j2k marking scheme peperiksaan percubaan spm … 2 peperiksaan percubaan spm (pkpsm-kedah) 2008...

Marking Scheme Peperiksaan Percubaan SPM (PKPSM - Kedah) 2008 Biology 1

1 D 26 A

2 C 27 B

3 A 28 B

4 C 29 D

5 A 30 A

6 B 31 A

7 C 32 D

8 C 33 B

9 D 34 B

10 D 35 A

11 D 36 D

12 A 37 C

13 D 38 B

14 C 39 D

15 D 40 C

16 B 41 C

17 D 42 B

18 A 43 C

19 D 44 B

20 D 45 B

21 C 46 A

22 C 47 B

23 B 48 C

24 B 49 C

25 D 50 C

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SMS MUZAFFAR SYAH , MELAKA

Biology 2 Peperiksaan Percubaan SPM (PKPSM-Kedah) 2008 Answer Scheme Question Sample Answer M1 M2

1 (a) (i) Organelle X : Chloroplast 1 (ii) Organelle Y : Mitochondrion 1 2 (b) P :

- P is the stroma - Dark reaction takes place in the stroma - Carbon dioxide is fixed and then reduced to form glucose Q : - Q is the granum - Light reaction takes place in the granum - granum trapped light energy to break down water molecule into hydrogen ion and hydroxyl ion. R : - Cellular respiration takes place in R - glucose is oxidized by oxygen to produce energy.

1 1 1

1 1 1

1 1

max 2

max 2

2

(c) - process in organelle X absorb carbon dioxide and release oxygen - process in organelle Y absorb oxygen and release carbon dioxide - process in organelle X use energy to synthesis glucose - process in organelle Y break down glucose to produce energy

1 1 1 1

4

Total 12

2 (a) (i) - semi-permeable membrane is the membrane which only permit some substances to move across it freely while other cannot.

1 1

(ii) - fluidity characteristics is caused by the protein molecules which are floating in the phospholipids bilayer. - the position of the molecules also keep on changing / not fixed in the position.

1

1

2

(b) (i) Fatty acids , glycerol , carbon dioxide, oxygen, water – any two 1 1 (ii) - molecule Q bind with the binding site of structure X.

- ATP provides energy to structure X and cause structure X to change its shape. - structure X push / pump molecule Q across it.

1 1

1

3 (c) (i) Solution X : salt solution

Solution Y : distilled water - both correct

1

1 (ii) - distilled water is hypotonic compare to the cytoplasmic fluid in the

red blood cells. - osmosis takes place - water molecules diffuse into red blood cell. - red blood cells swell and burst / haemolysis occur - cytoplasmic fluid of red blood cells cause the solution change into clear red

1

1 1 1 1

max 4

Total 12

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Question Sample Answer M1 M2

3 (a) (i) Organ X : small intestine Organ Y : liver

1 1

2

(ii) Molecule K : maltose Molecule M : glucose Enzyme L : maltase

1 1 1

3 (b) - enzyme L is highly specific, it means that it can only catalyse one

kind of enzyme. - enzyme L is not used up / destroyed at the end of the process.

1

1

2 (c) - muscle cells need to produce a lot of energy to carry out body

activities / movement. - muscle cells need molecule M to carry out cellular respiration to produce energy. - molecule M will be oxidised by oxygen to produce energy

1

1

1

3 (d) - excess blood sugar/glucose will be converted into glycogen and

stored in the liver. - excessive of blood sugar may lead a person to diabetes mellitus. - glycogen will be converted back into glucose when the glucose level in the body drops.

1

1 1

max 2

Total 12 4 (a) Factor X : Vitamin K

Factor Y : Ion Ca2+ 1 1

2

(b) Structure W: Fibrin 1 1 (c) Fibrinogen is soluble plasma protein whereas structure W is fibrin

which is insoluble protein. 1 1

(d) - to prevent serious blood loss when a person is injured - to prevent the entry of microorganisms and foreign particles into the blood/ body through the damage blood vessels - to maintain normal blood pressure - to maintain circulation of blood in a closed circulatory system

1 1

1 1

max 2

(e) (i) P : Thrombus/ Cholesterol / Plaque / deposit 1 1 (ii) - Excessive cholesterol, saturated fats and calcium are deposited on

the inner lining of the arteries. - Detached deposited cholesterol on the arteries wall can stimulate the agglutination of platelets which lead to the formation of blood clot / thrombus in the arteries to obstruct the blood flow

1

1

2

(f) (i) - Haemophilia 1 1 (ii) - Lack of certain blood clotting factors / factor VIII in Adam’s blood,. 1 1 (iii) - XhY 1 1 Total 12

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5 (a) - discontinuous variation - discontinuous variation - continuous variation

1 1 1

3 (b) - The differences between organism of the same species are known as

variation. 1 1

(c) Continuous variation Discontinuous variation - caused by genetic factor and

environmental factor. - has intermediate characteristics - shows gradual differences for a particular characteristics

- caused by genetic factor only - no intermediate characteristics - shows distinct differences for a particular characteristics.

1

1 1

max 2

(d) (i) No 1 1 (ii) Because they are not the same species. 1 1 (iii) Lepus alleni

- has bigger ear, to increase the ratio of TSA/V - to increase the rate of the heat loss from the body - to bring down the body temperature in the hot environment/ habitat Lepus articus - has smaller ear, to reduce the ratio of TSA/V - to slow down the rate of the heat loss from the body, - to maintain body temperature in the cold environment / habitat.

1 1 1

1 1 1

max 4

Total 12

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6 (a) (i) The moth with pale, speckled wings 1 1 (ii) - The moths with dark wings are more easily spotted/detected by their

predators.// the moths with pale, speckled wings, are well camouflaged by the pale bark of the trees.

- More moths with dark wings were hunted and eaten by the predators.

1

1

2

(iii) - use the capture, mark, release and recapture technique - capture as many moths as possible in the bush. - count the captured moths. - mark the moths with a small dot of Indian ink. - release the moths in the same place where they were captured. - after a few days, go back to the same place and capture again as many moths as possible. - count the recaptured moths, noted the number of moths which had been marked. - population size = (a x b) / c a = the number of the moths in the first capture. b = the number of the moths in the second capture. c = the number of marked moths in the second capture.

1 1 1 1 1 1

1

1

max 7

(b) (i) Abiotic component : gradient / steepness of the slope Biotic component : small insects/animals

1 1

2

(ii) - The steepness of the slope in Zone A is higher than Zone B. - In Zone A, steep slope cause rapid drainage and run-off. - the soil layer in Zone A is thinner and drier. - Zone A has less plants compare to Zone B - Zone B has higher population of small animals than zone A. - small animal like the earthworm feed on the rotten /dead/ plants / leaves - the activities of earthworm increase the organic substances/fertility in the soil - organic substances makes the soil in zone B more suitable/fertile for plants to growth.

1 1 1 1

1 1

1

1

8

Total 20

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7 (a) F1 Excess mineral salts can cause higher osmotic pressure in the blood 1m P1 Higher osmotic pressure can cause high blood pressure in the blood. 1m F2 Excess fats will be converted into cholesterol 1m P2 Cholesterol accumulates at the artery wall and cause arteriosclerosis 1m P3 High blood cholesterol levels are a risk factor for heart attack and stroke. 1m F3 Excess protein cause excess amino acids which lead to gout. 1m P4 and cause kidney failure. 1m F4 Low in roughage can cause constipation. 1m P5 Deficiency in roughage also leads to difficulties in peristalsis process

along digestive tract 1m

F5 Food preservatives, food colouring and food flavouring consists of carcinogenic substances.

1m max

P6 It may cause cancer. 1m 10m 7 (b) F1 The blood osmotic pressure of individual X is higher because of excess

taking of mineral salt 1m

P1 Pituitary gland is triggered to secrete more ADH (Antidiuretic Hormone) 1m P2 Cells lining of distal convoluted tubule and collecting duct become more

permeable to water 1m

P3 A lot of water is reabsorbed into surrounding blood capillaries 1m P4 through osmosis. 1m P5 Adrenal glands is not stimulated to release Aldosterone 1m P6 Less amount of salt be reabsorbed 1m P7 Hence, urine produced by individual X is concentrated and in a small

amount. 1m

F2 The blood osmotic pressure of individual Y is lower because of the meal

consumed is containing less salt 1m

P8 The osmoreceptor cells in the hypothalamus are less stimulated 1m P9 Less ADH is secreted 1m P10 Distal convoluted tubules and collecting duct is less permeable to water. 1m P11 Less water is reabsorbed from the filtrate 1m P12 Adrenal glands is stimulated to release Aldosterone 1m P13 Cells lining of distal convoluted tubule and collecting duct become more

permeable to salts. 1m

P14 Salts are actively reabsorbed into the blood capillaries 1m Max P15 Urine produced by individual Y is dilute and large in amount. 1m 10m Total 20 m

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NO MARKING CRITERIA MARKS

8(a) P1 P2 P3 P4 P5 P6

The formation of identical twins involves only one ovum and one sperm while fraternal twins involve two ova and two sperms.

Identical twins share the same placenta while fraternal twins have separate placentas. Identical twins have the same genetic information while fraternal twins have different genetic information. Identical twins have identical characteristics while fraternal twins do not have identical characteristics. Identical twins have the same sex while fraternal twins may not. Identical twins have the same blood group while fraternal twins do not have the same blood group.

[ P1 + 3 P ] = 4marks Max [Without P1 ] = 3 marks

1 m

1m

1m

1m

1m

1m _____

4 marks

( b) P1 P2 P3 P4 P5 P6

The sex in offspring is determined by the type of sperm which will fertilize the ovum. The sperm produced by the testis has 22 autosomes and one sex chromosomes /either the X chromosome or the Y chromosome// 22 + X or 22 + Y The ovum produced in the ovary has 22 autosomes and one X chromosome . If a sperm that contains the X chromosome fertilizes the ovum, a zygote that has two X /XX chromosome is produced, that is a girl. If a sperm that contains the Y chromosomes fertilizes the ovum, a zygote containing the XY chromosome is produced, that is a boy. As fertilization occurs at random, the probability of having a male child or female child is the same / 50 %

1m

1m

1m

1m

1m

1m

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P7 P8 P9 P10 P11 P12

Parents Phenotype Male X Female Genotype 44 + XY 44 + XX Meiosis

Gametes Random Fertilization Offspring Genotype 44 + XX 44 + XX 44 + XY 44 + XY Phenotype female female male male Ratio 1 female : 1 male There is an equal chance of the mother having a baby boy or a girl [Any 10 P Max

1m

1m

1m

1m

1m

1m

_____ 10 marks

P1 P2 P3 P4 P5 P6 P7

Insulin is produced by the langerhans cell in the pancreas. The genes that are responsible for the production of insulin are isolated from the DNA of langerhans cells / pancreatic cell. The genes are then inserted into the DNA molecule of bacteria (such as Escherichia coli . The bacterium contains a recombinant DNA with the human insulin gene The bacterium is then cultured in a suitable nutrient medium ( in laboratory). The bacterium has the gene for human insulin and able to produce human insulin The human insulin is extracted in large quantity

[any 6 P] Max

1m

1m

1m

1m

1m

1m

1m ___

6 marks

22 + X 22 + Y 22 + X 22 + X

Total 20

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9

(a) b)

– Ovulation releases a secondary oocyte , which enters the oviduct. - The secondary oocyte starts meiosis II which progresses until metaphase II. - The nuclei of a sperm cell (n) and the ovum (n) fuse and form a diploid zygote (2n). // A sperm fertilize the ovum to form a zygote.

- Zygote begins to divide repeatedly by mitosis as it travels along the fallopian tube towards the uterus. - Morula is form followed by blastula.

- Implantation occur / The blastocyst attaches itself to the endometrium.

Sample Answer : - Cigarette contain nicotine / DDT / lead particles. - The wall of maternal blood capillaries and the wall of foetal blood capillaries are semi-permeable. - Nicotine, drugs and alcohol are small in size.

- Nicotine, drugs and alcohol can diffuse from maternal blood capillaries to foetal blood capillaries - through the placenta - The substances carried by umbilical vein to the foetus.

- Nicotine, drugs or alcohol can affect the development of foetus - (example) cause disable / miscarriage . birth defect/ illness in the resulting baby.

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

max 4

Max 6

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(c) (i) (c) (ii)

- Method A is known as in-vitro fertilization (IVF) - Method A is use if the fallopian tubes of Mrs. Ali are blocked. - sperm cannot reach the ovum, fertilization fail to occur. - fertilization has to be done outside the body. - developed zygote/embryo then retransfer and implant in the uterus of Mrs. Ali. - the embryo then undergo normal development in the uterus of Mrs. Ali as normal pregnancy. - Method B is used if the uterus of Mrs. Ali fail to carry the implanted embryo because of damaged or abnormal uterus. - Madam X is the woman who is willing / hired to carry the implanted embryo to full term. - Madam X is known as surrogated mother. - Genetically the baby belongs to Mr. and Mrs. Ali. - Who is the real biological mother of the baby, Mrs. Ali or Madam X? - There are cases that the surrogated mother refuse to return

the baby to the couple after giving birth.

Total

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

Max 5 Max 5 20

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Marking Scheme Peperiksaan Percubaan SPM (PKPSM - Kedah) 2008 Biology 3 1(a) KB0603 – Measuring dan using numbers Score Criteria

3 Able to record all three distances with correct units Sample answer :

Air Speed Distance of PQ 1 2.5 cm 2 5.6 cm 3 10.3 cm

2 Able to record any two distances with the correct units 1 Able to record any one distance with the correct unit 0 No response or wrong response

1(b)(i) KB0601 - Observation Score Criteria

3 Able to state two different observations correctly [Observations must have values for MV and RV in table 1.1 or Comparison between two readings] Sample answer: 1. At air speed 1, the distance moved by air bubble in 5 minutes is 2.5 cm. 2. At air speed 3, the distance moved by air bubble in 5 minutes is 10.3 cm. 3. The distance moved by air bubble in 5 minutes at air speed of 3 is greater than

the distance moved by air bubble in 5 minutes at air speed of 1.

2 Able to state two different observations inaccurately Sample answer: 1. At air speed 1, the distance moved by air bubble in 5 minutes is the shortest. 2. At air speed 3, the distance moved by air bubble in 5 minutes is the longest. 3. Speed of fan influences the distance moved by air bubble in 5 minutes.

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

1. The air bubble moves 2. Distance the air bubble changes/ increases/ decreases

0 None of the above or No response

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Scoring For Observation and Inference

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

2

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

1

- 1 - 1 - - 1 1

0

1(b)(ii) KB0604 – Making inference Score Criteria

3 Able to state two different inferences correctly Sample answer:

1. In low speed of fan/ at air speed 1, low air movement causes low rate of transpiration.

2. In high speed of fan/ at air speed 3, high air movement causes high rate of transpiration.

3. Transpiration rate at air speed 3 is higher than air speed 1. 4. The higher the rate of air movement/ air speed, the higher the rate of

transpiration. 5. At air speed 1, the distance of 2.5 cm traveled by the bubble is caused by

low rate of transpiration in the shoot

2 Able to state two different inferences inaccurately Sample answer: 1. The differences in PQ is due to the different rate of transpiration 2. The rate of transpiration is influenced/ affected by the air movement/ air speed.

1 Able to state two different inferences at idea level Sample answer:

1. Transpiration occurs. 2. Rate of transpiration changes/ increases/ decreases 3. The rate of transpiration is different

0 None of the above or No response

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1(c) KB0610 – Controlling variables

Variables Method to handle the variable Manipulated v. Speed of air movement/ air speed/air movement

-Using fan with different air movement/air speed 1,2 and 3// -Change the air speed of fan, speed 1, 2 and 3

Responding v. Distance the air bubble moved in 5 minutes OR Rate of transpiration

-Measure the distance between initial level P and final level Q after 5 minutes by using a ruler/metre rule OR -Calculate the rate of transpiration by using

formula = Distance the air bubble moved cm/min time taken

Constant v. 1.Time taken for the air bubble moved/2.Type of plant/ 3.Temperature/ 4.Distance between the fan and balsam shoot

1. Fixed the time for 5 minutes/ 2. Fix/use the same type of plant, balsam/ 3. Fix the temperature at room temperature, 27 o C/

4. Fix the distance between the fan and balsam shoot at 30 cm

Score Criteria

3 6 ticks 2 4 – 5 ticks 1 2 -3 ticks 0 1 tick or No response

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1(d) KB0611 – State hypothesis

Score Criteria 3 Able to state a hypothesis relating the manipulated variable and the responding

variable correctly with the following aspects : P I = Manipulated variable (air speed/ air movement) P 2 = Responding variable (transpiration rate/ distance the air bubble moved) H = relationship Sample answer :

1. As the air speed increases, the rate of transpiration increases / distance the air bubble moved increases.// inversely.

2. The higher the air movement/ air speed, the higher the rate of transpiration. 3. When air movement/air speed is low, the transpiration rate will decrease //

vice versa

2 Able to state a hypothesis relating the manipulated variable and the responding variable inaccurately 1. Increasing the air movement/air speed increases transpiration 2. Air speed/air movement affect/ influence the rate of transpiration 3. Rate of transpiration is affected by air speed/air movement

1 Able to state a hypothesis relating the manipulated variable and the responding variable at idea level

1. Transpiration increase with air speed/ air movement 2. Air speed increases transpiration

0 No response or wrong response

1(e) KB0606 – Communicating data Score Criteria

3 Able to construct a table correctly with the following aspects: 1. Able to state 3 titles with units correctly 2. Able to record all the data correctly 3. Able to calculate and record rate of transpiration correctly

Sample answer : Speed of air movement 1 2 3 Distance of the air bubble moved in 5 minutes /cm

2.5 5.6 10.3

Rate of transpiration / cm min-1 0.5 1.12 2.06

2 Any 2 aspects correct 1 Any 1 aspects correct 0 None of the above OR no response

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1(f)(i) KB0607 – Correlating time and space Score Criteria

3 Able to draw the graph correctly with the following aspects : P (Axis) : Correct title with unit on both horizontal,vertical axis and uniform scale on the axis. T (Point) : All point plotted/ transferred correctly. B (Shape) : Able to join any two points to form a smooth graph, and positive gradient. All three aspects correct.

2 Any 2 correct 1 Any 1 correct 0 None of the above OR no response

1(f)(ii) KB0608 – Interpretating data Score Criteria

3

Able to interpret data correctly and explain with the following aspects : Relationship : P I = manipulated variable P 2= responding variable H = relationship Sample answer :

1. Transpiration rate increases when/with/as the air speed increases. 2. The higher the air speed, the higher the transpiration rate. 3. When the air movement/air speed increases, the transpiration rate will

increase.

2

Able to interpret data with 2 aspects correctly Sample answer : Transpiration rate is proportional to air speed

1

Able to interpret data with 1 aspect correctly

0 No response or wrong response ***

Reject 1. The higher the transpiration rate, the higher the air speed. 2. The air speed/air movement is proportional to the rate of transpiration.

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1(g) KB0605 – Predicting Score Criteria

3 Able to predict observation and the resulting rate of transpiration. Explain the outcome of the experiment correctly with the following aspects. P1 - Able to predict observation Sample answer for P1 :

1. Distance the air bubble moved shorter. P2 – Able to state the change in rate of transpiration Sample answer for P2 :

1. The rate of transpiration is lower P3 – Explanation Sample answer :

1. Distance the air bubble moved shorter and the rate of transpiration is lower, less than 0.50 cm min-1 because the air movement decreases when fan is not switched on during the experiment.

2 Able to predict any two criterias 1 Able to predict any one criteria 0 No response or wrong response

1(h) KB0609 – Defining by operation Score Criteria

3 Able to define operationally based on the result of the experiment with the following aspects. P1 - process losing water by balsam plant P2 – distance traveled by the air bubble in five minutes P3 – The rate of transpiration is affected by air speed/air movement //complete definition by theory // Hypothesis form Sample answer :

1. Transpiration is the process losing water in form of water vapour from the balsam plant through leaves and is affected by different air speed of fan that shown by the distance traveled by the air bubble in the capillary tube in 5 minutes..

2 Able to define operationally based on the result of the experiment with any two

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

aspect correctly 0 None of the above OR no response

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1(i) KB0602 – Classifying Score Criteria

3 Able to list all materials and apparatus used in the investigation Sample answer :

Materials (M) Apparatus (A) Vaseline Lamp

Plant shoot Metre rule Stop watch

Capillary tube Rubber tube Knife Basin Retort stand

All two materials and eight apparatus are correct.

2 Refer to scoring below 1 Refer to scoring below 0 Refer to scoring below

Scoring : Materials Apparatus Score 2M 8A 3 2M + 1A 7A 2M 6/7A 1M 8A

2

2M + 2A 6A 2M + 1A 6/7A 2M 4/5A 1M 5/6/7A

1

8A 2M 2M 1/2/3A 1M 1/2/3/4A

0

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Question 2 KB061201 Explanation Score

01 Able to state problem statement by relating P1, P2 and P3 in a question form correctly. P1- manipulated variable Distance of seedlings P2-responding variable The growth rate of plants (maize /paddy / any suitable plants) / the height of seedlings / mass of seedlings/ numbers of leaves P3-question form (How/does …? ) Sample answer: How / Does the distance of seedlings (P1) affects the growth rate of maize plants (P2) ? (P3) Note: - reject ‘mango plant’ as a named plant

3 P1+P2+P3

Able to state problem statement inaccurately Sample answer: 1. What is the effect of distance on plants ? (P1+P3) 2. The growth rate of plant is affected by the distance(no P3)

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

Able to state the idea Sample answer : 1. The distance affects the growth of plants ( no P2 + P3)

1 P1/P2/P3

No response or wrong response 0

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KB061202 Explanation Score

02 Able to state the hypothesis by relating two variables correctly (P1+P2+H) P1- manipulated variable Distance of seedlings P2-responding variable The growth rate of plants / the height of seedlings/ the mass of seedlings / the number of leaves H-relationship Sample answer: 1. The longer/shorter (H) the distance of seedlings (P1), the higher/lower (H) the growth rate of plants(P2) 2. The longer/shorter (H) the distance of seedlings (P1), the higher/lower the heights of seedlings(P2) 3. The longer / shorter the distance of seedlings the heavier / lighter the mass of seedlings 4. The longer /shorter the distance of seedlings, the more / lesser number of leaves. 5. As the distance of seedlings (P1) increase / decrease(H), the growth rate of plants (P2) increase/ decrease(H)

3 P1+P2+H

Able to state any two criteria correctly or inaccurate hypothesis Sample answer: 1. The distance of seedlings (P1) affect the growth rate of plants (P2) (no H) 2. The growth rate of plants in tray A is higher than the growth rate of plants in tray C ( no P1)

2 P1+P2/ P1+H/ P2+H

Able to draw the idea of hypothesis Sample answer: 1. The distance of seedlings affect the plants ( no P2 + H )

1 P1/P2/H

No response or wrong response 0

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04 Able to state K1, K2, K3, K4 and K5 (5K) correctly K1: The set up of apparatus (S1/ S2/S3/S4/S5) (any 3 ) K2: How to manipulate the variable (S3 ) K3: How to operate the responding variable ( S6 ) K4: How to fix the constant variable(S1 or S4) K5: Precautions (S4 or S5 ) S1- Three planting trays are prepared and filled with 3 kg of garden soil in each tray. S2- The trays are labeled as A, B and C with waterproof paint . S3- 30 numbers of maize seeds are planted in tray A at a distance of 10 cm intervals, 30 numbers of maize seeds in tray B at a distance of 5 cm intervals and 30 numbers of maize seeds in tray C at a distance of 2 cm intervals as shown below (not in correct scale) . 2cm γ-10cm- γ γ γ-5cm-γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ A B C S4- Each tray is watered daily with the same amount of water for 10 days S5- After 10 days, 10 maize seedlings are picked randomly from tray A and the root of seedlings are washed under running water S6- The height of maize seedlings are then measured by using metre rule. The average height are calculated by using formula = the total height of seedlings/cm 10 The growth rate is calculated by using formula = the average height of seedlings/cm time taken / day S7- Step 5-6 are repeated for seedlings from tray B and C. The average height and the growth rate of seedlings in tray B and C are measured and calculated separately. S8- The result are recorded in a table.

3 K1+K2+K3+

K4+K5 (5K)

Able to state any 3K – 4K correctly

2

Able to state any 1K – 2K correctly

1

Wrong response or no response

0

-10c

m-

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05 Able to list 3 materials and 3apparatus correctly to make a functional experiment and able to get the data MATERIALS (M) Maize seeds/ Paddy seeds /any suitable seeds Water Garden soil Note : reject ‘mango seed’ APPARATUS (A) Metre rule Tray / Basin / Container Waterproof paint /marker pen Spade * Beam / electronic / compression balance mass of * Oven seedlings

Notes : Score Material (M) Apparatus

(A) 3 3M 2A 2 3M

2M 1A 2A

1 2M 1M

1A 1A

3

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

2

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

1

Wrong response or no response 0

}

Explanation Score

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

The height of seedlings / cm The distance of seedlings/ cm(Tray)

1 2 3 4 5 6 7 8 9 10 Average heights of seedlings/ cm

The growth rate of plants cm/day

10(A) 5(B) 2(C)

B2 = 1 mark

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Construct Explanation Score

Able to state the correct technique with the following aspects Sample answer 1. Measure the height of seedling by using metre rule OR 2. Calculate the average height of seedlings by using formula = total heights of seedlings /cm OR 10 3.Calculate the growth rate of plants by using formula = the average heights of seedlings /cm time taken/day

B1 = 1 mark


03 Able to state 7-9 aspects of experimental planning correctly :

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

3

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

2

Able to state any 1 - 3 items correctly

1

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

0

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Sample Answer : √Problem Statement 01=3 Does the distance of seedlings affects the growth rate of plants? √Aim of experiment To study the effects of distance of seedlings on the growth rate of plants √Hypothesis

02=3 The longer the distance of seedlings , the higher the growth rate of plants √Variables Manipulated variable : The distance of seedlings Responding variable : The height of seedlings/ the growth rate of plants Constant variable : Number of seedlings / types of soil/ amount of water/ light intensity / time taken √Materials Maize seeds, water, garden soil ,

05=3 Apparatus Three planting trays / basins , metre rule , waterproof paint, √Techniques

B1=1 Measure the height of seedling by using metre rule OR Calculate the growth rate of seedlings by using formula = average heights of seedlings /cm time taken √Procedure 1- Three planting trays are prepared and filled with 3 kg of garden soil in each tray. Labeled the

trays as A, B and C with water proof paint. 04=3

2- 30 numbers of maize seeds are planted in tray A at a distance of 10 cm intervals, 30 numbers of maize seeds are planted in tray B at a distance of 5 cm intervals and 30 numbers of maize seeds are planted in tray C at a distance of 2 cm intervals a shown below (not in correct scale) . 2cm γ-10cm- γ γ γ-5cm-γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ

-10c

m-

γ γ γ γ γ γ γ γ γ γ γ γ A B C 3- Each tray is watered daily with the same amount of water for 10 days 4- After 10 days, 10 maize seedlings are picked randomly from tray A and the root of seedlings are washed under running water 5- The height of maize seedlings are then measured by using metre rule. The average

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height is calculated by using the formula = the total height of seedlings/cm 10 The growth rate of seedlings is calculated by using formula = the average height of seedlings/ cm Time taken / day 6- Step 4-5 are repeated for seedlings from tray B and C. The average height and the growth rate of seedlings in tray B and C are measured and calculated separately. 7- The result are recorded in a table. √Results

B2= 1 The height of seedlings / cm The distance of seedlings/ cm(Tray)

1 2 3 4 5 6 7 8 9 10 Average heights of seedlings/ cm

The growth rate of seedlings cm/day

10cm(A) 5cm(B) 2cm(C)

√Conclusion The longer the distance of seedlings , the higher the growth rate of plants . Hypothesis is accepted. Note:

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

03=3

17

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j2k marking scheme peperiksaan percubaan spm … 2 peperiksaan percubaan spm (pkpsm-kedah) 2008...

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