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SULIT 2 4551

4551/PP © 2016 Hak Cipta BPSBPSK [Lihat Halaman Sebelah]

SULIT

PAPER 1

No Answer No Answer No Answer No Answer No Answer

1 C 11 B 21 B 31 A 41 A

2 A 12 C 22 A 32 D 42 A

3 B 13 D 23 C 33 C 43 D

4 A 14 D 24 C 34 D 44 B

5 C 15 D 25 D 35 A 45 A

6 C 16 C 26 C 36 C 46 A

7 D 17 C 27 B 37 B 47 C

8 A 18 B 28 B 38 C 48 B

9 D 19 A 29 A 39 B 49 B

10 C 20 C 30 D 40 D 50 C

PAPER 2

Question 1

No Criteria Marks

(a) (i) Able to state the elements that make up the molecule X.

Answers:

- Carbon, hydrogen, oxygen (All correct) (Accept correct symbols)

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(ii) Able to write the chemical formula for molecule X.

Answer:

- C6H12O6

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(iii) Able to state the food class of substance X.

Answer:

- Carbohydrate

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(b) Able to explain what happens to molecule X after entering the body cells.

Sample answers:

P1 - (In body cell), glucose is oxidized

P2 - To produce energy

P3 - By cellular respiration

(Any 2)

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(c) Able to explain the effects of taking food high in glucose if the pancreas is

malfunctioned.

Sample answers:

P1 - No/less production of insulin

P2 - Excess glucose is not convert to glycogen

P3 - Causes high blood sugar/hyperglycemia

P4 - No/less uptake of glycogen by adipose tissue/muscle cells

P5 - He suffer diabetes mellitus

(Any 3)

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(d) (i) Able name molecule Y.

Answers:

- Maltose/Lactose/Sucrose

(Any 1)

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(ii) Able to explain the formation of molecule Y.

Sample answers:

P1 - By condensation

P2 - Eliminates/released one water molecule

P3 - Form a bond between two glucose molecules / glucose and galactose /

glucose and fructose

(Any 2)

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(e) Able to explain why molecule Y cannot be absorbed into the body cells.

Sample answers:

P1 - Molecule X / maltose is a disaccharide // Large (organic) molecule

P2 - Cannot pass through / cross the plasma membrane

P3 - (Because) there are no transport proteins to carry them across

(Any 1)

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TOTAL 12

Question 2

No Criteria Marks

(a) (i) Able to state the type of skeleton in earthworm.

Answer:

- Hydroststic skeleton

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(ii) Able to Eexplain the action of antagonistic muscles in the earthworm.

Sample answers:

P1 - A muscle opposes the action of another muscle

// One muscle contracts while another relaxes

P2 - When circular muscle contracts, the worm becomes longer/thinner

P3 - When longitudinal muscle relaxes, the worm shorter/fatter

(Any 2)

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(b) Able to explain why the earthworm dificult to move forward when place on

a white tile.

Sample answers:

P1 - The tile surface is smooth/slippery // No friction

P2 - The chaeta/bristles cannot grip/anchor on the tile/ground

P3 - The earthworm will slip backwards

P4 - The peristaltic locomotion is slowed down/difficult

(Any 2)

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(c) (i) Able to explain the adaptation of the scales that cover the body of the fish.

Sample answers:

P1 - Scales have a layer of slime/slimy mucus on them

E1 - To swim through water faster/smoother // Less friction

E2 - Slime protects fish from bacteria/parasites in the water (Any 2)

OR

P2 - Scales arranged backwards / in a head-to-tail configuration / like

roof tiles/overlapped

E3 - To allow smooth flow of water over the body// reduce friction with the

water.

E4 - To form a protective flexible armor to withstand blows and bumping /

protect fish from injuries (Any 2)

OR

P3 - Multiple/overlapping scales

E5 - That allows fish to move easily while swimming // Less friction

E6 - Provide a flexible covering // Scale protect from injuries

(Any 2)

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(ii) Able to name fin P.

Answer:

- Dorsal fin

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(iii) Able to explain the problem faced by the fish if fin Y is too small compared

to its body size.

Sample answers:

P1 - Fish becomes less stable // cannot balance as they move // cannot swim

through current / strong waves

P2 - Fish will not remain upright (as they swim through water)

P3 - Fish easily rolling // Tendency to roll // Rotate its belly up

P4 - Fish easily yawing // Tendency to yaw // Moving to the right or left

P5 - Fish has problem to make turns (in the water)

(Any 2)

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(iv) Able to explain how the fish uses its caudal fin to increase the speed of its

movement.

Sample answers:

P1 - Caudal fin is broad

P2 - It helps fish to propel forward through water quickly

// It provides forward thrust

P3 - It controls the fish direction

P4 - It provides a balance between speed and mobility/movement

// Fish can swim over long distances while maintaining high speeds.

(Any 2)

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TOTAL 12

Question 3

No Criteria Marks

(a) (i) Able to name of nerve cells.

Sample answers:

R - Afferent neurone // Sensory neurone

S - Efferent neurone // Motor neurone

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(ii) Able to state the difference in structure and function of R and S.

Sample answers:

Structure:

P1 R has long dendron S has long axon

P2 R is attached to receptor S is attached to effector

(Any 1)

Function:

F1 R sends nerve impulse from

receptor

S sends nerve impulse to effector

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(b) (i) Able to explain the meaning of reflex action.

Sample answers:

P1 - An automatic response/action (to a particular stimulus)

P2 - Involves involuntary contraction of skeletal muscles

P3 - Involves the spinal cord only // Does not involve central nervous system

(Any 2)

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(ii) Able to explain how the injury of the ventral root affect his action when he

accidentally pricked by a sharp pin.

Sample answers:

P1 - He is able to detect/feel the pain

P2 - Because the afferent neuron is not affected

// Impulse can be transmitted to the spinal cord / central nervous system

P3 - Impulse cannot be sent to effector

P4 - Muscle cannot be contracted

(Any 2)

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(c) Able to explain the transmission of nerve information across a synapse.

Sample answers:

P1 - (At the synaptic knob,) the nerve impulse is converted into chemical

signal/neurotransmitter.

P2 - Neurotransmitter diffuse through the synaptic knob to the adjacent/next

dendrite

P3 - (At the adjacent dendrite) the chemical signal/neurotransmitter is

converted back to nerve impulse // New impulse is triggered

(Any 2)

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(d) Able to state what organelles are abundant in nerve cells and explain why.

Sample answers:

P1 - Mitochondria

P2 - To generate more energy

P3 - For transmission of impulse // Ions pumps

(Any 2)

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TOTAL 12

Question 4

No Criteria Marks

(a) (i) Able to state two main organic food substances that contain in milk.

Sample answers:

- Protein / Lipid / Lactose

(Any 2)

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(ii) Able to explain the digestion of one organic food substance stated in (a)(i).

Sample answers:

P1 - Protein is digested to polypeptide

P2 - By enzyme pepsin

P3 - (Digestion takes place) in stomach

OR

P4 - Lipid is digested to fatty acid and glycerol

P5 - By enzyme lipase

P6 - (Digestion takes place) in duodenum /ileum

OR

P7 - Lactose is digested to glucose and galactose

P8 - By enzyme lactase

P9 - (Digestion takes place) in ileum

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(b) Able to explain why milk is good for pregnant mothers.

Sample answers:

P1 - Milk contain high calcium

P2 - Calcium is important to strengthen the mother’s bone

P3 - Calcium is important for formation of bone/teeth of baby/fetus

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(c) (i) Able to compare the absorption of digested food by R and S in villus

Sample answers:

S1 - Glucose, amino acids, fatty acid and glycerol (any 3 ) are small/simple

molecules

D2 - R absorbed glucose/amino acid while S absorbed fatty acid /glycerol

D3 - R absorbed water soluble substance while S absorbed fat soluble

substance

(Any 2)

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(ii) Able to describe the role of liver.

Sample answers:

P1 - Liver stores glycogen/vitamin A/ D/ iron

P2 - Regulate glucose level (in blood)

P3 - Excess glucose is converted to glycogen (by insulin hormone)

P4 - Deamination process

P5 - Excess amino acid is converted to urea

P6 - Produce bile

P7 - To emulsify lipid

P8 - Detoxification

// Convert toxic substance /examples, alcohol/drug into non-toxic form.

(Any 2)

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TOTAL 12

Question 5

No Criteria Marks

(a) (i) Able to explain role of light energy.

Sample answers:

P1 - To excited electrons from chlorophyll (molecules)

E1 - Energy is released // Energy is trapped in ATP molecules

P2 - To split water molecule

E2 - Into hydroxide ion and hydrogen ion

(Any 2)

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(ii) Able to explain the effects to the mechanism of dark reaction if the plant is

exposed to light for 24 hours everyday.

Sample answers:

P1 - More hydrogen (ions/atoms) are produced during light reaction

P2 - More carbon dioxide can be fix by hydrogen atom

P2 - More glucose / amino acid / fatty acid are produced

P3 - Rate of dark reaction increase

(Any 2)

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(b) Able to explain what happen to glucose in mitochondrion.

Sample answers:

P1 - Glucose is oxidised by oxygen

P2 - energy is released

P3 - ADP used energy released

P4 - To fix (one) phosphate molecule // Forms ATP/energy molecule

(Any 3)

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(c) Able to explain the difference in the production of CO2 gas by plant before

and after reaching the compensation point, as the light intensity increases.

Sample answers:

P1 - Before reaching compensation point, the carbon dioxide /CO2 produced

/released out from plant is decreasing but

after reaching compensation point, the CO2 released out is zero.

E1 - As light intensity increase, rate of respiration is decreasing while rate

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of photosynthesis is increasing // CO2 used in photosynthesis /dark

reaction is increased

E2 - After reaching compensation point, the rate of photosynthesis is higher

than rate of respiration (by plant)

E3 - CO2 produced / release out is zero

// More CO2 is taken from environment

(Any 2)

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(d) Able to suggest how to increase crop yield in the agricultural green house

and support the suggestion with suitable explanations.

Sample answers:

P1 - Increase light intensity / temperature / CO2 concentration / water supply

at optimum

E1 - High light intensity increase light reaction // more water molecules split

into ion hydrogen and hydroxide (by light energy)

E2 - More atom hydrogen produced // Fix more CO2 that supplied

E3 - (Thus) more glucose produced // starch/protein /lipid stored in plant

(tissue/ storage organ)

E4 - Rate of photosynthesis increase

E5 - Water supply / high humidity will reduce transpiration/water lost from

plant

P2 - Ventilation used to ensure air flow // To release excess O2 / and take in

more CO2 / control humidity

P3 - Condition of soil/water/temperature/humidity is automatically sensored

/controlled by computer

(Any 3)

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TOTAL 12

Question 6

No Criteria Marks

(a)

Able to explain how different concentration of sodium chloride solution

affect the condition of erythrocytes.

Sample answers:

At A

P1 - The percentage of erythrocytes hemolysed in hypotonic solution

depends on the degree of hypotonicity of the solution.

P2 - Between concentration of 0.0% and 0.3% of sodium chloride, the

percentage of hemolysis is high / 90%-100% of erythrocyte hemolysed

P3 - At 0% of sodium chloride solution, all / 100% of erythrocyte burst

/ hemolysed

P4 - The solution is hypotonic towards erythrocytes // in a stronger

hypotonic solution.

P5 - Water diffuse into the cell by osmosis

At B

P6 - From concentration 0.3% to 0.6% of sodium chloride, the percentage of

hemolysis is decreasing / decrease from 80% to 10% of erythrocyte

burst / hemolysed

P7 - The solution is less hypotonic towards erythrocytes

P8 - Less water diffuse into the cell by osmosis

At C

P9 - In solution above 0.6% sodium chloride, only a low percentage of

erythrocyte experience hemolysis / less than 20% of erythrocyte burst

P10 - At 0.9% sodium chloride solution is said to be isotonic to erythrocytes

// Osmotic equilibrium occurs

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P11 - Water movement into the cell is equal to the water movement out of

the cell.

(Any 8)

1

(b) Able to explain the formation of lymph in lymph capillary.

Sample answers:

P1 - Blood flows from arterial end of capillaries which has smaller diameter

P2 - Create higher hydrostatic pressure

P3 - Force water / small molecules substance filtered out of blood capillary

wall.

P4 - Large molecules such as erythrocytes/ platelet/ large protein molecules

are not filtered out.

P5 - Form interstitial fluid which fill the spaces between the cells

P6 - And constantly bathes the cells.

P7 - About 10-15% of interstitial fluid in the interstitial space diffuse into the

blunt end of lymph capillary.

P8 - To become lymph.

(Any 6)

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(c) Able to explain the health problem faced by a person whose lymphatic

vessels are blocked by filarial worms.

Sample answers:

P1 - (Filarial) worms lodge/fill in the lymphatic vessel

P2 - And lymph nodes

P3 - Reduce defence mechanism // Less/no phagocytisis/antibodes

P4 - Prevent the lymph from returning to the bloodstream

P5 - Interstitial fluid stay in the tissue

P6 - The worms producing millions of larvae // worm reproduce in lymphatic

vessel

P7 - (The larvae/worm) circulate in the blood

P8 - And cause the affected organ/tissue to become (massively) swollen //

oedema

P9 - (The disease is called) elephantiasis // skin/tissue thickening of limbs

(Any 6)

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TOTAL 20

Question 7

No Criteria Marks

(a)

Able to describe the energy transfer in the food chain.

Sample answers:

P1 - The food chain shows the flow of energy from one trophic level to

another

P2 - The paddy plant / producer received (the highest) energy from sun

P3 - The light energy is converted to chemical energy/food

P4 - Through photosynthesis

P5 - (When) rat eats/consumes paddy plant, energy is transferred to rats

P6 - 10% energy is transfer from plant to rats // 90% of energy loss

P7 - (90% of energy is lost) as respiration/reproduction/activity/any

suitable process

P8 - 10% of energy in rat /first consumer is transferred to snake /second

consumer when the snake consumes the rat

P9 - 10% of energy is transferred to eagle when the eagle consumes the

snake.

P10 - The eagle/third consumer obtain the lowest energy in the food chain.

(Any 6)

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(b) Able to explain the changes in population size of the rat and owl in the oil

palm plantation. Sample answers:

P1 - (The interaction between rat andowl is) prey-predator P2 - Rat is a prey // Owl is a predator

P3 - This is biological control P4 - When there are sufficient resources/food, the prey’s/rat population

increase

P5 - (When the population size of rat increase), population size of predator/

owl also increase

P6 - Because owl have sufficient food

P7 - (Thus) the population of rats decrease

P8 - Because rat/prey has been eaten by predator/owl

P9 - (Thus) the population of owl also decrease

P10 - Due to lack of food

P11 - The population of prey controls/determines the population of predator

(and vice versa)

P12 - The populations of both organisms is in a dynamic equilibrium.

(Any 8)

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(c) Able to describe the effect of competition on the growth rate of organism.

Sample answers:

P1 - Competition among members in the same species is called intra-

specific competition

P2 - Any example of intraspecific competition; Paddy plant and paddy

plant/ Bryophyllum with its offspring // male deer compete for mate

P3 - Competition between different species is called interspecific

competition

P4 - Any example of interspecific competition; Paddy plant with maize

/Paramecium caudatum with Paramecium aurelia // Leopard and lion

feed on the same prey

P5 - The organism competes with each other to obtain food /water/nutrient

/light/limited sources/space /mate

P6 - Dominant /strong species/individual obtain more food /water /space

/light /nutrient

P7 - (thus) the growth rate of dominant species/individual increase

P8 - Dominant /strong species /individual will dominate the habitat

P9 - The growth rate of weak species /individual decrease

P10 - Weak species will loss /extinct (from habitat)

(Any 6)

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TOTAL 20

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Question 8

No Criteria Marks

(a)

Able to describe how the levels of hormone secreted by pituitary gland and

ovary controls the menstrual cycle in a woman.

Sample answers:

Day 1-6

P1 - Pituitary gland secretes FSH

P2 - FSH will stimulate the development of (Graafian) follicle

P3 - Endometrium (wall) / Uterus lining disintegrate

P4 - Blood discharge // Menstruation

Day 7-13

P5 - Follicle cells/tissues secretes oestrogen (hormone)

P6 - Oestrogen helps to repair the endometrium (wall) / uterus lining

P7 - Oestrogen inhibit the secretion of FSH (by pituitary gland)

P8 - Oestrogen will stimulate the secretion of LH (by pituitary gland.)

Day 14

P9 - LH will stimulate ovulation

P10 - Graafian follicle will release the secondary oocyte

P11 - LH will stimulate the formation of corpus luteum

Day 15-28

P12 - Corpus luteum will secrete progesterone (and oestrogen)

P13 - Progesterone will maintain the thickness of endometrium (wall) /

uterus lining

P14 - Corpus luteum will degenerate

P15 - Reduces the secretion of progesterone

P16 - The endometrium (wall) / uterus lining will start to disintegrate

P17 - Menstrual cycle happens again

(Any 10)

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10

(b) Able to explain the advice and suggestions for both couples on how to

increase the chance to have babies. Sample answers:

Suggestion 1

P1 - Perform in vitro fertilization / IVF

P2 - Secondary oocyte / ovum is taken out from ovary

P3 - By using a laparoscopy

P4 - Sperm and secondary oocyte/ovum are fertilised

P5 - In culture medium/growth medium

P6 - The zygote is allow to grow to embryo // Zygote undergo mitosis to

form embryo (of eight cells)

P7 - The embryo is implanted into the uterus

P8 - Of biological mother (for couple 1)

P9 - Of surrogate mother (for couple 2)

Suggestion 2

P10 - Perform artificial insemination

P11 - Collect sperms (from husband) until certain concentration

P12 - (Concentrated sperm/semen) place in uterus tube of wife

P13 - Increase chances for fertilisation

Suggestion 3

P14 - Perform gamete intra-fallopion transfer/GIFT // zygote intra-fallopion

transfer/ZIFT technique

P15 - Gamete/zygote place in fallopion tube of wife

P16 - Increase chances for fertilisation

(Any 10)

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TOTAL 20

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Question 9

No Criteria Marks

(a)

Able to explain the factors of variation.

Sample answers:

P1 - Genetic factors / Environmental factors / Interaction between genetic

and environmental factors (Any 2)

P2 - Independent/random assortment of homologous chromosomes

P3 - During metaphase 1 (meiosis)

P4 - Crossing-over / exchange of genetic material

P5 - Between non-sister chromatids of homologous chromosomes

P6 - Recombination of gene (in chromosomes)

P7 - Occurs during prophase 1 (meiosis)

P8 - Random fertilization

P9 - All the sperms have the same chance to fertilise the ovum

P10 - Mutation // Sudden change in genetic materials

P11 - Due to mutagents // any two examples

P12 - Chromosomal mutation

P13 - Change in structure and number of chromosome

P14 - Example of change in structure of chromosome

P15 - Example of change in number of chromosome

P16 - Gene mutation

P17 - Change in the genetic codes / sequence of nitrogenous base

P18 - Example of change in the genetic codes / sequence of nitrogenous base

P19 - Example of environmental factors

P20 - Example of interaction between genetic and environmental factors

P21 - Example of continuous variation

P22 - Example of discontinuous variation (Any 10)

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10

(b) Able to explain the suitable husband or wife so all their children are Rhesus

positive.

Sample answers:

Husband for Susan

P1 - Susan can marry a man with Rhesus positive / Rhesus negative

P2 - Genotype for Susan is RhRh

P3 - Susan produced 100% gametes that carry Rh /dominant allele

P4 - Gamete with Rh/dominant allele (from Susan) fertilise with

Rh/dominant allele/ rh /recessive allele (from husband)

P5 - 100% /50% offspring has RhRh/Rhrh

Husband for Tina

P6 - Tina must marry a man that is homozigous Rhesus positive /RhRh

P7 - Genotype of Tina is heterozygous /Rhrh

P8 - Tina produce gamete with Rh/dominnat allele and rh /recessive allele

P9 - The man/husband produce 100% gamete with Rh/dominant allele

P10 - 50% /50% offspring has RhRh/Rhrh

Wife for Raul

P11 - Raul must marry a woman that is homozigous Rhesus positive /RhRh

P12 - Genotype of Raul is rhrh / homozygous recessive

P13 - Raul produced 100% gametes with rh

P14 - His wife produce 100% gamete with Rh

P15 - 100% of their child have (genotype with) Rhrh /heterozygous

(Any 10)

Note : Accept genetic diagrams with complete labels/explanations.

(Maximum 3 marks for each genetic diagram)

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TOTAL 20

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PAPER 3

Question 1

1 (a) [KB0603 - Measuring Using Number]

Score Criteria

3

Able to state all genotype and all phenotype of the offsprings correctly.

Answers:

Offspring Genotype Phenotype Offspring Genotype Phenotype

1 Bb Black 13 bb White

2 BB Black 14 Bb Black

3 Bb Black 15 BB Black

4 bb White 16 bb White

5 Bb Black 17 Bb Black

6 BB Black 18 BB Black

7 bb White 19 Bb Black


9 bb White 21 BB Black

10 Bb Black 22 bb White

11 BB Black 23 Bb Black


Note: If use different simbols, score 2 only.

2 Able to state any 20-23 of genotype and any 20-23 of the phenotype of the offsprings

correctly.

1 Able to state any 10-19 of genotype and any 10-19 of the phenotype of the offsprings

correctly.

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

Score Criteria

3 Able to state any two different observations correctly according to the criteria:

C1 Offspring

C2 Genotype / Phenotype

Sample answers:

Horizontal

1. Offspring 1/3/5/8/10/12/14/17/19/20/23/24 has genotype Bb / heterozygous black.

2. Offspring 1/3/5/8/10/12/14/17/19/20/23/24 has one black allele and one white allele.

3. Offspring 2/6/11/15/18/21 has genotype BB / homozygous black.

4. Offspring 2/6/11/15/18/21 has two black alleles /alleles for black.

5. Offspring 4/7/9/13/16/22 has genotype bb / homozygous white.

6. Offspring 4/7/9/13/16/22 has two white alleles.

7. Offspring 6/.. is black / has black fur / the phenotype is black.

8. Offspring 7/.. is white / has white fur / the phenotype is white.

Vertical

9. The number of offsprings with genotype Bb is double/twice/more than the number of

offsprings with genotype BB / bb.

10. The number of offsprings with genotype BB / bb is 6.

11. The number of offsprings with genotype Bb is the most.

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2 Able to state any one observation correctly and one incomplete observation.

or

Able to state any two incomplete observations.

Sample answers for incomplete observations:

Horizontal

1. Offspring 1/3/5/8/10/12/14/17/19/20/23/24 is heterozygous.

2. Offspring 2/6/11/15/18/21 is homozygous.

3. Offspring 4/7/9/13/16/22 is homozygous.

Vertical

4. The number of offsprings with genotype BB / bb is less.

5. The number of offsprings with genotype Bb is more.

6. The number of black offsprings is more.

1 Able to/state any one idea of observation (Any 1 criterion)

Sample answers:

1. Offspring 1/3/5/8/10/12/14/17/19/20/23/24 has one black button and one white button.

2. Offspring 1/3/5/8/10/12/14/17/19/20/23/24 inherit one black button from mother.

3. Offspring 1/3/5/8/10/12/14/17/19/20/23/24 inherit one white button from father.

4. Offspring 2/6/11/15/18/21 has two black buttons.

5. Offspring 4/7/9/13/16/22 has two white buttons.

6. Offspring 2/6/11/15/18/21 has same buttons.

7. Offspring 4/7/9/13/16/22 has same buttons.

1 (b) (ii) [KB0604 - Making inferences]

Score Criteria

3 Able to make one logical inference for each observation based on the criteria:

Horizontal

C1 Allele inherited (from mother / father)

C2 Dominant / recessive trait

Vertical

C3 Chance / Probability

Sample answers:

Horizontal (C1+C2)

1. Offspring (1/3/5/8/10/12/14/17/19/20/23/24) has (one) dominant allele and show

dominant trait // ..inherit (one) dominant allele and black (fur) is dominant (trait).

2. Offspring (2/6/11/15/18/21) has (two) dominant alleles and black (fur) is dominant.

3. Offspring (4/7/9/13/16/22) has (two) recessive alleles and white (fur) is recessive.

Vertical (C3)

4. The probability to get offsprings with Bb is double/twice/more than the offsprings with

genotype BB / bb.

5. The probability to get offsprings with genotype BB / bb is less/least.

6. The probability to get offsprings with genotype Bb is the most.

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2 Able to make one logical inference for any one observation and one inaccurate inference

for the other observation.

or

Able to make two inaccurate inferences base on one criterion (C1, C2 or C3) for each

observation.

Sample answers for incomplete inferences:

Horizontal (C1/C2)

1. Offspring (1/3/5/8/10/12/14/17/19/20/23/24) has one dominant allele.

2. Offspring (1/3/5/8/10/12/14/17/19/20/23/24) is black (fur) / has dominant trait.

3. Offspring (2/6/11/15/18/21) has (two) dominant alleles.

4. Offspring (4/7/9/13/16/22) is white (fur) / has recessive trait.

Vertical (incomplete C3)

5. The probability to get offsprings with Bb is double/twice/more.

1 Able to make an idea of inference with one criterion.

Sample answers:

Horizontal (idea C1/C2)

1. Offspring (1/3/5/8/10/12/14/17/19/20/23/24) is dominant.

2. Offspring (2/6/11/15/18/21) is dominant.

3. Offspring (4/7/9/13/16/22) is recessive.

Vertical (idea C3)

4. Easy to get offsprings with Bb.

For 1(b)(i) Observation and (ii) Inference:

Score Accurate Inaccurate Idea Wrong

3

2

1

0

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1 (c) [KB061001 - Controling Variables]

Score Criteria

3

Able to state all the variables and the method to handle the variables correctly.

Sample answers:

Variables Method to handle the variables

Manipulated variable:

Phenotype/fur colour of parents

// Genotype of parents

Two black rabbits are crossed/used

// Two rabbits (black parents), both with

genotype Bb are crossed / used

// Two (black) heterozygous/hybrid rabbits are

crossed/ used

// Parents genotype involved are Bb and Bb

// Use both Bb

Responding variable:

Phenotype of offsprings

(Count and) record the number of black and

white rabbits

// Calculate the percentage of black and white

rabbits : Number of offspring x 100%

All offsprings

// Calculate the ratio of black and white rabbits

as: X black : Y white

Controlled variable:

Type / species of organism

// Characteristic

// Traits

// Size of buttons

// Random fertilisation

Use/involve rabbits

// Use/involve fur colour

// Use/involve black and white fur

// Use same size (of button)

// Pick button randomly

2 Able to state 4 - 5 of the variables and the method to handle the variables correctly.

1 Able to state 1 - 3 of the variables and the method to handle the variables correctly.

1 (d) [KB0611 - Making Hypothesis]

Score Criteria

3 Able to state a hypothesis to show a relationship between the manipulated variable and

responding variable and the hypothesis can be validated, base on 3 criteria:

C1 Manipulated variable

C2 Responding variable

C3 Relationship (Accept if wrong number, percentage or ratio)

Sample answers:

1. When two black rabbits are crossed, the offsprings are 18 black and 6 white.

2. When black parents are crossed, the offsprings are 18 black and 6 white.

3. When two heterozygous/hybrid (black) rabbits are crossed, the ratio of offsprings is

3black : 1 white.

4. When two black rabbits are crossed, the offsprings are 75% black and 25% white.

5. When two rabbits with genotype Bb are crossed, the offsprings are 18 black and 6 white.

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2 Able to state less accurate hypothesis to show a relationship between manipulated variable

and responding variable base on 2 criteria.

Sample answers:

1. When two black rabbits are crossed, the offsprings are more black (than white). (C1, C2)

2. When two rabbits are crossed, the ratio of offsprings is 3black : 1 white. (C2, C3)

3. When two black rabbits are crossed, the offsprings are black and white. (C1, C2)

4. The offsprings are 18 black and 6 white. (C2, C3)

5. Two black rabbits are crossed, when/if the offsprings are 18 black and 6 white. (Reverse)

6. The phenotypic ratio of offsprings depends on the genotype of the parents.(Less accurate)

1 Able to state idea of hypothesis to show a relationship between manipulated variable and

responding variable base on 1 criterion.

Sample answers (Idea of C1/C2/C3):

1. Two rabbits are crossed. (Idea C1)

2. The offsprings are more black. (C2, Idea C3)

3. The ratio is 3 : 1. (Idea C3)

4. The offsprings are black and white. (C2, Idea C3)

5. The ratio of offsprings depends on the the parents. (Idea C1, C2)

1 (e) (i) [KB0606 - Communicating]

Score Criteria

3

Able to tabulate a table and fill in data accurately base on three criteria:

C1: Titles - Parents: Genotype, Phenotype

Offsprings: Genotype, BB, Bb, bb. Phenotype, Black, White.

C2: Recording data - Parents: BB, Bb, bb, Black, White

Offspring: Number of offspring

C3: Calculation: Percentage of offspring

Sample answer:

Parents:

Genotype BB Bb Bb

Phenotype Black Black White

Offsprings:

Genotype Phenotype

BB Bb bb Black White

Number 6 12 6 18 6

Percentage 25 50 25 75 25

OR

Genotype Phenotype

BB Bb bb Black White

6 12 6 18 6

25% 50% 25% 75% 25%

2 Able to tabulate a table base on two criteria.

1 Able to tabulate a table base on one criterion.

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1 (e) (ii) [KB0608 - Space and Time Relationship]

Score Criteria

3 Able to draw a bar chart based on three criteria:

C1: The x-axis and the y-axis are marked with appropriate values and constant scale.

C2: All points are transferred correctly

C3: Two bars with same width (Bar for black fur is taller than the bar for white fur)

2 Any two criteria

1 Any one criterion

1 (e) (iii) [KB0607 - Interpreting Data]

Score Criteria

3 Able to state the phenotypic ratio of fur colour among the rabbit offsprings and explain the

inheritance of fur colour.

C1: Relationship

(The phenotypic ratio of the offspring is) 3 Black : 1 White (If 3:1, idea level only)

C2: Any two Explanations:

E1 – Black (fur) is determine by a dominant allele/gene

Black (fur) is dominant trait // Black (fur) is dominant and white (fur) is recessive

// Black (fur) is dominant to white (fur)

E2 – (Both) parents have dominant allele / B

(Both) parents produced 50% gametes with B (/ 50% gametes with b)

E3 – The chance/probability of black offspring is 75% (/ white offspring is 25%)

Sample answer:

1. (The phenotypic ratio of the offspring is) 3 Black : 1 White.

Black (fur) is determine by a dominant allele. The parents have dominant allele.

2. 3 Black : 1 White. Black (fur) is a dominant trait. The parents produced 50% gametes

with dominant allele/B.

3. 3 Black : 1 White. Black (fur) is dominant and white (fur) is recessive. The probability

of black offspring is 75%.

2 Able to state the relationship and any one explanation (C1+1E),

OR idea of relationship and two explanations (IdeaC1+2E).

1 Able to state the relationship (C1),

OR idea of relationship and one explanation (IdeaC1+1E).

1 (f) [KB0609 - Define Operationally]

Score Criteria

3 Able to describe the term phenotype operationally, based on the experiment.

Criteria:

C1 Phenotype is (the trait inherited by rabbits, that are the) black fur and white fur in rabbits.

C2 Determined by the (random) pairing of black buttons and white button as the genotype.

C3 The (phenotypic ratio /) phenotype of offsprings depends on the genotype of parents.

Sample answer:

1. Phenotype is black fur and white fur in rabbits. It is determined by the pairing of black

buttons and white button as the genotype. The phenotype of offsprings depends on the

genotype of parents.

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2 Able to state any two criteria

Sample answers:

1. Phenotype is black fur and white fur in rabbits. It is determined by the pairing of black

buttons and white button as the genotype.

2. Phenotype is determined by the pairing of black buttons and white buttons as the

genotype. The phenotype of offsprings depends on the genotype of parents.

3. Phenotype is black fur and white fur in rabbits. The phenotype of offsprings depends on

the genotype of parents.

1 Able to state any one criterion or at idea level only.

Sample answers:

1. Phenotype is black fur / white fur.

2. Pairing of black buttons and white buttons is genotype.

3. The fur colour depends on the parents genes/alleles.

1 (g) [KB0605 - Predicting]

Score Criteria

3 Able to predict the number and the phenotypic ratio of the offsprings when the experiment is

repeated by crossing the same parents in a surounding with more food supply, and explain

the prediction based on three criteria.

C1 Prediction: Any number more than 24 // More than 24. (More, idea level only)

3 black : 1 white (If 3:1, idea level only)

(Either number or ratio, idea level only)

C2 Explanation 1: Higher growth rate // Grow faster // higher breeding rate

The parents are very reproductive.

C3 Explanation 2: Black is dominant trait/allele/gene // White is recessive trait/allele/gene

// The chance / probability of black fur is higher.

Sample answer:

1. 30 offsprings. Ratio 3 black : 1 white. The rabbits grow faster. Black is dominant trait.

2. More than 24 offsprings, 3 black : 1 white. The parents are very reproductive. The chance

of black fur is higher.

2 Able to predict less accurately (Prediction+1explanation//Prediction (idea)+ 2 Explanations)

Sample answer:

1. 30 offsprings. Ratio 3 black : 1 white. The rabbits grow faster.

2. 50 offsprings. The rabbits grow faster. Black is dominant trait.

3. More than 24 offsprings, 3 black : 1 white. The chance of black fur is higher.

1 Able to give idea of prediction. (Prediction // Prediction (idea) + 1 Explanation)

Sample answer:

1. 30 offsprings. Ratio 3 black : 1 white.

2. 50 offsprings. The rabbits grow faster..

3. 3 black : 1 white. The chance of black fur is higher.

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1 (h) [KB0602 - Classifying]

Score Criteria

3

Able to classify the characteristics and traits into three categories correctly, and in the

correct rows.

C1: Characteristic

C2: Dominant trait

C3: Recessive trait

Answer:

Characteristic Dominant trait Recessive trait Blood group Blood group A Blood group O

Height Tall Short

Eye iris colour Brown iris Blue iris

Type of hair Curly hair Straight hair

2 Able to classify the characteristics and traits into two categories correctly, and not

necessary in the correct rows.

1 Able to classify the characteristics and traits into one category correctly, and not necessary

in the correct rows.

Question 2

Problem Statement

Score Criteria

3

Able to state the problem statement of the experiment correctly that include criteria:

C1 Manipulate variables : Water samples // Water systems // Locations of water collected

C2 Responding variables : BOD level // Polution level

// Amount/level/content of (dissolved) oxygen

// Amount/level/content of organic substances

// Amount/level/content of microorganisms

C3 Relation in question form and question symbol [?]

Sample answers:

1. Which water sample has the highest BOD level?

2. Which water system has the least amount of oxygen?

3. In which location of water system the content of microorganisms is the highest?

2

Able to state the problem statement of the experiment with two criteria.

Sample answers:

1. Which water sample has the highest BOD level.

2. Which system has the least amount of oxygen?

3. In which location the content of microorganisms is the highest?

4. Which location has the most organic materials?

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1

Able to state the of problem statement with one criteria or at idea level.

Sample answers:

1. Which sample has high BOD level.

2. Which water has oxygen?

3. The location has more microorganisms.

4. What is polluted water?

Variables

Score Criteria

3

Able to state the three variables correctly

Sample answers:

Manipulate variables : Water samples // Water systems // Locations of water collected

(Reject: Water)

Responding variables : BOD level // Polution level

// Amount/level/content of (dissolved) oxygen

// Amount/level/content of organic substances

// Amount/level/content of microorganisms

// Time taken for the methylene blue solution to decolourise

Controlled variable: Temperature // light intensity // Volume of water (sample)

// Concentration/volume of methylene blue (solution)

2 Able to state any two variables correctly

1 Able to state any one variable correctly

Hypothesis

Score Criteria

3

Able to state the hypothesis correctly according to the criteria.

C1 Manipulate variables: Drain (water) in school canteen / boys hostel / girls hostel / dining

hall / school field // Fish pond in hostel / school garden // Fountain

in school / hostel // Tap water (mention once only)

(Reject: Drinking water)

C2 Responding variables

C3 Relationship of the variables

Sample answers:

1. Water sample taken from the drain near school canteen has the highest BOD level.

2. Water taken from the school fountain has the most amount of oxygen.

3. The content of microorganisms is the highest in hostel water system.

2

Able to state the hypothesis with two criteria.

Sample answers:

1. Different water samples has different pollution level.

2. Locations of water sample affects the BOD level.

3. The content of microorganisms in water depends on the location.

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1

Able to state the hypothesis with one criterion.

Sample answers:

1. Different location affect pollution.

2. Water sample decolourise the methylene blue solution.

Materials and Apparatus

Score Criteria

3

Able to state all functional materials and apparatus: 2*materials and 3*apparatus + 2 other

apparatus for the experiment.

Materials: *Water samples, *methylene blue solution

Apparatus: *(Reagent/sampling/dark) bottles, *syringe/measuring cylinder,

*stopwatch/timer, basin, black cloth/sugar paper, and glove.

(Must in the correct categories. Accept if not separated)

2 Able to state 2*materials and 3*apparatus + 1 other apparatus for the experiment.

1 Able to state 2*material and 3*apparatus for the experiment.

Procedure

Score Criteria

3

Able to state five procedures P1, P2, P3, P4 and P5 correctly.

P1 : How to Set Up The Apparatus (4P1)

P2 : How to Make Constant The Control Variable (1P2)

P3 : How to Manipulate The Manipulated Variable (1P3)

P4: How to Record The Responding Variable (2P4)

P5 : Precaution (1P5)

2 Able to state three of any procedures P1, P2, P3, P4 and P5 completely.

1 Able to state two of any procedures P1, P2, P3, P4 and P5 completely.

Example of Procedure:

1. Diagram with at least 4 labels. P1

2. Label 5 reagent bottles as A, B, C, D and E P1

3. Identify

five locations of water system in the school compound.

P1

P3

4. Fill the reagent bottles with

water samples,

fill to the top with no air trapped

cover with stoppers

tightly.

P1

P4

P5

P1

P5

5. Put in a basin

and cover with dark cloth and bring to the lab.

P1

P5

6. In the lab, add each water sample

with 1ml methylene blue solution,

with the needle immersed in the water,

and stir

P1

P2

P5

P1

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slowly.

Cover with dark cloth.

P5

P5

7. Start the stopwatch. P1

8. Record the time taken when the methylene blue solution decolourise. P4

9. Repeat the experiment to get average readings. P5

10. Tabulate the results in the table. P4

11. Record the scale of the BOD level as:

1 – most polluted

2

3

4

5 – least polluted

P4

Data

Score Criteria

2

Able to tabulate the correct table with units based on two criteria:

Heading with correct unit

Manipulated variable (at least 5 water systems)

Sample answers:

Locations of

water sample

Time the methylene blue decolourise

(minute) BOD level

Drain in school canteen

Drain in school school field

Drain in hostel

Fish pond in school garden

Fish pond in hostel

1 Able to tabulate the correct table with units based on any one criterion

END OF MARKING SCHEME

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