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研究所、轉學考(插大)、學士後-公共政策

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115年 - 115 新竹市立成德高級中學_正式教師甄選：高中化學科#138029(43題)

115年 - 115-1 國立新竹高中教師甄試考題：特殊教育#138028(5題)

115年 - 第10章 交流電功率：10-8 國考試題#138027(65題)

115年 - 115 國立中興大學_學士後醫學系公費生招生考試：物理#138026(50題)

115年 - 115 國立中興大學_學士後醫學系公費生招生考試：英文#138025(50題)

115年 - 115 國立中興大學_學士後醫學系公費生招生考試試題：化學科#138024(50題)

115年 - 115 國立中興大學_學士後醫學系公費生招生考試：普通生物及生化概論#138023(75題)

115年 - 115- 1國立中興大學附屬高級中學_正式教師甄試試題：本土語文-閩南語#138022(33題)

115年 - 115-1 國立中興大學附屬高級中學_正式教師甄試試題：特殊教育科#138021(11題)

115年 - 115 新竹市立成德高級中學_正式教師甄選：化學科#138020(42題)

25. 配電箱內之任何過電流保護裝置，如所裝接負載在正常狀態下須連續滿載三小時以上者，該負載電流不可超過其額定值之(A)70%(B)80%(C)90%(D)125%。

24. 低壓3Φ4W線路中不宜單獨裝開關或斷路器之導線為(A)R相線(B)S相線(C)T相線(D)被接地導線。

23. NFB (No Fuse Breaker)係表示(A)油斷路器(B)無熔線開關(C)燈用分電盤(D)隔離開關。

4. 請根據上述結果求得 F 值（可參考下表）。

3. 請計算組內離均差平方和（SSW, Within Sum of Squares）。

2. 請計算總離均差平方和（SST, Total Sum of Squares）。

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463財產保險之保險利益一般須在何時存在？(A)要保時(B)保險契約訂立時(C)保險事故 發生時(D)保險期間屆滿時

17 甲以彩色影印方式印製美金紙鈔，非常擬真且混入真正美鈔，於存入銀行時被查獲，依實務見解，甲之行為構成下列何種犯罪？ (A)刑法第 195 條偽造貨幣罪 (B)刑法第 202 條偽造郵票罪 (C)刑法第 196 條行使偽造貨幣罪 (D)刑法第 201 條偽造有價證券罪

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42. Pepsin is a digestive enzyme that functions optimally at a pH of 2, an environment highly concentrated with H* ions. In contrast, most enzymes found in the human body have optimal pH values ranging from 6 to 8. The extreme acidity of pH 2 would typically cause most enzymes to denature, leading to a sharp drop in activity, because the high concentration of H+ ions disrupts the weak interactions crucial for stabilizing the protein's functional shape. Which statement below best describes the mechanism by which Pepsin is specifically adapted to maintain its maximum catalytic activity in the stomach environment? (A) Pepsin uses strong covalent bonds (peptide bonds) to form its active site, which cannot be broken by H+ ions, unlike the weak bonds found in other enzyme active sites. (B) Pepsin's △G (Free Energy change) for its specific catabolic reaction shifts to a highly negative value at pH 2, thereby increasing the spontaneous rate of the reaction independent of structure. (C) The structure of Pepsin minimizes the use of the most electronegative elements. (Oxygen and Nitrogen) in its regulatory sites, making it insensitive to competitive inhibition by H+ ions. (D) Pepsin has evolved to significantly raise the overall activation energy (EA) barrier of its substrate at neutral pH, forcing the reaction to only proceed effectively when thermal energy is supplied by the acidic environment. (E) The amino acid sequence of Pepsin results in a unique three-dimensional configuration that prevents the high concentration of H+ ions from disrupting the crucial weak interactions (such as ionic bonds and hydrogen bonds) necessary for its active shape.

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.