biochem考试心得，希望对后来者有用

Deepocean1983UID：2272666

考生化有一段时间了，被一MM委托来写一点biochem考试心得，希望对后来者有用
先来简单介绍一下用的书， 绝对必要的书 英国人编的Instant notes生化和分子生物学， 非常之经典，特别是生化， 尽管分子生物学里面有一些知识有点陈旧， 看来在教育方面，英国还是比美国要经典一些：）。 我详细的看了三遍生化，结合POB， 就足以应付生化部分的考试。 POB太厚了，而且内容及其罗嗦， 有点看paper的感觉， 所以我只是辅助的看了一部分POB， 算作为instant  notes 的补充. 我选择了POB中酶， 硫的代谢， 核酸代谢，光合作用，乙醛酸循环途径 以及所有物化及酶的公式作为补充， 考试的时候基本上就没有遇到几道不会做的生化题。
分子生物学我用的是Instant notes结合waston编的molecular biology of the gene, 后者是一本相当权威经典的分子生物学教材， 是很多美国大学的首选教材，个人感觉应该比Gene VII好。 Instant notes 我也是从头到尾看了2，3遍， 虽然是99年的书， 但是还是有不少新的体会，太经典了， 流泪ing ：）。 waston的书是很好的补充， 有一点问题就是有时候太详细了， 把一些原理阐明的太清楚了。 所以读这本书时，要会把握重点。 我读的时候比较skipping， 所以大概一天可以看100页左右， 10天就可以看完。 这本书中比较重要的部分是honologous recombination, non-homologous recombination还有后面的真核及原核的转录调控。其他部分instant notes基本上可以覆盖。病毒学instant notes最后那一章基本可以覆盖， 需要补充的是HIV的相关知识。
遗传学我用的是大学的笔记， 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book大家购买的， 把老师气得半死，大呼版权。 对这本笔记我只能说，看了遗传学绝对是满分。
细胞生物学我用的是molecular biology of the cell, 1500多页的大厚书， 众多顶级科学家智慧的合成， 所有教科书中王者的感觉：）。 其实考sub感觉只用这本书考上90%都没有问题。 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。 neurology要考的知识点， 在cell membrane 那一章基本上可以覆盖。
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20多分钟的时间，再回过头来慢慢做那些题， 特别是一些遗传学的计算题， 直接跳过， 回头再作心情会平和很多， 也会更快做出来。 另外关于做题的顺序，我觉得先用50-60min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， 个人觉得如果你能排除一个以上答案，就猜， 总的得分的概率绝对高于空着， 毕竟人有时是要赌一下的：）
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。 其实细胞的内容是可以考的很多的很专业的， ETS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mechanism， 比如DNA repair， 他不会问你UVr 那几个分子的作用是什么， 一般是问你有几种DNA repair 方式 比如NER, BER。。。。。。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。
关于实验题， 是很灵活的一个部分， 需要有很好的背景知识基础， 所以大家准备时不要慌着准备这部分， 等基础部分牢了， 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， 比如要了解用flowcytometry测定细胞周期的原理， 还有一些试剂的原理要知道， 比如EDTA， EGTA是干嘛的。所以模考对这部分是extremly important，能让人熟悉题型。
关于模考，我觉得在你复习得差不多了， 留7-10天籁模考就足够了。 看完一遍书后可以模考一套来定位， 但是不要都早早的把三套题都模考掉了， 太浪费资源。。。。。。
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：




97-99
4. Absorbance=LgI0/I, 0.05 percent=0.05%X1000=0.5mg
12. Operon can contain more than one promoter for example the promoter for regulator genes
21. Although cellulose is synthesized on the membrane, but hemicellulose and pectin are synthesized in GA.
34. Prokaryotic ribosome (Large subunit contain about 30 proteins while small 20)
   Eukaryotes (Large 45, small 30)
40. There are least successful conjugation in Hfr2, and the largest region in this group is between His and Arg
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.
42. E because EGTA can inhibit VSV-G protein transportation.

94
11. partially dominance=incomplete dominance. This case also can be considered as codominance:
Alleles interaction: Incomplete dominance (1:2:1 when self-test), codominance, mosaic dominance
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)

16. PH=-lg(10-7+10-8)=7-lg1.1

24. Cytochrome always in mitochondria or SER

25.
29. Chloroplast inheritance is typical maternal inheritance (maze striped iojap trait)

30. RF=（1/2 T+ ND）/ total number, if PD=NPD, the genes are not linked, if PD>>NPD, they are linked.

35. When the reaction is S->P  this is the first-order reaction, v=κ[S], κ is the first-order constant.
When the reaction is S1+S2->P, this is a second-order reaction, κ is a second-order constant, v=κ[S1][S2], κ=KT/h e-σG++/RT
So the answer is first-order constant

49. avidin is a highly glycosylated protein with bound biotin, is relatively high non-specific binding, can be used for protein purification.
50. nitrate NO3-, nitrite NO2-, the elelctronegative potential of atoms F>O>N>S=C>P=H, so here the answer is phosphate.
57. Transposable element can transport within a replicon or between replicons.
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.
69. NADH is not truly cross mitochondria membrane, but via glycerol-3 phosphate shuttle and malate-aspartate shuttle.
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
73. The question asks the bases incorporated to products, not including primer.
76. endonucleolytic cleavage of 3’end is alternative splicing.
78. Fructose 1,6-bisphosphate is involved in the first step of glyceraldehyde 3-phosphate and dihydroxylacetone condensation.
80. AMP is not with a phosphoryl group that can be easily transferred.
85. a primary spermatocyte contain two complementary chromosomes and 4 chromatids, so 2n/4c.
88. Release of Ca2+ from SER in muscle through ryanodine channel depends on plasma membrane potential change which is conducted by transverse tubule.
96. POB P844
98. They are the amino group of Glu and amido group of Gln
102. nucleoplasmin is a chaperone in nucleus
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.
108......
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.
120. only enhancer can promote transcription both downstream and upstream.
123. Theta replication=bidirectional replication, so the answer is D
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.
126. Chiasma is cross-over, which marks recombination.
132. it contains UAA so it is a stop codon, so D
143. Mercaptoethanol and dithiothreitol can reduce and broke disulfide bond and acetoactate can form covalent bond with Cys. So C
146. D
149. Horizen line the number of bacteria is equal to no UV irradiation, so the DNA damage is repaired by photoactivation in strain 1.
151. Just like free radical OH, it is a chain of free-radical reaction,
152. The best way to terminate the reaction is to eliminate free-radical, so D.
156. phosphatase inhibitor can inhibit aggregation, besides cAMP can inhibit aggregation indicating cAMP-kinase inhibit aggregation. so protein dephosphorylation is important.
158. D contain “all”, so incorrect, so C is the most possible one.
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.
170. in order to study the study the relation between C4 acids, 3-PGA and Hexose phosphatase, the time should be decreased.
176. the enzyme hydrolyze X-P when Y is not present, but slower than Y is present.
Sub 2000
13. Retrotransposon may contain introns
22. Protein synthesis is before mRNA translation because mRNA of maternal-effect gene is translated before any zygotic-effect mRNA is transcribed.
28. All three are common pathway to regulate prokaryotes behavior
31. Stereoisomer=mirror image when the molecule contains only one chiral molecule. Only B contains one chiral molecule and are mirror image.
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.
61. TMamine+H2O= TMammonium+ +OH-   Kb=[OH-][TMammonium+]/[TMamine]
   TMammonium+ in water: TMammonium=TMammine+[H+]
So Ka=[TMammine][H+]/[TMammonium]=[TMammine][H+][OH-]/Kb[TMammine]
=10-14/7.4X10-5
   PKa=14-4.13=9.87
In other words, the PKa of acid and PKb of its corresponding base has the relationship:
KaKb=10-14 and PKa+PKb=14
74. The parental type is dominant than NP, so the genes are linked.
76. Bacterial and eukaryotes has the same translation intiation codon AUG
95. Cytosine has most double bond in its ring and thus the most plannar
100. Endoderm gives rise to: Digestive system and associated ligand, respiratory system and liver.
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.
Ectoderm: Hair, skin, sweat gland, mammary gland, sensory organ and nervous system.  
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.
118. Intron splicing occurs on mRNA so it is not DNA rearrangement. Intron splicing in cilia is self splicing.
127. Nuclear envelope can contain ribosome so it can be sites of protein synthesis.
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.
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.
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.
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.

Others: express sequence tag is a short cDNA fragments used in gene locating and mapping.
       Blast e-100 stands for


[ 本帖最后由 Deepocean1983 于 2008-4-22 23:19 编辑 ]

nkotbUID：2208953

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！

inertiaUID：2275366

明年我也准备考biochem了，多谢经验