尔雅Molecular Biology_3课后答案(学习通2023题目答案)
62 min read尔雅Molecular Biology_3课后答案(学习通2023题目答案)
1-1 The past and present journey of molecular biology随堂测验
1、The课后 HGP (Human Genome Project) was finished in:
A、1990
B、答案2001
C、学习2008
D、通题2006
2、目答Which of the following technique is 尔雅developed by Paul Berg who is named as “the father of modern genetic engineering”:
A、Technique of hybridization.
B、课后Technique of recombination DNA.
C、答案Technique of monoclonal antibody.
D、学习PCR
3、通题Which one in the following can be used for DNA sequencing:
A、目答Gersang method.
B、尔雅X-ray diffraction method.
C、课后Sanger dideoxy-mediated chain termination method.
D、答案Site-mutagenesis.
4、Two important experiments to prove that DNA is genetic material are: virulence transformation experiment of Diplococcus pneumonia and infection experiment of T2 phage to E.coli. The key point of these two experiments is:
A、DNA isolated from infected organisms can be used as pathogenic agents.
B、It is DNA mutation that cause the loss of virulence.
C、Foreign DNA (not protein) absorbed by organisms changed genetic potency.
D、DNA cannot be transferred between different organisms, so it must be a very conservative molecule.
5、In 1953, Watson and Crick found that:
A、Two polynucleotide chains form double helix structure of DNA by hydrogen bond linkage
B、DNA replication is semi-conservative and usually form heteroduplex DNA consisted of one parental chain and one daughter chain
C、Three continuous nucleic acids consist of a genetic code
D、Genetic material is usually DNA instead of RNA
6、The finding of Okazaki fragments proves that the replication of DNA is:
A、semi-conservative replication.
B、continuous replication.
C、indirect replication.
D、semi-continuous replication.
7、7.Based on “Central Dogma”, the direction to pass biological information is:
A、DNA>Protein>RNA
B、RNA>DNA> Protein
C、DNA>RNA> Protein
D、Protein >DNA>RNA
1-2 The nature of genetic material随堂测验
1、The basic component of nucleic acid is:
A、phosphate and ribose
B、nucleoside and nitrogen bases
C、mononucleotide
D、deoxyribose and nitrogen bases
2、To form a nucleic acid, the linkage between nucleotides is:
A、glycosidic bond
B、hydrogen bond
C、3’, 5’-phosphodiester
D、peptide bond
3、The content of which chemical elements is usually constant in a nucleic acid:
A、C
B、O
C、N
D、P
4、Which of the following is not included in a nucleic acid:
A、C
B、O
C、N
D、S
5、Which of the following bases is not included either in DNA nor RNA:
A、adenine
B、xanthine
C、thymine
D、uracil
6、In the free nucleotide, phosphate group is usually located in:
A、C2 of five-carbon sugar
B、C3 of five-carbon sugar
C、C4 of five-carbon sugar
D、C5 of five-carbon sugar
7、When DNA is completely hydrolyzed, the products are:
A、ribose and phosphate
B、deoxyribose and bases
C、ribose, phosphate and bases
D、deoxyribose, phosphate and bases
8、When DNA and RNA is completely hydrolyzed, the products are:
A、different sugar and partly different bases
B、different sugar but same bases
C、same sugar and same bases
D、different sugar and completely different bases
9、Which of the following is not included in DNA：
A、dAMP
B、dGMP
C、dCMP
D、dUMP
10、The linkage between five-carbon sugar (shorted as R), bases (shorted as N) and phosphate (shorted as P) in a nucleic acid is:
A、N-R-P
B、N-P-R
C、P-N-R
D、R-N-P
11、Basically, the primary structure of a nucleic acid is:
A、the sequence of bases in the polynucleotide chain
B、the way of bases pairing in the polynucleotide chain
C、the ratio of bases in the polynucleotide chain
D、the way of coiling and folding in the polynucleotide chain
12、What force holds complementary base pairs together in the nucleic acid:
A、hydrogen bond
B、glycosidic bond
C、phosphodiester
D、peptide bond
1-3 Structure of DNA double strain helix随堂测验
1、About the secondary structure of DNA, which one is false:
A、Three hydrogen bond are formed between A and T, two hydrogen bond are formed between G and C.
B、Nitrogen base groups face the inside of double helix in DNA.
C、There is base stacking force between base pairs.
D、Two strands are oppositely oriented.
2、Which is the main form of DNA naturally:
A、A-DNA
B、B-DNA
C、E-DNA
D、Z-DNA
3、The secondary structure of DNA is:
A、α- helix
B、β- sheet
C、β- turn
D、double helix
4、Which of the following is false about the content of bases in DNA:
A、A+T=C+G
B、A+G=C+T
C、G=C
D、A=T
5、Which of the following is not included in DNA：
A、phosphodiester
B、glycosidic bond
C、hydrogen bond
D、disulfide bond
6、Which of the following is false about the double helix structure of DNA:
A、Nitrogen bases are strictly paired in double helix.
B、The sugar and phosphate groups face outward of double helix.
C、Base plane is perpendicular to the central axis.
D、DNA is found in the right-handed configuration in all organisms.
7、Basically, the primary structure of a nucleic acid is:
A、the sequence of bases in the polynucleotide chain
B、the way of bases pairing in the polynucleotide chain
C、the way of coiling and folding in the polynucleotide chain
D、the linkage between groups in polynucleotide chain
8、Which of the following is false about the structure of nucleic acid:
A、There are major and minor grooves in the surface of double helix.
B、Base stacking force exists between neighboring base pairs in double helix DNA.
C、Double helix structure only exist in DNA.
D、Double helix structure can also exist in RNA.
9、The force contributing to DNA stability mainly include:
A、phosphodiester
B、hydrogen bond between base pairs
C、base stacking force
D、the repulsive force of negatively charged phosphate backbone
10、Which of the following is right about the double helix structure of DNA:
A、A=T
B、Nitrogen bases face the inside of double helix in DNA
C、10bp per turn, the pitch is 3.4nm in double helix.
D、Horizontally, DNA stability is maintained by hydrogen bond between base pairs.
11、Which of the following is correct about the nitrogen bases in DNA:
A、A=G, C=T
B、The composing and ratio of bases in DNA is different in different organisms.
C、The composing and ratio of bases in DNA is different in different organs of one organism.
D、The composing of bases in DNA is constant with the age increase.
1-4 Supercoiled DNA随堂测验
1、Which the following is false about the tertiary structure of DNA：
A、Supercoil
B、Loops formed within a single strand of DNA
C、Concatenated circle in circular DNA
D、Double helix of DNA
2、Which the following is false about the supercoil structure of DNA：
A、It’s the character of closed circular DNA (Once a single strand or double strands break, the supercoil of DNA is destroyed)
B、Topoisomerases in the cell can regulate the level of supercoil of DNA.
C、The formation of DNA supercoil needs DNA ligase.
D、Almost all DNA molecules in the cell is negatively supercoiled.
3、Which the following is correct about the topoisomerase II：
A、It can break the glycosidic bond in DNA double strands.
B、It doesn’t consume ATP when it works.
C、It can increase the number of negatively supercoiled DNA.
D、It can stabilize the topological structure of DNA molecule.
4、A circular DNA with B form reduces its crossing number by 1 with the help of topoisomerase I, then which the following is false:
A、The linking number of this DNA is reduced by 1.
B、No ATP is consumed in this process.
C、This DNA can maintain its relaxed B form.
D、A negatively supercoiled DNA would be formed if it maintains its relaxed B form.
1-5 DNA Denaturation, renaturation and hybridization随堂测验
1、Which is the maximum UV absorption wavelength of nucleic acids:
A、280nm
B、260nm
C、240nm
D、220nm
2、Which of the following statement is correct about the reason that nucleic acid has the ability to absorb UV:
A、The conjugated double bond between purine and pyrimidine
B、Ribose is connected to purine and pyrimidine
C、There are nitrogen atoms inside purine and pyrimidine
D、Ribose and phosphate is connected to purine and pyrimidine
3、DNA denaturation refer to：
A、Depolymeriztion of polynucleotide chains
B、DNA molecule turns to double helix from supercoil
C、The phosphodiester bond in the DNA molecule has been broken
D、The hydrogen bond between bases has been broken
4、The hybridization of nucleic acid could happen between DNA and RNA, DNA and DNA, then, which of the following RNA sequence could hybridize with single strand DNA with sequence 5’-CGGTA-3’:
A、5’-UACCG-3’
B、5’-GCCAU-3’
C、5’-GCCUU-3’
D、5’-AUCCG-3’
5、Which of the following DNA molecule has a relatively higher Tm value?
A、5’-ATATCATATGATATGTA-3’
B、5’-CGGTACTCGTGCAGGT-3’
C、5’-CGGTATTCGTGCAGGT-3’
D、5’-CCGTACTCGTGCAGGT-3’
6、Which of the following statements are correct about Tm value of DNA：
A、It is directly related to the bases sequence order of DNA
B、It is related to the length of DNA
C、It is directly proportional to the percentage of G-C base pairs
D、The higher G+C/A+T value is, the higher Tm value is
7、Under the melting temperature, what happened to the DNA double helix：
A、The double helix has been untied completely
B、The double helix has been untied about 50%
C、The UV absorption at 260nm increased
D、Part of hydrogen bonds between bases have been broken
8、Which of the following reactions are included in renaturation process：
A、The formation of the hydrogen bond
B、The formation of the phosphodiester bond
C、The formation of the phosphoester bond
D、The formation of the base stacking force
9、Which of the following statement about DNA renaturation is correct：
A、The renaturation takes place among two complementary strands of the denatured DNA molecule
B、The bigger of the DNA molecule is, the longer of the renaturation time will be
C、The heat-denatured DNA would renature only if it is cooled down slowly
D、The denaturation could happen among DNA and RNA strands
10、Which of the following statement about nucleic acid hybridization is correct:
A、It could happen between two DNA strands from different sources
B、It could happen between DNA and RNA strands from different sources
C、Hybridization process is based on the DNA denaturation and renaturation
D、The hybridization techniques could be used to study the structure and function of nucleic acids.
1-6 Nucleic acids extraction and gel electrophoresis随堂测验
1、What is the purpose of using liquid nitrogen to freeze and grind the sample during genome extraction of plants:
A、complete grinding of the sample
B、polysaccharide precipitation
C、protein denaturation
D、DNA precipitation
2、What is the function of CTAB in the genome extraction of plants:
A、complete grinding of the sample
B、polysaccharide precipitation
C、protein denaturation
D、DNA precipitation
3、What is the purpose of adding organic solvent mixture of phenol/chloroform/isoamyl alcohol during genome extraction:
A、complete grinding of the sample
B、polysaccharide precipitation
C、protein denaturation
D、DNA precipitation
4、What is the purpose of adding isopropanol:
A、complete grinding of the sample
B、polysaccharide precipitation
C、protein denaturation
D、DNA precipitation and elimination of small RNA molecules
5、Which of following is the best choice for storage of extracted DNA sample:
A、ultrapure water
B、TE buffer
C、75% ethanol
D、TBE buffer
6、Which of the following statement is correct about DNA agarose gel electrophoresis:
A、When melting the gel, it has to be heated thoroughly, with boiling for a while, then stop heating to transfer the melted gel into the gel tank.
B、The DNA bands could be observed directly after gel electrophoresis, since the loading buffer contains indicator.
C、Linear DNA molecules with different length hold different electrophoretic mobility.
D、DNA molecules with same length hold the same electrophoretic mobility.
Chapter 2 From gene to chromosomes
2-1 Genomes are not equal to chromosomes随堂测验
1、Which of the following statement about a gene is NOT true:
A、The whole DNA sequence responsible for synthesis of a functional protein or RNA molecule.
B、Polynucleotide sequence of DNA molecule that contains specific genetic information.
C、The minimal functional unit of genetic material.
D、The gene sequence consists of continuously arranged triplet codes.
2、Which is correct about organelle genome:
A、Several copies of the gene can be included in one organelle.
B、Consists of several chromosome.
C、Contains numerous short repeated DNA sequences.
D、DNA molecule in mitochondria and chloroplast is usually packaged with histones, to form chromatin structure.
3、Which of the following is NOT the characteristic of eukaryotic genome:
A、The genome of eukaryotes is much bigger than that of prokaryotes, with multiple replication origins, and the length of each replica is smaller in eukaryotes.
B、Somatic cell is usually the diploid cell with two copies of homologous genome.
C、The number of coding region is more than non-coding region.
D、The eukaryotic genes are usually discontinuous split genes consisting of exons and introns.
4、Which of the following is NOT the focus of human structural genomics:
A、Genetic map
B、Transcription map
C、Physical map
D、Analysis of gene function
5、Which of the following statement is the characteristic of prokaryotic genome:
A、Contains multiple DNA replication origins.
B、Single chromosome, always circular.
C、Chromosomal DNA is always fixed to binding proteins.
D、Consists of numerous non-coding sequences.
6、The genome of the virus can be:
A、dsDNA
B、ssDNA
C、dsRNA
D、ssRNA
7、A gene is the DNA sequence encoding proteins (enzymes).
8、The chemical basis of most genes in nature is DNA.
9、A split gene refers to that two genes are separated by some non-coding DNA sequences.
10、Plasmid contains origin of replication and can replicates independently in the host cell.
11、Chromosomal DNA of the eukaryotes is circular in cell.
2-2 Nucleosomes and their Assembly随堂测验
1、The length of DNA fragment in each nucleosome is about:
A、100bp
B、200bp
C、300bp
D、400bp
2、The histone with most species specificity in the nucleosome is:
A、H1
B、H2A
C、H3
D、H4
3、The basic unit of eukaryotic chromosome is:
A、30nm fiber
B、Nucleosome
C、Core particle of nucleosome
D、Histone octamer
4、Which of the following is NOT included in the core histone octamer:
A、H1
B、H2A
C、H3
D、H4
5、Which of the following statement about the nucleosome is NOT ture:
A、About 146bp of DNA sequence is included in each nucleosome.
B、Consists of chromatosome and linker DNA.
C、Core histones contain H2A, H3, H4 and H2B
D、H1is linker histone.
6、After DNA is packaged into nucloesome to form beads-on-a-string, the compression ratio is:
A、1/6
B、1/7
C、1/10
D、1/8400
2-3 Higher order chromatin structure随堂测验
1、The main component of chromosome scaffold is
A、histones
B、non-histone
C、DNA
D、RNA
2、The loop domains of chromosome consist of:
A、DNA fiber
B、10nm fiber
C、30nm fiber
D、Chromatid fiber
3、In solenoid model of chromosome, every ( ) nucleosomes per turn are included in one solenoid
A、2
B、4
C、6
D、8
4、Which of the following is correct about the assembly of chromosome:
A、nucleosome→core histone octamer→solenoid→microstrip→loop domain→chromatid
B、core histone octamer→nucleosome→solenoid→loop domain→microstrip→chromatid
C、core histone octamer→nucleosome→microstrip→solenoid→loop domain→chromatid
D、core histone octamer→nucleosome→solenoid→microstrip→loop domain→chromatid
5、The relationship of chromosome and chromatin is:
A、They are same substance in the same period of cell cycle.
B、They are different substances with different morphologies.
C、They are same substance with same morphology.
D、They are same substance in the different period of cell cycle.
6、DNA is compressed by ( ) folds from double strand molecule to 30nm fiber.
A、6
B、10
C、40
D、240
7、Which of the following statement is NOT true about the N terminal of histone and its modifications as well as its functions in nucleosome assembly:
A、The N terminal of core histone contains several amino acid residues with positive charge, which can interact with DNA molecule and stabilize the nucleosomes.
B、The Ser and Thr residues in the N terminal of core histone can be phosphorylated, to reduce the interaction of histone and DNA molecule.
C、The acetylation in the N terminal of core histone can stabilize the interaction of histone and DNA.
D、The N terminal of core histone can interact with histone octamer of the neighboring nucleosome, to stabilize 30nm fiber of chromosome.
8、Which of the following structures or components contribute to the process that nucleosome form stable 30nm fiber.
A、Four-helix bundle formed by core histone.
B、N terminal of core histone.
C、Histone H1
D、Core particle
2-4 Chromatin modification and remodeling随堂测验
1、Which of the following chromatin remodeling complex can make DNA molecule slide on core histone and shorten the distance of neighboring DNA.
A、NuRD family
B、ISWI family
C、SWR1
D、SWI/SNF family
2、Which of the following statement about the location of nucleosome is NOT true:
A、The interactions between nucleosome and DNA are dynamic.
B、The interactions of nucleosome and DNA are not sequence specific.
C、A nucleosome consists of DNA molecule with length of at least 150bp.
D、The DNA sequences don’t affect the location of nucleosome at all.
3、Which of the statement about the modification of histone is correct:
A、The acetylation of Lyc residue in the N terminal of histone can strengthen the interaction between different histones.
B、Chromo domain can recognize the acetylation site of histone.
C、The acetylation and phosphorylation of histone reduce the affinity between histone and DNA.
D、Bromo domain can recognize the methylation site of histone.
4、Which of the following statement about the chromatin remodeling is NOT true:
A、It is about the change of the structure and position of chromatin.
B、The location of nucleosome in DNA strands keeps unchanged.
C、It is closely related to the modifications in the N terminal of histone.
D、The complex for chromatin remodeling possess ATPase activity and can break the interaction of DNA and histone.
5、Which of the following statements about the chromatin remodeling complex are correct:
A、The chromatin remodeling complex consists of SWI/SNF, ISWI and SWR1 family.
B、The chromatin remodeling complex contains domains for recognition of modification site of histone.
C、SWR1 family can replace the H2A-H2B dimer of the nucleosome to H2A.Z-H2B dimer.
D、ISWI family can make DNA molecule slide on the core histone octamer, changing the arrangement of nucleosome and shortening the distance of neighboring DNA.
Chapter 3 Who is in pairs? DNA replication
3-1 Semiconservative DNA replication随堂测验
1、The technique that confirms semiconservative replication of DNA is:
A、Sanger method
B、density gradient centrifugation
C、α-complementation
D、Western blot
2、A double-stranded DNA fragment that is isotopically labeled with 15N, was cultured for n generations in the medium containing 14N. So the number of double-stranded DNA containing pure 14N is:
A、2^n+1
B、2^n-1
C、2^n-2
D、2^n
3、Which substrate in the following can be catalyzed by DNA polymerase to form DNA molecule:
A、dNTP
B、dNDP
C、dNMP
D、NTP
4、Which of the following fragment is complementary with TAGCAT during DNA replication:
A、TAGCAT
B、ATGCTA
C、ATCGTA
D、AUCGUA
5、The role of RNA primer during DNA replication is
A、to guide DNA polymerase to bind with DNA template
B、to provide 5’ phosphate end
C、to provide attachment sites for four kinds of NTP
D、to provide 3’-OH end to initiate synthesis of new DNA
6、The correct statement about DNA semiconservative replication is:
A、The substrate is four kinds of dNMP
B、The nucleotide sequences of two chains in the nascent DNA are identical
C、Both DNA polymerase and RNA polymerase are required during replication
D、The direction of synthesis for the new chain is 3’→5’ during replication
7、The direction of template during DNA replication is:
A、3’→5’
B、5’→3’
C、N-terminal → C-terminal
D、C-terminal → N-terminal
8、Semiconservative replication of DNA refers to synthesis of nascent DNA using parent DNA as template, and the nascent DNA would contain one parental strand as well as a newly-synthesized strand.
9、The nucleotide sequences of nascent DNA stand are identical to the parent DNA when read from 5’→3’ direction.
3-2 The mechanism of DNA polymerase随堂测验
1、The DNA polymerase responsible for chromosomal DNA replication in E. coli is:
A、DNA polymerase I
B、DNA polymerase II and DNA polymerase III
C、DNA polymerase V
D、DNA polymerase I and DNA polymerase III
2、The correct statement about DNA polymerase III of prokaryotes is:
A、with the activity of 3’→5’ exonuclease
B、with the activity of 5’→3’ exonuclease
C、with the activity of 3’→5’ polymerase
D、only add substrate in the 3’ end of deoxyribonucleic acid
3、In prokaryotes, filling the gaps generated by DNA replication or DNA repair mainly depends on
A、nuclease H
B、DNA polymerase I
C、DNA polymerase II
D、DNA polymerase α
4、Which of the following statement about DNA polymerase of eukaryotes is NOT true:
A、DNA pol α/primase are involved in initial synthesis of the nascent strand.
B、DNA polδ and ε are mainly responsible for the synthesis of DNA chains.
C、Replacement of DNA polymerase could happen during DNA synthesis process.
D、There are five kinds of DNA pol：α、β、γ、δ and ε in eukaryotes
5、The fidelity of DNA replication do NOT include:
A、The instant proofreading of DNA pol δ in eukaryotes
B、The instant proofreading of primase
C、The selection of DNA pol for different bases
D、The strict rule of base complementary pairing
6、In prokaryotes, which of the following enzymes may be functional both in the processes of removing RNA primer and adding deoxyribonuleotides:
A、DNA polymerase III
B、DNA polymerase II
C、DNA polymerase I
D、exonuclease MFI
7、Which of the following are not functions of the core of DNA pol:
A、Synthesis of DNA in 5’→3’ direction
B、Synthesis of DNA in 3’→5’ direction
C、Proofreading
D、Ligation of Okazaki fragments
8、8. If the DNA polymerase in E.coli lost its function of 3’→5’ exonuclease, the synthesis of DNA would be slowed down, while the fidelity of DNA would not be affected.
9、9. In eukaryotes, DNA pol α and primase can form a complex, to initiate primer synthesis and replacement of polymerase, then DNA pol α continue to synthesize the leading strand.
3-3 Process of DNA replication随堂测验
1、The enzyme that releases the positive supercoiled structure of DNA is:
A、Primase
B、Helicase
C、Gyrase
D、Telomerase
2、Which of the following statement about the role of helicase in DNA replication is NOT true:
A、to open hydrogen bond between the base pair
B、will consume ATP to provide energy
C、It will bind to the DNA double helix
D、Its activity is activated by DNA replication complex
3、The function of DNA topoisomerase is:
A、Identifying the origin of replication
B、Straightening out the DNA strand during replication
C、Stabling DNA molecule topological conformation
D、Opening the hydrogen bond between the DNA double helix
4、The wrong statement about the semi-discontinuous synthesis of DNA is:
A、Leading strand is synthesized continuously
B、Lagging strand is synthesized discontinuously
C、The discontinuously synthetic fragments are Okazaki fragments
D、Half of the leading strand as well as half of the lagging strand is discontinuously synthesized
5、The enzyme catalyzing the formation of phosphodiester bond between 5’phosphate group and the 3’hydroxyl in DNA is:
A、methyltransferase
B、ligase
C、DNA pol I
D、terminal transferase
6、DNA replication requires: (1) DNA polymerase III, (2) helicase, (3) DNA polymerase I, (4) DNA-directed RNA polymerase, (5) DNA ligase, the correct order of their participation is:
A、2,3,4,1,5
B、4,2,1,5,3
C、4,2,1,3,5
D、2,4,1,3,5
7、Which of the following statement is correct about the DNA replication in E.coli:
A、One strand of DNA double helix is synthesized discontinuously.
B、Okazaki fragments are generated in this process.
C、RNA primer is needed.
D、Single-strand binding protein can prevent the unwinding of DNA double helix in the replication process.
8、Which of the following statement is correct about DNA topoisomerase in E.coli:
A、to catalyze the ligation of gaps in DNA double strands.
B、to release the supercoiled structure of DNA ahead of replication fork.
C、to break hydrogen bond and base stacking force and improve the extension activity of DNA polymerase.
D、to break phosphodiester bond of DNA and separate two strands of DNA.
9、Which of the following statements is correct about the synthesis of DNA lagging strands in the replication:
A、The synthesis of lagging strand is semi-conservative.
B、The synthesis of lagging strand is discontinuous.
C、Lagging strand is synthesized in the form of Okazaki fragment.
D、Several DNA polymerases are participated in the synthesis of lagging strand.
10、Which of the following statement is correct about DNA replication:
A、in the direction of 3’→5’
B、need DNA ligase.
C、need RNA primer
D、need DNA polymerases
11、Which of the following is correct about the Okazaki fragment:
A、It is synthesized by DNA polymerase I.
B、It is the discontinuous fragment in the lagging strand.
C、It is synthesized in the direction of 3’→5’.
D、It has free 3’-OH end.
12、When DNA topoisomerase II in yeast is mutated, DNA replication process would not be affected, but its chromosome cannot be separated in the process of mitoses.
13、In the process of DNA replication, the leading strand is synthesized in the direction of 5’→3’, while the lagging strand in the direction of 3’→5’.
14、Single-strand binding protein in the replication fork can separate two strands of DNA and prevent the complementary pairing of bases.
3-4 Regulation of DNA replication initiation随堂测验
1、A replicon is:
A、Whole DNA fragments replicated in the cell division phase.
B、Replicated DNA fragments and enzymes and proteins required in this process
C、DNA sequences replicated from one replication origin.
D、DNA fragments between replication origin and replication fork.
2、How many replication forks can be generated from one replication origin？
A、1
B、2
C、3
D、4
3、The recognition of replication origin by the origin recognition complex (ORC) take place in the () of cell cycle in eukaryotes:
A、G1 phase
B、G2 phase
C、M phase
D、S phase
4、The protein that recognizes DNA replication origin of E.coli is:
A、DnaA
B、DnaB
C、DnaC
D、ORC
5、The correct description about the assembly of pre-replication complex in eukaryotes is:
A、It couples with activation of replication complex.
B、Protein Cdc6 recognizes the origin of replication.
C、Assembled pre-replication complex may not be activated for a long time
D、The pre-replication complex can be assembled in any stages of cell cycle except S phase.
6、The characteristics of all replication origins in prokaryotes, eukaryotes and virus are:
A、It is unique DNA sequence consisting of several short tandem repeats.
B、It is palindromic sequence that can form stable secondary structure.
C、DNA binding proteins can specific recognize these short tandem repeats.
D、It is next to the DNA sequence rich of A-T and easy to unwind.
7、The methylation of GATC sequence near the OriC site of E.coli is closely related to the initiation of replication.
8、SeqA protein and methyltransferase Dam in E.coli would competitively bind to methylated GATC sequence.
9、After DnaA protein binds to the 9bp short tandem repeats of oriC, it would recruit helicase DnaB to bind to 13bp repeats and unwind DNA.
10、The binding of DnaA to the replicator is sequence specific.
11、In E.coli, yeast and virus, DNA replication is initiate at a specific origin site, which consists of several short tandem repeats and binds to initiate proteins.
3 5 End replication problem of DNA in eukaryotic cell随堂测验
1、Telomerase is a:
A、DNA-dependent DNA polymerase
B、DNA-dependent RNA polymerase
C、RNA-dependent DNA polymerase
D、RNA-dependent RNA polymerase
2、Which of the following statements about telomere is correct?
A、Telomeres are special structure composing of nucleosomes in the end of eukaryotic chromosomes
B、Telomeres can stable DNA secondary structure
C、Telomerases act as template as well as reverse transcriptase
D、Cell aging is related to the abnormal extend of telomere
3、The correct description about the telomerases is:
A、They are abound in eukaryotes
B、C, H, O, N, P are included in the telomerases
C、They belong to ribozyme, catalyzing the reaction which RNA is involved
D、They can carry genetic information
4、Which of the following will not happen when telomerase loses its function in cells?
A、Telomere was gradually shortened with every cell division
B、Cells exhibit aging or even death after 30 to 50 times of cell division
C、Immune system loses some defense ability gradually
D、Numerous somatic cells start to proliferate indefinitely
5、A special structure (telomere) is the “clock” that controls the frequency of cell division, which will become shorter with cell division. There are some enzymes extending telomere both in cancer cells and normal germ cells. According the information, the reason why somatic cells can not divide indefinitely is:
A、Lack of amino acid required to synthesize telomere
B、Lack of genes to control synthesis of telomerase
C、The mutagenesis of genes that control synthesis of telomerase
D、The expression level and activity of telomerase are strictly controlled
6、After sheep Dolly was born, it is doubtable that Dolly was not from somatic cell of an adult sheep, but from polluted embryonic cell in the experiment. The evidence that supports Dolly is from somatic cell rather than embryonic cell is:
A、DNA fingerprint of Dolly
B、DNA melting temperature of Dolly
C、Telomere length of Dolly
D、Biological clock of Dolly
3-6 Let’s “cook” DNA随堂测验
1、The necessary conditions for PCR to amplify DNA is ①target gene ②primer ③four deoxynucleotide ④DNA polymerase ⑤mRNA ⑥ribosome(A)
A、①②③④
B、②③④⑤
C、①③④⑤
D、①②③⑥
2、PCR is an enzymatic reaction that depends on DNA polymerase, with the presence of primer, template and four kinds of deoxyribonucleotides. The key factor to determine its specificity is:
A、template
B、primer
C、dNTP
D、Mg2+
3、Polymerase chain reaction namely PCR is a technique that amplifies DNA fragments rapidly in vitro. Generally, the process of PCR contains the following 30 times cycles: denaturation of template DNA at 95℃ → annealing (primer binds to DNA template) at 55℃ → extension to form the nascent DNA chain at 72℃. Which of the following statement is NOT true about PCR process:
A、In PCR process, both of the two strands of DNA are synthesized continuously, the same as synthesis of leading strand in DNA replication.
B、The binding of primer with DNA template strand depends on the principle of complementary base pairing in melting process.
C、The process of extension requires DNA polymerase, ATP and four kinds of ribonucleotides
D、The optimal temperature of DNA polymerase in PCR process is higher compared with DNA replication in cells.
4、The wrong statement about PCR is:
A、PCR contains three steps including denaturation, annealing and extension
B、PCR technique can be used for gene diagnosis, determination of the genetic relationship etc.
C、The DNA polymerase used in PCR can be stored at room temperature due to its tolerance of higher temperature.
D、Sanger method for DNA sequencing is based on PCR technique.
5、In genetic engineering, the target genes (1000bp, including 460 adenine deoxynucleotides) are amplified for 4 generation in the PCR instrument, the number of cytosine deoxynucleotides in PCR buffer should be at least:
A、640
B、8100
C、600
D、8640
6、The denaturation temperature of PCR is generally:
A、95℃
B、85℃
C、72℃
D、100℃
7、An archaeologist found the frozen muscle of an extinct giant animal in the ice of Siberian taiga region. He wanted to know the similarity of DNA of the giant animal and that of modern Indian elephant, and following detection work was done. The right steps and order is:①Lower the temperature and detect hybrid DNA region with double strand ②amplify DNA from giant animal and elephant by PCR ③ transducer DNA from giant animal into elephant cells ④ DNA from giant animal and elephant are heated together in the water bath under a certain temperature(D)
A、②④③
B、②④③①
C、③④①
D、②④①
8、Which of the following enzymes can tolerant high temperature:
A、Taq DNA polymerase
B、Hind III
C、T4 ligase
D、RNA enzyme
9、Which of the following may cause non-specific amplification of DNA in PCR reaction?
A、excessive TaqDNA polymerase
B、excessive primer
C、excessive Mg2+ in buffer
D、either of A, B or C
Chapter 4 DNA mutation and repair
4-1 Replication errors and the mismatch repair system随堂测验
1、Frameshift mutation results from base-pairs:
A、Transition
B、Transversion
C、Hydrolysis
D、Insertion
2、In nucleic acid, the point mutation is:
A、One base-pair is replaced by another is called point mutation
B、Only the mutation that leads to the change of amino acid can be called point mutation
C、The mutation that don’t change phenotype is called point mutation
D、The mutation that causes dysfunction of a single gene is called point mutation
3、A man suffers from fragile X syndrome <CGG>n. The cause is that n turns to 300 from 60. Then which the following is the cause of this disease:
A、missense mutation
B、nonsense mutation
C、frameshift mutation
D、dynamic mutation
4、The protein MutH in E. coli can recognize:
A、Twisty double-strands of DNA
B、Hemi-methylated GATC
C、The insertion site of the intercalator
D、The gap between Okazaki fragments
5、The Dam methylase in E. coli can methylate the A of GATC sequence in DNA. Which of the following is unrelated to this modification:
A、Protein SeqA can inhibit the methylation of GATC by Dam
B、The marker of DNA parental strand after replication
C、The full-methylated sequence GATC is the binding site of initiator protein DnaA
D、The methylation site is recognition site of MutH in mismatch repair system
6、Sickle cell anemia is the homozygote genotype of abnormal hemoglobin. β-strand aberrance is caused by which of the following mutations:
A、mononucleotide insertion
B、mononucleotide loss
C、chromosome nondisjunction
D、base substitution
7、Knocking out the kinase A of mouse can lead to its death in embryonic period, then which of the following mutation of the sequence encoding kinase A most likely lead to the death of mouse:
A、cytosine replaces guanine
B、methylcystein replaces cytosine
C、loss of three nucleotides
D、insertion of one nucleotide
8、Which of the following change is transversion:
A、T → G
B、A → T
C、C → G
D、C → T
9、Which of the following point mutation can affect the sequence of polypeptide:
A、missense mutation
B、non-sense mutation
C、Insertion
D、deletion
10、MutS in mismatch repair system can recognize mismatch sites of nascent strands by specific sequences.
11、In mismatch repair system of eukaryotes, the gap between Okazaki fragments can be used for MutH binding and its hydrolysis of mismatch site.
12、MutL can hydrolyse the sequences containing mismatch base-pairs in nascent strands of DNA, with consumption of ATP.
13、The gap generated by hydrolysis of mismatch site in replication can be repaired by DNA polymerase I in E. coli.
4-2 DNA damage随堂测验
1、Which of the following belongs to spontaneous damage of DNA:
A、Single-base mismatch during DNA replication
B、Formation of thymine dimer
C、Cytosine deoxidation
D、DNA cross-linking
2、The damage of ultraviolet irradiation to DNA mainly is:
A、Base substitution
B、Broken of phosphodiester bond
C、Deletion of base
D、Formation of covalent pyrimidine dimer
3、Single-base substitution is a kind of DNA damage, they can:
A、affect transcription process instead of replication
B、affect DNA sequence but not DNA secondary structure
C、change DNA double helix structure but won’t affect replication process
D、be induced by ultraviolet irradiation（such as pyrimidine dimer）or the formation of additive compound (such as alkylation
4、The most common form of ultraviolet irradiation-induced DNA damage is the generation of thymine dimer. Which of the following statement is correct about changes of DNA structure in this process:
A、It is the covalent linkage between thymines of the two complementary polynucleotide chain
B、It can be repaired by the relevant enzyme system including nucleotide excision repair system
C、It is catalyzed by thymine dimeric enzyme
D、It will not affect DNA replication process
5、During cultivation of E.coli, the spontaneous point mutations are largely caused by:
A、The tautomeric shift of hydrogen atom
B、The broken of DNA sugar-phosphate backbone
C、The insertion of one base-pair
D、The cross-linking of inter-strands
6、Insertion or deletion of base-pairs will cause frameshift mutation. Which of the following compounds can most likely to result in this kind of mutation:
A、Acridine derivatives
B、5-bromouracil
C、Azathioprine
D、Ethyl ethanesulfonic acid
7、Which kind of the mutations is most irreversible:
A、Deletion or insertion of nucleotide
B、Hydrolytic deamination
C、Octaoxa-guanine
D、Pyrimidine dimer
8、The cytosine in nucleic acid can be easily hydrolyzed to become its deaminized form, so there are many CTPs in the genome.
9、The hydrolytic deamination of 5-methylcytosine is the hotpot of mutations in the genome.
10、The oxidative damage of bases in DNA can occur spontaneously.
11、In the process of depurination sites repairing and nucleotide excision repairing, the glycosidase is needed to catalyze the ligation of correct bases to pentose.
12、Nitrite is a common food additives. Itself as well as the nitrosamine generated by heating reaction of nitrite with amino acid are both mutagenic agents of DNA.
4-3 Repair of DNA damage随堂测验
1、Which of the following regarding to DNA repair is NOT correct:
A、UV irradiation can induce crosslinking between the adjacent thymines
B、DNA polymerase III is needed in the nucleotide excision repair system to repair the nick of single strand.
C、DNA repairing process needs DNA ligase
D、Specific damaged bases can be removed by different glycosylase in the mammal cells.
2、The UvrAB complex in E.coli can recognize:
A、Gaps between Okazaki fragments
B、Normal DNA molecule
C、Hemi-methylated GATC
D、Twisty DNA double helix
3、The DNA strand containing pyrimidine dimer can still functions as a template during replication. When replication fork meets damaged site, the synthesis of nascent strand will be completed by which of the following enzyme:
A、DNA polymerase I
B、DNA polymerase III
C、RecA
D、Damage-removing synthase
4、Which enzyme is not needed during base excision repair:
A、DNA polymerase
B、Phosphodiesterase
C、Exonuclease
D、Ligase
5、In photoreactivation repair, which of the following enzyme binds to pyrimidine dimer?
A、Photolyase
B、Exonuclease
C、Endonuclease
D、Ligase
6、Four steps are involved in most of DNA repairs. The correct order of the four steps is:
A、recognition、excision、re-synthesis、re-ligation
B、re-ligation、re-synthesis、excision、recognition
C、excision、re-synthesis、re-ligation、recognition
D、recognition、re-synthesis、re-ligation、excision
7、Among the responses of cells to DNA damage, which is most likely to lead to high mutation rate:
A、Photoreactivation repair
B、Base excision repair
C、Recombination repair
D、Damage-removing synthesis
8、Mismatch repair is based on recognition of the mismatched base during replication. Which of the following is correct about this:
A、UvrABC system recognizes the mismatched bases and add the correct one with the help of DNA polymerase I.
B、If recognition happens before methylation of DNA parental strand, the sequence of repair tends to be wild type.
C、Mismatch site is usually repaired by single-strand exchange.
D、Mismatch repair relies on the repairing function which is repressed under normal conditions by the protein LexA.
9、Which of the following damages can be repaired by nucleotide excision repair system:
A、The broken of DNA double strands
B、Base insertion
C、Alkylation of bases
D、The formation of thymine dimer
10、The prokaryotic cell is susceptible to environment and DNA damage can be easily generated, therefore its damage-removing synthase is actively expressed as a result of transcriptional activation of RecA.
11、The depurination sites of DNA strand can be repaired by directly addition of bases to pentose by specific glycosylase based on complementary base pairing.
12、Nucleotide excision repair system is similar with mismatch system in replication process, both of which recognize the twisty DNA double helix as damaged site. So the same protein complex can be used to find damaged sites in these two system.
4-4 DNA Homologous Recombination随堂测验
1、Which of the following repair cannot fundamentally eliminate DNA structural damage:
A、nucleotide excision repair
B、mismatch repair
C、photoreactivation repair
D、recombination repair
2、In prokaryotic cell，the main function of protein RecA in genetic recombination is:
A、To promote the autosynapsis and single-strand exchange between DNA molecules
B、To disassociate the single strand from the duplex DNA molecule
C、Possessing activity of cutting the single strand DNA at specific site
D、To promote the formation of the single strand DNA
3、Which of the following statement is NOT correct about the non-homology end joining repair after the break of the double strands in mammal cells?
A、The terminal of DNA needs the protection of proteins such as Ku70, Ku86 etc.
B、The repair can happen at any period of cell cycle
C、Independent of DNA homology, the repair forces the two DNA broken ends to link together
D、The repair can result in exchanges between the chromosome arms but usually will not influence the expression of the relevant genes
4、What are the common characteristics of nucleotide excision repair and homologous recombination repair:
A、The endonuclease is needed to cut one strand of DNA molecule.
B、The dysfunction of DNA caused by pyrimidine dimer can be avoided.
C、DNA polymerase is needed to synthetize single-strand DNA, repairing gaps in DNA strands.
D、Ligase is needed to make covalent bond between new fragment and the old one.
5、Protein RecBCD is a multi-functional enzyme which？
A、is a single-strand and double-strand exonuclease that consumes ATP
B、is a helicase
C、is a sequence-specific endonuclease that acts on single strand
D、is a polymerase
6、RecA protein is a sequence-specific endonuclease that acts on single strand, as well as a helicase that can dissociate single-strand DNA from double helix, the latter of which is ATP-dependent.
7、The exchange of DNA fragments is included in DNA recombination, which involving the break and re-formation of phosphodiester bond in DNA backbone.
8、No matter how Holliday junction is divided, heteroduplex DNA would be generated in both of two DNA molecules.
9、RecA protein can bind to both single-strand DNA and double-strand DNA, to promote the autosynapsis between them.
Chapter 5 Transcription in prokaryotes
5-1 Transcription and RNA polymerase随堂测验
1、The recognition sequence of RNA polymerase in prokaryotic cell is usually located in:
A、-35,-10 region upstream transcription start site.
B、-35,-10 region downstream transcription start site.
C、TATA box upstream transcription start site.
D、no specific region.
2、Which of the following descriptions about DNA and RNA polymerase are correct:
A、RNA polymerase synthesize RNA chain using nucleoside diphosphate instead of nucleoside triphosphates.
B、RNA polymerase can initiate transcription without a primer, and nucleotides are added to 5’ end of the growing RNA chain.
C、RNA polymerase can add nucleotides in both 3’ or 5’ end of RNA chain.
D、Both RNA polymerase and DNA polymerase can only add nucleotides to 3’ end of the polynucleotides.
3、Which is the substrate of RNA synthesis during transcription process?
A、dATP
B、NTP
C、dNTP
D、AMP
4、Which of the following statement about the promoter in prokaryotes is correct:
A、The sequences which DNA polymerase can recognize and bind to
B、The sequences that contains TATA box
C、The sequences which is a 40-60bp RNA fragment
D、Its -10 region has a consensus sequence of 5'-TATAAT-3’
5、Which of the following statement about DNA replication and transcription is NOT correct:
A、Both of them use DNA as template.
B、The product of DNA replication is double-strand DNA, while the product of transcription is single-stand RNA.
C、Both of them are based on the complementary base pairing of A-T and G-C.
D、The products are polynucleotides.
6、The core enzyme of RNA polymerase in prokaryotic cell consists of:
A、αββ′
B、ααββ′
C、αββ′ω
D、ααββ′ω
7、The sense strand is also called:
A、template strand
B、leading strand
C、coding strand
D、lagging strand
8、If the template sequence is ”5’-AGCATCTA”, which of the following is the sequence of its transcription product:
A、5’TCGTAGAT
B、5’UCGUAGAU
C、5’UAGAUGCU
D、5’TAGATGCT
5-2 Transcription process in prokaryotes随堂测验
1、A RNA transcription process including:
A、Initiation, elongation and termination
B、entrance, transpeptidation and translocation
C、Formation of primosome, chain elongation and termination
D、formation of closed complex, formation of open complex and promoter escape
2、Which of the following statement about Rho-independent transcription termination is NOT correct?
A、The terminator intensively contains A-T pair and G-C pair
B、5’ end of transcription product is rich of U.
C、The transcription product could form stem-loop structure
D、The stem-loop structure could stop the transcription of RNA polymerase.
3、The ternary complex in the elongation step of transcription consists of:
A、Complex ofσfactor, core enzyme and DNA strand which are formed in the promoter region.
B、Complex of core enzyme, DNA strand and nascent RNA chain
C、Complex of three holoenzymes formed in transcription start site
D、Complex ofσfactor, core enzyme and helicase
4、Which of the following factor is related to transcription termination in prokaryotic cell:
A、σ
B、α
C、β
D、ρ
5、The signal of transcription termination prokaryotic cell is:
A、Termination codon
B、Terminator
C、The end of DNA chain
D、The inactivation of RNA polymerase
6、Which of the following is the necessary and sufficient condition(充要条件) of the effective initiation of bacterial transcription:
A、Helicase
B、Primase
C、The holoenzyme of RNA polymerase
D、The core enzyme of RNA polymerase
7、Which of the following descriptions about transcription termination in prokaryotic cell is correct:
A、occurs randomly
B、depends on theρfactor of the holoenzyme of RNA polymerase
C、If the transcription terminator sequence contains palindromic sequence, ρfactor is not necessary for the termination.
D、It can effectively terminate the transcription process if the end of the gene is rich of A-T.
8、When transcription is terminated in prokaryotes,σfactor can recognize the termination signal.
9、The ρ factor possesses both ATPase and helicase activity.
10、Transcription products in prokaryotes must be polycistron.
11、The core enzyme of RNA polymerase slide along the template strand in the direction of 5'→3'.
5-3 Transcriptional regulation in bacterial I: Lac operon随堂测验
1、In the lac operon model, the trans-acting factor is___, and the cis-acting element is___
A、Repressor protein, lac O sequence
B、lac I gene, lac O sequence
C、Repressor protein, lac I gene
D、lac I gene, lac P gene
2、The lac operon regulates gene expression on:
A、Replication level
B、Transcription level
C、Post-transcription level
D、Translation level
3、In lac operon the role of allolactose is:
A、Repress the negative regulation
B、Induce the negative regulation
C、Repress the positive regulation
D、Induce the positive regulation
4、In lac operon, ___binds to the repressor protein could initiate the transcription of structural genes.
A、cAMP
B、Lactose
C、allolactose
D、CAP
5、The repressor protein can binds to ___ and inhibit the transcription of structural genes.
A、lac O
B、lac P
C、CAP binding site
D、lac I
6、Bacteria is intended to consume glucose at first, and enzymes to metabolize other sugars would be produced after exhaustion of glucose. This phenomenon can be called:
A、Repression
B、Induction
C、Coordinate regulation
D、Catabolite repression
7、Which of the following conditions lead to the lowest expression of lac operon:
A、Both glucose and lactose exist
B、Only glucose exists
C、Neither glucose nor galactose exists
D、Only lactose exists
8、The role of glucose in catabolite repression is:
A、Improving the activity of adenylate cyclase, to increase the concentration of intracellular cAMP.
B、Inhibiting the activity of adenylate cyclase, to decrease the concentration of intracellular cAMP.
C、Improving the activity of phosphodiesterase, to increase the concentration of intracellular cAMP.
D、Inhibiting the activity of phosphodiesterase, to decrease the concentration of intracellular cAMP.
9、An operon containing three structural genes can be regulated by negative transcriptional regulation, then which of the following statements is true:
A、The operon contains three binding sites for repressor proteins.
B、The operon contains three promotors.
C、The inhibitor of the regulator protein of the operon would inhibit the transcription of structural genes.
D、The regulator gene of the operon can be located downstream of this operon.
5-4 Transcriptional regulation in bacterial II: Tryptophan operon随堂测验
1、Tryptophan operon belongs to:
A、negative regulation-induction system
B、negative regulation-repression system
C、positive regulation-induction system
D、positive regulation-repression system
2、Which of the following conditions lead to the highest expression of tryptophan operon:
A、High concentration of Trp while low concentration of other amino acids.
B、High concentration of other amino acids while low concentration of Trp.
C、Both Trp and other amino acids are high concentration.
D、Both Trp and other amino acids are low concentration.
3、Besides the negative regulation of repressor protein, what’s the unique regulation of tryptophan operon:
A、Repression
B、Induction
C、Attenuation
D、Catabolite repression
4、Which of the following descriptions regarding tryptophan operon in E.coli is correct:
A、it contains regulations in both transcription level and translation level
B、It contains regulations in both initiation step of transcription as well as termination step of transcription.
C、It contains both positive regulation and negative regulation.
D、It contains regulations for enzyme induction as well as enzyme repression.
5、Which of the following is NOT correct regarding tryptophan operon:
A、Ribosome are involved in transcription termination
B、Attenuator is a key regulatory element
C、When Trp is deficient, transcription termination will be initiated in advance
D、Coupled transcription and translation is the molecular basis of transcription regulation
6、The transcriptional regulation of tryptophan operon is closely related to：
A、repressor protein
B、transcriptional factor
C、cAMP-CAP
D、Ribosome
Chapter 6 Transcription in eukaryotes
6-1 Eukaryotic RNA polymerase and promoter随堂测验
1、In gene expression regulation of eukaryotes, which of the following regulatory element can inhibit the transcription:
A、Enhancer
B、Silencer
C、Promoter
D、Upstream activation sequences
2、Which is NOT true about RNA polymerases in eukaryotic cell:
A、There are at least three RNA polymerases including I, II, III in eukaryotes.
B、RNA polymerase I has the highest transcriptional activity in eukaryotic cells
C、RNA polymerase I is located in the nucleolus, while RNA polymerase II, III are located in nucleoplasm
D、The RNA polymerase II is responsible for the transcription of rRNA genes
3、Which of the following statement about transcriptional regulation element is NOT correct:
A、Enhancer can be recognized and combined by some transcriptional activation factors.
B、Silencer which also belongs to transcriptional regulatory factors can be combined by transcription inhibitors.
C、TFIID factor can recognize several elements of the core promoter at the same time.
D、The elements in the core promoter functions independently, without synergistic activation between each other.
4、Which is wrong description about promoter:
A、It is located upstream a gene.
B、The promoter of prokaryotic cells is recognized and combined by RNA polymerase, while the promoter of eukaryotic cells is recognized by general transcription factors.
C、There are some conservative sequences in the promoter region.
D、Genes encoding proteins all have the same core promoter factors.
5、Which of the following catalyze the synthesis of mRNA in eukaryotic cell:
A、RNA polymerase I
B、RNA polymerase II
C、RNA polymerase III
D、RNA polymerase core enzyme
6、Which factors in the following are included in the core promoter of eukaryotic cells:
A、TATA box
B、Inr
C、DPE
D、BRE
7、Which of the following belong to long-distance control elements:
A、Enhancer
B、Silencer
C、Upstream activation sequences
D、Insulator
8、Which of the following products are transcribed by RNA polymerase III in eukaryotic cells:
A、hnRNA
B、tRNA
C、mRNA
D、5SrRNA
9、RNA polymerase in eukaryotic cells only exists in the nucleus.
10、Promoter of all higher eukaryotes have TATA box.
11、Bacteria use only one RNA polymerase for transcription of all RNA in the cell, while eukaryotic cells have various RNA polymerases.
12、Both the enhancer and promoter are factors that promot transcription.
13、Enhancer must be located upstream the target gene, but its effects are not related to its sequence direction.
6-2 Transcription process in eukaryotic cell随堂测验
1、Which is NOT correct about the transcription process of eukaryotic cells:
A、RNA polymerase can synthesis RNA only when DNA exists.
B、NTPs are the substrates.
C、The direction of RNA chain is 3' -5'.
D、The bases of nascent RNA form complementary base-pairing with bases of DNA molecule.
2、What is the transcriptional elongation factor of eukaryotic cells:
A、TFIID
B、TFIIH
C、GRE
D、TFIIS
3、Which of the following statement is NOT true about the transcription initiation of eukaryotes:
A、General transcription factor are located at promoter by recruiting RNA polymerase II, which can effective initiate most of gene transcription.
B、The components of mediator complex that are necessary for the transcription initiation in eukaryotes may be different.
C、The CTD domain of RNA polymerase II would be phosphorylated in multiple sites in the initiation of transcription.
D、The combination of TFIIB factor and DNA determines the direction of transcription initiation complex.
4、Which is NOT true about the mediator complex:
A、It is a necessary component of basal apparatus.
B、Mediator complexe can be connected with many transcription factors, while it does not directly connected with RNA polymerase.
C、The inactivation of a subunit of mediator complex can lead to the abortion of some genes expression.
D、Different subunits in mediator complex can response to different kinds of transcription factors.
5、Which is correct about DNA polymerase and RNA polymerase:
A、RNA polymerases use nucleoside diphosphate instead of nucleoside triphosphate for synthesis of polynucleotide chains
B、RNA polymerases need primer and add nucleotides in the 5 'end of the polynucleotide chain
C、DNA polymerase can add nucleotides in both ends of the polynucleotide chain
D、All of the RNA polymerases and DNA polymerases can only add nucleotides in the 3’ end of the polynucleotide chain
6、Which of the following process is associated with the phosphorylation in CTD domain of RNA polymerase II:
A、Degradation of mRNA
B、Splicing of mRNA
C、Termination of mRNA synthesis
D、Modification of the mRNA 5’ end
7、Which is correct about the mRNA transcription in eukaryotic cells:
A、The mRNA can contain coding sequences of several proteins.
B、The generation of the 3 'end depends on polyadenylation signal sequence.
C、The 5 'end of mRNA will be capped after the termination of transcription.
D、After mRNA is synthesized and released, RNA polymerase II will still be able to continuously move forward on the DNA template and synthesize another RNA molecule.
8、The TBP protein can recognize the sequence in the major groove of DNA molecule and bind to the TATA box by its folding structure.
9、In the transcription process of eukaryotes, the combination of general transcription factors with the promoter is prior to the location of RNA polymerase.
10、The transcription initiation in eukaryotic cells doesn't consume ATP. It can use the interaction of general transcription factors and DNA to unwind of double-stranded DNA in the start point.
11、TFIIH factor has kinase activity, which can promote RNA polymerase II escaping from the promoter.
12、Transcription activator has a DNA-binding domain as well as transcription activation domain.
13、Because RNA polymerase does not have exonuclease activity, the transcription process have no proofreading function.
14、TATA box only exists in the promoter of genes transcribed by RNA polymerase II.
6-3 Transcription factors of eukaryotic cell随堂测验
1、Activator protein 1 (AP-1) consists of c-jun protein and c-fos proteins. The type of protein motif that make it bind to DNA is:
A、Leucine zipper
B、HLH
C、HTH
D、Zif
2、Which is NOT the mechanism of the activating transcription factors:
A、Combined with general transcription factors to promote the assembly of transcription initiation complex
B、Changing protein conformation to activate the enzyme in basal apparatus.
C、Recruiting chromatin modification complex to change chromatin morphology.
D、Generation of accessible and free DNA by chromatin remodeling.
3、Which is NOT true about the transcription factor with Zinc finger structure:
A、It usually contains multiple zinc finger arrangement.
B
学习通Molecular Biology_3
学习通Molecular Biology_3是一门关于分子生物学的在线课程。这门课程旨在为学生提供关于DNA、RNA、蛋白质、基因组等内容的深入认识，帮助学生了解生命的基本单位，为生物科学的学习打下基础。
该课程由教育部赞助，由多位专业教授授课。课程内容涵盖了分子生物学的各个方面，包括DNA复制、转录、翻译、基因表达、基因组编辑等等。这些知识点是生物学、医学和生物工程等学科的基础，对于理解生命的基本机理至关重要。
该课程还将介绍一些实用的技术，例如PCR、基因工程、蛋白质纯化、基因测序等。这些技术在研究生命科学、医学和生物工程方面有着广泛的应用，对于学生未来的职业发展也具有重要的意义。
课程内容
该课程共分为十个模块，每个模块的内容如下：
- 模块1：DNA的结构和功能
- 模块2：DNA复制
- 模块3：基因表达和转录
- 模块4：翻译
- 模块5：基因调控
- 模块6：基因组和基因组学
- 模块7：基因组编辑和CRISPR-Cas技术
- 模块8：蛋白质的结构和功能
- 模块9：蛋白质纯化和分析
- 模块10：DNA测序和分析
每个模块包含多个课程视频、阅读材料、测验以及讨论论坛。通过这些方式，学生可以充分了解每个主题，提高对生命科学的理解和应用能力。
课程收获
通过学习该课程，学生可以收获以下知识和技能：
- 了解DNA、RNA、蛋白质等生命基本单位的结构和功能；
- 了解基因的表达和调控机制，以及如何使用基因编辑技术进行研究；
- 学习基本的实验技术和数据分析方法，例如PCR、基因测序、蛋白质纯化等；
- 了解当前分子生物学领域的研究进展和应用前景。
这些知识和技能对于生命科学、医学和生物工程等领域的研究和工作都具有重要的意义。通过学习该课程，学生可以为未来的职业发展打下坚实的基础。
适合人群
该课程适合以下人群：
- 生命科学、医学或生物工程专业的本科生或研究生；
- 对分子生物学感兴趣，希望了解生命基本单位的人群；
- 希望提高实验技术和数据分析能力的人群。
无论你是否有生物学或化学背景，只要你感兴趣并愿意学习，该课程都适合你。
学习方式
该课程是一门在线课程，学生可以在任何时间、任何地点进行学习。学习内容包括视频课程、阅读材料、测验和讨论论坛。学生可以根据自己的兴趣和时间安排，自由选择课程内容和学习进度。
学生还可以通过讨论论坛与其他学生和教授交流，分享自己的学习体验和思考。这不仅可以帮助学生更好地理解课程内容，还可以拓展学生的思维和视野。
结语
学习通Molecular Biology_3是一门非常有价值的在线课程。通过学习该课程，学生可以深入了解生命的基本单位，提高实验技术和数据分析能力，为生命科学、医学和生物工程领域的职业发展打下坚实的基础。
如果你对生命科学和分子生物学感兴趣，想要学习相关知识和技能，不妨来试试这门课程。相信你一定会有所收获。
