细胞生物学习题

３、 染色质纤维上非组蛋白的主要功能是 （a） 组装核小体 （b） 调控基因表达 （c） 组成异染色质

（d） 协助DNA卷曲成染色体

４、 染色体的核小体是由四种组蛋白分子构成的： （a） 异二聚体 （b） 四聚体 （c） 八聚体

５、 真核细胞中编码蛋白质的基因是由（ ）所转录的 （a） RNA多聚酶I （b） RNA多聚酶II （c） RNA多聚酶III （d） Klenow大片段 4、是非题：

1. 不论转录活性是高是低，细胞中核孔的密度和总数都是一样的………………………………………………………（ ）

2. 肌球蛋白分子的头部具ATP酶活性。………………………………（ ） 3. 染色体中DNA 与非组蛋白的重量之比总是1：1。……………（ ） 4. 染色质上不活跃转录的基因DNA的甲基化程度普遍高于活跃转录的基因。………………………………………（ ）

5. 常染色体的所有基因都具有转录活性。………………………（ ）

6. 染色体上由于“位置效应”形成的非活性区在所有细胞后代中都能稳定的遗传下去。………………………………（ ）

7. 外层核膜与内质网相连续。……………………………………（ ） 8. 多线染色体上的胀泡是翻译水平较高的地区。………………（ ） 9. 细胞核是关键的细胞器之一，没有细胞核的细胞是不能存活的。……………………………………………………（ ）

10. 核仁在蛋白质合成旺盛、活跃生长的细胞中很大，在不具蛋白质合成能力的细胞中很小。…………………..…（ ） 5、 答题：

1. 细胞核的基本结构及其主要功能。（尽量简要）

2.. 孔复合体的结构模型是什么？（注意：陈建国认为P251，栓可能是正在运入/运出核的物质，而非核孔复合体，但出在试题中，还是要按书上答） 3. 概述染色质的类型及其特征。

4. 比较组蛋白与非组蛋白的特点及其作用？（要尽量简要，前者参与构成染色体，后者参与基因表达）

5. 试述核小体的结构要点及其实验证据？ 6. 请叙述染色质包装的多级螺旋模型？ 7. 概述核仁的结构及其功能。

8. 已有一瓶长成致密单层的Hela细胞和质粒中含有人的rRNA基因克隆的E.Coli，试设计一个实验证明rRNA存在于Hela细胞的哪几条染色体上？（北大考研细胞，1992）

9. 已有一只受精的鸡卵和质粒中含有rRNA基因克隆的E.Coli菌株，试设计一个突验显示rDNA存在于鸡8.12的细胞时哪些染色体上。（北大考研细胞，1994）

10. 核孔复合物的结构和功能是什么？（北大考研，细胞2001） 11. 细胞核是由哪几部分组成？说明核孔复合体的结构和功能。 12. 核层是怎样构成的？有什么作用？ 13. 根据含珠模型阐明染色质纤维的结构。

14. 组蛋白和非组蛋白在染色质中的作用是什么？有何实验根据？ 15. 阐明核仁基本结构和功能。

16. 什么叫核基质？它对核结构和基因表达有何作用？

17. 试述染色体结构与基因活性的关系？在个体发育过程中染色体结构发生哪些变化，影响了基因活动。

18. 通过对基因组大小及基因结构差异的比较，试述这些差异对物种进化的意义。 19. 主题:请教甲基化的表示问题

请问mCGmm1G中，甲基化m在前面与在后面个各代表什么意思?

RE:mCGmm1G中，甲基化m在前面代表碱基上的甲基应有数字上下标,在后面个代表核糖环2'-OH位上的甲基,无数字上下标.你的题大概未打全. 19…有声无名主题:ＤＮＡ复制时端部为何会缩短？

原核生物的ＤＮＡ是环装的，无所谓端部。可真核生物中，ＤＮＡ是线形的，其末端复制时主要是由于不够一个冈奇片段舍弃复制还是由于ＲＮＡ引物的部位由于没有引物而不能起始复制？

RE: 端粒酶中有RNA,可以碱基配对方式接在一条SSWDNA的5'末端附近,以端粒酶中的RNA作模板使一条SSDNA原先作引物的部分变为DNA延伸部分合成SSDNA,端粒酶是反转录酶. 参见:

翟中和等,细胞生物学,高教出版社,2000. 20. 去又看了一下细胞

cmh79觉得各位没有明白我的意思：

对于端粒，发生复制是可以产生两个子代双链---------一条（就是后随连）就如书上所说，倒也行的通；可是另一条（也就是先导链）它应该是没有3’突出端了，那么它的telemere怎么产生？（肯定不是有端粒酶来延长-------正常细胞没有该酶活性！）本来画了一个图，可是贴不上来，没办法！

21. Where would you expect to find phenylalanine residues in a folded protein? What amino acid residues would

you expect to find in the loop regions connecting different α helices?

Since phenylalanine residues are hydrophobic, they would probably be located in α helices or β sheets within the interior of the protein. The loop regions connecting these elements of secondary structure would be expected to contain hydrophilic amino acids. 22．Repetitive DNA sequences were first identified by studies of rates of DNA reassociation. What relative rates of reassociation are expected for sequences repeated 1000 times in the genome compared to genes with only a single copy?

DNA reassociation is a bimolecular reaction, so the reaction rate is proportional to the square of the DNA concentration. A sequence repeated 1000 times in the genome would therefore reassociate with a rate 106-fold greater than that of single-copy DNA.

23、 ow many histone molecules are associated with the chromosomes of a human cell?

Assuming that a typical cell contains about 1 ng of protein, what fraction of total cellular protein

corresponds to histones? Given one nucleosome every 200 base pairs, there are 3 × 107

nucleosomes in a diploid human cell. Each nucleosome contains two molecules of the four core histones and one molecule of histone H1, so there are nearly 3 × 108 histone molecules in the cell. Taking the average molecular weight of the histones as approximately 15,000 daltons, the total mass of histones is approximately 0.008 ng—nearly 1% of total cell protein.

24、Why are YAC vectors useful for analysis of complex genomes? What is the role of telomeres in these vectors?

Because they can accommodate inserts containing thousands of kilobases, YACs can be used for mapping large genomes, which may contain thousands of megabases of DNA. Telomeres maintain the ends of YACs, allowing them to replicate as linear, chromosome-like molecules in yeasts.

25、Assuming that yeast and C. elegans utilize a similar number of proteins to carry out basic cellular functions, what fraction of C. elegans genes are expected to encode proteins involved in aspects of cell regulation specific to multicellular organisms?

C. elegans and yeast contain about 19,000 and 6,000 genes, respectively, so about two-thirds of C. elegans genes are expected to specifically function in multicellular organisms.

26、Approximately 30,000 cDNAs have been localized on the human genome map. What is the average distance between these markers? Approximately 100 kilobases.

27、How is the fidelity of DNA replication affected by the fact that DNA polymerase adds nucleotides only to a primer strand that is hydrogen-bonded to the template?

This property of DNA polymerases is necessary for proofreading because it enables the polymerase to recognize and excise mismatched bases that are not hydrogen-bonded to the template strand.

28、Would you expect the RNA fragments synthesized by primase to be accurate copies of the template DNA? How does this affect the overall accuracy of DNA replication?

Primase, like other RNA polymerases, is not capable of proofreading, so errors occur at a

comparatively high frequency in RNA primers. However, since the primers are later removed, the overall fidelity of replication is not compromised.

Patients with xeroderma pigmentosum suffer an extremely high incidence of skin cancer but have not been found to have correspondingly high incidences of cancers of internal organs (e.g., colon cancer). What might this suggest about the kinds of DNA damage responsible for most internal cancers? The wild-type gene will be regulated normally, but the temperature-sensitive gene will be expressed constitutively. β-Galactosidase will therefore be produced at the permissive but not the nonpermissive temperature.

29、Patients with xeroderma pigmentosum suffer an extremely high incidence of skin cancer but have not been found to have correspondingly high incidences of cancers of internal organs (e.g., colon cancer). What might this suggest about the kinds of DNA damage responsible for most internal cancers?

The high frequency of skin cancer results from DNA damage induced by solar UV irradiation,

which is subject to repair by the nucleotide-excision repair system. The lack of elevated incidence of other cancers may suggest that similar types of damage are not frequent in internal organs and that most cancers of these organs result from other types of mutations (e.g., the incorporation of mismatched bases during DNA replication).

30、RecA mutants of E. coli are sensitive to UV irradiation in addition to being recombination-deficient. Why?

RecA mutants are sensitive to UV irradiation because they are deficient in recombinational repair of DNA damage.

29、 What phenotype would you predict for a mutant mouse lacking one of the genes required for site-specific recombination in lymphocytes?

The mouse would be immunodeficient, lacking both B and T lymphocytes, as a result of being unable to rearrange its immunoglobulin and T cell receptor genes.

31、The consensus sequence of the E. coli -10 promoter element is TATAAT. You are comparing two promoters that have -10 element sequences of TATGAT and CATGAT, respectively. Which would you expect to be transcribed more efficiently?

The promoter containing the sequence TATGAT, which more closely resembles the consensus sequence, will be transcribed more efficiently. 32、You are working with two strains of E. coli. One contains a wild-type β-galactosidase gene and an i- mutation; the other contains a temperature-sensitive β-galactosidase gene and an oc mutation. After mating these strains, you assay for the production of β-galactosidase at both permissive and nonpermissive temperatures in the absence of lactose. What do you expect to find?

The wild-type gene will be regulated normally, but the temperature-sensitive gene will be

expressed constitutively. β-Galactosidase will therefore be produced at the permissive but not the nonpermissive temperature. 33、You are comparing the requirements for in vitro basal transcription of two polymerase II genes, one containing a TATA box and the other containing only an Inr sequence. Does transcription from these promoters require TBP or TFIID?

The promoter containing the TATA box can be transcribed in vitro in the presence of either TBP or TFIID. However, the Inr promoter requires TFIID, since the Inr sequence is recognized by TAFs rather than by TBP. 34、You are studying the enhancer of a gene that normally is expressed only in neurons. Constructs in which this en-hancer is linked to a reporter gene are expressed in neuronal cells but not in fibroblasts. However, if you mutate a specific sequence element within the enhancer, you find expression in both fibroblasts and neuronal cells. What type of regulatory protein would you expect to bind to that enhancer element?

The sequence element would be a candidate binding site for a tissue-specific repressor

35、A transcription factor is found to activate transcription by binding to different DNA sequences in muscle cells and liver cells. How might alternative splicing be involved in determining this tissue specificity of activator function?

Two different DNA-binding domains could be joined to the factor's activation domain by