2016年6月13日 星期一

中五生物下學期EXP02 實驗22.X: 探究酵母進行的缺氧呼吸














































結果:                                                                                                                                         (8 )
彎液面刻度隨時間的變化  詳盡數據 (1)
時間 (min)
0







彎液面刻度 (ml)








CO2釋出體積 (ml)









選軸 (1), 軸標註和單位 (1), 標題 (1), 畫線 (曲線圖) (1)

        酵母缺氧呼吸的最高速率      =     最高體積轉變 / 時間  
=      最高斜率 = (y2 – y1) / (x2 – x1) (1)
                                                                計算結果 (1)
        (列出計算過程)

碳酸氫鹽指示劑檢驗釋出氣體的結果 (1)
碳酸氫鹽指示劑原先顏色
紅色
排進氣體後碳酸氫鹽指示劑的顏色
黃色
















































結果闡釋:(6 )
描述並解釋實驗結。                                                                                                                      (6)
描述 (2), 解釋 (4). 建議答案:
實驗裝置的針筒內並不含有空氣和氣泡, 以維持缺氧狀態, 酵母只能進行缺氧呼吸以產生能量維持其代謝活動.
透過缺氧呼吸, 酵母分解混合物中的葡萄糖, 釋出二氧化碳和乙醇 (1). 二氧化碳氣體釋出後會佔據溶液的空間, 產生的壓力會把移液管內溶液的彎液面壓向下 (1)
根據線圖所示, 缺氧呼吸速率在0 – 15分鐘時較慢, 但卻隨時間上升 (1). 這是因為酵母需要時間適應新環境, 所以其代謝活動及呼吸速率並不穩定 (1).
15分鐘後, 酵母的缺氧呼吸速率變得穩定及達至最高速率 (1). 酵母在15分鐘後完全適應的環境, 所有代謝活動包括缺氧呼吸也達至平衡狀態 (1).

討論:(6 )
1     解釋為何需要在實驗開始前將針筒內的空氣排走。                                                                       (1)
以肯定針筒內是缺氧環境 (1)

2     你可如何改變本實驗的裝置以探究葡萄糖濃度對缺氧呼吸的影響。試繪畫標註圖加以說明。(5)
自變項: 不同濃度葡萄糖溶液 (1); 控制變項: 相同體積葡萄糖溶液與相同體積酵母混合 (1);
應變項: 缺氧呼吸在某葡萄糖濃度的速率 (1);


繪圖 (1)顯示重覆所有裝置但只轉變葡萄糖濃度 (自變項)

標註 (1)















結論:(2)                                                                                                                          

酵母在缺氧環境下會進行缺氧呼吸並釋出二氧化碳 (1), 其最高速率是… (1)

中五生物下學期EXP01 實驗21.X: 探究水生植物的光合作用
















































結果:                                                                                                                                         (8 )
開始時彎液面的刻度 (ml)
0.9
結束時彎液而的刻度 (ml)
-0.25
彎液面刻度的變化 (ml)
1.15 (1)
實驗進行時間 (min)
4 X 60 = 240 (1)

        光合作用速率                 = 1.15 / 240 (1) = 0.0048 (1) ml min-1 (1)
        (列出計算過程)

描述餘燼木條的變化。這結果有何啟示?(2)
重燃 (1); 顯示釋出的氣體是氧 (1)

結果闡釋:                                                                                                                                     (2 )
實驗結果顯示植物在有光情況下進行了甚麼過程?這過程的產物是甚麼?
光合作用 (1); (1)

討論:                                                                                                                                         (6 )
1     實驗中利用碳酸氫鈉溶液而用水甚麼好處?(1)
提供更多的二氧化碳供黑藻進行光合作用之用 (1)

2     你可如何改變本實驗的裝置以探究光強度對光合作用的影響。試繪畫標註圖加以說明。(5)              
重複實驗 (1), 唯每次改變光強度 / 光的距離 (1), 其他控制變項要保持不變 (1)


(1)
標註 (1)
















結論:(2)                                                                                                                           (2 )

黑藻能進行光合作用釋出氧 (1); 而光合作用速率為 0.0048 ml min-1 (1)


















F4 2nd Term EXP03: Practical 10.X

Practical 10.X
Investigation of the difference in
transpiration rate between the two leaf
surfaces

Objective:       To measure the transpiration rate of a leafy shoot by using bubble photometer (1), and
(2 marks)         to deduct the difference in stomatal density between the upper and the lower leaf surface by comparing the transpiration rate between the two surfaces of the leaf (1)

Biological principle and experiment design:                                                                          (8 marks)
Most of the water absorbed by a plant is eventually lost in the form of water vapour (1) from
the leaf surface. This process is called transpiration (1), which creates a pulling force
to transport water and minerals (1) in the xylem vessels upwards to the leaves. This is the main transport mechanism in plants.

To measure the rate of transpiration, a (bubble) potometer (1) can be used. Since water is continuously absorbed and lost through transpiration, the air bubble inside the connected capillary tube moves towards the direction of the leafy shoot. Transpiration rate can be calculated as shown below:

                Transpiration rate = Volume of absorbed water (1)
                                                Time of measurement (1)
The abundance of the stomata (1) on both surfaces of a leaf is different. The difference can be deduced indirectly by comparing the transpiration rates between both surfaces. To measure the transpiration rate of one surface of a leaf, surface of the other side should be smeared with Vaseline that is impermeable to water and air.

To ensure a valid experimental result, we have to make an assumption that all the water uptake by the

leafy shoot is lost through transpiration (2)

Procedures:                                                                                                                            (6 marks)
1.      Smear the upper surface of a leaf from a plant with Vaseline.
2.      Cut the leafy shoot from the plant and fit it tightly into the potometer tightly. This should be done under
so as to prevent the air bubble from entering the xylem vessles (1) that may cause blockage of water
transport and absorption (1).

3.      Set up the apparatus as shown below. Seal off all connections with vaseline (1).

1.      Introduce an air bubble (1) into the capillary tube so that the change in water absorption can be observed.
2.      Allow the setup to equilibrate for 10 minutes, so that the transpiration rate would be stabilized.
3.     Record the distance (1) travelled by the bubble in (actual time) (1) minutes.
4.      Open the tap of the reservoir to re-adjustment the position of the air bubble. Repeat steps 4 to 5 to take two more readings (if time allowed).
5.      Repeat steps 1 to 7, but use another leaf with similar size and smear the lower surface with Vaseline.

Results:                                                                                                                                  (7 marks)
Leafy shoots
Upper surface smeared
Lower surface smeared
Initial position of the air bubble (cm)
1.3 ml
4.3 ml (1)
Final position of the air bubble (cm)
7.87 ml
2.7 ml (1)
Distance of the bubble movement (cm)
6.57 ml
1.6 ml (2)
Time of the bubble movement (min)
54
48 (1)
Rate of the bubble movement (cm / min)
0.085 ml min-1
0.033 ml min-1 (2)

Result Interpretation:                                                                                                           (7 marks)
1          According to the experimental result, deduce the difference in the abundance of stomata on both surfaces of a leaf.                                                                                                  (3 marks)
The transpiration rate of the leaf with lower surface smeared is much lower than that with upper surface smeared (1). This shows that the transpiration rate is much more reduced by smearing the lower leaf surface than smearing the upper surface (1).
It can be deduced that the lower leaf surface may have higher stomatal density than the upper surface (1).

2          Describe and explain the movement of the air bubble inside the capillary tube.             (2 marks)
The air bubble moves towards the leafy shoot along the capillary tube (1).
During the process of transpiration, water is absorbed into the leafy shoot by transpiration pull, leading the air bubble to moving towards the leafy shoot (1)

3          Explain why the rate of the movement of the air bubble cannot directly show the rate of transpiration?                                                                                                             (2 marks)
The rate of the movement of the air bubble can only show the rate of water absorption directly (1). Actually, not all the water absorbed is lost through transpiration (1)

Discussion:                                                                                                                              (2 marks)
4     State the limitations of this experimental setup or the source of errors of this experiment.      (1 mark)
The friction between the air bubble and the capillary wall may hinder the movement of the air bubble (OR any other reasonable answers) (1)

5     Suggest improvements for the limitations or the errors as stated in question (4).         (1 mark)
To reduce the volume of the air bubble introduced so as to reduce the friction during movement (1)

Conclusion:                                                                                                                            (2 marks)
The transpiration rates of the upper leaf surface and the lower leaf surface are 0.033 ml min-1 and 0.085 ml min-1 respectively (1).

By comparing the transpiration rates between the two leaf surfaces, it can be deduced that the lower leaf surface may have a higher stomatal density than the upper leaf surface (1).