SPM Trial 2009 Bio Ans (Pahang)

8/14/2019 SPM Trial 2009 Bio Ans (Pahang)

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MARKING SCHEME

BIOLOGY 1 (4551/1)

SPM TRIAL EXAMINATION

2009

1. A 26. B

2. B 27. C

3. B 28. B

4. A 29 D

5. D 30. B

6. D 31. D

7. B 32. B

8. D 33. A

9. C 34. C

10. C 35. C

11. D 36. B

12. B 37. B

13. B 38. B

14. A 39. A

15. C 40. C

16. B 41. C

17. D 42. B

18. D 43. C

19. C 44. D

20. A 45. D

21. C 46. A

22. A 47. B23. C 48. D

24. A 49. D

25. A 50 A

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MARKING SCHEME

BIOLOGY 2 (4551/2)


2009

No. Marking Criteria Mark1(a)

1(b)(i)

(ii)

1(c)

1 (d)(i)

1(d)(ii)

Able to name the parts labelled Q and R.

Sample answer :

Q : Carrier proteinR : Channel protein / pore protein

Able to state the component of structure P.

Sample answer :

It is composed of two layers of phospholipids

Able to explain the main function of P.

Sample answer:

Acts as a barrier between the internal and external

environment of the cell// Allows only specific molecules to pass through it// provide the structural basis for all cell membrane.

Able to give the meaning of semi-permeable.Sample answer :

A semi-permeable plasma membrane is a membrane thatallows only certain substances to move freely across it.

Able to state the concentration which is isotonic toblood plasma.

Sample answer :

0.45 g/100 cm3

Able to explain the answer in (d)(i).

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2

2

1

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1(d)(iii)

1(e)(i)

(ii)

SampleAnswer :

Both percentage of haemolysis of red blood cells andpercentage of crenation of red blood cells are zero (0%).

Able to comment on the osmotic pressure at Q.

SampleAnswer :

F : The osmotic pressure inside the red blood cells isequivalent to its environment.

P2 : Amount of water moving in and out of the cells arethe same,

P3 : therefore the size and structure of the red blood cellsdoes not change.

( F + Any P2/P3 )

Able to define active transport.

Sample answer :

Active transport is a movement of substances / molecules /ions against the concentration gradient / from low to highconcentration across the plasma membrane with the helpof carrier protein and energy / ATP.

Able to explain what will happen to the uptake of theions by root cells if the roots are immersed in asolution containing metabolic poisons such ascyanide.

Sample answer :

P1 there is no uptake of ions by root cellsP2 metabolic poisons kill/ damaged the (root) cellsP3 no energy/ ATP is producedP4 active transport does not occur

(Any three)

1

1

1

1

3

TOTAL

4

4

13 marks

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2(a)(i)

(ii)

(b)

(c)

(d)

Able to name organ X and organ Y.

Sample answer :

Organ X : Ileum // small intestineOrgan Y : Liver

Able to name molecule K, molecule M and enzyme L.Sample answer :

Molecule K : Starch

Molecule M : GlucoseEnzyme L : (Pancreatic) Amylase

Able to state two characteristics of enzyme L based onDiagram 2.1

Sample answer :

1. Enzyme remains unchanged at the end of thereaction (and can be used again).

2. Enzyme is substrate specific / reaction is very

specific

Able to explain why molecule M is needed in musclecells.

Sample answer :

Pt. 1 Molecule M / glucose is the substrate for respirationPt. 2 As the muscle cells contract and relax, energy is

needed for activitiesPt. 3 therefore, molecule M is needed in muscle cells to

provide energy from respiration process.

Able to explain the importance of forming glycogen.

Sample answer :

Pt.1 : Glycogen is the main reserve of carbohydrates inanimals

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1

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1

1

1

1

1

5

2

3

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Pt. 2 It can be converted back to glucose when energy isneeded from respiration process 1

TOTAL

2

12 marks

3(a(i))

(ii)

(b)(i)

Able to name stage X and Y.

Sample answer :

X : Prophase I

Y : Metaphase I

Able to Able to state two differences betweenchromosomal behaviour at X and Y.

Sample Answer:

Prophase I Metaphase I(Paired homologouschromosomes) arearranged randomly.

(Paired homologouschromosomes) arearranged on the

metaphase plate /equatorial plane.Spindle fibre does not holdon the centromere of thechromosomes .

Spindle fibre holds on thecentromere of thechromosomes.

(The homologouschromosomes paired and)crossing over take place.

(The homologouschromosomes paired)crossing over does nottake place.

( Any 2 )

Able to state the occurrence at Z.

Sample Answer:

P1 : Four daughter cells formedP2 : Each daughter cell has two chromosomes / haploid / n

1

1

1

1

1

11

2

2

2

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(ii)

(c)(i)

(ii)

(iii)

Able to state the chromosome number in each of thedaughter cell in Z and able to give reason.

Sample answer :

P1 : 6 (chromosomes).P2 :(During meiosis) the daughter cell receives halfthe number of chromosome from the parent cell / 2n

// Daughter cell haploid / n, parent cell diploid / 2n

Able to state either cell A, cell B and cell C aregenetically identical and explain.

Sample answer :

F : Cell A is similar to cell B but is different from cell C.P : Cell A and cell B are products of mitosis whereas cellC is a product of meiosis.

Able to state the number of chromosome in Cell if CellB undergoes an improper cell division.

Sample answer :

24 (chromosomes)

Able to state the syndrome of the individual.Sample answer :

Downs syndrome// Klinefelters syndrome

1

1

1

1

1

1

1

TOTAL

2

2

2

12 marks

4(a)

4(b)

Able to state the function of the eosin solution.

Sample answer:

To stain the xylem (vessels) (with red dye)

Able to name the parts labelled K and M.

Sample answer:

1 1

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4(c)

4(d)

(e)(i)

(ii)

(iii)

K : XylemM : Phloem

Able to name the tissue which is responsible fortransporting water and mineral ions from the roots to

the upper parts of the plant.

Sample answer:

Xylem

Able to draw and label the observation of the root cutacross.

Sample answer:

Drawing 1 mAny 2 labels 2 m

Able to state the type of transport involved in

Diagram 4.3.

Sample answer :

Translocation

Able to explain why does the part above the ringbecome swollen after two weeks.

Sample answer :

F : The products of photosynthesis cannot be transportedto the parts below the ring

P : as tissue M / phloem is removed

Able to explain why have the leaves not wiltedaftertwo weeks.

Sample answer :

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2

1

3

Xylem

PhloemPericycle

Cortex// ground tissue

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F : Water can still be transported to the leavesP : as tissue K / xylem is not removed from the stem 1

1

TOTAL

5

12 marks

5(a)

5(b)

Able to complete the drawing the appropriate neuronsinvolved in the reflex action.

Sample answer :

3 neurones 2 m2 neurones - 1 m

Able to explain the transmission of impulse from oneneurone to another neurone.

Sample answer :

Pt..1 When an impulses arrives in the axon terminalPt. 2 it stimulates (synaptic) vesicles to move towards and

bind with the presynaptic membranePt. 3 The vesicles fuse / release the neurotransmitter into

the synapsePt. 4 The neurotransmitter molecules across the synapse

to the dendrite of another neuronePt. 5 Stimulated to trigger a new impulses which travels

along the neurone( Max 4 )

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1

1

1

1

1

2

4

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5(c)

5(d)

5(e)

5 (f)

Able to name the structure M and N.

Sample answer :

M : Sensory reseptor // finger tipN : Effector // muscles tissues

Able to differentiate the reflex action with thevoluntary action.

Sample answer :

The reflex action is governed by the spinal chord whereasthe voluntary action is governed by the cerebrum.

Able to state the importance of reflex action to us.

Sample answer :

To protect the body against injuries

Able to predict the effect on O if it is injured ordamaged.

Sample answer:

1. The nerve impuls will be sent from afferent neurone tothe effector

2. The effector / muscles will not contract3. The hand will not be removed immediately from the

needle.(Any one )

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2

1

1

1

11 marks

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6(a)

(b)

Able to relate the tissues involved in producing therunning movement

Sample Answer:

Pl- Tendons, ligaments, bones, muscles and joints areimportant features in a movement,P2- Tendons connect muscles to bonesP3- Tendons are strong and non elasticP4- Force is transferred to bones through tendons.P5- Movement at the joint is possible with the aid ofligaments.P6- Ligaments connect two bones together

P7-to give support and strength to the joint.P8- Ligaments are strong and elastic.P9- The quadriceps / extensor muscles contract while thebiceps femoris muscles relax and the leg is straightened.P10- The biceps femoris muscles contract while thequadriceps / extensor muscles relax and the leg is bent.P11- Calf muscles contract to lift up the heels.P12-Feet push downward and backwardP13-Repeated contraction and relaxation of muscles result inthe running movement.

MAXIMUM: 10 marks

Able to give example and explain how the supportsystem in woody plants differs from that of non-woodyplants.

Examples 2 marks ,Facts 8 marks

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Sample answers:

Non-woody plants (herbaceous plants)Example: Balsam plant/ any suitable answer

P1: (Support in herbaceous plants is) provided by the turgidityof the parenchyma / collenchyma cellsP2: (When there is enough warm in the ground). the cells takein water by osmosis and become turgid.P3: The turgor pressure of the fluids in the vacuoles pushesthe cell contents / plasma membrane against the cell wallP4: creating support for it stem/ roots /leavesP5: The thin thickening die cell walls with cellulose /collenchyma cells gives support to herbacous plantsWoody plants :

Example : Rambutan tree/ hibiscus/ any suitable example

P6: Woody plants have specialised tissues/ sclerenchymatissues/ xylem vessels / tracheids. to give them support;P7: These tissues have cellulose walls which have depositsof lignin for added strength.P8: Sclerenchyma cells have very thick walls (which do notallow water to pass through).P9: (These cells are dead cells and) their function is to pro-vide support for the plant.P10: Xylem vessels have thick walls of lignin which are

deposited during the plant's secondary growth.P11: The lignified xylem vessels form the woody tissues of thestem.P12: This makes the plant stronger and also provides supportfor the plant.P13: Tracheids are also dead cells with thick walls and verysmalldiameters.P14: They are found with the xylem vessels and together theysupport the plants.

MAXIMUM: 10 marks

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TOTAL

7(a) (i) Able to explain how the transport of oxygen andcarbon dioxide takes place in the body cells

Sample answers:

P1: The blood circulatory system transport oxygen from thealveoli to the body cells.P2: Oxygen combines with the haemoglobin in the red bloodcellsP3: to form oxyhaemoglobin (which is unstable.)P4: Oxygen is carried (in form of oxyhaemoglobin) to thetissues (which have a low partial pressure of oxygen.)P5: The (unstable) oxyhaemoglobin breaks down into oxygenand haemoglobin again.P6: Oxygen (molecules are) transferred to the body cellsP7: Carbon dioxide binds (itself) to the haemoglobin

P8: (and is) transported in the form of carbaminohaemoglobin.P9: Carbon dioxide is (also) transported as dissolved carbondioxide (in the blood plasma.)P10: Most of carbon dioxide is carried as bicarbonate ions(dissolved in the blood plasma.)P11: When the blood carrying carbon dioxide reaches thebody cells, the carbon dioxide diffuses into the blood plasmaand combines with the red blood cells.P12:Carbon dioxide reacts with water to form carbonic acid.P13:Carbonic anhydrase in the red blood cells catalyse theformation of carbonic acid.P14: The carbonic acid then dissociates into a hydrogen ionsand bicarbonate ions.

MAXIMUM: 6 marks

(ii) Able to describe the adaptations of the alveolus forgaseous exchange

Sample answer:

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20 marks

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(b)

F1: The millions of alveoliP1: provide a large surface area for gaseous exchange.F2: The walls of the alveoli are moistP2: and this allows respiratory gases to dissolve easily tothem.

F3: The walls of the alveoli are very thin (one-cell thick)P3: for quick / easy diffusion of gases.F4: The alveoli are richly supplied with blood capillariesP4: to increase the rate of diffusion / the rate of thetransportation of gases

MAXIMUM: 4 marks

Able to explain how an oxygen debt is built up when anathlete is running and how it is settled after he stopsrunning.

Sample answer:

P1: During a vigorous exercise /running, the breathing rate isincreased.P2:This is to supply more oxygen (quickly to the muscles)P3:for rapid muscular contraction).P4: However, the supply of oxygen to muscles is stillinsufficient P5: and the muscles have to carry out anaerobicrespiration (to release energy).P6: The glucose is converted into lactic acid,P7: with only a limited amount of energy being produced

P8: An oxygen debt builds up in the body as shown in thegraphP9: High levels of lactic acid in the musclesP10: cause them to ache.P11: After running, the athlete breathes more rapidly / deeplythan normal for 20 minutes (shown in the graph)P12: There is a recovery period (from the 10th minute until the20th

P13:when oxygen is paid back (during aerobic respiration)minute)

P14: About 1/6 lactic acid is oxidised to carbon dioxide, waterand energy.

MAXIMUM: 10 marks

4

10

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TOTAL

8(a)

(b)

Able to explain how the inheritance happen

Answer:

P1: The situation involved is monohybrid inheritance.P2: The genotype of blood group A can be IAIA /1A1P3: while the genotype of blood group B can be I

0

BIB or IBIO

P4: Blood group 0 has a genotype, I.

OIO (while thegenotype of blood group AB is IAIB

P5: Alleles 1.

A and IB

P6: Iare codominant

O

P7: Mr. Nick is heterozygous dominant/Iallele is recessive.

AIO

P8: while his wife is heterozygous dominant/ I

(for his bloodgroup A)

BI0

P9: Mr. Nick and his wife produce haploidgametes/sperm/ovum (as a r e s u l t o f m e i o s i s )

(for bloodgroup B)

P10: Mr. Nick produces (gametes with) genotypes IA/IP11: (while) his wife (will) produce (gametes with) genotypes1

O

A/ lP12: The gamete (I

OO) of Mr. Nick fuses with his wife's gamete

(10

P13: to produce a zygote with genotype II)

o

P14: (Thus, they will) produce an offspring with blood group 0..

MAXIMUM: 10 marks

(i) Able to explain how DNA fingerprinting is carried out.

Answer:

P1: Tissue samples are taken from the scene of a crime and

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20 marks

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DNA is extracted.P2: An enzyme breaks down the DNA into fragments.

P3: The DNA fragments are classified according to size.

P4: An alkali is added to separate the double-stranded D N A

i n t o s i n g l e s t r a n d s .

P 5 : Each single strand is laid on a nylon membrane and

radioactive matter is added to it. A banding pattern appears.

P6: An X-ray film is produced and the positions of black bands

are compared with the part of DNA treated with radioactive

matter.

MAXIMUM: 4 marks

(ii) Able to state the advantages and disadvantages of

DNA fingerprinting

Sample answer:

Advantages:

P1:DNA fingerprinting is more accurate than common

fingerprinting as no two people have the same DNA

fingerprints.

P2: DNA fingerprinting is more efficient than blood-type

identification because many people have the s a m e

b l o o d t y p e

P3: DNA fingerprinting requires only a small amount of DNA

to obtain a highly accurate result

P4: DNA samples last longer than fingerprints.

P5: Mixed DNA samples can still be used.

P6: DNA evidence is harder to clean up compared to

fingerprints.

Disadvantages:

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P7:DNA samples may be degraded by adding

chemicals, and this will affect the accuracy of the technique.

P8: Human errors are possible when different procedures and

standards are used in DNA fingerprinting.

MAXIMUM: 6 marksTOTAL

9(a) (i) Able to explain why most plants cannot colonise andgrow in the swamps.

Sample answer:

P1: The ground is too soft and unable to support plants,P2: The water-logged / muddy swamps provide very littleoxygen for root respiration.P3: The swamp water has a high concentration of salt and ishypertonic.P4: The plants growing in swamp will have the problem ofdehydration.P5: Seeds that fall into the muddy swamp will die ofdehydration / insufficiency of oxygen.P6: The swamp is exposed to strong sunlight and intenseheat.

P7: As a result, the plants growing there will lose water veryfast by transpiration.

MAXIMUM: 5 marks

(ii) Able to explain how the mangrove trees adaptthemselves to the harsh living conditions

Sample answer:

P1: Root system which is highly branched and spreads over a

big area to give good support to the plants.P2: Pneumatophores (breathing roots) which grow protrudingupwards above the ground.P3: The plant cells have high concentration of cell sap.P4: Hence, the cells are able to withstand the high saltcontent of theswamp.P5: Excess salt is eliminated through hydatodes found at the

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(b)

lower epidermis of leaves.P6: Viviparous seeds which germinate while stillattached to the parent plant.P7: The long radical produced will let the seedling stick intothe ground and not submerge or drift away.

P8:Thick cuticle and sunken stomata which help to reduce therate of transpiration.

MAXIMUM: 5 marks

Able to describe the effects of unplanned developmentand improper management of the ecosystem.

P1: The leave canopy in the forest protects the soil fromexcess rain water.

P2: When the forest is cleared, the soil is exposed to rain

(water) / wind.

P3: this will cause soil erosion

P4:The soil that is exposed to wind will be blown to

another area,

P5: while soil that is exposed to rain water will be eroded

and deposited at the bottom of the river / pond /lake.

P6:The soil at the hill slopes can (also) be washed away by

P7: resulting in land slides.

heavy rain water

P8: (The deposited soil will) cause the water level to increase

rapidly when it rains and

P9: this will in turn cause flash floods.

P10:Wild life species will also be threatened

P11: when their habitat is destroyed.

P12: Global warming will occur

P13:due to an increase in the Earth's temperature,P14:which is caused by excess emissions of carbon

dioxide/ methane/ CFC /nitrogen dioxide (into the

atmosphere).

P15:

P16:

These gases trap the heat that is reflected by the Earth.

The thinning of the ozone layer occurs

P17: when the ozone layer (that protects the Earth from

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ultraviolet radiation) is destroyed by chlorofluorocarbons

(CFC).

MAXIMUM: 10 marks

TOTAL

10

20 marks

20 marks

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1

MARKING SCHEME

BIOLOGY 3 (4551/3)


2009

Question1

1 (a)(i) (ii) KB0603 Measuring Using Numbers

Score Explanation

3

Able to record 1 reading for theinitial mass andall 4 readings for thefinal mass of potato strips correctly .

Sample Answer:Initial mass : 50 gm

Type ofsolution

Final mass of potato strip after 30minutes

( gm)

0.2M 580.4M 520.6M 46

0.8M 42

2Able to record 1 reading for the initial mass and all 3 readings for thefinal mass of potato strips correctly .

1Able to record any 1 reading for the initial mass and all 2 readings forthe final mass of potato strips correctly .

0No response or 1 reading for the initial mass and 1 reading for the finalmass .

1 (b) (i) [KB0601 - Observation]

Score Explanation

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2

3

Able to state two correct observations based on the following criteria :[Observation must have values for MV and RV from Table 1 orcomparison between two readings.]

MV : Concentration of sucrose solution

RV : The final mass of potato strip

Sample answer:1. In 0.2M sucrose solution, the final mass of potato strip is 58 gm.2. The final mass of potato strip immersed in 0.8M sucrose solution is

42 gm.3. The final mass of potato strip immersed in 0.8M sucrose solution is

less than the final mass of potato strip immersed in 0.2M sucrosesolution // inversely

2

Able to state two different observations inaccurately OR without values.

Sample Answer:1. At concentration 0.8 M, the final mass is the lowest // inversely.2. The highest concentration of sucrose solution, the final mass is 58

gm // inversely.3. The concentration of sucrose solution influences the final mass of

potato strip .

1

Able to state two different observations at idea level.

Sample Answer:1. The concentration changes / increases / decreases.2. The final mass of potato strip changes /increases /decreases.

0No response or wrong response.

ScoringScore Correct Inaccurate Idea Wrong

3 2 - - -2 1 1 - -

- 2 - -

11 - 1 -- - 2 -- 1 1 -

0 - 1 - 1- - 1 1

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3

1 (b) (ii) [KB0604 Making inference]

Score Explanation

3

Able to state two inferences for each observation made correctly and

accurately for each observation and equivalent in 1(b)(i).

Sample answers:1. (At concentration of 0.2M),2. (At concentration of 0.8M),

water molecule diffuse into the cell.

3.water molecule diffuse out of the cell.

More water molecule diffuse out of the cell at 0.8M but more waterdiffuse into the cel

l at 0.2M.

2

Able to state any two inferences inaccurately .

Sample answers:1. More water molecule diffuse .2. The diffusion of water is influenced by concentration

1

Able to state two inferences at idea level.

Sample Answer:1. Osmosis occurs.

0 No response OR wrong response.

Scoring:Score Correct Inaccurate Idea Wrong3 2 - - -2 1 1 - -

- 2 - -

11 - 1 -- - 2 -- 1 1 -

0 - 1 - 1- - 1 1

1 (c) [KB0610 Controlling Variables]

Score Explanation

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4

3

Able to state all 3 variables and the methods to handle the variablecorrectly.Sample Answer :

Variables Method to handle the variable correctlyManipulated variable

Concentration of sucrosesolution

:

Use 0.2M,0.4M,0.6M and 0.8M / differentconcentration of sucrose solution

Responding variableFinal mass of potato strip //percentage change in massof potato strip

:Using a triple beam balance measureand record the final mass of potato strip

// Calculate the percentage change inmass of potato using formula:Final mass-Initial mass

Initial massX 100%

Constant variable1. Duration of immersion /:

2. Length / mass of thepotato strip / Volume ofsucrose solution

1. Fix the time of 30 minutes toimmersedthe potato strip.

2. Used the length of 5mm / Fix themass at50 gm.

3. Fix / used the volume at 20ml

Reject way how to handle the variable if variable is wrong

2Able to state 4-5 correctly.

1Able to state 2-3 correctly

0No response or only one criteria correct.

1 (d) [KB0611 Making Hypothesis]

Score Explanation

3

Able to state a hypothesis by relating the manipulated variable andresponding variable correctly with following aspects:

P1 : Stating manipulated variable.P2: Stating responding variableH : Showing a specific relationship/ showing direction of relationship

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5

Sample Answer :1. As the concentration of sucrose solution increases, the final mass

of potato strip decreases // the percentage change in massdecreases.

2. As the concentration of sucrose solution increases / decreases,the final mass of potato strip increases /decreases // thepercentage change in mass increases // decreases

2

Able to state a hypothesis relating the manipulated variable inaccurately.

Sample Answer:1. The increase of the concentration of sucrose solution influences /

affects the final mass of potato strip.2. The percentage change in mass of potato strip is affected by

concentration of sucrose solution.

1

Able to state a hypothesis relating the manipulated variable at idea level.

Sample Answer :1. Final mass of potato strip / concentration of sucrose solution

changes.2. As the final mass of potato strip increases the percentage change

increases.

0No response or wrong response if no P1 or P2 no mark for each.

1 (e) (i) [KB0606 Communication]

Score Explanation

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6

3

Able to construct a table correctly with the following aspects:

1 : Able to state the 4 titles with units correctly.2 : Able to record all the data correctly.3 : Able to calculate and record percentage change correctly

Sample answer :Concentration ofsucrose solution

(M)

Initialmass(gm)

Finalmass(gm)

Percentage change inmassFinal mass-Initial mass

Initial mass

X100

(%)0.2 50 58 16.00.4 50 52 4.00.6 50 46 -8.0

0.8 50 42 -16.02

Any two aspects correct

1Any one aspect correct


1 (e)(ii) [KB0612 Relationship between space and time]

Score Explanation

3

Able to draw the graph correctly with the following aspects:

P(paksi) : Corrected title with unit on both horizontal, vertical axis anduniform scale on the axis.

T(titik) : All points plotted / transferred correctly.B(bentuk) : Able to join the points to form a smooth graph / line

2Able to state any two correct.

1Able to state any one correct


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7

1 (f) [KB0608 Interpreting Data]

Score Explanation

3

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

Relationship :

P1 = Able to state the concentration of sucrose solution which is isotonicto the cell sap

ExplanationP2 = K Linmacam tak lengkap aje..P3 = Able to state

Sample answer :

Produced (P3).

2Able to interpret data with two aspects correctly.

1Able to interpret data only one aspect correctly.


1 (g) [KB0609 Defining by Operation ]

Score Explanation

3

Able to define operationally based on the result of the experiment with thefollowing aspects:

P1 : Movement of water in / outP2 : Plasma membrane of the potato cellsP3 : Difference in concentration gradient between the sucrose solution

and the cell sap.

Sample answer:

1. Osmosis is a process in which water molecules entering / leaving thepotato strips (P1) across the plasma membrane of the potato strip (P2)when there is a difference in concentration gradient between the

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sucrose solution and the cell sap (P3).

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

1Able to define operationally based on the result of the experiment with

only one aspects correctly.0 No response or wrong response

1 (h) [KB0605 Predicting]

Score Explanation

3

Able to predict and explain the outcome of the experiment correctly withthe following aspects:

Prediction :P1 : Able to predict the mass of the potato strip correctly.

Explanation :P2 : Able to state distilled water is hypotonicP3 : Able to state more water molecules diffuse into the potato strip

Sample answer:1. The mass of the potato strip increases more than 46 gm // any

values more than 46 gm .Water molecules diffused into the potato

strip because distilled water is hypotonic.

** P1 must be correct to get P2 & P3, if P1 wrong automaticallyreject P2 & P3 - for score 3, 2, 1

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

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


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9

1(i) [KB0602 Classifying]

3

Able to classify all 3 solutions concentration and types of solutioncorrectly:

Solutionconcentrations

Kepekatan larutan(%)

Types of solution compared tothe osmotic concentration of cellsapJenis larutan dibandingkandengan kepekatan osmotik sapsel

0.25% natrium chloridesolution Hypotonic

0.8% natrium chloride

solutionIsotonic

1.10% natrium chloride

solution.Hypertonic

2Able to classify 2 solutions concentration and types of solution correctly

1Able to classify 1 solution concentration and types of solution correctly

0 No response or wrong response

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10

Question 2

KB061201 ( Problem statement)

Score Criteria

3

Able to state the problem statement correctly :

C1 : Manipulated variable

C2 : Responding variable

R : Question form and have relationship

Sample Answer :

1. How does air movement affect the rate of transpiration ?2. How does ( wind) affect the rate of transpiration ?** Without question mark (?) score 2

2

Able to give a statement of identified problem but incomplete.

Sample Answer:

1. Does air movement have a relationship with the rate of

transpiration ?

2. Does wind have a relationship with the rate of

transpiration ?

1

Able to give idea of a statement of identified problem.

Sample Answer:

1. What is the effect of air movement / wind?

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KB061202( Making Hypothesis )

Score Criteria

3

Able to state the hypothesis correctly by relating two variables.

Criteria set:

C1 : State the manipulated variable

C2 : State the responding variable

R : Show the specific relationship and direction between the

manipulated variable and the responding variable.

Answer must have C1, C2 and R

Sample Answer :

1. The higher the air movement , the higher the rate of

transpiration.

2. The higher the wind , the higher the rate of

transpiration

2Able to make a statement of hypothesis which relates the

manipulated variable to the responding variable.

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12

Answer must have C1 and C2 but without correct relationship

Sample Answer :

1. The air movement affect the rate of transpiration2. The wind affect the rate of transpiration

1

Able to state an idea of a statement of hypothesis.

Sample Answer:

1. The air movement / wind increases.


KB061203 - Planning ( Planning for investigation)

Score Criteria

3

Scoring Criteria :

Able to state 7- 9 planning investigation of experiment following :

Problem statement (PS) idea Aim of investigation / Objective (Ob) Relation between C1

and C2

Sample answer

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1. To investigate the effect of air movement / wind on the rate of

transpiration.

Statement of hypothesis (Hp) idea States variables (Vr)

All threevariables must be correct :

Manipulated variable : Moving air / wind

Responding Variable : Time taken by the air bubble to move

a distance of 2 cm.

Constant Variable : Temperature / Relative humidity / light

intensity

List of materials and apparatus (AM) Technique (Tq) Correctly and accurately

Bonus 1/B1 = 1 mark

Sample Answer:

Record the time taken for the bubble to move a distance of 2 cm

using a stopwatch.

Procedure / Method of investigation (K) must have at least onecriteria either K1 / K2 / K3 /K4 /K5

Data presentation // presentation of result (RD) Have tablewith 3 titles with correct units and no data is required

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Sample Answer :

External condition Time taken for the air bubble

to move a distance of 2 cm

(seconds)

Rate of

transpiration

(mms-1)

1 2 3 Average

Using the fan / air

movement ( wind)

Without fan / no

air movement

Bonus 2/B2: 1 mark

Conclusion (Cn) Must be the same with hypothesis. If

hypothesis is wrong, rejectconclusion.

Sample answer :

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

( Hypothesis is accepted)

If students only write hypothesis accepted in conclusion, reject

conclusion.

2Scoring Criteria :

State 4 - 6 items

1 Scoring Criteria:

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State 1 - 3 items


KB061204 ( Method / procedure of investigation)

Score Criteria

3

Able to state all five criteria K1,K2, K3, K4 and K5 :

Criteria :

K1 : Preparation of materials & apparatus

- Cut a leafy balsam / named plant shoot

(any 5)

- The capillary tube and rubber tubing are filled with water.

- The leafy shoot is inserted into the rubber tubing under water.

- The capillary tube is lifted up above the water surface to trap an air

bubble.

- The capillary tube is placed back into the beaker of water and kept

upright using a retort stand.

- A fan is used to create condition air movement .

- Steps 5-9 is repeated twice to obtained an average reading.

- The experiment is repeated without using the fan .

- The results are recorded in a table.

- The rate of transpiration is calculated.

Remark :

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Able to state any five (K) steps to get P1.

K2 : Operating Fixed variable

- The potometer is placed at room temperature ,37 o

C

K3 :

- Record the time taken by the air bubble to move a distance

Operating responding variable

of 2 cm using a stopwatch

K4 :

- Repeat the experiment without using fan.

Operating manipulated variable

K5 : Precaution / Accuracy of experiment

State one precaution steps in the experiment.

(Any 1)

Sample Answer:

- Cut a leafy balsam / named plant shoot slantly under water

- The leaves are wiped dry .

- Some vaseline Is smeared around the rubber tubing

Sample Answer:

Method / Procedure :

1. Cut a leafy balsam / named plant shoot slantly under water to

prevent air from entering the xylem.

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1. The capillary tube and rubber tubing are filled with water.2. The leafy shoot is inserted into the rubber tubing under water.3. The leaves are wiped dry. Some vaseline is smeared around the

rubber tubing to make the apparatus airtight.4. The potometer is placed in a beaker of water for 5 minutes at

room temperature to allow water to move up the capillary tube.5. The capillary tube is lifted up above the water surface to trap anair bubble.

6. The capillary tube is placed back into the beaker of water andkept upright using a retort stand.

7. Mark two points , P and Q at a distance of 2 cm apart on thecapillary tube.

8. A fan is used to give air movement.9. A stopwatch is activated and the distance travel by the air

bubble from P to Q point is recorded.10. Steps 5-9 is repeated twice to obtained an average reading.

11. The experiment is repeated without using fan.12. The results are recorded in a table.13. The rate of transpiration is calculated.

2 Able to state 4 criteria

1 Able to state 2-3 criteria


KB061205 (Listing of Materials and Apparatus)

Skor Perkara

3

Abble to state all the materials and apparatus:

Sample Answer:

Materials :

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*Leafy balsam shoot,

*distilled water

vaselin

Apparatus :

*capillary tube,* rubber tubing,

beaker,*stopwatch,*fan

marker pen/thread, tissue

paper/cloth , retort stand

2 Able to state two of the * materials and four * apparatus

1 Able to state two of the * materials and two *apparatus including


Mark:

3 X 5 = 15 marks

B1 = 1 mark( technique)

B2 = 1 mark( Data presentation)

TOTAL = 17 marks

END OF MARKNG SCHEME

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SPM Trial 2009 Bio Ans (Pahang)

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