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 PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM TAHUN 2013 SET B PERATURAN PEMARKAHAN Set B Biologi  Kertas 1, 2 dan 3 Peratur an pemarkahan ini mengandungi 23 halaman bercetak  w w w . m  y  s  c h  o  o l   c h i  l   d r  e n .  c  o  w w w . m  y  s  c h  o  o l   c h i  l   d r  e n .  c  o  u  a  t   t   u r  u n  (   p  e r  c  u m  a  )   s  o  a l   a n l   a i   d i  : M  u  a  t   t   u r  u  (   p  e r  c  u m  a  )   s k  e m  a l   a i   d i  www.myschoolchildren.com

Skema 2013 Kedah

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PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM

TAHUN 2013

SET B

PERATURAN PEMARKAHAN

Set B

Biologi Kertas 1, 2 dan 3

Peraturan pemarkahan ini mengandungi 23  halaman bercetak 

 

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1

SULIT 4551/2

Skema Modul Peningkatan Prestasi SPM Biologi 2013: Kertas 2/ Set 2.

Question No

Mark ing Criteria Submarks

Totalmarks

1(a) (i) Tonoplast1 1

(b) (i)Semi permeable 1 1

(ii) P1: store chemicals such as organic acids/ mineral salts/ oxygen/carbon dioxides/ waste substances/ pigments / metabolic by- products

P2: contribute to plants elongations by absorbing water and causingcells to expands

P3: create turgidity to provide mechanical support for the plants.P4: regulates water balance in plants

Any 2

1

1

1

2

(c) (i) cellulose  1 1

(ii) P1 : maintains the shape of plant cells

P2 : provides mechanical strength and supports to plant cells

P3 : protect plant cells from ruptured when excess water moves

into the cells Any 2 

111

2

(d) (i)

Plasma membrane shrink and pull away from cell wall

Size o vacuole must be smaller

1

1 2(ii) P1: the cell plasmolysed / plasmolysis

P2:the concentrated sugar is hypertonic to the cell sap

P3:water diffuse out of the cellP4: by osmosis

P5:the cytoplasm and vacuole shrink

P6:the plasma membrane pulls away from the cell wall

Any 3

1

1

11

1

1

3

12

cytoplasm

vacoule

nucleus

Cell wall

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3

SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

3. (a) A: vascular bundle 1 2

B: aerenchyma cell / air sac 1

(b) A: Xylem vessel has a thick wall / lignified wall provide

support to the plant.

1

2B: Aerenchyma tissue/ air sac in the stem and leaves ( helps

maintain buoyancy in the floating plants)

1

(c) (i) To provide mechanical support for growing plants //

prevent the cells from collapsing. 1 1

(ii) P1: to create capillary action

P2: by forming cohesion force and adhesion force

P3: To form a continuous water column from root to the

leaves.

P4: which enable the water move up along the narrow xylem

vessel

Any 2

1

1

1

2

(d) (i) Xylem tissue // wood/ woody tissue. 1 1

(ii) P1: Produce hard and strong timber // wood can be used in

construction / furniture / bridges.

P2: Fibres are used in paper and textile industriesP3: Wood is used as a source of fuel

Any 2

1

1

1

2

(e) P1: Tying the wire on the stem for a long period of time

causes the phloem to be cut off

P2: Food from the upper part cannot be transported to the

area below the wire.

P3: Organic substances accumulated above the wire causing

the area to swell up.

Any 2

1

1

1 2

4. (a) (i) Simple diffusion 1 1

(ii) F: Partial pressure of oxygen in the alveolus is higher than in

blood capillary.

P: Oxygen diffuses from alveolus into blood capillary which

follow the concentration gradient.

1+1 2

(b) (i) Pulmonary vein 1 1

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4

SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

(ii) P1: In the blood, oxygen (from alveolus) combines with

respiratory pigment/ haemoglobin to form

oxyhaemoglobin/ oxygenated blood.

P2: oxygenated blood // oxyhaemoglobin is transported tothe heart

P3: the heart pumps the oxygenated blood to muscle cells

via the aorta.

 Any 2

1

1

1

2

(c) (i) P: Aerobic respiration

Q: Anaerobic respiration

1

1 2

(ii)P Q

D1 Complete breakdownof glucose.

Incomplete breakdownof glucose.

E1 Release large amount

of energy // 2898 kJ

per molecule glucose

// 38 molecules of

ATP

Releases less amount of

energy // 150 kJ per

molecule glucose // 2

molecules of ATP.

D2 Produce carbon

dioxide and water.

Produce lactic acid.

E2 Produced as waste

product.

Caused muscular

cramps/ fatigue

D3 Occurs in the

mitochondria.

Occurs in the cytoplasm.

E3 Oxygen is required. Oxygen is not required.

Any one pair.

1+1

2

(d) F1: (During the vigorous activity) the muscle cells are in a state ofoxygen deficiency/ oxygen debt // the blood cannot supply

oxygen fast enough to meet the demand for energy/ ATP.

P1: (The increase in heartbeat rate) is to deliver more oxygen/

glucose to muscle cells.

P2: to produce extra energy (from aerobic respiration) // to

increase the rate of cellular respiration.

P3: to remove more carbon dioxide from the muscle cells.

 Any 2

1+1

2

Total 12

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SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

5. (a) (i) P : Primary Follicle Q: Corpus luteum 1 + 1 2

(ii) FSH stimulates the growth/development of the primary

follicleStructure P will grow/develop becomes secondary

follicle/Graafian follicle

1

1 2

(b) Ovulation will not occur

Graafian follicle will not release the secondary oocyte ( into

the fallopian tube)

1

1 2

(c) (i) Able to complete the graph that shows the thickness of

Endometrium wall is decreasing/thinning

1 1

(ii) Corpus luteum degenerateNo progesterone is secreted (to thicken the endometrium

wall)

11

2(d) P1: Q secrete progesterone

P2: (Progesterone will thickened and) maintained the

thickness of endometrium wall

P3: Endometrium wall ready for of implantation of

embryo//

P4: Prevent miscarriage/abortion

1

1

13

Total12

6 (a) (i)  Able to calculate the ratios of phenotype dominant to

 phenotype recessive correctly

Sample answer

C1 : Round yellow : round green : wrinkled yellow : wrinkled

green

315 : 108 : 101 : 32

556 556 556 556

0.9 : 0.3 : 0.3 : 0.1

or9 : 3 : 3 : 1

1

1

1 3

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6

SULIT 4551/2

Marking Criteria Sub

(ii) Able to describe the cross based on Mendel’s Second Law by using Punnet

Square

Sample answer:

P1: A pure breeding with dominant characteristic round yellow (RRYY) with

recessive characteristic of pea plant.

Parental Pure breeding plant Pure breeding plant 

phenotypes Round and Yellow seeds  Wrinkled and Green 

Parental genotypes RRYY X rryy

Meiosis

Gametes RY r y

F1 genotypes  RRyY

F1 phenotypes All round and yellow seeds

F1 x F1  RrYy X RrYy

Gametes  RY Ry rY ry RY Ry rY ry

Male Gamete

Female

Gamete

RY Ry rY ry

RY RRYY RRYy RrYY RrYy

Ry RRYy RRyy RrYy Rryy

rY RrYY RrYy rrYY rrYy

ry RrYy Rryy rrYy rryy

F2 genotypes RRYY : RRYy, : RRyy, : rrYY, : rryy

RrYY : RrYy Rryy rrYy

F2 phenotypes Round, : Round : Wrinkled : Wrinkled,Yellow Green Yellow Green

F2 genotypic ratio 1:2:2:4 1:2 1:2 1

F2 phenotypic ratio 9 : 3 : 3 : 1

Any 10

1

1

1

1

1

1

1

1

1

1

1

Max

10

1

1

1

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SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

(b)

 Able to explain the possible genetic combination a blind colour man

marries a carrier woman through schematic diagram

Sample answer:

Phenotype : Colour blind man X Female colourblind carrier 

Genotype : Xb

Y X B X 

Meiosis

Gamete Xb

Y X X B

 X b 

Fertilisation

F1 Genotype X B X 

b X 

bY 

  X 

BY X 

b X 

Phenotypes Female Male Male Female

carrier colour blind normal colour blind

The children of the couple will have the following possible genetics

combinations :

25% carrier female : 25% colour blind : 25% normal male : 25% femalecolour blind

1

1

1

1

1

1

1

7

Total 20

7. (a) P1: The diagram shows a saprophytic fungus.

P2: It obtain its nutrient by secreting digestive enzymes onto the

substrate

P3: The enzymes digest the complex substances into simple forms.

P4: The simple forms are then absorbed by hypha.

1

1

1

1 4

F1- In the month of January until April, the increase in the prey's is

followed by an increase in predator population

P1 - due to abundance of food

F2 - However from April to August, when the number of predator

increases the number of preys will then decrease

P1 - This is because the high number of predators will easily

consume the prey.

1+1

1+1

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SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

(b)

P2 - When the prey reduces, the predators will have less to eat.

P3 - There is intraspesific competition

F3 - The number of predators also reduces in the following months

from August to December.

P1 - The prey - predator relationship takes place in cycle.

F4 - This keeps the population of both organisms in a dynamic

equilibrium

Any 6

1+1

1+1

1 Max

6

(c) F1 – To build proteins, plants need the element nitrogen 1

P1 - The nitrogen gas in the atmosphere is about 78% 1

P2 - but plants are not able to utilize this nitrogen 1

P3 - The nitrogen has to be fixed before it can be absorbed by the

plants 1

F2 - Nitrogen fixing bacteria / Noctoc sp / Azotobacter / bacteria A

which lives freely in the soil 1

and Rhizobium sp / bacteria B which lives in the root nodule of

Leguminous plant. 1

P1 - can convert nitrogen in the atmosphere into ammonia. 1

F3 - Nitrosomonas / bacteria C oxidizes/ converted ammonia into

nitrites 1

F4 - Nitrobacter  oxidizes / converted nitrites into nitrates. 1

P1 - Nitrites and Nitrate can be absorbed by plants as their

nitrogen source. 1

P2 - The plants protein is transferred to the animals when

consumed by the animals 1

F5 - The animals and plants proteins are transferred to human

when consumed. 1

Any 10 Max

10

Total 20

8. (a)

P1

P2

Able to state the function of tissue R and tissue S.

Sample answer :

Tissue R : Xylem

Function : transports water and dissolved mineral salt from the

roots to the stem, leaves, flowers and fruits or// it also provide

mechanical support to the plant

1+1

2

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SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

Able to state the differences in the function and structure of tissue R and

tissue S.

Tissue R Tissue S

P1 : Xylem P2 :Phloem

P3: Function

transports water and dissolved

mineral salt from the roots to the

stem, leaves, flowers and fruits

or// it also provide mechanical

support to the plant

P4: Function:

transports organic food

substances, for example sucrose

and amino acids, from the

leaves both upwards and

downwards to various plant

parts to the growing regions or

storage structureP5: Consists of sieve tubes and/ or

companion cells

P6: Consists of xylem vessels

and/ or tracheids

P7: Has sieve plate on the sieve

tube

P8: Hollow tube

P9: The cell wall is not thickened

with lignin/ not lignified

P10: The cell wall is thickened

with lignin/ lignified

P11: Has thin layer of protoplasm P12: No protoplasm

Any 8

1+1

1+1

1+1

1+1

1+1

1+1

8

(b)

P1

P2

P3

P4

Able to explain adaptation of tissue R to carry out its function

in water transport efficiently .

Sample answer

Structure of xylem Explanation

P1: Tissue R consists of

(long) cells joined end to end

//hollow tube

E1: Allows water to flow in a

continuous column.

P2: The end walls of xylem

vessels have broken down

E2: To give an uninterrupted

flow of water from roots to the

leave.// allows water to flow in

a continuous columnP3: There are pits (at

particular points in the

lignified wall )

E3: To permit lateral flow of

water

P4: The walls are

impregnated/

thickened with lignin //

lignified

E4: To prevent them

collapsing (under the large

tension forces set up by the

transpiration pull.)

Any 8

1+1

1+1

1+1

1+1

8

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SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

(c) Able to explain how this process helps to prevent the occurrence

of overheating in plants.

Sample answerP1: (when the water reach the leaves,) water is convert to water

vapour

P2: water vapour diffuse out through stoma

P3: thus, the plant loses heat

P4: causes the cooling effect on the plant // reduces temperature

of the plant

1

1

1

1 4

Total 20

9. (a)  Able to explain the three processes which occur to the fat before it

can be used by body cells correctly

Sample answer

Diagram 9.1(a)

P1: Process P is digestion

F1: breakdown of fats occurs at duodenum

F2: bile produced by the liver helps to emulsify fat

F4: pancreas secretes lipase

F5: which digest lipid into fatty acid and glycerol

Diagram 9.1(b)

P2: Process Q is absorptionF1: take place in the ileum.

F2: fatty acid and glycerol absorb in the lacteal

F3: the lipid then is transported via lymphatic system into

bloodstream

Diagram 9.1(c)

P3: Process R is assimilation

F1: take place in the cell

F2: lipid is the major components to form complex compound such

as plasma membrane

1

1

1

1

1

11

1

1

1

1

1

Max

4

4

Max

2

Max

10

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SULIT 4551/2

Question No

Mark ing Criteria Submarks

Totalmarks

9 (b)  Able to explain the relationship between the eating habit and

 physical condition of each person and able to suggest the ways to

improve their daily food intake

Sample answer:

Diagram 9.2(a)

F1:Anorexia nervosa

P1:experience an intense fear of gaining weight

P2: skips meals for long periods of time/lose appetite to eat / do not

want to eat at all

P3:cause drastic lost in body weight (which is 15% below than

normal weight)

P4: cause tissue repair cannot take place due to lack of

protein//heart malfunction//fatal

Ways to improve their health condition

P5: counselling correction any distorted belief about their body

P6: take right amount of carbohydrates to provides energy//protein

to produce new cells//fats to provide energy//vitamin

/mineral/fibres/drink lot of water to maintain healthy/ practise

balanced diet

Diagram 9.2(b)

F2:ObesityP1: Body weight exceed by 25% of normal weight  //excessive

storage of fat

P2: which result s from imbalance between food intake and energy

expenditure.

P3: eating too much fat/oily foods/excess carbohydrates/excess

sugar

P4: lack of exercise

Ways to improve daily food intake

P5:practised balanced diet

P6:eating not more than what is required by the body

P7:eat more fruits/vegetables//drink a lot of water to prevent

constipation

P8:eat less fat/oily foods/carbohydrates/excess sugar

P9:practise more exercise

1

1

1

1

1

1

1

11

1

1

1

1

1

1

1

1

Max

3

2

Max

3

Max

2

Max

10

Total 20

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SULIT 4551/3

[Lihat sebelah

4551/3 SULIT

MARKING SCHEME : PAPER THREE  – set 1

Question 1 : 1(a)

Score Mark Scheme

3

Able to record all readings of the number of gas bubbles produce in the table

provided

Concentration of sodium

hydrogen carbonate

solution (%)

Number of gas bubbles

released in 5 minutes

0.2 2

0.4 5

0.6 8

0.8 10

2 Able to record any 3 data

1 Able to record any 1- 2 data

0 No response or wrong response

1(b)(i)

Score Mark Scheme

3

Able to state two correct observations based on following criteria.

C1 –  Concentration of sodium hydrogen carbonate solution (%)

C2 –  Number of gas bubbles released in 5 minutes

Sample Answer:

1.  At the concentration of 0.2% sodium hydrogen carbonate, the number of gas

bubbles released are 2.

2. At the concentration of 0.8% sodium hydrogen carbonate, the number of gas

bubbles released are 10 .

2 Able to state one correct observation and one inaccurate response.

Sample answer:

1.  At the concentration of 0.2% sodium hydrogen carbonate, the number of gas

bubbles released least.

1 Able to state one correct observation or two inaccurate response or idea.

Sample answer:

1. Gas bubbles is produced

0 No response or wrong response (response like hypothesis)

1(b) (ii)

Score Mark Scheme

3

Able to state two reasonable inferences for the observation.

Sample answer:

1. When the concentration of carbon dioxide is low caused more gas to be produced,

 photosynthesis process is less // the rate of photosynthesis decrease /low  

2. When the concentration of carbon dioxide is more caused more gas to be

 produced, photosynthesis process is more// the rate of photosynthesis increase/

high

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2 Able to state one correct inference and one inaccurate inference.

Sample answer:

1. When the concentration of carbon dioxide is low caused more gas to be

 produced// the rate of photosynthesis  is low. 

1 Able to state one correct inference or two inaccurate inference or idea.

Sample answer:

1. The number of gas bubble is influenced by concentration of carbon dioxide.

0 No response or wrong response (inference like hypothesis)

Score Mark Scheme

3 Able to state all the variables and the method to handle variable correctly

(√) for each variable and method

Able to get 6 √ (with the correct key words) 

Variable Method to handle the variable

Manipulated variable

Concentration of sodium hydrogen

carbonate// concentration of carbon

dioxide

Use different concentration of sodium

hydrogen carbonate

Responding variable

1. The number of gas bubble released//

2. The rate of photosynthesis 

Count and record the number of gas

bubble//

Calculate the rate of photosynthesis

by using formula(number of bubble

 per minute) 

Fixed variable

Temperature of the water// distance of

the lamp// light intensity

Fixed water temperature at 28 0 C//

Fixed the distance of lamp at 20cm//

Fixed the 10 W lamp bulb  

2 Able to get 4 – 5 √

1 Able to get 2 – 3 √ 

0 No response or wrong response

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1(d)

Score Mark Scheme

3

Able to state the hypothesis correctly based on the following criteria:

V1 – the concentration of sodium hydrogen carbonate

V2 – the number of gas bubble

R - State the relationship between V1 and V2.

 As the concentration of sodium hydrogen carbonate/ the concentration of carbon

dioxide increases, the number of gas bubble increases // the rate of photosynthesis

increases as the concentration of sodium hydrogen carbonate increases.

2

Able to state the hypothesis but less accurate.

Number of bubbles gas depends on the concentration of carbon dioxide

1

Able to state the idea of the hypothesis

Concentration of carbon dioxide affects photosynthesis

0 No response or wrong response

1(e)(i)

Score Mark Scheme

3

Able to construct a table and record the result of the experiment with the following

criteria:

If without unit (x)

Concentration ofsodium hydrogen

carbonate(%)

Number of gasbubbles(unit)

Rate of photosynthesis(unit/minute)

0.2 2 0.4

0.4 5 1.0

0.6 8 1.6

0.8 10 2.0

2 Able to construct a table and record any two criteria1 Able to construct a table and record any one criteria

0 No response or wrong response

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1(e)(ii)

Score Mark Scheme

3

Able to draw a line graph of concentration of sodium carbonate solution against the

rate of photosynthesis with three criteria.

Axes (A)  – both axis are labeled with units, uniform scales, independent

variable on horizontal axis. (√) Point (P)  – All points are correctly plotted. (√) 

Curve (C) - smooth curve

2 Graph with any two criteria.

1 Graph with any one criteria.

0 No response or wrong response.

1(f)

Score Mark Scheme

3

Explain the relationship between the rate of photosynthesis and the concentration of

sodium hydrogen carbonate solution based on the graph in 1(e)(ii).

The higher the concentration of sodium hydrogen carbonate, the higher the rate of

 photosynthesis. More gas/ oxygen is produced because more carbon dioxide will

increase the number of bubble gas produced  

2

Able to explain briefly the rate of photosynthesis and the concentration of sodium

hydrogen carbonate solution

The higher the concentration of sodium hydrogen carbonate, the higher the rate of

 photosynthesis because more gas/ oxygen is produced

1

Able to explain the idea the rate of photosynthesis and the concentration of sodiumhydrogen carbonate solution

The higher the concentration of sodium hydrogen carbonate, the higher the rate of

 photosynthesis. 

0 No response or wrong response

Rate of photosynthesis

(unit/minute)

Concentration of sodium

carbonate solution (%)

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1(g)

Score Mark Scheme

3

Able to predict correctly and explain the prediction based on the following item:

C1 – the rate of photosynthesis

C2 – light intensityC3 - concentration of sodium carbonate solution

The rate of photosynthesis will increase more then 2.0 unit/minute because the light

intensity has increased and the concentration of carbon dioxide is the limiting

 factor.

2 Able to predict based on any two criteria.

1 Able to predict based on any one criteria.

0 No response or wrong response

1(h)

Score Mark Scheme

3

 Able to state the definition of photosynthesis correctly, based on the following criteria.

C1  – Hydrilla sp in sodium carbonate solutionC2  – produces gas bubblesC3  – the rate of photosynthesis is influence by the concentration of sodium carbonate

solution

Photosynthesis is a process when an aquatic plant/ Hydrilla sp release gas

bubbles/oxygen and is affected by different concentration of sodium hydrogen

carbonate // carbon dioxide.

2 Able to state the definition of urine based one of the two criteria.

1 Able to state the idea of photosynthesis

0 No response or wrong response

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1(i)

Score Mark Scheme

3

Able to classify the apparatus and material used in the experiment

Materials Apparatus

0.3% sodium hydrogen

carbonate solution

 An aquatic plant

Water bath

Lamp

Stopwatch

thermometer

2Able to classify two apparatus and two material correctly.

1Able to classify two apparatus and one material correctly.

0 No response or wrong response

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QUESTION 2

Question Mark scheme Marks

2(i) Able to state a problem statement relating the MV to RV correctly.

P1 : the concentration of enzyme

P2: the rate of enzyme reaction//the time taken for the hydrolysis

of starch to be completed

P3: relationship with question form

Sample answers:

1. What is the effects of enzyme concentration on the activity of

salivary amylase on starch//the time taken for the hydrolysis of

starch to be completed ?

2. How does the enzyme concentration affects the activity of the

salivary amylase//the time taken for the hydrolysis of starch to be

completed?

3

Able to state a problem statement less accurately / less one

criteria.

Sample answers:

1.  Enzyme concentration affects the activity of salivary amylase

2. What is the effect of enzyme on the activity of salivary amylase?

2

Able to state a problem statement at idea level / less two criteria.

Sample answer:

1. Enzyme affects the activity of salivary amylase

1

No response or incorrect response 0

2(ii) Able to state a hypothesis based on the following criteria correctly.

P1 : the concentration of enzyme (Manipulated Variable)

P2: the rate of enzyme reaction//activity of salivary amylase//

the time taken for the hydrolysis of starch to be completed

(Responding Variable)

H : relationship between MV and RV

Sample answers:

1.  The higher the enzyme concentration, the higher the rate of

enzyme reaction

2.  When the enzyme concentration increase, the rate of enzyme

reaction increase

Note: Accept hypothesis with wrong conclusion

3

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Able to state a hypothesis less accurately / less one criteria.

Sample answers:

1.The enzyme concentration affects the rate of enzyme reaction

2

Able to state a hypothesis at idea level / less two criteria.

Sample answer:

1.  Enzyme affects the time taken for hydrolysis of starch to be

completed

1

No response or incorrect response 0

2(iii) Able to state all 3 variables correctly.

Sample answers:

1.  Manipulated variable:

Enzyme concentration

2.  Responding variable :

The rate of enzyme reaction//the time taken for hydrolysis of

starch

3.  Constant variable :

Temperature // volume of enzyme//concentration of

substrate/starch

3

Any 2 variables correct 2

Any 1 variable correct 1

No response or incorrect response 0

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Question Mark scheme Marks

2(iv) Able to list all the important apparatus and materials correctly

-6 apparatus and 4 materials = 6A + 4 M

Sample answers:

Apparatus: (A)

1.Water bath // (Beakers +distilled water+ thermometer)

2.Tile with grooves

3.Test tube

4.Thermometer

5.Syringe

6.Stopwatch

Note: the use of water bath is equal to( beaker + water +

thermometer )

Materials: (M)

1.starch suspension

2.saliva

3. distilled water

4.iodine solution0

3

Any 4 A + 3M

2

Any 2A + 2M

1

No response / wrong response

Sample answer:

Any 3/2/1 A only

0

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2(v) Able to describe the steps of the experiment correctly.

Sample answer:

1.  Prepare three test tubes and label with A- C.(K1)

2.  Fill test tube A with 10% amylase enzyme concentration

solution (K1)3.  Immerse the test tube in water bath set at 37C throughout the

experiment (K1,K5)

4.  Meanwhile , add few drops of iodine solution onto the grooves

of white tile. (K1)

5.  Add 4 ml of 1% starch suspension to the test tube by using a

syringe.(K1,K2)

6.  Immediately, start the stopwatch (K1)

7.  Stir the mixture in the test tube by using a glass rod (K1,K5)

8.  Remove a small amount of the mixture and test with iodine

solution on the tile(K1)

9.  Repeat iodine test at 30-second intervals until the mixturedoes not turn blue-black when tested with iodine

solution(K1,K2)

10. Record the time taken for the hydrolysis of starch to be

completed (K3).

11. Calculate the rate of enzyme reaction by using formula

= 1 ( min –1

 ) (K3)

time

12. Repeat step 2 to 11 with 15% and 20% of amylase enzyme

concentration.(K1,K4)

13. Record the results in a table. (K3)

K1 : Steps 1, 2, 3,4,5,6, 7,8,9 and 12 ( any 4 K1)

K2 : Steps 5,9 ( any one step)

K3 : Steps 10,11 (any one step)

K4 : Steps 12

K5 : Steps 3,7 ( any one step)

5 K 3

3- 4 K 2

1-2 K 1

0

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SULIT 4551/3

[Lih t b l h

Question Mark scheme Marks

2(vi) Able to construct a table to present data with the following criteria:

T : Title with correct units

M: Manipulated variableSample answers:

Concentration of

amylase (%)

Time taken for

the hydrolysis of

starch to be

completed (min)

Rate of enzyme

reaction

( 1/ t ) (min-1

)

10

15

20

2

No response or incorrect response 0

END OF MARKING SCHEME

 PERATURAN PEMARKAHAN TAMAT

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