Q-addmaths-2008

SOALAN ULANGKAJI SPM 2008

SPM 2008

Biology

Kertas 1 1. C 2. B 3. A 4. B 5. B 6. C 7. B 8. C 9. D 10. C 11. B 12. C 13. A 14. C 15. D 16. B 17. D 18. C 19. B 20. D 21. C 22. C 23. B 24. B 25. D

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Soalan 26. C 27. C 28. B 29. D 30. A 31. D 32. C 33. B 34. D 35. B 36. D 37. B 38. B 39. B 40. B 41. A 42. D 43. A 44. A 45. D 46. D 47. D 48. B 49. C 50. B

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SOALAN ULANGKAJI SPM 2008

SPM 2008

Jawap

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Biology

Soalan

Kertas 1

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Soalan

Butiran SECTION A

Markah

1(a)

An involuntary action that occurs automatically and spontaneously without conscious control towards a stimulus.

2 marks

(b)

P : Sensory receptor Q : Afferent neurone R : Interneurone S : Efferent neurone T : Effector (biceps)

5 marks

(c)

• A sharp pin pierces the skin, causing the sensory receptors in the skin to generate nerve impulses. • The nerve impulses are transmitted along an afferent neurone toward the spinal cord. • In the spinal cord, the nerve impulses are transmitted from the afferent neurone to an interneurone. • From the interneurone, the nerve impulses are transmitted to an efferent neurone. • The efferent neurone carries the nerve impulses from the spinal cord to the effector (biceps muscle) so that the pin can be pulled out from the skin immediately.

5 marks

2 (a)

U : V : W: X :

Avicennia sp. Bruguiera sp. Rhizophora sp. Sonneratia sp.

4 : 2 marks 2 –3 : 1 mark 0 –1 : 0 mark

(b)

• The pioneer species of a mangrove swamp are the Sonneratia sp. and Avicennia sp. • As the time passes, the soil becomes more compact and firm. This conditions favours the growth of the Rhizophora sp. The root system of the Rhizophora sp. and Bruguiera sp. traps silt and mud, creating a firmer soil structure over time • As more sediments are deposited, the shore extends futher to the sea. Over time, terrestrial plants like nipah palm begins to replace the Bruguiera sp. • The gradual transition and succession from a mangrove swamp to a terrestrial forest and eventually to a tropical rainforest, which is a climax community, takes a long time.

3 marks

(c)

• Soft muddy soil • Waterlogged soil which lacks oxygen • Seawater with high salinity (high salt content) • Strong sunlight and extreme heat.

2 marks

(d)

1. 2. 3. 4.

3 marks

3 (a) (b)

Pneumatophores – breathing roots. Hydathode (pores) in the epidermis of leaves. Thick cuticles and sunken stomata. Viviparity seeds.

K : Tracheal system L : Gills filaments

2 marks

Q : Air enters Q through tiny opening. T : To prevent the air tubes from collapsing or become deflated.

4 marks

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Biology

Jawap

Soalan

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Kertas 2

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Soalan Butiran (c) • Air enters the tracheae through the spiracles and travels through the tracheoles to the fluid-filled tips. • Oxygen in the fluid diffuses directly from the tracheoles into the cells, and carbon dioxide diffuses out from the cells into the tracheoles. (d)

(e)

• When an insect inhales, the abdominal muscles relax and P open. The increased air pressure forces air out through P. • The tracheal tubes carry oxygen from the air directly to body cells. K The large number of tracheoles provides a large surface area for the diffusion of gases

L The large surface area of the filaments and lamellae increases the efficiency of gaseous exchange in fish.

Markah 2 marks

2 marks

2 marks

4 (a)

A : Bone tissue B : Muscle tissue

2 marks

(b)(i)

Tendon

1 mark

(ii)

1 mark Tendon

(c)

(d)(i) (ii)

• The movement of the forelimb is brought about by the contraction and relaxation of a pair of antagonistic muscles A and B. • The arm is flexed by the contraction of the B muscle. The C muscle relax as the B muscle contract causing the arm bent. The B muscle become thicker and shorter. Contraction of the B muscle pulls the ulna up thus bending the arm at the elbow joint. • To straighten the arm, the C muscle contract while the B muscle relax. Contraction of the C muscle pulls the ulna down and the arm is straighten.

3 marks

More effort is required.

1 mark

So that a larger force is produced to lift the thing.

2 marks

SOALAN ULANGKAJI SPM 2008

SPM 2008

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Biology

Soalan

Kertas 2 Soalan 5 (a) P is produced by an anther Q is produced by a testis.

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Butiran

Markah 2 marks

(b)

M : pollen tube N : tail of the aperm K : the generative nucleus

3 marks

(c)

M : The pollen tube grows down through the tissues of the style and enters the ovule through the micropyle. N : The tail helps the sperm to propel towards the ovum.

2 marks

(d)

The generative nucleus (K) divides by mitosis to form two male gamete nuclei while the tube nucleus (L) disintegrates.

2 marks

(e)(i)

(ii)

2 marks

Double fertilization.

1 mark

• The body needs a balanced diet which contains carbohydrates, proteins, fats, vitamins, mineral salts, fibres and water in the correct proportion. • A balanced diet sustains the body and maintains health and for growth and repair. • Malnutrition is caused by an insufficient intake o one or more nutrients in our diet. • Malnutrition may be caused by eating too little, or overeating of certain types of food, which does not supply a balanced diet.

8 Marks

(b)

• Eating habits refer to what a person eats, the frequency a person eats and the amount a person eats automatically without thinking and without control.

2 Marks

(c)

• The meat-based fast food is not suitable for a teenager when consumed frequently for a long period because it contains excessive amount of salts, fats, proteins and food additives. • A balanced diet should contain seven classes of food in the correct proportions according to the age, sex and the level of activities. • Excess fats caused the cholesterol content in the blood to rise. This increase is likely to cause the cholesterol to be deposited in the arteries. This can lead to cardiovascular diseases. • Excess mineral salts increase the blood osmotic pressure. It can also lead to the formation of kidney stones. • Excess proteins can increase the uric acids content of the blood and lead to gout and kidney failure. • Food flavorings and colorings contain harmful chemicals that can cause cancer. • Insufficient fibre can lead to constipation and haemorrhoids.

10 Marks

6 (a)

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Biology

Jawap

Soalan

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Kertas 2

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Soalan Butiran 7 (a) • Distilled water is hypotonic to the cytoplasm of the red blood cells and the cell sap of the onion cells. • During osmosis there is a net flow of water from the surroundings into the cell. • The red blood cells will swell and eventually burst. • The vacuoles in the onion cells expands and the plant cells become turgid. • However, the cell wall prevents the onion cells from bursting because it is tough and rigid. • Vegetables which are soaked too long in a salt solution will become soft because the salt solution is hypertonic to the cytoplasm of the vegetable cells. • Water flows out of the cells, causing the cytoplasm and the vacuoles to shrink. • The plasma membranes pull away from the cell walls. • In this condition, the cells are said to be plasmolysed. • To make the vegetables crisp again, Puan Mashita can soak the drooped vegetables in water. SECTION C 8 (a)

• Osmoregulation is the process of regulating the blood osmotic pressure by regulating the water content and the concentration of salts in the body. • The normal blood glucose concentration in humans is about 90 mg of glucose in 100 cm3 of blood. • This level of glucose is regulated by the negative feedback mechanism controlled by hormones. • Two organs are involved : (i) Pancreas secretes the insulin directly into the blood. (ii) Liver is the main target organ of insulin and glucagon. • Insulin converts the excessglucose in the blood to glycogen. The conversion of glucose to glycogen lowers the blood glucose concentration to its optimum level. • In liver cells, the excess glucose in the blood will be converted to lipids. • Glucagon converts the stored glycogen in the liver to glucose. • The glucose then diffuse out of the liver cells into the blood. • Glucagon also increases the conversion of glucose from amino acids in the liver cells. • This increase the blood glucose concentration to its optimum level. • Tropic response or tropism is controlled by auxins. Auxins are produced in the root tip in the cell division zone. • Response of root tip towards light: i. When a root tip is exposed to light from one direction, the concentration of auxins is higher in the shaded region. ii. The high concentration of auxins in the shaded region of the root tip inhibits cell elongation. iii. Cells in shaded region grow slower than cells in region exposed to light. iv. Hence, the root bends away from the light showing negative phototropism. Objectives of Environmental Education : • Raising the awareness of society towards the environment and its problems. • Helping society to change the attitude so that there will be more motivation in playing the respective roles of shouldering the problems of the environment. • Helping individuals to attain skills for solving environmental problems. • Helping individuals to evaluate environmental education action and programmes.

Markah

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Biology

Jawap

Soalan

Kertas 2

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Soalan Butiran 9 (a) Objectives of Environmental Education : • Raising the awareness of society towards the environment and its problems. • Helping society to change the attitude so that there will be more motivation in playing the respective roles of shouldering the problems of the environment. • Helping individuals to attain skills for solving environmental problems. • Helping individuals to evaluate environmental education action and programmes.

Markah 10 marks

Formal environmental education: • Students of primary and secondary schools be exposed through the Science and Biology syllabus. Informal environmental education : • Distribution of pamphlet and poster. • Producing films and magazines. • Sponsoring exhibitions, talks and competitions on the environmental. • Celebrating World Environmental Day. • Collect and bury rubbish in a consistent and safe way. • Treat human and animal waste before disposing them into the sea or river. • Using biodegradable detergents. • Recycle and reuse plastic, tin, aluminium materials and old newspaper. • Educating and encouraging society on the importance of controlling pollution.

10 marks

SOALAN ULANGKAJI SPM 2008

SPM 2008

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Biology

Soalan

Ulang

Kertas 3 Soalan

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Butiran

Markah

SECTION A 1(a)

(i) 29°C (ii) W X Y Z

(b)

5√ = 3m 4 – 3√ = 2m 2√ = 1m 0 – 1 √ = 0m

= 43°C = 54°C = 78°C = 60°C

(i) 1. The water temperature increase when it is heated/water becomes hot. 2. The food sample will become smaller and darker.

3m

(ii) 1. Because heat from the food sample is transferred to the water. 2. Because the food sample will be burnt and chemical energy is converted to heat energy.

3m

(c)

Variables Food sample

Particulars to be implemented The type of food sample which is used is different. High temperature of water is recorded. Difference between highest temperatures with initial temperature will give temperature change. -mass of food sample weighed accurately 5.0 g -Volume of distilled water is 20 ml -Distance of burnt food sample from the base of test tube is the same.

Temperature changes

-Mass of sample food/ -volume of distilled water/ -distance of food sample from test tube. (d)

Content of chemical energy in food can be determined by burning method.

(e)

(i) Food sample W

Increase in water temperature 14

X

25

Y

49

Z

31

1√ = 1m Total = 6 m

3m

Energy value 20 x 4.2 x 14 J 5g = 235.2 J/g 20 x 4.2 x 25 J 5g = 420.0 J/g 20 x 4.2 x 49 J 5g = 823.2 J/g 20 x 4.2 x 31 J 5g = 520.8 J/g

4√ = 3m 3√ = 2m 2√ = 1m 0 - 1√ = 0m

SOALAN ULANGKAJI SPM 2008

SPM 2008

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Biology

Soalan

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Kertas 3 Soalan

Butiran (ii) Food Y has high energy value so that food Y may contains a lot of fats. // Food W contains high carbohydrate but less fats.

Markah 3m

(f)

1. The water in the boiling tube must be stirred during the experiment to ensure even distribution of heat.

3m

2. A shield must be used to enclose the boiling tube to minimize the heat loss to the surroundings. (g)

The experimental energy value is different from the actual energy value of the food sample because of the surroundings as heat energy and light energy.

3m

(h)

Energy value is the amount of heat generated from the combustion of one gram of food sample. The heat energy is absorbed by water thus increasing the temperature of the water.

3m

(i)

Title 1m Each food class = 1m

Food samples with equivalent energy values Carbohydrates Fats Rice Butter Bread Palm oil Steamed potatoes Groundnut SECTION B

Problem Statement : Does yeast carry out anaerobic respiration?

3m 3m

Aim : To investigate the process of anaerobic respiration in yeast. Hypothesis : Yeast carries out anaerobic respiration and produces carbon dioxide, ethanol and heat. Variables : Manipulated variables : Presence of yeast Responding variables : Carbon dioxide, ethanol and Controlled variables : Absence of oxygen

heat.

Technique used : Yeast cells are left to ferment in the absence of oxygen. The products of fermentation, namely carbon dioxide, alcohol and heat are then detected using simple tests. Materials/Apparatus: Yeast suspension, glucose solution, limewater and paraffin oil, thermometers, boiling tubes, test tubes, beakers, Bunsen burner, a tripod, wire gauze and stoppers with delivery tubes.

3m

SOALAN ULANGKAJI SPM 2008

SPM 2008

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Biology

Soalan

Ulang

Kertas 3 Soalan

an kaji

Butiran Procedure : 1. Boil 100ml o water in a beaker, cool it and use it to prepare a 5% glucose solution (add boiled water to 2.5 g glucose and bring it to 50 ml) and a 5% yeast suspension (add boiled water to 2.5 g yeast and bring it to 50 ml).

Markah 3m

2. Label the tubes as A and B. 3. Pour the 5% glucose solution to tubes A and B until 1/3 full. Add 5 ml of yeast suspension to tubes A. 4. Add sufficient paraffin to tubes A and B, to form a layer covering their contents. 5. Connect both tubes with stoppers that have attached U-shaped delivery tubes and thermometer. 6. For each tube, dip the other free end of the U-shaped delivery tube into a test tube containing limewater. 7. For each tube, record its initial temperature. Set both setups A and B aside at room temperature and observe them again after one hour. Results : Boiling tube

Initial temp (ºC)

Final temp (ºC)

Lime water

A

30

33

Chalky

B

30

30

colourless

Smell and appearence in boiling tube - smell of ethanol - the solution in A is foamy - no smell - solution looks clear

Conclusion : Experimental results show that yeast carries out anaerobic respiration in the absence of oxygen, producing heat and ethanol. Hypothesis accepted.

3m

1m

1m – complete 9 criteria include variables, hypothesis and technique.