SPM Bio Ans (Kedah)

8/14/2019 SPM Bio Ans (Kedah)

1/24

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

j2kMOZ@C

SMS MUZAFFAR SYAH , MELAKA


2/24

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 solutionSolution 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

j2kMOZ@C



3/24


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 glucoselevel 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 Adams blood,. 1 1

(iii) - Xh

Y 1 1Total 12

j2kMOZ@C



4/24


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/ habitatLepus 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

j2kMOZ@C



5/24


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

j2kMOZ@C



6/24

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. 1mF3 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) 1mP2 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

j2kMOZ@C



7/24

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 donot 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

j2kMOZ@C



8/24

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

asEscherichia 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 producehuman 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

j2kMOZ@C



9/24

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

j2kMOZ@C



10/24

(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 orMadam 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

j2kMOZ@C



11/24

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 - ObservationScore 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 moves2. Distance the air bubble changes/ increases/ decreases

0 None of the above or No response

j2kMOZ@C



12/24

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 oftranspiration.

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

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 bylow 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/ decreases3. The rate of transpiration is different

0 None of the above or No response

j2kMOZ@C



13/24

1(c) KB0610 Controlling variables

Variables Method to handle the variable

Manipulated v.

Speed of air movement/ air speed/airmovement

-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/mintime 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

j2kMOZ@C



14/24

1(d) KB0611 State hypothesis

Score Criteria3 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 airbubble 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 movement2. Air speed increases transpiration


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 correctly2. 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

j2kMOZ@C



15/24

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 uniformscale 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 correct0 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 variableP 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

1Able to interpret data with 1 aspect correctly

0 No response or wrong response***

Reject1. The higher the transpiration rate, the higher the air speed.2. The air speed/air movement is proportional to the rate of transpiration.

j2kMOZ@C



16/24

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 observationSample 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


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 thebalsam plant through leaves and is affected by different air speed of fan thatshown 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

j2kMOZ@C



17/24

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 LampPlant 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



Scoring :

Materials Apparatus Score

2M 8A 3

2M + 1A 7A

2M 6/7A

1M 8A

2

2M + 2A 6A2M + 1A 6/7A

2M 4/5A

1M 5/6/7A

1

8A 2M

2M 1/2/3A

1M 1/2/3/4A

0

j2kMOZ@C



18/24

Question 2

KB061201 Explanation Score

01 Able to state problem statement by relating P1, P2 andP3 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 thegrowth 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

j2kMOZ@C



19/24


02 Able to state the hypothesis by relating two variablescorrectly (P1+P2+H)

P1- manipulated variable

Distance of seedlings

P2-responding variableThe 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 seedlings4. 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

j2kMOZ@C



20/24


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

-10cm- -5cm-

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

-10cm-

j2kMOZ@C



21/24


05 Able to list 3 materials and 3apparatus correctly tomake a functional experiment and able to get the data

MATERIALS (M)

Maize seeds/ Paddy seeds /any suitable seeds

WaterGarden 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 ofseedlings/

cm

The growth

rate of plantscm/day

10(A)

5(B)

2(C)

B2 = 1

mark

j2kMOZ@C



22/24

Construct Explanation Score

Able to state the correct technique with the following

aspects

Sample answer

1. Measure the height of seedling by using metre ruleOR2. 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 problemObjectiveHypothesisVariables ( The three variables are correct)List of materials and apparatusTechnique used

ProcedurePresentation of dataConclusion

Note:

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

3

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

experimental planning correctly

2

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

j2kMOZ@C



23/24

Sample Answer :

Problem Statement01=3

Does the distance of seedlings affects the growth rate of plants?

Aim of experimentTo study the effects of distance of seedlings on the growth rate of plants

Hypothesis

02=3The longer the distance of seedlings , the higher the growth rate of plants

VariablesManipulated 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

MaterialsMaize seeds, water, garden soil ,

05=3

Apparatus

Three planting trays / basins , metre rule , waterproof paint,

TechniquesB1=1Measure 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

Procedure1- 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, 30numbers 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

-10cm- -5cm-

-10cm-

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 ofseedlings are washed under running water

5- The height of maize seedlings are then measured by using metre rule. The average

j2kMOZ@C



24/24

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= 1The height of seedlings / cmThe distanceof seedlings/

cm(Tray)

1 2 3 4 5 6 7 8 9 10 Averageheights of

seedlings/

cm

The growth rateof seedlings

cm/day

10cm(A)

5cm(B)

2cm(C)

ConclusionThe longer the distance of seedlings , the higher the growth rate of plants . Hypothesis

is accepted.

Note:

7-9

- 3 marks4-6 - 2 marks1-3 - 1 mark 03=3

17

j2kMOZ@C

SPM Bio Ans (Kedah)

Documents

Transcript of SPM Bio Ans (Kedah)