细胞生物学资料：00级期末答案.doc

中山大学生科院细胞生物学期末试卷参考答案 (2000级生物科学、生物技术、药学专业, 共164人) 2002年1月6日 姓 名∶ 班 级∶ 一 二 三 四 五 六 七 总 分一、填空题(每空0.5分，共10分)1. 寡聚糖，内质网，高尔基体，出芽，蛋白包被小泡，笼形蛋白包被小泡，调节型，包被蛋白复合体，组成型，辅基蛋白B-100，受体介导的内吞作用2. START 3. 螺线管压缩成超螺线管， 40倍4. 基因调节作用5 ①铁硫蛋白；②辅酶Q；③黄素蛋白；④细胞色素二、判断题(正确的标T，错误的标F,或写出必要的答案，共15分)1.Answer: (a)S ; (b)M ; (c)M; (d)G1 ; (e)M; (f)Gl;(g)Gl, G2, S;``(h)M, G2, S; (i)Gl, S, G2, M;(j)Gl, G2, M2. 答：T, 正确3. 答：错误,粉末状的染色体4. 答：T,正确。

5.答： F,错误，核纤层蛋白B与内核膜相连6 答：F,错误，被阻止在M期7. 答：不正确，因为二者的膜蛋白不同8. 答：正确9. 答：错误, 8个小亚基是核基因编码;10. 答：错误, 给胞质分裂传递信号11.答: 错误, 同源染色体间的分子重组是随机发生的三、选择题(请将正确答案的代号填入括号，每题1分，共15分)1. Answer: b2. Answer: d 3. Answer: b4.Answer: c5. Answer: b6. Answer :D7. Answer: a8. Answer: a9. Answer1.D10. Answer2.D11. Answer: C12. ( d ) 13. ( d )14. ( D ) 15. ( A ) 四、简答题(选做4题，每题5分，20分)1. Answer. Regulated secretion occurs only in response to a signal. The proteins to be secreted are stored in special secretory vesicles. Sorting into the regulated secretory pathway is controlled by selective protein aggregation. Constitutive secretion appears to occur by default with secretory proteins, which do not selectively aggregate being included in transport vesicles.2. Answer If GTP is present but cannot be hydrolyzed, microtubules will continue to grow until all free tubulin subunits have been used up.3. Answer: The centrosome serves as a microtubule-organizing center in vivo, andall of the microtubules radiating from the centrosome apparently have the same polarity.4. 答：加在于粗面内质网上合成的蛋白质上的糖基可由两种途径连接：通过天冬氨酸残基的N原子或通过丝氨酸和苏氨酸残基的O原子。

N-连结糖蛋白合成的第一步在粗面内质网上进行，糖链是从磷酸多萜醇转移至新生肽链上这种糖基化在高尔基体中继续被修饰O-连结的糖基化是在高尔基体中进行的5.答: 过氧化物酶体中的氧化酶都是利用分子氧作为氧化剂, 催化下面的化学反应: RH2 + O2 ---------→ R + H2O2这一反应对细胞内氧的水平有很大的影响例如在肝细胞中,有20%的氧是由过氧化物酶体消耗的,其余的粒体中消耗在过氧化物酶体中氧化产生的能量以产热的方式消耗掉, 而粒体中氧化产生的能量贮存在ATP中线粒体与过氧化物酶体对氧的敏感性是不一样的,线粒体氧化所需的最佳氧浓度为2%左右,增加氧浓度,并不提高线粒体的氧化能力过氧化物酶体与线粒体不同, 它的氧化率是随氧张力增强而成正比地提高(图7-44)因此,在低浓度氧的条件下,线粒体利用氧的能力比过氧化物酶体强,但在高浓度氧的情况下,过氧化物酶体的氧化反应占主导地位,这种特性使过氧化物酶体具有使细胞免受高浓度氧的毒性作用五、计算与推理(第1题必做，2、3选一题，每题5分，共10分)1. Answer： There are six nucleosomes per helical turn of the solenoid structure, and one helical turn of the solenoid corresponds to slightly less than 30nm along the length of a chromatin thick fiber.Assuming, for simplicity of calculation, one helical turn per 30 nm, then there are 6 nucleosomes per 30-nm stretch of thick fiber. A 900-nm-long, thick fiber thus has 30 solenoid turns (900 nm divided by 30 nm/turn) and contains 180 nucleosomes (6 nucleosomes/turn 30 turns).The DNA content of each human nucleosome plus the linker DNA connecting it to adjacent nucleosomes is about 200 bp. This thick fiber thus contains 36,000 bp of DNA: (200 bp/nucleosome) (180 nucleosomes/900-nm thick fiber).2. Answer：Loss of cyclin leads to inactivation of the mitotic Cdk. As the result, its target proteins become dephosphorylated by phosphatases, and the cells exit mitosis--they disassemble the mitotic spindle, reassemble the nuclear envelope, decondense their chromosomes, and so on. Cyclin is degraded by ubiq-uitin-dependent destruction in proteosomes, and the activation of the mitotic Cdk most likely causes the ubiquitination of the cyclin, but with a substantial delay. As discussed in Chapter 5, ubiquitination tags proteins for degradation in proteasomes.3. Answer：The actual explanation is that the single amino acid change causes the protein to misfold slightly so that, although it is still active as a pro-tease inhibitor, it is prevented by chaperone proteins in the ER from exiting this organelle. It therefore accumulates in the ER lumen and is eventually degraded. Alternative interpretations might have been: (1) the mutation affects the stability of the protein in the bloodstream so that it is degraded much faster in the blood than the normal protein, or (2) the mutation inactivates the ER signal sequence and prevents the protein from entering the ER. (3) Another explanation could have been that the mutation altered the sequence to create an ER retention signal, which would have retained the mutant protein in the ER. One could distinguish between these possibilities by using fluorescent-tagged antibodies against the protein to follow its transport in the cells (see Panel 5-3, pp. 158-159).六、比较题（每题5分，共10分）1. Answer: Both dynein and kinesin are large motor proteins that convert the chemical energy of ATP into movement. Both are found affiliated with microtubules, although only dynein occurs on the microtubules of cilia and flagella. Kinesin is a plus-end directed microtubular motor, and dynein, among its other roles, is a minus-end directed microtubular motor. In spite of their similarities in function, they are not homologous proteins, and they a。