第二章基因的概念.pdf

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1、第二章 基因的概念 一 选择题 1.A DNA nucleotide may consist of 一分子DNA核苷酸可能有下列哪一项组成 A a ribose sugar,a phosphate group,and adenine.B a phosphate group,deoxyribose,and cytosine.C uracil,deoxyribose,and a phosphate group.D deoxyribose,thymine,and a hydroxyl group.2.Which is the most accurate representation of the o

2、rganization levels of the genetic information in cells?下列哪一项最准确地代表了细胞中遗传信息的组织水平 A genes nucleotidechromosomes genome.B genome genesnucleotideschromosomes.C chromosomesgenes nucleotides genome.D nucleotidesgeneschromosomesgenome.3.Eukaryotic cells differ from prokaryotic cells in that only the former

3、 contain A ribosomes.B cytoplasm.C DNA.D a nucleus.4.The prokaryotic organism that has been the subject of many genetic studies is A Saccharomyces cerevisiae.B Neurospora crassa.C E.coli.D Drosophila melanogaster.5.Which element is not found in a molecule of DNA?下列哪一元素不会在DNA分子中发现？A carbon.B sulfur.C

4、 nitrogen.D oxygen.6.Which is found in RNA,but not DNA?下列哪一项只在RNA中存在，而不会在DNA分子中出现？A Phosphate.B Adenine.C Ribose.D Cytosine.7.The two polynucleotide chains in a molecule of DNA are held together by what type of bond?DNA分子中的两条多核苷酸链是依赖于哪一种类型的化学键结合在一起的？A Phosphodiester.B Phosphate.C Peptide.D hydrogen.

5、8.The DNA that makes up bacterial chromosomes is A single-stranded.B circular and supercoiled.C complexed with histones.D all the above.9.Which of the following is a DNA base pair?A A-T.B T-C.C A-U.D G-T.10.Which of the following is a RNA base pair?A A-T.B A-C.C U-C.D U-A.11.The DNA and histone prot

6、eins in a eukaryotic chromosome are compacted into structures called 在真核生物的染色体中，DNA和组蛋白组成的结构称为 A proteosomes.蛋白体 B nucleosomes.核小体 C telomeres.端粒 D centromeres.着丝粒 12.In his transformation experiments,Frederick Griffith observed that virulent strains of Streptococcus pneumoniae produced _ colonies.在

7、转化试验过程中，弗雷德里克格里菲斯观察到致病的肺炎荚膜链球菌菌株形成_菌落 A smooth,shiny B rough,dry C exceptionally large D unusually colored 13.Viral genomes may be composed of 病毒基因组的可能组成是 A RNA only.B DNA only.C either RNA or DNA.D both RNA and DNA.14.In the Hershey-Chase blender experiments,their major conclusion was that Hershey

8、和 Chase利用混合实验得出的主要结论是 A a single gene directed the synthesis of a single polypeptide.B DNA was the genetic material.C DNA was a double helix.D the genetic material was located in the nucleus of cells.15.The type of DNA found most commonly in living cells is the _ form.在活细胞中最普遍存在的DNA构型是 A A.B B.C C.D

9、 Z 16.Genes located in which region of a eukaryotic chromosome are most likely to be transcribed?基因位于真核生物染色体_区域时最有可能被转录 A centromere 着丝粒.B telomere端粒.C euchromatin 常染色质 D heterochromatin 异染色质 17.A nucleoside consists of a 核苷的基本组成是 A pentose sugar and a nitrogenous base.B phosphate group and a nitrog

10、enous base.C pentose sugar and a phosphate group.D pentose sugar,a phosphate group,and a nitrogenous base 18.Prokaryotic chromosomes consist mostly of A unique sequence DNA only.B repetitive sequence DNA only.C either unique sequence or repetitive sequence DNA.D both unique sequence and repetitive s

11、equence 19.To create a karyotype 染色体组,_ chromosomes are spread on a slide and stained.A interphase.B telophase.C metaphase.D anaphase 20 A chromosome from an unknown microscopic organism is examined and found to contain only unique sequence DNA.This organism is most likely A virus.B bacterium.C fung

12、us.D protozoan.21.In eukaryotic cells,the genetic material is found in the A ribosomes.B nucleus.C endoplasmic reticulum.D cytoplasm.22.Griffiths experiment injecting a mixture of dead and live bacteria into mice demonstrated that(choose the correct answer):格里菲斯通过实验将死细菌与活细菌提取物混合后注射到老鼠体内是为了证明（选择正确的答案

13、）A DNA is double-stranded B mRNA of eukaryotes differs from mRNA of prokaryotes C A factor was capable of transforming one bacterial cell type to another D Bacteria can recover from heat treatment if live helper cells are present 23.The X-ray diffraction data obtained by Rosalind Franklin suggested(

14、choose the correct answer):罗莎琳德富兰克林获得的X射线衍射图像的数据表明（选择正确的答案）A DNA is a helix with a pattern that repeats every 3.4 nanometers B Purines are hydrogen bonded to pyrimidines C DNA is a left-handed helix D DNA is organized into nucleosomes 24.Griffith injected mice with different types of bacteria.For ea

15、ch of the following bacteria types injected,indicate whether the mice lived or died:格里菲斯将不同类型的细菌注射到老鼠体内，请根据注射的细菌类型，判断老鼠是存活还是死亡？a.type R b.type S c.heat-killed S d.type R+heat-killed S A lived lived died died B died lived died lived C lived died lived died D died died lived lived 二 填空题 1基因敲除（Gene kno

16、ck-out）即是（将特定基因失活的过程），它是研究（基因功能）的一种反向遗传学方法。2（假基因pseudogene）是与有功能的基因在基因结构的组成上非常相似，却不具表达功能的基因。3在原核生物的基因表达调控中，因为没有核膜，（转录）和（翻译）是耦联的。4A nucleosome is comprised of two copies of histones（H2A），（H2B），（H3），（H4），one copy of histone（H1），and（200）bp of DNA.三 简答题 1From a chemical（bonding）view,why is double-strand

17、ed DNA so stable?从化学（键）的角度，说明双链DNA比较稳定的原因 答案：Hydrogen bonds between bases and hydrophobic bonds due to base stacking 碱基之间的氢键以及疏水的碱基堆积力。2What is“Chargaffs Rule”?Chargaff规则的内容是什么？答案：A%=T%and G%=C%in dsDNA.腺嘌呤和胸腺嘧啶的摩尔数相等，即A=T；鸟嘌呤和胞嘧啶的摩尔数相等，即G=C；含氨基的碱基（A和C）总数等于含酮基的碱基（G和T）总数，即A+C=G+T；嘌呤的总数等于嘧啶的总数，即A+G=C+

18、T。3Supercoiling DNA requires energy and allows work to be done.What are two biochemical functions that this pent-up energy is used for?使DNA超螺旋化需要能量，并可使生物学事件有序进行。超螺旋中的能量的主要用于哪两种生化作用？答案：Energy can cause strand separation used for:（1）DNA replication；（2）Transcription 4You have isolated a plasmid DNA tha

19、t is a closed circular molecular 1050 bp in length with 5 negative supercoils.What are the linking number,helical turns,writhe,and superhelical density?你分离出的质粒 DNA 为 1050 bp 长度的共价闭合环状分子，具 5 个负超螺旋。请问这个 DNA 分子的连接数（L）、螺旋数（缠绕数 T）、扭曲数（W）、以及超螺旋密度是多少？答案：Writhe=W=-5；helical turns=T=1050/10.5=100；L=T+W=100+（

20、-5）=95 Superhelical density=W/T=-5/100=-0.05 5What is a pseudogene,and how do you recognize one?什么是假基因？你如何识别出假基因？答案：Psuedogenes are are defined by their possession of sequences that are related to those of the functional genes,but that cannot be translated into a functional protein.假基因是指与正常基因结构相似，但丧

21、失正常功能的DNA序列，即不能翻译出有功能蛋白质的基因。can be recognized by the occurrence of one or more mutations that obviously render them non-functional.可以利用补偿其功能的突变体而区分出假基因。6.In eukaryotic cells,DNA is packaged into chromatin.The repeating unit of chromatin is called nucleosome.在真核细胞中，DNA 被包装成染色质，其重复单元为核小体。（1）What const

22、itute a mononucleosome?每个核小体单元是由什么组成的？（2）What is the consequence of DNA packaging on transcription?DNA 被包装后，对转录会产生怎样的影响后果？答案：（1）A mononucleosome is composed of 146 bp of core DNA wrapped around a histone octamer of 2 copies of each of H2A,H2B,H3 and H4.每个核小体单元由 146 bp 核心 DNA 和各两分子的组蛋白 H2A、H2B、H3、H4

23、组成的八聚体构成。（2）The consequence of DNA packaging is to inhibit transcription due to the inability of transcription machinery and transcription factors to gain access to DNA.DNA 被包装成染色质结构后，会抑制转录。因为这一结构能够阻止转录机构及转录因子与 DNA 之间的相互接近。7Which combinations of histones form“histone-fold”dimers with each other in t

24、he nucleosome?在核小体中，每个组蛋白二聚体分别由哪几种组蛋白组成？答案：H3 and H4 form one dimer pair,H2A and H2B form the other.组蛋白 H3 和 H4 形成一个二聚体；组蛋白 H2A 和 H2B 形成另一个二聚体。8Through X-ray diffraction analysis of crystallized DNA oligomers,different forms of DNA have been identified.These forms include A-DNA,B-DNA,and Z-DNA,and e

25、ach has unique molecular attributes.通过对 DNA 晶体的 X-射线衍射分析，鉴定出了 DNA 的不同形式。DNA 分子的二级结构存在三种不同的构型即 A-DNA、B-DNA、Z-DNA，每一种构型各有其独特的分子特征。（1）Which of these forms is the most common form in living cells?在活的生物细胞中，普遍存在的分子构型是哪一种？（2）Z-DNA has an unusual conformation resulting in more base pairs per helical turn th

26、an B-DNA.What is the conformation?Does this molecule have any function in living cells?Z-DNA 与 B-DNA 相比，每圈螺旋含有较多的碱基对数，请进一步说明 Z-DNA 的结构特点，在活的生物细胞中，Z-DNA 具有怎样的功能？（3）Which of these forms is never found in living cells?在活的生物细胞中，哪一种构型没被发现。答案：（1）B-DNA is the form most common to living cells.B-DNA 是活细胞中 DN

27、A 的普遍存在形式。（2）Z-DNA is a long and thin(about 2 nm wide)double helix,like B-DNA.Unlike the right-handed B-DNA however,Z-DNA is left-handed.It has an axis that runs through the minor groove,and has 12 base pairs per helical turn that are inclined 8.8 from a plane perpendicular to the axis.In contrast,B

28、-DNA has an axis that runs through the base pairs,and has 10 base pairs per helical turn that are inclined 2 from a plane perpendicular to the axis.The major groove of Z-DNA is not very distinct,as it is thin and flattened out along the helix surface,while the major groove of B-DNA is wide and inter

29、mediate in depth(between that of A-DNA and Z-DNA).The minor groove of Z-DNA is extremely narrow and very deep,while that of B-DNA is narrow and of intermediate depth.It has been proposed that regions with Z-DNA provide a stretch of left-handed helical turns that are involved in replication,recombina

30、tion and transcription.Stretches of left-handed turns may aid in unwinding right-handed helical turns in B-DNA during these processes.Z-DNA may also be more stable under extreme environmental conditions.与 B-DNA 相比，Z-DNA 比较 A-DNA is found only when the DNA is dehydrated,so it is unlikely that lengthy

31、 sections of A-DNA would be found in living cells.9Benzer 用一般遗传学方法测出 T4 噬菌体 r突变型的一个 cistron 中含有许多个突变子（或重组子）并且指出一个突变子的大小是 1-3 个核苷酸，后来证明这是科学上的一个惊人的预见。请回答：（1）他的工作对基因概念的发展有何意义?（2）你能说出在当时的条件下（没有 DNA 序列分析技术，遗传密码还没发现）用什么样的研究方案能得出一个突变子是 13 个核苷酸的结论。答案：基因是 DNA 分子中含有特定遗传信息的一段核苷酸序列，是遗传物质的最小功能单位。基因内可以较低频率发生基因内的重

32、组，交换。通过大量的成对突变型的杂交，测得其最小的重组频率为 0.02%，即 0.02 个遗传图距。已知 T4噬菌体的基因组是 1.8105bp，其长度为 1500 个图距单位。因此可求算出 0.02 个遗传图距相当于多少核苷酸：1.810515000.02=2.4 bp 因此指出一个突变子是 1-3 个核苷酸，暗示了三联体密码的存在。10比较基因组的大小和基因组复杂性的不同：一个基因组有两个序列，一个是 A，另一个是 B，各有 2000 bp 长。其中一个是由 400 bp 的序列重复 5 次而组成，另一个则由 50 bp 的序列重复 40 次而组成，请问：（1）这个基因组的大小怎样？（2）

33、这个基因组的复杂性如何？答案：基因组的大小是指在基因组中 DNA 的总量。复杂性是指基因组中所有单一序列的总长度。（1）这个基因组的大小为 4000bp；（2）这个基因组的复杂性为 450bp 11 What evidence do we have that in the helical form of the DNA molecule the base pairs are composed of one purine and one pyrimidine?我们有什么证据表明在 DNA 分子的螺旋构型中，在每一个碱基对中都含有一个嘌呤碱基和一个嘧啶碱基？答案：Two different lin

34、es of evidence support the view that a base pair is composed of one purine and one pyrimidine.有两方面的证据支持在 DNA 分子的螺旋构型中，每一个碱基对中都含有一个嘌呤碱基和一个嘧啶碱基（1）When the chemical components of double-stranded DNA from a wide variety of organisms were analyzed quantitatively by Chargaff,it was found that the amount o

35、f purines equaled the amount of pyrimidines.More specifically,it was found that the amount of adenine equaled the amount of thymine,and that the amount of cytosine equaled the amount of guanine.The simplest hypothesis to explain these observations was the existence of complementary base pairing,A on

36、 one strand paired with T on the other strand,and G paired with C.（2）More direct physical evidence was provided by X-ray diffraction studies.These established the dimensions of the DNA double helix and allowed for comparison with the known sizes of the bases.The diameter of the double helix is const

37、ant throughout its length at 2 nm.This is the right size to accommodate a purine paired with a pyrimidine,but too small for a purine-purine pair,and too large for a pyrimidine-pyrimidine pair.12The double-helix model of DNA,as suggested by Watson and Crick,was based on data gathered on DNA by other

38、researchers.The facts fell into the following two general categories;give two examples of each:(1)chemical composition;(2)physical structure.Watson 和 Crick 在其他研究者成果的基础上提出了 DNA 双螺旋模型，这些研究成果可以划分为两类即 DNA 的化学组成与 DNA 的物理结构。请详细阐述这两类研究成果的具体结论。答案：The double-helix model of DNA suggested by Watson and Crick h

39、ad to incorporate existing information about its chemical composition and physical structure.In terms of its chemical composition,it was known that DNA is composed of polynucleotides,that(A)=(T)and(G)=(C)(Chargaffs rules),and that while the percent GC varies between organisms,the A/T and G/C ratios

40、do not.In terms of its physical structure,the structure and molecular dimensions of the component molecules(the bases,sugars,and phosphates)were known.It was also known from studies of Franklin and Wilkins that the molecule is organized in a highly ordered,helical structure,and that there are two di

41、stinctive regularities at 0.34 and 3.4 nm along the molecules axis.化学组成即 DNA 碱基组成的 Chargaff 规则：腺嘌呤和胸腺嘧啶的摩尔数相等，即 A=T；鸟嘌呤和胞嘧啶的摩尔数相等，即 G=C；含氨基的碱基（A 和 C）总数等于含酮基的碱基（G 和 T）总数，即 A+C=G+T；嘌呤的总数等于嘧啶的总数，即 A+G=C+T。DNA 的碱基组成具有生物种的特异性。物理结构：DNA 分子是由核苷酸组成，核苷酸有含氮的碱基、戊糖、磷酸构成。根据富兰克林和威尔金斯的 x 射线衍射图像表明：DNA 分子是一个十分有序的双螺旋结

42、构，每两个相邻碱基平面的垂直距离是 3.4，每个螺旋包含 10 个碱基对。13Hershey-Chase 所做的噬菌体转染实验中，为什么32 P 只标记在 DNA 分子中，而35S 只标记在蛋白质的外壳上？如果用35S 标记的噬菌体去感染细菌，那么在子代病毒中是否会出现带35S 标记的病毒？如果是用32 P 标记的噬菌体重复实验，那么在子代病毒中是否可以找到带32 P 标记的病毒？答案：因为 DNA 分子中含大量磷酸基团，不含硫；而蛋白质中部分氨基酸含硫，但都不含磷，所以32 P 只标记在 DNA 分子中，而35S 只标记在蛋白质的外壳上。如果用35S 标记的噬菌体去感染细菌，那么在子代病毒中

43、不会出现带35S 标记的病毒。因为噬菌体感染细菌时其蛋白质外壳并不进入细菌体内，子代噬菌体的蛋白质外壳是在细菌体内新合成的，因此不带标记。如果用32 P 标记的噬菌体去感染细菌，那么在子代病毒中会出现带35P 标记的病毒。因为噬菌体感染细菌时其核酸部分进入细菌体内，并作为遗传物质传给子代噬菌体，因此在子代噬菌体可以找到带35P 标记的病毒。14The co-crystal structure of ORC（origin recognition complex）bound to DNA shows that 200 bp of DNA is wrapped in 3 full right-han

44、ded turns around the yeast ORC complex.You assemble ORC onto a circular 8 kb plasmid containing an ARS（autonomously replicating sequence）element.Before assembly,the plasmid had an average of 10 negative supercoils.After assembly,you extract the reaction with phenol-chloroform,recover the DNA,and run

45、 it on a gel.(Hint:DNA is wrapped in a left-handed turn around nucleosomes).Please answer：（1）What is the linking number,twist,and writhe of the plasmid after extraction(assume 10.5 bp/turn of relaxed B DNA).答案：W=-10（ten negative supercoils）T=8000bp/10.5 bp per turn=761 or 762 L=T+W=751 or 752 （2）Whe

46、re does the extracted DNA migrate on an agarose gel lacking ethidium bromide?答案：At the position of supercoiled DNA.（3）What will happen to the mobility of the plasmid as you add increasing amounts of ethidium bromide to the gel during electrophoresis?Explain briefly.答案：Since ETBr locally untwists DNA

47、(twist goes down),the plasmid will acquire positive supercoils(writhe goes up).At first,this will result in the loss of negative supercoils and the plasmid will migrate less rapidly.If you add enough ETBr,the plasmid will become positively supercoiled and again migrate at the position of supercoiled

48、 DNA.（4）Now you repeat the experiment,but you include Topoisomerase I in the assembly reaction.Under these conditions,what is the linking number,twist,and writhe of the plasmid after extraction?答案：W=+3 T=8000bp/10.5 bp per turn=761 or 762 L=764 or 765 The 10 negative supercoils present in the plasmi

49、d will be instantly removed by topoisomerase I,but the solenoidal supercoils,constrained by ORC,will not.Assuming that each ORC will generate 3 positive plectonemic supercoils,after protein extraction,W=+3,T=761(since DNA always adopts the most favorable value),L=T+W=764.In addition to just stating the name of the strain,we were looking for at least some explanation of how it would work.