分子生物学(双语)1 Genes Are DNA
分子生物学中英文对照

acetyl CoA / 乙酰辅酶A 一种小分子的水溶性代谢产物,由与辅酶A 相连的乙酰基组成,产生于丙酮酸、脂肪酸及氨基酸的氧化过程;其乙酰基在柠檬酸循环中被转移到柠檬酸。
actin / 肌动蛋白,肌纤蛋白富含于真核细胞中的结构蛋白,与许多其他蛋白相互作用。
其球形单体( G2肌动蛋白) 聚合形成肌动蛋白纤丝( F2肌动蛋白) 。
在肌肉细胞收缩时F2肌动蛋白与肌球蛋白相互作用。
activation energy / 活化能(克服障碍以) 启动化学反应所需的能量投入。
降低活化能,可增加酶的反应速率。
active site / 活性中心,活性部位酶分子上与底物结合及进行催化反应的区域。
active transport / 主动转运离子或小分子逆浓度梯度或电化学梯度的耗能跨膜运动。
由ATP 耦联水解或另一分子顺其电化学梯度的转运提供能量。
adenylyl cyclase / 酰苷酸环化酶催化由ATP 生成环化腺苷酸(cAMP) 的膜附着酶。
特定配体与细胞表面的相应受体结合引发该酶的激活并使胞内的cAMP 升高。
allele / 等位基因位于同源染色体上对应部位的基因的两种或多种可能形式之一。
allosteric transition / 变构转换小分子与蛋白质上特定调节部位相结合所引起的蛋白质之三级及(或) 四级结构的改变,其活性随之发生变化。
多亚单位酶的变构调节很普遍。
alpha(α) helix /α螺旋常见的蛋白质二级结构,其氨基酸线性序列叠为右旋螺旋,借助主链上的羧基与酰胺基间的氢键维持稳定。
aminoacyl2tRNA / 氨酰转移核糖核酸用于蛋白合成的氨基酸的激活形式,含有借高能酯键与tRNA 分子上3’2羟基相结合的氨基酸。
amphipathic / 两亲的,兼性的指既有亲水性部分又有疏水性部分的分子或结构。
anaphase / ( 细胞分裂) 后期姐妹染色体(或有丝分裂期的成对同源物) 裂开并分别(分离) 朝纺锤体两极移动的有丝分裂期。
分子生物学---蛋白组学整理英文

proteomicsProteome: 细胞或组织或机体在特定时间和空间上表达的所有蛋白质。
Proteomics: 分析细胞内动态变化的蛋白质组成成分,表达水平于修饰状态,了解蛋白质之间的相互作用于联系,在整体水平上研究蛋白的组成与调控的活动规律。
研究蛋白组学希望达到的目标:By studying global patterns of protein content and activity and how these change during development or in response to disease, proteomics research is poised to boost our understanding of systems-level cellular behaviors. Clinical research also hopes to benefit from proteomics by both the identification of new drug targets and the development of new diagnostic markers.蛋白质组学研究内容:蛋白鉴定,蛋白定量,蛋白相互作用,蛋白修饰。
Why proteomics(为什么研究蛋白组学)•Proteins distinguish various types of cells, since all cells have essentially the same “Genome” their differences are dictated by which genes are active and the corresponding proteins that are made.•Similarly, diseased cells may produce dissimilar proteins to healthy cells.•Post-translational modifications can dramatically alter protein function - the task of studying proteins is often more difficult than genes.What’s MS(mass spectrometry),即质谱的工作原理1.The basic principle of MS is to generate ions from either inorganic or organiccompounds by suitable method, to separate these ions by their mass-to-charge ratio (m/z) and to detect them qualitatively and quantitatively by their respective m/z and abundance.即质谱能够实现不同质量离子的分离和相对定量,m/z(谱图中的x轴)值可以区分出不同的离子,intensity(谱图中的y轴)表示离子的相对丰度。
分子生物学题库 (3)

生 命 科 学 学院 2xxx —2xxx 学年第1学期考试 A 卷答案考生 信 息 栏 ______学院______系______ 专业 ______年级姓名______学号___ 装订线2.What’s alternative mRNA processing? List the four types of alternative mRNAprocessing.同一个mRNA前体,由于加工方式不同,能形成不同的成熟mRNA。
可变mRNA 剪切的方式主要有:利用不同的polyA位点,部分内含子的保留,部分外显子的去除,以及RNA编辑等。
3.How did Meselson and Stahl prove that DNA replication is semi-conservative?用15N 标记DNA(即让细菌在15N的液体中生长到所有的DNA均含15N),然后移到14N的培养基中生长,每一代提取DNA,用氯化铯密度梯度离心,检测样品中DNA的浮力密度,发现得出的结果只能用半保留复制机制才能解释。
4.What’s DNA cloning? What are the major steps of DNA cloning?DNA克隆是将一特定的DNA片段插入具有自我复制能力的载体中,然后转化入可以大量繁殖的宿主中的过程或技术5.How to screen a cDNA expression library to find target genes? List all the possiblemethods.免疫筛选、DNA筛选、菌落PCR筛选,EST测序等6.How does transcription in prokaryotes initiate?RNA聚合全酶-正确识别DNA模板上的启动子--由酶、DNA和核苷三磷酸(NTP)构成的闭合三元起始复合物—开放起始复合物—数次合成流产—RNA合成成功,转录即自此开始。
分子生物学双语复习知识点

生命基本特征(本质):生命是生物体所表现出来的复合现象,包括:自身繁殖,生长发育,新陈代谢,遗传变异,对刺激产生反应等生物学Biology:是一门研究生命的现象与本质及活动规律的科学。
它包揽了生命的各个方面,从生命的化学组成,细胞的结构与功能,个体生物学,生物的多样性,到生物的遗传、进化及生态等方面的完整知识体系。
分子生物学Molecular Biology:它是研究核酸、蛋白质等生物大分子的结构与功能,并从分子水平上阐述蛋白质与核酸、蛋白质与蛋白质之间相互作用的关系及其基因表达调控机制的学科,是人类从分子水平上真正揭开生物世界的奥秘,由被动地适应自然界转向主动地改造和重组自然界的基础学科。
Is a subject to understand the five basic cell behavior patterns (growth, division, specialization, movement, and interaction) in terms of the various molecules that are responsible for them.That is, molecular biology wants to generate a complete description of the structure, function, and interrelationships of the cell’s macromolecules, and thereby to understand why living cells behave the way they do.分子生物学的研究内容Research Contents of Molecular Biology :生物大分子的结构功能研究(结构部分,又称结构分子生物学):包括基因、基因组的结构;DNA 复制、转录、翻译(功能部分);基因表达调控研究(调控部分);DNA重组技术(又称基因工程)Structure and Function of Macromolecules (Structural Part, Also Known as Structural & Molecular Biology);DNA Replication, Transcription, Translation (Functional Part);Regulation of Gene Expression (Control Section);Recombinant DNA Technology (Genetics)F.Miescher就发现了核素(nuclein);Boyer 和Berg等发展了重组DNA技术,完成了第一个细菌基因的克隆;Sanger 等发明了DNA测序技术;Sanger、Maxam和Gilbert先后发明了三种DNA序列的快速测定法;Mullis等发明的聚合酶链式反应(PCR);90年代全自动核酸序列测定仪问世;生物芯片技术是生命科学研究中继基因克隆技术、PCR技术、基因自动测序技术后的又一次革命性技术突破;分子遗传学基本理论建立者Jacob和Monod最早提出的操纵元学说分子生物学的3条基本原理:构成生物体各类有机大分子的单体在不同生物中都是相同的;生物体内一切有机大分子的建成都遵循共同的规则;某一特定生物体所拥有的核酸及蛋白质分子决定了它的属性。
2024年《分子生物学》全册配套完整教学课件pptx

运输功能
如载体蛋白,血红蛋白等 ,在生物体内运输各种物 质。
免疫功能
如抗体蛋白,参与生物体 的免疫应答。
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蛋白质的功能与调控
调节功能
如激素,生长因子等,调节生物 体的生长发育和代谢过程。
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储存功能
如植物种子中的贮藏蛋白,动物体 内的肌红蛋白等,储存能量和营养 物质。
个性化医疗
根据患者的基因信息,制定个 性化的治疗方案。
药物基因组学
预测患者对药物的反应和副作 用,指导合理用药。
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基因治疗的原理与应用
基因治疗的原理
通过导入正常基因或修复缺陷基因, 从而治疗由基因突变引起的疾病。
遗传性疾病的治疗
如视网膜色素变性、腺苷脱氨酶缺乏 症等。
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癌症治疗
利用基因编辑技术,修复或敲除癌症 相关基因,抑制肿瘤生长。
基因表达调控的层次
基因表达调控可分为转录前调控、转录水平调控、转录后调控和翻 译水平调控等多个层次。
基因表达调控的意义
基因表达调控对于生物体的生长发育、代谢、免疫应答等生理过程具 有重要意义,同时也是疾病发生发展的重要因素。
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原核生物的基因表达调控
1 2 3
原核生物基因表达调控的特点
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DNA损伤的修复机制
直接修复
针对某些简单的DNA损伤,如碱 基错配,可通过特定的酶直接进行 修复。
碱基切除修复
通过识别并切除受损碱基,再合成 新的DNA片段进行修复。
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核苷酸切除修复
针对较严重的DNA损伤,如嘧啶 二聚体,通过切除一段包含受损部
分子生物学英文试题

Multiple Choice(1) The attachment site for RNA polymerase in bacteria is called the:a. Initiatorb. Operatorc. Promoterd. Start codon(2) The specificity of bacterial RNA polymerase for their promoters is due to which subunit?a. αb. βc. γd. σ(3) The first protein complex to bind to the core promoter for a protein-coding gene in eukaryotes is;a. RNA polymerase IIb. General transcription factor TFIIBc. General transcription factor TFIIDd. General transcription factor TFIIE(4) Which modification must be made to RNA polymerase II in order to activate the preinitiation complex?a. Acetylationb. Methylationc. Phosphorylationd. Ubiquitination(5) What is the name of the DNA sequence that is located near the promoter of the lactose operon, and which regulates expression of the operon in E. coli?a. Activatorb. Inducerc. Operatord. Repressor(6) Which of the following types of sequence module enables transcription to respond to general signals from outside of the cell?a. Cell-specific modulesb. Developmental modulesc. Repression modulesd. Response modules(7) Which of the following is NOT a type of activation domain?a. Acidic domainsb. Glutamine-rich domainsc. Leucine-zipper domainsd. Proline-rich domains(8) Which of the following is NOT a experiment used to define the site on a DNA molecule to which a protein binds?a. Gel retardation assayb. DNA footprinting assayc. Modification interference assayd. Y east two hybrid assay(9) Which of the following DNA sequences can increase the rate of transcription initiation of more than one gene/promoter?a. Activatorsb. Enhancersc. Silencersd. T erminators(10) Approximately how many base pairs form the attachment between the DNA template and RNA transcript during transcription?a. 8b. 12-14c. 30d. The entire RNA molecule remains base-paired to the template until transcription is finished.(11) Which factor is thought to be most important in determining whether a bacterial RNA polymerase continues or terminates transcription?a. Nucleotide concentrationb. Structure of the polymerasec. Methylation of termination sequencesd. Thermodynamic events(12) What is the role of the Rho protein in termination of transcription?a. It is a helicase that actively breaks base pairs between the template and transcript.b. It id s DNA-binding protein that blocks the movement of RNA polymerase along the template.c. It is a subunit of RNA polymerase that binds to RNA hairpins and stalls transcription.d. It is a nuclease that degrades the 3’ ends of RNA transcripts.(13) Antitermination is involved in regulation of which of the following?a. Operons encoding enzymes involved in the biosynthesis of amino acids with regulation dependent on the concentration of the amino acids.b. Operons encoding enzymes involved in the degradation of metabolites, regulation dependent on the presence of the metabolitec. Genes present in the upstream region of the operond. Genes present in the downstream region of the operon.(14) What is the major transcriptional change that occurs during the Stringent Response in E. coli?a. Transcription rates are increased for most genes.b. Transcription rates are increased only for the amino acid biosynthesis operons.c. Transcription rates are decreased for most genes.d. Transcription rates are decreased only for the amino acid biosynthesis operons.(15) Which of the following is necessary for the RNA endonuclease activity of RNA polymerase that occurs when RNA polymerase is stalled during transcription?a. Rhob. RelAc. GreAd. RNAse H(16) How is the lariat structure formed during splicing of a GU-AG intron?a. After cleavage of the 5’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of the nucleotide at the 3’ splice site.b. After cleavage of the 5’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of an internal adenosine.c. After cleavage of the 5’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of the nucleotide at the 5’ splice site.d. After cleavage of the 3’ splice site, a new phosphodiester bond is formed between the 5’ nucleotide and the 2’ carbon of an internal adenosine.(17) What are cryptic splice sites?a. These are splice sites that are used in some cells, but not in others.b. These are splice sites that are always used.c. These are splice sites that are involved in alternative splicing, resulting in the removal of exons from some mRNA molecules.d. These are sequences within exons or introns that resemble consensus splicing signals, but are not true splice sites.(18) What statement correctly describes trans-splicing?a. The order of exons within an mRNA transcript is rearranged to yield a different mRNA sequence.b. Exons are deleted from some mRNA transcripts but not others.c. Intron sequences are not removed from RNA transcripts and are translated into proteins.d. Exons from different RNA transcripts are joined together.(19) The chemical modification of eukaryotic rRNA molecules takes place in the:a. Cytoplasm.b. Endoplasmic reticulum.c. Nuclear envelope.d. Nucleolus.(20) Which of the following is an example of RNA editing?a. Removal of introns from an RNA transcript.b. Degradation of an RNA molecule by nucleases.c. Alteration of the nucleotide sequence of an RNA molecule.d. Capping of the 5’ end of an RNA transcript.(21) Nonsense-mediated RNA decay (NMD) is a system for the degradation of eukaryotic mRNA molecules with what features?a. NMD degrades mRNA molecules with stop codons at incorrect positions.b. NMD degrades mRNA molecules that encode nonfunctional proteins.c. NMD degrades mRNA molecules that lack a start codon.d. NMD degrades mRNA molecules that lack a stop codon.(22) Which of the following describes RNA interference?a. Antisense RNA molecules block translation of mRNA molecules.b. Double-stranded RNA molecules are bound by proteins that block their translation.c. Double-stranded RNA molecules are cleaved by a nuclease into short interfering RNA molecules.d. Short interfering RNA molecules bind to the ribosome to prevent the translation of viral mRNAs.(23) How are RNA molecules transported out of the nucleus?a. Passive diffusion through the membrane.b. Through the membrane pores in an energy-dependent process.c. Through membrane pores in an energy independent process.d. Through a channel in the membrane that leads to the endoplasmic reticulum.(24) Match protein/RNA with its function (answers can be used more than once or not at all)Spliceosome a. small nuclear ribonucleoproteins (snRNP)microRNAs b. guanylyl transferasemRNA capping c. ribozymeautocatalytic RNA splicing d. dicere. poly-A polymerase(25) Match protein with its function (answers can be used more than once or not at all)JAK a. G-proteinGlucocorticoid receptor b. EndonucleaseRAS c. DNA-binding proteinIF-2 d. RNA binding proteine. Kinase(26) Match lambda gene with its function (answers can be used more than once or not at all)cI a. Anti-terminationN b. Transcriptional repressorCRO c. Transcriptional activatorcII d. Transcriptional terminatore. Translation factorAnswers to practice exam #3How is it possible for microRNAs to regulate eukaryotic gene expression by binding to the 3’ untranslated end of an mRNA ?Binding to the 3’-UTR initiates an RNA cleavage event that removes the polyA tail and begins the mRNA degradation processWhy is attenuation absent in eukaryotic organisms ?Attenuation is the mechanism whereby amino acid biosynthesis operons are regulated by the cellular concentration of the amino acid that is the product of the genes in the operon by transcription termination. The attenuation mechanism requires that translation by ribosomes and transcription occur in the same subcellular compartment. In eukaryotes transcription and translation are carried out in different compartments, so attenuation would not be possible in eukaryo tes.What are the differences between activator and coactivator proteins ?An activator is a DNA binding protein that stabilizes construction of the RNA polymerase II transcription initiation complex. A coactivator is a protein that stimulates transcription initiation by binding nonspecifically to DNA or via protein-protein interactions.Explain what a “modification protection assay” is intended to discover and how it is carried out .Modification protection is a technique used to identify nucleotides i nvolved in interactions with a DNA-binding proteinHow are Caenorhabditis elegans and Drosophila melanogaster good model organisms for development in higher eukaryotes ?Developmental pathways in animals utilize similar regulators, therefore discovery of regulators in lower animals can reveal how development is controlled in higher animals.How does the anchor cell of C. elegans induce the vulva progenitor cells to differentiate into vulva cells? Why do the vulva progenitor cells follow different pathways upon receiving the signal from the anchor cell ?The anchor cell produces a diffusible signal that stimulates differentiation of vulva cells. Different vulva cells undergo different differentiation pathways because they are exposed to differing concentr ations of the signal molecule, and the vulva cells themselves produce secondary signaling molecules that control differentiation in nearby vulva cells.The process of excision of a GU-AG intron and splicing of exons is defined as requiring two transesterification reactions. What does this mean ?A transesterification reaction is the simultaneous cleavage and reformation of a phosphodiester bond. During intron splicing the donor site phosphodiester bond is cleaved and then reformed with the branchpoint nucleotide within the intron, forming a lariat structure. In the second transesterification, the branch point phospodiester bond I cleaved and simultaneously formed between the donor and acceptor sites. The net effect is that there is no change in the number of phosphodiester bonds. During sporulation in Bacillus σE and σF are present in both the prespore and mother cells. How is σF activated in the prespore?Sigma F is activated in the prespore by when it is released from protein-protein interaction with AB. Sigma F is inactive when it is bound to AB.Explain how the iron response protein (IRP) functions to activate expression of Ferritin and at the same time inhibit expression of Transferrin.The iron response protein can bind to iron response elements in RNA only when it is not bound to iron. In the case of ferritin, binding of IRP to the 5’-IRE blocks translation of the ferritin mRNA, so when it is not bound ferritin protein is produced. In the case of transferrin, binding to the 3’-IRE blocks degradation of the transferrin mRNA thereby increasing half life of the mRNA and stimulating transferrin protein production.。
分子生物学试题及答案1

(A) always expressed ; (B) usually expressed unless a signal turns them "off";(C) usually not expressed unless a signal turns them "on"; (D) never expressed12、The X-ray diffraction data obtained by Rosalind Franklin suggested (choose the correct answer):(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.13、To begin DNA replication, a short ___ primer must first be produced.(A) DNA ;(B) RNA ;(C) polypeptide ;(D) histone14、Through their experiments with DNA from the bacterium Escherichia coli , Meselson and Stahl showed that DNA replication is(A) conservative.; (B) dispersive ;(C) duplicative.;(D) semi-conservative15、A mutation changes a CG base pair to an AT base pair. This is a ___ mutation.(A ) transversion ;(B )transition ;(C )transpositional ;(D )translocation16、Which of the following is an example of a nonsense mutation?(A )ACG to ACC ;(B )AUG to UUG ;(C )UAC to UAG ;(D )AAA to UUU17. The genetic code is(A) triplet.;(B) noncontinuous.;(C) overlapping.;(D) all of the above18、Mutations are(choose the correct answer)(A )caused by genetic recombination ; (B )heritably changes in genetic information(C )caused by faulty transcription of the genetic code ; (D ) usually but not always benefical to the development of the individuals in which they occur19、Which of the following is not a class of mutation?(A )frameshift ; (B )missense ; (C )transition ; (D )transversion ;(E )none of the above (meaning all are classes of mutation )20、Utraviolet light usually causes mutations by by a mechanism involving(choose the correct answer)(A )one-strand breakage in DNA ;(B )light-induced change of thymine to alkylated guanine ;(C )inversion of DNA segments ;(D )induction of thymine dimmers ; (E )deletion of DNA segments21、亚硝酸能够诱导DNA 分子中碱基的 反应(A )嘧啶二聚体;(B )脱氨基;(C )烷基化;(D )羟基化作用22、在真核基因表达的调控中, ______调控元件能促进转录的速率。
双语分子生物学

老阚班双语分子生物学题型:填空、名解、选择、判断改错、简答、论述一、绪论1.molecular biology(分子生物学):分子生物学是研究核酸、蛋白质等所有生物大分子的形态、结构特征及其重要性、规律性和相互关系的科学。
2.DNA发现者:James Watson and Francis CrickDNA发现的实验:图1-1(P7)、图1-2(P8)3.DNA全称——Deoxyribonucleic AcidRNA全称——Ribonucleic Acid4.分子生物学主要研究内容:◆DNA重组技术◆基因表达调控研究◆生物大分子的结构功能研究—结构分子生物学◆基因组、功能基因组与生物信息学研究5.DNA recombinant technique(DNA重组技术):是20世纪70年代初兴起的技术科学,目的是将不同DNA片段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二、染色体与DNA1.Nucleosome(核小体):核小体是染色体结构的基本单位、由H2A、H2B、H3、H4各两个分子生成的八聚体和由大约200bp的DNA组成的。
八聚体在中间,DNA分子盘绕在外,而H1则在核小体的外面。
每个核小体只有一个H1。
2.C值反常现象(C-value paradox):C值往往与种系进化的复杂程度不一致,某些低等生物却具有较大的C值。
3.关于DNA结构的词汇:double–helix(双螺旋)、base(碱基)sugar-phosphate backbonts(糖-磷酸骨架)、base pairs(碱基对)right/left-handed double-helix(右/左手双螺旋)4.DNA类型:右手螺旋:A-DNA,B-DNA左手螺旋:Z-DNA5.组蛋白:Octameric core (H2A、H2B、H3、H4)+外围(H1)6.真核生物重复序列类型:单拷贝、轻度(或低度)重复、中度重复、高度重复(四个空)不重复序列、中度重复序列、高度重复序列(三个空)7.真核生物基因组的特征:1)真核基因组庞大2)存在大量的重复序列3)大部分为非编码序列4)转录产物为单顺反子5)断裂基因,有内含子结构6)存在大量的顺式作用原件7)具有端粒结构8)细胞器基因, 转录和翻译在时间和空间上是不偶联的;9)有假基因化现象8.Z-DNA的特点:1)left-handed double helix with a zig-zag conformation of thebackbone (less smooth than B-DNA)2)Narrower, more elongated helix than A or B.3)Only one groove is observed, resembling the minor groove4) A high G-C content favours Z conformation.5)Z-DNA formation occurs during transcription of genes, attranscription start sites near promoters of actively transcribed.6)Base pairs nearly perpendicular to helix axis9.DNA复制所需要的元素(Requirements for DNA replication):1.原料(Precursor):四种脱氧核苷三磷酸(dATP、dGTP、dCTP、dTTP)2.模板(Template):以DNA的两条链为模板链,合成子代DNA3.引物(Primer):DNA的合成需要一段RNA链作为引物4.酶:a)引物合成酶(引发酶)(Primases)、b)DNA聚合酶(DNA polymerase)c)DNA连接酶(Ligases)d)DNA 拓扑异构酶(DNA Topisomerase)e)DNA 解螺旋酶/解链酶(DNA helicase)10.复制的几种方式(Modes of DNA replication):a)线性DNA双链的复制b)环状DNA双链的复制:1)Theta replication2)Rolling Circle Replication3) D loop11.The steps of BERa)Damaged baseb)DNA glycosylase recognize the damaged basec)Remove the base by hydrolysis of the N-glycosidic bondd)The AP endonuclease removes the AP site and neighboringnucleotidese)The gap is filled by DNA polymerase I and DNA ligase12.转座子(transposon, Tn):是存在于染色体DNA上可自主复制和位移的基因单位13.转坐(transposition):DNA转座是由可移位因子介导的遗传物质重排现象, 又称移位。
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FIGURE 07: Eukaryotic cells can acquire a new phenotype as the result of transfection
Basic DNA Structure
• Nucleotide – phosphate, sugar, base • Bases – pyrimidines and purines • Phosphodiester bonds – phosphate/sugar
FIGURE 20: Nucleic acids all replicate by synthesis of complementary strands
FIGURE 21: The amount of nucleic acid in the genome varies over an enormous range
– proofreading – mutations
FIGURE 15: Replication of DNA is semiconservative
FIGURE 16: The replication fork
FIGURE 17: An endonuclease cleaves a bond within a nucleic acid
FIGURE 13: The two strands of DNA form a double helix
© Photodisc
DNA Replication Basics
• Origins of replication • Semi-conservative • Denaturation, annealing, extension • DNA Polymerases
FIGURE 03: Griffith experiment
FIGURE 04: The DNA of S-type bacteria can transform R-type bacteria into the same S-type
FIGURE 06: The genetic material of phage T2 is DNA
backbone • 5’ to 3’ direction, antiparallel double helix • Base pairing • Minor versus major groove in B-DNA • 3.4 nm and 10.4 base pairs per turn • Alternative DNA structures
– Depends on complementarity
• Intermolecular or intramolecular • Hybridization is the key to many methods
FIGURE 23: Denatured single strands of DNA renature to give the duplex form
FIGURE 22: Base pairing
Hybridization
• Nucleic acids hybridize by base pairing • Denaturation is ds to ss
– Tm depends on sequence
• Renaturation is ds to ss
分子生物学(双语)1 Genes Are DNA
Figure CO: X chromosome in blue
© Science VU/DOE/Visuals Unlimited
FIGURE SO: Part I: Genes and Chromosomes
Photo courtesy of S. V. Flores, A. Mena, and B. F. McAllister. Used with permission of Bryant McAllister, Department of Biology, University of Iowa.
DNA as the Genetic Material
• Pneumococcus pathogenicity tranferred to normal strain by transformation
• T2 bacteriophage genetic material is DNA • Transfection of eukaryotic cells with TK+
FIGURE 19: The central dogma
Control of Gene Expression
• Processes affected by control • Complexity from control • Regulatory elements
cis-acting trans-acting
FIGURE 08: A polynucleotide chain
FIGURE 12: Flat base pairs lie perpendicular to the sugar-phosphate backbone
FIGURE 14: Base pairing provides the mechanism for replicating DNA
FIGURE 18: An exonuclease removes bases one at a time
Gene Expression
• Transcription
– Reverse transcription
• RNA processing • Translation • Post-translational modifications
© Mopic/ShutterStock, Inc.
FIGURE CO: Genes are DNA
FIGURE 01: A brief history of genetics
FIGURE 02: A gene codes for an RNA, which may code for protein