Part III 1.3’ TACGGGCATACGTAAG 5’ 2.The double-helical structure 可藉由hydrogen bond 来稳定, the base stacking between adjacent bases on the same strand. Base stacking in nucleic acid 可以 减少UV的吸收。而Denaturation of DNA, 使的Bases 失去Base stacking的效果, 增加UV 的吸收。 3.每一个amino acid 对应a triplet of three nucleotide pairs, so the 192 amino acids are encoded by 576 nucleotide pairs, 但真正在gene上的有1440 nucleotide pairs. 多余的864 nucleotide pairs 也许可能是introns 或者转译出polypeptide, 但 经过post-translational modification, 把一些不需要的peptide enzymatic cleavage. 5.Leading strands and lagging strands synthesis all need a. template b. primer c. 5’->3’ direction. Leading strands：continuous replication and need dATP , dGTP , dCTP , dTTP , for precursors and need enzymes： a.DNA helicase : required ATP b.Single-strand DNA binding proteins c.DNA gyrase d.polymerase III : carries out the elongation by addition of nucleotide units e.pyrophosphatase : hydrolyzes PPi released as each new nucleotide units is added. Lagging strands：discontinuous replication and need dATP , dGTP , dCTP , dTTP , for the source of nucleotide in the new strand. UTP, ATP, CTP, and GTP are required for formation of the RNA primer. Enzyme： a.leading strands enzyme b.primase c.DNA polymerase I d.DNA ligase 6.Spontaneous deamination of 5-methylcytosine 得到了thymine, 所以有G-T mismatched repair. 而G-T mismatches 是最常发生的mismatch 在 Eukaryotes DNA. 因 此, 有一套特别的repair system 是回复G≡C pair. 7.(Ans: please see chapter 12 page 337) 8.既然很多DNA nucleotides 是由每一个gene 所得到的, 有一个error in any one molecule 会导致protein 上有不正确的amino acid. In quality control, 不正确的 protein 会很快地被degrade, 而有问题的mRNA 也不会存活太久, 所以问题不大. 但DNA replication, errors would be transmitted to the next generations of cells. 9.a. 皆需template b. 发生方向 5’-> 3’ c. 所有enzymes 都藉由phosphodiester bond 来增加nucleotide. d. All use (deoxy) ribonucleoside triphosphates as substrate, and release pyrophosphates as a product. 10.Telomerase 是protein+RNA, 可以再telomere 的 G-rich strand 上, 以它自己的 RNA 当template, 合成DNA 在telomere上, 它可以同时加很多nucleotide 在telomere 上 , 然後再translocation 去进行下一次的合成, 而Reverse transcriptase, RNA polymerase, DNA polymerase 一次只能增加一个nucleotide, 然後translocation, 去进 行下一次。 11.Polymerase chain reaction (PCR)：利用双股DNA重复地denature 复制, 而在试管 中快速且大量复制任何DNA fragment的技术。我们可以利用PCR来大量生产特定DNA sequence, 因此可用来大量生产并分析存在於单一细胞的DNA, 可作为疾病的诊断。 Taq polymerase ( 耐高温的polymerase ) used in PCR reaction. 目前要再加入m RNA → DNA by reverse transcription. 13. Lactose operon 是一种inducible negative control, 也就是平常在repressor的存 在下, 此gene operator 是处於不表现的状态, 而在有Inducer ( lactose) 存在下, 此 operon 才可被induced 而表现。 → lactose 可和repressor 结合以改变repressor的结构形状, 使它无法接在operator 上而破坏其repression, 因为lactose 就是Inducer. 14. Lys and Arg side chain of histones are modified. 15. (Ans: please see chapter 10 page 267) 16. (Ans: please see chapter 11 page 214-215) --

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