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2008年国际生物奥赛试题

THEORETICAL TEST – PART A

19th INTERNATIONAL BIOLOGY OLYMPIAD

13th – 20th July, 2008

Mumbai, INDIA

THEORETICAL TEST – PART A

理論題–第 A 部分

Write all answers in the ANSWER SHEET.

所有的答案都必須回答於答案卷上

THEORETICAL TEST – PART A

Dear Participants 親愛的參賽者

?You have a total of 120 minutes for answering Part A.

你有120 分鐘作答。

?The questions in Part A have only one correct answer. Mark the correct answer with …X? on the Answer Sheet, which is provided separately. The correct way of marking the cross is shown below. Use a dark pencil to mark your answers.

本詴題全部為單選。答題時在正確的空格內以深色鉛筆畫‘X’。可參考下圖範例。

?The answers written in the Question Paper will not be evaluated.

作答於題目紙上不計分。

?Mark your answers clearly. Avoid any corrections in the Answer Sheet.

作答應清楚,並避免汙染非答案區。

IBO – 2008

INDIA

THEORETICAL TEST – PART A

?NOTE: Some of the questions may be marked “S kipped” / “D elete d”. DO NOT attempt these questions. Also, read the question completely before

attempting it as some questions may continue from one page to the next.

注意:如有遇到跳過或刪除的題目,請勿作答。題目可能跨頁,請詳細讀完。

?The maximum number of points is 61.

總題數為 61。

?Your Answer Sheets will be collected at the end of the examination.

答案卷在考完後將會被收回。

Good Luck!! 祝好運

Country:

First name:

Middle name:

Family name:

Student Code:

THEORETICAL TEST – PART A

PART A

CELL BIOLOGY (13 points) 細胞學(13 分)

1. (1 point) The central dogma originally proposed by Francis Crick has seen

changes reflecting new insights obtained from time to time. Which one of the following schematics correctly depicts our current understanding of the

replication of genetic material and the “flow of informati on” in biological

systems?

(1 分) 中心教條(central dogma) 最早由Francis Crick 提出,下者何者能正確

的代表現今遺傳物質的複製與訊息傳達?

THEORETICAL TEST – PART A

2. (1 point) In an experiment, mice were injected intravenously with uniformly

labeled [14C] – glucose. The molecules in the body where the 14C would be found are:

(1 分) 實驗中,小鼠自靜脈內注射含有14C 標定的葡萄糖。請問14C 會在下列

何種分子中被發現:

a. essential amino acids and proteins.

必需胺基酸與蛋白質

b. lipids and all vitamins.

脂質與維他命

c. proteins and lipids.

蛋白質與脂質

d. proteins and all vitamins.

蛋白質與維他命

THEORETICAL TEST – PART A

3. (1 point) The following schematics depict the orientation of F1F O-ATPase

along with the direction of H+-transport and ATP synthesis/hydrolysis.

(1 分) 下圖中所表示的是有關F1F O-ATPase 中H+ 運輸方向與ATP 合成/ 水

解,請問下列何者正確:

Of the above schematics,

a.Only I is correct.只有I 正確

b.Only II is correct.只有II 正確

c.Only III is correct.只有III 正確

d. Both I and III are correct. I 與III 是正確

THEORETICAL TEST – PART A

4. (1 point) A given DNA sample has 60% purines. The source of this DNA is

most likely to be:

(1 分) 在一種未知的DNA 樣本中含有60% 的嘌呤,此DNA 可能源自於下列

何種生物:

a. a eukaryotic cell.

真核生物

b. a bacterial cell.

細菌

c. a bacteriophage with double-stranded DNA.

雙股DNA 噬菌體

d. a bacteriophage with single-stranded DNA.

單股DNA 噬菌體

THEORETICAL TEST – PART A

5. (1 point) The stage of cell division shown in the figure below represents:

(1 分) 此圖為細胞分裂的那一時期及染色體套數為何:

a. Meiotic metaphase I with n = 4

減數分裂中期 I ,n = 4

b. Meiotic metaphase II with n = 4

減數分裂中期 II ,n = 4

c. Meiotic metaphase II with n = 8

減數分裂中期 II ,n = 8

d. Meiotic metaphase I with n = 2

減數分裂中期 I ,n = 2

THEORETICAL TEST – PART A

6. (1 point) Polymerase Chain Reaction (PCR) is a technique for rapid

amplification of DNA segments. If you are given double-stranded DNA with appropriate forward and reverse primers as shown in the figure below, the minimum number of cycles you will require to obtain at least one copy of the desired fragment PQ, as dsDNA without overhangs, will be:

(1 分) PCR 是一種能快速倍增DNA 片段的技術。實驗中添加雙股DNA 與適

當的正、反向引子(如下圖),在沒有添加過量dsDNA 情況下,至少要有多少的PCR 週期數才可獲得一個全新的PQ 片段:

a. 1

b. 3

c.

4 d.

40

THEORETICAL TEST – PART A

7. (1 point) Which of the primer pairs is the correct one to amplify the gene

sequence below with PCR?

(1 分) 以PCR 實驗擴增下列序列時,下列引子組何者正確?

5?-GCGTTGACGGTATCAAAACGTT AT……TT TACCTGGTGGGCTGTTCTAATC-3?

a. 5?-GCGTTGACGGTATCA-3?and 5?-TGGGCTGTTCTAATC-3?

b. 5?-CGCAACTGCCATAGT-3?and 5?-TGGGCTGTTCTAATC-3?

c.

5?-GCGTTGACGGTATCA-3?and 5?-GATTAGAACAGCCCA-3? d.

5?-TGATACCGTCAACGC-3?and 5?-GATTAGAACAGCCCA-3?

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