，使细胞从G1期进入S期，引发细胞分裂。

1·下列有关TATA盒(Hognessbox)的叙述,哪个是正确的:B

A.它位于第一个结构基因处

B.它和RNA聚合酶结合

C.它编码阻遏蛋白

D.它和反密码子结合

`2转录需要的原料是:A.DNTP B. dNDP C. dNMP D. NTP E . NMP (D)

3DNA模板链为5’-ATTCAG-3 ’ , 其转录产物是:D

A. 5 ’ -GACTTA-3 ’

B. 5 ’ -CTGAAT-3 ’

C. 5 ’ -UAAGUC-3 ’

D. 5 ’ -CUGAAU-3 ’

4DNA复制和转录过程有许多相同点,下列描述哪项是错误的? D

A.转录以DNA一条链为模板,而以DNA两条链为模板进行复制

B. 在这两个过程中合成均为5`-3`方向

C. 复制的产物通常情况下大于转录的产物

D. 两过程均需RNA引物

5.真核生物的mRNA加工过程中,5’端加上（），在3’端加上（），后者由（）催化。如果被转录基因是不连续的，那么，（）一定要被切除，并通过（）过程将（）连接在一起。(帽子结构、多聚腺苷酸尾巴、poly(A)聚合酶、内含子、剪接、外显子)

6.–10位的（TATA ）区和–35位的（TTGACA ）区是RNA聚合酶与启动子的结合位点，能与σ因子相互识别而具有很高的亲和力。

7·决定基因转录基础频率的DNA 元件是（启动子）,它是（RNA聚合酶）的结合位点8·原核生物RNA 聚合酶核心酶由（2αββ′ω）组成，全酶由（2αββ′ωσ）组成。

9·下面那一项不属于原核生物mRNA的特征（ C ）

A：半衰期短B：存在多顺反子的形式

C：5’端有帽子结构D：3’端没有或只有较短的多聚（A）结构

10·真核细胞中的mRNA帽子结构是（A ）

A. 7-甲基鸟嘌呤核苷三磷酸

B. 7-甲基尿嘧啶核苷三磷酸

C. 7-甲基腺嘌呤核苷三磷酸

D. 7-甲基胞嘧啶核苷三磷酸

11·下面哪一项是对三元转录复合物的正确描述（ B ）

（A）σ因子、核心酶和双链DNA在启动子形成的复合物

（B）全酶、模板DNA和新生RNA形成的复合物

（C）三个全酶在转录起始点形成的复合物（D）σ因子、核心酶和促旋酶形成的复合物

四

1．多数氨基酸都有两个以上密码子，下列哪组氨基酸只有一个密码子？D A．苏氨酸、甘氨酸B．脯氨酸、精氨酸

C．丝氨酸、亮氨酸D．色氨酸、甲硫氨酸E．天冬氨酸和天冬酰胺

2．tRNA分子上结合氨基酸的序列是B

A．CAA-3′B．CCA-3′C．AAC-3′D．ACA-3′E．AAC-3′

3．遗传密码BCE

A．20种氨基酸共有64个密码子B．碱基缺失、插入可致框移突变

C．AUG是起始密码D．UUU是终止密码E．一个氨基酸可有多达6个密码子

4、tRNA能够成为氨基酸的转运体、是因为其分子上有AD

A．-CCA-OH 3′末端B．3个核苷酸为一组的结构

C．稀有碱基D．反密码环E．假腺嘌吟环

5、蛋白质生物合成中的终止密码是( ADE )。

（A）UAA （B）UAU （C）UAC （D）UAG （E）UGA

6、Shine-Dalgarno顺序(SD-顺序)是指:（A ）

A.在mRNA分子的起始码上游8-13个核苷酸处的顺序

B.在DNA分子上转录起始点前8-13个核苷酸处的顺序

C.16srRNA3‘端富含嘧啶的互补顺序

D.启动基因的顺序特征

7、“同工tRNA”是：( C )

(A)识别同义mRNA密码子(具有第三碱基简并性)的多个tRNA

(B)识别相同密码子的多个tRNA

(C)代表相同氨基酸的多个tRNA

(D)由相同的氨酰tRNA合成酶识别的多个tRNA

8、反密码子中哪个碱基对参与了密码子的简并性(摇摆)。（A ）

（A）第—个(B)第二个(C)第三个(D) 第一个与第二个

9、与mRNA的GCU密码子对应的tRNA的反密码子是（ B ）

（A）CGA（B）IGC （C）CIG（D）CGI

10、真核与原核细胞蛋白质合成的相同点是（ C ）

A翻译与转录偶联进行B模板都是多顺反子C 都需要GTP D甲酰蛋氨酸是第一个氨基酸

11、是翻译延长所必需的B 12、氨基酸与tRNA连接D 13、遗传密码的摆动性A

A、mRNA上的密码子与tRNA上的反密码子不一定严格配对

B、转肽酶

C、酯键

D、磷酸化酶

E、N-C糖甘键

七、

1、基因表达调控主要表现在两种水平：（转录水平）和（转录后水平）。其中，后者又包括mRNA加工成熟水平上的调控和（翻译水平上的调控）。

2、不同生物使用不同的信号来指挥基因调控。原核生物中，（营养状况）和（环境因素）对基因表达起主要影响。在高等真核生物中，（激素水平）和（发育阶段）是基因表达调控的最主要手段。

3、原核生物的基因调控主要发生在转录水平上，根据调控机制的不同可分为（正转录调控）和（负转录调控）两大类。在负转录调控系统中，调节基因的产物是（阻遏蛋白）。根据其性质可分为（负控诱导）和（负控阻遏）系统。

4、在葡萄糖存在的情况下，即使在细菌中加入乳糖、半乳糖或其他的诱导物，与其相应的操纵子也不会启动，不会产生出代谢这些糖的酶来，这种现象称为（葡萄糖效应）或称为降解物抑制作用。

5、细菌实施应急反应的信号是（鸟苷四磷酸）和（鸟苷五磷酸）。产生这两种物质的诱导物是（空载tRNA）。

6、原核生物调节基因表达的意义是为适应环境，维持（A）

A、细胞分裂

B、细胞分化 C 器官分化 D 组织分化

7、原核生物的基因调控主要发生在（A）。

A 转录水平

B 转译水平

C 转录后水平

D 转译后水平

1、关于管家基因叙述错误的是D

(A) 在生物个体的几乎各生长阶段持续表达(B) 在生物个体的几乎所有细胞中持续表达

(C) 在生物个体全生命过程的几乎所有细胞中表达

(D) 在生物个体的某一生长阶段持续表达(E) 在一个物种的几乎所有个体中持续表达

2、一个操纵子（元）通常含有B

(A) 数个启动序列和一个编码基因

(B) 一个启动序列和数个编码基因(C) 一个启动序列和一个编码基因

(D) 两个启动序列和数个编码基因(E) 数个启动序列和数个编码基因

3、下列情况不属于基因表达阶段特异性的是，一个基因在A

(A) 分化的骨骼肌细胞表达，在未分化的心肌细胞不表达

(B) 胚胎发育过程不表达，出生后表达(C) 胚胎发育过程表达，在出生后不表达

(D) 分化的骨骼肌细胞表达，在未分化的骨骼肌细胞不表达

(E) 分化的骨骼肌细胞不表达，在未分化的骨骼肌细胞表达

4、乳糖操纵子（元）的直接诱导剂是E

(A) 葡萄糖(B) 乳糖(C) β一半乳糖苷酶(D) 透酶(E)异构乳糖

5、Lac阻遏蛋白结合乳糖操纵子（元）的 B

(A) CAP结合位点(B) O序列(C) P序列(D) Z基因(E) I基因

6、cAMP与CAP结合、CAP介导正性调节发生在C

A)葡萄糖及cAMP浓度极高时(B) 没有葡萄糖及cAMP较低时

C没有葡萄糖及cAMP较高时(D有葡萄糖及cAMP较低时(E) 有葡萄糖及CAMP较高时

7、Lac阻遏蛋白由D

(A) Z基因编码(B) Y基因编码(C) A基因编码(D) I基因编码(E) 以上都不是

8、色氨酸操纵子（元）调节过程涉及E

(A)转录水平调节(B)转录延长调节(C)转录激活调节D翻译水平调节(E)转录／翻译调节

9、与O序列结合 A 10、与P序列结合 B

11、与CAP结合 C 12、与CAP位点结合 D

(A) Lac阻遏蛋白(B) RNA聚合酶(C) 环一磷酸腺苷(D) CAP-cAMP (E)异构乳糖

13、D

A B DNA结合影响模板活性E合

C RNA聚合酶结合影响其活性D影响该蛋白质结合DNA

14、DNA损伤修复的SOS系统B

A B LexA蛋白是一系列操纵子的阻遏物

C RecA蛋白是一系列操纵子的阻遏物D体

15、cAMP对原核基因转录的调控作用的叙述错误的是D

A cAMP可与分解代谢基因活化蛋白(CAP)结合成复合物

B cAMP-CAP复合物结合在启动子前方

C cAMP水平不高

D葡萄糖和乳糖并存时，细菌优先利用乳糖E葡萄糖和乳糖并存时，细菌优先利用葡萄糖

21、Lac 阻遏蛋白由__I__ 基因编码，结合__O__ 序列对Lac 操纵子（元）起阻遏作用。

22、Trp 操纵子的精细调节包括__阻遏机制_ 及_弱化机制_ 两种机制。

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Do you think “we all have a responsibility to be safe”? There is no denying that safety plays an important part in our daily life. Recently our school has carried out a program aiming at increasing our safety awareness. I consider this as indeed a good practice, which will remind us that safety should always come first in our daily life, whether when we stay at school or outside it. For example, when we are doing sports, we should attach great importance to safety and avoid being hurt. Besides, we should follow the traffic rules when we are going to school or when we are on our way back home. I firmly believe that we should develop a good sense of safety and do everything safely. What measures have been taken to prevent driving infractions like drunk driving and speeding in China? Driving infractions cases have been increasing and have caused numerous deaths. A more severe punishment should be given to those who drive after they have drunk, especially to those who are drunk and have caused serious injuries or deaths while driving.In order to crack downdriving infractions, the government should introduce a more severe law. Besides, more measures should be taken to cultivate the drivers’ awareness of traffic laws and respect others’ lives as well as their own when they are driving on the road, which I think should be the key to decrease drunken driving cases. Can you tell us how to escape a fire, esp. when you are in a high-rise building? As we know, no one knows for sure when a fire will happen, so it is extremely necessary to make preparations before a fire starts. To beginwith, checking the fire escape in advance and make sure that you can find it when lights have failed.What’s more, taking care not to be overcome by smoke, which containing monoxide gas, can kill you quickly. At last, cover your mouth with a wet towel or a wet cloth, and avoid getting into the smoke. In short, if you make preparations before and take precautions during the fire, chances are that you will survive in case of a fire if it really breaks out some day. What are the things you have to defend yourself so that you can have a peaceful and wonderful life?