生物讨论专区

Jec

23. D
--- Hormones are grouped into three classes based on their structure:

• steroids
• peptides
• amines

--- The endocrine system acts by releasing hormones that in turn trigger actions in specific target cells. Receptors on target cell membranes bind only to one type of hormone. More than fifty human hormones have been identified; all act by binding to receptor molecules. The binding hormone changes the shape of the receptor causing the response to the hormone. There are two mechanisms of hormone action on all target cells. Here, we mention only non-steroid hormone mechanism so that it is sufficient to understand the nature of the question no. 23.

--- Non-steroid hormones (water soluble) do not enter the cell but bind to plasma membrane receptors, generating a chemical signal (second messenger) inside the target cell. Five different second messenger chemicals, including cyclic AMP have been identified. Second messengers activate other intracellular chemicals to produce the target cell response.

  

Jec

24. A
--- 这道题目又是另一题跟上题有关的。这题是讲steroid hormone。除了option number A,其它都是叙述non-steroid hormone。 如果已经全面知道the mechanism of hormone,你会觉得非常容易。

Jec

25. A

26. C

以下的解说会更详细一点。

[ 本帖最后由 Jec 于 23-9-2007 01:14 PM 编辑 ]

Jec

27. C
--- Non-specific immune response is matched with these structures: skin and phagocytic white blood cells.The relavant defences are phagocytosis, Natural-Killer Cells (NK cells), inflammation and fever attack.

Specific immune response includes antibodies and lymphocytes. Antibodies are specific proteins,known as immunoglobulin, that recognises and binds to specific antigens. Lymphocytes consists of two types, namely B-cells and T-cells (TC, TH and TS).

我们来谈谈免疫学的一些重要的概念. "免疫学"起源于拉丁文 "immune", 意为"安全". 当一个活的生物被外部物质攻击的时候, 防御功能立刻反应以抵抗. 免疫反应是防御功能之一.

第一道防线有皮肤, 胃(胃酸: pH2), 黏膜, 泪液, salivary amylase唾液(溶菌酶, IgA)等可以抵御病毒, 细菌和霉. 当第一道防御线被穿越, 下一道就是免疫反应, 有先天免疫和后天免疫之分.

补体, macrophage巨噬细胞, 溶解酵素, 和干扰素通过非特异性反应参与先天免疫. 所以, 溶解酵素和干扰素被称作天然抗体, i.e. antibody.

后天免疫是一种细胞防卫, 能够识别几种特定的对身体有害的外部物质. 这种能识别多种抗原的强大反应, 得益于三种类型的细胞表面感受器的进化. 每一类都有很多, 分别是T细胞, 抗体和MHC分子. 考虑到它们的蛋白质结构类似, 有人推测他们起源于相同的基因. 由于难以判断一种外部刺激物是否有害,

我们的身体总是假定未知物为有害的入侵者. 如果免疫反应成功, 身体复原并产生特定记忆, 引起后天免疫, 所以它下次能处理相同的物质. 然而, 免疫反应并非总是好事. 极端免疫反应能引起自身免疫性, 移植物对抗宿主疾病和allergic reaction(过敏).

因此, 在自身免疫, 过敏症和移植手术时必须抑制免疫反应. 尤其是如果一个器官接受者在手术之前已输入了器官捐赠人的血, 移植将引起敏锐的排斥反应, 因为接受者已被免疫.

[ 本帖最后由 Jec 于 24-9-2007 09:54 PM 编辑 ]

Jec

28. C
--- One example of Bryophyte is Marchantia sp.
Fertilisation:
During fertilisation, the antheridium will burst when matured and the male gametes swim to the archegonia within films of water (either dew or rain).

Spore dispersal:
The capsule in the sporogium or sporophyte discharges its spores with aids of the elaters which undergo a twisting movement to push spores out of it.

[ 本帖最后由 Jec 于 23-9-2007 06:12 PM 编辑 ]

Jec

29. A
--- The following steps are the correct sequence of the early development of vertebrate embryos.
1. The cleavage
   o Rapid cell division but the overall size does not increase.

2. Blastula formation
   o Blastula is a hollow ball of cells. Due to the secretion of blastocoel by  the dividing blastomeres in the centre of morula, the blastula is formed.

3. Grastula formation
   o Another name for this process is grastulation. Three primary germ layers are formed, namely ectoderm, mesoderm and endoderm.

4. Organogenesis
   o The germinal layers rearrange to differentiate and form organs.The complexity increases  at the peak in this last stage of the embryonic development.

muse

wow..真的太详细了。。
谢谢jec..

Jec

原帖由 muse 于 23-9-2007 06:01 PM 发表
wow..真的太详细了。。
谢谢jec..

我快做到吐血了。。。 谢谢你的鼓励，我越做越开心。很大的鼓舞和满足感！

Jec

30. A
--- Option number IV definitely is incorrect.

Jec

31. D
--- Relative growth rate curve = the gain as percentage of previous mass over time.

32. A
--- R phase is also called decelerating phase. S phase is also called stationary phase. S phase does not have any growth observed as the overall growth has ceased.

Jec

33. A
--- Calculate gene map:
(46+48)/500  X 100%  = 18.8 %  = 18.8 map units

This case is a recombinant case.

In test-cross, the genotype we used must be recessive, vg vg b b.

Jec

34. A
--- Test-cross is the cross between the progeny and a genotypically recessive plant.
    Back cross is the cross between the progeny and one of its parent plant.
    Reciprocal cross is the cross between the plants which are reversal genotype or phenotype compared to the previous cross.

Jec

35. B
--- Tetraploid means four sets of chromosomes.

The origin of polyploids:

1. Two models for the origin of allopolyploids have been proposed. The one-step model suggests that an allotetraploid is formed by fusion of unreduced male and female gametes from two diploid species or by direct hybridization between two autotetraploid species, because almost every plant species produces a variable but small amount of unreduced gametes and many plant species are autotetraploids.

2. The two-step model proposes that an allotetraploid is formed through hybridization between two diploid species followed by chromosome doubling of the F 1 hybrid.

The two-step process, usually using mitotic inhibitors such as colchicine, has been used to produce artificial allotetraploids such as in Brassica and wheat.

[ 本帖最后由 Jec 于 23-9-2007 06:51 PM 编辑 ]

Jec

36. A
--- Addition (or insertion) as well as deletion of a nucleotide can cause frameshift mutation.Addition and deletion  of a nucleotide must happen after the start codon, not before the start codon to see the frameshift mutation take its consequence. Inversion does not cause frameshift mutation because only the base (or bases) which are inversed are involved in the gene level changes, the rest of the genes are still intact. This explanation also works for substitution.

Frameshift mutations:
Since they occur in a region which is translated into protein, such an addition or deletion of the base will put the downstream region out of proper reading frame.

我的周末都化为乌有啦。若还有空闲的话，我将会回来完成未完成的工作。这些答案都是辛苦搜罗回来的喔， 哈哈！

[ 本帖最后由 Jec 于 24-9-2007 05:03 PM 编辑 ]

Jec

37. D
--- 所谓 albino gene 就是暗示frequency of homologous recessive individual可以等于q2 = 100/360.
那么， q = squared root of (100/360)

我们知道Hardy-Weinberg equation：p+q = 1
也就是说， p = 1 - q
           p = 1 - (10/squared root of 360)
             = 0.47295
p2 + 2pq + q2 = 1

既然已查出 p 和 q 的数值， 我们便可以找到 2pq 了。
0.22369 + ２pq + 0.27778 = 1
                    2pq  = 1 - 0.22369 - 0.27778
                    2pq  = 0.49850


Number of heterozygous individuals:
  0.49850 X 360 = 179

Jec

38. C
--- Mutation I is a substitution. Reason: Base A in the 2nd triplet code ACG is substituted with C becoming triplet code CCG.

--- Mutation II is an inversion. Reason: The 5th triplet code CCT is inverted into the triplet code CTC.

[ 本帖最后由 Jec 于 24-9-2007 05:26 PM 编辑 ]

Jec

39. D
--- 如果要回答这道题目，首先要明白什么会导致frequency of dominant and recessive alleles 保持不变。有关factors是：
1. Large population size
2. Random mating
3. No mutation
4. No migration
5. No natural selection

因此，采取逆向思维便能告诉我们如何寻获答案了。

Option I is wrong. Mutation causes changes in the allele frequency.
Option II is wrong. Natural selection causes changes in the allele frequency.
Only Option III and IV are correct.

Jec

40. A
--- The regulator gene codes for the synthesis of a mRNA which is translated into a repressor protein. When lactose is absent, the active repressor molecule which has a strong affinity for the operator binding site, binds to the operator and covers parts of the promoter. The promoter region is the binding site for the RNA polymerase.

Jec

41. B
--- Three main components of lactose operon: structural gene, operator & promoter.

Lac Operon: Section of DNA with a promoter/operator in bacteria (one kind of prokaryitic cell) that acts as an on/off switch for several genes that control the breakdown of the sugar lactose into glucose + galactose.

Promoter: DNA section where RNA polymerase binds.

Operator: DNA section where repressor binds

Structural gene: Gene that makes an enzyme or other protein

Regulator gene: A gene not found next to the operon that makes a repressor protein.

Repressor: protein that shuts off operon. If bound to operator, it prevents RNA polymerase from initiating transcription.

图像跟上题一样。不过，我亦看上了这个插图，顺便post上来共享。

Jec

42. A
--- Bacteria carrying plasmids with an antibiotic resistance gene can grow in a medium containing ampicillin. Bacteria without ampicillin resistance gene in its plasmid will be killed. As a result, this is a very important procedure in screening the recombinant bacteria.