04年细胞生物学思考题.docx

1、04年细胞生物学思考题Testing & Thinking of Cell Biology思考题（04-03-09）1填空：无论是真核细胞还是原核细胞，都具有以下共性： ； ； ； 。答：都有DNA，都有核糖体，都是分裂法繁殖，都有细胞膜 。2判断：原核生物和真核生物细胞膜内都含有胆固醇。（）答：错误。 3选择：影响膜脂流动性的重要因素是磷脂分子脂肪酸链的不饱和程度。不饱和性越高， 流动性越（ ），其原因是（ B ）。 A小，双键多、折曲小； B大，双键多、折曲多； C小，分子排列疏松； D大，分子排列紧密答：B思考题（04-03-16）1缬氨霉素是一种（ A ）。 A可动离子载体； B电位门

2、控通道； C配体门控通道； D离子门控通道答：A2简述Na+/葡萄糖协同运输的主要特点。答：无须直接消耗ATP，但需要依赖电化学梯度。载体蛋白有两种结合位点，分别结合Na+与葡萄糖；载体蛋白借助Na+/K+泵建立的电位梯度，将Na+与葡萄糖同时转运到胞内；胞内释放的Na+又被Na+/K+泵泵出细胞外建立Na+梯度。3为什么带3蛋白又叫阴离子传递蛋白?答：具有阴离子转运的功能。4说明细胞外基质的主要组成及它们的主要功能。答：细胞外基质的组成可分为3大类：（1）蛋白聚糖，是由糖胺聚糖以共价形式与线性多肽链连接而成的复合物，能形成水性的胶状物。（2）结构蛋白，如胶原和弹性蛋白，赋予胞外基质一定的强度

3、和韧性。（3）黏着蛋白，如纤连蛋白与层粘连蛋白，促使细胞基质结合。其中胶原和蛋白聚糖为基本骨架在细胞表面形成纤维网状复合物，再通过纤连蛋白或层粘连蛋白，以及其他连接分子直接与细胞表面受体连接或附着在受体上。由于多数受体是膜整合蛋白，并与胞内骨架相连，因此胞外基质通过膜整合蛋白将胞外、胞内连成一个整体。细胞外基质对于细胞的活性及形态起关键作用，一些动物培养细胞的ECM对于细胞的合成与分泌活动具有影响。除了决定器官、组织的形态及保护作用外，还可参与信号转导，细胞分化以及形态建成等。思考题（04-03-23）1. G-蛋白的-亚基上有3个活性位点，分别是： 、 和 。答：GTP结合位点，GTP酶活性

4、位点，ADP核糖基化位点2. 判断：参与信号转导的受体都是膜蛋白。答：错误3. 肾上腺素与胰高血糖素均与G蛋白受体结合并激活糖原分解。因此，肾上腺素与胰高血糖素一定是：A具有十分相似的结构并且与相同的受体结合;B与含有不同的配体结合位点但是有相似功能的受体相结合；C与不同的受体结合并且激活不同的第二信使;D与相同的受体结合，一个是在细胞内而另一个在细胞外。答：B思考题（04-03-29）1简答题： PKA和PKC系统在信号放大中的根本区别是什么？答：PKA途径激活的是蛋白激酶A；PKC途径激活的是蛋白激酶C。 G-蛋白偶联受体与酶联受体的主要不同点是什么？答：G蛋白偶联受体都是7次跨膜的结构，

5、在信号转导中全部与G蛋白偶联；酶联受体都属于单次跨膜受体。2. 选择题： 下列通讯系统中，受体可进行自我磷酸化的是（ B ）。 A鸟苷酸环化酶系统 B酪氨酸蛋白激酶系统 C腺苷酸环化酶系统 D肌醇磷脂系统答: B 动员细胞内贮存Ca2+ 的第二信使分子是（C） Ac-AMP BDG CIP3 DPKC 答：C思考题（04-04-06）1线粒体是一种具有双层膜结构的细胞器，内外膜的结构相似，功能相同。（ ）答：错误2. 过氧化物酶体对细胞的氧张力具有调节作用。( )答：正确。3糖的氧化分3个阶段进行： ； ； 答： 糖氧化为丙酮酸； 丙酮酸脱羧形成乙酰coA； 乙酰coA进入三羧酸循环，彻底氧化

6、4在线粒体电子传递链中，只有（ B ）不是质子移位体。 A复合物 B复合物 C复合物 D复合物答：B思考题（04-04-13）1SRP受体又称停靠蛋白（Docking Protein），它的的功能是使 。答：SRP-信号序列-新生肽-核糖体复合物结合到ER膜上2 判断题：只有在翻译后，被分泌到胞外空间或整合到膜上的真核生物蛋白才被转运到ER的腔中。（ ）答：错误3比较顺面高尔基体内的蛋白和反面高尔基体内的蛋白，可发现：（ ）A两个部分的蛋白是一样的；B顺面高尔基体蛋白是糖基化的，含有修饰过的氨基酸，而反面高尔基体内的蛋白未经修饰；C顺面高尔基体内的蛋白是糖基化的，而反面高尔基体内的蛋白也糖基化

7、，含有修饰过的氨基酸；D顺面高尔基体的蛋白比反面高尔基体的蛋白短答：C思考题（04-04-20）1细胞松弛素B虽然是与微丝特异结合的药物，但并不影响肌肉收缩，原因是肌纤维中的 。答：肌动蛋白纤维是稳定的结构，不存在聚合和解聚过程2肌肉收缩的基本单位是 ，由粗肌丝和细肌丝组成。构成粗肌似的主要成分是 ，它的头部具有 的活性。构成细肌丝的主要成分是肌动蛋白纤维，其上结合有肌钙蛋白和 。肌钙蛋白由3个亚基所组成，其中 亚基与Ca2+ 结合，对肌肉的收缩具有 作用。答：肌原纤维，肌球蛋白，ATP酶，原肌球蛋白，TnC，调节3试说明一个肌节收缩的原理。答：肌节的缩短不是由于纤丝的缩短，而是纤丝间互相滑动

8、所致。细肌丝向肌节中央滑动，导致重叠部分增加，缩短了肌节。粗肌丝与细肌丝之间的滑动必然涉及肌球蛋白头部与肌动蛋白细肌丝的接触，产生粗、细肌丝间的交联桥才能产生滑动。思考题（04-04-27）1核质蛋白协助组蛋白与DNA形成正常的核小体，机理主要是降低 ，从而阻止了错误的装配。答：组蛋白的表面正电荷2判断题：组蛋白和非组蛋白一样都是碱性蛋白质。（）答：错误3常染色质是（ C ）。 A经常存在的染色质； B染色很深的染色质； C不呈异固缩的染色质； D呈现异固缩的染色质答：C4简答 比较导肽与入核信号的区别。答：入核信号与导肽的区别在于入核信号可以反复利用，即永久性；入核信号由核孔复合物识别。 H

9、1组蛋白的作用。答：H1组蛋白维持染色质的30nm纤维结构，没有H1组蛋白，染色质纤维将展开。2004Testing and Thinking(8-30-2004)1. Assume you were given a mixture consisting of one molecule each of all possible sequences of a smallish protein of molecular weight 4800. How big a container would you need to hold this sample? Assume that the avera

10、ge molecular weight of an amino acid is 120.Answer1. A protein with a molecular weight of 4800 is made of about 40 amino acids; thus there are 1.1 1052 (= 2040) different ways to make such a protein. Each individual protein molecule weighs 8 10-21 g (= 4800/6 1023); thus a mixture of one molecule ea

11、ch weighs 9 1031 g (= 8 10-21 g 1.1 1052), which is 15,000 times the total weight of the planet earth, 1.1weighing 6 1024 kg. You would need a quite large container, indeed.Testing and Thinking(9-6-2004)1. What does the p in pH stand for?2. The proteins in a mammalian cell account for 18% of its net

12、 weight. If the density of a typical mammalian cell is about 1.1 g/mL and the volume of the cell is 4 x 10-9 mL, what is the concentration of protein in mg/mL?3. Each of the following statements is an answer; indicate in each case what the question is.(b) This property of water makes it possible for

13、 land animals to cool themselves by surface evaporation with minimum loss of body fluid.Answers for (9-6-2004)1. Although the symbol p in common usage denotes the negative logarithm of what it stands for is unclear. In the original 1909 paper in which the concept of pH was developed, the author-Dani

14、sh chemist Soren EL. Sorensen-was not explicit. In textbooks where it is commented on at all, it is most commonly reputed to stand for the French or German words for power or potential. Close examination of the original paper reveals that the p in pH is likely a consequence of the authors arbitrary

15、choice to call two solutions by the letters p and q. The q solution had the known H+ concentration of 1, the p solution had the unknown H+ concentration. If the solutions had been switched, do you think qH would ever have caught on?2. The concentration of protein is about 200 mg/mL (0.18 x 1.1 g/mL

16、= 198 mg/ mL). Given the density of the cell, you dont need to know its volume to calculate the concentration of protein.3. (b) What is the temperature-stabilizing capacity of water? (Or, alternatively, what is the specific heat of water?)Testing and Thinking(9-15-2004)10-33 Proteins that form a -ba

17、rrel pore in the membrane have several strands that span the membrane. The pore-facing side of each strand carries hydrophilic amino acid side chains, whereas the bilayer-facing sides carry hydrophobic amino acid side chains. Which of the three 10-amino acid sequences listed below is the most likely

18、 candidate for a transmembrane strand in a -barrel pore? Explain the reasons for your choice. (See inside back cover for one-letter amino acid code.) A. A D F K L S V E L T B. A F L V L D K S E TC. A F D K L V S E L T10-25 (True/False) The basic structure of biological membranes is determined by the

19、 lipid bilayer, but their specific functions are carried out largely by proteins. Explain your answer.Answer（9-15-2004）10-33 A. In a strand, adjacent amino acid side chains protrude from opposite sides of the strand; thus, every other amino acid side chain will face the same side of the strand. If a

20、 strand is part of a -barrel pore, its amino acid side chains will alternate between hydrophobic and hydrophilic, so that one side of the strand will be hydrophobic and the other side will be hydrophilic. Only choice A has alternating hydrophobic and hydrophilic amino acids.10-25 True. The lipid bil

21、ayer defines the structure of the membrane and provides a permeability barrier that separates the inside from the outside of the cell. Specific membrane proteins allow particular solutes to enter and leave the cell, bind signaling molecules, and mediate attachment to the extracellular matrix.Testing

22、 and Thinking(9-20-2004)11-3 If a frog egg and a red blood cell are placed in pure water, the red blood cell will swell and burst, but the frog egg will remain intact. Although a frog egg is about one million times larger than a red cell, they both have nearly identical internal concentrations of io

23、ns so that the same osmotic forces are at work in each. Why do you suppose red blood cells burst in water, while frog eggs do not?11-3 Red blood cells have water channels-aquaporins-that make them about 10-fold more permeable to water than a lipid bilayer. Frog eggs do not express aquaporins and thu

24、s their permeability to water is roughly that of a lipid bilayer. However, if there were only a 10-fold difference in permeability to water, wouldnt a frog egg still burst, but just take ten times as long to do so? A part of the answer lies in the enormous volume difference between a red cell and a

25、frog egg, more specifically in the surface-to-volume ratio. Assuming both are spheres, which of course red cells are not, the 106-fold difference in volume translates into a 100-fold lower surface-to-volume ratio in the egg. (This is an underestimate of the difference in the surface-to-volume ratio

26、because the red cell biconcave disc has a larger surface-to-volume ratio than does a sphere of the same volume.) Thus, in the absence of aquaporins the egg should take up water at more than a lO00-fold lower rate than a red cell. If aquaporins are engineered to be expressed into frog eggs, they also

27、 swell and burst when placed in water.19. When present in the culture medium, ouabain inhibits the Na+/K+ ATPase in cultured cells, but when the drug is microinjected into cells, it exerts no inhibitory effect. Although ouabain itself possesses no negative charge, ouabain treatment causes the a subu

28、nits of the Na+/K+ ATPase to become more negatively charged but does not alter the charge properties of the subunits.a. On which side of the plasma membrane are the ouabain and ATP-binding sites of the enzyme located?b. During the transport process, the Na+/K+ ATPase becomes transiently phosphorylat

29、ed. What do these observations suggest about a likely mechanism for ouabain inhibition of this ion pump?c. Which of the two subunits, or , is most likely to be phosphorylated?Answer：19a. The ouabain-binding site of the Na+/K+ ATPase must be located on the exoplasmic face of the plasma membrane, as m

30、icroinjected ouabain does not inhibit the ATPase. The ATP-binding site is located on the cytoplasmic face of the plasma membrane, where it has access to ATP in the cytosol.19b. The addition of a phosphate group to the enzyme would cause it to become more negatively charged. In the presence of extrac

31、ellular ouabain, this negative charge, which normally is temporary, becomes permanent. Most likely, ouabain treatment inhibits the Na+/K+ ATPase by blocking the reaction at the point when phosphate has been added to the enzyme. Thus the phosphorylated form of the enzyme accumulates, blocking further ion transport. 19c. The observation that ouabain treatment results in the subunit becoming more negatively charged but has no effect on the charge of the subunit suggests that the subunit is phosphorylated. This hypothesis could be tested by investigating the effect of a phosphatase