Ecology 整合2018-2021
21 Q9 Interaction: competition
20 Q2 Trophic levels biomass
20 Q6 nitrogen cycle
19 Q7 ecological succession: secondary
19 Q8c natural selection
19 Q9 interaction: competition+food chain
18 Q10 natural selection
Ecology marking 整合
Q1 Why X is more competitive than Y?
A1 X has a higher growth rate (21Q9a,19Q9a). X occupied a better and more favourable habitat (16Q5a).
Q2 3 reasons of energy loss across trophic level.
A2 loss as heat in respiration, indigestible parts are digested, uneaten body parts (20Q2a).
Q3 Food chain relationship given top consumer decrease in population.
A3 When top consumer decrease, secondary consumer increase, primary consumer decrease, and thus producer increase (20Q2b,19Q9b).
Q4 3 roles of oxygen in nitrogen cycle.
A4 Nitrification N2 to No3-, respiration of organism, aerobic decomposition (20Q6b).
Q5 Change in soil nitrogen content during succession.
A5 High N to low N: Rapid growth of herbaceous plant; Low N to high N: growth of plants increases population of nitrogen fixing bacteria and decomposer, thus increase soil N content (19Q7bi).
Q6 Principle of natural selection
A6 4 分罐頭答法 (19Q8c, 18Q10b)
Q7 Why cut and glue back its tail in an experiment?
Q7 Ensure cutting doesn’t affect DV, so DV is due to IV only (19Q8bi).
Q8 Why an imported species become dominant species in a local habitat. (2 reasons)
A8 higher reproduction rate than local species. Lack of natural predator in local habitat (19Q9c).
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台主有白卡屬性 冇事
Experiment 整合2018-2021
21 Q9 Interaction of plants with different leaf sizes
21 Q10 In vitro digestion of breast milk using enzymes in digestive juice
20 Q5 Transpiration rate of a leaf through upper and lower epidermis
20 Q7 Killing mosquitoes using parasitism.
19 Q6 The change in mass when 3 potato cylinders are immersed in sucrose solution with different concentration
18 Q8 Digestion of lipid in milk using lipase
15 Q7 Breakdown of hydrogen peroxide by catalase in different temperature
Experiment marking 整合
Q1 4 Control variables of transpiration experiments
A1 rate of water evaporation by adding oil, surface area of the leaf epidermis, wind speed by using enclosed system, volume of air containing water vapour by using identical flasks, leaves of the same potted plant (20Q5b). Temperature (by monitoring by a thermometer) (17Q7b)
Q2 種族滅絕原理(落藥) (3分)
A2 less survival rate of the treated, population of the treated is lower, amplified effect after several generations (20Q7cii).
Q3 Importance of immersing 3 cylinders instead of 1 cylinders in each concentration of solution.
A3 Increase reliability (accuracy 冇分, reliability=low sample variations, accuracy= low measuring error, by using different measuring method)(19Q6c).
Q4 Why increase the temperature (from 37C to 85C) at the end of an enzymatic experiment before measurement?
A4 To denature all the digestive enzymes to stop the reaction (21Q10aii).
Q5 Why pH indicator could be used to determine the lipid digestion?
A5 digestion of lipid by lipase produce fatty acids, which is acidic that could change the colour of the pH indicator by lowing pH (18Q8biii).
Q6 The importance of incubation before reactions
A6 Ensure the reactants have reached the set temperature at the beginning of the reaction (15Q7b).
Q7 How to compare amylase activities. (2 experiments)
A7 Iodine test: drill two holes on a starch agar plate, add 2 amylase samples to the holes respectively, incubate the plate, add iodine solution, the larger the dark blue zone the higher the amylase activities (14Q6c).
Reducing sugar test: Mix 2 amylase samples with 5 mL of starch solution, incubate the mixture, boil with excess Benedict’s solution, the larger the amount of brick red precipitate, the higher the amylase activities.
Q8 How to measure stomatal density on a leaf in the upper and lower epidermis.
A8 microscopic examination of epidermal peel, count the number of stomata per unit area in the upper and lower epidermis.
Immerse the leaf in hot water, count the number of bubbles per unit area in the upper and lower epidermis (20Q5ciii).
21 Q9 Interaction of plants with different leaf sizes
21 Q10 In vitro digestion of breast milk using enzymes in digestive juice
20 Q5 Transpiration rate of a leaf through upper and lower epidermis
20 Q7 Killing mosquitoes using parasitism.
19 Q6 The change in mass when 3 potato cylinders are immersed in sucrose solution with different concentration
18 Q8 Digestion of lipid in milk using lipase
15 Q7 Breakdown of hydrogen peroxide by catalase in different temperature
Experiment marking 整合
Q1 4 Control variables of transpiration experiments
A1 rate of water evaporation by adding oil, surface area of the leaf epidermis, wind speed by using enclosed system, volume of air containing water vapour by using identical flasks, leaves of the same potted plant (20Q5b). Temperature (by monitoring by a thermometer) (17Q7b)
Q2 種族滅絕原理(落藥) (3分)
A2 less survival rate of the treated, population of the treated is lower, amplified effect after several generations (20Q7cii).
Q3 Importance of immersing 3 cylinders instead of 1 cylinders in each concentration of solution.
A3 Increase reliability (accuracy 冇分, reliability=low sample variations, accuracy= low measuring error, by using different measuring method)(19Q6c).
Q4 Why increase the temperature (from 37C to 85C) at the end of an enzymatic experiment before measurement?
A4 To denature all the digestive enzymes to stop the reaction (21Q10aii).
Q5 Why pH indicator could be used to determine the lipid digestion?
A5 digestion of lipid by lipase produce fatty acids, which is acidic that could change the colour of the pH indicator by lowing pH (18Q8biii).
Q6 The importance of incubation before reactions
A6 Ensure the reactants have reached the set temperature at the beginning of the reaction (15Q7b).
Q7 How to compare amylase activities. (2 experiments)
A7 Iodine test: drill two holes on a starch agar plate, add 2 amylase samples to the holes respectively, incubate the plate, add iodine solution, the larger the dark blue zone the higher the amylase activities (14Q6c).
Reducing sugar test: Mix 2 amylase samples with 5 mL of starch solution, incubate the mixture, boil with excess Benedict’s solution, the larger the amount of brick red precipitate, the higher the amylase activities.
Q8 How to measure stomatal density on a leaf in the upper and lower epidermis.
A8 microscopic examination of epidermal peel, count the number of stomata per unit area in the upper and lower epidermis.
Immerse the leaf in hot water, count the number of bubbles per unit area in the upper and lower epidermis (20Q5ciii).
呀肥巴！！！原來你改咗名
唔知厹幾時蒲頭
唔知厹幾時蒲頭Grad左 想second degree
過幾日考留名，aim lvl 4
基本上唔係神科都可以無限combine cert
唔自爆了 總之係考評局未開始玩實驗題目嘅年份
覺得當年未夠俾心機溫
都唔係因為冇咗SBA所以多experiment 題
2018 已經開始多啦
2018 已經開始多啦
有冇邊題想深入討論？
2020 Q6aii
考生表現極差劣(very poor)
同通識卷一樣 收到題目搵晒所有key words 出黎
Describe how plants can obtain nitrate from waste water and make use of it for protein synthesis in their leaves.
1) how 題 (3分）
所有步驟或過程背晒出嚟
2) obtain nitrate
好多考生寫nitrate製造過程 例如
Nitrogen fixing bacteria convert ammonium into nitrate.
冇分
佢問你點吸(obtain/absorb)唔係點整(make/produce)
Nitrate is absorbed by root though diffusion and active transport (1分) transported into xylem.
3) to protein synthesis in leaves
由於題目問how 所有步驟都要講
而家d nitrate喺xylem到 佢要去leaves
nitrate in xylem is transported upward to leaves by transpiration pull (1分).
最後答 nitrate to amino acid to protein synthesis
可以諗係deamination 嘅相反
in leaves, nitrate combines with intermediate products from photosynthesis to form amino acid for protein synthesis (1分).
以下Out syll:
請勿詳細討論 問返dse問題
第3分準確d係 intermediates from photosynthesis, glycolysis and Kerbs cycle.
20隻amino acids 有20種 carbon skeletons
每隻amino acid 都有自己嘅intermediate 原型(precursor)
但有例外 形狀太類似嘅 amino acid 就來自同一intermediate
自己Google plant amino acid synthesis
有曬d biochemical pathways
千祈唔好記
Out syll part 完
考生表現極差劣(very poor)
同通識卷一樣 收到題目搵晒所有key words 出黎
Describe how plants can obtain nitrate from waste water and make use of it for protein synthesis in their leaves.
1) how 題 (3分）
所有步驟或過程背晒出嚟
2) obtain nitrate
好多考生寫nitrate製造過程 例如
Nitrogen fixing bacteria convert ammonium into nitrate.
冇分
佢問你點吸(obtain/absorb)唔係點整(make/produce)
Nitrate is absorbed by root though diffusion and active transport (1分) transported into xylem.
3) to protein synthesis in leaves
由於題目問how 所有步驟都要講
而家d nitrate喺xylem到 佢要去leaves
nitrate in xylem is transported upward to leaves by transpiration pull (1分).
最後答 nitrate to amino acid to protein synthesis
可以諗係deamination 嘅相反
in leaves, nitrate combines with intermediate products from photosynthesis to form amino acid for protein synthesis (1分).
以下Out syll:
請勿詳細討論 問返dse問題
第3分準確d係 intermediates from photosynthesis, glycolysis and Kerbs cycle.
20隻amino acids 有20種 carbon skeletons
每隻amino acid 都有自己嘅intermediate 原型(precursor)
但有例外 形狀太類似嘅 amino acid 就來自同一intermediate
自己Google plant amino acid synthesis
有曬d biochemical pathways
千祈唔好記
Out syll part 完
突然想起當年日校有句
我寧願做一條mRNA 因為咁 我就可以擁有U
Sequences in Gene transcription:
Coding strand (ATCG) = mRNA (AUCG)
Template strand (ATCG) = anti-codon tRNA (AUCG)
題目出玩genetic 呢個字 答案出現mitosis/meiosis機會率極高
0/10
我寧願做一條mRNA 因為咁 我就可以擁有U
Sequences in Gene transcription:
Coding strand (ATCG) = mRNA (AUCG)
Template strand (ATCG) = anti-codon tRNA (AUCG)
題目出玩genetic 呢個字 答案出現mitosis/meiosis機會率極高
0/10
補充少少：寫題目比咗你嘅資訊作為答案，係唔會有分
aii之所以寫ammonia converted to nitrate冇分，係因為題目已經寫咗呢句。如果題目冇比你，咁你就要寫，而題目分數亦好可能變成4分，因為的確係要寫呢個point, 即使字眼係"obtain"都好。
所以前年寫親呢句嘅人，應該都係唔識答題常識，唔關Bio知識or扣字眼(obtain/make)事
aii之所以寫ammonia converted to nitrate冇分，係因為題目已經寫咗呢句。如果題目冇比你，咁你就要寫，而題目分數亦好可能變成4分，因為的確係要寫呢個point, 即使字眼係"obtain"都好。
所以前年寫親呢句嘅人，應該都係唔識答題常識，唔關Bio知識or扣字眼(obtain/make)事
炒左喇呢科
肥巴覺得出auxin long question機率高唔高 好多年冇出了
仲係咁肥嗎？
Aim lvl 4要集中温咩topics？