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PEPERIKSAAN PERCUBAAN SPM 2010 4551/1

Peraturan Pemarkahan

( Jawapan )

BIOLOGY Paper 1 29 Ogos 2010

1 4

1 jam

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH

NEGERI KEDAH DARULAMAN

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MARKING SCHEME PAPER 1 TRIAL KEDAH 2010 1. C 26. A 2. A 27. A 3. C 28. B 4. D 29. C 5. C 30. C 6. B 31. C 7. A 32. B 8. D 33. A 9. D 34. C

10. B 35. D 11. C 36. A 12. D 37. A 13. D 38. A 14. D 39. D 15. B 40. A 16. D 41. A 17. C 42. C 18. A 43. B 19. D 44. B 20. C 45. A 21. B 46. B 22. D 47 D 23. A 48. A 24. B 49. C 25. C 50. B

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PEPERIKSAAN PERCUBAAN SPM 2010

BIOLOGY PAPER 2

Two hours and thirty minutes


( Jawapan )

4551/2 Biology Kertas 2 29 Ogos 2010

2

12 hours


NEGERI KEDAH DARUL AMAN

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2

BIOLOGY SECTION A

PAPER 2 [4551/2] No.

Marking Criteria / Sample Answers Marks

1 (a) (i) Gills

1

(ii) Tracheal system

1

(b) P : Filaments Q: Spiracles

1 1

2

(c) (R is ring of chitin which) support the tracheal / prevent the tracheal from collapsing.

1

(d) Diagram 1.1(b): P1: The filament have numerous thin-walled lamellae to maximise the surface area for gaseous exchange. P2: The gill filaments have thin membrane and covered by a net work of capillaries to transport respiratory gases. P3: The surface of the gills is moist which allows the gases to be dissolved. Any 1P

1

Diagram 1.2(b) P1: The large number of tracheoles provides a large surface for the diffusion of gases. P2: Tip of tracheoles have thin permeable walls and contain fluid in which respiratory gases can dissolve. P3:Terminal ends of the tracheol remains moist which allows the gases to be dissolved. Any 1P

1

(e) (i) P1:( The gaseous exchange process occurs over the whole body surface in an Amoeba sp) through simple diffusion. P2:Higher concentration of oxygen in the water surrounding causes oxygen to diffuse into the Amoeba. P3:Higher concentration of carbon dioxide in the cell causes carbon dioxide to diffuse out of the Amoeba. Any 2P

1 1 1

2

(ii) S: Contractile vacuole

1

(iii) P1: Freshwater is hypotonic to the cytoplasmic fluid of Amoeba sp . P2: Water diffuses into the cell and fill the contractile vacuole by osmosis P3: When the contractile vacuole is filled with water to its maximum size, it contracts to expel its content from time to time. Any 2P

1 1 1

2

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3

No.

Marking criteria/ Sample answers Mark

2 (a) (i) Osmosis

1

(ii) P1 : Sucrose solution is hypertonic / more concentrated. P2 : Water diffuse from distilled water into the sucrose solution P3 : The level of sucrose solution in the capillary tube stop rising at the equilibrium stage / the concentration inside and outside of the visking tubing is the same / the amount of water diffuse into and out from the visking tubing is the same. Any 2 Ps

1 1 1

2

(b)

F- Sucrose molecules are too large E- The visking tubing is a semi permeable membrane/ which only allows certain substances to pass through.

1 1

2

(c) (i) Y : crenation Z : haemolysis

1 1

2

(ii) P1- Solution Z is hypotonic compare to red blood cell. P2- Osmosis occur P3- water leaves/ diffuses into the cell P4- Red blood cell expand/ swell and burst. Any 3P

1 1 1 1

3

(iii) F : No P1 : Plant cell consists of cell wall P2 : Cell wall is made up of cellulose // Cell wall able to withstand the pressure. Any 2

1 1 1

2

Total

12

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4

No.


3 (a) (i) Absorption / Simple diffusion / facilitated diffusion

1

(ii) F1 thin wall/ one cell thick E1 increase rate of diffusion of digested food/ nutrients F2 large surface area/ has microvilli E2 increase rate of absorption of digested food/ nutrient F3 has a network of capillaries/ blood vessels E3 to transport the absorbed nutrients Any F + E

1 1 1 1 1 1

2

(b)

P: hepatic portal vein Q: lymphatic/lymph vessel/ duct

1 1

2

(c) P1: Deamination.// The amino group is removed (from amino acid)/ converted to ammonia . P2: (Ammonia) is converted to urea. P3: urea will be excreted through the kidneys. Any 2 Ps

1 1 1

2

(d) L1: A major energy reserve in the body// L2: (phospholipids are) components of the plasma membrane// L3: Lipids is used as a respiratory substrate// L4: Excess fats are stored in adipose tissues (under the skin, around internal organs) Any 1L A1:Amino acids are used in protein synthesis// A2:For repair and production of new protoplasm/growth and repair// A3:Used in the formation of enzymes/ some hormones/protein part of haemoglobin/ antibodies Any 1A G1:Glucose is used as the main respiratory substrate// It is oxidised to release energy (water and carbon dioxide)// G2:Excessive glucose is converted to glycogen // Blood glucose level rise / increase. Any 1 G

1 1 1 1 1 1 1 1 1

3

(e) P1: Diabetes mellitus // Blood sugar level increases// Hyperglycemia P2: Excess glucose cannot be converted to glycogen.

1 1

2

Total 12

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5

No.


4 (a)

Both arrows correct

1

(b) A – Pulmonary artery B – Pulmonary vein

1 1

2

(c)

F : Contraction of ventricle / heart E1: generates a (high) pressure E2 : (to) propel/ force / pump the blood flow from the heart/ ventricle to vessel A Any two

1 1 1

2

(d)(i)

Coronary artery

1

1

(ii) P1: Cut the supply of O2/ nutrients to the heart muscle P2: causing chest pain / angina / heart attack / myocardial infarction Reject ‘Heart problem’

1 1

2

(e) (i)

(ii)

P1: platelets break down and release chemicals P2: to cause platelets to stick to each other P3: platelets clump together to form a plug to prevent blood loss . P4: released thrombokinase and other clotting factors Any 2P P1 : Fibrinogen is soluble, fibrin is insoluble / not soluble P2 : Fibrin able to form fibres / meshwork / thread to trap blood cells, fibrinogen is not able to do so.

1 1 1 1 1 1

2 2

Total 12

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6

No.

Marking Criteria / Sample Answers Marks

5 (a) (i) (Transfer/flow of) energy

1

(a) (ii) F : Phytoplankton is an autotrophic organism. P1 : Able to absorb light energy / consists of chloroplast. P2 : synthesis their own food / carry out photosynthesis Any 2

1 1 1

2

(b) F1 : population of small fish increases P1 : no shark feed on small fish // shark is the predator F2 : population of plankton decreases P2 : more small fish feed on the plankton F3 : Eventually the population of small fish decreases Any 3

1 1 1 1 1

3

(c) F : Commensalism P1 : Shark is the host / neither gain any benefit nor harmed. P2 : Remora benefits P3 : Remora obtain protection / food / transport from the shark. Any 3

1 1 1 1

3

(d) P1 : Fertilizer washed away by rain water into the lake P2 : Nutrient / minerals content in the lake increase. P3 : alga bloom / alga grow rapidly in the lake. P4 : eutrophication occur. P5 : Oxygen content in the lake decrease / drop P6 : Fishes die / population decrease Any 3 P

1 1 1 1 1 1

3

Total

12

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7

BIOLOGY SECTION A

PAPER 2 [4551/2] - ESSAY

No.

Marking Scheme Mark

6(a) (i)

Continuous variation : body weight, height Discontinuous variation : types of earlobe, types of finger print.

1 1

2

(a)(ii) Continuous Variation Discontinuous variation P1 The changes of

characteristics among individual are gradual

The differences among individuals are distinct.

P2 Continuous variation is quantitative // characteristics can be measured.

Discontinuous variation is qualitative // characteristic is either present or absent.

P3 The graph shows the normal distribution curve.

The graph shows the discrete distribution.

P4 The character is determined by many genes

The character is determined by a single genes

P5 The characteristic is influenced by the environmental factor and genetic factor.

The characteristic is influenced by the genetic factor.

P6 Exhibits a range of phenotype with intermediate characters.

There are no intermediate groups.

Any 4 pair

Max 4 m

(b) Albinisme F : Albinisme is caused by the change in gene // mutation P1 : Body / skin unable to produce black pigment (melanin) P2 : The skin and hair of albinos are white // their eyes are pink. Any 2 Sickle cell anaemia F : Sickle cell anaemia is caused by the change in the genes // mutation. P1 : haemoglobin produced is not normal / abnormal P2 : Abnormal haemoglobin unable to bind / transport / carries with oxygen efficiently. P3 : The patient will always feel weak / cannot carries out vigorous activities. Any 2

1 1 1 1 1 1 1

Max 2 m

Max 2m

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8

6(c) (i) Abiotic factors that cause variation between the two sets of ginger plants are: F1: Sun light P1: Plants need light energy to carry out photosynthesis for growth P2: Set A, plants are obtain more / exposed to sunlight // Plants in set B obtain less sunlight / not exposed to Sunlight. P3 : Growth rate of plants in Set A is higher than plants in Set B.

1 1 1 1

F2: Space P4: Plants need (space) to grow a large root system / leaves P5: Plants able to absorb sufficient water and minerals/sunlight. P6: Set A, plants have larger space for the root and leaves to Grow // Plants in set B have smaller space for the root and leaves to grow.

1 1 1 1

F3: Soil / minerals P7: Plants need mineral for (healthy) growth. P8: Loam soil provides more minerals in Set A. // Sandy loam soil in Set B contains less minerals. P9: Loam soil able to trap / store water better than sandy loam soil. Any 8

1 1 1 1

max 8

6(c) (ii) F1 : Plantlets from tissue culture have the same genetic material. P1 : This is to show /ensure/proof the differences of plants in Set A and Set B are not caused by genetic factor / have the same genetic material. // This is to show /ensure/proof the differences of plants in Set A and Set B are caused by abiotic factors.

1 1

2

Total 20 No.

Marking Scheme Mark

7(a) P1 : Nerve impulses arrive at the axon terminal of (presynaptic) neurone. P2 : Causes the synaptic vesicles to move towards the (presynaptic) membrane and fuse with the membrane. P3 : Neurotransmiters /acetylcoline (examples) molecules are released from synaptic vesicles. P4 : (The neurotransmitter molecules) diffuse across the synaptic cleft into the postsynaptic knob / dendrite / cell body of neighbouring neurone.. P5 : The neurotransmitter molecules bind to specific receptor sites in the postsynaptic knob. P6 : The binding triggers / generates new nerve Impulses. P7 : The impulses then move along the postsynaptic neurone. P8 : The release of neurotransmitter is in one direction, from the synaptic knob to the postsynaptic neurone. P9 : Mitochondria in the synaptic knob generate ATP / energy to synthesis neurotransmitter molecules. Any 6

1 1 1 1 1 1 1 1 1

Max 6

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9

No.

Marking Scheme Mark

7 (b) P1 : The receptor at the terminal of X stimulated by the heat. P2 : The receptor generates a nerve impulse. P3 : The nerve impulse travels along X / afferent neurone To the spinal cord. P4 : In the spinal cord, the nerve impulse is transmitted to an interneurone. P5 : From the interneurone, the nerve impulse is transmitted to an efferent neurone/ neurone Y. P6 : Nerve impulse travels along efferent neurone / Y and reach the effector / muscle tissue / fingers. P7 : Muscles contract to withdraw the hand / finger. Any 4

1 1 1 1 1 1 1

Max 4 7 (c) P1 : The receptors in the eyes detect the dog.

P2 : Nerve impulses are generated and transmitted to the brain via the afferent neurone. P3 : The hypothalamus in the brain is stimulated. P4 : It actives the sympathetic nervous system to generate nerve impulses. P5 : Nerve impulses are transmitted to the adrenal medulla to stimulate secretion of adrenaline. P6 : Adrenaline carried / transported by blood circulatory system to the targeted organs. P7 : Adrenaline promotes the breakdown of glycogen to glucose. P8 : (Adrenaline) increases the breathing rate. P9 : More oxygen will be taken into the body P10 : (Adrenaline) increases the rate of heartbeat/ blood pressure. P11 : Rate of the blood flow increase. P12 : More glucose and oxygen will be supplied to the muscles. P13 : More energy produced by the muscles. // metabolic rate increase. P14 : Body has enough energy to face the ‘fight or flight’ situation. Any 10

1 1 1 1 1 1 1 1 1 1 1 1 1 1

Max 10

Total 20

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10

No. Marking Scheme Mark 8 (a)(i) P1 fish have streamline shapes // the anterior of the fish is

smooth and rounded // the body is long and tapers towards the end.

P2 the body of a fish is covered with scales that have a slimy coating

1 1

2

(a)(ii) P1 myotomes muscles are arranged in both side of the body P2 the vertebral column of the fish is flexible and can bent

from side to side P3 myotome muscles act antagonistically in fish./ carry out

opposite action in a fish P4 when the muscles on right side contract, the muscle on

the left side relax P5 the tail/body will be bent to the right. P6 when the muscles on left side contract, the muscle on

the right side relax P7 the tail/body will be bent to the left. P8 alternate contraction of the right and left myotome block

enable its tail to move left and right P9 to produce a force that propel the fish forward.

[ any 6]

1 1 1 1 1 1 1 1 1

Max 6

(b)(i) Similarities: F1 Both Joint S and Joint T has a cavity filled with svnovial fluid // lined with synovial membrane El Synovial fluid acts as lubricant to reduce friction between bones // absorbs shock of the movement. F2 The end surfaces of the humerus bone of Joint S and Joint T are covered with cartilage E2 To protect the bone / reduce friction between the bones F3 Both Joint S and T are connected with ligaments E3 to absorb shock // strengthen the articulation of bones/ joint. Differences: D1 Joint S is hinge joint E4 Joint S allows the movement of bones in one plane / direction D2 Joint T is ball-and-socket joint. E5 Joint T allows rotational movement of bones in all directions.

[ any 8 ]

1 1 1 1 1 1 1 1 1 1

Max 8

8 (b)(ii) Osteoporosis P1 : the bone become thinner / more brittle / porous / fragile. P2 : Loss of bone mass. P3 : Lack of calcium / phosphorus / vitamin D Arthritis P4 : Cartilage between bones become thinner. P5 : Ligaments become shorter / loss elasticity P6 : Less production of synovial fluid. P7 : The joints become swollen / stiff / painful

[ any 4 ]

1 1 1 1 1 1 1

Max 4

Total 20

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11

No.

Marking Scheme Mark

9 (a)

(b)

The tree F1 : Less tree will be chopped / felled P1 : More CO2 absorbed by the trees for photosynthesis P2 : Avoid the increasing of CO2 in the atmosphere. P3 : Reduce the impact of Green house effect // global warming P4 : Less habitat of fauna and flora will be destroyed. P5 : Reduce / avoid the extinction of fauna and flora. P6 : To maintain / preserve the biodiversity. The oil / fuel // Save Energy F2 : Reduce the burning of oil / fuel P7 : More fuel/energy can be preserved for future. P8 : Less green house gases / acidic gases released. P9 : Reduce / avoid the impact of green house effect / acid rain. The Landfill F3 : Less landfill will be opened P10 : Landfill cause leaching / ground water pollution. P11 : Less diseases / health problem caused by the improper managed landfill. The Water F4 : Less used water / effluent / untreated sewage released into river. P12 : Reduce / avoid the impact of water pollution / avoid the extinction of aquatic organisms. Any 10 Good Effect G1 : Generate hydropower electricity G2 : As reservoir / to store water / supply fresh water G3 : Supply water for agricultural / industries. G4 : Place/site for recreation / tourism G5 : Reduce the flood problem at the downstream. Bad Effect B1 : Flooded / submerge trees / habitat of the fauna and flora B2 : Less tree / plants to carry out photosynthesis // Less CO2 absorbed for photosynthesis B3 : Amount of CO2 in the atmosphere increase B4 : Increase the impact of green house effect / global warming. B5 : Many species of fauna and flora extinct // Reduce the biodiversity. B6 : Reduce the flow of water at the downstream. B7 : Cause the population of aquatic life at the downstream reduce. B8 : Reduce the land used for residential / agricultural B9 : Flooded / destroy / loss of historical building / site. Any 10

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1

max 10

max 10

Total 20

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( Jawapan )


NEGERI KEDAH DARUL AMAN

PEPERIKSAAN PERCUBAAN SPM 2010 4551/3 BIOLOGY Kertas 3 23 September 2010 1 ½ jam

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Skema Biology P3

Percubaan SPM 2010 PKPSM Kedah

2

1 (a) Date / Tarikh Set A Set B Set C

Green / Hijau

Green / Hijau Green / Hijau

Green / Hijau

Yellowish green

Hijau kekuningan Yellow Kuning

Yellowish Green

Hijau kekuningan Yellow Kuning

Yellow Kuning

Yellow

Kuning Yellow Kuning

Yellow Kuning

3 days …………………..

2 days ………………………

1 day …………………….

Time Taken for the bananas to turn yellow / day Masa yang diambil untuk pisang menjadi kuning / hari

Score 3 : 3 ticks Score 2 : 2 ticks Score 1 : 1 tick Score 0 : 0 tick / no answer

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Skema Biology P3


3

Scoring

Correct Inaccurate Idea Wrong Score 2 - - - 3

1 1 - - - 2 - -

2

1 - 1 - - - 2 - 1 - - 1 - 1 1 -

1

- 1 - 1 1 1

0

QUESTION SCORE MARK SCHEME NOTE

KB0601 – Observation

3

Able to state two different observations correctly Sample Answers : Vertical:

1. Bananas in set C took 1 day to ripen / turn yellow

2. Bananas in set A took 3 days to ripen / turn yellow.

Horizontal: 3. In day 1/ 16 Nov, bananas in set A have turn yellow ,

bananas in Set B and Set C still in green and yellowish green colour.

4. In day 2/ 17 Nov, bananas in set A still in yellowish green, bananas in set B and set C have been turn yellow.

2 Able to state two different observations inaccurately. Sample Answers :

1. Bananas in set C turn yellow the fastest. 2. Bananas in set A turn yellow the slowest.

1 Able to state two different observations at idea level. Sample Answers :

1. Ripening process of the bananas is affected by the number of ripe mangoes.

2. Ripening process of the bananas become faster if the number of ripe mangoes increase.

1 (b) (i)

0 None of the above OR No response

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Skema Biology P3


4

Scoring

Correct Inaccurate Idea Wrong Score 2 - - - 3

1 1 - - - 2 - -

2

1 - 1 - - - 2 - 1 - - 1 - 1 1 -

1

- 1 - 1 1 1

0


KB0604 – Making inference

3

Able to state two inferences correctly Sample answers : 1. In Set C, the concentration of ethylene produced by

the mangoes is the highest. Ethylene induced the bananas to ripen faster.

2. in Set A, no ethylene induces the ripening process of the bananas. The ripening process is the slowest.

3. In day 1, bananas in Set C turn yellow/ ripen the fastest because the concentration of ethylene in Set C is the highest, ethylene induce the ripening of bananas.

4. In day 2, bananas in Set A is still unripe because no ethylene induce the ripening process.

Must have the concept I . ripe mangoes produce ethylene. II. ‘ethylene induce the ripening process’

2 Able to state two inferences inaccurately Sample answers : 1. In Set C, the concentration of ethylene produced by

the mangoes is the highest. 2. In Set A, no ethylene produced, the ripening process

of the bananas is the slowest.

Does not have concept II

1 Able to state two inferences at idea level Sample answers : 1. In Set C, the number of the mangoes is the most.

Mangoes induce bananas ripe faster. 2. In Set A, no mango induce the ripening of bananas.

Does not have concept I and concept II

1 (b) (ii)


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Skema Biology P3


5


KB0610 – Controlling variables

3

Able to state all 3 variables and the methods to handle the variable. Sample answer :

Variables Method to handle the variable Manipulated variable The number of ripe mango.

Use different number of mangoes in different set of experiment.

Responding variable Time / Day taken by the unripe bananas to ripe / turn yellow.

Count the day for the unripe bananas to turn yellow by referring to the calendar.

Constant variable 1. The size of the plastic container. 2. The unripe bananas

1. Use the same size plastic containers.. 2. Use the bananas from the same stalk of bananas.

All 6 ticks

2 4 to 5 ticks

1 2 to 3 ticks

1 (c)


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6


KB0611 – State hypothesis

3 Able to state a hypothesis relating the manipulated variable and the responding variable correctly with the following aspects : P1 = manipulated variable (the number of ripe mango / concentration of ethylene) P2 = Responding variable ( time/ day taken by the bananas to ripe / turn yellow) H = relationship (Higher/increases or inversely) Sample answers :

1. The more the number of ripe mangoes / the higher the concentration of ethylene, the faster the bananas to ripen/ become ripe / turn yellow.

Has all P1, P2 and H.

2 Able to state a hypothesis relating the manipulated variable and the responding variable inaccurately Sample answers :

1. Time / day taken by the unripe bananas to ripen is affected by / depend on the number of mangoes / concentration of ethylene. ( No H / relationship )

Has any 2 aspect

1 Able to state a hypothesis relating the manipulated variable and the responding variable at idea level Sample answer : 1. Ripe mangoes induce the unripe bananas to ripen / turn yellow. ( No P1, P2 and H )

1 (d)

0

None of the above OR No response

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Skema Biology P3


7

1(e)(i) Able to construct a table correctly with the following aspects :

1. Able to construct the table with 3 columns 2. Able to state the manipulated variable and responding variable in the table. 3. Able to record all the data in the table correctly.

Score 3 : All the 3 aspects correct Score 2 : Any 2 aspects correct Score 1 : Any 1 aspect correct Score 0 : None of the above OR no response. Sample Answer :

Set of Experiment Number of Mangoes Used

Time taken for the unripe bananas to turn yellow

Set A 0 3 Set B 1 2 Set C 2 1


KB0607 –Correlating time and space

3 Able to draw the graph correctly with the following aspects: P (paksi) : Correct title with unit on both horizontal, vertical axis and uniform scale on the axis. T(titik) : All points plotted/transferred correctly. B(bentuk): Able to join all the points to form the graph

All three aspects correct.

2 Any two correct.

1 Any one correct.

1 (e)(ii)

0


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8


KB0608 – Interpretating data

3

Able to interpret data correctly and explain with the following aspects ;

Relationship : P1 = Able to state the relationship between the manipulated variable and responding variable. Explanation : P2 = Able to state mangoes produce / release ethylene. P3 = Ethylene induce the ripening of bananas. Sample answer : 1. As the number of ripe mangoes increase, the

bananas ripen faster. This is because more ethylene is produced by the mangoes. Bananas induced by ethylene to ripen / turn yellow faster

2 Able to interpret data correctly with two aspects correctly.

1 Able to interpret data correctly with the only one aspect correctly.

1 (f)

0


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9


KB0609 – Defining by operation

3

Able to deduce about ripening process of bananas based on the results of the experiment with the following aspects. P1 : The condition of ripening process. P2 : Induced by ethylene / ripe mangoes. P3 : The higher the concentration of the ethylene , the faster the ripening process occur. Sample answer :

1. Ripening process of bananas occur when the bananas turn yellow. The process induced by ethylene / number of mangoes present. The higher the concentration of the ethylene, the faster the ripening process occur.

2 Able to define operationally based on the result of the experiment with two aspects correctly.

1 Able to define operationally based on the result of the experiment with only one aspect correctly.

1 (g)

0


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Skema Biology P3


10


KB0605 – Predicting

3

Able to predict and explain the outcome of the experiment correctly with the following aspects : Prediction : P1 : Able to predict the time taken for the unripe bananas to turn yellow / ripen. Explanation : P2 : Able to state that without the cover, ethylene diffuse into the surrounding. P3 : Able to state that less ethylene to induce the unripe bananas to ripen.

Sample answer : 1. The time / day taken for the unripe bananas in Set C

to ripen will become more than one day. This is because without the plastic cover, ethylene diffuse to the surrounding, less ethylene induces the unripe bananas to ripen.

2 Able to predict and explain the outcome of the experiment correctly with the two aspects correctly.

1 Able to predict and explain the outcome of the experiment correctly with one aspect correctly.

1 (h)

0


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11

1 (i) Able to classify the factors in Table 3 correctly.

Plant Hormones

Function / Usage

Auxin

Used to produce seedless fruits

Gibberellins

Used to promote the growth of main stem

Cytokinin

Used in storage of green vegetable.

Score 3 : 3 ticks Score 2 : 2 ticks Score 1 : 1 tick Score 0 : 0 tick

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Skema Biology P3

12

12

Question 2 Explanation Score

01 Able to state problem statement by relating P1, P2 and P3 in a question form correctly. P1- manipulated variable The deficiencies of nitrogen in culture solution/types of culture solution P2-responding variable The height of seedling/growth rate of seedling P3-question form (What …? ) Sample answer: 1. What is the effect of nitrogen deficiencies in culture solution (P1) on the height / the growth rate of seedling (P2)? (P3) 2. How does the deficiencies of nitrogen in culture solution (P1) affects the height / the growth rate of seedling (P2) ? (P3)

3 P1+P2+P3

Able to state problem statement inaccurately Sample answer: 1. What is the effect of deficiencies of nitrogen in culture solution on plants ? (P1+P3) 2. The height / growth rate of seedling is affected by the deficiencies of nitrogen in culture solution (no P3)

2 P1+P2/ P1+P3/ P2+P3

Able to state the idea Sample answer : 1. The deficiencies of nitrogen in culture solution affects the plants ( no P2 + P3)

1 P1/P2/P3

No response or wrong response 0 Explanation Score

02

Able to state the hypothesis by relating two variables correctly (P1+P2+H) P1- manipulated variable The deficiencies of nitrogen in culture solution/ the types of culture solution P2-responding variable The height of seedling/ the growth rate of seedling H-relationship Sample answer: 1. The height / growth rate of seedling (P2) is lower / slower (H) in nitrogen deficiencies of culture solution.(P1) 2. In complete culture solution (P1), the higher/ slower (H) , the height / the growth rate of seedling (P2) 3. The height / the growth rate of seedling (P2) is higher (H) in complete Knop’s solution (P1) 4. In complete Knop’s solution (P1), the height of seedling / the growth rate (P2) is higher (H)

3 P1+P2+H

Able to state any two criteria correctly or inaccurate hypothesis Sample answer: 1. The deficiencies of culture solution (P1) affect the height /growth

2 P1+P2/ P1+H/

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Skema Biology P3

13

13

rate of seedling (P2). (no H) 2. The height of seedling is higher (no P1)

P2+H

Able to draw the idea of hypothesis Sample answer: 1. The deficiencies of nitrogen in culture solution affect the plants (noP2+H)

1 P1/P2/H

No response or wrong response 0 KB061204 Explanation Score

04 Able to state K1, K2, K3, K4 and K5 (5K) correctly K1: The set up of apparatus (S1/ S2/S3/S4/S5/S6/S7/S8) (any 3 ) K2: How to manipulate the variable (S2/S3/S4 /S11) K3: How to operate the responding variable ( S10/S12) ( any 1 ) K4: How to fix the constant variable(S5/S6/S10) ( any 1 ) K5: Precautions ( S5/S6/S7/S8/S9)

S1- Three glass jars labelled A, B and C are prepared S2- In glass jar A, distilled water is fulfilled which serves as a control experiment. S3- In glass jar B, a complete culture solution is prepared using the composition of the Knop’s solution as a guide. S4- In glass jar C , a culture solution deficient in nitrogen is prepared by replacing calcium nitrate with calcium chloride and potassium nitrate is replaced by potassium chloride. . S5- Each jar is wrapped with black paper to prevent light from penetrating into the culture solution which will cause the growth of green algae. S6-Three maize seedlings of the same height are chosen and put into each jars. S7- Keep the roots of seedlings are fully immersed in each solutions. The culture solution is aerated using an air pump to ensure the root of the seedling obtain enough oxygen for respiration. S8- All set of apparatus are exposed to light so the seedling are able to carry put photosynthesis S9- The culture solution in each jar is replaced every week to ensure that the nutrients which are supposed to be available are not depleted. S10- After one month , seedling in jar A is taken out and the height of seedling is measured by using a ruler . The growth rate of the seedling is calculated and is recorded in a table . (Any abnormal

3 K1+K2+

K3+ K4+K5

(5K)

Seedling /anak benih To air pump/pam Cotton wool / kapas Culture solution / Larutan kultur Glass jar/ balang kaca Black paper/kertas hitam

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Skema Biology P3

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characteristics are not to be observed.) S11- Step S10 is repeated with seedling in glass jar B and glass jar C are observed . S12- Record the result in a table and plot a bar chart showing the growth rate of seedlings ( cm/day) against the types of solution.

Able to state any 3K – 4K correctly

2

Able to state any 1K – 2K correctly

1

Wrong response or no response

0

KB061205 Explanation Score

05 Able to list all materials and apparatus correctly to make a functional experiment and able to get the data MATERIALS (M) √ Tomato seedling/ maize seedling, Calcium nitrate Potasium nitrate √ Potasium dihydrogen phosphate Magnesium sulphate Iron (III) phosphate(trace) √ Calcium chloride √ Potasium chloride √ Distilled water √ Cotton wool √ Black paper APPARATUS (A) √ Glass jar √ Glass tubing √ L – shaped delivery tubes √ Air pump √ Rubber bung √ Ruler Notes : Score Material (M) Apparatus

(A) 3 7M 6A 2 5M

3M 3A 2A

1 2M 1M

1A 1A

3

Able to list any 5 materials and any 3 apparatus related to the experiment ( 5M + 3A / 3M + 2A )

2

Able to list any 2 material and any 1 apparatus related to the experiment (2M + 1A / 1M + 1A)

1

Wrong response or no response 0

}Notes: Accept if written as Knop’s Solution (√) only. If solutions are listed, reject if list out are incomplete

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Skema Biology P3

15

15

Explanation Score

Able to construct a table to record data with the following aspects - Titles - Data is not required

The height of seedling /(cm)

Glass Jar

Types of solution

Initial height

Final height

The growth rate of seedling / (cm/day)

A Distilled water

B Complete Knop’s Solution

C Nitrogen Deficient in culture solution

B2 = 1 mark

Construct Explanation Score Able to state the correct technique with the following aspects

Sample answer Measure the height of seedling from the tip of the shoot to the root by using ruler OR Calculate the growth rate of seedling by using formula : The growth rate of seedlings= The height of seedling (cm) Time taken (days)

B1 = 1 mark

Explanation Score

03 Able to state 7-9 aspects of experimental planning correctly : √Statement of problem √Objective √Hypothesis √Variables ( The three variables are correct) √List of materials and apparatus √Technique used √Procedure √Presentation of data √Conclusion Note: 7-9 √- 3 marks 4-6 √- 2 marks 1-3 √- 1 mark

3

Able to state any 4 - 6 items/aspects in the experimental planning correctly

2

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Skema Biology P3

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16

Able to state any 1 - 3 items correctly

1

Wrong response or no response Example: The report is in the form of explanation without planning item

0

Sample Answer : √Problem Statement What is the effect of nitrogen deficiencies in culture solution on the height /growth rate of seedling ? √Aim of experiment To study the effect of nitrogen deficiencies in culture solution on the height/ growth rate of seedling √Hypothesis The height / growth rate of seedling is lower / slower in nitrogen deficiencies of culture solution. √Variables Manipulated variable : The types of culture solution Responding variable : The height of seedling/ growth rate of seedling Constant variable : The initial height of seedling / the type of seedling √Materials Tomato seedling/ maize seedling, calcium nitrate*, potassium nitrate*, potassium dihydrogen phosphate*, magnesium sulphate*, iron (III) phosphate*, calcium chloride, potassium chloride,distilled water, cotton wool, black paper Notes: accept 5 * if it is written as Knop’s solution . Apparatus Glass jar , Glass tubing , L – shaped delivery tubes, Air pump, Rubber bung , Ruler √Techniques Measure the height of seedling from the tip of the shoot to the root by using ruler OR Calculate the growth rate of seedling by using formula : The growth rate of seedlings= The height of seedling (cm) Time taken (days) √Procedure 1. Three glass jars labelled A, B and C are prepared 2. In glass jar A, distilled water is fulfilled which serves as a control experiment. 3. In glass jar B, a complete culture solution is prepared using the composition of the Knop’s

01=3

02=3

B1=1

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Skema Biology P3

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solution as a guide. 4. In glass jar C , a culture solution deficient in nitrogen is prepared by replacing calcium nitrate with calcium chloride and potassium nitrate is replaced by potassium chloride. . 5. Each jar is wrapped with black paper to prevent light from penetrating into the culture solution which will cause the growth of green algae. 6. Three maize seedlings of the same height are chosen and put into each jars. 7. Keep the roots of seedlings are fully immersed in each solutions. The culture solution is aerated using an air pump to ensure the root of the seedling obtain enough oxygen for respiration. 8. All set of apparatus are exposed to light so the seedling are able to carry put photosynthesis 9. The culture solution in each jar is replaced every week to ensure that the nutrients which are supposed to be available are not depleted. 10. After one month , seedling in jar A is taken out and the height of seedling is measured by using a ruler . The growth rate of the seedling is calculated and then is recorded in a table . (Any abnormal characteristics on the leaves are not to be observed.) 11. Step S10 is repeated with seedling in glass jar B and glass jar C are observed . 12. Record the result in a table and plot a bar chart showing the growth rate of seedlings ( cm/day) against the types of solution. √Results

The height of seedling /(cm) Glass Jar

Types of solution

Initial height Final height

The growth rate of seedling / (cm/day)

A Distilled water

B Complete Knop’s Solution

C Nitrogen Deficient in culture solution

√Conclusion The height/ the growth rate of seedling is lower/slower in nitrogen deficiencies of culture solution. The hypothesis is accepted. Note:

7-9√ - 3 marks 4-6√ - 2 marks 1-3√ - 1 mark

03=3

17

B2= 1

j*k

j*k - trial paper collection · ( jawapan ) biology paper 1 29 ogos 2010 1 4 1 jam persidangan...

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