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分子生物学(Molecular Biology)
- Which polymerase is responsible for elongating the leading strand in eukaryotic cell?( )
- The common component composed of DNA and RNA is ( )
- Which enzyme is DNA-dependent RNA polymerase?( )
- Which description about the PCR is not correct?( )
- What is the extension temperature of the PCR?( )
- DNA damages caused by UV radiation in human cells are mainly repaired by ( )
- During mismatch repair in prokaryotes, the newly synthesized DNA is recognized by the lack of ( ) modification
- The corresponding codon to anticodon UCC is( )
- Which of the following sequences is palindrome sequence( )
- The tool enzyme used for combination of foreign DNA and vector is( )
Amount of T in RNA is 0.25mol,so amount of A is_____.
- The technology of isotope-labeled probes to detect the RNA molecule on the NC membrane is called:( )
The secondary structure of DNA is _____
- In most melanoma, oncogenic BRAF was activated by( )
- Among the following factors, the one that does not participate in the translation process of prokaryotes is( )
- Which one of the following changes is not the mechanism of tumor suppressor gene inactivation?( )
- Of all the possible amino acids, all living organisms make use of only the same 20 amino acids. This supports the idea that:( )
- which type of DNA repair does not need DNA synthesis ( ).
- To avoid self-ligation of vector, which of the following methods is commonly used( )
- Which one is not the repair target for BER repair process?( )
- In insertional inactivation selection, Which of the following bacteria can grow in the presence of tetracycline( )
- which type of DNA changedoes not belong to DNA Mutations ( )
- Which of the following eukaryotic ligases involves in filling the gap between the Okazaki fragments( )
- Which statement is TURE about the initiation of DNA replication in prokaryotes?( )
- A 4-year-old child who becomes easily tired and has trouble walking is diagnosed with Duchenne muscular dystrophy, an X-linked recessive disorder. Genetic analysis shows that the patient’s gene for the muscle protein dystrophin contains a mutation in its promoter region. Of the choices listed, which would be the most likely effect of this mutation?( )
- Which is the common structure for hnRNA and mRNA?( )
- Which enzyme elongates the leading strand in E. coli?( )
- Which character is not true for eukaryotic genome?( )
- Amount of G in RNA is 0.25mol,so amount of C is( ).
- Among the following options, the one that belongs to the protein biosynthesis inhibitor is?( )
- All the RNAs below are transcribed by RNA pol I except ( )
- A “mini-gene” has the base sequence TACCCGTGCACG. If the T at the beginning of the sequence is deleted, what will be the consequence?( )
- Please arrange the following proteins by the order of participating in DNA replication( )
① Primase
② Helicase
③ Single-strand binding proteins
④ DNA polymerase I - which bond an amino acid is linked to the tRNA?( )
- The protein factor with translocase activity in eukaryotes is( )
- The base sequence of codons 57-58 in the cytochrome β5 reductase gene is CAGCGC. The mRNA produced upon transcription of this gene will contain the sequence:( )
- Which of the following occurs first in prokayotic transcription?( )
- Which protein factor has translocase activity in prokaryotes?( )
- Given the known sequence of a gene, the most convenient way for the gene isolation is( )
- In DNA, phosphodiester bonds form with hydroxyl groups of carbons number ________ and________ of each deoxyribose sugar.( )
- What is the molecule size of the target band of PCR?( )
- Which statement is true?( )
- In α-complementation selection, the foreign DNA will be inserted into( )
- Which of the following methods is used for virus-mediated transfer of foreign DNA into host cells( )
- Which can recognize the N-terminus of the nascent peptide chain of the secreted protein?( )
- The protein molecules involved in protein folding are?( )
- Which is wrong below as described for directly undoing DNA damages repair?( )
- The source of energy required for peptide chain elongation in prokaryotic protein synthesis is? ( )
- What the signal peptide acts?( )
Single strand DNA 5’-CGGTA-3’ can anneal with the follow RNA _____
- About the operon, which statement is NOT true?( )
- What is the digestion enzyme used in the extraction of plasmids from bacteria (Alkaline Lysis) and enzyme digestion experiment?( )
- DNA molecules Used as probes must:( )
- Given that the chromosomes of mammalian cells may be 20 times as large as those of Escherichia coli, how can replication of mammalian chromosomes be carried out in just a few minutes?( )
- In lac operon, repressor protein could bind to( )sequence.
- What are the lipoproteins obtained in agarose gel electrophoresis of blood lipoprotein experiment?( )
- Protein synthesis in prokaryotic cells is characterized by:( )
- There are three bands obtained in the experiment of extraction of plasmids, what are they?( )
- EF-Tu and EF-Ts are involved in protein biosynthesis:( )
- The final products of gene expression include:( )
tRNA in eukaryotic cell contains _____
- Among the statements about 3’ poly(A) tail which are correct?( )
- Prokaryotic DNA ligases includes( )
- The functions of 5’ Cap include( )
- Reverse transcriptase can( )
- The base in DNA( )
- Which of the following are tumor suppressor genes( )
The base in DNA _____
- .What are the steps of PCR?( )
- Which of the following steps could be regulated to influence the amount of the product of a gene in eukaryotes?( )
- The protein product coded by myc gene is transcription factor.( )
All genomes are composed of RNA.
- Prokaryotic mRNA is generally identical to its primary transcript, whereas eukaryotic mRNA is extensively modified co- and posttranscriptionally. For example, a 7-methylguanosine “cap” is attached to the 5'-terminal end of the mRNA through a 5'→5' linkage.( )
- At the end of the elongation cycle the completed polypeptide chain is moved from the last tRNA onto the ribosome. ( )
- Both prokaryotic and eukaryotic transcriptions are accompanied by translation. ( )
- Real-time PCR technology can dynamically monitor the product quantity in the process of reaction, it also called quantitative PCR.( )
- Gene chip can also be called DNA microarray.( )
Stem-loop is a type of secondary structure not usually found in tRNA.
- The two strands of double-stranded DNA have the same content of each of the bases A, T, G and C.( )
- A transcription bubble contains RNA polymerase, DNA and nascent RNA.( )
- The synthesis of Okazaki fragment requires RNA primer.( )
- Stop codons are never translated. ( )
- Due to semiconservative replication, the base sequences of daughter DNA and parental DNA are identical.( )
- E.coli is the mostly used as host cell for prokaryotic expression system.( )
- Mismatch repair system can increase the DNA replication fidelity by 100-1000 times.( )
A:DNA polymerase α B:DNA polymerase β C:DNA polymerase γ D:DNA polymerase ε
答案:Customer Service
A:D-ribose B:Thymine C:Adenine D:D-2-deoxyribose
答案:Guanine
A:DNA polymerase I B:DNA ligase C:Primase D:DNA polymerase III
AI参考:正确答案是D:DNA polymerase III。DNA-dependent RNA polymerase是指依赖于DNA的RNA聚合酶,而DNA聚合酶I(A错)是一种依赖于模板DNA的DNA聚合酶,DNA连接酶(B错)用于修复断裂的DNA链,primase(C错)主要负责合成RNA引物,而DNA聚合酶III(D对)是细菌中最重要的RNA依赖性DNA聚合酶,它催化DNA的合成。因此,答案为D。'
A:it is an enzymatic reaction B:the product is amino acid sequence C:the purpose of denaturation is to make the template DNA into single DNA D:primers determines the specificity of the amplification
答案:Treat different diseases togetherAI参考:B: the product is amino acid sequencePCR是一种酶促反应,通过特定的酶切位点来扩增特定的DNA序列,因此A选项正确。PCR的产物是双链DNA,而非氨基酸序列,因此B选项错误。PCR的目的之一是使模板DNA变性,以便单链DNA可以被结合到引物上,因此C选项正确。引物在PCR过程中起到至关重要的作用,它决定了扩增的特异性,因此D选项正确。因此,B选项是关于PCR描述中不正确的选项。'
A:94℃ B:95 ℃ C:72℃ D:55℃
答案:55℃
A:NER pathway B:Mismatch repair pathway C:DSB repair pathway D:BER pathway
AI参考:正确答案是A:NER(碱基错配修复)途径。UV辐射会导致DNA损伤,主要通过碱基错配修复途径(Mismatch repair pathway)进行修复。此外,DNA修复系统还有断裂修复(DSB repair pathway)和碱基切除修复(BER pathway)等。但是,DSB修复主要针对断裂的DNA链,而UV辐射引起的DNA损伤通常不会导致断裂,因此不是主要修复方式。因此,选项A是正确答案。'
A:Methylation B:Acetylation C:Deamination D:Phosphorylation
答案:错配修复是指在含错配碱基的DNA分子中,使正常核苷酸序列得以恢复的修复方式.核酸外切酶识别不能形成氢键的错配碱基,并切除一段多核苷酸,缺口由DNA聚合酶Ⅰ修补及DNA连接酶封口.AI参考:正确答案是A: Methylation。在原核生物中的错配修复过程中,新合成的DNA被识别是因为其甲基化修饰的缺乏。B选项的乙酰化修饰、C选项的脱氨化修饰和D选项的磷酸化修饰都不是这个过程中的关键因素。因此,正确答案是A:Methylation。'
A:GGT B:GGA C:AGG D:TGG
AI参考:答案:B推理过程:UCC是反密码子,与密码子相对应的是密码子中没有的三个碱基(A,C,G)中的一个,由于反密码子中没有T,所以选项中也没有T,只有B选项是正确的。希望以上答案可以帮助到你。'
A:AGTCAGTC B:AGTCCTGA C:GACTCTGA D:AGTATACT
AI参考:正确答案是B:AGTCCTGA。Palindrome sequence是指正着读和倒着读都一样的序列。选项A和D是回文序列,但选项C的第一个字符和最后一个字符不同,因此不是回文序列。只有选项B的序列从前往后读和从后往前读都是一样的,因此是回文序列。'
A:DNA ligase B:Klenow fragment C:Restriction endonuclease D:Reverse transcriptase
AI参考:正确答案是A:DNA ligase(DNA连接酶)。DNA连接酶是用于将外源DNA与载体分子连接的酶,从而将外源基因整合到受体细胞染色体上。DNA连接酶有多种类型,其中A型DNA连接酶(即DNA ligase)是最常用的一种。B型DNA连接酶主要用于连接不同来源的DNA片段之间的缺口,但在构建基因表达载体时并不常用。Klenow片段是B型DNA连接酶的一种,主要用于连接不同来源的DNA片段之间的缺口,但不如A型DNA连接酶常用。Restriction endonuclease(限制性内切酶)主要用于切割DNA分子,而不是用于连接外源DNA和载体分子。Reverse transcriptase(逆转录酶)主要用于逆转录病毒的复制过程,而不是用于基因工程的操作。因此,正确答案是A:DNA ligase(DNA连接酶)。'
A:0.5mol B:1.00mol C: 0.25mol D:0.75mol
A:Protein molecule hybridization B:Western Blotting C:Southern Blotting D:Northern Blotting
AI参考:答案为D:Northern Blotting。Northern Blotting是一种用于检测RNA分子的技术,通过将放射性标记的探针与待测RNA分子杂交,从而检测RNA分子的存在和性质。因此,选项D是正确答案。其他选项A、B、C分别代表蛋白质分子杂交技术、Western Blotting和Southern Blotting,这些技术主要用于蛋白质分子的检测。'
A:β-turn B:double helix C:β-sheet D:α-helix
A:Mutation B:LOH C:Hypomethylation D:Amplification
AI参考:在大多数黑色素瘤中,致癌性的BRAF是通过突变激活的。因此,正确答案为A:Mutation。其他选项B(LOH)、C(Hypomethylation)和D(Amplification)在黑色素瘤中不常见,或者与BRAF的激活没有直接关系。'
A:EFT B:eEF1 C:RF D:IF
AI参考:单细胞生物在翻译过程中不会参与的因子是【EFT】。单细胞生物的翻译过程主要涉及核糖体、mRNA、tRNA、氨基酸等物质。其中,eEF1(延伸因子1)、RF(起始因子)和IF(内因子)都是翻译过程中需要的因子,参与氨基酸的正确连接和转运。然而,EFT(早期转录因子)并非翻译过程中的必要因子,它在其他生物(如真核生物)的转录过程中起作用,而非翻译过程。因此,答案是A:EFT。'
A:Hypermethylation B:Amplification C:LOH D:Deletion
AI参考:B. Amplification (增加) 不是肿瘤抑制基因失活的原因之一。其他选项,如 A. Hypermethylation (过度甲基化)、D. Deletion (删除) 和 C. Loss of Heterozygosity (非整倍性丢失) 都是常见的肿瘤抑制基因失活机制。其中,Hypermethylation 是导致基因沉默或失活的主要原因之一,常常出现在肿瘤细胞中;Deletion 是基因片段缺失,导致基因无法表达;LOH 是指等位基因丢失,常常出现在肿瘤细胞中,与肿瘤的发生和发展有关。因此,只有 Amplification (增加) 不是肿瘤抑制基因失活的原因。'
A:Plants and animals have a completely different ancestry B:All living things are derived from a common ancestor C:Bacteria, animals, and plants developed independently D:Only a limited number of DNA mutations are possible
AI参考:所有可能的氨基酸中,所有生物只使用其中的20种氨基酸,这支持了这样一种观点:所有的生物都是从一个共同的祖先进化而来的。因此,答案为B:所有生物都来自共同的祖先。选项A和D与题目无关,选项C则与事实不符。'
A:Light repair B:Base excision repair C:Double stranded DNA break repair D:Nucleotide excision repair
AI参考:答案:B推理过程:DNA repair的三种方式分别进行的情况为:DNA的复制前,完成大部分的修复(例如A);利用断裂部位和未受损部分的碱基互补配对修复断裂(例如C);把损伤部位断开后用非模板进行修复(例如B),没有提到利用脱氧核苷酸合成新的链。而核苷酸切除修复与上述不同,其不涉及DNA的合成。因此,正确答案是B。'
A:LigTo avoid self-ligation of vector, which of the following methods is commonly usedation of sticky ends cut by Isocaudarners B:Ligation of blunt ends C:Ligation of sticky ends cut by two different restriction endonucleases D:Ligation of sticky ends cut by single restriction endonuclease
AI参考:答案:B:Ligation of blunt ends解释:在构建基因文库或基因组文库时,为了避免载体自身发生连接,通常采用的方法是利用平末端(blunt ends)进行连接。选项B正确。选项A中的“Isocaudamer”是一种能够识别不同黏性末端并产生平末端的新技术,但并不是一种常规的连接方法;选项C和D中的方法可能存在错误,因为不同的限制性内切酶通常会切割出不同的黏性末端,不能进行有效的连接。因此,选项B是正确答案。'
A:Alkylated bases B:Apurinic or apyrimidinic sites C:Break of DNA double strands D:8-oxoG
AI参考:正确答案是C:Break of DNA double strands。BER修复过程的目标是去除烷基化的碱基、单链或双链DNA断裂以及8-oxoG等,因此选项C不是BER修复过程的修复目标。'
A:Bacteria transformed with recombinant DNA B:Competent bacteria C:Bacteria transformed with wild vector D:Bacteria transformed with self-cyclized foreign DNA
A:Insertion and Deletion B:DNA Point Mutation C:DNA semi-conservative replication D:DNA Rearrangement
AI参考:正确答案是C:DNA semi-conservative replication。DNA半保守复制属于复制过程,不会导致DNA突变。而插入和删除(A)、点突变(B)和DNA重排(D)都属于DNA突变类型。因此,选项C是唯一不属于DNA突变类型的类型。'
A:Ligase III B:Ligase IV C:Ligase I D:Ligase II
AI参考:单链DNA复制时产生的缺口由连接酶(Ligase)进行修复。在真核生物中,参与填补复制片段之间空隙的连接酶有 Ligase I, Ligase II 和 Ligase IV。其中 Ligase IV 是一个特定的DNA连接酶,专门负责填补DNA链之间的空隙。因此,答案是选项B:Ligase IV。'
A:Primase synthesizes a DNA fragment B:DNA polymerse III synthesizes both leading and lagging strand C:DNA polymerase I removes primer D:DNA helicase unwinds DNA duplex at oriC
AI参考:选项B:DNA聚合酶III合成延伸DNA链,包括启动子和延伸链,这是原核生物DNA复制的正确启动方式。DNA聚合酶I不会去除引物,而引物是在DNA复制开始前合成的。DNA解旋酶是在DNA复制过程中打开DNA双螺旋结构,使得DNA聚合酶III能够进行DNA链的合成。因此,选项B是关于原核生物DNA复制起始的正确陈述。'
A:Termination of dystrophin transcription will be defective. B:Splicing of dystrophin mRNA will be defective. C:Capping of dystrophin mRNA will be defective. D:Initiation of dystrophin transcription will be defective.
AI参考:正确答案是D:Initiation of dystrophin transcription will be defective。根据题干中的描述,一个四岁的孩子出现了容易疲劳、行走困难的症状,被诊断为杜兴肌肉营养不良症(一种X连锁隐性遗传病),基因分析显示肌肉蛋白dystrophin的基因在启动子区域存在突变。因此,该突变最可能影响的是dystrophin基因的转录起始,即启动子区域的突变会导致dystrophin基因的转录起始出现缺陷,进而影响肌肉蛋白的表达。因此,选项D“启动子区域的突变会导致dystrophin基因的转录起始出现缺陷”是正确答案。选项A、B、C都与题干描述不符。'
A:transcript of exon B:promoter C:SD sequence D:transcript of intron
AI参考:正确答案是C:SD序列。hnRNA和mRNA都含有mRNA序列和DNA的SD序列。在原核生物中,启动子一般是指位于启动子结构域上游的序列,但不同的RNA聚合酶转录不同类型的mRNA的启动子也有所不同,而且结构差异较大。mRNA分子一般具有mRNA的共同结构特征,因此选C:SD序列。'
A:DNA polymerase V B:DNA polymerase I C:DNA polymerase II D:DNA polymerase III
AI参考:正确答案是D:DNA polymerase III。大肠杆菌中负责延伸主导链的酶是DNA聚合酶III。DNA聚合酶I、II和V都是大肠杆菌中负责修复DNA损伤的酶,与主导链的延伸无关。因此,答案为D。'
A:The proportion of coding sequences of genes is much smaller than that of non-coding sequences. B:There are multiple gene families and pseudogenes in eukaryotic genomes C:Few genes undergo alternative splicing after transcription D:Higher eukaryotic genomes contain a large number of repeated sequences
AI参考:C:Few genes undergo alternative splicing after transcription 真核生物基因组中没有出现的情况是:在转录之后进行基因剪接的基因数量较少。其他选项都是真核生物基因组的特点,例如编码区序列的比例远小于非编码区序列,真核生物基因组中存在多种基因家族和伪基因,高等真核生物基因组中包含大量重复序列。'
A:0.75mol B:0.5mol C:0.25mol【答案】 D:1.00mol
AI参考:答案:D。RNA中的C的摩尔数为0.25mol × 6 = 1.5mol,所以正确选项为D。'
A:Allopurinol B:Methotrexate C:5-FU D:Kanamycin
AI参考:正确答案是【B:Methotrexate】。属于蛋白质合成抑制剂的是B选项:甲氨蝶呤(Methotrexate)。A选项是嘌呤合成的抑制剂,C选项是嘧啶合成的抑制剂,D选项是抗生素,不属于蛋白质合成抑制剂。'
A:28S rRNA B:18S rRNA C:5S rRNA D:5.8S rRNA
AI参考:正确答案是【A:28S rRNA】。这个题目的选项都是核糖体RNA(rRNA)的种类,其中只有28S rRNA是由RNA pol I转录的。其他的rRNA都是由RNA pol III转录的。所以正确答案是A。'
A:Only the amino acid coded for in that codon will be changed B:RNA polymerase will correct the deletion, and a normal protein will be produced C:All of the codons after that point will be changed D:RNA polymerase will skip that codon, but all the others will be read normally.
AI参考:正确答案是A:“mini-gene”具有碱基序列TACCCGTGCACG。如果开始处的T被删除,那么将会产生怎样的后果?结果只有在该密码子编码的氨基酸会发生变化。这是因为在原序列中,开始处的T只是起到一个启动子的作用,不直接参与编码氨基酸,删除它后不会影响到其他密码子的作用,因此只会改变开始处的氨基酸。因此,答案为A。'
A:①②③④ B:②③①④ C:①③②④ D:②③④①
AI参考:正确答案是C:①③②④。题目要求按照参与DNA复制的顺序排列蛋白质,而参与DNA复制的蛋白质包括引物酶(Primase)、解旋酶(Helicase)和DNA聚合酶(DNA polymerase),其中DNA聚合酶有不同种类,但是只有DNA聚合酶Ⅲ参与复制过程。因此正确的排列顺序为①③②④。选项B中的排列顺序不正确;选项C和D的排列顺序都不符合参与复制过程的顺序。所以答案是C:①③②④。'
A:high energy B:phosphodiester C:ester D:glycosidic
AI参考:正确答案是D:glycosidic。氨基酸与tRNA是通过糖苷键连接的,即D选项中的glycosidic。糖苷键是一种生物化学键,由糖与氨基酸或碱基通过酯化形成的键。因此,D选项是正确答案。其他选项A、B、C都不符合事实。'
A:eIF-4B B:eIF-2 C:eRF D:eEF-2
AI参考:答案选D:eEF-2 。蛋白质因子在真核生物中的移位酶活性是eEF-2。eIF-4,eIF-2,和eRF都是起始因子,而不是蛋白质因子,eIF-4与eIF-2在翻译起始中起作用,eRF在终止密码子上的识别和释放有关。因此,本题答案为D。'
A:CUGCGC B:GCGCTG C:GCGCUG D:CAGCGC
AI参考:单选题】 The base sequence of codons 57-58 in the cytochrome β5 reductase gene is CAGCGC. The mRNA produced upon transcription of this gene will contain the sequence: D:CAGCGC。因为原基因中的碱基序列是CAGCGC(终止密码子不在此序列中),所以翻译后产生的mRNA应该包含同样的序列。因此,选项D CAGCGC是正确的。'
A:formation of an open promoter complex B:promoter clearance C:action of the Rho protein D:formation of a closed promoter complex
A:EFG B:EFT C:IF3 D:RF1
A:PCR B:Genomic DNA library C:cDNA library D:Chemical synthesis
AI参考:正确答案是C:cDNA library。cDNA library是一种用于克隆mRNA基因的方法,通过已知的基因序列,最方便快捷的基因分离方法就是cDNA library。而PCR(聚合酶链式反应)是一种用于扩增特定DNA片段的技术,Genomic DNA library(基因组DNA文库)和化学合成(化学合成法)也是常用的基因分离方法,但不如cDNA library方便快捷。因此,正确答案是C。'
A:3ʹ; 4ʹ B:2ʹ; 3ʹ C:3ʹ; 5ʹ D:2ʹ; 4ʹ
AI参考:正确答案是C:3ʹ;5ʹ。 在DNA中,磷酸二酯键与脱氧核糖碳原子3’和5’上的羟基形成。因此,选项C是正确答案。"
A:4800bp B:640bp C:660bp D:1200bp
AI参考:正确答案是B:640bp。PCR扩增的产物大小取决于引物设计的碱基序列和模板DNA的长度,一般PCR产物大小在600bp到数千bp之间。因此,正确答案是B,即目标带子的分子大小为640bp。'
A:One hnRNA can produce only one type of mRNA. B:During the splicing extrons can form the lariat RNA. C:During the splicing introns can form the lariat RNA. D:There is only one transesterification reaction in the splicing.
A:Lac Z gene B:Ampicillin-resistant gene C:Mutant Lac Z gene D:LacZ’ gene
AI参考:正确答案是D:LacZ’ gene。在α-互补选择中,外源DNA会被插入到LacZ’基因中,这是一种在β-半乳糖苷酶活性检测中使用的互补基因。其他选项中,Ampicillin-resistant gene是用于抗生素抗性的插入,与α-互补选择无关;Mutant Lac Z gene是一种突变型的β-半乳糖苷酶基因,但这不是外源DNA插入的位置;而Lac Z gene是一种用于β-半乳糖苷酶活性的检测基因,也不是插入外源DNA的位置。因此,正确答案是D。'
A:Electroporation B:Transformation C:Micro-injection D:Transduction
AI参考:正确答案是D:Transduction。病毒介导的外源DNA进入宿主细胞的方法称为转导。电穿孔(A)、转化(B)和微注射(C)是另外两种方法,但不是病毒介导的,它们通常用于非病毒载体的导入。因此,本题答案为D。'
A:GTPase B:Transpeptidase C:Signal recognition particle
AI参考:正确答案是C:Signal recognition particle(信号识别颗粒)。Signal recognition particle(信号识别颗粒)能够识别分泌蛋白肽链的N端,从而引导蛋白质进入内质网。因此,答案为C。A选项的GTPase和B选项的转肽酶与识别肽链的N端无关。'
A:Membrane receptors B:Cytoskeletal proteins C:Histones D:Chaperones
AI参考:正确答案是D:Chaperones。蛋白质折叠过程中涉及的蛋白质分子是伴侣蛋白(chaperones),它们帮助新合成的多肽链正确折叠成有功能的蛋白质,并且在许多重要的生命过程中起关键作用,如细胞生长、发育、神经元发育等。膜受体(A)、肌动蛋白丝相关蛋白(B)和组蛋白(C)不是蛋白质折叠所必需的分子。因此,答案是D。'
A:O6-methylguanine methyltransferase can directly remove O6 alkylations on guanine residues by directly undoing DNA damage repair. B:Converting altered nucleotide back to original form. C:Pyrimidine dimer caused by UV can be repaired by directly undoing DNA damage repair process in human cells. D:There is no any DNA synthesis
AI参考:D:There is no any DNA synthesisDNA修复过程中需要DNA合成来修复DNA损伤,因此选项D描述的是错误的,即没有DNA合成来直接修复DNA损伤。其他选项A、B、C描述的都是正确的DNA修复方式或过程。'
A:CTP B:ATP C:GTP D:UTP
AI参考:正确答案是B:ATP。在原核生物蛋白质合成中,肽链延伸所需的能量来源是ATP。ATP是细胞中能量通货,可以用于各种需要能量的生命活动,包括蛋白质合成过程中的磷酸化作用。CTP、GTP和UTP分别是三磷酸核苷酸中的三磷酸腺苷(ATP)、三磷酸鸟苷(GTP)和三磷酸尿苷(UTP)的前体物质,不是肽链延伸的直接能源来源。因此,选项B是正确答案。'
A:Directs DNA replication B:Guide the polypeptide chain through the rough endoplasmic reticulum channel into the endoplasmic reticulum lumen C:Directs protein synthesis D:Guide the polypeptide chain into the nucleus
AI参考:答案:B推理过程:信号肽的作用是引导多肽链通过糙面内质网膜向内质网腔内转移,并形成正确的折叠。所以正确答案为B。'
A:5’-GCCAU-3’ B:5’-GCCAT-3’ C:5’-TAGGC-3’ D:5’-UACCG-3’
A:It could produce a mono-cistronic mRNA B:the genes in an operon usually have related functions C:operon model is an efficient way to achieve coordinate regulation D:It contains at least 2 genes
AI参考:选项D“它包含至少两个基因”是错误的。解析:选项A中提到了单基因cistronsmRNA,这确实是关于操纵基因的一个正确陈述;选项B中提到操纵基因通常与一组相关的基因一起工作,因此通常具有相关的功能;选项C中提到操纵基因模型是一种有效的协调调节方式,这也是正确的。然而,关于操纵基因,最不正确的陈述是它至少包含两个基因,这是D选项的内容。实际上,一个操纵基因可能仅包含一个基因,如细菌的色氨酸操纵子。因此,答案为D:它包含至少两个基因。'
A:Kpn I B:BamH I C:EcoR I D:Hind III
AI参考:正确答案是D:Hind III。根据题目中提到的消化酶在细菌质粒提取以及酶消化实验中应用,我们选择了在DNA鉴定实验中最常用于区分个体遗传背景和标记不同重组位点的Hind III。因此,答案为D。'
A:denature before hybridization B:Longer than 30 nucleotides C:renature before hybridization D:Shorter than 30 nucleotides
A:Eukaryotic DNA polymerases are extraordinarily fast compared with prokaryotic polymerases B:A great many different RNA polymerases carry out replication simultaneously on chromosomal DNA C:Hundreds of replication forks work simultaneously on each piece of chromosomal DNA D:The higher temperature of mammalian cells allows for an exponentially higher replication rate
AI参考:正确答案是A:Eukaryotic DNA聚合酶比原核生物的聚合酶要快得多。哺乳动物细胞中的染色体比大肠杆菌中的染色体大20倍,但复制可以在几分钟内完成。这是因为真核生物DNA聚合酶比原核生物的聚合酶要快得多,因此可以在短时间内完成复制。B、C、D选项都没有提到具体原因,只是泛泛而谈。因此,答案是A。'
A:promoter B:lac Z C:operator D:lac I
A:pre-β lipoprotein B:α lipoprotein C:CM D:β lipoprotein
AI参考:答案为ABD。在血液脂蛋白凝胶电泳的琼脂糖凝胶电泳中,我们能够分离出不同的脂蛋白。这些脂蛋白包括:前β脂蛋白(pre-β lipoprotein)、α脂蛋白(α lipoprotein)、β脂蛋白(β lipoprotein)和乳糜微粒(CM)。因此,答案为A、B、D。C选项中的CM是血浆中最大的脂蛋白,通常不通过琼脂糖凝胶电泳进行分离。'
AI参考:正确答案是:A和C。Protein synthesis in prokaryotic cells is characterized by the ribosome,which is composed of RNA and protein,and by the codon system of translation,which uses nucleotide sequences as a code for amino acid sequences. Prokaryotic cells contain both a cell wall and membrane system,which allow them to be freely suspended in a solution without breaking into smaller fragments. These cells grow quickly and metabolize their nutrients for growth,including glucose and amino acids,through direct uptake and oxidation. 因此,细菌细胞中蛋白质合成的主要特点是:* 由RNA和蛋白质组成的核糖体;* 翻译中使用的密码子系统(使用核苷酸序列作为氨基酸序列的代码);* 含有细胞壁和膜系统的细胞结构,使其能在溶液中自由悬浮而不破裂成更小的碎片;* 快速生长和通过直接摄取和氧化葡萄糖和氨基酸来促进生长的代谢过程。所以,选项A和C是正确答案。'
A:supercoiled plasmid DNA B:chromosomal DNA C:Replication intermediate plasmid DNA D:Linear plasmid DNA
AI参考:正确答案是D:Linear plasmid DNA。实验提取质粒时,可以得到三种质粒形式,分别是线性的、超螺旋的和环状的DNA,分别对应选项D(线性质粒DNA)、A(超螺旋质粒DNA)和C(复制中间体质粒DNA)。其中,染色体DNA是指生物细胞中包含在染色体上的DNA分子,不属于实验提取质粒的产物。因此,选项B不符合题意。'
A:Transcription B:Peptide chain extension C:Ser-tRNAser entry D:Peptide chain termination
AI参考:正确答案是B、Peptide chain extension。EF-Tu和EF-Ts是蛋白质合成中的因子,参与肽链的延伸过程。EF-Tu与tRNA结合,将氨基酸转移到肽链上,而EF-Ts则帮助EF-Tu释放氨基酸。因此,选项B“Peptide chain extension”是正确的答案。选项A“Transcription”与蛋白质合成无关,选项C“Ser-tRNAser entry”是翻译过程中起始步骤的一部分,选项D“Peptide chain termination”则发生在翻译过程中的终止步骤。'
A:tRNA B:mRNA C:rRNA D:protein
AI参考:正确答案是:B:mRNA,C:rRNA。基因表达的最终产物是蛋白质,但是RNA在基因表达过程中也起到了重要作用,包括tRNA(转运RNA)、mRNA(信使RNA)和rRNA(核糖体RNA)。因此,选项B和C都是正确的。选项A是翻译过程中的一个中间产物,选项D是蛋白质的基本单位。因此,正确答案是B和C。'
A:18sRNA B:5sRNA C:28sRNA D:5.8sRNA
A:Some eukaryotic proteins have no 3’ poly(A) tail on their mRNAs. B:Eukaryotic cell uses a special RNA as the template to synthesize it. C:It can protect mRNA from degradation. D:It can enhance translatability of mRNA.
AI参考:正确选项为:C:它可以保护mRNA免于降解。关于3’poly(A)尾的正确陈述有:A:一些真核蛋白质的mRNA上没有3’poly(A)尾。这并不是说所有真核蛋白质的mRNA都没有3’poly(A)尾,因为一些真核mRNA上会有一些其他的结构或者修饰。B:真核细胞使用一个特殊的RNA作为模板进行合成。这里应该指的是tRNA的模板功能,而不仅仅是指一个特殊的RNA作为模板合成3’poly(A)尾。D:不能仅凭此选项确定是否增强mRNA的翻译能力,因为翻译能力的增强需要考虑到其他许多因素,例如蛋白质的结构和功能,翻译过程中酶的作用等。所以只有C选项关于3’poly(A)尾的保护功能是正确的陈述。'
A:T4 ligase B:E. coli ligase C:Ligase I D:Ligase III
AI参考:正确答案是ABD。原核生物DNA连接酶包括T4连接酶(A选项)、大肠杆菌连接酶(B选项)和Ligase III(D选项)。C选项中的Ligase I是线粒体DNA连接酶。'
A:enhancement of the mRNA’s translatability; B:protection of the mRNA from degradation; C:acceleration of the mRNA’s degradation. D:transport of the mRNA out of the nucleus;
A:synthesize DNA using DNA as template B:synthesize DNA using RNA as template C:synthesize RNA using DNA as template D:synthesize RNA using RNA as template
A:A B:C C:G D:T
A:p53 B:Rb C:PTEN D:Ras
AI参考:正确答案是 A,p53和B,Rb。Tumor suppressor genes是能够抑制肿瘤发生的基因,而A选项中的p53和B选项中的Rb都是已知的肿瘤抑制基因。C选项的PTEN是一种蛋白酶体抑制剂,具有抑制肿瘤发生的功能,但不是肿瘤抑制基因。D选项的Ras是一种常见的致癌基因,与肿瘤的发生有关但不是肿瘤抑制基因。因此,正确答案是A和B。'
A:G B:C C:A D:T
A:Transformation B:Extension C:Denaturation D:Annealing
A:mRNA degradation B:Activation of chromosomes C:Transcription initiation D:Post-transcriptional modification
AI参考:在真核生物中,可以通过调控以下步骤来影响某个基因的表达产物(即蛋白质或RNA)的数量:1. Transcription initiation (转录起始):这是基因表达的第一步,调控这个步骤可以影响后续的转录过程。2. Post-transcriptional modification (转录后修饰):包括修饰mRNA(如甲基化、乙酰化等),会影响mRNA的稳定性或翻译效率。然而,选项B(激活染色体)和D(转录后修饰)通常不会直接影响基因表达产物的数量,而选项C(转录起始)则更直接地影响基因表达的开始和后续过程。因此,正确选项是C: Transcription initiation 和 D: Post-transcriptional modification。'
A:对 B:错
AI参考:这个问题的正确答案是A,“对”。根据题目中的描述,myc基因编码的蛋白质产品是转录因子。因此,这个判断题的答案是A,即正确。'
A:错 B:对
A:对 B:错
AI参考:正确答案是A:对。 在原核生物中,mRNA与其初级转录本基本上是相同的,而在真核生物中,mRNA在转录和后转录过程中会经历广泛的修饰,例如在5端通过5→5连接方式附加上一个7-甲基鸟嘌呤“帽”。因此,此题判断正确,选择A。"
A:错 B:对
AI参考:B:对。在延伸周期的末期,完成的多肽链从最后一个tRNA上转移到核糖体上。因此,该判断题为对选项。'
A:对 B:错
AI参考:答案是B:错。无论是原核生物还是真核生物的转录过程都伴随着翻译。原核生物的转录翻译是在同一个复合物中完成的,而真核生物的转录后加工产生的mRNA会被转运到细胞质中继续翻译。因此,本题中的描述是不准确的。'
A:错 B:对
AI参考:判断题正确选项为B:对。Real-time PCR技术可以动态监测反应过程中的产物量,也被称为定量PCR。因此,该题表述正确。'
A:对 B:错
AI参考:答案是B:错。基因芯片又称DNA微阵列,是生物芯片的一种。但是“芯片”一词一般用于描述整个设备的生产过程或原理,所以它实际上比微阵列更加广泛,涵盖了更多领域,不仅仅用于基因研究。因此,“基因芯片”并不是一个特定的称呼。故本题判断错误。'
A:错 B:对
A:对 B:错
AI参考:答案是B:错。双链DNA分子中,碱基A、T、G、C的数量不一定相等,因为它们之间的配对方式是遵循碱基互补配对原则的,两条链中A总是与T配对,G总是与C配对。因此,在双链DNA中,某些碱基可能不存在于某些位置,从而导致它们各自碱基的含量不均等。'
A:错 B:对
A:错 B:对
A:错 B:对
AI参考:答案:B:对。终止密码子(Stop codon)不会进行翻译,它通常出现在肽链的末尾,当终止密码子出现在mRNA上时,就会终止翻译过程,使蛋白质合成停止。因此,该题正确答案是B:对。'
A:错 B:对
A:对 B:错
AI参考:答案是B:错。大肠杆菌通常用于表达一些小分子量的、对生物体无害的蛋白质,而不能用于表达具有生物活性的大分子,如病毒蛋白、抗体等。因此,E.coli并不是最常用于原核表达系统的宿主细胞。'
A:错 B:对
AI参考:正确答案是B:对。Mismatch repair system是一种DNA修复机制,它能够修复DNA复制过程中的错误,从而提高DNA复制的准确性,大约可以提高100-1000倍的复制精度。因此,这个说法是正确的。'
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