2010 PSPM Kedah Biology 123 w Ans

8/8/2019 2010 PSPM Kedah Biology 123 w Ans

1/93

*k


2/93

*k


3/93

*k


4/93

*k


5/93

*k


6/93

*k


7/93

*k


8/93

*k


9/93

*k


10/93

*k


11/93

*k


12/93

*k


13/93

*k


14/93

*k


15/93

*k


16/93

*k


17/93

*k


18/93

*k


19/93

*k


20/93

*k


21/93

*k


22/93

*k


23/93

*k


24/93

*k


25/93

*k


26/93

*k


27/93

*k


28/93

*k


29/93

*k


30/93

*k


31/93

*k


32/93

*k


33/93

*k


34/93

*k


35/93

*k


36/93

*k


37/93

*k


38/93

*k


39/93

*k


40/93

*k


41/93

*k


42/93

*k


43/93

*k


44/93

*k


45/93

*k


46/93

*k


47/93

*k


48/93

*k


49/93

*k


50/93

*k


51/93

*k


52/93

*k


53/93

*k


54/93

*k


55/93

*k


56/93

*k


57/93

*k


58/93

*k


59/93

*k


60/93

*k


61/93

*k


62/93

*k


63/93

*k


64/93

PEPERIKSAAN PERCUBAAN SPM 2010 4551/1

Peraturan Pemarkahan( Jawapan )

BIOLOGYPaper 1

29 Ogos 2010

1

4

1jam

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

SEKOLAH MENENGAH

NEGERI KEDAH DARULAMAN

*k


65/93

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. C10. 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. B19. 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

*k


66/93

PEPERIKSAAN PERCUBAAN SPM 2010

BIOLOGY

PAPER 2

Two hours and thirty minutes

Peraturan Pemarkahan

( Jawapan )

4551/2

Biology

Kertas 2

29 Ogos 2010

2

12 hours


SEKOLAH MENENGAH

NEGERI KEDAH DARUL AMAN

*k


67/93

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 carbondioxide 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

*k


68/93

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 solutionP3 : 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

11 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 wallP2 : Cell wall is made up of cellulose

// Cell wall able to withstand the pressure.

Any 2

1

11 2

Total 12

*k


69/93

4


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, aroundinternal 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

*k


70/93

5


4 (a)

Both arrows correct 1

(b) A Pulmonary artery

B Pulmonary vein

1

1 2

(c) F : Contraction of ventricle / heartE1: generates a (high) pressure

E2 : (to) propel/ force / pump the blood flow from the heart/ ventricle to

vessel A

Any two

11

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 infarctionReject 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

*k


71/93

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

11

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

11

1 3

Total 12

*k


72/93

7

BIOLOGY

SECTION A

PAPER 2 [4551/2] - ESSAYNo. 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) AlbinismeF : 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 anaemiaF : 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

11

1

1

1

1

Max2 m

Max

2m

*k


73/93

8

6(c) (i) Abiotic factors that cause variation between the two sets of ginger plantsare:

F1: Sun light

P1: Plants need light energy to carry out photosynthesis for

growthP2: 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

11

F3: Soil / mineralsP7: 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

11

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 thesynaptic 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

*k


74/93

9


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 neuroneTo 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

11

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

11

1

1

1

1

1

11

1

1

1

1

1 Max 10

Total 20

*k


75/93

10


8 (a)(i) P1 fish have streamline shapes // the anterior of the fish issmooth and rounded // the body is long and tapers

towards the end.

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

1

1 2

(a)(ii) P1 myotomes muscles are arranged in both side of the bodyP2 the vertebral column of the fish is flexible and can bent

from side to side

P3 myotome muscles act antagonistically in fish./ carry outopposite action in a fish

P4 when the muscles on right side contract, the muscle onthe 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 blockenable 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

11

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 ofJoint S and

Joint T are covered with cartilage

E2 To protect the bone / reduce friction between the bonesF3 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 jointE4 Joint S allows the movement of bones in one plane /

directionD2 Joint T is ball-and-socket joint.

E5 Joint T allows rotational movement of bones in

all directions.

[ any 8 ]

1

1

1

11

1

1

1

1

1

Max

8

8 (b)(ii) OsteoporosisP1 : 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

*k


76/93

11


9 (a)

(b)

The tree

F1 : Less tree will be chopped / felledP1 : 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 LandfillF3 : 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 electricityG2 : As reservoir / to store water / supply fresh water

G3 : Supply water for agricultural / industries.

G4 : Place/site for recreation / tourismG5 : 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

11

11

1

1

1

1

1

1

1

1

1

1

1

1

11

1

11

1

1

1

1

1

1

1

1

1

max

10

max

10

Total 20

*k


77/93

Peraturan Pemarkahan

( Jawapan )


SEKOLAH MENENGAH

NEGERI KEDAH DARUL AMAN

PEPERIKSAAN PERCUBAAN SPM 2010 4551/3BIOLOGYKertas 3

23 September 20101 jam

*k


78/93

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 greenHijau kekuningan

YellowKuning

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 tickScore 0 : 0 tick / no answer

*k


79/93

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 yellow2. 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 yellowishgreen, 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 thenumber of ripe mangoes.

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

1 (b) (i)

0 None of the above OR No response

*k


80/93

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 thebananas to ripen faster.

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

3. In day 1, bananas in Set C turn yellow/ ripen thefastest 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 bythe mangoes is the highest.

2. In Set A, no ethylene produced, the ripening processof 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

haveconcept I

and

concept II

1 (b) (ii)


*k


81/93

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.

Respondingvariable

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 theplastic container.

2. The unripe

bananas

1. Use the same size plasticcontainers..

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)


*k


82/93

Skema Biology P3


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 higherthe 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 variableand 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)


*k


83/93

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 columns2.

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)


*k


84/93

Skema Biology P3


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, thebananas ripen faster. This is because more ethyleneis 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)


*k


85/93

Skema Biology P3


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 thebananas turn yellow. The process induced by

ethylene / number of mangoes present. The higher

the concentration of the ethylene, the faster theripening 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)


*k


86/93

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 Cto 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 experimentcorrectly with one aspect correctly.

1 (h)


*k


87/93

Skema Biology P3


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 tickScore 0 : 0 tick

*k


88/93

Skema Biology P3 12

12

Question 2

Explanation Score

01 Able to state problem statement by relating P1, P2 and P3 in aquestion form correctly.

P1- manipulated variableThe 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 inaccuratelySample 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)

2P1+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 ofculture solution

P2-responding variableThe 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 Knops solution (P1)

4. In complete Knops 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/

*k


89/93

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) correctlyK1: 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 Knops 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 ensurethat 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 theseedling 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

*k


90/93

Skema Biology P3 14

14

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 dataMATERIALS (M)

Tomato seedling/ maize seedling,Calcium nitrate

Potasium nitrate

Potasium dihydrogen phosphateMagnesium sulphate

Iron (III) phosphate(trace)

Calcium chloride Potasium chloride Distilled water

Cotton wool Black paper

APPARATUS (A)

Glass jar Glass tubingL shaped delivery tubesAir pump Rubber bung Ruler

Notes :Score Material (M) Apparatus

(A)

3 7M 6A

2 5M

3M

3A

2A

1 2M1M

1A1A

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

Knops Solution ()only.

If solutions are listed,

reject if list out are

incomplete

*k


91/93

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)

GlassJar

Types of solution

Initialheight

Finalheight

The growth rate ofseedling / (cm/day)

A Distilled water

B Complete Knops

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 problemObjectiveHypothesisVariables ( The three variables are correct)List of materials and apparatus

Technique usedProcedurePresentation of dataConclusion

Note:

7-9 - 3 marks4-6 - 2 marks1-3 - 1 mark

3

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

correctly

2

*k


92/93

Skema Biology P3 16

16

Able to state any 1 - 3 items correctly 1

Wrong response or no responseExample:

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 ofseedling

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 seedlingConstant variable : The initial height of seedling / the type of seedling

MaterialsTomato 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 Knops 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 prepared2. 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 Knops

01=3

02=3

B1=1

*k


93/93

Skema Biology P3 17

solution as a guide.

4. In glass jar C , a culture solution deficient in nitrogen is prepared by replacing calcium nitratewith 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 isaerated 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 CompleteKnops 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

B2= 1

*k

2010 PSPM Kedah Biology 123 w Ans

Text of 2010 PSPM Kedah Biology 123 w Ans

2011 PSPM Kedah Physics 3 w Ans

2012 PSPM Kedah Sains 1 w Ans

2011 PSPM Kedah BM 2 w Ans

2015 PSPM Kedah BC2 w Ans

2015 PSPM Kedah BI2 w Ans

2015 PSPM Kedah Fizik1 w Ans

2010 PSPM Kedah PM w Ans

2015 PSPM Kedah Fizik3 w Ans

2011 PSPM Kedah PI 1 w Ans

2012 PSPM Kedah Moral w Ans

2010 Pspm Kedah Mm12 w Ans

2011 PSPM Kedah Biologi 3 w Ans

2015 PSPM Kedah Kimia2 w Ans

2011 PSPM Kedah PI 2 w Ans

2011 PSPM Kedah Chemistry 2 w Ans

2015 PSPM Kedah Fizik2 w Ans

2012 PSPM Kedah Kimia 3 w Ans

2015 PSPM Kedah Kimia1 w Ans

2011 PSPM Kedah Perdagangan 2 w Ans

2015 PSPM Kedah Kimia3 w Ans

2011 PSPM Kedah Math 1 w Ans

2011 PSPM Kedah BI 2 w Ans

2011 PSPM Kedah Math 2 w Ans

2011 PSPM Kedah Chemistry 1 w Ans

2010 PSPM Kedah Perdagangan12 w Ans

2011 PSPM Kedah Biologi 1 w Ans

2015 PSPM Kedah BM2 w Ans

2010 PSPM Kedah PAkaun12 w Ans

2010 Pspm Kedah Bm12 w Ans

2010 PSPM Kedah Sejarah12 w Ans

Documents

2010 PSPM Kedah Biology 123 w Ans

Text of 2010 PSPM Kedah Biology 123 w Ans