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2010国际生物奥赛B卷


IBO2010 K REA

THEORETICAL TEST Part B

ENVELOPE COVER SHEET

Student Code:

The 21st INTERNATIONAL BIOLOGY OLYMPIAD Changwon, KOREA 11th – 18th July,

2010

THEORETICAL TEST: PART B 理论题 B 卷 Time available: 150 minutes 考试时间 150 分钟
GENERAL INSTRUCTIONS 说明 1. Open the envelope after the start bell rings. 铃声响后才能打开试卷封套 2. A set of questions and an answer sheet are in the envelope. 试卷封套内含试题与答案卷 3. Write your 4-digit student code in every student code box. 写上四位数的学生编号 4. The questions in Part B may have more than one correct answer. Fill the Answer Sheet with checkmarks (√), numbers, or characters to answer each question. 可有多个正确答案,依指示注记作答 5. Use pencils and erasers. You can use a ruler and a calculator provided. 利用铅笔与橡皮擦作答,可以使用尺与大会提供的计算器 6. Some of the questions may be crossed-out. DO NOT answer these questions. 被删除的题目,不要 作答 7. Stop answering and put down your pencil IMMEDIATELY after the end bell rings. 铃声响后 立即 停止作答 8. At the end of the test session you should leave all papers at your table. It is not allowed to take anything out. 结束后,把试卷留在桌上,勿携出教室

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IBO2010 KOREA THEORETICAL TEST Part B

Country:

Student Code:

The 21st INTERNATIONAL BIOLOGY OLYMPIAD
Changwon, KOREA 11th – 18th July, 2010

THEORETICAL TEST: PART B 理论题 B 卷
Time available: 150 minutes 考试时间 150 分钟
GENERAL INSTRUCTIONS 说明 1. Write your 4-digit student code in every student code box. 写上四位数的学生编号 2. The questions in Part B may have more than one correct answer. Fill the Answer Sheet with
checkmarks (√), numbers, or characters to answer each question.

可有多个正确答案,依指示注记作答 3. Use pencils and erasers. You can use a scale and a calculator provided. 利用铅笔与橡皮擦作答,可以使用尺与大会提供的计算器 4. Some of the questions may be crossed-out. Do not answer these questions. 被删除的题目,不要 作答

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IBO2010 KOREA THEORETICAL TEST Part B

5. The maximal point of Part B is 107.1. 满分为 107.1 分 6. Stop answering and put down your pencil immediately after the end bell rings. 铃声响后 立即 停止作答 7. At the end of the test session you should leave all papers at your table. It is not allowed to take anything out. 结束后,把试卷留在桌上,勿携出教室

GOOD LUCK!!

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IBO2010 KOREA THEORETICAL TEST Part B

CELL BIOLOGY

B1. (2.7 points) The Western blot below shows migration distances of five signal molecules (a~e) involved in a growth hormone-regulated cell-signaling pathway. 下图西方墨点法的结果显示五种参与激素调控的细胞讯息途径的 信息分子 (a~e),其之间的移动 距离。

To determine the order of molecules (a~e) in the signal cascade that occurs upon the growth hormone treatment, cells were treated with different inhibitors (I~IV) of cell signaling. The following blots show the changes in signal molecule expression patterns resulting from inhibitor treatment. 为了了解这 五种信息分子 (a~e) 参与激素调控的细胞讯息途径,分别加入 四种不同的抑制剂 (I~IV) 进行信息传导之研究。下图为分别加 四种抑制剂 后所得到的西方墨点实验结果。

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IBO2010 KOREA THEORETICAL TEST Part B

B1.1. (1.5 points) Fill in the boxes in the answer sheet to show the order of proteins (a~e) in the signaling cascade. 在答案纸的方块中,写上 蛋白质 (a~e) 的代号。

B1.2. (1.2 points) Fill in the circles in the answer sheet to show the site where each inhibitor (I~IV) exerts its action. 在答案纸的方块中,写上 抑制剂 (I~IV) 的代号。

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IBO2010 KOREA THEORETICAL TEST Part B

B2. (2.4 points) Match the molecular constituents (a~f) on the right with the cellular structures (A~D) that maintain cell morphology on the left. Each cellular structure can have more than one molecular constituent. 配合题,右边 (a~f) 代表 组成分子,左边 (A~D) 代表 细胞结构。每一种细胞结构可能会有 多种组成分子参与。
a. Cadherin 钙黏着蛋白 A. Cytoskeleton 细胞骨架 B. Cell wall 细胞壁 C. Desmosome junction 胞桥体 D. Extracellular matrix 胞外基质 b. Cellulose 纤维素 c. Collagen 胶原蛋白 d. Actin 肌动蛋白 e. Keratin 角蛋白 f. Lignin 木质素

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IBO2010 KOREA THEORETICAL TEST Part B

B3. (1.5 points) In the figure, the letter in each box represents an organ or tissue. Match each listed organ or tissue in the answer sheet to the correct box in the figure. 配合题,下图中,方块分别代表不同的 器官或组织。在答案纸中将 正确 的英文字母 与 器官 或组织 名称进行配对。

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IBO2010 KOREA THEORETICAL TEST Part B

B4. (2 points) When E. coli is grown on a medium containing a mixture of glucose and lactose, it shows complex growth kinetics, as shown in the graph below. 下图为大肠菌在含有葡萄糖与乳糖混合的培养基中之生长曲线。

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IBO2010 KOREA THEORETICAL TEST Part B

B4.1. (1 point) Which pair of graphs correctly shows the changes in glucose concentrations in the medium and β-galactosidase activity within the cells? 下列有关培养基中的葡萄糖浓度与细菌内 β- 半乳糖酶活性的关系配对,何者 正确?

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IBO2010 KOREA THEORETICAL TEST Part B

B4.2. (1.2 points) The graph below shows the expression pattern of lac mRNA in wild-type and mutant E. coli cells after lactose is added to a glucose-depleted medium. 下图为野生型与突变株的大肠菌在葡萄糖 用尽 、 改添加乳糖的培养基中 lac mRNA 的表现趋 势。

Indicate with a checkmark (√) in the answer sheet whether each mutant is able or unable to show the mutant expression pattern. 有关下列四种突变株 是否 能正确表现 突变株 趋势,请在答案纸上 正确处 以 √ 作答。 Mutant 突变株

I. An E. coli mutant in which the repressor is not expressed. 抑制子不表现 II. An E. coli mutant in which the repressor can bind to the operator, but not to lactose. 抑制子会与操作子结合,但是无法与乳糖结合 III. An E. coli mutant in which the operator is mutated so that the repressor cannot bind to the operator. 操作子突变,抑制子无法与之结合 IV. An E. coli mutant in which RNA polymerase cannot bind to the promoter of the lac operon. RNA 聚合酶无法与 乳糖操纵组的启动子结合

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IBO2010 KOREA THEORETICAL TEST Part B

B5. (1.5 points) Transcription and translation of a gene in a prokaryote cell are depicted in the picture below.

下图为原核细胞中基因 转录与转译的 简图

Indicate with a checkmark (√) in the answer sheet whether each description is true or false. 有关下列三种叙述,请在答案纸上 正确 或 错误 处 以 √ 作答。 Description I. The direction of transcription is from (B) to (A). 转录方向由 (B) 向 (A) II. Location (C) of the mRNA is the 5' - end. (C) 位在 mRNA 的 5? 端 III. The polypeptide on ribosome (D) is longer than the polypeptide on ribosome (E). 在 (D) 处核糖体 多肽链比 (E) 处核糖体 长 叙述

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IBO2010 KOREA THEORETICAL TEST Part B

B6. (2 points) A part of the nucleotide sequence of one strand of a double-stranded DNA molecule and the corresponding amino acid sequence are shown. The table shows a portion of the genetic code. 下图为双股 DNA 中的单股核苷酸序列与胺基酸对应简图,表中为部分的遗传码与胺基酸对应序列 Codon position DNA strand polypeptide 密码子位置 DNA 股 胺基酸 5'........ ........ a TTT Phe b AAG Lys c TTA Leu d AGC Ser .......3' .......

Codon 密码子 UUU UUA AAG AGC

Amino acid 胺基酸 Phe Leu Lys Ser

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IBO2010 KOREA THEORETICAL TEST Part B

Indicate with a checkmark (√) in the answer sheet whether each description in true or false. 有关下列四种叙述,请在答案纸上 正确 或 错误 处 以 √ 作答。 (Assume that the length of the DNA is the same as that of its primary transcript.) (本提前题为各条件下 DNA 在它们初转录时的长度是相同的)

Description I. The DNA strand shown is a template strand. 该股 DNA是为进行转录之模板

叙述

II. If the G+C content of the DNA strand shown is 40%, then the A+T content of its complementary DNA strand is 60%. 若该股 DNA 的 G+C 含量为 40%,则它的互补 DNA 中,A+T 的比例为 60% III. If the G+C content of the DNA strand shown is 40%, then the A+U content of the primary transcript is 60%. 若该股 DNA 的 G+C 含量为 40%,则初转录的mRNA中 A+U 的比例为 60% IV. The nucleotide sequence of mRNA is 5' ....... UUU AAG UUA AGC ....... 3'. mRNA 的核苷酸序列为 5' ....... UUU AAG UUA AGC ....... 3'

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IBO2010 KOREA THEORETICAL TEST Part B

B7. (2 points) The picture below shows the process of generating a transgenic plant harboring gene X using the Agrobacteria Ti-plasmid. 下图为利用 农杆菌 Ti 质体 产生具有 X 转殖基因的植物步骤。

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IBO2010 KOREA THEORETICAL TEST Part B

B7.1. (1 point) Which explanation about this process is true or false? 有关下列五种解释,请在答案纸上 正确 或 错误 处 以 √ 作答。 Explanation 解释

I. Restriction enzymes and ligase are used to make the recombinant DNA. 限制酶与接合酶会用来重组 DNA II. Plant tissue culture techniques are used to differentiate the leaf discs into a plant. 植物组织培养技术可应用在将一小块叶片分化转成植株 III. The whole recombinant Ti-plasmid harboring gene X gets integrated into the plant genome. 具有 X 基因的整个重组 Ti 质体可以插入植物的基因体中 IV. The introduction of gene X into the transgenic plant genome can be confirmed by using genomic PCR or genomic Southern blot analysis. 可以利用基因体 PCR 与南方墨点法,检测转植基因植物基因体中是否具有 X 基因 V. The expression of the introduced gene X in the plant cell can be checked by using RT (reverse transcriptase) -PCR, Northern blot analysis, or Western blot analysis. 可以利用 RT (反转录)-PCR、北方墨点法或西方墨点法,检测植物细胞中是 否具有 X 基因的表现

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IBO2010 KOREA THEORETICAL TEST Part B

B7.2. (1 point) Evaluate whether the following description is true or false for a plant expression vector in general? 有关下列五种常见的植物表现载体叙述,请在答案纸上 正确 或 错误 处 以 √ 作答。

Description I.

叙述

It should include the selection marker gene that is needed for selecting the transformed cell. 需要具有筛选基因来筛选转型过的细胞

II.

It should include a promoter that can express the introduced gene within the plant cell. 需要启动子来让基因在植物细胞中表现 It usually contains a multiple cloning site used for insertion of the foreign gene. 需要一个多重选殖位点来插入外来基因 It should have the same nucleotide sequence with the specific part of the plant genome because the foreign gene is inserted by homologous recombination. 须要有一段与植 物基因体中完全相同的核苷酸序列,因为外来基因需要置段特殊 的基因以利同源 重组法进行基因插入

III.

IV.

V.

It should have the replication origin needed for cloning during the process of making the recombinant vector. 选植过程中,载体必须含有复制起点才能复制重组载体

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IBO2010 KOREA THEORETICAL TEST Part B

B8. (1.5 points) Caulobacter bacteria undergo a special cell division. Division of the mother cell results in two different daughter cells: a ?roaming? (r) cell and a ?pedicle? (p) cell. Roaming cells permit Caulobacter to spread out. Pedicle cells stay and use the pedicle to stick at that place. The picture below shows how roaming and pedicle cells divide. 茎菌 有一种特别的细胞分裂方法,分裂母细胞会形成两种子细胞:游动细胞 (r) 与茎着细胞 (p)。游动细胞会让茎菌散布出去,茎着细胞会产生茎的构造让细胞留在原地。游动细胞与茎着细 胞的分裂方法如下图所示。

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IBO2010 KOREA THEORETICAL TEST Part B

The division cycle period when starting with a roaming cell (r = 90 min) is longer than when starting with a pedicle cell (p = 60 min). The extended length of period (r) is because the roaming cell 细胞分裂周期的长短不一,游动细胞需要 90 分钟,远大于由茎着细胞所需的 60 分钟。 游动细胞需要较长的分裂时间是因为:

A. produces more DNA than the pedicle cell. 与茎着细胞相比,需要合成较多的 DNA B. produces a pedicle before division. 在分列前要先产生茎柄 C. produces a flagellum during division 分裂中需要产生鞭毛. For each of the above explanations, indicate with a checkmark (√) on the answer sheet whether it is true or false. 有关有关上述三种答案,请在答案纸上 正确 或 错误 处 以 √ 作答。

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IBO2010 KOREA THEORETICAL TEST Part B

B9. (2 points) In the experiment described below, cells (1) were put in a medium with a salt concentration lower than the cytoplasm, causing them to swell and rupture at one location (2). Ruptured cells were then washed out and resealed to form ?ghosts? (3). This process also produced smaller vesicles whose membrane was either right-side-out (4) or inside-out (5), depending on the ionic conditions of the solution used for the disruption procedure.

如下图所示,细胞培养时 (1),当培养基盐的浓度低过细胞,所以细胞会有膨胀与破裂现象 (2),破裂的细胞经过清洗后会重新愈合,形成鬼细胞 (3),产生鬼细胞过程中会出现许多小囊 泡,这些小囊泡在胀破过程中,会因为离子的条件不同而产生正确面向外 (4) 与内面外翻 (5) 两种。

Prepared ghosts/vesicles were then mixed with a radioactive labeling reagent that is water-soluble, and could covalently attached to protein (3~5). The proteins embedded in the membrane were then solubilized with detergent and analyzed by SDS polyacrylamide-gel electrophoresis. Segregated proteins were visualized by Coomassie Blue staining (I) and autoradiography (II). 在准备鬼细胞或是囊泡的过程中,一种水溶性的放射物质被添加到溶液中,藉以跟蛋白质产生 共价附着 (3~5)。接着将这些细胞膜蛋白利用清洁剂溶解萃取后,以 SDS 电泳分离,并藉由蛋 白质染剂 (Coomassie Blue) 染色 (I) 与自动放射显影 (II) 得到下图。

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IBO2010 KOREA THEORETICAL TEST Part B

Which of the proteins (a~e) is/are transmembrane protein(s)? 蛋白质 (a~e) 中,何者是穿膜蛋白? A. Protein b B. Protein c C. Protein d D. Proteins a~e

b
c d a~e a 与 e

E. Protein a and protein e

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IBO2010 KOREA THEORETICAL TEST Part B

B10. (1.5 points) Subcellular organelles and their cellular components can be easily separated by the size-fractionating differential centrifugation method, as depicted below. During the process, four pellets (nucleus and 1~3) are formed. 使用离心机,利用大小差异梯度分离法可以将胞器与细胞成分进行区分。操作方法如下图所示, 操作后会得到四个沉淀物 (细胞核与沉淀物 1~3)

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IBO2010 KOREA THEORETICAL TEST Part B

The table below shows descriptions about subcellular organelles collected in different centrifugation pellets. 下表为有关细胞核与沉淀物 1~3 的特性描述 Pallets 沉淀物 Nucleus 细胞核 Pellet 1 沉淀物 1 Pellet 2 沉淀物 2 Pellet 3 沉淀物 3 Description 描述

An organelle containing a linear DNA harboring telomeric sequences. 具有端粒序列的线状 DNA An organelle inheriting its genetic information by maternal inheritance. 具有自有的遗传物质并有母系遗传特性的的胞器 An organelle performing glycosylation of most proteins. 含有大多数糖解作用的蛋白质的胞器 An organelle composed of two subunits and synthesizing proteins. 具有两个次单位且参与蛋白质合成的胞器

Provided that the subcellular structures are not disrupted during the centrifugation process, determine whether descriptions A, B and C of different subcellular structure in the pellet are true or false taking above information as a reference. Mark the appropriate box with a checkmark (√) in the answer sheet. 假设在操作过程中并没有影响酵素活性,下表为有关 沉淀物 1~3 的功能说明。请在答案纸上 参照上表,针对 A, B 与 C 的描述,判断答案是 正确 或 错误,并以 √ 作答。 Pellet 沉淀物 A Pellet 1 沉淀物 1 Pellet 2 沉淀物 2 Description 描述

An organelle containing a bunch of proteases, lipases, and nucleases. 含有蛋白酶,脂肪酶与核酸酶 An organelle carrying an enzyme catalyzing the conversion of hydrogen peroxide (H 2 O2 ) to water and oxygen 含有可催化过氧化氢转化成水的酵素之胞器. The infected intracellular virus covered with viral coat. 含有病毒外套蛋白的细胞内病毒颗粒

B

C

Pellet 3 沉淀物 3

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IBO2010 KOREA THEORETICAL TEST Part B

B11. (2 points) The SalI and XhoI restriction map of a 5 kb linear DNA molecule is shown below. 一段 5 kb 线状 DNA 的限制酶舆图。

The 3.5 kb DNA fragments obtained from a XhoI digestion were ligated with the 1.0 kb DNA fragments obtained from a SalI digestion. The resulting 4.5 kb DNA molecules were digested with SalI. Write down all the different lengths of DNA fragments you can get from this digestion. (Assume that restriction enzymes completely cut all the DNA molecules, and ignore blunt-end ligation.) 一段且以 XhoI 剪切的 3.5 kb DNA 片段和另一段且以 SalI 剪切的 1.0 kb DNA 片段接合。 此 4.5 kb 的 DNA 片段经由 SalI 完全作用后,所能得到的 DNA 片段大小分别为何? (假设限制酶将所有的 DNA 作用完全,同时不考虑平端接合)

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IBO2010 KOREA THEORETICAL TEST Part B

B12. (1.5 points) The following graphs show the quantitative change in DNA content at each of four stages in the cell cycle (G1, S, G2, M).

下图分别是四个不同细胞周期 (G1, S, G2, M) 阶段的 DNA 含量的改变

Select the graph (A~D) representing the stages described in I~III. 请将下列描述 I~III 对应 图 A~D 中,在 正确 处打 √。 Cellular activity and response 细胞活性与反应

I. Taxol treatment, which prevents microtubule deploymerization, arrests the cell at this stage. 使用微管去聚合化药剂 Taxol 处理,细胞会停留在这个阶段 II. With a mitogen treatment, such as an epidermal growth factor, an arrested cell at this stage proceeds to the next stage of the cell cycle. 使用促进细胞分裂素,例如上皮细胞生长因子,细胞会停留在这个阶段并准备进入下一个阶 段 III. The cell cycle check point at this stage confirms that DNA duplication is complete before the cell proceeds to the next stage. 这个阶段具有细胞周期校正点,以确认在进入下一个阶段前,会完成 DNA 已完成复制

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IBO2010 KOREA THEORETICAL TEST Part B

PLANT ANATOMY AND PHYSIOLOGY
植物解剖与生理 B13. (2 points) A transgenic Arabidopsis plant (2n) has a total of two copies of a kanamycin-resistant gene in its nuclear genome, one on chromosome 1 and the other on chromosome 3. For each description of this plant, indicate with a checkmark (√) in the answer sheet whether the description is true or false. B13. (2 points) 一基 因 转殖阿拉 伯芥植物 的 细胞核基 因组中总 共 有 2 份拷备 的抗 抗生 素基 因 (kanamycin-resistant gene),一在染色体 1 、另一在染色体 3 上。下列为有关此植物的描述, 判断其真或伪,并在答案纸上适当空格中打勾(√)。 Description描述 I. All pollen grains of this plant have kanamycin-resistant genes. 此植物的所有花粉粒都有抗抗生素基因 II. Endosperms formed by self-fertilization of this plant have 0~6 copies of the kanamycin-resistant gene. 此植物经自体受精所产生的胚乳中有0~6份拷备的抗抗生素基因 III. If seeds from self-fertilization of this plant are germinated, the ratio of kanamycin-resistant to kanamycin-sensitive seedlings is 3 to 1. 若此植物经自体受精所产生的种子萌发了,其可抗抗生素与对抗生素敏感的幼苗比例 为3:1 IV. A cell containing 4 copies of the kanamycin-resistant gene exists among root cells at prophase of mitosis in this plant. 在此植物根部细胞之有丝分裂前期时,可见细胞中具有4份拷备的抗抗生素基因

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IBO2010 KOREA THEORETICAL TEST Part B

B14. (1.5 points) Figure a shows an ABA signal transduction pathway in a guard cell. Figure b shows changes occurring after ABA treatment in (1) the cytoplasmic Ca2+ concentrations of guard cell and (2) stomata aperture size.

图a显示离层素ABA在保卫细胞中的讯号传递路径,图b是在ABA处理之后(1)为保卫细胞的
细胞质之 Ca2+ 浓度变化; (2) 为气孔大小变化。

For each description about ABA action, indicate with a checkmark (√) in the answer sheet whether the description is true or false. 下列为有关ABA作用的描述,判断其真或伪,并在答案纸上适当空格中打勾(√)。 Description描述 I. With ABA treatment, Ca2+ is moved from outside of the guard cell into the cell interior. ABA处理后,Ca2+从保卫细胞外送至细胞内 II. With ABA treatment, the concentration of K+ is increased in the cytoplasm of guard cells. ABA处理后,保卫细胞内K+浓度上升 III. The K+ channel (I) is outward, and the K+ channel (II) is inward. K+ 信道 (I)是向外,K+ 通道 (II)是向内

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IBO2010 KOREA THEORETICAL TEST Part B

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IBO2010 KOREA THEORETICAL TEST Part B

B15. (3 points) The chloroplast, a plant organelle, originated from ancestors of the cyanobacteria; however, many proteins in the chloroplast are encoded from genes in the nuclear genome. 植物的叶绿体源自蓝绿菌;然而叶绿体内的许多蛋白质是由细胞核基因所编译的。

B15.1. (1.2 points) For each property of chloroplast DNA, indicate a checkmark (√) in the answer sheet whether the property is similar to that of prokaryote or eukaryote genomic DNA. 以下叶绿体DNA特性与哪类生物基因组DNA特性相似?在答案纸上原核生物或真核生物之 适当空格中打勾(√)。 Property特性 I. The DNA is a circular double strand. II. Introns are found. 可以找到内含子 III. Component of 70S ribosome is encoded. 70S 核糖体的组成可被编译出来 DNA为环状双股

IV. Usually, polycistronic mRNA is transcribed. 多顺反子的mRNA通常可被转录

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IBO2010 KOREA THEORETICAL TEST Part B

B15.2. (1.8 points) Protein X, a thylakoid lumen protein, is transcribed in the nucleus and translated in the cytoplasm. Next, the protein is translocated into the stroma of the chloroplast by signal peptide I. In the stroma, signal peptide I is cleaved, and the remaining protein is targeted to the thylakoid lumen by signal peptide II. In the thylakoid lumen, signal peptide II is cleaved, and the remaining polypeptide III is usually observed. 一种类囊体腔室蛋白X 在细胞核被转录,并在 细胞质中转译。然后此蛋白被讯息胜肽 I 转 送到叶绿体的基质中。在基质中讯息胜肽 I 会被切截,剩下的蛋白质会被讯息胜肽 II 标记 至类囊体腔室中 。 在类囊体腔室中讯息胜肽 II 会被切截 , 而剩下最后所被看到的多肽链 III 。

Several recombinant vectors of protein X are transformed into the nuclear genome and expressed. For each recombinant vector, fill the blanks in the 2nd column with the cellular location (A~D) where the expressed proteins are mainly observed. Fill the blanks in the 3rd column with the polypeptides (E~H) observed in that location. 蛋白质 X 的许多重组载体被转型至细胞核基因组,并表现之。对每个重组载体,在第二栏 中填入此蛋白质在细胞中主要的表现位置(如下所列A~D);在第三栏中填入在此位置可观察 到的多肽链(如下所列E~H)。 < Cellular location of expressed proteins > 蛋白质表现的细胞位置 A. Cytoplasm 细胞质 B. Stroma 基质 C. Thylakoid membrane 类囊体膜

D. Thylakoid lumen 类囊体腔室 < Observed polypeptides >观察到的多肽链 E. I-II-III F. I-III G. II-III H. III

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IBO2010 KOREA THEORETICAL TEST Part B

B16. (1.5 points) Figure a shows organogenesis of plant calluses incubated on media containing different concentrations of IAA (an auxin) and kinetin (a cytokinin). In nature, Agrobacterium, a soil bacterium, induces crown gall tumors on the roots of legume plants. The bacterium induces these tumors by integrating its T-DNA into the plant genome and by expressing a group of genes necessary for gall formation (Figure b). 图 a 显示植物愈伤组织在不同浓度的 IAA(植物生长素)及激动素(细胞分裂素)组成的培养基中培养 后,所特化出之器官。自然情况下,农杆菌诱导豆类植物的根形成冠状瘿,此细菌藉由将其 T-DNA 加入植物基因组,并表现一群与诱导产生瘤的相关基因如图 b 所示。

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IBO2010 KOREA THEORETICAL TEST Part B

If an infecting Agrobacterium lacks or over-expresses the auxin-biosynthetic genes or cytokinin-biosynthetic genes, determine the most expected callus phenotype (A~D) for mutations (I, II, and III) described in the table below. Indicate with a checkmark (√) in the appropriate box in the answer sheet. 若受农杆菌感染者 缺乏或过度表现 植物生长素合成基因或细胞分裂素合成基因,其愈伤组织 最有可能特化出的表现型(A~D) 和各突变(I, II, and III) 间的关系如下表所描述。在答案纸上适 当空格中打勾(√)。

< Expected callus phenotypes > 愈伤组织最有可能特化出的表现型 A. Shooty callus 茎 C. Undifferentiated callus 未特化 B. Rooty callus 根 D. Propagation-deficient callus 增殖不全

Gene mutation 基因突变 I. Deletion of iaaH, overexpression of ipt. iaaH 缺失 ipt 过度表现

II. Overexpression of iaaH, deletion of ipt. iaaH 过度表现 III. Deletion of iaaH and ipt. iaaH 及 ipt 缺失 ipt 缺失

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IBO2010 KOREA THEORETICAL TEST Part B

B17. (2.4 points) Plant root cell type is determined by the division and differentiation of a particular stem cell (meristematic cell). Figure a shows the whole microscopic structure of a longitudinallysectioned Arabidopsis primary root. Figure b is an enlarged diagram corresponding to a region of the inset in Figure a, showing the arrangement of root primordia (stem cells).

植物根细胞型态是藉由特定干细胞(分生能力旺盛的细胞)之分裂与分化来决定,图 a 是阿拉伯 芥主根纵切面的显微构造。图 b 是图 a 小方格对应区域的放大图,显示根的始原细胞(干细 胞)的排列情形。

Fill in the table to best match the listed function with the correct root cell type (1~6 in Figure a) and with the corresponding initial cell (7~11) in Figure b. 在表格中填入最适当对应者--将功能 的 7~11)。 对应正确根细胞类型(图 a 的 1~6);及对应始原细胞(图 b

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IBO2010 KOREA THEORETICAL TEST Part B

B18. (1.5 points) The figures below show the inner structures of pine and persimmon seeds. 下图为松树及柿子的种子构造

Indicate with a checkmark (√) whether the following statements are true or false. 指出下列叙述的真或伪,并在适当空格中打勾(√)。

I. Structures a and b are the same in ploidy, but they differ in genetic composition. 构造a, b 之染色体倍数相同,但其基因组成不同 II. Structures a, b, and c consist of two different sporophytic structures and one gametophytic structure. 构造a, b, and c包含两种不同孢子体构造及一种配子体构造 III. Structures x and y are the same in both ploidy and genetic composition. 构造x and y在染色体倍数及基因组成均相同 IV. Structure z is three-times higher in ploidy than structure c. 构造z的染色体倍数是构造 c 的3倍 V. Structures a and x are both surrounded by the ovary. 构造a and x均被子房包被住

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IBO2010 KOREA THEORETICAL TEST Part B

ANIMAL ANATOMY AND PHYSIOLOGY

B19. (1.8 points) Human blood can be separated into three parts using a table top centrifuge, as shown in the following figure. (1.8 分)用桌上型离心机可将人的血液分离成 3 个部份,如下图所示。

Of these blood parts (a~c), select the part that contributes most to the listed functions of blood. Answer by placing a checkmark (√) in the appropriate box in the answer sheet. 由此三部份的血液(a~c),勾选(√)与下方所列与其提供的功能相当者,填入答案卷上。 Function I. Antibody production. 生产抗体功能。 II. Transport of carbon dioxide. 运输二氧化碳。 III. Transport of iron. 运输铁。 IV. Transport of oxygen. 运输氧。 V. Blood clotting. 血液凝固。 VI. Neutralizing snake venom. 中和蛇毒。 功能

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IBO2010 KOREA THEORETICAL TEST Part B

B20. (2.2 points) The picture depicts the adult human skeleton and the table lists different types of joints.

本题不作答
B20.1. (1.2 points) Choose the type of each joint by placing a checkmark (√) in the appropriate box in the answer sheet. B20.2. (1 point) For each statement concerning the function of joints and bones, indicate with a

checkmark (√) whether the statement is true or false. Function I. The joint between the skull and the first cervical vertebra enables the rotation of the head. II. The fibula, as well as the tibia, plays an important role in supporting the body weight.
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B21. (2.4 points) Chordates are distinguished from other animals by 4 distinctive key morphological characters. (2.4 分)脊索动物与其他动物不同于 4 种关键的型态特征。 B21.1. (1.2 points) Choose the 4 key morphological characters from the following list and write their member in the left-hand column of the table in the answer sheet. (1.2 分)由下表中选择此 4 种关键的型态构造,填在答案卷上所提供的左表列中。 Morphological character 1. Cirri, 须 2. Brain, 脑 型态特征 4. Gills, 鳃

3. Pharyngeal slits, 咽裂

5. Notochord, 脊索 8. Anus, 肛门

6. Intestine, 小肠 9. Tail. 尾

7. Tubular dorsal nerve cord, 管状背神经索

B21.2. (1.2 points) The morphological characters of a lancelet (Branchiostoma) are shown in the illustration below. Find each of the morphological characters that you listed in the table (from B21.1) - write the corresponding letter code in the right-hand column of the table in the answer sheet. (1.2 分)下图为文昌鱼的型态,由 B21.1 表中所列的特征选出对应者,将其图中所列之 英文代号填入答案卷上右表列中。

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IBO2010 KOREA THEORETICAL TEST Part B

B22. (2 points) The graph below depicts the pressure changes in an aorta, left ventricle, and left atrium that occur concurrently during the mammalian cardiac cycle. Below the graph are sketches of the heart illustrating blood flow and valve state (opened/closed). (2 分)下图显示在哺乳动物心搏周期间,主动脉左心室及右心房血压的同步变化。图下方为 显示当时心脏的血流方向及办膜(开/闭)的状态之简图。

Match each numbered event in the cardiac cycle graph with the letter of its corresponding heart sketch. Write the corresponding letter code in the right-hand column of the table in the answer sheet. 请将心搏周期中各数字所相对应下图(A~E)之状况, 将答案填写于答案卷上所提供的表内右列中。

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IBO2010 KOREA THEORETICAL TEST Part B

B23. (1.5 points)

Fig. I shows the relationship between weight and the specific metabolic rate of the

indicated animal species, and Fig. II shows the O 2 consumption rate of the indicated species as a function of running speed (on a treadmill machine). (1.5 分)图 I 显示所列动物之体重与代谢率的关系,图 II 显示所列动物在奔跑时(在跑步机 上的结果)的耗氧率。

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IBO2010 KOREA THEORETICAL TEST Part B

Read each of the following explanations, and indicate with a checkmark (√) in the answer sheet whether the explanation is true of false. 阅读下列解释,勾选(√)其是否正确或错误于答案上。 Explanation 解释

A. At rest, smaller animals consume more energy per weight than the bigger animals consume. 在休息时,小型动物单位体重的代谢率高于大型动物。 B. Using the same amount of food per body weight, a smaller animal can travel a longer distance than a bigger animal can travel. 以单位体重所得之相同食物量来看,小型动物比大型动物走得远。 C. Using the same amount of food, bigger animals generate more ATP than the smaller ones generate. 使用相同量的食物,大型动物产生的 ATP 较小型动物多。

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IBO2010 KOREA THEORETICAL TEST Part B

B24. (1.8 points) If an astronaut lived on a heavier and larger planet than Earth, he would experience stronger gravitational forces. In that case, what would you expect to happen in this astronaut?s body? For each symptom listed below, indicate with a checkmark (√) whether the symptom is whether the symptom is expected or inexpected. (Assume that the composition of the atmosphere of the planet is the same as that of Earth.) (1.8 分)若航天员住在一个比地球更大更重的星球,他会感受到更大的地心引力,在此状况 下,航天员的身体会发生什么变化?由下列表中列出的症状,指出哪些是被预料到的,哪些不 是预期的(假设该星球之大气组成与地球相同。)

Symptom

症状

A. Increase in blood pressure. 血压增加 B. Decrease in the respiration rate. 呼吸率降低 C. Increase in muscle mass. 肌肉量增加 D. Increase in bone density. 骨骼密度增加

E. Decrease in the number of red blood cell. 红血球数量减少 F. Increase in oxygen content in the blood. 血中含氧量增加

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IBO2010 KOREA THEORETICAL TEST Part B

B25. (1.5 points) The following dissection figure shows the blood vessels in liver tissue. The three main blood vessels are indicated by capital letters (A~C). (1.5 分)下图为血管在肝组织中的解剖图,A~C 为其内三条主要的血管。

Following statements describes properties of blood that flows through particular blood vessels. For each description, indicate with a checkmark ( √ ) in the appropriate box with matching vessel where that blood would be found. 下列叙述系血液在不同血管中流动时的特性。根据每个叙述勾选(√) 出血液与血管的关联。 Description I. Blood with the highest oxygen content. 血液内含最高的含氧量。 II. The blood shows the first increase in lipid content after the meal. 在用餐后最先显示其脂肪含量增加血液。 III. The blood shows the first increase in glucose content after the meal. 在用餐后最先显示其血液葡萄糖含量增加。 叙述

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IBO2010 KOREA THEORETICAL TEST Part B

B26. (3 points) A Korean professor, Charlie Shin, was bilingual, such that he is fluent in Korean and English. He was also good at communicating using sign language. Unfortunately, he had a stroke while taking part in discussions at the 2010 IBO International Jury meeting. Dr. Oliver diagnosed that Charlie had damage in his left cerebral cortex which controls some part of his language output area and whole arm areas. (3 分)查理辛是一位韩国教授,能说韩语及英语的双语学者。他也懂得用手语来沟通。很不 幸地,他在 2010 IBO 国际审查委员讨论时中风。奥利佛医师诊断出查理的大脑皮质左侧受损。 此部份系控制(语言输出/口说)及整个手臂的活动。

B26.1. (1 point) A novice nurse examined Charlie?s language ability. Select a correct diagnosis among below. (1 分)一位新手护士检查查理的语言能力,选择下列何者为正确的诊断。 A. Charlie had difficulty in understanding Dr. Oliver ?s talk. 查理对奥利佛医师的谈话产生了解上的困难。 B. Charlie had difficulty in understandings of the 2010 IBO theoretical questions written on a paper. 查理对写在纸上有关 2010 IBO 理论题有了解上的困难。 C. Charlie had a hard time to understand a word “LOVE” written on his back by Dr. YT Kim. 查理对金外踢伯市在他背上写“LOVE”这个自有了解上的困难。 D. Charlie?s ability to speak Korean fluently had disappeared. 查理能流利说韩语的能力消失了。 E. Charlie?s ability to write Korean poems with his right hand remained intact. 查理用他的右手来写韩语的能力仍被完整的保持。

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IBO2010 KOREA THEORETICAL TEST Part B

B26.2. (1 point) The ability of Charlie?s sign language and the movement of upper extremity were also carefully examined by Dr. Oliver. The results showed that he was also incapable of proper execution of sign language expression in either way and of moving his right arm. What can we conclude from this? (1 分)查理的手语能力及上肢的活动亦受到奥利佛医师细心的诊断,结果显示他不能有 效地用任一手来执行手语,右手亦不能移动。根据这些观察,我们可下何种结论? A. The damaged language area is responsible for both sign as well as spoken language. 脑部受损的语言区掌控手语及口语。 B. Motor neurons in the right cerebral cortex govern the muscles of the right side. 在右脑皮质部的运动神经先掌控右侧的肌肉。 C. The language comprehension region is located in the right hemisphere. 语言理解区是位在右脑半球。 D. His visual system is also damaged. 他的视觉系统也受到损害。 E. His sign language expression with left arm is also abnormal. 他的左手对手语的表示仍是正常的。

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IBO2010 KOREA THEORETICAL TEST Part B

B26.3. (1 point) A brain-machine interface (BMI) study using a monkey was reported in the Science journal. An array of micro-wire recording electrodes was implanted in associative, arm movement planning area in the frontal cortex of a normal monkey. During upper arm movements electromyogram (EMG), recordings were taken from upper extremities, and at the same time neural recordings were made from implanted recording electrodes in the frontal cortex. Correlations between EMG and neural signals were obtained every 200 msec and used as commands for robot arm movement. The monkey intentionally controlled the robot arm with almost 100% success rate, without using arm muscles. Evaluate whether the following would be true or false if this BMI technology is used for human. (1 分)科学期刊上有篇利用猴子来建立脑与机械链接接口的研究,该研究中科学家将一 微电极值入正常猴子大脑皮质前叶中的联合区与手臂运动意图区。当上肢在运动时,肌电图 (EMG)会被记录下来,同时脑中的神经元活动状况也会被所植入的电极记录下来。EMG 和 神经讯号间的相关性会以每 200 msec(毫秒)收讯一次,并作为机械手臂动作上指令。这样 一来,猴子就可以在不运动手臂的状况下,用意识控制机械手臂的移动。如将此 BMI 科技 用于人类时,在下列情况何者正确,何者 错误。 Description 叙述

I Immunological reaction is one of obstacles to overcome for future development of a prosthetic device for patients such as Charlie. 生物的免疫反应是未来发展给如查理的病人等使用时的人工义肢时等所需克服的障碍。 II For accurate decoding of motor planning information, the number of simultaneously recorded neurons should be increased. 为能正确地解读运动意图的信息,进行同步记录的神经元细胞数量应该增加。 III It is more difficult to design prosthetic robot fingers than a robot arm using this kind of BMI technology. 用此种(BMI)科技,要发展出人工机械手指要比人工机械手臂更困难。 IV This BMI technology is applicable to overcome Charlie?s language disability by decoding motor production information. 此 BMI 科技可用于解决查理的语言障碍问题,即将运动产出的信念加以解读后使用。 V The described BMI technology can be classified as a motor (output) BMI, while artificial cochlea can be classified as an sensory (input) BMI. 上述之 BMI 科技可被归为是一种运动(输出)型之 BMI,而人工耳蜗(电子耳)则可 视为是一种感觉(输入)型之 BMI。

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IBO2010 KOREA THEORETICAL TEST Part B

B27. (3 points) A spinal nerve has four different kinds of axons carrying out physiological functions like muscle contractions and cutaneous sensory, thermal and pain sensations. Myelinated, large-diameter axons carry motor information, while unmyelinated, small-diameter axons carry pain information. An electrophysiological experiment was carried out using an isolated rat spinal nerve. Four different intensities of electrical stimulation were delivered to the nerve. Since the stimulation caused simultaneous activation of all axons in the nerve, including both small and large diameter axons, we observed different peaks (a to d) in the compound action potential (CAP) traces on an oscilloscope. The averaged post-stimulus time delays of these CAP peaks were: a, 2 ms; b, 2.5 ms; c, 12 ms; and d, 55 ms. The length of the spinal nerve was 10 cm. (3 分)一条脊髓神经有 4 种不同种类的轴突来执行不同的生理功能,如:肌肉收缩及皮肤的 感觉、温觉和痛觉。具髓鞘粗径的轴突传递运动信息,而不具髓鞘细径的轴突传递痛觉信息。 今有一个生理实验用分离出的老鼠脊髓神经来进行,以 4 种不同程度的电讯号刺激此神经上, 由于刺激使此神经上所有的轴突同时被活化,包括细径及粗径轴突。我们可以在示波器上观察 由复合的动作电位所形成的不同的高峰(a to d)系这些 CAP 波峰的平均后刺激时间延迟为 a, 2 ms; b, 2.5 ms; c, 12 ms; and d, 55 ms。此一脊髓神经的长度为 10 cm。

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IBO2010 KOREA THEORETICAL TEST Part B

B27.1. (1 point) Calculate the conduction velocity (m/sec) of the CAP peak a. (1 分)计算此 CAP 波峰 a 的传导速度(米/秒) 。

B27.2. (1 point) After the middle part of the nerve is exposed to a local anesthetic that blocks Na+ channels, which of the following is expected to occur? (1 分)在此神经的中间部份经由会将Na+的通道阻隔的局部麻醉后则下列何者会发生? A. The height of all CAP peaks is reduced. CAP 的各种波峰皆会降低。 B. The post-stimulus time delays of all CAP peaks are shortened. 此种所有 CAP 波峰的后刺激时间延迟皆会变短。 C. Peaks are reduced and delays are shortened selectively in CAP peaks c and d. 在 CAP 波峰 c 及 d 中,其波峰会降低且延迟会变短。 D. Peaks are reduced and delays are shortened selectively in CAP peaks a and c. 在 CAP 波峰 a 及 c 中,其波峰会降低且延迟会变短。 E. Peaks are reduced and delays are shortened selectively in CAP peaks b and c. 在 CAP 波峰 b 及 c 中,其波峰会降低且延迟会变短。

B27.3. (0.5 point) Which CAP peak is response to painful stimulation? (0.5 分)哪一个 CAP 波峰是与痛的刺激有关?

B27.4. (0.5 point) Which CAP peak is responsible for muscle contractions? (0.5分)哪一个CAP波峰是与肌肉收缩有关?

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IBO2010 KOREA THEORETICAL TEST Part B

B28. (2.7 points) The figures below present the skeletal structures of Tetrapod anterior limbs. In the figure, (a) corresponds to an early amphibian limb. The numbers and letter codes with each limb represent different bones as indicated in the legend under the figure. (2.7分)下图为四足动物的前肢骨骼构造。在图中(a)相当于一只早期的两生类的肢,图中的 数字及英文字母代号表示在每一肢上的不同骨骼,其简写说明如下表中。

H: Humerus, 肱骨

U: Ulna, 尺骨

R: Radius, 桡骨

C: Carpals, 腕骨 1~5: Phalanges.趾骨

M: Metacarpals, 掌骨

S: Sesamoid bone, 种子骨

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IBO2010 KOREA THEORETICAL TEST Part B

B28.1. (1.8 points) Among the following statements, decide which statements are true or false? (1.8分)下列的叙述,何者 正确,何者 错误? I. (c) and (e) show loss or fusion in the skeletons as compared to the ancestral condition. 与祖先的情况比较,(c) 及 (e)显示骨骼有消失或愈合之现象。 (b) and (g) show adaptation for life in the ocean. (b) 及 (g)显示在海洋中生活的适应。 (b) and (d) shows convergent evolution of the skeleton. (b) 及 (d)显示骨骼的趋同演化。 (i) shows adaptation for grasping. (i)显示对抓取的适应。 The sesamoid bones in (f) and (g) are evolutionary reversals. 在(f) 及 (g)的(种子)骨在演化上是相反的结果。 The figures show homologous characteristics of Tetrapod anterior limbs. 此图显示四足动物前肢的同源特征。

II.

III.

IV.

V.

VI.

B28.2. (0.9 point) Which of the anterior limbs in the figure above show adaptation for flight or no adaptation for flight? Indicate with checkmarks (√) in the appropriate box the answer sheet. (0.9分)在上图哪一个前肢显示其适应飞行或不适飞行,请将答案勾选(√)于答案卷上。

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IBO2010 KOREA THEORETICAL TEST Part B

ETHOLOGY

B29. (3 points) In matriphagy, a spider female is cannibalized by her offspring who attack and eat her body, when they reach a specific age. The young then live in a group for a short time period and disperse from the nest individually after the third molt. However, some mothers avoid matriphagy. If a mother is not eaten by the first clutch, there is a 30% probability that she will be able to produce a second clutch. The table presents demographic data for this species. (3分)在以母亲为食物的蜘蛛中,子代杀死母亲并吃牠的身体,直到一定的年龄为止。幼体 会成群在一起度过短暂的时间,而后在个体经历第三次蜕皮后个别由巢区向外扩散。然而有些 母蜘蛛会避免被其子女捕食,若母蜘蛛在生产第一窝时不被子代吃掉,则有 30%的机会繁殖第 二窝子代。下表显示此种蜘蛛的族群增长的资料。 Clutch size at emergence 出生时窝巢的 子代数 1st clutch with matriphagy 第一窝有食用 母亲者 1st clutch without matriphagy 第一窝未食用 母亲者 2nd clutch with matriphagy 第二窝有食用 母亲者 40 95% 3.5 mg 20% 100 70% 2.0 mg 10% 100 95% 3.5 mg 20% Survival rate at the 3rd molting 小蜘蛛在达第 三次蜕皮时之 存活率 Body mass at dispersal 子代开始扩散 时的体重 Survival rate from emergence until adulthood 出生到成蛛时 的存活率

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IBO2010 KOREA THEORETICAL TEST Part B

B29.1. (1 point) If spiders avoid matriphagy and attempt to produce a second clutch, what is the total clutch size, on average, that these spiders would produce? (1分)若母蜘蛛躲过子女的攻击而进入第二窝繁殖,则此类母蜘蛛平均所能产下的总窝 卵(子代)数为何?

B29.2. (1 point) Calculate and write down the reproductive success of the two strategies in which a female spider (1分)计算并写出母蜘蛛在此二种不同的策略下,其繁殖成功的结果。 (i) produces only a single clutch and is cannibalized, or 在产下第一窝子代后即被子女吃掉,或 (ii) avoids being eaten and attempts to produce a second clutch? 避免被第一窝子代吃掉又尝试繁殖第二窝? (Reproductive success refers to the mean number of reproductively viable offspring on individual produces.) (繁殖成功系指母蜘蛛所产生之所有子代能达到成体,即可以繁殖的年龄的平均数量。)

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IBO2010 KOREA THEORETICAL TEST Part B

B29.3. (1 point) From an evolutionary perspective and given the constraints above, which behavior would be selected for? (1分) 由上例所显示的限制条件,试由演化的观点来看,哪一种行为会被天择保存下来? A. The female does not allow matriphagy because the behavior decreases her survivorship. 雌体不让子代把牠吃掉,因为此种行为降低牠的存活机会。 B. The female leaves the nest before emergence of the young from the egg sac. 雌体在子代从卵块孵化前即离开窝巢。 C. The female is eaten by its second clutch after leaving the first clutch just before matriphagy. 雌体在第一窝子代准备吃牠前离开窝巢被第二窝子代吃掉。 D. The female is eaten by its first clutch. 雌体被牠第一窝的子代吃掉。 E. The female does not produce the offspring which cannibalize the mother. 雌体不生产会吃掉母亲的子代。

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IBO2010 KOREA THEORETICAL TEST Part B

B30. (2.6 points) Honeybee workers (Apis species) perform dances to transmit information about the distance and direction of the food source. (2.6分)蜜蜂工蜂(蜜蜂属)会表演舞蹈来传递有关食物资源离巢的距离及方向的信息。

B30.1. (1 point) What is the primary sensory mechanism involved in this communication between colony members in the nest? (1分)在巢中的蜜蜂族群中,下列何者为最主要的沟通机制?

A. Acoustic 听觉 B. Gustatory 味 觉 C. Olfactory 嗅 觉 D. Tactile 触觉 E. Visual 视觉

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IBO2010 KOREA THEORETICAL TEST Part B

B30.2. (1.6 points) The previous figure below shows the location of 8 food sources (1~8) relative to the hive. The next figure shows a waggle dance pattern for food source 1. The dotted line indicates the direction of gravity. (1.6分)下图显示蜂巢附近有8处食物资源(1~8),再下图显示蜜蜂针对食物资源1.所表演 摇摆舞蹈。虚线代表地心引力(重力)的方向。

Match each food source direction with its corresponding waggle dance pattern in the following figures. 用下列4种不同的摇摆舞型来表明其蜂巢与食物资源所在方向的关联。

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IBO2010 KOREA THEORETICAL TEST Part B

B31. (1.5 points) Over a number of generations, two strains of rats were selected in a normal environment for their increased or decreased maze-learning ability: ?Maze-bright? rats vs. 'Maze-dull? rats. For the experimental test, rats from each strain were reared in three environments that differed in the amount of visual stimuli present: restricted, normal, and enriched. The graph below shows the behavioral performance of adults in terms of the number of errors committed in running a maze for the maze-bright and maze-dull rats. (1.5分)用老鼠来进行探索迷宫能力的实验,经过好几代的繁殖筛选后:产生两种品系,对 走迷宫聪明的老鼠及对走迷宫笨拙的老鼠。在所要进行的实验中,两个不同品系的老鼠分别养 在3种不同的环境中,提供3种不同程度的视觉刺激:受限的、正常的及丰富的。下图显示之品 系(聪明及笨拙)成体在走迷宫时行为(犯错次数)的表现。

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IBO2010 KOREA THEORETICAL TEST Part B

Mark whether each of the conclusions below is true or false by putting a checkmark (√) in the appropriate box in the answer sheet. 在答案卷上勾选(√)下列的结论何者为正确,何者为错误。 Conclusion I. 结论

This experiment proves that selection for a behavioral trait leads to genetic differences between strains. 此一实验证明藉由行为上表现的筛选可导致其品系间遗传上的差异。

II.

If the two strains of rats are raised in a normal environment, the two strains make a similar number of errors. 若此二品系的老鼠养在正常的环境中 , 则此二品系的老鼠走迷宫时会犯同样程度的错误。

III.

This experiment shows that exposure to visual cues during early development influences behavioral performance in adult rats. 此一实验显示在老鼠早期发育过程中给予视觉上的刺激会影响其在成体时行为的表现。

IV.

The threshold amount of visual stimuli that markedly improves adult behavioral performance is different for maze-dull and maze-bright rats. 改善成体行为表现的视觉刺激量在两品系的老鼠(聪明及笨拙)中是有差异的。

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IBO2010 KOREA THEORETICAL TEST Part B

GENETICS AND EVOLUTION 遗传与演化

B32. (2 points) The fruit fly Drosophila melanogaster has a XX -XY system of sex determination. The Y chromosome determines maleness in humans, but not in Drosophila. Instead, sex determination in Drosophila depends on the ratio of the number of X chromosomes to the number of autosomal haploid sets in an individual fly. 果蝇是以 XX-XY 系统来决定性别。Y 染色体在人类可决定雄性,但在果蝇则不能;果蝇是以 X 染色体数目与单套体染体套数的比例,来决定性别。 The table below describes five mutants whose sex-chromosome complements and haploid sets of autosomes differ from the normal condition. 下表记载五个果蝇突变株,分别具有与正常不同的性染色体组成及单套体染色体套数

Sex-chromosome complement 性染色体组成 A B C D E X XXY XXX XXXY XX

Haploid sets of autosomes 单套体染色体套数 2 2 3 3 4

Indicate with a checkmark (√) the sex phenotype of all the mutant flies.

判断五种突变果蝇的性别,在答案卷的正确位置打勾 (√)

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IBO2010 KOREA THEORETICAL TEST Part B

B33. (2.4 points) The following statements concern evolutionary patterns of animal morphological traits. Mark whether each statement is true or false by putting a checkmark (√) in the appropriate box on the answer sheet.

以下叙述与动物形态特征的演化模式有关。判断下列六种陈述的对错,在答案卷的 正
确 及 错误 的适当格子中打勾(√)。

Statements

叙述

I. Evolution is invariably a phenomenon with direction; therefore, morphological complexities evolved from simplicities. 演化是具方向性的不可变现象;因此,在形态上皆由简单到复杂 II. Genetic mutations always lead to morphological changes. 遗传突变都会导致型态学上的变化 III. Increases in animal body size are not universal within evolutionary lineages. 动物体型的增大,不是演化系谱中的通例 IV. Morphological changes of individuals do not result from allometric growth, the differential growth of body parts. 个体形态学上的改变不是由体型的变异生长造成,变异生长是指身体各部不等速生 长的意思 V. Chordate species are more similar in the embryonic stages rather than in the adult stages. 脊索动物于胚胎期时的种间相似度,多半大于成体期 VI. Phylogenetic analyses have revealed trends of morphological evolution in some lineages. 系统发生学上的分析可显示出某些族系中形态演化的趋势

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IBO2010 KOREA THEORETICAL TEST Part B

B34. (3 points) The following tables present results of plant crosses involving three linked genes: F is a flower-color gene, S is a seed-color gene, and L is a plant-height gene. Each gene has two alleles with one allele exhibiting complete dominance over the other allele. Dominant phenotypes are red flowers, yellow seeds, and tall plants; recessive phenotypes are white, green, and short, respectively. Assume that crossing-over between two genes occurs once.

下列各表显示包含三个联锁基因的植物杂交结果。F 是花色基因,S 是种皮颜色基因,L 是 植株高度基因。每个基因的两个对偶基因中一个对另一个是完全显性。显性表型分别为红 花、黄种皮及高茎,隐性表型为白花、绿种皮及矮茎。假设两基因间发生一次互换。

Parents 亲代 F1 phenotypes F1 表型 Frequency of F 1 F1 频率

Red flower / Yellow seed (FfSs) X White flower / Green seed (ffss) 红花 / 黄种皮 (Ff / Ss) X 白花 / 绿种皮 (ff / ss) Red flower / Yellow seed White flower / Green seed Red flower / Green seed White flower / Yellow seed

红花 / 黄种皮 白花 / 绿种皮 0.49 0.49

红花 / 绿种皮 白花 / 黄种皮 0.01 0.01

Parents 亲代 F1 phenotypes F1 表型 Frequency of F 1 F1 频率

Tall height / Yellow seed (LlSs) : self fertilization 高茎 / 黄种皮 (LlSs) 自体受精 Tall height / Yellow seed 高茎 / 黄种皮 0.51 Tall height / Green seed 高茎 / 绿种皮 0.24 Short height / Yellow seed 矮茎 / 黄种皮 0.24 Short height / Green seed 矮茎 / 绿种皮 0.01

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IBO2010 KOREA THEORETICAL TEST Part B

B34.1. (0.9 point) Indicate with a checkmark (√) in the answer sheet whether each description is true of false.

判断下列描述的 对 或 错,在答案卷的适当格子中打勾(√)。
Description 描述 I. S is closer to L than to F. S 距离 L 比距离 F 近 II. Some of F1 plants with tall height / green seed are due to crossing-over. 部份F1 植物具有高茎 / 绿种皮是由于发生互换 III. Crossing-over occurs at prophase of meiosis I.

互换发生于第一次减数分裂的前期

B34.2. (0.8 point) How many genotypes can be observed in F1 plants having tall height/yellow seed?

在F1 植物中能表现 高茎 / 黄种皮 的基因型有多少种?

B34.3. (1.3 points) Calculate the map unit between gene L and gene S. (One map unit = distance of 1% recombination) 计算基因 L 与基因 S 间的舆图单位 ( 1 个舆图单位 = 发生 1% 重组的距离)

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IBO2010 KOREA THEORETICAL TEST Part B

B35. (2 points)

Shown below is a pedigree for the genetic trait PKU (phenylketonuria) that is caused by

a recessive mutation of the PAH gene (that encodes phenylalanine hydroxylase). Under the pedigree is the RFLP (Restriction fragment length polymorphism) pattern of each individual for the PAH gene. II-2 individual has the PKU. 下图显示一带有PKU (苯丙酮尿症)遗传特征的谱系 , PKU由一PAH基因(苯丙氨酸羟化酶之编码 基因)的隐性突变所引起。谱系图下方为各个人PAH基因的RFLP(限制酶片段长度多型性分析) 之结果。II-2 个体为PKU患者。

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IBO2010 KOREA THEORETICAL TEST Part B

B35.1. (1 point) The RFLP phenotype for individual II-2 is not given. From the gel shown below (A~D), choose all the patterns that would be a correct match for II-2. 图中没有II-2 个体的RFLP表型 , 由下面A~D的电泳胶片结果图中 , 选出最能正确配合 II-2 的电泳模式

B35.2. (1 point) The RFLP phenotype for individual II-4 is not given. From the gel shown below, determine whether each molecular phenotype (A~D) could be a possible match for II-4. 图中没有 II-4 个体的RFLP表型,由下面的电泳胶片结果图中,决定A~D的每个分子表型 是否可能与 II-4 配合

<Example>

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IBO2010 KOREA THEORETICAL TEST Part B

B36. (2 points) 105 cells of a triple-mutant yeast strain (leu

-

his

-

trp ) were spread either on minimal

-

medium or on minimal medium supplemented with various combinations of histidine, leucine, or tryptophan. The cultures were grown at either 25℃ or 37℃ for 3 days. Colony numbers in each plate were counted, and the data are listed in the following table. 将带三重突变(leu his trp )之酵母菌株的 105 个细胞,涂在基本培养基或添加有组氨酸(his)、 白氨酸(leu)、色氨酸(trp)的基本培养基上;在 25℃ 或 37℃培养 3 天后,数算每盘的菌落数目 记载于下表: Supplements added on minimal medium 加入基本培养基之添加物 25℃ 37℃ Number of colonies 菌落数目
-

None 无 His, Trp Leu, His Leu, Trp Leu, His, Trp

None 无 None 无 8 Confluent Confluent 极多

None 无 None 无 7 11 Confluent 极多

B36.1. (1 point) What kind of mutation most probably causes the his phenotype? 能引起his 的表型,最有可能发生的是何种突变? A. Conditional mutation B. Deletion mutation C. Point mutation D. Missense mutation E. Nonsense mutation 条件突变 缺失突变 点突变 错义突变 无意义突变
-

-

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IBO2010 KOREA THEORETICAL TEST Part B

B36.2. (1 point) What type of mutation most probably causes the leu phenotype? 能引起leu 的表型,最有可能发生的是何种突变? A. Conditional mutation 条件突变 B. Deletion mutation C. Point mutation D. Missense mutation E. Nonsense mutation 缺失突变 点突变 错义突变 无意义突变
-

-

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IBO2010 KOREA THEORETICAL TEST Part B

B37. (2 points) Human ABO blood type is determined by two genes (H and I). First, the H gene codes for the antigen precursor. The dominant allele (H) leads to expression of the precursor; the recessive allele (h) does not. Second, the I gene has three allele forms, IA, IB and i, and determines blood type (A, B, O or AB).

人类ABO血型由二基因(H 及 I)决定。首先,H基因为抗原先驱物质的编码区,显性对偶基因(H)
会使此先驱物表现,隐性对偶基因(h)则不会。其次、I基因以IA、 IB 、及 i三个对偶型决定血 型(A, B, O 或 AB)。

A male with blood type A and a female with blood type B marry. Each of them is heterozygous for both the H gene and the I gene. What is the probability of having a son with blood type O? Give your answer as a percentage (%) rounded to an integer (without any decimals). 某男子血型为 A 型,与一 B 型女子结婚,两人之 H 基因及 I 基因均为异型合子。他们儿子为 O 型的机率为何?答案请以整数(不要小数点)的百分比(%)表示。

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IBO2010 KOREA THEORETICAL TEST Part B

B38. (2 points) The presence of a beard on some goats is determined by the B (beard) gene, which has two alleles: beardless (B+) and bearded (Bb). The Bb allele is dominant in males but recessive in females. F1 progeny were born from a cross between a beardless male and a bearded female; F2 progeny were produced by crossing two F1 individuals.

某些山羊是否有胡子是由B基因决定,B基因有不长胡子(B+)及长胡子(Bb)两个对偶基因,Bb对偶
基因在雄羊为显性,在雌羊为隐性。将一无胡子雄羊与一有胡子雌羊交配得到F1 子代。

A beardless male (♂) × A beard female (♀) 无胡雄羊 x 有胡雌羊 ↓ F1 × F1 ↓ F2

Mark whether each statement is true or false by putting a checkmark (√) in the appropriate box on the answer sheet.

判断下列描述的 正确 或 错误,在答案卷的适当格子中打勾(√)。

Description 描述 A. F1 females have beards. F1 雌羊有胡子

B. One half of F2 progeny have beards. 在 F2 子代中一半有胡子 C. One fourth of F2 females have beards. 在 F2 子代雌羊中, 1/4 具有胡子 D. The beard gene is sex-linked. 胡子基因是性联遗传的

E. The beard gene is inherited according to Mendel's principles. 胡子基因的遗传是根据门德尔定律

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IBO2010 KOREA THEORETICAL TEST Part B

B39. (3 points) You sequence a 16 bp DNA molecule with the Sanger DNA sequencing procedure. Shown below is the high resolution electrophoretic pattern of the fragments. As you can see, the ddCTP lane was damaged.

你用 Sanger 的 DNA 定序法将一 16 bp 的 DNA 分子定序。下图显示各片段的高分辨率电泳分析 图型,但其中 ddCTP 一栏受损了。

B39 .1. (1 point) Indicate with a checkmark (√) which of the following components are required in the reaction mixture containing ddGTP? 在含有 ddGTP 的反应混合物中,下列成分是 需要 或 不需要,在答案卷适当格子中打勾(√)。

Component 成分 A. DNA polymerase B. Primer C. dATP D. dGTP DNA 聚合酶 引子 dATP dGTP

E. Template DNA to be sequenced 要定序的模板 DNA

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IBO2010 KOREA THEORETICAL TEST Part B

B39.2. (1 point) How does the absence of a 3?-OH group in ddNTPs affect DNA synthesis? 在 ddNTPs 中缺少了 3’-OH,会如何影响 DNA 的合成?

A. It promotes DNA breakage. B. It prevents the proper base pairing. C. It destabilizes the phosphodiester bond. It activates nucleases.

它会促使 DNA 断裂 它会妨碍碱基正确的配对 它使磷酸双酯键不稳定 D. 它会活化核酸酶

E. It prevents phosphodiester bond formation. 它会停止磷酸双酯键的生成

B39.3. (1 point) What would be the correct DNA sequence? 下列何者是正确的 DNA 序列? A. 5'-AGGCTACCAGAAATCC-3' B. 5'-CCTAAAGACCATCGGA-3' C. 5'-GGATTTCTGGTAGCCT-3' D. 5'-TCCGATGGTCTTTAGG-3' E. 5'-TGATGGTTTTAGG-3'

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IBO2010 KOREA THEORETICAL TEST Part B

B40. (2 points) Answer the next two questions using the genetic code table provided below. 用下面的遗传密码表来回答后面的两个问题

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B40.1. (1 point) Which of the following mutations would create new template DNA from which the shortest peptide would be translated? 下列何种突变会产生新的模本DNA,而导致生成最短的多肽链。

5’Template strand DNA Sequence 模板DNA序列 Nucleotide number核苷酸数 3’-

ATG

GCT

GGC

AAT

CAA

CTA

TAT

TAG

-3’

CGA

CCG

TTA

GTT

GAT

ATA

ATC

-5’

1

4

7

10

13

16

19

22

A. a deletion of nucleotide number 7.

第 7 核苷酸的缺失

B. a G→C transversion of nucleotide number 9. 第 9 核苷酸G→C的易位 C. a G→A transition of nucleotide number 13.
第 13 核苷酸G→A的转换

D. insertion of -GGT- after nucleotide number 5. 第 5 核苷酸后方-GGT-的插入 E. a T→A transversion at nucleotide number 18. 第 18 核苷酸T→A的易位

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IBO2010 KOREA THEORETICAL TEST Part B

B40.2. (1 point) A series of point mutations occurred in a bacterial gene, resulting in the substitution of amino acid residues in the order shown in the diagram below. 在细菌基因发生一系列点突变,造成胺基酸残迹的置换,其次序如下图所示。

Which amino acid in the diagram can have more than one option for its codon given this particular process of point mutation? 图中的哪个胺基酸,由于在此特殊的点突变过程中,其编码子会导致其可有超过一个以 上的选择? A. Gly B. Arg C. Ile D. Leu E. Lys

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IBO2010 KOREA THEORETICAL TEST Part B

B41. (2 points) Suppose you have a population of flour beetles with 1,000 individuals. Normally the beetles are a red color; however, this population is polymorphic for a mutant autosomal body color, black, designated by b/b. Red is dominant to black, so B/B and B/b genotypes are red. Assume the population is in Hardy–Weinberg equilibrium, with f(B) = p = 0.5 and f(b) = q = 0.5. 假设面粉象鼻虫族群有1000只个体,正常象鼻虫为红色;然而此族群是多型性,具有体色为 黑色标示为b/b的体染色体突变, ,红色相对于黑色是显性,故基因型B/B and B/b皆表现为红色。 假设此族群符合哈温平衡,其中B的频率f(B) = p = 0.5而b的频率f(b) = q = 0.5。

B41.1. (1 point) What would be the expected B and b allele frequencies, respectively, if 1,000 black individuals migrated into the population? (Assume that all other Hardy–Weinberg conditions were met.) 若1000只黑色个体移入此族群中,则B and b的预期频率分别为多少?(假设皆符合其他哈 温平衡状况)

B41.2. (1 point) What would be the frequencies of B and b alleles respectively, if a population bottleneck occurred and only four individuals survived: one female red heterozygote and three black males? 若族群发生瓶颈效应,且仅有4只个体存活,其中1只为雌的红色异型合子、3只为黑色的雄 个体,则B and b的频率分别为多少?

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IBO2010 KOREA THEORETICAL TEST Part B

ECOLOGY 生态学

B42. (2 points) Island biogeography theory states that the number of species on an island is determined by immigration rates of new species to the island and extinction rates of species on the island. Immigration rates to an island decline as its distance from the mainland increases, and extinction rates decrease with increasing island size. When the immigration and extinction rates on an island are equal, the number of species on the island reaches equilibrium. (2分)岛屿生物学的理论显示岛屿上的物种数受新种移入此岛的速率与岛屿上既有物种的灭 绝率所影响。新种移入岛屿的速率会随着岛屿与大陆间的距离增加而减少,而灭绝率则会随着 岛屿面积的增加而减少。当一岛上的移入率与灭绝率相等时,则岛上的物种数达到平衡状态。

Give the correct equilibrium number of species (S 1 ~S 4 ), in the answer sheet, for each of four islands with different combinations of distance (near and far) and area (small and large) as shown in the figure above. 在答案卷上写下平衡时4岛(S 1 ~S 4 )相对应各岛的距离(近及远)与面积(小集大)的关系。

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IBO2010 KOREA THEORETICAL TEST Part B

B43. (2 points) The contents of 3 soils (a, b, and c) were examined for soil pH and amounts of acidic cations (H+, Al3+) and other cations (Ca2+, Mg2+, K+, Na+). The figure below shows the results of that examination: the white and shaded portions of each column represent the amount of acidic and other cations, respectively. (Values given are in units of centimoles/kg.)
(2 分)检验 3 种土壤(a, b, and c)中pH及其内之酸性阳离子(H , Al )及其他阳离子(Ca , Mg , K , + Na )之数量。下图显示检验结果:柱状图中,白色表示酸性阳离子,有色表示其他阳离子(单位 系分摩耳/公斤)
+ 3+ 2+ 2+ +

For each description below, indicate with a checkmark (√) whether it is true or false. 下列叙述中,勾选(√)何者正确,何者错误。 Description 叙述

I. Aluminum toxicity tends to be most severe in soil a. 铝的毒性在土壤a中最严重。 II. Soil b contains the most nutrient minerals plants can use. 土壤b具有最多可被植物利用的营养盐。 III. Anions such as NO 3 - and PO 4 - tend to be retained in soil more than cations are retained. 阴离子如NO 3 - 及 PO 4 - 较阳离子似在土壤中被保存的更多。 IV. As more H+ displaces other cations, the soil becomes more acidic. 当更多的H+取代其他的阳离子,土壤变得更酸。

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IBO2010 KOREA THEORETICAL TEST Part B

B44. (2.2 points) The figure below shows standing biomass pyramids of two ecosystems, each with four tropic levels.
(2.2 分)下图显示二个生态系的生物量塔,每个生态系各有 4 个营养阶层。

B44.1. (1.2 point) Which of the following explanations are true or false? Indicate with a checkmark (√) in the answer sheet. (1.2 分)下列叙述何者正确,何者错误,请在答案卷上勾选(√) Explanation 叙述

I. Pyramid a reflects energy losses due to respiration within trophic levels and energy losses during energy transfer between trophic levels. 塔a显示能量的损失来自各阶层间的呼吸作用及能量在不同营养阶层中的传递。 II. Pyramid b represents an ecosystem with fast turnover in the primary producer level. 塔b显示在此生态系中初级生产者的转换率很快。 III. For each ecosystem, its energy pyramid is opposite to its biomass pyramid. 在每个生态戏中,其各自所有之能量塔与其生物量塔呈相反分布。 IV. For both ecosystems, production efficiency becomes higher as the trophic level increases. 在此二个生态系中,当营养阶层升高时,其生产效率亦提高。

B44.2. (1 point)

Assuming an ecological efficiency of 10% between trophic levels, how much net

primary productivity is required to harvest 2 g C/m2 annually from the tertiary consumer level? (1分)假设在不同的营养阶层间,其生态的效率为10%,今若要由第三级消费者的阶层 产生每年2 g C/m2间(每平方米2克的碳),则需有多少的初级生产量为基础?

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IBO2010 KOREA THEORETICAL TEST Part B

B45. (2.8 points) Recent changes in the mean global temperature are largely attributed to increases in levels of some atmospheric gases and aerosols(small panticheosuspended in air), many of which have been generated by human activities. (2.8 分)近年全球平均气温的改变主要来自某些大气中的气体及喷雾剂(在空气中悬浮的小
颗粒) ,其中皆是来自人的各种活动。

B45.1. (0.8 point) Evaluate whether the following statements are true or false in relation to the role of these gases and aerosols changing global temperature. (0.8分)判断下列对气体与喷雾剂与全球气温改变之关系的叙述,何者正确,何者错误。 I. These gases scatters short-wave radiation emitted from the sun. 这些气体会分散阳光中的短波辐射。 II. These gases absorb and re-radiate infrared radiation emitted from the earth?s surface. 这些气体吸收并再辐射由地表产生的红外线辐射。

III. Aerosols prevent heat convection into space. 喷雾剂阻止热对流到太空中。 IV. Regardless of the presence of gases or aerosols, solar radiation itself has increased recently. 太阳本身的辐射在最近增加,无关乎这些气体及喷雾剂的存在。

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IBO2010 KOREA THEORETICAL TEST Part B

B45.2. (2 points) For each statement below, choose from the following list of gases the one that is most likely to be related to that description.
(2 分)由下表中所提供的气体种类选取其与下列叙述最为相关者。 <List of gases> a. Hydrofluorocarbons (HFCs) 氢氟碳化物 d. N 2 b. CH 4 e. O 3 c. CO 2 f. N 2 O

Description

叙述

I.The gas largely derived from fossil fuels and clearing of forests that contributes the most to global warming. 此气体主要来自化石燃料及砍阀森林的结果是造成全球暖化的主角。 II. The gas with the highest global warming potential (compared to CO 2 ). 此气体与CO 2 比较具有造成全球暖化最大的潜能。 III. A gas that in the stratosphere is essential to support human life on earth, while in the troposphere it exerts harmful effects on humans. 一种气体在同温层中,对地球人类生命的保护相当重要,但在对流层其对人则产生 有害效果。 IV.A gas that is not thought to contribute to global warming. 一种不被认为会对地球暖化产生影响的气体。 V.A gas derived from landfills and the livestock sector that has increased most rapidly in the past 200 years. 一种气体由土地掩埋及饲养牲口所产生,在过去200年间快速增加。

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IBO2010 KOREA THEORETICAL TEST Part B

B46. (2 points) The picture shows a schematic representation of the production of three well-known trees of a deciduous forest. The production is indicated in kg dry mass per hectare per year. (2 分) 下图显示在落叶森林中 3 种为人熟知的树种概要,其生产立为每年每公顷的植物干中 (公斤) 。

Calculate how much of the total production comes from above ground woody parts. 计算地表以上所有木质部份的总产量。 Give your answer as a percentage (%) rounded to an integer (without any decimals). 答案请以百分比(%)显示,四舍五入以整数表示。

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IBO2010 KOREA THEORETICAL TEST Part B

BIOSYSTEMATICS
生物系统分类 B47. (2 points) Figures a and b show the characteristics of a cactus from the American desert and a spurge from the African desert, respectively. An evolutionary mechanism has been proposed to explain the morphological similarities between these nonrelated species. That same evolutionary mechanism has been reported to operate at the DNA sequence level. 图 a 为美洲沙漠的仙人掌,图b为非洲沙漠的大戟。某演化机制可用来解释此两种亲缘相距甚远 的物种为何具有相似形态。而其DNA序列也有相同的演化机制。

a. Cactus仙人掌

b. Spurge大戟

c. Molecular evolution model分子演化模式

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IBO2010 KOREA THEORETICAL TEST Part B

Which of the molecular evolutionary trees shown in Figure c is the best molecular model of the morphological evolutionary mechanism that we observe in the cactus and spurge? The symbols A, C, G, and T on the molecular evolutionary trees represent DNA bases. 上页的图 c 中,哪一个分子演化树是仙人掌与大戟之形态演化机制的最佳分子模式?分子演化 树上的符号A, C, G, and T代表的盐基。 A. Tree (a), 1 with 4. B. Tree (b), 1 with 2. C. Tree (c), 1 with 5. D. Tree (c), 2 with 3.

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IBO2010 KOREA THEORETICAL TEST Part B

B48. (2 points) The following figure represents a recent phylogenetic tree for the animal kingdom. Carefully observe the tree topology, and answer the following questions. 下图为动物界的亲缘演化树,仔细检视树型分枝,并回答以下问题。

B48.1. (1 point) What are the most appropriate synapomorphic characters for the numbers (1) and (2), respectively? Mark appropriate boxes with a checkmark (√). 下列何者分别是演化树中的(1) 及 (2)最适当的共衍征?在适当空格中打勾(√) A. Segmented body 身体分节

B. True tissue differentiation 真的组织分化 C. Embryogenesis D. Bilateral symmetry 胚胎发生 两侧对称 外骨骼发育

E. Exoskeleton development

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IBO2010 KOREA THEORETICAL TEST Part B

B48.2. (1 point) Which of the following groups are members of Deuterostomia (taxon number (3))? 下列哪些群是后口类成员(演化树中的分类群(3)) A. Echinodermata, Arthropoda. 棘皮动物、节肢动物 B. Echinodermata, Chordata. 棘皮动物、脊索动物 C. Mollusca, Arthropoda. 软件动物、节肢动物 D. Annelida, Mollusca. 环节动物、软件动物 E. Chordata, Mollusca. 脊索动物、软件动物

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IBO2010 KOREA THEORETICAL TEST Part B

B49. (2 points) The Influenza A virus is responsible for annual flu epidemics and for occasional flu pandemics. Influenza A has a genome composed of eight RNA strands that encode a total of 11 proteins. Influenza A strains can be classified based on the combination of two coat proteins, Hemagglutinin (H1~H13) and Neuraminidase (N1~N9). In this way, various flu types such as H1N1, H3N1, H7N2, etc. can be recognized. The virus strains also can be classified by the host animal. The following figure represents the phylogeny of flu viruses based on the nucleoprotein gene of the flu virus genome. Indicated for each viral strain is the host species from which it was isolated, the year, and the type of Hemagglutinin and Neuraminidase it carries. Indicate with checkmarks (√). whether the following statements are true or false. A型流行性感冒病毒是每年流行感冒的元凶,且偶而也会造成广泛感染。A型流行性感冒病毒基因 组 具 有 8 股 RNA 片 段 , 总 共 编 译 11 个 蛋 白 质 。 其 品 系 可 依 据 两 种 蛋 白 质 鞘 血 凝 集 素 (Hemagglutinin) (H1~H13) 及 神经胺酸酶 (Neuraminidase) (N1~N9)的组合来分类,因此可分出 多种感冒类型例如H1N1, H3N1, H7N2等。这些病毒品系也可用寄主动物来分类,下图代表根据 流 行 性 感 冒 病 毒 基 因 组 的 核 酸 蛋 白 基 因 所 得 之 亲 缘 关 系 树 , 以 及 Hemagglutinin and Neuraminidase 所带有的类型。判断下列叙述的真或伪,并在适当空格中打勾(√)。

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IBO2010 KOREA THEORETICAL TEST Part B

I.

The avian flu virus consists of the most diverse types, and some avian flu types also are found in some mammalian species such as whales and dolphins. Therefore, the avian flu virus represents the most archaic type of flu virus. 禽流感病毒包括最多样的类型,且有些类型也出现在哺乳类物种如鲸豚类上。因此禽流 感病毒代表流感病毒中最古老的类群。

II.

The phylogenetic tree suggests that the host shift and genetic recombination of flu virus have occurred between birds and pigs. 亲缘关系树显示寄主转移及流感病毒的遗传重组已发生在鸟与猪上

III.

The virulence of virus can be changed rapidly by host shifts and mutations. Therefore, vaccine developments are relatively difficult compared to other common diseases. 病毒的毒性可因寄主转移及突变而快速改变,因此其疫苗发展比其他常见疾病困难

IV.

Swine flu strains are phylogenetically more closely related to the human flu strains than to other strains. 新流感品系与人类流感品系之亲缘关系较其他品系还近

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IBO2010 KOREA THEORETICAL TEST Part B

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IBO2010 KOREA THEORETICAL TEST Part B

B50. (1.5 points) Following table summarizes the main characteristics of the four major phyla of seed plants. Check (√) in the answer sheet whether each characteristics is absent (─) or present (+) for A~E. 下列表格是种子植物的四个主要门,判断各门特征之有(+)、无(─),将答案填在答案纸上。

Character 特征 Phylum 门 Cycadophyta 苏铁门 Ginkgophyta 银杏门 Pinophyta 松门 Magnoliophyta 被子植物门

Flagellated sperm 精细胞具鞭 毛

Double fertilization 双重受精

Vessel in xylem 木质部导 管 ─

Flowers and fruits 花及果实

Development of the secondary xylem 次生木质部发育 ─

+
A

B









E







D

+ +



+

C

+

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IBO2010 KOREA THEORETICAL TEST Part B

B51. (2.4 points) All organisms use carbon as well as energy in order to live and function. Organisms can be divided into four nutrition modes based upon the species? main sources of energy and of carbon.

所有生物皆需利用碳及能量以维持生存及生理功能。生物根据物种其主要能量与碳的来源,可 分为四种营养方式。

B51.1. (1.2 points) From the following list of nutrition modes, fill in the answer sheet with the correct term corresponding to each combination of carbon and energy source. 下列为各种营养方式,在答案纸上填入对应每种碳与能量来源的组合。

<Nutrition mode> I. Photoautotroph, 光自营 学自营 III. Photoheterotroph, 光异营 学异营 IV. Chemoheterotroph. 化 II. Chemoautotroph, 化

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IBO2010 KOREA THEORETICAL TEST Part B

B51.2. (1.2 points) From the list of organisms provided, choose two organisms belonging to each nutrition mode. 从下列生物中选出两种属于每种营养方式

<Organisms> a. Cyanobacteria, 蓝绿菌 菌

生物

b. Green nonsulfur bacteria, 绿色非硫化细

c. Purple nonsulfur bacteria, 紫色非硫化细菌 e. Most archaebacteria, 大部分古细菌 植物 g. Animals, 动物

d. Fungi, 真菌 f. Most plants, 大部分

h. Nitrifying bacteria.

硝化细菌

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IBO2010 KOREA THEORETICAL TEST Part B

B52. (2 points) The following figure represents a recent phylogeny for the plant kingdom. 下图代表植物界的亲缘关系

For each number (1)~(4), select the appropriate apomorphic trait from the list provided. 由下列衍生特征中选出适当者。

< Apomorphic traits >衍生特征 A. Leaves with well-developed vascular bundles, 叶具完整维管束 B. Embryos, 胚胎

C. Seeds, 种子 D. Vascular tissues, 维管束组织 E. Phragmoplast. 细胞壁成膜体

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