2017 dse生物試題目及答案
2017 dse生物答案，名師解答生物dse past paper中文答案。包括2017 dse生物卷一、卷二答案，2017 dse bio paper 1b、paper 2 answer。
2017 dse生物答案
2017 DSE BIO Suggested Answers (by James Fong)
01. C 02. B 03. A 04. C 05. A
06. C 07. C 08. B 09. A 10. D
11. D 12. B 13. A 14. B 15. A
16. D 17. C 18. D 19. B 20. B
21. B 22. C 23. D 24. C 25. A
26. D 27. C 28. A 29. C 30. D
31. D 32. D 33. C 34. B 35. A
36. A
2017 DSE Bio Suggested Solution
(by James Fong)
[Key:- (T)︰正確；(X)︰錯誤]
1. C
C (T) Protein synthesis is a building up process (anabolic)
A (X) Digestion is a breaking down process (catabolic)
B (X) Formation of glycogen is a building up process (anabolic)
D (X) Absorption is not a metabolism
C (T) 蛋白質的合成是組成過程 (合成代謝)
A (X) 消化是分解過程 (分解代謝)
B (X) 形成糖原是組成過程 (合成代謝)
D (X) 吸收並不是代謝作用
2. B
For starch digestion, enzyme X is amylase
(1) Mouth cavity contains salivary amylase
(2) Stomach contains proteases
(3) small intestine contains amylase secreted by pancreas
在澱粉的消化中酶 X 是澱粉酶
(1) 口腔含有唾液澱粉酶
(2) 胃部含有蛋白酶
(1) 小腸含有由胰臟分泌的澱粉酶
3. A
At the beginning, there are Starch & amylase in R
(─) Benedict’s test: tests for reducing sugar (maltose)
(+) Iodine test: tests for starch
(+) Test for proteins: all enzymes are proteins (amylase)
在開始時 R 含有澱粉及澱粉酶
(─) 本立德溶液測試：測試還原糖 (麥芽糖)
(+) 碘液測試：測試澱粉
(+) 測試蛋白質：所有酶都是蛋白造的 (澱粉酶)
4. C
P (T): showing the result of iodine test if starch is present
Q (T): showing enzyme X alone cannot give positive result in Benedict's test
S (X): showing enzyme X is the causing agent for the reaction (control) rather than showing enzyme X is denatured after boiling
Water bath (T): all the tubes are under the same condition same as human body temperature (controlled variable, temperature)
P (T)︰顯示若澱粉存在，碘液測試的結果
Q (T)︰顯示若只有酶 X，本立德溶液測試沒有正反應
S (X)︰顯示酶是引致這反應的原因 (對照)，而非煮沸會將酶變性
水浴 (T)：所有試管處於與人體體溫相同下進行實驗 (控制變項，溫度)
5. A
Different molecular size means the length and the amino acid sequence of the polypeptides are different. They will have different conformation. However they should have similar shape at the active site to bind with starch molecule to form the enzyme substrate complex.
不同的分子大小，代表多肽的長度及所載的氨基酸排序有所不同，會形成不同的構象。但他們會有結構相似的活性區，便可與澱粉分子結合成酶受質複合物。
6. C
A (X) regeneration of carbon dioxide acceptor (Calvin cycle in stroma)
B (X) reduction of 3-C compound (Calvin cycle in stroma)
C (T) photolysis of water (Photochemical reaction in thylakoid space)
D (X) carbon dioxide fixation (Calvin cycle in stroma)
A (X) 二氧化碳受體的再生 (基質內︰卡爾文循環)
B (X) 三碳化合物的還原 (基質內︰卡爾文循環)
C (T) 水的光解 (類囊體腔內︰光化反應)
D (X) 二氧化碳固定 (基質內︰卡爾文循環)
7. C
A (X) regeneration of NAD (on inner membrane of mitochondrion)
B (X) production of carbon dioxide (in matrix of mitochondrion)
C (T) conversion of pyruvate to acetyl-CoA (in matrix of mitochondrion)
D (X) conversion of triose phosphate to pyruvate (in cytoplasm)
A (X) NAD的再生 (線粒體的內膜上)
B (X) 二氧化碳的產生 (線粒體的基質上內)
C (T) 轉化丙酮酸鹽成乙酰輔酶A (線粒體的基質內)
D (X) 轉化丙糖磷酸成為丙酮酸鹽 (在細胞質內)
8. B
Mother may be homozygous B (IBIB) or heterozygous (IBi). Even if the mother is heterozygous, each child has a 50% chance to be of blood group B.
母親可以是純合 B 型 (I^B I^B) 或是雜合型 (I^B i)，儘管母親是雜合型，每個孩子都有 50% 機會是血型 B。
9. A
AAC: coding strand of the DNA (code)
TTG: non-coding strand of the DNA (template strand: anticode)
AAC: mRNA (codon)
UUG: tRNA (anticodon)
[* code ~ codon; *anticode ~ anticodon]
AAC：DNA的編碼股 密碼
TTG：DNA的非編碼股 模板 反密碼
AAC：密碼子
UUG：反密碼子
[*密碼與密碼子相似；*反密碼與反密碼子相似]
10. D
The process of speciation:
(2) isolation: the ancestor is divided into two groups to stop gene flow in between
(1) mutation: produces new character (new gene)
(3) natural selection: each isolated group was subjected to a different set of environmental conditions. As a result, adaptive traits specific to those particular environmental conditions were selected by natural selection
A new species is formed until their genetic compositions were so different that they could not interbreed again
物種形成的過程
(2) 隔離：始祖會被分隔成兩組，阻止它們會有基因交流
(1) 突變：產生新特徵 (出現新的基因)
(3) 自然選擇：被隔離的始祖面對不同環境情況，結果，不同適應該獨特環境的特徵，在自然選擇下被選擇保留下來
直至它們的基因組差異過大而不能進行雜交，新物種便形成。
11. D
A (X) Q is more adaptive than P at area II.
B (X) P grows better in areas I.
C (X) Areas II and I are different in environmental conditions. So a species Q is evolved.
D (T) P and Q belong to the same Family in the classification system because Q are evolved from P.
A (X) 在區域 II，Q 會較 P 適應。
B (X) 在區域 I，P 的生長較佳。
C (X) 區域 II 與區域 I 的環境條件不同，所以進化出 Q。
D (T) 在分類系統中，因為 Q 進化自P，P 和 Q 是屬於同一科。
12. B
At the beginning, the biodiversity at I is greater than II. Therefore I is secondary succession
在開始時，I 的生物多樣性較 II 高， 所以 I 是次級演替。
13. A
Large amount of water is lost by transpiration.
有大量的水分經蒸騰流失。
14. B
The transpiration rate of a leafy shoot is measured by the length of water moved up in xylem per unit time. The leafy shoot should be cut the lower end under the red-coloured solution (3) to avoid the forming of air bubbles in the xylem.
帶葉片支條的蒸騰率，可以量度某時段內，水分在木質導管內上升的距離來計算。支條的下端需要在紅色溶液中切割 (3)，用以防止氣泡在木質導管內形成。
15. A
Both hot and bright conditions increase the rate of transpiration.
炎熱和光亮兩者都會增加蒸騰速率。
16. D
In stem, the vascular bundles is distributed at the periphery and xylem is pointed to the pith side.
在植物的莖部，維管組織會分佈於外圍， 而木質部則對內指向髓部。
17. C
(2) xylem vessels: in mid rib and veins provide rigidity for support.
(3) mesophyll cells: large proportion and provide turgidity for support.
(2) 木質導管：在中脈及葉脈提供剛度支持。
(3) 葉肉細胞：比例佔大多數，提供膨脹度支持。
18. D
A (X) Veins have a large lumen: blood flows through with less resistance
B (X) Arteries have a thick layer of elastic tissue: withstands the high blood pressure inside
C (X) The aorta has the highest blood pressure: pumping action of the heart.
A (X) 靜脈的管腔大：血液承受較少阻力
B (X) 動脈有層厚的彈性組織：抵禦管腔內的高血量壓
C (X) 大動脈內的血壓最高：心臟產生的泵壓引致
19. B
Ovum can still alive for 1 to 2 days after released in reproductive tract. It is still at the upper part of the oviduct.
卵子釋出在生殖道仍能存活一至兩天，期間仍在輸卵管的上部。
20. B
The blockage prevents the meeting of sperm and ovum.
堵塞使得精子不能與卵子相遇。
21. B
(1) (T) The cell number is increased from stages I to IV.
(2) (X) As the photomicrographs are at the same magnification, the size of the embryo developed during the early stages do not change. The active dividing cells becomes smaller and smaller and do not have time for enlargement.
(3) (T) In cell stage IV, it differentiates with outer layer of cells surrounded a fluid-filled cavity and an inner mass of cells.
(1) (T) 細胞的數目由階段 I 至IV 是不斷的增加
(2) (X) 若顯微照片以相同倍數放大，胚胎在早期發育階段的大小沒有改變，細胞只在不斷分裂下變得越來越細小，未有時間進行細胞增大。
(3) (T) 細胞階段 IV 已進行分化，形成外層細胞及內層細胞團。
22. C
2 x 2 x 2 x 2 = 16 (2^n = 16; n = 4)
23. D
The cell number at stage I to III are not enough for implantation.
Stage IV is blastocyst. The cell number is enough for implantation.
(The zygote (stage I) develops by mitosis into 16 cells morula (stage III) and then develops into blastula. It differentiates with outer layer of cells surrounded a fluid-filled cavity and an inner mass of cells. This is then known as the blastocyst. Its formation begins about 5 days after fertilization. The blastocyst is about 0.1-0.2 mm in diameter and is composed of 200-300 cells following rapid cell division. About 5–6 days after fertilization, it reaches the uterus and begins implantation. The process is finished only 11–12 days after fertilization. )
階段 I 至階段 III 的細胞量不足以進行植入。而階段四 IV 是胚泡，有足夠細胞進行植入。
(階段 I 的合子藉有絲分裂發展成為 16 個細胞階段的桑葚胚，再發展成囊胚。及後分化出外層細胞，包圍著一個液體腔及內細胞團，此乃胚泡。胚泡大約在受精後第五天開始形成，直徑約0.1至0.2 mm，經快速分裂後約有 200 至 300 個細胞。受精後第五至第六日，胚泡到達子宮並開始植入。植入的階段要在受精後第11至第12日才完成。)
24. C
I (head): brain develops fast at childhood (0-8 years old)
II (whole body): develops rapidly at both childhood and puberty
III (reproductive system): develops rapidly at puberty (12-20 years old)
I (頭部)：小童階段腦部發展較快 (0至8歲)
II (整個身體)：小童階段及青春期的發展較快
III (生殖系統)：青春期的發展較快 (12至20歲)
25. A
1 is semi-circular canal which detects head movement and sends nerve impulse to the brain (Out of syllabus)
1 是半規管，可偵測頭部運動並送出神經脈衝到腦部 (課程不用考)
26. D
Ear drum (4): converts sound wave to vibration.
Ear bones (5): amplifies and transmits vibration to oval window (2)
耳膜 (4)：轉換聲波成震動。
聽小骨 (5)：擴大及傳送震動至卵圓窗 (2)
27. C
Cause of short-sighted eye: eyeball is too long / lens is too thick.
Problem: the image of distant object is focused in front of the retina. A blurred image is formed onto the retina.
Correction: a concave lens can shift the focus point back onto the retina. A sharp image will be detected.
近視成因：眼球過長或晶體過厚。
毛病：遠物的影像聚焦在視網膜前，視網膜上偵測到的影像模糊不清。
矯正：用凹透鏡將焦點移後至視網膜 形成清晰影像。
28. A
Reading a book is a short distance vision; a more convex lens is required for focusing. Therefore, the ciliary muscles need to contract and our eyes may feel tired after a long time reading.
以近視覺長時間看書，需要凸度較高的晶體作對焦。因此睫狀肌需要長期收縮，所以感覺疲倦。
29. C
A (X) “Y”
B (X) “X to Y”
C (T) Neuromuscular junction is also a synapse.
D (X) “Neurotransmitters”
A (X) “Y”
B (X) “X to Y”
C (T) 神經肌肉接點是突觸的一種。
D (X) “神經遞質”
30. D
Eye is a sense organ and connected by peripheral nervous system (optic nerve)
眼睛是感覺器官，與外周神經元系統連接 (視神經)
31. D
P: region of cell differentiation
Q: region of cell elongation
P：細胞分化區
Q： 細胞延長區
32. D
Fungi are decomposers
真菌是分解者。
33. C
4 is absorption
4 是吸收作用。
34. B
(1) (X) S is producer (autotroph).
(3) (X) “S”
(1) (X) S 是生產者 (自養生物)。
(3) (X) “S”
35. A
All the dead bodies will be decomposed by Q
所有屍體都會被 Q 分解。
36. A
Type I: pancreas is defective; it cannot produce enough insulin upon stimulation
Type II: Their pancreas is functioning normally but their body cells are insensitive to insulin.
B (X): Blood insulin level are “low”.
C (X): remains “high” even after meal.
D (X): will “not” drop significantly
一型糖尿病：因胰臟的缺陷，受刺激下不能產生足夠胰島素。
二型糖尿病：胰臟的功能正常，但體細胞對胰島素的敏感度不足。
B (X): 血液胰島素的水平很低
C (X): 用餐後血液胰島素的水平會升高
D (X): 注射胰島素後血糖的水平不會顯著下降
