尔雅Medical Molecular Biology章节答案(学习通2023题目答案)

点击查看答案

Chapter 1 The尔雅 structure and function of nucleic acid

The structure and function of nucleic acid

1、Which of the following are pyrimidines?节答
A、adenine and cytosine
B、案学adenine and guanine
C、习通adenine and thymine
D、题目cytosine and thymine
E、答案guanine and thymine

2、尔雅Which kind of nucleotide does not exist in DNA？
A、节答dAMP
B、案学dTMP
C、习通dUMP
D、题目dGMP
E、答案dCMP

3、尔雅Which of the following is 节答the connection of primary structure of nucleic acid molecule?
A、2 ′,案学 3 ′ phosphodiester bond
B、3 ′, 5 ′ phosphodiester bond
C、2 ′, 5 ′ phosphodiester bond
D、glycosidic bond
E、hydrogen bond

4、In nucleotide molecule, which carbon atom of ribose and the purine N9 linked by glycosidic bond?
A、C-1 '
B、C-2 '
C、C-3 '
D、C-4 '
E、C-5 '

5、Which of the following may be right if a nucleic acid has the following base composition: A = 16%, G= 39%, C = 22%, T= 23% (C)?
A、double strand DNA
B、tRNA
C、single strand DNA
D、single strand RNA
E、double strand RNA

6、Which of the following refers to melting temperature of DNA?
A、The temperature when A260 reached the maximum
B、The temperature when A260 reaches the maximum value of 50%
C、DNA begin unwinding of temperature
D、The complete unwinding DNA strands of the temperature
E、The temperature when A280 reaches the maximum value of 50%

7、In nucleic acid molecules, which of the following is the key part that the genetic information is stored in?
A、pentose conformation
B、The rotation angle of bases
C、The base sequence
D、pentose phosphate backbone
E、phosphate ester bond

8、Which are correct about the descriptions of nucleic acid molecular hybridization?
A、It refers to renaturation of denatured DNA
B、DNA can be hybridized with RNA to form a double helix
C、RNA can be hybridized with its encoded polypeptide chain
D、It can detect the homology between hybrid nucleic acids
E、?DNA or RNA from different sources can be hybridized as long as some bases can pair

9、Which are correct about the descriptions of mRNA?
A、Among the three RNAs, the metabolic rate is the fastest?
B、The secondary structure contains local double helix
C、It consists of two subunits
D、There are m7Gppp cap and polyA tail
E、It contains genetic information

10、The sequence of a DNA strand is 5 '- A-T-T-C-3', which can be its complementary strand?
A、5′-T-A-A-G-3′
B、5′-T-C-C-G-3′
C、5′-G-A-A-T-3′
D、5′-U-A-A-G-3′
E、5′-G-A-A-U-3′

The structure and function of nucleic acid

1、What is the background and significance of DNA double helix structure? What inspiration can you get from it?

The structure and functio of nucleic acid

1、What is the structure of nucleosome? What is the key points of DNA double helix?

Chapter 2 DNA replication

DNA replication

1、Which statement is correct about DNA pol in prokaryotes
A、DNA-pol I is the major DNA replicase
B、DNA-pol II has highest activity, is the major enzyme in DNA replication
C、DNA-pol III is the major DNA replicase, and has the 3′→5′ exonuclease activity
D、DNA-pol III seal the nick during replication and repair
E、DNA-pol II has 5′→3′ exonuclease activity

2、The function of DNA topoisomerase is
A、to stable unwound DNA duplex
B、to recognize replication origin site
C、to synthesize RNA primer
D、to unwind DNA duplex, in order to replicate
E、to unwind DNA super coil, avoid tangling together

3、Which process doesn’t require DNA ligase
A、DNA damage repairing
B、reverse transcription
C、DNA replication
D、DNA recombination
E、genetic engineering

4、The function of RNA primer is
A、to separate DNA double helix
B、to provide 3′-OH to synthesize DNA molecule
C、to provide 5′-P to synthesize DNA molecule
D、to provide 5′-P to synthesize RNA molecule
E、to release supercoil of DNA

5、Which enzyme can use RNA as template to synthesize DNA molecule
A、RNA pol
B、reverse transcriptase
C、DNA topoisomerase
D、DNA helicase
E、DNA ligase

6、About DNA replication origin site, which one is correct
A、has one origin site only in prokaryotes
B、has multi-origin site in prokaryotes
C、has one origin site in eukaryotes
D、recognize and bind by β subunit of DNA-pol III
E、recognize and bind by primase

7、Which statement about DNA replication is WRONE
A、require primer
B、is bidirectional replication
C、is semiconservative replication
D、is semidiscontinuous replication
E、the orientation of newly synthesized chain is 3’→5’

8、The biological activity of prokaryotic DNA pol III involve:
A、5′→3′ exonuclease activity
B、5′→3′ polymerization activity
C、3′→5′ exonuclease activity
D、3′→5′ polymerization activity
E、ligase activity

9、Which statements about telomere and telomerase are correct:
A、Telomerase consists of DNA and proteins
B、Telomerase consists of RNA and proteins
C、Telomere is specialized structure at the end of a linear eukaryotic chromosome
D、Telomere consists of DNA and proteins
E、Telomerase is special reverse transcriptase

10、Which statements about reverse transcription are correct:
A、require reverse transcriptase
B、only appear in prokaryotes
C、dNTP act as substrate
D、use RNA as template to synthesize DNA
E、use RNA as template to synthesize RNA

DNA replication

1、Please list all materials that involved in whole process of DNA replication and discribe their functions.

Chapter 3 RNA transcription

Chapter 3 RNA transcription

1、Pribnow box refers to
A、AATAAA
B、AAUAAA
C、TAAGGC
D、TTGACA
E、TATAAT

2、RNA polymerase holoenzyme initiates transcription which involves
A、recognition of the -10 sequence and -35 sequence
B、recognition of the -10 sequence onl
C、recognition of the -35 sequence only
D、recognition of the -100 sequence only
E、none of the above

3、In eukaryotes, the RNA polymerase responsible for mRNA transcription is
A、RNA polymerase I
B、RNA polymerase II
C、RNA polymerase III
D、RNA holoenzyme
E、RNA core enzyme

4、Pre-initiation complex refers to
A、RNA polymerase binds to TATAAT sequence
B、RNA polymerase binds to TATA sequence
C、mutiple transcription factors bind to RNA polymerase and DNA template
D、σ factor binds to RNA polymerase
E、repressor dissociates from operator

5、Exons are
A、result of gene mutation
B、digested DNA fragments
C、un-transcribed DNA
D、coding sequence of eukaryotic gene
E、noncoding sequence of eukaryotic gene

6、Rho-independent termination of transcription in E. coli
A、requires ATP
B、requires about 20 nucleotides of hairpin loop
C、?requires Rho protein
D、Rho protein binds with RNA
E、Rho protein has ATP helicase activity?

7、Transcription factors are
A、promoters
B、TATA box
C、proteins binding to DNA and regulating transcription
D、CATT box
E、GC box

8、Which of the following enzyme can bind to DNA?
A、RNA-pol
B、DNA-pol
C、capping enzyme
D、DNA topoisomerase
E、DNA helicase

9、Post-transcriptional modification of eukaryotic mRNA include
A、addition of5’-cap
B、addition of3’polyA
C、splicing
D、mRNA editing
E、chemical modification to get rare base

10、Which of the following statement regarding the poly A tail of eukaryotic mRNA is correct?
A、The poly A tail is located at the 3′ end of RNA
B、The poly A tail is transcribed from the DNA template
C、Poly A tails can protect mRNA
D、Poly A polymerase is a DNA-independent RNA polymerase involved in polyadenylation of RNA
E、Most of eukaryotic mRNA have a ploy A tail

RNA transcription

1、Please list the difference between DNA replication and RNA transcription.

Chapter 4 Protein Synthesis

Protein synthesis

1、How many genetic codons exist in the cell?
A、20
B、4
C、32
D、64
E、96

2、Which of the following is not the properties of genetic codes
A、Commaless
B、Universality
C、Degeneracy
D、Exclusion
E、Wobble

3、What is the amino acid that the start codon encode?
A、Asparagine
B、Cysteine
C、Serine
D、Leucine
E、Methionine

4、Amino acid activation in the process of translation is:
A、Binding of amino acids to corresponding rRNA
B、Binding of amino acids to small ribosome subunits
C、Binding of amino acids to corresponding tRNA
D、Binding of amino acids to corresponding mRNA
E、Binding of amino acids to large ribosomal subunits

5、The function of rRNA is to
A、Synthesize DNA
B、Synthesize mRNA
C、Form ribosomes
D、Transfer amino acids to ribosomes
E、Templet for translation

6、The Shine-Dalgarno sequence can be found in
A、30S ribosome
B、mRNA
C、5S RNA
D、16S RNA
E、tRNA

7、Put the following steps of prokaryotic translation initiation in the correct order (1) Binding of mRNA to 30S subunit (2)binding of fMet-tRNAfMet to P site (3) Dissociation of 50S subunit from 30S subunit (4) Binding of 50S subunit to 30S subunit
A、3214
B、3124
C、1423
D、1234
E、2341

8、The 3 important functional parts of the prokaryotic ribosome are
A、A site -aminoacyl site: binding aminoacyl -tRNA
B、P site - peptidyl site: binding peptidyl -tRNA
C、E site - exit site: release the unloaded tRNA
D、alone in the large subunit
E、is formed by 60s and 40S subunits of the ribosome

9、Peptide chain biosynthesis needs:
A、ATP or GTP for energy
B、Mg2+, transpeptidase, aminoacyl -tRNAsynthetase
C、initiation factor
D、elongation factor
E、release factor

10、Post-translational processing includes
A、folding
B、disulfide bond formation
C、peptide hydrolysis.
D、Chemical modification of amino acid residues
E、subunit polymerization

protein synthesis

1、What kind of specific amino acid can initiates protein synthesis in bacteria? How does that specific amino acid formed?

Chapter 5 The Regulation of Gene Expression

The Regulation of Gene Expression

1、The lac operon is shut off when
A、Lactose is present
B、Lactose is absent
C、Fructose is present
D、Glucose is absent
E、Galactose is present

2、Prokaryotic RNA polymerase specific for some genes transcription is governed by
A、σ subunit
B、α subunit
C、γ subunit
D、δ subunit
E、β subunit

3、The repressor in lactose (lac) operon is encoded by
A、Z gene
B、Y gene
C、I gene
D、A gene
E、X gene

4、Proteins that can bind to DNA and prevent transcription are called:
A、positive regulatory protein
B、trans- acting factor
C、inducer
D、repressor protein
E、catabolic gene activating protein

5、The positive regulation mediated by the combination of cAMP and CAP occurs in
A、has glucose and high cAMP
B、has glucose and low cAMP
C、without glucose and high cAMP
D、without glucose and low cAMP
E、high concentration of glucose and cAM

6、Which location of the operon does RNA polymerase bind to
A、initiation region of structure gene
B、CAP binding site
C、regulatory gene
D、promoter sequence
E、S-D sequence

7、House-keeping gene
A、Its expression can be induced under some environmental signals
B、Its expression can be repressed under some environmental signals
C、Belongs to regulatable gene
D、Express only in specific cells and specific stage of life
E、Express in almost all cells and in all life period

8、Which of the following is NOT involved in the Lacoperon?
A、Z gene
B、Y gene
C、Enhancer
D、Operator
E、Promoter

9、Which of the following is the importance of Gene expression regulation?
A、Maintaining the growth, development and reproduction of organisms
B、Adapting to the environmental changes
C、Maintaining the development and cellular differentiation of the organism
D、Keeping all the gene expressing all the time
E、The expression of genes can be regulated at transcriptional level only

10、The DNA sequence that regulates the transcriptional activity of the gene include
A、promoter
B、enhancer
C、silencer
D、exon
E、CAP binding site

The Regulation of Gene Expression

1、Describe the regulation of the lac operon by lac repressor and CAP.

Chapter 6 Recombinant DNA technology

Recombinant DNA Technology

1、Which of the following is the basic conditions for gene cloning vector ?
A、It contains resistance genes
B、Expression regulatory elements
C、Transcription ability
D、Translation ability
E、Self replication ability

2、Which of the following is not a characteristic of a plasmid?
A、It exists in the cytoplasm of most bacteria
B、Double stranded linear molecule
C、About several thousand base pairs
D、Easy to transfer from one bacterium to another
E、Often carrying drug resistance gene

3、What is correct about the characteristics of ideal plasmid vector?
A、It is linear single stranded DNA
B、It contains single site of multiple restriction enzyme
C、It contains multiple sites of the same restriction enzyme
D、Its replication is controlled by the host
E、It does not contain drug resistance gene

4、In gene cloning, Which of the following refers to DNA recombinants?
A、Renaturation of two DNA single strands from different sources
B、The linker formed between target gene and vector
C、The linker formed between prokaryotic DNA and eukaryotic DNA
D、The linker formed by two different structural genes
E、The linker formed between DNA and RNA

5、Which of the ligation mode is directional and the efficiency is the highest between the target gene and the vector?
A、The sticky end produced by one restriction endonuclease
B、The blunt end produced by one restriction endonuclease
C、Sticky ends produced by two different restriction endonucleases
D、The blunt end produced by two restriction endonucleases
E、Two blunt ends produced by the filling of sticky ends

6、In DNA recombination, Which of the enzyme catalyze the formation of recombinant DNA molecules?
A、Lyase
B、Reverse transcriptase
C、DNA polymerase
D、DNA ligase
E、Topoisomerase

7、Which of the following can be simply summarized as operation procedure of genetic engineering?
A、Application of restriction endonuclease
B、Isolation, cut, ligation, transformation, screening and express
C、The recombinant plasmid was introduced into the host cells and the positive cell lines were screened out
D、The vector was ligated with the target gene to form recombinant DNA
E、Isolation of vector and target gene

8、Which of the following are required in genetic engineering?
A、Gene vector
B、Restriction endonuclease
C、DNA ligase
D、Terminal nucleotide transferase
E、Ribozyme

9、The plasmid used for gene cloning does not exist in
A、Viral genome
B、Bacterial extra chromosome
C、Bacterial chromosome
D、Yeast chromosome
E、Mitochondrial genome of mammalian cells

10、Which of the following is wrong about competent cells?
A、Host cells after ingestion of recombinant DNA
B、Antibiotic sensitive host cells
C、Host cells in optimal uptake and tolerance of recombinants
D、Host cells recombined with exogenous DNA
E、Host cells in their natural state

Recombinant DNA Technology

1、What are the basic mechanisms of genetic engineering?

Chapter 7 Cellular Signal Transduction

Cellular Signal Transduction

1、Which enzyme could be directly activated by NO in smooth muscle relaxation?
A、Protein kinase A
B、Protein kinase C
C、Adenylyl cyclase
D、Guanylyl cyclase
E、MAPK

2、Which enzyme is responsible for the degradation of cAMP?
A、Adenylyl cyclase
B、Guanylyl cyclase
C、Protein kinase A
D、Protein kinase C
E、Phosphodiesterase

3、Which second messenger(s) will be produced after the hydrolysis of PIP2 (phosphatidylinositol 4,5 bisphosphate)?
A、cAMP
B、cGMP
C、DAG and IP3
D、DAG only
E、IP3 only

4、In EGFR-Ras pathway, the adaptor protein Grb2 binds to phosphorylated Tyr of EGFR by which domain?
A、SH2
B、SH3
C、Proline-rich domain
D、PH
E、None of the above

5、Which of the following signaling pathway is correct?
A、Gs-cAMP -PKA
B、Gs-cAMP-PKC
C、Gq-cAMP-PKA
D、Gq-Ca2+-PKA
E、Gs-cGMP-PKC

6、Which of the following is NOT extracellular signaling molecules?
A、Glucagon
B、Epinephrine
C、G protein
D、Dopamine
E、Angiotensin II

7、Which of the following signaling molecules could be directly activated by Ca ion binding?
A、cAMP
B、Calmodulin
C、Protein kinase A
D、Protein kinase B
E、Protein kinase G

8、Which of the followings is Ser/Thr kinase?
A、Protein kinase A
B、MAPK
C、Protein kinase C
D、Protein kinase G
E、MEK

9、Which of the following statements regarding to signaling transduction is correct?
A、One extracellular molecule might initiate different signaling pathways
B、One receptor might mediate different signaling pathways
C、One signaling molecule might participate in multiple pathways
D、Different signaling pathways always regulate different biological effects
E、One ligand might activate one or more receptors

10、Which of the followings is the common feature of GPCR?
A、7 transmembrane helixes
B、Extracellular N terminus
C、3 intracellular loops for G protein coupling
D、Multiple polypeptide chains
E、3 extracellular loops for ligand binding

Cellular Signal Transduction

1、What is the activating mechanism of PKA by cAMP?

学习通Medical Molecular Biology

Medical Molecular Biology是一门研究生物分子结构、生物分子功能及其调控、基因表达等相关内容的课程。在这门课程中，学习者将探索细胞遗传学、信号转导、基因调控等重要领域，并了解最新的医学分子生物学研究进展。

课程安排

Medical Molecular Biology课程共分为14个单元，每个单元的内容如下：

• 单元一：分子生物学基础概念
• 单元二：细胞遗传学
• 单元三：DNA复制与修复
• 单元四：转录和RNA加工
• 单元五：翻译和蛋白质后修饰
• 单元六：信号转导概论
• 单元七：细胞生长因子信号转导
• 单元八：细胞凋亡和坏死
• 单元九：细胞周期和细胞增殖
• 单元十：内源性信号调节机制
• 单元十一：免疫应答和炎症反应
• 单元十二：基因调控及其在发育和疾病中的作用
• 单元十三：疾病基因组学
• 单元十四：新药研发

知识点概述

以下为Medical Molecular Biology课程涵盖的主要知识点：

细胞遗传学

细胞遗传学是研究细胞遗传性质和遗传变异的学科。在这门课程中，你将学习以下内容：

• 染色体结构和功能
• 有丝分裂和减数分裂
• 染色体不平衡和异常
• 基因组编辑技术

分子生物学基础概念

分子生物学是研究生命分子结构、功能及其相互关系的学科，是现代生物学的基础。在这门课程中，你将学习以下内容：

• 生物大分子的结构和功能
• 生物分子的互作和调控
• 基本的分子生物学实验技术

信号转导

信号转导是指细胞外信号分子的传递过程，从而调控细胞内的相关分子活动。在这门课程中，你将学习以下内容：

• 信号转导的基本概念和分类
• 常见的信号转导途径和信号转导通路
• 常见的信号分子和信号受体及其作用机制
• 信号转导的调控和疾病相关性

基因调控

基因调控是指细胞内基因表达的调节过程，是生物体发生和发展的基础。在这门课程中，你将学习以下内容：

• 基因表达的基本原理和调控机制
• 转录因子和核心启动子的作用
• 表观遗传调控的作用机制和调控方式
• 基因调控在发育和疾病中的作用

课程特色

Medical Molecular Biology课程具有以下特色：

• 课程内容全面，涵盖分子生物学的基本知识和最新进展
• 课程教学方法灵活多样，采用视频讲解、实验演示、案例分析等方式
• 课程难度适中，适合具有一定生物基础的学习者
• 学习过程中有许多练习题和案例分析，帮助学习者更好地理解和应用所学知识

学习方法

以下是学习Medical Molecular Biology的建议方法：

• 认真阅读课程大纲和每个单元的内容概述，了解课程的结构和学习目标
• 观看课程视频，理解每个单元的核心概念和知识点
• 完成每个单元的练习题和案例分析，加深对所学知识的理解和应用
• 参考课程中推荐的相关论文或书籍，了解最新的研究进展

结语

Medical Molecular Biology是一门涉及广泛、内容丰富的课程，是研究生物分子生物学的重要基础。通过学习这门课程，你将掌握最新的分子生物学知识和应用技术，为未来的科学研究和职业发展打下坚实的基础。

学习通Medical Molecular Biology

Medical Molecular Biology是一门研究生物分子结构、生物分子功能及其调控、基因表达等相关内容的课程。在这门课程中，学习者将探索细胞遗传学、信号转导、基因调控等重要领域，并了解最新的医学分子生物学研究进展。

课程安排

Medical Molecular Biology课程共分为14个单元，每个单元的内容如下：

• 单元一：分子生物学基础概念
• 单元二：细胞遗传学
• 单元三：DNA复制与修复
• 单元四：转录和RNA加工
• 单元五：翻译和蛋白质后修饰
• 单元六：信号转导概论
• 单元七：细胞生长因子信号转导
• 单元八：细胞凋亡和坏死
• 单元九：细胞周期和细胞增殖
• 单元十：内源性信号调节机制
• 单元十一：免疫应答和炎症反应
• 单元十二：基因调控及其在发育和疾病中的作用
• 单元十三：疾病基因组学
• 单元十四：新药研发

知识点概述

以下为Medical Molecular Biology课程涵盖的主要知识点：

细胞遗传学

细胞遗传学是研究细胞遗传性质和遗传变异的学科。在这门课程中，你将学习以下内容：

• 染色体结构和功能
• 有丝分裂和减数分裂
• 染色体不平衡和异常
• 基因组编辑技术

分子生物学基础概念

分子生物学是研究生命分子结构、功能及其相互关系的学科，是现代生物学的基础。在这门课程中，你将学习以下内容：

• 生物大分子的结构和功能
• 生物分子的互作和调控
• 基本的分子生物学实验技术

信号转导

信号转导是指细胞外信号分子的传递过程，从而调控细胞内的相关分子活动。在这门课程中，你将学习以下内容：

• 信号转导的基本概念和分类
• 常见的信号转导途径和信号转导通路
• 常见的信号分子和信号受体及其作用机制
• 信号转导的调控和疾病相关性

基因调控

基因调控是指细胞内基因表达的调节过程，是生物体发生和发展的基础。在这门课程中，你将学习以下内容：

• 基因表达的基本原理和调控机制
• 转录因子和核心启动子的作用
• 表观遗传调控的作用机制和调控方式
• 基因调控在发育和疾病中的作用

课程特色

Medical Molecular Biology课程具有以下特色：

• 课程内容全面，涵盖分子生物学的基本知识和最新进展
• 课程教学方法灵活多样，采用视频讲解、实验演示、案例分析等方式
• 课程难度适中，适合具有一定生物基础的学习者
• 学习过程中有许多练习题和案例分析，帮助学习者更好地理解和应用所学知识

学习方法

以下是学习Medical Molecular Biology的建议方法：

• 认真阅读课程大纲和每个单元的内容概述，了解课程的结构和学习目标
• 观看课程视频，理解每个单元的核心概念和知识点
• 完成每个单元的练习题和案例分析，加深对所学知识的理解和应用
• 参考课程中推荐的相关论文或书籍，了解最新的研究进展

结语

Medical Molecular Biology是一门涉及广泛、内容丰富的课程，是研究生物分子生物学的重要基础。通过学习这门课程，你将掌握最新的分子生物学知识和应用技术，为未来的科学研究和职业发展打下坚实的基础。