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    PEPERIKSAAN PERCUBAAN SPM 2010 4551/1

    Peraturan Pemarkahan( Jawapan )

    BIOLOGYPaper 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. 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

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    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 

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

    SEKOLAH MENENGAH

    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 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

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    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

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    No. Marking criteria/ Sample answers Mark

    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

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    No. Marking criteria/ Sample answers Mark

    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

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    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

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    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) 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

    11

    1

    1

    1

    1

    Max2 m

    Max

    2m

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    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

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    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 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

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    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 aslimy 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

    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 of Joint 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 /

    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

    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

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    No. Marking Scheme Mark

    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

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    Peraturan Pemarkahan

    ( Jawapan )

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

    SEKOLAH MENENGAH

    NEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2010 4551/3BIOLOGYKertas 3

    23 September 20101 ½ jam

    *k 

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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

    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 

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    Percubaan SPM 2010 PKPSM Kedah

    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 thenumber 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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    Percubaan SPM 2010 PKPSM Kedah

    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

    QUESTION SCORE MARK SCHEME NOTE

    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 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

    haveconcept I

    and

    concept II

    1 (b) (ii)

    0 None of the above OR No response

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    Percubaan SPM 2010 PKPSM Kedah

    5

    QUESTION SCORE MARK SCHEME NOTE

    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)

    0 None of the above OR No response

    *k 

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    6

    QUESTION SCORE MARK SCHEME NOTE

    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 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)

    0 None of the above OR No response

    *k 

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    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

    QUESTION SCORE MARK SCHEME NOTE

    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 None of the above OR No response

    *k 

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    8

    QUESTION SCORE MARK SCHEME NOTE

    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)

    0 None of the above OR No response

    *k 

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    Percubaan SPM 2010 PKPSM Kedah

    9

    QUESTION SCORE MARK SCHEME  NOTE

    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 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)

    0 None of the above OR No response

    *k 

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    Percubaan SPM 2010 PKPSM Kedah

    10

    QUESTION SCORE MARK SCHEME NOTE

    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)

    0 None of the above OR No response

    *k 

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    Percubaan SPM 2010 PKPSM Kedah

    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 

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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 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 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) 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 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 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

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    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 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 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

    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)

    GlassJar

    Types of solution

    Initialheight

    Finalheight

    The growth rate ofseedling / (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 marks4-6  √- 2 marks1-3  √- 1 mark

    3

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

    correctly

    2

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    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 seedling Constant 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 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 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 Knop’s

    01=3

    02=3

    B1=1

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    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 CompleteKnop’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

    B2= 1

    *k