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. B

23. C 48. D

24. A 49. D

25. A 50 A

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MARKING SCHEME BIOLOGY 2 (4551/2)

SPM TRIAL EXAMINATION 2009

No. Marking Criteria Mark 1(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 protein R : 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 that allows only certain substances to move freely across it. Able to state the concentration which is isotonic to blood plasma. Sample answer : 0.45 g/100 cm3

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

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1(d)(iii) 1(e)(i) (ii)

SampleAnswer : Both percentage of haemolysis of red blood cells and percentage 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 is equivalent to its environment. P2 : Amount of water moving in and out of the cells are the same, P3 : therefore the size and structure of the red blood cells does 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 high concentration across the plasma membrane with the help of carrier protein and energy / ATP. Able to explain what will happen to the uptake of the ions by root cells if the roots are immersed in a solution containing metabolic poisons such as cyanide. Sample answer : P1 – there is no uptake of ions by root cells P2 – metabolic poisons kill/ damaged the (root) cells P3 – no energy/ ATP is produced P4 – active transport does not occur (Any three)

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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 intestine Organ Y : Liver Able to name molecule K, molecule M and enzyme L. Sample answer : Molecule K : Starch Molecule M : Glucose Enzyme L : (Pancreatic) Amylase Able to state two characteristics of enzyme L based on Diagram 2.1 Sample answer :

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

2. Enzyme is substrate specific / reaction is very specific

Able to explain why molecule M is needed in muscle cells. Sample answer : Pt. 1 Molecule M / glucose is the substrate for respiration Pt. 2 As the muscle cells contract and relax, energy is needed for activities Pt. 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 in animals

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Pt. 2 It can be converted back to glucose when energy is needed from respiration process

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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 between chromosomal behaviour at X and Y. Sample Answer: Prophase I Metaphase I (Paired homologous chromosomes) are arranged randomly.

(Paired homologous chromosomes) are arranged on the metaphase plate / equatorial plane.

Spindle fibre does not hold on the centromere of the chromosomes .

Spindle fibre holds on the centromere of the chromosomes.

(The homologous chromosomes paired and) crossing over take place.

(The homologous chromosomes paired) crossing over does not take place.

( Any 2 ) Able to state the occurrence at Z. Sample Answer: P1 : Four daughter cells formed P2 : Each daughter cell has two chromosomes / haploid / n

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(ii)

(c)(i)

(ii)

(iii)

Able to state the chromosome number in each of the daughter cell in Z and able to give reason. Sample answer : P1 : 6 (chromosomes). P2 :(During meiosis) the daughter cell receives half the 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 are genetically 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 cell C is a product of meiosis. Able to state the number of chromosome in Cell if Cell B undergoes an improper cell division. Sample answer : 24 (chromosomes) Able to state the syndrome of the individual. Sample answer : Down’s syndrome // Klinefelter’s syndrome

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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 :

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4(c)

4(d)

(e)(i)

(ii) (iii)

K : Xylem M : Phloem Able to name the tissue which is responsible for transporting 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 cut across. Sample answer :

Drawing – 1 m Any 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 ring become swollen after two weeks. Sample answer : F : The products of photosynthesis cannot be transported

to the parts below the ring P : as tissue M / phloem is removed Able to explain why have the leaves not wilted after two weeks. Sample answer :

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Xylem

Phloem Pericycle

Cortex // ground tissue

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

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12 marks

5(a)

5(b)

Able to complete the drawing the appropriate neurons involved in the reflex action. Sample answer :

3 neurones – 2 m 2 neurones - 1 m Able to explain the transmission of impulse from one neurone to another neurone. Sample answer : Pt..1 When an impulses arrives in the axon terminal Pt. 2 it stimulates (synaptic) vesicles to move towards and bind with the presynaptic membrane Pt. 3 The vesicles fuse / release the neurotransmitter into the synapse Pt. 4 The neurotransmitter molecules across the synapse to the dendrite of another neurone Pt. 5 Stimulated to trigger a new impulses which travels along the neurone ( Max 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 tip N : Effector // muscles tissues

Able to differentiate the reflex action with the voluntary action. Sample answer : The reflex action is governed by the spinal chord whereas the 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 or damaged. Sample answer: 1. The nerve impuls will be sent from afferent neurone to the effector 2. The effector / muscles will not contract 3. The hand will not be removed immediately from the needle.

(Any one )

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11 marks

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6(a)

(b)

Able to relate the tissues involved in producing the running movement Sample Answer: Pl- Tendons, ligaments, bones, muscles and joints are important features in a movement, P2- Tendons connect muscles to bones P3- Tendons are strong and non elastic P4- Force is transferred to bones through tendons. P5- Movement at the joint is possible with the aid of ligaments. 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 the biceps femoris muscles relax and the leg is straightened. P10- The biceps femoris muscles contract while the quadriceps / extensor muscles relax and the leg is bent. P11- Calf muscles contract to lift up the heels. P12-Feet push downward and backward P13-Repeated contraction and relaxation of muscles result in the running movement.

MAXIMUM: 10 marks Able to give example and explain how the support system in woody plants differs from that of non-woody plants. 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 turgidity of the parenchyma / collenchyma cells P2: (When there is enough warm in the ground). the cells take in water by osmosis and become turgid. P3: The turgor pressure of the fluids in the vacuoles pushes the cell contents / plasma membrane against the cell wall P4: creating support for it stem/ roots /leaves P5: The thin thickening die cell walls with cellulose / collenchyma cells gives support to herbacous plants Woody plants : Example : Rambutan tree/ hibiscus/ any suitable example P6: Woody plants have specialised tissues/ sclerenchyma tissues/ xylem vessels / tracheids. to give them support; P7: These tissues have cellulose walls which have deposits of lignin for added strength. P8: Sclerenchyma cells have very thick walls (which do not allow 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 the stem. P12: This makes the plant stronger and also provides support for the plant. P13: Tracheids are also dead cells with thick walls and very small diameters. P14: They are found with the xylem vessels and together they support the plants.

MAXIMUM: 10 marks

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7(a)

(i) Able to explain how the transport of oxygen and carbon dioxide takes place in the body cells Sample answers: P1: The blood circulatory system transport oxygen from the alveoli to the body cells. P2: Oxygen combines with the haemoglobin in the red blood cells P3: to form oxyhaemoglobin (which is unstable.) P4: Oxygen is carried (in form of oxyhaemoglobin) to the tissues (which have a low partial pressure of oxygen.) P5: The (unstable) oxyhaemoglobin breaks down into oxygen and haemoglobin again. P6: Oxygen (molecules are) transferred to the body cells P7: Carbon dioxide binds (itself) to the haemoglobin P8: (and is) transported in the form of carbaminohaemoglobin. P9: Carbon dioxide is (also) transported as dissolved carbon dioxide (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 the body cells, the carbon dioxide diffuses into the blood plasma and 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 the formation of carbonic acid. P14: The carbonic acid then dissociates into a hydrogen ions and bicarbonate ions.

MAXIMUM: 6 marks (ii) Able to describe the adaptations of the alveolus for gaseous exchange Sample answer:

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(b)

F1: The millions of alveoli P1: provide a large surface area for gaseous exchange. F2: The walls of the alveoli are moist P2: and this allows respiratory gases to dissolve easily to them. 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 capillaries P4: to increase the rate of diffusion / the rate of the transportation of gases

MAXIMUM: 4 marks Able to explain how an oxygen debt is built up when an athlete is running and how it is settled after he stops running. Sample answer: P1: During a vigorous exercise /running, the breathing rate is increased. 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 still insufficient P5: and the muscles have to carry out anaerobic respiration (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 the graph P9: High levels of lactic acid in the muscles P10: cause them to ache. P11: After running, the athlete breathes more rapidly / deeply than normal for 20 minutes (shown in the graph) P12: There is a recovery period (from the 10th minute until the 20th

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

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

MAXIMUM: 10 marks

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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 the genotype of blood group AB is IAIB

P5: Alleles 1.

A and IBP6: I

are codominant O

P7: Mr. Nick is heterozygous dominant/I allele is recessive.

AIO

P8: while his wife is heterozygous dominant/ I

(for his blood group A)

BI0

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

(for blood group B)

P10: Mr. Nick produces (gametes with) genotypes IA /IP11: (while) his wife (will) produce (gametes with) genotypes 1

O

A/ lP12: The gamete (I

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

(10

P13: to produce a zygote with genotype I°I)

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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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 marks

TOTAL

9(a)

(i) Able to explain why most plants cannot colonise and grow in the swamps. Sample answer: P1: The ground is too soft and unable to support plants, P2: The water-logged / muddy swamps provide very little oxygen for root respiration. P3: The swamp water has a high concentration of salt and is hypertonic. P4: The plants growing in swamp will have the problem of dehydration. P5: Seeds that fall into the muddy swamp will die of dehydration / insufficiency of oxygen. P6: The swamp is exposed to strong sunlight and intense heat. P7: As a result, the plants growing there will lose water very fast by transpiration.

MAXIMUM: 5 marks (ii) Able to explain how the mangrove trees adapt themselves 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 protruding upwards above the ground. P3: The plant cells have high concentration of cell sap. P4: Hence, the cells are able to withstand the high salt content of the swamp. 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 still attached to the parent plant. P7: The long radical produced will let the seedling stick into the ground and not submerge or drift away. P8:Thick cuticle and sunken stomata which help to reduce the rate of transpiration.

MAXIMUM: 5 marks Able to describe the effects of unplanned development and improper management of the ecosystem. P1: The leave canopy in the forest protects the soil from excess 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

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20 marks

20 marks

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MARKING SCHEME

BIOLOGY 3 (4551/3)

SPM TRIAL EXAMINATION

2009

Question1

1 (a)(i) (ii) KB0603 – Measuring Using Numbers Score Explanation

3

Able to record 1 reading for the initial mass and all 4 readings for the final mass of potato strips correctly . Sample Answer: Initial mass : 50 gm Type of solution

Final mass of potato strip after 30 minutes

( gm)

0.2M 58 0.4M 52 0.6M 46 0.8M 42

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

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

0

No response or 1 reading for the initial mass and 1 reading for the final mass .

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 or comparison 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 sucrose solution // 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.

0 No response or wrong response.

Scoring Score Correct Inaccurate Idea Wrong 3 2 - - - 2 1 1 - -

- 2 - - 1

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

0 - 1 - 1 - - 1 1

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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 water diffuse 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 Wrong

3 2 - - - 2 1 1 - -

- 2 - - 1

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

0 - 1 - 1 - - 1 1

1 (c) [KB0610 – Controlling Variables] Score Explanation

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3

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

Variables Method to handle the variable correctly Manipulated variableConcentration of sucrose solution

:

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

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

:

Using a triple beam balance measure and record the final mass of potato strip // Calculate the percentage change in mass of potato using formula: Final mass-Initial mass Initial mass

X 100%

Constant variable1. Duration of immersion /

:

2. Length / mass of the potato strip / Volume of sucrose solution

1. Fix the time of 30 minutes to

immersed the potato strip.

2. Used the length of 5mm / Fix the mass at 50 gm.

3. Fix / used the volume at 20ml

Reject way how to handle the variable if variable is wrong

2 Able to state 4-5 correctly.

1 Able to state 2-3 correctly

0 No response or only one criteria correct.

1 (d) [KB0611 – Making Hypothesis] Score Explanation

3

Able to state a hypothesis by relating the manipulated variable and responding variable correctly with following aspects: P1 : Stating manipulated variable. P2: Stating responding variable H : 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 mass decreases.

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

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

1 (e) (i) [KB0606 – Communication] Score Explanation

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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 of sucrose solution

(M)

Initial mass (gm)

Final mass (gm)

Percentage change in mass Final mass-Initial mass

Initial mass

X 100

(%) 0.2 50 58 16.0 0.4 50 52 4.0 0.6 50 46 -8.0 0.8 50 42 -16.0

2 Any two aspects correct

1 Any one aspect correct

0 No response or wrong response.

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

2 Able to state any two correct.

1 Able to state any one correct

0

No response or wrong response.

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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 isotonic to the cell sap Explanation P2 = K Linmacam tak lengkap aje……………………………….. P3 = Able to state Sample answer : Produced (P3).

2

Able to interpret data with two aspects correctly.

1

Able to interpret data only one aspect correctly.

0

No response or wrong response.

1 (g) [KB0609 –Defining by Operation ] Score Explanation

3

Able to define operationally based on the result of the experiment with the following aspects: P1 : Movement of water in / out P2 : Plasma membrane of the potato cells P3 : 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 the

potato 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).

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 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 with the following aspects: Prediction : P1 : Able to predict the mass of the potato strip

correctly.

Explanation : P2 : Able to state distilled water is hypotonic P3 : 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 automatically reject P2 & P3 - for score 3, 2, 1

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

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

0

No response or wrong response.

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1(i) [KB0602 – Classifying]

3

Able to classify all 3 solutions concentration and types of solution correctly:

Solution concentrations

Kepekatan larutan (%)

Types of solution compared to the osmotic concentration of cell sap Jenis larutan dibandingkan dengan kepekatan osmotik sap sel

0.25% natrium chloride solution

Hypotonic

0.8% natrium chloride solution

Isotonic

1.10% natrium chloride solution.

Hypertonic

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

1

Able to classify 1 solution concentration and types of solution correctly

0 No response or wrong response

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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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0 No response or wrong response

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

2 Able to make a statement of hypothesis which relates the manipulated variable to the responding variable.

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Answer must have C1 and C2 but without correct relationship

Sample Answer :

1. The air movement affect the rate of transpiration 2. 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.

0 No response or wrong response

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 three variables 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 one criteria either K1 / K2 / K3 /K4 /K5

• Data presentation // presentation of result (RD) – Have table with 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, reject conclusion.

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.

2 Scoring Criteria :

State 4 - 6 items

1 Scoring Criteria:

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State 1 - 3 items

0 No response or wrong response

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 an

air bubble. 6. The capillary tube is placed back into the beaker of water and

kept upright using a retort stand. 7. Mark two points , P and Q at a distance of 2 cm apart on the

capillary 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

0 No response or wrong response

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

0 No response or wrong response

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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