hce_nthu
115年
生物與生化
第 44 題
In an $E. coli$ cell undergoing rapid replication, a newly identified chemical agent completely and specifically inhibits the function of DNA ligase. Assuming all other replication proteins (Helicase, Primase, DNA pol I, and DNA pol III) are fully functional, what structural consequence would be immediately observable following the completion of the first round of DNA synthesis?
- A Both parental DNA strands would remain permanently associated, halting the replication process before the replication fork could open fully.
- B The leading strand would fail to elongate beyond the initial RNA primer because DNA pol III requires DNA ligase to begin continuous synthesis.
- C The lagging strand would consist of multiple Okazaki fragments that are fully synthesized DNA segments but lack covalent bonds between them.
- D The ends of the circular bacterial chromosome would shorten significantly because the replication machinery cannot replace the terminal RNA primers.
- E The concentration of thymine dimers would increase dramatically due to the inability of the DNA replication complex to proofread mismatched bases.
思路引導 VIP
請試著想像落後股（lagging strand）的合成過程：當所有的「建築材料」（核苷酸）都已經由聚合酶按照模板擺放就位，且 RNA 引子也都被置換成了 DNA，那麼這條新合成的長鏈在微觀結構上，是已經連成一氣的完整絲線，還是像一排緊密靠在一起、但彼此沒有黏死的積木？是什麼化學鍵負責這最後的「黏合」動作？
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AI 詳解
AI 專屬家教
太棒了！你能精準判斷出 DNA 接合酶 (DNA ligase) 的缺失對複製過程的具體影響，代表你對原核生物 DNA 複製的動態過程有著非常紮實且清晰的理解。
DNA 複製中的分工與接合
在 $E. coli$ 的複製過程中，雖然 DNA 聚合酶 III (DNA pol III) 負責主要延伸，DNA 聚合酶 I (DNA pol I) 負責移除 RNA 引子並填補 DNA 缺口，但這兩者都只能將核苷酸接到已有的 $3'-OH$ 末端。當落後股 (lagging strand) 的各個片段合成完畢後，相鄰的片段之間雖然緊密相連，卻仍缺乏最後一個磷酸二酯鍵 (phosphodiester bond)。由於題目設定只有 DNA 接合酶受到專一性抑制，因此 岡崎片段 (Okazaki fragments) 會被完整地合成出來，卻無法像正常狀態下那樣被「縫合」成一條連續的長鏈。
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