Question 1i. P- Mitochondria

R- Choloroplast

ii. P- Site of energy production

R- Has chorophyll to trap sunlight and undergo photosynthesis.

a) Q is where the ribosome attached to. Protein that is synthesised by the ribosome cannot be transported when Q is absent. Enzyme production cannot be occured as the protein is not transported to golgi apparatus.Question 2

Essay Section B

a) Nitrates and phosphates that contain in fertilizers flow to the rivers and lakes. This will cause algal bloom that prevent the penetration of sunlight to the lake. The submerged plants receive less amount of sunlight and cannot undergo photosynthesis. As a result, the plants will die and oxygen will be less leading to the death of other aquatic organism such as fishes. The fertiliser will be used up that causes the alga to die by itself. Dead substance will lead to high amount of bacteria to undergo decomposition on death substance. Much bacteria will cause less amout of oxygen that lead to high BOD level.this will cause dead zone as no life can survive. The example is at Gulf of Mexico Dead Zone.b) i) Treating sewage can remove pathogens and bacteria that carry various diseases such as malaria. It can also improve the environment through proper drainage and disposal of waste water. Treating sewage can maintain water ecosystem ingtergrity so that aquatic organism can stay alive. It can prevent flood as no sediments and materials deposited in lakes so that water can move easily. Treating sewage can preserving water quality for human uses.ii) Organic fertilizer is beneficial to water environment rather than inorganic fertilizer because it wont build up harmful residues or cause pollution due to run-off from irrigation or rain. Organic fertilizer have adequate amount of phosphorus but inorganic fertilizers have excess amount of phosphorus that lead to eutrophication. Organic substance have neutral pH value rather than inorganic substance that is acidic so, it can mantain the pH value of water. Organic substance does not contain any chemical so aquatic plants can live and absorb dissolved carbon dioxide in water.Question 2

a) Yes. Milk contains high amount of calcium and phophorus that is needed for the formation of strong teeth and bones. It also contains protein that function for the formation of new tissues. Wholemeal bread contains high amount of carbohydrate that provide energy for the daily activities. It also contains minerals such as iodine, zinc and iron that is important in body regulation. Butter contains protein to help the adult to grow and lipid to provide stored energy.b) When the food is chewed in the mouth, salivary galnd screted salivary amylase. Amylase will digest carbohydarate into maltose. The food enter the oesophagus and it is called as bolus. In the stomah, hydrocloric acid kills the bacteria in the food. The food enter the small intestine, amylase secreted by pancreas digest starch to maltose. At the end of the ileum, the intestinal juice that contains maltase, sucrase and lactase enzymes will digest the carbohydrates into its constituent monosaccharides. The monosaccharides will be absorbed at the villi to the bloodstream and transported to all parts of the body.

QUESTION 3

In X, antibodies are produced by the body while in Y, antibodies are obtained directly. The reaction of X is slow while the reaction of Y is immediate. X is injected with vaccine while Y is injected with antiserum that contains specific antibody. X is long lasting immunity while y is short lasting immunity. In X, it is injected before a person is infected with disease while in Y, it is injected when person is infected or have high risk of getting the disease. In x, it needs second injection or booster dose because first injection ussually is slow and low level of antibody while in Y, it does not need second injestion.PAPER 3

1. Diagram 1 shows an experiment that had been carried out to investigate the effect of air movement on the rate of transpiration hibiscus plant by using a potometer. Time is taken for an air bubble to move from X to Y (10 cm distance) by using stopwatch.

Rajah 1 menunjukkan satu eksperimen yang dijalankan untuk mengkaji kesan pergerakan udara terhadap kadar transpirasi pokok bunga raya menggunakan satu potometer. Masa di ambil untuk satu gelembung udara untuk bergerak dari X ke Y [jarak 10 cm] dengan menggunakan jam randik

The potometer is placed near a fan with air speed adjusted at different velocity in the same room with temperature 30oC. The result of the experiment is shown in the table 1.

Potometer ini diletakkan dalam bilik yang sama, berhampiran dengan kipas angin dimana halaju angin berbeza. Keputusan eksperimen ditunjukkan dalam Jadual 1Fan speedStopwatch readingTime taken by air bubble to move from X to Y (minute)

136

228

3

18

415

Base on the Diagram 1 and Tables 1 answer all questions:1(a)(i) State two observations on the time of air bubble moves.

Nyatakan dua pemerhatian tentang masa dalam eksperimen ini

Observation 1

When the fan speed is 1, the time taken by air bubble to move from X to Yis 36 minutes.

Observation 2

When the fan speed is 4, the time taken by air bubble to move from X to Y is 15 minutes.

[3 marks]

(a)(ii)State one inference for each observation made in (a) (i).

Nyatakan satu inferen bagi setiap pemerhatian yang dibuat pada 1(a)(i)Inference for observation 1 The time taken by air bubble to move from X to Y is the longest because water molecule evaporate at surface of the leaves slowly under slow air movement causing the air bubble to move in long time.Inference for observation 2

The time taken by air bubble to move from X to Y is the shortest because water molecule evaporate at surface of the leaves rapidly under fast air movement causes it to move in short time.

[3 marks]

(b)(i)Draw a table based on the following criteria:

Bina satu jadual berdasarkan criteria berikut

Fan speed Time taken

Transpiration rate[Transpiration rate = Distance [cm/min] Time Fan speed

Time taken for air bubbles to move (min)

Transpiration rate(cm /min)

1

36

0.28

2

28

0.36

3

18

0.56

4

15

0.67

[3 marks](c)

State the relationship between the increase of air movement and the rate of transpiration. Explain your answer

Nyatakan hubungan antara pertambahan keamatan cahaya dengan kadar transpirasi. Terangkan jawapan anda

When the air movement increases, the rate of transpiration increases. This is because the water

Molecules evaporate at surface of the leaves rapidly under high air movement causing the rate of transpiration increases.

[3 marks]

(d)

State the variables in the experiment and explain how the variables are operated

Nyatakan pembolehubah dan cara mengendalikan pembolehubahVariables in the experiment

How the variables are operated

Manipulated variable

Air movementUse different speed of fan that is 1, 2, 3 and4

Responding variable:

Rate of transpiration

Calculate and record the rate of transpiration by using formula distance/time (cm/min)

Fixed variables :

Temperature

Fix the temperature at 30 degree celcius.

[3 mar(e)

State the hypothesis for this experiment.

Nyatakan hipotesis eksperimen ini

The faster the air movement, the higher the rate of transpiration.

[3 marks]

(f)

Based on the experiment, define transpiration operationally

Berdasarkan eksperimen, beri definisi transpirasi secara operasi

Transpiration is a process of loss of water vapour shown by the time taken by air bubble to travel from X to Y affected by the air movement.

[3 marks](g)

The experiment is repeated by placing the set-up in the dark without fan. Predict transpiration rate of the plant shoot under this condition. Explain your prediction.

Jika eksperimen diulang dengan meletakkan radas dalam gelap, ramalkan kadar transpirasi tumbuhan ini. Terangkan ramalan anda

0.1 cm/min.The rate of transpiration is lower due to low light intensity and lower air movement.

[3 marks](f)

Classify the apparatus and materials used in this experiment. Kelaskan alat radas dan bahan yang digunakan dalam eksperimen ini.ApparatusMaterialsThermometer

Stopwatch

RulerWater

Hibiscus plant

Capillary tube

[3 marks]

[33 marks]

Paper 3(Sec B)Problem statement:

Which sources of water sample has higher BOD level?Variables:

MV: Type of water sample

RV:BOD value

CV: Amount of water sample

Hypothesis:

C has higher BOD level compared to A and B.

Apparatus and material:

Reagent bottle with stoppers, cupboard, syringe, stopwatch, label paper, measuring cylinder, beaker, water sources from A, B and C, methylene blue solution.

Procedure:

1. Collect 200ml of water sample from A, B and C.2. Label 3 reagents of bottle as A, B and C respectively.

3. Measure 100ml of water sample A using measuring cylinder and put in the reagent bottle.

4. Add 1ml of 0.1% methylene blue solution at the base of water sample using a syringe.

5. Close the bottles quickly without shaken it.

6. Repeat step 3-5 by using water sample from B and C.

7. Place all bottles in a cupboard and start the stopwatch.

8. Examine the bottles from time to time.

9. Record the time taken for the water samples to decolourise the methylene blue solution.

10. Record the results in a table.

Data presentation

Water sampleTime taken to decolourise(hour)BOD value

A

B

C

10 cm

Air bubble

X

Y

Fresh plant shoot

Diagram 1