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15. (A) acknowledged by (B) allocated to (C) confined to (D) distinguished from

1. 根據刑事訴訟法規定,告訴乃論之罪,應在知悉犯人之時起,多久內提出告訴? (A)3個月 (B)4個月 (C)5個月 (D)6個月

13.適應體育的 STEP 原則有利於幫助教師構思動作難易度的調整,並有利於規劃差異化教學。請問下列變化何者錯誤? (A)Space:調整投籃距離的遠近 (B)Task:調整投籃的動作 (如: 單手、雙手) (C)Equipment:調整活動進行的速度、對抗強度 (D)People:調整比賽場上的對抗人數

24 依實務見解,下列何種行為會構成刑法第 315 條之 1 之窺視竊聽竊錄罪? (A)在捷運車廂內拿密錄器偷拍乘客大腿外側、膝蓋及小腿部位 (B)在咖啡廳裡用肉耳偷聽隔壁桌客人的對談 (C)在某人車上裝設 GPS 追蹤器,長時間蒐集駕駛使用車輛之移動軌跡 (D)拿望遠鏡在遠處偷看在公園裡開心玩耍的小女孩

43. During the mitotic phase, the enzyme separase is responsible for cleaving the cohesin proteins that hold sister chromatids together, a step crucial for the initiation of anaphase. If a cell possesseda dominant mutation causing separase to be constitutively active and begin cleaving all cohesins prematurely during prometaphase (before the M checkpoint criteria were met), which outcome is the most likely and immediate consequence? (A) The resulting daughter cells would skip the G1 phase and immediately enter the S phase due to premature activation of maturation-promoting factor (MPF). (B) The cell would successfully complete mitosis, but the spindle poles would fail to move apart due to inactive non-kinetochore microtubules. (C) The M checkpoint would fall, and the liberated chromatids would segregate randomly, resulting in genetically unequal daughter cells. (D) The nuclear envelope would reform immediately, triggering early telophase and preventing the remaining spindle microtubules from attaching to kinetochores. (E) Cytokinesis would initiate during prophase, leading to the formation of multiple, small nuclei within a single parent cell.

44. 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. NTHU115 (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.