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发表于 24-9-2007 06:16 PM
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43. D
--- By using DNA fingerprinting technique (Restriction Fragment Length Polymorphism or RFLP), the victim can be identified.
通常，一個獨立樣本的DNA首先被提取和純化。純化後的DNA可以用PCR反應擴增。隨後，用restriction endonuclease 切成“限制性片段”，每種内切酶只能切除能夠識別的特殊序列。隨後，限制性片段通過 gel electrophoresis (瓊脂糖凝膠電泳)將不同長度的片段分開。結果可以通過Southern blotting(南方墨點法)被加強。
[ 本帖最后由 Jec 于 24-9-2007 06:34 PM 编辑 ] |
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发表于 24-9-2007 06:46 PM
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44. D
--- Fungi (singular fungus) are a kingdom of eukaryotic organisms.
The fungi are heterotrophic organisms characterized by
o non-motile
o a chitinous cell wall, and
o in the majority of species,
o filamentous growth as multicellular hyphae forming a mycelium;
o some fungal species also grow as single cells.
Sexual and asexual reproduction is via spores, often produced on specialized structures or in fruiting bodies.
Yeasts, moulds, and mushrooms are examples of fungi.
The discipline of biology devoted to the study of fungi is known as mycology.
[ 本帖最后由 Jec 于 24-9-2007 09:31 PM 编辑 ] |
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发表于 24-9-2007 06:56 PM
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45. A
--- 当然是gene bank咯！因为ex situ conservation的意思是人为的，不在原来的地方做conservation. Ex-situ conservation means literally, "off-site conservation". It is the process of protecting an endangered species of plant or animal by removing part of the population from a threatened habitat and placing it in a new location, which may be a wild area or within the care of humans.  |
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发表于 24-9-2007 07:02 PM
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46. B
--- This is a very famous example for polymorphism in natural selection!
There are five types of natural selection:
1. Polymorphism
2. Stabilising selection
3. Directional selection
4. Disruptive selection
5. Sexual selection
在英国，这种小蛾子的翅膀，本来是有许多黑白相间的斑斑点点，在正常情况下，其翅膀上的斑点使蛾子能混入环境中，不易被捕食它们的鸟所发现。但是，当环境因为工业污染而发生变化，使环境变得黑的时候，原来蛾子翅膀上的黑白相间的斑点反而更易让鸟发现。只有那些翅膀全部是黑色的蛾子，不易被鸟发现。久而久之，翅膀是斑点的蛾子越来越少，翅膀是黑色的蛾子越来越多。反之，如果环境没有被污染，黑色蛾子越来越少。
这个例子是教科书里面讲解 Darwinian evolutionary theory (达尔文进化论) 的“自然选择理论”时的经典故事。
其实，自然选择，性选择或人工选择，会导致生物群体内部的不同个体比例发生变化，没有什么稀奇或不可理解的。现在许多家养的物种，都是“选择力”作用下的结果，最明显的是中国人培养的金鱼。 |
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发表于 24-9-2007 07:11 PM
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47. C
--- Allopatric speciation results in new species to develop because the populations (demes) become physically isolated by occupying different geographical areas. Allopatric speciation也叫异域成种, 在生物演化中占据了主要地位，认为物种形成首先是由于在祖种分布域中出现地理隔离而产生的.
Option A is wrong. Speciation does take place.
Option B is wrong. Genetic drift also takes place.
Option C is correct.
Option D is wrong. Sympatric speciation (同域成种) refers to intrinsic isolation where populations (demes) are reproductively isolated within the SAME geographical area. |
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发表于 24-9-2007 07:24 PM
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48. D
--- This is a graph about Type II of the Survivorship Curve. It indicates a constant and continuous loss of the rate of the population growth when the population size increases. At point X, the rate of population growth is zero. This means the population size reaches at peak end. Therefore, X is the carrying capacity. And what is carrying capacity? Okay, here is the meaning of carrying capacity. Carrying capacity refers to the largest population that a habitat can support and sustain for an indefinite period, assuming there are no changes in the environment.
请看Option A.这个graph不可能是K-strategy action.原因？因为要当k-strategist,必须符合以下特性：
1. stable habitat
2. low rate of reproduction
3. usually reproduce later in life
4. highly specialised
5. size of offspring is big
6. low mortality rate
再看Option B & Option C. 这些statements应该改写成这样才对。
B. Minimum rate of natality.
C. Maximum rate of mortality.
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发表于 24-9-2007 08:55 PM
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49. D
--- In an ecosystem, as enery is transferred along a food chain, large losses of energy occur in every transfer. Energy transfer (能量转移) 是生物体内能量交接系统和状态的总称。生物界总体接收能量的窗口是利用光能的光合作用 or photosynthesis，由此产生了保存于化合物中的化学能，它可通过在同一个体内或向其它个体的传送，变为其它化合物的化学能，或者变为运动的力学能，变为光、电、物质输运以及热等等能量。起核心的关键作用的物质是ATP，核心系统是光合作用以外的呼吸和糖酵解。Each trophic level receives less energy than its previous trophic level.
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发表于 24-9-2007 09:10 PM
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50. D
--- 标记再捕法 marking-and-recapture method or capture-recapture method:
o 放出许多个标记的个体，再根据捕回的资料推算种群的迁移、分散的过程和范围，或者直接
计数有困难、流动性大的动物个体数的推算方法。
o 推算个体数最简单的方法是林肯法（Lincoln index）。例如对s个体作标记后放走，在以后
某个时间内捕得的n个体中，如有m个体是标记后再捕到的个体数，则总个体数N＝s·n/m。
o 这种方法只适用于没有出生、死亡、迁出、迁入等变动的种群。但通过几次反复捕捉、再捕
捉，即使个体数有变化，也能推算个体数、减少（迁出、死亡）率、增加（出生、迁入）
率，而提出了各种模式。
o 用标记再捕法来获得无偏差的个体数的推算值，必须满足下列条件：作有标记的个体和未作
标记的个体在捕捉的难易上必须是无差异的；标志不得消失等等。
Population size = (72 X 85)
---------
8
= 765 wood mice
[ 本帖最后由 Jec 于 24-9-2007 09:13 PM 编辑 ] |
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发表于 24-9-2007 09:27 PM
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恭喜恭喜 ..
终于完成了.. !!! |
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发表于 24-9-2007 09:33 PM
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发表于 27-9-2007 10:34 PM
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哈哈，我也看完了啦。。
恭喜我吧。。
哈哈！！！
还有题目吗？？ |
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发表于 27-9-2007 10:54 PM
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噢， 我好久没来咯！原来这里还有这么一个帖子。。。
在这，有什么特别活动吗？ |
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发表于 29-9-2007 11:35 AM
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发表于 30-9-2007 11:32 AM
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Hey, 2006年的STPM Bio试卷 paper 2来咯！
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发表于 30-9-2007 11:34 AM
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发表于 30-9-2007 11:35 AM
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发表于 30-9-2007 11:41 AM
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发表于 30-9-2007 11:44 AM
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回复 #173 zfc 的帖子
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我推荐你买6-YEAR SERIES 的STPM BIOLOGY (Longman)哦！里面有你要的资料！ |
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发表于 30-9-2007 07:57 PM
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回复 #177 Jec 的帖子
谢谢你啦。。。
是你太好人了。。。
对了，不好意识，向你教教我关于heridity and genetic 那课的， 我真的blur blur...
先谢谢你啦。。。 |
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发表于 30-9-2007 09:14 PM
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原帖由 Jec 于 30-9-2007 11:44 AM 发表 
我推荐你买6-YEAR SERIES 的STPM BIOLOGY (Longman)哦！里面有你要的资料！
其实那些考卷也是从longman pre-u biology的cd拿来的。他们只给试卷，却没给答案，一点诚意也没有！还说什么complimentary cd！:@
我还是lower 6，老师说年尾的考试会从往年的stpm试题来出题，所以临时临急才来这里问看谁有答案。 |
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