biochem考试心得，希望对后来者有用
[size=3][font=Times New Roman]考生化有一段时间了，被一MM委托来写一点biochem[/font][font=宋体]考试心得，希望对后来者有用[/font][/size][size=3][font=宋体]先来简单介绍一下用的书，[/font][font=Times New Roman] [/font][font=宋体]绝对必要的书[/font][font=Times New Roman] [/font][font=宋体]英国人编的[/font][font=Times New Roman]Instant notes[/font][font=宋体]生化和分子生物学，[/font][font=Times New Roman] [/font][font=宋体]非常之经典，特别是生化，[/font][font=Times New Roman] [/font][font=宋体]尽管分子生物学里面有一些知识有点陈旧，[/font][font=Times New Roman] [/font][font=宋体]看来在教育方面，英国还是比美国要经典一些：）。[/font][font=Times New Roman] [/font][font=宋体]我详细的看了三遍生化，结合[/font][font=Times New Roman]POB[/font][font=宋体]，[/font][font=Times New Roman] [/font][font=宋体]就足以应付生化部分的考试。[/font][font=Times New Roman] POB[/font][font=宋体]太厚了，而且内容及其罗嗦，[/font][font=Times New Roman] [/font][font=宋体]有点看[/font][font=Times New Roman]paper[/font][font=宋体]的感觉，[/font][font=Times New Roman] [/font][font=宋体]所以我只是辅助的看了一部分[/font][font=Times New Roman]POB[/font][font=宋体]，[/font][font=Times New Roman] [/font][font=宋体]算作为[/font][font=Times New Roman]instant notes [/font][font=宋体]的补充[/font][font=Times New Roman]. [/font][font=宋体]我选择了[/font][font=Times New Roman]POB[/font][font=宋体]中酶，[/font][font=Times New Roman] [/font][font=宋体]硫的代谢，[/font][font=Times New Roman] [/font][font=宋体]核酸代谢，光合作用，乙醛酸循环途径[/font][font=Times New Roman] [/font][font=宋体]以及所有物化及酶的公式作为补充，[/font][font=Times New Roman] [/font][font=宋体]考试的时候基本上就没有遇到几道不会做的生化题。[/font][font=Times New Roman] [/font][/size]
[size=3][font=宋体]分子生物学我用的是[/font][font=Times New Roman]Instant notes[/font][font=宋体]结合[/font][font=Times New Roman]waston[/font][font=宋体]编的[/font][font=Times New Roman]molecular biology of the gene, [/font][font=宋体]后者是一本相当权威经典的分子生物学教材，[/font][font=Times New Roman] [/font][font=宋体]是很多美国大学的首选教材，个人感觉应该比[/font][font=Times New Roman]Gene VII[/font][font=宋体]好。[/font][font=Times New Roman] Instant notes [/font][font=宋体]我也是从头到尾看了[/font][font=Times New Roman]2[/font][font=宋体]，[/font][font=Times New Roman]3[/font][font=宋体]遍，[/font][font=Times New Roman] [/font][font=宋体]虽然是[/font][font=Times New Roman]99[/font][font=宋体]年的书，[/font][font=Times New Roman] [/font][font=宋体]但是还是有不少新的体会，太经典了，[/font][font=Times New Roman] [/font][font=宋体]流泪[/font][font=Times New Roman]ing [/font][font=宋体]：）。[/font][font=Times New Roman] waston[/font][font=宋体]的书是很好的补充，[/font][font=Times New Roman] [/font][font=宋体]有一点问题就是有时候太详细了，[/font][font=Times New Roman] [/font][font=宋体]把一些原理阐明的太清楚了。[/font][font=Times New Roman] [/font][font=宋体]所以读这本书时，要会把握重点。[/font][font=Times New Roman] [/font][font=宋体]我读的时候比较[/font][font=Times New Roman]skipping[/font][font=宋体]，[/font][font=Times New Roman] [/font][font=宋体]所以大概一天可以看[/font][font=Times New Roman]100[/font][font=宋体]页左右，[/font][font=Times New Roman] 10[/font][font=宋体]天就可以看完。[/font][font=Times New Roman] [/font][font=宋体]这本书中比较重要的部分是[/font][font=Times New Roman]honologous recombination, non-homologous recombination[/font][font=宋体]还有后面的真核及原核的转录调控。其他部分[/font][font=Times New Roman]instant notes[/font][font=宋体]基本上可以覆盖。病毒学[/font][font=Times New Roman]instant notes[/font][font=宋体]最后那一章基本可以覆盖，[/font][font=Times New Roman] [/font][font=宋体]需要补充的是[/font][font=Times New Roman]HIV[/font][font=宋体]的相关知识。[/font][/size]
[size=3][font=宋体]遗传学我用的是大学的笔记，[/font][font=Times New Roman] [/font][font=宋体]是一很牛的前辈做的大概两百页的老师的笔记，后来影印成[/font][font=Times New Roman]book[/font][font=宋体]大家购买的， 把老师气得半死，大呼版权。[/font][font=Times New Roman] [/font][font=宋体]对这本笔记我只能说，看了遗传学绝对是满分。[/font][/size]
[size=3][font=宋体]细胞生物学我用的是[/font][font=Times New Roman]molecular biology of the cell, 1500[/font][font=宋体]多页的大厚书，[/font][font=Times New Roman] [/font][font=宋体]众多顶级科学家智慧的合成，[/font][font=Times New Roman] [/font][font=宋体]所有教科书中王者的感觉：）。[/font][font=Times New Roman] [/font][font=宋体]其实考[/font][font=Times New Roman]sub[/font][font=宋体]感觉只用这本书考上[/font][font=Times New Roman]90%[/font][font=宋体]都没有问题。[/font][font=Times New Roman] [/font][font=宋体]此书的毛病还是太详细，需要会把握重点。发育那一章基本上可以包括发育要考的所有知识点。[/font][font=Times New Roman] neurology[/font][font=宋体]要考的知识点，[/font][font=Times New Roman] [/font][font=宋体]在[/font][font=Times New Roman]cell membrane [/font][font=宋体]那一章基本上可以覆盖。[/font][/size]
[size=3][font=宋体]考试的时候，不会做的题目一定要跳过去，等所有题都做完了，一般会有[/font][font=Times New Roman]20[/font][font=宋体]多分钟的时间，再回过头来慢慢做那些题，[/font][font=Times New Roman] [/font][font=宋体]特别是一些遗传学的计算题，[/font][font=Times New Roman] [/font][font=宋体]直接跳过，[/font][font=Times New Roman] [/font][font=宋体]回头再作心情会平和很多，[/font][font=Times New Roman] [/font][font=宋体]也会更快做出来。[/font][font=Times New Roman] [/font][font=宋体]另外关于做题的顺序，我觉得先用[/font][font=Times New Roman]50-60min[/font][font=宋体]完成实验题部分比较好，因为最后考试时间紧张再做实验题的话可能会非常慌张。另外关于猜题还是空题，[/font][font=Times New Roman] [/font][font=宋体]个人觉得如果你能排除一个以上答案，就猜，[/font][font=Times New Roman] [/font][font=宋体]总的得分的概率绝对高于空着，[/font][font=Times New Roman] [/font][font=宋体]毕竟人有时是要赌一下的：）[/font][/size]
[size=3][font=宋体]考试内容我的感觉是生化的统治地位在下降，细胞生物学的地位在上升，特别是最后的实验题部分。[/font][font=Times New Roman] [/font][font=宋体]其实细胞的内容是可以考的很多的很专业的，[/font][font=Times New Roman] ETS[/font][font=宋体]算是比较人性了，没有太过刁难我们。而且他不会考很具体的[/font][font=Times New Roman]mechanism[/font][font=宋体]，[/font][font=Times New Roman] [/font][font=宋体]比如[/font][font=Times New Roman]DNA repair[/font][font=宋体]，[/font][font=Times New Roman] [/font][font=宋体]他不会问你[/font][font=Times New Roman]UVr [/font][font=宋体]那几个分子的作用是什么，[/font][font=Times New Roman] [/font][font=宋体]一般是问你有几种[/font][font=Times New Roman]DNA repair [/font][font=宋体]方式[/font][font=Times New Roman] [/font][font=宋体]比如[/font][font=Times New Roman]NER, BER[/font][font=宋体]。。。。。。所以把握重点是很必要的。把前人的回忆题总结下来， 根据回忆题的知识点来看书是很必要的。[/font][/size]
[size=3][font=宋体]关于实验题，[/font][font=Times New Roman] [/font][font=宋体]是很灵活的一个部分，[/font][font=Times New Roman] [/font][font=宋体]需要有很好的背景知识基础，[/font][font=Times New Roman] [/font][font=宋体]所以大家准备时不要慌着准备这部分，[/font][font=Times New Roman] [/font][font=宋体]等基础部分牢了，[/font][font=Times New Roman] [/font][font=宋体]再来准备会效率很高。这部分也要求很强的实验技术的背景，[/font][font=Times New Roman] [/font][font=宋体]比如要了解用[/font][font=Times New Roman]flowcytometry[/font][font=宋体]测定细胞周期的原理，[/font][font=Times New Roman] [/font][font=宋体]还有一些试剂的原理要知道，[/font][font=Times New Roman] [/font][font=宋体]比如[/font][font=Times New Roman]EDTA[/font][font=宋体]，[/font][font=Times New Roman] EGTA[/font][font=宋体]是干嘛的。所以模考对这部分是extremly important，能让人熟悉题型。[/font][/size]
[size=3][font=宋体]关于模考，我觉得在你复习得差不多了，[/font][font=Times New Roman] [/font][font=宋体]留[/font][font=Times New Roman]7-10[/font][font=宋体]天籁模考就足够了。[/font][font=Times New Roman] [/font][font=宋体]看完一遍书后可以模考一套来定位，[/font][font=Times New Roman] [/font][font=宋体]但是不要都早早的把三套题都模考掉了，[/font][font=Times New Roman] [/font][font=宋体]太浪费资源。。。。。。[/font][/size]
[font=宋体][size=3]最后送给大家的是当初考试时我自己总结的三套真题里部分题目的详细解答，希望有用：[/size][/font]
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[font=宋体][font=Times New Roman][size=3]97-99[/size][/font]
[font=Times New Roman][size=3]4. Absorbance=LgI0/I, 0.05 percent=0.05%X1000=0.5mg[/size][/font]
[font=Times New Roman][size=3]12. Operon can contain more than one promoter for example the promoter for regulator genes[/size][/font]
[font=Times New Roman][size=3]21. Although cellulose is synthesized on the membrane, but hemicellulose and pectin are synthesized in GA.[/size][/font]
[font=Times New Roman][size=3]34. Prokaryotic ribosome (Large subunit contain about 30 proteins while small 20)[/size][/font]
[size=3][font=Times New Roman] Eukaryotes (Large 45, small 30)[/font][/size]
[font=Times New Roman][size=3]40. There are least successful conjugation in Hfr2, and the largest region in this group is between His and Arg[/size][/font]
[font=Times New Roman][size=3]41. EGTA is related to EDTA, but with a much higher affinity for Ca2+ than Mg2+. There are still some VSV-G proteins be transported to cis-Golgi, so C is incorrect, the most likely is A.[/size][/font]
[font=Times New Roman][size=3]42. E because EGTA can inhibit VSV-G protein transportation.[/size][/font]
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[font=Times New Roman][size=3]94[/size][/font]
[font=Times New Roman][size=3]11. partially dominance=incomplete dominance. This case also can be considered as codominance:[/size][/font]
[font=Times New Roman][size=3]Alleles interaction: Incomplete dominance (1:2:1 when self-test), codominance, mosaic dominance[/size][/font]
[font=Times New Roman][size=3]Nonallelic interaction: Complementary (9:6:1), dominant epistasis (12:3:1), double recessive epistasis (9:7), recessive epistasis (9:4:3), inhibitor gene (13:3) [/size][/font]
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[font=Times New Roman][size=3]16. PH=-lg(10-7+10-8)=7-lg1.1[/size][/font]
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[font=Times New Roman][size=3]24. Cytochrome always in mitochondria or SER[/size][/font]
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[font=Times New Roman][size=3]25. [/size][/font]
[font=Times New Roman][size=3]29. Chloroplast inheritance is typical maternal inheritance (maze striped iojap trait)[/size][/font]
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[size=3][font=Times New Roman]30. RF=[/font][font=宋体]（[/font][font=Times New Roman]1/2 T+ ND[/font][font=宋体]）[/font][font=Times New Roman]/ total number, if PD=NPD, the genes are not linked, if PD>>NPD, they are linked.[/font][/size]
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[size=3][font=Times New Roman]35. When the reaction is S->P this is the first-order reaction, v=[/font][font=宋体]κ[/font][font=Times New Roman][S], [/font][font=宋体]κ[/font][font=Times New Roman] is the first-order constant.[/font][/size]
[size=3][font=Times New Roman]When the reaction is S1+S2->P, this is a second-order reaction, [/font][font=宋体]κ[/font][font=Times New Roman] is a second-order constant, v=[/font][font=宋体]κ[/font][font=Times New Roman][S1][S2], [/font][font=宋体]κ[/font][font=Times New Roman]=KT/h e-[/font][font=宋体]σ[/font][font=Times New Roman]G++/RT[/font][/size]
[font=Times New Roman][size=3]So the answer is first-order constant[/size][/font]
[font=Times New Roman][size=3][/size][/font]
[font=Times New Roman][size=3]49. avidin is a highly glycosylated protein with bound biotin, is relatively high non-specific binding, can be used for protein purification.[/size][/font]
[font=Times New Roman][size=3]50. nitrate NO3-, nitrite NO2-, the elelctronegative potential of atoms F>O>N>S=C>P=H, so here the answer is phosphate.[/size][/font]
[font=Times New Roman][size=3]57. Transposable element can transport within a replicon or between replicons.[/size][/font]
[font=Times New Roman][size=3]66. Testosterone is steroid hormone, so it has nuclear receptor as transcription activator, once bound the receptor can be released from sequestration such as Hsp90, and dimerize to bind to DNA and activate transcription. Vassopressin and adrenocorticotropic hormone are both peptide hormone.[/size][/font]
[font=Times New Roman][size=3]69. NADH is not truly cross mitochondria membrane, but via glycerol-3 phosphate shuttle and malate-aspartate shuttle.[/size][/font]
[font=Times New Roman][size=3]70. G-oligosaccharides are formed on dolichol in RER with the composition of (GlcNAc)2 (Man)9 (Glc)3, and undergoes core glycosylation to proteins. Galactose is added only in terminal glycosylation in GA. So the answer is GA[/size][/font]
[font=Times New Roman][size=3]73. The question asks the bases incorporated to products, not including primer.[/size][/font]
[font=Times New Roman][size=3]76. endonucleolytic cleavage of 3’end is alternative splicing.[/size][/font]
[font=Times New Roman][size=3]78. Fructose 1,6-bisphosphate is involved in the first step of glyceraldehyde 3-phosphate and dihydroxylacetone condensation.[/size][/font]
[font=Times New Roman][size=3]80. AMP is not with a phosphoryl group that can be easily transferred.[/size][/font]
[font=Times New Roman][size=3]85. a primary spermatocyte contain two complementary chromosomes and 4 chromatids, so 2n/4c.[/size][/font]
[font=Times New Roman][size=3]88. Release of Ca2+ from SER in muscle through ryanodine channel depends on plasma membrane potential change which is conducted by transverse tubule.[/size][/font]
[font=Times New Roman][size=3]96. POB P844[/size][/font]
[font=Times New Roman][size=3]98. They are the amino group of Glu and amido group of Gln[/size][/font]
[font=Times New Roman][size=3]102. nucleoplasmin is a chaperone in nucleus[/size][/font]
[font=Times New Roman][size=3]103. Rhodopsin accept light energy and conformational change to activate G protein transducin which in turn activate cGMP phosphdiesterase to decrease level of cGMP which leads to closure of Na+ channel.[/size][/font]
[font=Times New Roman][size=3]108......[/size][/font]
[font=Times New Roman][size=3]119. Because AZT is thymidine analog, so it is not likely that it can not enter nucleus, and because mismatch repair enzyme does not distinguish HIV cDNA and host DNA, so it is also not likely that MMR play a role. The most likely reason is DNA Pol does not accept this analog just as it does not accept a mismatch nucleotide or oxyribonucleotide while reverse transcriptase accepts.[/size][/font]
[font=Times New Roman][size=3]120. only enhancer can promote transcription both downstream and upstream.[/size][/font]
[font=Times New Roman][size=3]123. Theta replication=bidirectional replication, so the answer is D[/size][/font]
[font=Times New Roman][size=3]125. phase of meiosis prophase I: leptotene: introduce DSB and single strand invasion to initiate recombination. Zygotene: Formation of synapsis and synaptonemal complex. Patchtene: recombination occurs. Diplotene: Synaptonemal complex dissemble and homologous chromosome remain at chiasma. Diakinesis: Condensation of chromosome. Synaptonemal complex consists of chromatin, lateral element and central element, between central and lateral element are traverse filament. [/size][/font]
[font=Times New Roman][size=3]126. Chiasma is cross-over, which marks recombination.[/size][/font]
[font=Times New Roman][size=3]132. it contains UAA so it is a stop codon, so D[/size][/font]
[font=Times New Roman][size=3]143. Mercaptoethanol and dithiothreitol can reduce and broke disulfide bond and acetoactate can form covalent bond with Cys. So C[/size][/font]
[font=Times New Roman][size=3]146. D[/size][/font]
[font=Times New Roman][size=3]149. Horizen line the number of bacteria is equal to no UV irradiation, so the DNA damage is repaired by photoactivation in strain 1.[/size][/font]
[font=Times New Roman][size=3]151. Just like free radical OH, it is a chain of free-radical reaction,[/size][/font]
[font=Times New Roman][size=3]152. The best way to terminate the reaction is to eliminate free-radical, so D.[/size][/font]
[font=Times New Roman][size=3]156. phosphatase inhibitor can inhibit aggregation, besides cAMP can inhibit aggregation indicating cAMP-kinase inhibit aggregation. so protein dephosphorylation is important.[/size][/font]
[font=Times New Roman][size=3]158. D contain “all”, so incorrect, so C is the most possible one.[/size][/font]
[font=Times New Roman][size=3]163. They still have T cells, so can induce the clonal selection of memory and other T cells to mount immune response to previously met or not met antigens.[/size][/font]
[font=Times New Roman][size=3]170. in order to study the study the relation between C4 acids, 3-PGA and Hexose phosphatase, the time should be decreased.[/size][/font]
[font=Times New Roman][size=3]176. the enzyme hydrolyze X-P when Y is not present, but slower than Y is present.[/size][/font]
[font=Times New Roman][size=3]Sub 2000[/size][/font]
[font=Times New Roman][size=3]13. Retrotransposon may contain introns[/size][/font]
[font=Times New Roman][size=3]22. Protein synthesis is before mRNA translation because mRNA of maternal-effect gene is translated before any zygotic-effect mRNA is transcribed.[/size][/font]
[font=Times New Roman][size=3]28. All three are common pathway to regulate prokaryotes behavior[/size][/font]
[font=Times New Roman][size=3]31. Stereoisomer=mirror image when the molecule contains only one chiral molecule. Only B contains one chiral molecule and are mirror image.[/size][/font]
[font=Times New Roman][size=3]42. when the two mutations do not belong two same cistron, in heterozygous status, they can complement each other. If not, they are very likely to be in the same cistron. B is correct if they are in the different complementation group.[/size][/font]
[font=Times New Roman][size=3]61. TMamine+H2O= TMammonium+ +OH- Kb=[OH-][TMammonium+]/[TMamine][/size][/font]
[size=3][font=Times New Roman] TMammonium+ in water: TMammonium=TMammine+[H+][/font][/size]
[font=Times New Roman][size=3]So Ka=[TMammine][H+]/[TMammonium]=[TMammine][H+][OH-]/Kb[TMammine][/size][/font]
[font=Times New Roman][size=3]=10-14/7.4X10-5[/size][/font]
[size=3][font=Times New Roman] PKa=14-4.13=9.87[/font][/size]
[font=Times New Roman][size=3]In other words, the PKa of acid and PKb of its corresponding base has the relationship:[/size][/font]
[font=Times New Roman][size=3]KaKb=10-14 and PKa+PKb=14[/size][/font]
[font=Times New Roman][size=3]74. The parental type is dominant than NP, so the genes are linked.[/size][/font]
[font=Times New Roman][size=3]76. Bacterial and eukaryotes has the same translation intiation codon AUG[/size][/font]
[font=Times New Roman][size=3]95. Cytosine has most double bond in its ring and thus the most plannar[/size][/font]
[font=Times New Roman][size=3]100. Endoderm gives rise to: Digestive system and associated ligand, respiratory system and liver.[/size][/font]
[font=Times New Roman][size=3]Mesoderm: Chordalmesoderm gives rise to notochord. Paraxial endoderm gives rise to somite and head, somite gives rises to cartilage, bone, vertebral column and associated connective tissue, skeletal muscle, dermis. Intermediate mesoderm gives rise to gonad and kidney. Lateral plate mesoderm gives rise to circulatory system including heart, blood vessel and blood cells, body cavity lining, mesoderm (connective tissue) of limb and extraembryonic membrane for import nutrients for embryo.[/size][/font]
[size=3][font=Times New Roman]Ectoderm: Hair, skin, sweat gland, mammary gland, sensory organ and nervous system. [/font][/size]
[font=Times New Roman][size=3]108. 5000,000dpm corresponds to 1000 micromole in 2 hours, so 5000dpm corresponds to 10 micromole in 2 hours, so corresponds to 5 micromole per hour.[/size][/font]
[font=Times New Roman][size=3]118. Intron splicing occurs on mRNA so it is not DNA rearrangement. Intron splicing in cilia is self splicing.[/size][/font]
[font=Times New Roman][size=3]127. Nuclear envelope can contain ribosome so it can be sites of protein synthesis.[/size][/font]
[font=Times New Roman][size=3]139. Arginase is a urea cycle enzyme catalyzing the argine to orthonine and urea. The formation of arginosuccinate is catalyzed by arginosuccinate synthetase and its convertion to succinate and argine is catalyzed by arginosuccinase.[/size][/font]
[font=Times New Roman][size=3]158. disintegration of radioactive molecules when exposure is corresponding to its amount in the labeled molecules. Disintegration higher means the radioactive molecules in that labeled molecules are higher. The disintegration of A,B, C, D, E corresponds respectively to their length, so they show same specificity.[/size][/font]
[font=Times New Roman][size=3]171. Glucose 6-phosphate level decrease after O2 is removed because no ATP (Pasteur Effect) is generated through TCA, so the only ATP source is glycolysis, so the rate of glycolysis increases, and glucose 6-phosphate is used for glycolysis.[/size][/font]
[font=Times New Roman][size=3]176. Hormone-independent regulatory sequence are upstream and downstream of -315 sequence because the CAT activity show changes in the absence of hormone in both deletion of upstream and downstream sequence of -315. [/size][/font]
[font=Times New Roman][size=3][/size][/font]
[font=Times New Roman][size=3]Others: express sequence tag is a short cDNA fragments used in gene locating and mapping.[/size][/font]
[size=3][font=Times New Roman] Blast e-100 stands for[/font][/size]
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[[i] 本帖最后由 Deepocean1983 于 2008-4-22 23:19 编辑 [/i]] cong下大牛:loveliness:
偶就买过一本POB看了两页改翻沈同三了。。。没得比啊！ 明年我也准备考biochem了，多谢经验
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