中国大学Molecular Biology_5答案(mooc完整答案)
61 min read中国大学Molecular Biology_5答案(mooc完整答案)
1-1 The past and present journey of molecular biology随堂测验
1、The大学答案答案 HGP (Human Genome Project) was finished in:
A、1990
B、完整2001
C、中国2008
D、大学答案答案2006
2、完整Which of the following technique is 中国developed by Paul Berg who is named as “the father of modern genetic engineering”:
A、Technique of hybridization.
B、大学答案答案Technique of recombination DNA.
C、完整Technique of monoclonal antibody.
D、中国PCR
3、大学答案答案Which one in the following can be used for DNA sequencing:
A、完整Gersang method.
B、中国X-ray diffraction method.
C、大学答案答案Sanger dideoxy-mediated chain termination method.
D、完整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 1 Tests
1、What are the basic components of a nucleic acid:
A、phosphate and ribose
B、nucleoside and nitrogenous base
C、mononucleotide
D、deoxynucleoside and base
2、What is the direction of genetic information flow of reverse transcription:
A、DNA→DNA
B、DNA→RNA
C、RNA→DNA
D、RNA→RNA
3、Which of the following structure causes the UV absorption of nucleic acids:
A、The hydrogen bond between purine and pyrimidine
B、The glycosidic bond between base and pentose
C、The conjugated double bond between purine and pyrimidine ring
D、The phosphodiester bond between nucleotides
4、What is the key component for storage and transmission of genetic information in nucleic acids:
A、nucleoside
B、pentose
C、phosphate group
D、base sequence
5、DNA molecule does not include:
A、phosphodiester bond
B、hydrogen bond
C、disulphide bond
D、Van der Waals' force
6、Below are the two molecules that has been heat-denatured, when temperature of the solution was dropping slowly, which one would be easier to recover:
A、5’-ATATCATATGATATGTA-3’
B、5’-CGGTACTCGTGCAGGT-3’
C、None
D、None
7、Which of the following statements about the structure of nucleic acid is NOT true:
A、There are major and minor grooves on the surface of double helix
B、There is base stacking force between upper and lower layers of bases
C、The double helix structure only exits in DNA molecule
D、There is complementary base pairing existing in the double helix structure
8、DNA denaturation means:
A、Supercoiled DNA molecule turn to double helix structure
B、The phosphodiester bond of the molecule has been broken
C、The hydrogen bond between bases has been broken
D、The complete hydrolysis of the nucleic acid molecule
9、When running DNA agarose gel electrophoresis, which of the following is NOT true:
A、After the gel was ready, add electrophoretic buffer and then pull out the comb
B、Put the sample well on the positive side of the electrophoresis chamber and load the samples
C、Mix the sample solution with loading buffer and then load them in the sample well
D、Use ethidium bromide (EB) to stain the gel after running and check the bands under UV light
10、Under the melting temperature, what happened to the DNA double helix:
A、All of the G-C base pairs disappeared
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
11、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 covalent bond
D、The formation of the base stacking force
12、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
13、Which of the following statement is NOT true about the virulence transformation experiment of Diplococcus pneumonia carried out by Griffith and Avery:
A、Only S type of Diplococcus pneumonia has the capsule which is similar to cell wall.
B、In Griffith’s experiment, if S type strain and heat-denatured R type strain was injected into mice, the mice would not die
C、The virulence transformation of bacterial strain is mainly related to the capsule of the S type strain
D、The transformation of alive R type strain to S type strain could also happen in the tube besides happening in vivo in the mice.
14、The UV absorption value of nucleic acid is unrelated to the pH value of solution.
15、The component of bases in DNA from different biological tissues are different.
16、The Tm value of DNA decreased with the increasing of (A+T) /(G+C) value.
17、The DNA renaturation normally take place at a temperature which is about 20 degree lower than its Tm value.
18、Most of the DNA molecules are negatively supercoiled naturally in organisms.
19、For purified DNA sample, the value of OD260/OD280 lower than 1.8 indicating the sample contains RNA.
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 2 Tests
1、Which of the following statement about genome is NOT correct:
A、The whole nucleic acid information that responsible for the survival, proliferation, development and activities of an organism.
B、Both chromosomal genome and organelle genome can be included in a eukaryotic cell.
C、Some regulation sequences for gene expression are included in genome.
D、The most sequences of eukaryotic genome encode protein product.
2、HIV that causes AIDS is:
A、Double-strand DNA virus
B、Single-strand DNA virus
C、RNA virus
D、Phage
3、Which of the following is NOT included in the histone core octamer of nucleosome:
A、H1
B、H2A
C、H3
D、H4
4、DNA fragment included in a nucleosome core particle is:
A、55bp
B、146bp
C、166bp
D、200bp
5、5.What’s the following statement about the effect of modification in histone terminal on nucleosome stability is NOT correct:
A、The positive charged amino acid residues in histone terminal benefit for the stable connection of histone and DNA.
B、The acetylation of Lys residue in histone terminal make the packaging of 10nm fiber more closer.
C、The phosphorylation of histone terminal reduces its positive charges and weakens the interaction of DNA and histone.
D、The acetylation and methylation site of histone can be used to recruit the chromatin remodeling complex, which would affect the interaction of DNA and histone.
6、What’s the following is correct about the loop model:
A、It is formed by the further folding of 10nm solenoidal fiber.
B、The central axis of loop model consists of non-histones.
C、16 loops are connected with non-histone and radially arranged in a plane, which forms a microstrip.
D、About 108 micro-strips are included in a chromatid.
7、In the modification of histone, Bromo domain can recognize:
A、The acetylation site of histone.
B、The methylation site of histone.
C、The histone terminal that are not modified.
D、Non-histone.
8、The differences of prokaryotic genome compared to eukaryotic genome are:
A、It contains several DNA replication origins.
B、Single chromosome and always circular.
C、Chromosomal DNA is not connected with histone to form nucleosome.
D、It contains numerous non-coding sequences.
9、Which of the following histones are highly conserved:
A、H1
B、H2A
C、H3
D、H4
10、Which is included in a chromatosome:
A、Histone core octamer
B、H1
C、55bp linker DNA
D、146bp DNA
11、Which of the following structures are included in the chromatin:
A、Nucleosome beads
B、Solenoid
C、Loops
D、Chromatid
12、A gene is the DNA sequences encoding a protein(enzyme).
13、The genome of DNA virus is double-strand DNA.
14、The chromosome of bacteria only consists of a single circular double-strand DNA.
15、Most of DNA sequences in high eukaryotes don’t encode proteins.
16、DNA is genetic material but RNA is not.
17、Four kinds of histones in nucleosome core particle as well as H1 is highly conserved, without species and tissue specificity.
18、The main difference of chromosome and chromatin is the compactness, while their chemical composition is exactly the same.
19、The diameter of beads-on-a-string structure of nucleosome is 30nm, to which linker DNA is connected to form 30nm fiber.
20、The N terminal of core histone is necessary to form 30nm fiber of chromotin.
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 3 Tests
1、Okazaki fragment refers to:
A、DNA fragments in DNA template
B、DNA fragments synthesized in lagging strand
C、DNA fragments synthesized in leading strand
D、DNA fragments synthesized by DNA ligase
2、Which of the following statement about DNA replication of E.coli is NOT true:
A、Four kinds of dNMP are needed.
B、It requires DNA ligase.
C、RNA primer will be formed in the process.
D、DNA polymerase I is needed
3、Which of the following statements about prokaryotic and eukaryotic DNA replication is NOT true:
A、The replication direction of two nascent strand is 5'→3'
B、For one DNA template strand, the replication is continuous on one side of replication origin and discontinuous on the other side.
C、Okazaki fragments formation and RNA primer removal would happen in both eukaryotic and prokaryotic cells.
D、The whole process of DNA replication can occur spontaneously without consumption of ATP.
4、Which of the following enzyme is NOT required during DNA replication?
A、ligase and gyrase
B、RNA polymerase which uses DNA as template
C、DNA polymerase which uses RNA as template
D、helicase and topoisomerase
5、Which of the following statements about DNA polymerase III is NOT true:
A、Deoxyribonucleotide are connected to 5’ hydroxyl of synthesized DNA by DNA pol III
B、Requires four kinds of 5’-dNTP
C、To catalyze deoxyribonucleotide connected to 3’ terminal of RNA primer
D、Pyrophosphate is one of the product
6、The enzyme that releases the DNA supercoiled structure is:
A、DNA polymerase
B、helicase
C、topoisomerase
D、telomerase
7、Which of the following statement about DNA polymerases in E.coli is NOT true:
A、DNA Pol I can be divided into two subunits.
B、DNA Pol III is very important for the extension of DNA strands during replication
C、Use four kinds of deoxyribonucleotides as substrate
D、DNA Pol III consists of four subunits
8、The wrong explanation for low error rate of DNA replication in prokaryotic cell is:
A、DNA Pol I have 3’ →5’ exonuclease activity
B、Both DNA Pol I and III have proofreading function
C、DNA Pol III have 5’ →3’ exonuclease activity
D、DNA Pol III have 3’ →5’ exonuclease activity
9、The common characteristic of DNA replication in eukaryotes to prokaryotes is:
A、The direction of synthesis is 5’ →3’
B、The length of Okazaki fragments is similar
C、Both of them have several replication origins
D、The speed of DNA replication is similar
10、Telomerase is:
A、DNA ligase
B、RNA polymerase
C、DNA exonuclease
D、Reverse transcriptase
11、Which of the following enzyme has the function of proofreading, repairing and filling the gap?
A、DNA pol III
B、DNA pol V
C、DNA pol δ
D、DNA pol I
12、The enzyme closely related to the synthesis of RNA primer is:
A、DNA pol α
B、DNA pol γ
C、DNA pol I
D、DNA pol III
13、The correct description about the assembly of pre-replication complex in eukaryotes is:
A、It is coupled with activation of replication complex
B、Protein Cdc6 recognizes the origin of replication
C、The assembly of pre-replication complex may not be activated for a long time
D、The complex can be assembled in any stages of cell cycle except S phase
14、The sequence of DNA template is ATAGC, and the sequences of synthetic nascent DNA is:
A、5′-GCTAG-3′
B、5′-GCTAT-3′
C、5′-GCATA-3′
D、5′-GATAT-3′
15、Semi-conservative replication process means that DNA parental strands are used as template of daughter strands, and the nascent DNA molecule consists of an old parental strand as well as a new daughter strand.
16、In DNA replication, the sequence of nascent DNA strand is the same as the template when reading in 5'→3' direction.
17、Both two strands in the telomere can be synthetized by its telomerase using its RNA as template.
18、The deficiency of 3’→5’ exonuclease activity of DNA polymerase in E. coli would slow down the synthesis of DNA, while the fidelity of DNA would not be affected.
19、Both DNA polymerase and RNA polymerase are needed in DNA replication process.
20、The DNA replication in E.coli, yeast, virus are all initiated at a specific origin site, which consists of several short tandem repeats and they are the targets of initiator proteins.
21、Topoisomerase I need to break the phosphodiester bond in sugar-phosphate backbone of DNA to unwind double helix, which would consume ATP.
22、DNA replication process needs DNA polymerase, which need free 3’-OH to initiate the polymerization. The free 3’-OH can come from a RNA primer, or synthetized DNA strand.
23、DNA replication machinery consists of a large replication complex which contains one DNA polymerase core enzyme, primase, helicase, slide clamp and assembler etc.
24、In Sanger method for DNA sequencing, the substrate should be four double de-oxyribonucleotide, including ddATP、ddGTP、ddCTP and ddTTP.
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 4 Tests
1、According to the reversibility of gene mutation, the gene A can turn to gene a, meanwhile gene a can also return to gene A. If the mutation rate of A→a is U, and a→A is V, then in most cases:
A、U<V
B、U>V
C、U is equal to V
D、U is not equal to V
2、Which of the following is NOT true about the excision repair system after DNA damage:
A、Repairs by homologous recombination are included in the excision repair system
B、Excision repair system can repair various DNA damages
C、Excision repair can be initiated by glycosylase
D、Repairs initiated by UVrABC are included in excision repair system.
3、The mechanism that thymine dimer inhibits DNA synthesis is:
A、It makes slip clamp of DNA polymerase cannot get through DNA template strand.
B、It breaks DNA template strand.
C、It inactivates DNA polymerase.
D、dNTP cannot be synthesized into DNA daughter strands by DNA polymerase III.
4、The mutation which could be induced by nitrite is:
A、pyrimidine dimer
B、Glu→Gln
C、C→U
D、T→A
5、The mutation relevant to β chain in sickle-cell disease is:
A、insertion
B、loss
C、crosslinking
D、point mutation
6、Xeroderma pigmentosum is a kind of human genodermatosis. After sun exposure, the DNA of epidermic cells in patient’s skin is damaged under ultraviolet light and it tends to develop to carcinoma cutaneum. The most possible molecular mechanism of this disease is:
A、The deficiency of permeability of cell membrane induces rapid water loss
B、The deficiency of DNA excision repair system.
C、The expression of methyltransferases in the patients is not enough
D、The cytochrome synthetic pathway is inhibited.
7、Ionization radiation can increase the canceration rate of the cell, so which of the following DNA damages are closely related to Ionization radiation:
A、The breaks of DNA double strands
B、The oxidation of base
C、The formation of pyrimidine dimer
D、The deamination of cytosine
8、Which of the following are NOT true about DNA polymerase I in E. coli:
A、It can repair the gap in the single-strand DNA.
B、It can replace DNA polymerase III to add dNTP into DNA chain when DNA template is damaged.
C、Using dNTP as substrate.
D、It possesses activity of 5’→3’ endonuclease.
9、Which of the following can lead to frameshift:
A、Alkylation
B、Insertion of intercalator
C、Loss of single base
D、Oxidative deamination
10、UvrD is a kind of exonuclease, it can hydrolyze single-strand DNA harboring mismatch site in mismatch repair system, and also has same function in nucleic acid excision repair system.
11、The reason for the deamination of both 5-methylcytosine and cytosine is the same, while the mutagenesis rate of the former is higher than the latter.
Chapter 5 Transcription in prokaryotes
5-1 Transcription and RNA polymerase随堂测验
1、Th
中国大学Molecular Biology_5
中国大学的分子生物学专业是近年来备受关注的一个专业，随着生物科技的迅猛发展，分子生物学在生命科学领域的重要性越来越显著。
课程设置
中国大学Molecular Biology_5的课程设置涵盖了分子生物学领域的主流研究方向，如基因结构与调控、细胞信号转导、蛋白质结构与功能、分子遗传学等。其中，基因结构与调控课程着重介绍基因组的组成、组织、结构及其调控机制；细胞信号转导课程介绍细胞内的信号传导网络及其调控；蛋白质结构与功能课程讲解蛋白质的结构、功能及其调控；分子遗传学课程研究基因的转录、翻译及其调控。通过这些课程的学习，学生能够全面掌握分子生物学的基础知识和研究方法。
实验课程
实验课程是中国大学Molecular Biology_5专业的重要组成部分。实验内容包括PCR扩增、基因克隆、细胞培养、蛋白质分离纯化等。学生在实验中能够深入了解实验原理和步骤，熟悉实验操作技能，提高科学思维和实验设计能力。同时，实验课程也为学生开展科研工作打下了良好的基础。
学科竞赛
中国大学Molecular Biology_5专业鼓励学生参加各种学科竞赛，如全国大学生生命科学竞赛、国际分子生物学奥林匹克竞赛等。参加学科竞赛能够锻炼学生的综合能力，提高实验技能和科研水平，并为日后的科研和就业提供更多机会。
就业前景
中国大学Molecular Biology_5专业的毕业生可以在生物医药、生物科技、食品安全等领域从事基础研究、新药研发、质量控制、技术开发等工作。毕业生可以选择去大型生物医药企业、科研院所、教育机构等单位工作，也可以选择创业或者继续深造攻读硕士、博士等学位。
总结
中国大学Molecular Biology_5专业以其扎实的分子生物学理论基础、丰富的实验课程及广泛的学科竞赛，培养出了大量优秀的毕业生，为推进我国生命科学事业的发展做出了重要贡献。
