I want to know the answers to some molecular biology questions.

Short answer: 1. Please classify eukaryotic DNA into different categories according to its complexity and describe its properties. A: There are three types: 1. Highly repetitive sequences with high copy number and low cot value. 2. Non-repetitive sequences with single copy number and high cot value. 3. The sequence of moderate repetition between them. Features: 1. It accounts for about 10% of the total DNA and can be divided into three categories: satellite DNA, microsatellite DNA and microsatellite DNA. 2. It accounts for 20%~80% of the total DNA and can be divided into coding sequence and non-coding sequence. 3. Most structural genes and single copy sequences. In the genome, a single copy sequence stores huge genetic information and codes protein with different functions. Please describe the process of retrovirus reverse transcription and integration. Please describe the process of retrovirus and retrovirus integration. A: Reverse transcription is divided into two steps. The first step is the process of synthesizing DNA negative strand with RNA as template under the action of reverse transcriptase. The host-unloaded tRNA will appear in the virus particles, and the 18bp sequence at the 3' end of the trna can be base-paired with the site at the 5' end of the virus RNA molecule about 100~200bp. When it reaches the end point, the synthesis will stop temporarily. The R region at the 5' end is degraded, so that the R region at the 3' end can be base-paired with the newly synthesized DNA, and then the whole RNA is transcribed into DNA from the 3' end by reverse transcriptase. The result of template transformation and extension is to add a U3 sequence at the 5' end to form the first LTR. The second step is the process of synthesizing DNA positive strand using DNA negative strand as template. First, the tRNA primer is degraded, then the RNA is degraded, and the remaining fragments are used as primers for DNA synthesis. The process of integration into host DNA is the process from linear DNA to protovirus. Mainly catalyzed by integrase. Please describe the structure and function of DNA pol I and pol II. Please describe the structure and function of DNA polymerase I and II. Answer: Pol I is a single subunit protein, encoded by Pol A gene, with a molecular weight of 65,438+million. Besides the ability to synthesize DNA, POLI has 3'5' nucleic acid exonuclease activity and 5'3' nucleic acid exonuclease activity. Its main functions are DNA repair and RNA primer excision during DNA replication. Pol II has DNA polymerase activity and 3'5' nucleic acid exonuclease activity, but does not have 5'3' nucleic acid exonuclease activity. Its main function is DNA repair. 4. Please describe the DNA replication of prokaryotes. Please describe the beginning of DNA replication in prokaryotes. A: First of all, DnaA protein is used to find the replication initiation site. It can recognize and bind the 9bp repeats of OriC. Once these four repeats are occupied, more than 20 additional DnaA will combine with OriC in a synergistic manner to form the initial complex. Then DnaA untied three AT-rich repeats on the left side of the starting site to form an open complex. With the help of DnaC, DnaB was bound to this open starting site in the form of hexamer, forming a pre-initiated complex. DnaB is a kind of helicase, which can be used to further untie DNA double strands, and the formed DNA unrequited love is quickly combined by SSB tetramer to prevent DNA chain recombination. Then, with the help of other protein, primer enzymes with the function of producing RNA primers are also combined to form primers, and then RNA primers are synthesized on the leading strand and the lagging chain respectively using DNA as a template. Once the primer is synthesized, the primer enzyme falls off and waits for the next combination. Then DNA polymerase III binds to replication fork, and the first dNTP is attached to the 3-OH of the RNA primer, and the initial process ends. 5. Please describe the role, mechanism and process of mismatch repair. Please describe the function, mechanism and process of mismatch repair. A: One aspect of keeping the error rate low in DNA replication comes from mismatch repair. Mismatch repair is a kind of repair system, which distinguishes the parent and offspring DNA chains after DNA replication according to the different methylation status, and then uses the mother chain as a template to correct the mismatched bases in the daughter chains. Before DNA methylation, the repair system checks DNA. When mismatched bases are identified, the corresponding enzymes always remove and replace nucleotides from unmethylated chains to ensure the recovery of the original base pairs. After the mismatch repair mechanism is checked correctly, the newly formed sub-chain is methylated, just like labeling it as qualified. 6. Please describe the mechanism, process and consequences of transcnl 7. How to remove the characteristics of introns in pre-tRNA and pre-rRNA of eukaryotes? The characteristics of introns in eukaryotic precursor tRNA and precursor rRNA, how to remove them? Answer: the precursor rRNA belongs to intron I of self-splicing, and it can self-splicing itself from the RNA precursor without enzyme catalysis. It cuts itself off by a two-step transesterification reaction. In the first step, guanine nucleotide is used as a cofactor to provide a free 3'- hydroxyl group attached to the 5' end of intron. In the second step, the 3-OH of exon A attacks the 5' end of exon B, thus releasing the intron and connecting the two exons together. Intron in eukaryotic tRNA precursor is a single type of intron. Taking yeast as an example, there is a sequence complementary to the anti-codon of tRNA, which is located in the downstream region of the anti-codon, and there is no conservative sequence in the intron. The combination of intron excision enzyme and tRNA precursor molecule depends on the identification of secondary structural characteristics of tRNA precursor molecule, not the identification of intron primary sequence characteristics. The process of intron excision is firstly the recognition and cleavage of substrate, and no energy is needed. A special endonuclease cuts both ends of intron in the precursor of tRNA to remove intron. Then there is the ligation reaction. In yeast and plants, cyclic phosphorus groups are opened under the action of cyclic phosphotriesterase, and the products formed are 2'- phosphate groups and 3' hydroxyl groups. Finally, in the presence of ATP, ligase connects two tRNA molecules together. For mammals, ligase can directly connect the 2', 3'- cyclic phosphate group of RNA with the 5'- hydroxyl terminal to form a normal 5', 3' phosphodiester bond instead of generating two additional phosphate groups. 8. Characteristics of eukaryotic precursor mRNA introns and how to remove them? The characteristics of eukaryotic precursor mRNA introns, how to remove them? A: Most introns in mRNA precursors are GU-AG introns. They can't splice RNA only through their own sequences. They complete the RNA splicing process with the catalytic help of splicer. His 5- splicing site contains a conserved GU sequence, and his 3- splicing site contains a conserved AG sequence. The process of intron excision is divided into three parts. In the first stage, the 5' end of intron is cut to form a free left exon and a right intron-exon molecule, forming a lasso structure. In the second stage, the 3-terminal splicing site is cut off and then released in the form of lasso. The third stage is that the exon on the right is connected with the exon on the left. 9. what is the function of prokaryotic RNA pol subunit? What is the function of prokaryotic RNA polymerase subunit? A: The α subunit is necessary to equip ribozymes and plays a certain role in promoter recognition. It also plays a role in the interaction between RNA polymerase and other regulatory factors. β' subunit is the most abundant subunit in RNA polymerase, and its function is to bind to DNA. The function of β subunit is to combine with NTP and has catalytic polymerization activity. β' and β * * * together constitute the active center of RNA synthesis. Sigma factor can recognize the promoter. The function of ω subunit is to promote the assembly of RNA polymerase. 10. What is the role of CTP of RNA pol II in RNA transcription and processing? The role of RNA polymerase Ⅱ CTD in RNA transcription and processing A: CTD is the carboxyl terminal binding domain of the largest subunit of RNA polymerase 2. It consists of 52 repeated heptapeptides, and the phosphorylation of CTD is related to the beginning of transcription extension stage. Phosphorylation of polymerase promotes the separation of enzyme from GTF and promoter, and the enzyme moves from the initial complex along the DNA template. CTD can also be used as a platform to combine with many protein related to RNA processing. 1 1. Transcription termination of eukaryotic mRMA and generation of 3'poly(A) tail? Transcription termination of eukaryotic mRNA and formation of 3'- terminal polyA A: There are two models of transcription termination of eukaryotic mRNA. The first model is the allosteric model. Phosphorylated CTD can bind protein factors related to RNA cleavage and polyadenylation. These protein factors can recognize the signal of polyadenylation loaded on the transcribed RNA, and then cut the RNA by endonuclease, and then they all fall off from the RNA polymerase complex, thus changing the conformation of RNAP and weakening the continuous synthesis of RNA. The second model is a torpedo model. CTD combined with guanylate transferase which catalyzes the formation of 5- terminal cap can add 5- terminal cap structure to newly synthesized RNA. This structure can be recognized by exonuclease, and the exonuclease of nucleic acid is digested continuously from 5 to 3 ends until it catches up with RNA polymerase, so that RNA polymerase falls off the DNA template and transcription stops. Poly(A) polymerase can continuously load ATP to the 3- terminal of RNA, resulting in poly(A) tail. The terminal A can be used as a template for poly (a). 12. Transcription initiation of eukaryotic RNA pol ii iii? Eukaryotic RNA polymerase Ⅱ Ⅱ Ⅲ A: POL Ⅱ: Initiator and TATABOX constitute the core promoter, which is a necessary and sufficient condition for transcription initiation, but its transcription efficiency is relatively low, which requires the participation of activating proteins. A few of them have no TATBOX promoter, and their core promoters are composed of promoters and DPE elements. Upstream of the core promoter are two regulatory elements, CAATBOX and GCBOX. Transcription initiation requires GTF, which is a universal transcription factor. TBP can recognize and bind TATABOX, TAF is the linking factor of TBP, TF2D of TBP binds TATABOX, and TF2A can stimulate the transcription process by stabilizing the interaction between D and TATABOX. TF2E and TF2H combine to form a preinitiated complex. TF2H phosphorylates CTD, and transcription begins. Pol I: Only the rRNA gene is transcribed, and the promoter consists of the core promoter and the upstream UCE. The core binding factor is mainly responsible for ensuring that RNA polymerase is located at the starting point correctly, and the auxiliary factor UBF increases the transcription frequency, so that the core binding factor can bind to the core promoter more effectively. Pol III: According to the different positions of promoters, it can be divided into internal promoters and upstream promoters. Internal promoters can be divided into 1 type and type 2, both of which need the participation of TF3ABC. Type 1: firstly, TF3A and BOXA combine, resulting in the combination of TF3C and BOXC, then TF3B and the starting point combine, and then polymerase combines. Type2: Different from type 1, TF3C binds to cassettes A and B at the same time, so that TF3B binds to the starting point and polymerase binds. Type3: firstly, snRNA activating protein complex SNAPc binds to PSE, then SNAPc can help TF3B bind to TATABOX, and finally, with the help of TF3B, RANP3 binds to the transcription start point. 13. Transcription termination in prokaryotes A: Transcription termination can be divided into Rho factor-dependent and Rho factor-independent types. Independence: First, when RNA polymerase transcribes two GC-rich reverse repeats, it enters the oligo-U synthesis region; Because these two GC-rich sequences are complementary, these two sequences and the non-repetitive sequence in the middle form a neck ring structure; The formation of this neck ring structure or hairpin structure will make RNA polymerase slow down RNA synthesis, or simply stop RNA synthesis, so that RNA polymerase stays in the polyU synthesis region. At this time, the RNA-DNA hybrid chain will be melted at the weak rU:dA base pairing in the termination region, and the RNA chain will fall off from the DNA template chain, and the transcription will be terminated. Dependent: RNA polymerase is transcribed first, and then Rho factor recognizes and binds to the rut site on the transcribed RNA. Then, driven by ATP hydrolysis, RNA polymerase is chased along RNA. Rho may catch up with RNA polymerase when RNA binding enzyme stops when it encounters terminator. Rhp factor melts RNA-DNA hybrid chains in transcription vesicles, and the interaction between Rho and RNA polymerase may participate in the melting process. Then the transcription stopped. 14. Conservative characteristics of bacterial promoter 15. Fidelity of AA-trna synthesis.