Part III
1.3’ TACGGGCATACGTAAG 5’
2.The double-helical structure 可藉由hydrogen bond 來穩定, the base stacking
between adjacent bases on the same strand. Base stacking in nucleic acid 可以
減少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.每一個amino acid 對應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. 多餘的864 nucleotide pairs 也許可能是introns 或者轉譯出polypeptide, 但
經過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 是最常發生的mismatch 在 Eukaryotes DNA. 因
此, 有一套特別的repair system 是回復G≡C pair.
7.(Ans: please see chapter 12 page 337)
8.既然很多DNA nucleotides 是由每一個gene 所得到的, 有一個error in any one
molecule 會導致protein 上有不正確的amino acid. In quality control, 不正確的
protein 會很快地被degrade, 而有問題的mRNA 也不會存活太久, 所以問題不大. 但DNA
replication, errors would be transmitted to the next generations of cells.
9.a. 皆需template
b. 發生方向 5’-> 3’
c. 所有enzymes 都藉由phosphodiester bond 來增加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 當template, 合成DNA 在telomere上, 它可以同時加很多nucleotide 在telomere 上
, 然後再translocation 去進行下一次的合成, 而Reverse transcriptase, RNA
polymerase, DNA polymerase 一次只能增加一個nucleotide, 然後translocation, 去進
行下一次。
11.Polymerase chain reaction (PCR)：利用雙股DNA重複地denature 複製, 而在試管
中快速且大量複製任何DNA fragment的技術。我們可以利用PCR來大量生產特定DNA
sequence, 因此可用來大量生產並分析存在於單一細胞的DNA, 可作為疾病的診斷。
Taq polymerase ( 耐高溫的polymerase ) used in PCR reaction.
目前要再加入m RNA → DNA by reverse transcription.
13. Lactose operon 是一種inducible negative control, 也就是平常在repressor的存
在下, 此gene operator 是處於不表現的狀態, 而在有Inducer ( lactose) 存在下, 此
operon 才可被induced 而表現。
→ lactose 可和repressor 結合以改變repressor的結構形狀, 使它無法接在operator
上而破壞其repression, 因為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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