101年 高普考 高考三級 藥事 藥物分析與生藥學 試卷

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101
年公務人員高等考試三級考試試題 代號37140
科: 藥事
目: 藥物分析與生藥學(包括中藥學)
考試時間: 2小時
※注意:
禁止使用電子計算器。
不必抄題,作答時請將試題題號及答案依照順序寫在試卷上,於本試題上作答者,不予計分。
(請接背面)
全一張
正面
一、馬兜鈴酸之一的馬兜鈴酸甲(AAI)之結構如下:
列舉二種含馬兜鈴酸類成分之中藥。(6分)
為何含馬兜鈴酸類成分之中藥被行政院衛生署公告禁用?(6分)
二、Ginsenoside Rg1Ginsenoside Rb1之結構如下:
此二成分含於何植物科別及何種藥材中?(6分)
本結構在化學分類上屬於何種類別(如 indole alkaloid)?(4分)
此二成分在酸水解去醣基後之產物之分子式分別為C30H52O4C30H52O3,則其化
學結構各為何?(10 分)
此二成分之分析宜利用何種層析技術及檢示器?(12 分)
三、生合成 Morphine 之部分途徑示於下圖:
9
(R)
試寫出 IIII 之反應機轉(含參與之酵素)。(12 分)
101
年公務人員高等考試三級考試試題 代號37140
科: 藥事
目: 藥物分析與生藥學(包括中藥學)
全一張
背面
四、下列九個類黃酮化合物(19)係分離自菲律賓楠(Machilus philippinensis),其
分離方法/步驟如下:
The EtOH extract (160 g) of the dried leaves of M. philippinensis (774 g) was partitioned
into fractions soluble in CH2Cl2 (38.30 g), EtOAc (10.51 g), n-BuOH (31.02 g) and H2O
(74.12 g) by a liquid-liquid partitioning process. Part of the EtOAc-soluble fraction (6.03 g
out of 10.51 g) was fractionated on a Sephadex LH-20 column (815 × 35 mm, MeOH) to
give six fractions (E1~6). An aliquot of fraction E2 (45.0 mg out of 500.7 mg) was separated
by a semi-preparative RP-18 HPLC column (Phenomenex® Prodigy ODS-3, 250 × 10 mm,
5 μm), each run 5.0 mg, delivered by 16% MeCN in H2O for 30 min, to 25% MeCN in 40 min
by a linear gradient mode, and then MeCN for 15 min, with a flow rate of 2.5 mL/min and
detection at 300 nm. After nine runs, the fractions containing pure compounds were evaporated
under reduced pressure to give 3 (2.9 mg, tR = 21.39 min), 4 (10.0 mg, tR = 22.61 min), 6
(7.3 mg, tR = 30.86 min), 7 (4.8 mg, tR = 36.36 min), and 9 (1.9 mg, tR = 42.33 min),
respectively. An aliquot of fraction E3 (273.8 mg out of 1.20 g) was separated on a Lobar
(low-pressure) RP-18 column (LiChroprep RP-18, size B, 310 × 25 mm; 40-63 μm, Merck),
delivered by a stepwise gradient of MeOHH2O from 30:70 to 100:0, to give six
subfractions (E3-1~6). Fractions E3-3 (24.7 mg) and -5 (2.0 mg) were pure 5 and 8,
respectively. An aliquot of fraction E6 (201.0 mg out of 756.0 mg) was separated on the same
Lobar RP-18 column, delivered by a stepwise gradient of MeOHH2O from 35:65 to 85:15,
to give five subfractions (E6-1~5). Fractions E6-2 (14.4 mg) and -4 (2.3 mg) were pure 1 and
2, respectively.
以上圖右之方式表達化合物 19之分離流程。(10 分)
試由分離之結果簡述所使用的層析法之分離原理與特性。(16 分)
OH
MeO
Eugenol
5
6
2
78
9
3
五、Eugenol 之結構如下:
此成分含於何植物科別及何種生藥材中?(6分)
此成分之1H NMR在芳香氫區(aromatic proton region)與烯氫區(olefinic proton
region)呈現那些訊號(含偶合及偶合常數之分析)?(12 分)
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