1

BIOLOGYSPM Forecast Paper

Instructions: For Questions 1 to 50, each question is followed by four options, A, B, C and D. Choose the best option for each question.

(50 marks) Time: 1 hour 15 minutes PAPER 1

1 In which organelle is the genetic material of a cell located?A The vacuoleB The chloroplastsC The cytoplasmD The chromosomes

2 Diagram 1 shows the organelle of a cell.

P

Diagram 1

What is the function of structure P ?A Synthesises proteins such as enzymesB Transports the proteins synthesised at the

ribosomesC Transports and synthesises fatty acids and

glycerolD Sorts and packages proteins according to their

destinations

3 Which of the following tissues contains the most number of cells at different stages of cell division?A Ground tissueB Vascular tissueC Meristematic tissueD Epidermal tissue

4 Diagram 2 shows the fluid mosaic model of the plasma membrane.

X

Y

Z

Diagram 2

What do X, Y and Z represent?

X Y Z

A Lipid Carbohydrate Protein

B Lipid Protein Carbohydrate

C Protein Carbohydrate Lipid

D Carbohydrate Lipid Protein

5 Diagram 3 shows the conditions of some cells after an experiment.

solution P solution Q solution R

Diagram 3

Three types of cells, cheek cells, red blood cells and plant cells are soaked in solutions P, Q and R respectively for 10 minutes. Which of the following shows the osmotic concentrations of P, Q and R compared to the initial osmotic concentration of the cells?

Solution P Solution Q Solution R

A Hypotonic Isotonic Hypertonic

B Isotonic Hypotonic Hypertonic

C Hypertonic Isotonic Hypotonic

D Hypotonic Hypotonic Isotonic

6 Which substance is not an organic molecule in the human body?A LipidB ProteinC WaterD Carbohydrate

2

7 Why does maltase hydrolyse maltose but not starch?A Maltose is a smaller molecule than starch.B Starch denatures the tertiary structure of

maltase.C The active sites of maltase are complementary

to those of maltose but not starch.D Maltase becomes inactive when it binds to

starch.

8 Diagram 4 shows a graph of the rate of reaction catalysed by enzymes P, R and S at different pH values.

P

1 2 3 4 5 6 7 8 9 10pH

R S

Di agram 4

What are enzymes P, R and S ?

P R S

A Rennin Salivary amylase Lipase

B Pepsin Trypsin Salivary amylase

C Rennin Trypsin Lipase

D Pepsin Lipase Trypsin

9 Which sub-phase of the interphase results in the doubling of the DNA content?A G1 phaseB S phaseC G2 phaseD M phase

10 Which of the following processes involve both mitosis and meiosis?

P : Formation of pollen in the antherQ : Formation of ova in the ovaryR : Regeneration of flatwormsS : Production of clones through tissue culture

technique

A P and Q C P, R and SB R and S D Q, R and S

11 Which of the following cells is a product of meiosis?A C

B D

12 The following shows four types of food classes and their energy contents.Which of the food classes represents lipids?

Food class Energy content (kJ per 100 g)

A 40

B 3000

C 1650

D 1400

13 Which process converts starch into glucose?A CondensationB PhotolysisC HydrolysisD Assimilation

14 What happens to excess proteins in the body?A Stored in the liverB Excreted through the kidneysC Broken down by the kidneysD Broken down by the liver

15 Diagram 5 shows the longitudinal section of a leaf.

A

B

CD

Diagram 5

Which region, A, B, C or D, is the main site for carbon fixation?

16 Chlorosis in plant leaves is a result of the deficiency ofA magnesium and calciumB magnesium and nitrogenC iron and calciumD nitrogen and iron

17 What changes take place during a period of intensive exercise?

Breathing rate

Concentration of lactic acid in the

blood

Body temperature

A Increases Decreases Decreases

B Increases Increases Decreases

C Increases Increases Increases

D Decreases Increases Increases

3

18 The following are some of the activities that occur during the processes of inhalation and exhalation in fish.

K : Water flows through the lamellae.L : The mouth closes.M : The floor of the buccal cavity lowers.N : The pressure in the buccal cavity increases.O : The operculum remains closed.

Which activities occur during the exhalation process in fish?A K, L and N C L, M and NB K, M and O D M, N and O

19 During vigorous activities, I the rate of heartbeat increases II the rate of ventilation increases III the rate of respiration increases IV the energy requirement decreases

A I, II and III C II, III and IVB I, III and IV D I, II, III and IV

20 Three organisms, P, Q and R, are cultured together in a nutrient solution for three weeks. The changes in their populations are shown in the graph in Diagram 6.

population of organisms

time (days)

R

QP

Dia gram 6

Which relationship between organisms P, Q and R is correct?

P Q R

A Producer Primary consumer

Secondary consumer

B Predator Prey Producer

C Fourth trophic level

Third trophic level

Second trophic level

D Carnivore Herbivore Producer

21 In a prey-predator relationship, A there are always more predators than preyB the numbers of predators and prey are always

the sameC the predator population increases when prey

are not presentD the prey population increases when predators

are not present

22 Diagram 7 shows part of the nitrogen cycle.

Diagr am 7

What are the organisms in groups X, Y and Z ?

X Y Z

A Fungi Nitrosomonas sp.

Nitrobacter sp.

B Rhizobium sp. Saprophytic bacteria

Nostoc sp.

C Nostoc sp. Nitrobacter sp. Nitrosomonas sp.

D Saprophytic bacteria

Nitrobacter sp. Nitrosomonas sp.

23 Carbon monoxide is poisonous to the respiratory system because this gasA decreases the efficiency of the nervous system

in controlling the respiration rateB competes with oxygen to bind with

haemoglobinC inhibits the formation of red blood cellsD decreases the efficiency of diffusion of gases

across the alveoli

24 Which of the following steps help to solve the problem of acid rain?

I Control the sulphur content in fuel II Control the emissions of carbon monoxide from

factories III Encourage the consumption of unleaded petrol IV Use catalytic converters to clean up emissions

from car exhaustsA I and IIB I and IVC I, II and IIID II, III and IV

25 What are the effects of global warming? I Deterioration of the human immune system II Destruction of corals III Increase in sea levels IV Zonal changes on hillsides

A I, II and IIIB I, III and IVC II, III and IVD I, II, III and IV

Nitrogen in animal proteins

Ammonium compounds

Nitrite compounds

Nitrate compounds

X

Z

Y

4

26 Diagram 8 shows a cross section through the mammalian heart.

CDA B

Diagra m 8

Which blood vessel, A, B, C or D, carries blood directly to the lungs?

27 Blood samples from three veins in the body are tested for the concentrations of glucose, oxygen and urea. The results, in arbitrary units, are shown in Table 1.

VeinGlucose

concentrationOxygen

concentrationUrea

concentration

P 0.9 39 1.4

Q 1.3 40 0.6

R 0.8 91 4.1

Table 1

Which of the following are true of veins P, Q and R ?

Hepatic portal vein

Pulmonary vein

Renal vein

A P Q R

B Q P R

C R P Q

D Q R P

28 A foetus who receives antibodies from her mother through the placenta obtainsA a naturally acquired active immunityB an artificially acquired immunityC a naturally acquired passive immunityD an artificially acquired passive immunity

29 Which of the following does not result in the opening of a stoma?A A higher number of ATP molecules which is

available to the guard cellsB An increase in light intensityC A decrease in the amount of water in the guard

cellsD The accumulation of ions in the guard cells

from adjacent cells

30 Diagram 9 shows two types of cells found in a plant.

cell R

cell S

Diagram 9

What are the functions of cells R and S ?

Cell R Cell S

A Transports water Provides energy

B Transports organic substances

Provides energy

C Provides energy Transports water

D Provides energy Transports organic substances

31 Which of the following bones are parts of the axial skeleton?

P

Q

R

S

A P and QB P and SC Q and RD R and S

32 Which pair of tissues have cell walls thickened with lignin?A Collenchyma and tracheidsB Collenchyma and sclerenchymaC Sclerenchyma and sieve tubesD Sclerenchyma and tracheids

33 Diagram 10 shows a type of plant.

Diagram 10

5

The plant obtains support I by having woody tissues II with the help of water buoyancy III by having aerenchyma tissues IV by having air sacs in the leaves

A I and IIB II and IIIC II and IVD III and IV

34 Diagram 11 shows a longitudinal section of the brain and the spinal cord.

W

ZX

Y

Diagram 11

Which labelled parts carry out the following functions?

Control body movements and

posture

Regulates internal body processes

A W X

B Z Y

C X Z

D Z X

35 A boy is paralysed because of an injury to his head. Which part of his brain is damaged?A CerebrumB CerebellumC HypothalamusD Medulla oblongata

36 What changes are likely to occur when a person walks into a cold room?

Sweating ShiveringSkin blood

vessels

A Decreases Increases Dilate

B Decreases Increases Constrict

C Increases Decreases Dilate

D Decreases Decreases Constrict

37 Which hormone is secreted when a person faces a threatening situation?A InsulinB AdrenalineC AldosteroneD Antidiuretic hormone

38 A farmer wants to induce the development of fruit without fertilisation. Which suitable hormone can he use?A CytokininB AuxinC EthyleneD Growth hormone

39 Where does spermatogenesis occur?A Interstitial cellsB Seminal vesiclesC Seminiferous tubulesD Vas deferens

40 During meiosis in the human ovary, which cell is the first to be haploid?A The oogoniumB The primary oocyteC The secondary oocyteD The ovum

41 Which contraceptive method prevents the release of ova from the ovaries?A VasectomyB Contraceptive pillsC DiaphragmD Depo-Provera injection

42 What are the products of double fertilisation in flowering plants?A One triploid zygote and one diploid nucleusB Two diploid zygotes and one triploid nucleusC One diploid zygote and one triploid nucleusD Two triploid zygotes and one diploid nucleus

43 During secondary growth in plants, which tissue gives rise to secondary phloem and xylem?A CorkB Cork cambiumC Apical meristemD Vascular cambium

44 Which of the following statements is a correct definition of growth?A Growth is an irreversible increase in the weight

and size of an organism.B Growth is an irreversible increase in the height

and size of an organism.C Growth is an irreversible increase in the weight

and height of an organism.D Growth is an irreversible increase in the number

of cells in an organism.

45 A man has blood group A while his wife has blood group B. The couple has a child of blood group O. What is the genotype of both the man and his wife?

6

Man Wife

A AA BB

B AO BB

C AA BO

D AO BO

46 What are the possible genotypes of a person with type B blood group?

I IAIA III IBIO

II IBIB IV IAIB

A I and II C II and IIIB I and IV D III and IV

47 Diagram 12 shows how the sex of offspring can be determined.

FatherMother

Male baby Female baby

Parents :

Gametes : P Q R

Fertilisation :

Diagram 12

What are the genotypes of P, Q, and R?

P Q R

A 44 + X 44 + Y 44 + Y

B 44 + XX 44 + XY 44 + XX

C 22 + X 22 + Y 22 + X

D 22 + X 22 + X 22 + Y

48 A pair of identical twins shares the following characters.

P – IntelligenceQ – Attached ear lobeR – Skin colourS – Weight

Which characters will most likely be different when the pair of identical twins grow up?A P and Q C P, R and SB P, Q and R D Q, R and S

49 Diagram 13 shows a type of chromosomal mutation.

abc

defg

bc

defg

Diagram 13

What type of mutation is this?A Translocation C DeletionB Duplication D Inversion

50 Which environmental factors result in variations in plants?

I Humidity II Temperature III Soil fertility IV Nutrition

A I and IIB I, III and IVC II, III and IVD I, II, III and IV

7

1 In an enzyme-catalysed reaction, the amount of substrate used was measured and the rate of product formation was determined. The results are shown in Table 1.

Amount of substrate (g) Rate of product formation (g min−1)

0.2 0.2

0.6 0.6

1.0 1.2

1.4 1.2

1.8 1.2

Table 1

(a) (i) Plot a graph of the rate of product formation against the amount of substrate used.

[3 marks]

(ii) Explain the results based on the graph plotted in (a)(i).

[2 marks] (iii) Suggest one way to increase the rate of product formation when the amounts of substrate used are

greater than 3.5 g.

[2 marks]

Instructions: Answer all questions in this section. Write your answers in the spaces provided.

Section A (60 marks) Time: 2 hours 30 minutes PAPER 2

8

The graph in Diagram 1 shows the changes in the rate of an enzyme-catalysed reaction over time.

reaction rate

progress ofreaction

X Y

A

B

C

Diagram 1

(b) Explain why the reaction rate becomes constant at point X.

[2 marks](c) Which labelled line correctly illustrates what would happen if more enzyme is added at point Y ? Explain your

answer.

[3 marks]

2 Rats are pests on oil palm plantations. In order to control the population of rats, a plantation manager decides to release 500 owls on his plantation. The changes in the population size of rats and owls over the years are shown in Diagram 2.

populationsize

time (years)

ratowl

Key:

1000

500

Diagram 2

(a) State the relationship between the owls and the rats.

[1 mark](b) (i) What is the method used to control the population of rats on the oil palm plantation?

[1 mark] (ii) Explain how this method controls the rat population.

[2 marks]

(c) State one advantage of using the method in (b)(i) over chemical pesticides.

[2 marks]

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(d) Explain one method which can be used to estimate the number of rats on the plantation.

[4 marks](e) Based on the graph in Diagram 2, the populations of rats and owls are said to have reached a dynamic

equilibrium. Explain the meaning of this statement.

[2 marks]

3 Diagram 3 shows some blood cells as observed under a microscope.

A B

X

C

Diagram 3

(a) (i) What is cell A?

[1 mark]

(ii) How does the structure of cell A allow it to carry out its function?

[2 marks]

(b) (i) Name cell C.

[2 marks]

(ii) State the substance produced by cell C. How does the substance destroy pathogens which enter the body?

[2 marks]

(c) Explain how cell B destroys a pathogen.

[2 marks]

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(d) The mammalian circulatory system is described as a closed double circulatory system. (i) Explain briefly what is meant by a ‘double circulatory system’.

[1 mark]

(ii) Suggest why having a double circulatory system is advantageous to a mammal.

[2 marks]

4 Diagram 4.1 shows the structure of a nephron in the mammalian kidney.

P

R

T5

T2

T3

T1

T4

Q

Diagram 4.1

(a) State two substances in the blood that are filtered into T5.

[2 marks]

(b) Name the process which allows the movement of substances in (a) into T5. Explain how structures P, Q and R are adapted to carry out this process.

[2 marks](c) What does the process of secretion normally occur?

[1 mark]

(d) Table 2 shows the concentrations of glucose and urea at T4 and T3.

Substance Concentration at T4 (g l −1) Concentration at T3 (g l −1)

Glucose 1.5 0

Urea 0.5 2.5

Table 2

Explain the changes in the concentration of the two substances in T3.

[2 marks]

11

(e) Patients who are in the later stages of chronic kidney disease have to undergo treatments using a dialysis machine as shown in Diagram 4.2.

dialysis solution

dialysis solution

to the patient

blood from patient tube blood

dialysis solution

Diagram 4.2

(i) How does the dialysis machine help a kidney patient?

[1 mark] (ii) State one characteristic of the tube which allows the dialysis machine to function.

[1 mark] (iii) How can useful substances like mineral salts be retained in the blood when the patient is treated using

a dialysis machine?

[1 mark] (iv) Explain why the dialysis solution has to be replaced constantly with fresh dialysis solution during the

course of treatment.

[2 marks]

5 Each DNA nucleotide consists of a phosphate, a sugar and a base. Diagram 5 shows the arrangement of these components on one side of a DNA strand. The nitrogenous bases are adenine (A), cytosine (C), thymine (T) and guanine (G).

C

C

T

G

G

A

Diagram 5

12

(a) (i) Fill in the complementary bases of the strand in Diagram 5.[3 marks]

(ii) Label each of the following parts on Diagram 5: nucleotide; phosphate and sugar.[3 marks]

In green pea plants, yellow seeds (Y) are dominant over green seeds (y) which are recessive.(b) Describe what is meant by the terms dominant and recessive alleles.

Dominant alleles:

Recessive alleles: [2 marks]

(c) A homozygous green-seeded pea plant is crossed with a heterozygous yellow-seeded pea plant. The allele for yellow seed is a dominant allele.

Complete the following Punnet square to show the genotypes and phenotypes of the offspring.

Heterozygous yellow seed

Homozygous green seed

[2 marks](d) Based on your answers in (c), explain the difference between genotype and phenotype.

[2 marks]

Section B (40 marks)Instructions: Answer any two questions in this section.

6 The chemical process of photosynthesis can be summarised as in the schematic diagram in Diagram 6.

ATP hydrogen atoms

H2O

C6H12O6

O2

light

ADP

NADP+

CO2

light reactions

light-independent reactions

Diagram 6

(a) State the meaning of photosynthesis based on the schematic diagram. [2 marks]

(b) Based on Diagram 6, describe how a green plant synthesises starch molecules. [10 marks]

(c) Leela’s mother is a vegetarian whose main diet includes beans, high fibre food and citrus fruits. Discuss and evaluate how these foods contribute to her health.

[8 marks]

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7 (a) Explain the benefits of having good postures. [4 marks](b) Suggest other ways to care for the musculoskeletal system apart from maintaining good postures. [6 marks](c) Explain how these animals move in their respective habitats. (i) An earthworm (ii) A grasshopper

[10 marks]

8 (a) The following statements describe the lymphatic system and blood circulatory system in humans.

• The fluid that diffuses from the blood plasma into the interstitial space will re-enter the blood circulatory system eventually.

• The lymphatic system also functions in transporting digested food products.

Explain the above statements to justify that the lymphatic system acts as a complementary system to the blood circulatory system. [10 marks]

(b) • Multicellular organisms such as humans need a specific transport system.• Unicellular organisms such as Amoeba sp. do not need a transport system.

Evaluate the above statements based on the physical characteristics of the organisms with regards to the

following processes: (i) Respiration (ii) Nutrition (iii) Excretion

[10 marks]

9 (a) Variation is the result of the interaction between genetic and environmental factors.

Explain by giving suitable examples, how both factors influence one another to produce variations among the organisms of the same species. Explain why variation is important to the survival of a species.

[10 marks](b) Many types of crop plants and farm animals have been genetically modified to increase their yields and

produce. Describe the benefits of producing genetically modified (GM) food. Discuss the controversial issues of consuming GM food.

[10 marks]

Time: 1 hour 30 minutes PAPER 3Instructions: Answer all questions.

1 Four different cubes, with side dimensions measuring 1 cm, 2 cm, 3 cm and 4 cm respectively are prepared. The cubes are prepared from 1.0% agar-phenolphthalein. Each cube is placed in a Petri dish. 100 ml of 0.2 M sodium hydroxide solution is poured into each Petri dish. Immediately, the stopwatch is activated. After 5 minutes, each cube is removed and dried with filter paper. Diagram 1 shows the portion of the cube that has changed colour.

Diagram 1

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(a) Measure the length that has changed colour for each cube. [3 marks](b) (i) State two different observations based on the diagram. Observation 1:

Observation 2:

[3 marks] (ii) State the inferences from the observations in (b)(i). Inference from observation 1:

Inference from observation 2:

[3 marks](c) Complete Table 1 based on the experiment that was carried out.

Variable Method to handle the variable

Manipulated variable: How to alter the manipulated variable:

Responding variable: How to determine the responding variable:

Constant variables: How to maintain the constant variables:

Table 1

[3 marks](d) State the hypothesis for this experiment.

[3 marks](e) Construct a table to record the results of this experiment. Your table should contain the following titles: • Side dimension of the cube (cm) • Total surface area (cm2) • Volume of the cube (cm3) • Total surface area/volume ratio • Length of diffusion (cm) • Percentage portion of the cube which has changed colour

Use the following formula to calculate the percentage portion of the cube which has changed colour.

Percentage portion of the cube length of diffusion (cm) = —————————————————————— × 100% which has changed colour total length of diffusion pathway

[3 marks]

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(f) Based on the results, state the relationship between (i) the size of the cube and the total surface area/volume ratio

(ii) the total surface area/volume ratio and the percentage portion which has changed colour after 5 minutes

[3 marks](g) If the cube with a side dimension of 4 cm is immersed in the sodium hydroxide solution for another

15 minutes, predict the change in the percentage portion of the cube which will change colour with time. Explain why.

[3 marks](h) Predict the percentage portion which will change colour if the experiment is repeated using a cube that has

a side dimension of 5 cm.

[3 marks](i) Based on the results of the experiment, what can you deduce about the size of an organism and the total

surface area/volume ratio?

[3 marks]( j) Classify the following organisms according to the rate of diffusion (slow, medium and fast) of substances to

its internal cells.

Bird, Amoeba sp., Hydra sp., Paramecium sp., frog, Hydrilla sp.

Rate of diffusion

Slow Medium Fast

[3 marks]

2 The phloem tissue transports organic substances from the leaves downwards to storage organs and from the storage organs upwards to growing regions such as shoots and buds.

Based on the above information, plan a laboratory experiment to study the role of phloem tissue as a continuous tube system that transports organic substances. The planning of the experiment should cover the following aspects:

• Proble statement• Hypothesis• Variables• List of apparatus and materials• Experimental procedure• Presentation of data

[17 marks]

16

ANSWERS

Paper 1 1 D 2 B 3 C 4 B 5 A 6 C 7 C 8 A 9 B 10 A11 C 12 B 13 C 14 D 15 B16 B 17 C 18 A 19 A 20 B21 D 22 A 23 B 24 B 25 C26 A 27 D 28 C 29 C 30 B31 B 32 D 33 C 34 B 35 A36 B 37 B 38 B 39 C 40 C41 B 42 C 43 D 44 D 45 D46 C 47 C 48 C 49 C 50 D

Paper 2Section A 1 (a) (i) Rate of product formation against the amount

of substrate

rate

of p

rodu

ct p

rodu

ctio

n (g

min

–1)

amount of substrate (g)

0 1.0

0.5

1.0

1.5

2.0

(ii) As the amount of substrate increases, the rate of product formation increases since there are more collisions between the substrate and the enzyme molecules.

(iii) The graph levels off at 1.2 g min–1 as the number of active sites available becomes limited.

All active sites are engaged in the reaction. To increase the rate of reaction, add more

enzyme or increase the temperature to the optimum level.

(b) All active sites are occupied by substrate molecules (saturated). The reaction is at the maximum rate.

(c) Line A. By adding more enzyme, more active sites are made available to bind with substrate molecules.

2 (a) Prey-predator(b) (i) Biological control (ii) Biological control is the use of natural

predators (owls) to control the population of the pests (rats).

(c) Biological control is a better method because the natural predator used is very specific to a particular pest.

Hence, other organisms are not affected by its presence.

(d) The method which can be used is the capture, mark, release and recapture method.• Initially, a number of rats is captured (x) and

tagged with a ring.• The marked rats are then released on the oil

palm plantation.• After a duration of one week, a second

sample of rats (y) is captured and the number of marked rats in the second sample (z) is recorded.

• Based on the data obtained, the estimated population size of rats in the area can be calculated by using the following formula:

number of rats in the first sample (x) × number of rats in the second sample (y) Population size = ————————————–————— number of marked rats recaptured (z)

(e) This means the population of owls and rats are stable and the changes in numbers occur in a cycle that keeps the populations of both organisms within a certain range.

3 (a) (i) Red blood cell (ii) It is shaped like a biconcave disc and does

not have a nucleus. This provides a large surface area to volume ratio for the rapid diffusion of oxygen across its plasma membrane.

(b) (i) Lymphocyte (ii) Cell C produces antibodies. Antibodies bind

to specific antigens on the pathogens and destroy them.

(c) Cell B extends its pseudopodia towards the pathogen to engulf it.

Ingestion of the bacterium forms a phagosome. The phagosome then combines with a

lysosome which releases lysozyme into the phagosome to kill the pathogen.

(d) (i) Blood passes through the heart twice during every circuit.

(ii) There is no mixing of oxygenated and

17

deoxygenated blood; therefore, more oxygen can reach the tissues.

It also helps maintain a high blood pressure. 4 (a) Glucose and amino acids

(b) Ultrafiltration. P (afferent arteriole) has a larger diameter than Q (efferent arteriole).

As a result, the blood reaching R is under high pressure. This pressure forces the blood through the filtration membrane into T5. Moreover, the wall of R is highly permeable to small molecules and allows them to pass through easily.

(c) At T2.(d) No glucose is found in T3 because all glucose is

reabsorbed into the blood at T4.

The concentration of urea is higher in T3 because a large amount of water is reabsorbed at the loop of Henle and at T2.

At the same time, more urea is secreted at T2.(e) (i) The dialysis machine helps to remove waste

products in the blood of the patient. (ii) The dialysis tube is semi-permeable and

allows small molecules to pass through easily.

(iii) The dialysis solution contains solutes like mineral salts that are of the same concentration as those in the blood plasma.

(iv) As the waste products in the patient’s blood diffuses into the dialysis solution, the concentration gradient of the waste products between the blood and the dialysis solution decreases. Therefore, the dialysis solution needs to be changed constantly to maintain a high concentration gradient.

5 (a) (i) and (ii)

phosphate

sugar

nucleotide

C G

T A

G

G C

C

A T

C G

(b) Dominant allele: The allele that is expressed in the phenotype even when a recessive allele is present.

Recessive allele: The allele that is only expressed in the phenotype when the dominant allele is absent.

(c) Let Y represent the dominant allele for yellow seeds.

Let y represent the recessive allele for green seeds.

Heterozygous yellow

seedHomozygous green seed

Y y

yYyYellow seed

yyGreen seed

yYyYellow seed

yyGreen seed

(d) Genotype is the genetic make-up of an organism. For example, a heterozygous yellow-seeded

plant carries the genotype Yy. Phenotype is the observable characters of an

organism. For example, the phenotype of a seed which

carries the genotype Yy is yellow.

Section B 6 (a) Photosynthesis is a process which involves two

main stages:• Light reaction which occurs only in the

presence of sunlight. [1]• Dark reaction which occurs in the presence or

absence of sunlight and results in the production of starch molecules. [1]

[2](b) Light reaction

• During the light reaction, chlorophyll traps light energy which excites the electrons of chlorophyll molecules to higher levels. [1]

• In the excited state, the electrons can leave the chlorophyll molecules. [1]

• Light energy is also used to split water molecules into hydrogen ions (H+) and hydroxyl ions (OH–). This reaction is known as photolysis of water. [1]

• The hydrogen ions then combine with electrons released by chlorophyll to form hydrogen atoms. [1]

• At the same time, each hydroxyl ion loses an electron to form a hydroxyl group. This electron is received by chlorophyll. [1]

• The hydroxyl groups then combine to form water and gaseous oxygen. [1]

Dark reaction• During the dark reaction, the hydrogen atoms

are used to fix carbon dioxide in a series of reactions catalysed by photosynthetic enzymes. [1]

• The overall reaction results in the reduction of carbon dioxide into glucose. [1]

6CO2 + 24H 6(CH2O) + 6H2Ocarbon dioxide hydrogen atoms glucose water

6(CH2O) C6H12O6

18

• (CH2O) is the basic unit of glucose. Six units of (CH2O) combine to form one molecule of glucose. [1]

• The glucose monomers then undergo condensation to form starch which is temporarily stored as starch granules in the chloroplasts. [1]

[10](c) • High fibre food such as wholemeal bread and

cereals provide roughage, vitamins and minerals. [1]

• The indigestible cellulose fibre provides bulk to food and assists in peristaltic movement along the alimentary canal. [1]

• The dietary fibre also aids bowel movement. This prevents constipation and other disorders of the large intestine. [1]

• Vitamins and minerals do not provide energy, but are essential for the maintenance of good health, and for various functions of the body. [1]

• Beans provide proteins, minerals and vitamins. For example, beans contain vitamin B1 which is needed for the formation of coenzymes required for cellular respiration. [1]

• Even though beans provide the essential amino acids, plant proteins usually lack one or more essential amino acids. Thus, her diet should include a variety of beans to ensure she obtains all the essential amino acids. [2]

• Citrus fruits contain vitamin C, water and roughage. Vitamin C is needed for maintaining healthy epithelial tissue. [1]

• The food consumed by Leela’s mother contributes to her good health.

[8] 7 (a) • Posture is the way a person holds his or her

body. [1]• Having a good posture prevents injuries to

the musculoskeletal system. [1]• A good posture allows free circulation of

blood and does not cramp some of the vital organs. [1]

• A good sitting posture, for example, distributes the weight of the entire body to the thigh muscles and this avoids straining the back. [1]

[4](b) • Eating food rich in calcium. For example, milk

and yoghurt. [1]• Wear comfortable shoes with strong, flexible

soles and cushioned insoles that protect the feet from hard surfaces. [1]

• Do not wear stiletto for walking because the weight of the body is on the front part of the feet. This can injure soft structures, joints and bones. [1]

• Wear loose clothing that are not too tight around the waist to allow for proper functioning of the stomach muscles. [1]

• Always warm up before doing vigorous activities. [1]

• Practise correct and safe exercise techniques to prevent injuries during physical activities or while performing daily chores. [1]

[6](c) (i) Locomotion in an earthworm

• The earthworm moves with the help of antagonistic muscles. [1]

• The force of contraction is applied to a fluid-filled chamber, the coelom. [1]

• When circular muscles contract, the longitudinal muscles relax and the body of the earthworm becomes thinner and longer. [1]

• When the longitudinal muscles contract, the circular muscles relax and the earthworm becomes shorter and thicker. [1]

• The rhythmic contractions and relaxations of both muscles produce peristaltic waves along the body which help the earthworm to move in its habitat. [1]

[5] (ii) Locomotion in a grasshopper

• The grasshopper also moves with the help of antagonistic muscles. [1]

• The flexor and extensor muscles are attached to the internal surface of the exoskeleton. [1]

• When the flexor muscles in the upper part of a grasshopper’s leg contract, the lower leg is pulled toward the body. [1]

• In this sitting position, the hind leg is folded in a Z shape and the grasshopper is ready to jump. [1]

• When the extensor muscle contracts, the leg jerks backwards, propelling the grasshopper forward and upwards into the air. [1]

[5] 8 (a) (i) • Blood that enters the arteriole end of a

capillary is under high pressure. [1]• The high pressure causes fluid to diffuse

continuously from the blood plasma into the interstitial space. [1]

• Most of this fluid re-enters the blood circulatory system.

• However, the remaining fluid that still remains in the interstitial fluid is collected by the lymphatic system. [1]

• Once inside the lymph capillaries, the fluid is known as the lymph. [1]

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• The lymph re-enters the blood circulatory system through the right lymphatic duct which empties its lymph into the right subclavian vein. [1]

• Lacteals, which are found in the villi, are lymph capillaries which contain droplets of lipids and fat-soluble vitamins. [2]

• Lacteals are joined to larger lymphatic vessels which eventually pass into the thoracic duct and into the left subclavian vein. [2]

• Thus, the lymphatic system serves as a complementary system to the blood circulatory system because it returns interstitial fluid which contains water, nutrients and other molecules which continuously diffuse out of blood capillaries back into the blood circulatory system. [1]

[10]

(b) • Multicelluar organisms like humans are larger in size and have a smaller total surface area/volume ratio. Diffusion of substances from the surroundings into the body cells is slow. [1]

• Therefore, they need a specific transport system because their body cells are not in direct contact with the surrounding environment. [1]

• On the other hand, unicellular organisms such as Amoeba sp. are smaller in size (only one cell) and have a large total surface area/volume ratio. [1]

• The exchange of substances between the organisms and the surroundings occurs easily through diffusion. Therefore, they do not need a specific transport system or face any problem in obtaining their cellular requirements because their body cells are in direct contact with the surrounding environment. [1]

(i) Respiration• Multicellular organisms, such as humans,

need a specific respiratory structure or lungs for the exchange of respiratory gases. [1]

• Unicellular organisms do not need a specific organ for the exchange of respiratory gases because this process occurs through the plasma membrane by diffusion. [1]

(ii) Nutrition• Multicellular organisms, such as humans,

need a digestive system for the digestion of food substances and absorption of nutrients. [1]

• Unicellular organisms do not need a digestive system because they can engulf food particles through phagocyctosis. The food particles are then digested by lysozyme and the nutrients are absorbed. [1]

(iii) Excretion• Multicelluar organisms need an excretory

system to excrete materials from their bodies. [1]

• Unicellular organisms do not need an excretory system because waste materials, such as carbon dioxide, are eliminated by diffusion through the plasma membrane. [1]

• Amoeba sp. has contractile vacuoles which contract rhythmically to eliminate excess water and waste materials. [1]

[Maximum: 10] 9 (a) • Variations are the phenotypic differences in

individuals of the same species. [1]• Genetic and environmental factors interact

with one another at different levels in different individuals to control and produce various characters. [1]

• As a result of this interaction, no two organisms have the same phenotype. [1]

• For example, a person’s build can be affected by his nutrition. When a child suffers from an imbalanced diet, his growth is likely to be stunted and he will not reach his full height although he may carry the gene for tallness. [1]

• On the other hand, a medium-sized man who eats a right diet and does weight training can alter his phenotype and achieve the physique of a bodybuilder. [1]

• Individuals with the same genotypes such as identical twins may have different phenotypes in different environments. For example, if one identical twin is brought up in a well-nourished environment, while the second in a poorly-nourished environment, the former will be heavier and has a larger build than his twin. [1]

• Infertile soil can result in unhealthy plants. A limited space with a dense population of organisms can retard the growth of an organism. [1]

• However, environmental factors cannot influence the phenotype more than what is already predetermined by the genotype. [1]

• Variation is important to the survival of a species because the environment is a dynamic, constantly changing entity. This poses new challenges for organisms living within the environment. [1]

20

• If there is no variation among individuals of a species, an unfavourable change in the environment to a species may cause an entire species to be wiped out or become extinct as they do not possess the necessary characters to help them adapt to the changing environment. Hence, variation enables an organism to live more successfully when compared to organisms that do not have variations. [1]

[10]

(b) • Genetically modified organisms are organisms whose genetic materials have been deliberately altered. [1]

• Plants have been genetically modified to improve their flavours, to resist diseases and improve their yields. Examples of GM plants are wheat, soya beans, maize and tomatoes. [1]

• GM maize, for example, is able to resist attacks by insects while GM soya beans have been altered to produce a healthier plant oil for human consumption. [1]

• Farm animals are engineered to increase the quantity and quality of their meat and milk produce. [1]

• Some examples of genetically modified animals are tilapia fish which has an increased growth rate and cows which are able to secrete lactoferrin (a protein normally found in human breast milk) in its milk, thus making cow’s milk more nutritious for human babies. [1]

• Despite its benefits, there are many controversies concerning the consumption of GM food. For example, GM organisms could change the natural evolution process and ecological balance in various environments. [1]

• These organisms could mutate into other types of organisms over which man has no control. [1]

• There is also a possibility that scientists may transfer the wrong genes when engineering an organism, inadvertently creating something unintended. [1]

• Genetically engineered organisms may pass their genes on to other similar organisms or to organisms with which they have close contact. [1]

• The long-term effects of eating genetically modified food are still unknown. [1]

[10]

Paper 3 1 (a)

Side dimension (cm)

Length that has changed colour (cm)

1 1

2 1

3 0.9

4 0.6

(b) (i) Observation 1: The le ngth that has changed colour for the

cube with the side dimension of 1 cm is 1 cm.

Observation 2: The length that has changed colour for the

cube with the side dimension of 4 cm is 0.6 cm.

(ii) Inference from observation 1: The rate of diffusion is faster when the side

dimension is shorter. Inference from observation 2: The rate of diffusion is slower when the

side dimension is longer.

(c)Variable

Method to handle the variable

Manipulated vari able:The side dimension of the cube

How to alter the manipulated variable:Prepare four different dimensions of the cubes

Responding variable:The length that has changed colour

How to determine t he responding variable:Measure the length that has changed colour for each cube with a ruler

Constant variables: Immersion time, 1.0% agar-phenolpht halein cubes

How to maintain the constant variables:Fix 5 minutes as the immersion time and use only cubes prepared from 1.0% agar-phenolphthalein cubes

(d) The higher the total surface area to volume

ratio, the larger the percentage portion of the cube which has changed colour.

21

(e)

Side dimension

of the cube (cm)

Total surface area (cm2)

Volume of the cube

(cm3)

Total surface area/

volume ratio

Length of

diffusion (cm)

Percentage portion of the cube which has changed colour

1 6 1 6 : 1 1.0 100%

2 24 8 3 : 1 1.0 50%

3 54 27 2 : 1 0.9 30%

4 96 64 1.5 : 1 0.6 15%

(f) (i) The larger the size of the cube, the smaller the total surface area/volume ratio.

(ii) Th e higher the total surface area/volume ratio, the larger the percentage portion which has changed colour after 5 minutes.

(g) The percentage portion of the cube which has changed colour increases with time because a longer time is allowed for diffusion to take place.

(h) The percentage portion that will change colour if the experiment is repeated using a cube with a side dimension of 5 cm will be lower than 15% because the smaller the total surface area/volume ratio, the slower the rate of diffusion.

(i) The larger the size of an organism, the smaller the total surface area/volume ratio.

(j) Slow rate of diffusion

Medium rate of diffusion

Fast rate of diffusion

Bird, frog Hydra sp.Hydrilla sp.

Amoeba sp.Paramecium sp.

2 Problem statement:What is the effect of removing a ring of phloem tissue from the stem of a plant?Hypothesis:The tissue just above the ring swells, whereas the tissue below the ring shrinks.The diameter of the stem above the ring is larger that the diameter of the stem below the ring.

Variables:• Manipulated variable: A stem that is ringed and a

stem that is not ringed• Responding variable: The condition/diameter of

the stem above and below the ring after one month

• Constant variable: Stems taken from the same tree

List of apparatus and materials:A sharp knife, a healthy plant with small stems, Vaseline

Experimental procedure:1. Use a knife to remove a complete ring of bark

from a stem.2. Apply Vaseline on the exposed surface.3. Observe the condition of the stem above and

below the ring.4. Draw the condition of the stem.5. Measure and record the diameters of the stem

above and below the ring.

Presentation of data:

Diameter of the stem abo ve the ring (cm)

Diameter of the stem below the ring (cm)