Name: .................................................................................. Form:.................................

964/1 BIOLOGY (BIOLOGI) PAPER 1 (KERTAS 1)

MULTIPLE-CHOICE (ANEKA PILIHAN)

One hour and forty-five minutes (Satu jam empat puluh lima minit)

PEJABAT PELAJARAN DAERAH GOMBAK

(GOMBAK DISTRICT EDUCATION OFFICE)

PEPERIKSAAN PERCUBAAN SIJIL TINGGI PELAJARAN MALAYSIA 2008

(MALAYSIA HIGHER SCHOOL CERTIFICATE TRIAL EXAMINATIONS 2008)

Instructions to candidates:

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO.

There are fifty questions in this paper. For each question, four suggested answers are given.

Choose one correct answer and indicate it on the multiple-choice answer sheet provided.

Read the instructions on the multiple-choice answer sheet carefully.

Answer all questions. Marks will not be deducted for wrong answers.

This question paper consists of 17 printed pages and 1 blank page.

[Turn over

2

1. Which of the following is made up of the condensation of β-glucose monomers?

A. Sucrose

B. Cellulose

C. Starch

D. Glycogen

2. The following is the formula for an amino acid.

This amino acid is

I Hydrophilic

II Acidic

III Negatively charged

IV Insoluble in water

A. I only

B. I and II

C. I, II and III

D. II, III and IV

3. The water potential of three adjacent plant cells is shown in below. In which direction

does water move?3

A. From cell X to cells Y and Z

B. From cell Y to cells X and Z

C. From cell Z to cell X

D. From cell Z to cell Y

3

4. The following diagram shows the sequence of movement of a molecule across the

membrane.

The movement of the molecule in the diagram is called

A. Phagocytosis

B. Pinocytosis

C. Diffussion

D. Facilitated diffusion

4

5. Which of the following diagrams is prokaryote?

A. I only

B. I and II

C. II and III

D. III and IV

6. Which of the following is not true of a mitochondrion?

A. It is the site of Krebs cycle

B. It contains DNA

C. It possesses two layers of membranes

D. The folded inner membrane is called thylakoid

7. Which of the following cells has not undergone specialisation?

A. Sieve tube

B. Endodermis

C. Tracheid

D. Meristem

5

8. The Diagram below shows the types of xylem thickening P, Q, R, S and T found in the

stem of a dicotyledon plant.

What are the names of the xylem thickening shown in the above diagram?

P Q R S T

A Helix Reticulate Annulus Pitted scalarform

B Helix Annulus Pitted Reticulate Scalariform

C Annulus Helix Helix Reticulate Pitted

D Pitted Helix Helix Scalariform Annulus

9. The following reaction occurs with the help of an enzyme.

Enzyme that helps in the reaction is classified as

A. Lyase

B. Isomerase

C. Hydrolase

D. Transferase

6

10. The graph below shows Lineweawer-Burke plot for a reaction catalyzed by enzyme.

The value on the graph which is used to calculate Michaelis constant Km is

A. I

B. II

C. III

D. IV

11. In chloroplasts, chemiosmosis translocates photons from

A. the matrix to the stroma.

B. the stroma to the chlorophyll.

C. the intermembrane space to the matrix.

D. the stroma into the thylakoid compartment.

7

12. Which of the following statements best describe the relationship between

photosynthesis and respiration?

A. Respiration is anabolic and photosynthesis is catabolic.

B. ATP molecules are produced in photosynthesis and used up in respiration.

C. Respiration is the exact reversal of the biochemical pathways of photosynthesis.

D. Photosynthesis stores energy in complex organic molecules, while respiration

releases it.

13. Which of the following are products of the light reactions of photosynthesis that are

utilised in the Calvin cycle?

A. Electrons and H+

B. CO2 and glucose

C. ATP and NADPH

D. ADP, Pi and NADPH

14. Photorespiration lowers the efficiency of photosynthesis by removing which of the

following from the Calvin cycle?

A. ATP molecules

B. RuBP carboxylase

C. Ribulose bisphosphate molecules

D. Glyceraldehyde phosphate molecules

15.. In a plant cell, where is ATP synthase located?

I Plasma membrane

II Thylakoid membrane

III Inner mitochondrial membrane

A. I only

B. I and II only

C. II and III only

D. I, II and III

16. Which substance enters and leaves in mitochondrion during aerobiosis respiration?

Enters leavers

A. Acetyl-CoA Carbon dioxide

B. Pyruvate ATP

C. Acetyl-CoA ATP

D. Pyruvate NADH

8

17. In a food chain, the decomposer organism acts as

A. an epiphyte

B. a parasite

C. a saprophyte

D. a holozoic

18. Oxygen can easily dissociate from oxyhaemoglobin in condition of

A. Low carbon dioxide concentration

B. Low blood pH

C. Low H+ ion concentration

D. High oxygen content

19. Which of the following sequences is correct for the flow of impulse through the

structures listed below in the heart?

I. Atrioventricular node

II. Purkinje tissue

III. His bundle

IV. Sinoatrial node

A. I , IV , III , II

B. III , IV , II and I

C. IV , III , I and II

D. IV , I , III and II

20. The two-directional translocations in the sieve tube can be explained by

A. Mass flow

B. Munch model

C. Root pressure

D. Cytoplasmic streaming

21. Which of the following reaction is not true in ornithine cycle for the formation of urea?

A. Arginine to urea

B. Arginine to Ornithine

C. Ornithine to Citruline

D. Citruline to Ornithine

9

22. Low water potential in our body caused b y excessive sweating will result in

A. Secretion of ADH inhibited and then the rate of water reabsorption increases

B. Secretion of ADH stopped and then more water is excreted

C. Secretion of ADH stimulated and then less water is excreted

D. Secretion of aldosteron stimulated and then excretion of sodium increases

23. Which of the following occur in the myofibrils when a muscle contracts?

Sarcomere H Zone A band I Band

A Shortens Shortens

No change in

length

Shortens

B Lengthens Shortens

No change in

length

Shortens

C Shortens Lengthens Shortens Lengthens

D Shortens

No change in

length

Shortens Shortens

24. The diagram below shows the sequence of events occurring as an action potential

arrives at a synapse. The arrows indicate the direction of the movement of substances

across the membranes.

Which of the following are represented by 1, 2, 3, 4 and 5?

1 2 3 4 5

A K+ Na

+ Acetylcholine Ca

2+ K

+

B K+ Na

+ K

+ Ca

2+ Acetylcholine

C Na+ K

+ Ca

2+ Acetylcholine Na

+

D Na+ K

+ Na

+ Acetylcholine Ca

2+

10

25. The diagram below shows the control of hormones in a female reproductive system.

Higher Brain

Centre

Hypothalamus

Anterior

pituitary

I II

Ovary

Follicles

Corpus

luteum

III

IV

Uterus and

secondary sex

characteristics

Which of the following are represented by I, II, III and IV?

I II III IV

A FSH LH Progesterone Oestrogen

B LH FSH Oestrogen Progesterone

C FSH LH Oestrogen Progesterone

D LH FSH Progesterone Oestrogen

11

26. The diagram below shows four different light treatments given to a short-day plant

which has a critical night length of 11 hours.

Which of the following treatments can induce flowering in the plant?

A. P and R

B. P and S

C. Q and R

D. Q and S

27. What shuts off the immune response in T cell and B cell after an infection has been

conquered?

A. Cytozoic T cell

B. Pyrogens

C. Natural killer cells

D. Suppressor T cells

28. Oestrogen and progesterone are used in contraceptive pills. What is the effect of these

hormones on the menstrual cycle?

A. To maintain the endometrium in the uterus

B. To stimulate the release of luteinising hormone

C. To stimulate the release of follicle stimulating hormone

D. To inhibit the production of gonadotropin releasing hormone

12

29. Which of the following shows the structures of a plant and their number of

chromosomes?

Pollen

nucleus

Pollen

grain

Pollen

nucleus

Cell in the

testa

Cell in the

pericarp

A Diploid Diploid Diploid Diploid Diploid

B Diploid Haploid Haploid Haploid Haploid

C Haploid Diploid Haploid Haploid Diploid

D Haploid Haploid Haploid Diploid Diploid

30. Four types of growth patterns are shown in the table below.

Growth Pattern Example

I Isometric growth (a) Human organs

II Allometric growth (b) Beans

III Limited growth (c) Fish

IV Intermittent growth (d) Woody plants

Which growth pattern corresponds to its example?

I II III IV

A (a) (c) (d) (b)

B (b) (c) (d) (a)

C (c) (a) (b) (d)

D (d) (a) (d) (b)

31. Polygenic inheritance forms the basis of

A codominace

B continuous variation

C discontinuous variation

D linkage gene

13

32. Which of the following statement is incorrect?

A A dominant allele exerts its dominance on an alternative allele of the same gene.

B Dihydbrid inheritance involves the transmission of two alleles independently of

each other at the same time.

C Codominance occurs when both alleles of gene express themselves equally in the

same phenotype.

D Epistasis occurs whwn one gene suppresses or masks the effects of another gene.

33. In radish, the elongates shape has the genotype of LL, the round ll or oval Ll. While its

colour can be red of genotype RR, white genotype rr and the purple genotype Rr. If a

cross is made between white oval radish plant wit that of are oval, which phenotype is

correct?

Red

Elongated

Purple

Elongate

Red

Oval

Purple

Round

Red

Round

Purple

Round

A 1 - 2 - 1 -

B - 1 - 2 - 1

C - 1 - 1 - 1

D 1 - - 2 1 -

34. A small proportion of men have the genotype XYY. Such a genotype is possible if one

contributory gamete to the zygote is

A an ovum produced by non-disjunction in meiosis

B an ovum containing an X and a Y chromosome

C a sperm produced by non-disjunction at meiosis 1

D a sperm produced by non-disjunction at meiosis 11

35. In the population that is in Hardy-Weinberg equilibrium, 16% of the individuals show

the recessive trait. What is the frequency of the dominant allele in the population?

A 0.84 B 0.36 C 0.6 D 0.4

14

36. An mRNA molecule transcribed from the lactose operon contains nucleotide sequences

complementary to

A structural genes coding for enzymes

B the operator region

C the promoter region

D the repressor gene

37. Which of the following are the uses of recombinant DNA technology?

I. to poduce crops and farm animals of high content

II. to produce bacteria which can clean up oil spills

III. to clones of the sheep

A I and II

B. I and III

C. II and III

D. I, II and III

38. Which of the following techniques can be used to identify a criminal that left a strand of

hair at the crime scene?

A Southern blotting

B DNA finger printing

C DNA sequencing

D DNA clonning

39. cDNA library consist of

A clones that have the complete genomic information of an organism

B clones that have introns and exons of an organism

C clones that have only exons of an organism

D clones that have only introns of an organism

15

40. The table below shows the taxonomic hierarchy from kingdom to genus for three

different organisms.

Taxonomic hierarchy

Examples of organisms

Kingdom

Plantae

Animalia

Animalia

Phylum

X

Annelida

Z

Class

Dicotyledoneae

Y

Mammalia

Order

Renales

Terricolae

Primates

Family

Ranunculacae

Lumbricidae

Hominidae

Genus

Ranunculus

Lumbricus

Homo

What do X, Y and Z represent? X Y Z

A. Angiospermmophyta Anthozoa Echinodermata

B. Angiospermmophyta Oligochaeta Chordata

C. Bryophyata Nematoda Mollusca

D. Coniferophyta Cestoda Cnidaria

41. Which of the following is an advantage of the coelom formation in the body of some

annelids?

I. The digestive system can move independently of the body wall; therefore,

peristalsis can occur every time.

II. The coelom can act as the main excretory organ because its fluid is able to

eliminate waste products.

III. The coelom is able to function as a hydrostatic skeleton that is able to change its

shape without changing its volume.

A. I only C. II and III only

B. I and II only D. I, II and III

16

42. Based on the table below, match the phyla of organisms to their characteristics.

Phylum

Characteristic

I. Cnidaria

P Body divided into head, muscular foot and visceral mass

II. Arthropoda

Q Diploblastic body, polymorphism

III. Mollusca

R Segmented legs, chitinous exoskeleton

IV. Nematoda

S Body covered with thin and elastic cuticle pseudocoelom

I II III IV

A. P Q S R

B. Q R P S

C. R S Q P

D. S R P Q

43. Which of the following are examples of continuous variation?

I. Human skin colour.

II. Human fingerprints.

III. Human MN blood group.

IV. Human height.

A. I, II and III C. II and III

B. I and IV D. II, III and IV

44. A species of organism is separated into two populations by a mountain range. Mating

between individuals of the two populations produces sterile progeny. Which of the

following explains this phenomenon?

A. Speciation does not take place.

B. Genetic drift does not take place.

C. Allopatric speciation takes place.

D. Sympatric speciation takes place.

45. Industrial melanism shown by the Biston betularia moth is an example of

A. mutation C. reproductive isolation

B. natural selection D. geographical isolation

17

46. What are the causes of genetic variation?

I. Mutation of chromosomes.

II. Random assortment of chromosomes.

III. Light intensity.

IV. Dominant and recessive alleles.

A. I, II and III C. I, II and IV

B. II, III and IV D. I, II, III and IV

47. Which of the following energy flows in an ecosystem involves the transfer of the

greater amount of energy?

A. Plant →Herbivore

B. Plant →Decomposer

C. Herbivore→Carnivore

D. Carnivore→Decomposer

48. The canopy of the tropical rain forest is very dense. Which of the following statements

are the effects of the dense condition?

I Little sunlight is able to pass through the forest.

II Detritus is constantly present due to a high rate of decomposition activity.

III The area under the canopy is dry, with low light intensity.

IV Many species of lower level plants grow there but the forest is unable to

accommodate too many plants.

A. I and II

B. I and IV

C. III and III

D. I,II and IV

18

49. A plant sampling is done in an area that covers 2500m². The number of plant in 25

quadrats, each with the area of 1 m², is as shown in the table.

Number of plants in each

quadrat

5 4 6 4 5

6 5 4 6 6

7 7 5 4 3

2 1 8 3 2

3 8 9 7 5

What is the total number of plants in an area of 10 000 m²?

A. 5

B. 125

C. 500

D. 50 000

50. In an investigation to estimate the population size of a species of beetle by capture –

recapture method, the lower surface of the beetle is marked with a small spot of paint.

Which is the reason why a beetle is not marked on its back?

A. The paint will be more visible on the surface of the beetle.

B. The paint will make the beetle more noticeable to the predators.

C. The paint will damage the epidermis of the beetle

D. The paint will hinder the movement of the beetle

19

Marking Scheme

Paper 1

1. B 2. C 3. A 4. D 5. A

6. D 7. D 8. C 9. D 10. B

11. D 12. D 13. C 14. C 15. D

16. B 17.C 18. B 19. D 20. D

21. D 22. C 23. A 24. C 25. C

26. D 27. D 28. D 29. D 30. C

31. B 32. B 33. D 34. D 35. C

36. A 37. A 38. B 39. C 40. B

41. D 42. B 43. B 44. C 45. B

46. C 47. A 48. A 49. D 50. B

20

Name: ........................................................................................... Form: ............................

964/2 BIOLOGY (BIOLOGI)

PAPER 2 (KERTAS 2)

STRUCTURE AND ESSAY (STRUKTUR DAN ESEI)

Two and a half hours (Dua jam setengah)

PEJABAT PELAJARAN DAERAH GOMBAK

(GOMBAK DISTRICT EDUCATION OFFICE)

PEPERIKSAAN PERCUBAAN SIJIL TINGGI PELAJARAN MALAYSIA 2010

(MALAYSIA HIGHER SCHOOL CERTIFICATE TRIAL EXAMINATIONS 2010)

Instructions to candidates:

DO NOT OPEN THIS QUESTION PAPER UNTIL

YOU ARE TOLD TO DO SO.

Answer all questions in Section A. Write your answers

in the spaces provided

Answer any four questions in Section B. Write your

answers on the answer sheets provided. Begin each

answer on a fresh sheet of paper. Answers should be

illustrated by large and clearly labelled diagrams

whenever suitable.

Answers may be written in either English or Bahasa

Malaysia.

Arrange your answers in numerical order and tie the

answer sheets to this question paper.

For examiner’s use

(Untuk kegunaan

pemeriksa)

1

2

3

4

5

6

7

8

9

10

Total

(Jumlah)

___________________________________________________________________________

This question paper consists of 11 printed pages and 1 blank page.

[Turn over

21

21

Structure Question

1. The below shows the ‘lock and key’ hypothesis for enzyme activity.

(a) Define the term enzyme.

_____________________________________________________________

_____________________________________________________________

[1 mark]

(b) Briefly explain the ‘lock and key’ hypothesis.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[3 marks]

(c) Molecule A must attach permanently to the enzyme before the enzyme can

function as a catalyst. Identify molecule A.

_____________________________________________________________

_____________________________________________________________

[1 mark]

22

22

(d) Molecule B is an organic molecule that must attach temporarily to the same

enzyme before the enzyme can function. Identify molecule B.

_____________________________________________________________

_____________________________________________________________

[1 mark]

(e) Name site X.

_____________________________________________________________

_____________________________________________________________

[1 mark]

(f) Explain the effect of non-competitive inhibitor that attaches to side X.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[3 marks]

23

23

2. The diagram below shows the nephron of a mammalian kidney.

a) Name the process that takes place in the Malphigian body

_____________________________________________________________

_____________________________________________________________

[1 mark]

b) Give one feature of the blood vessels associated with the glomerulus and

explain how this features facilitates the process that takes place in (a)

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[2 marks]

24

24

c) Describe the process that takes place at X and Y to produce concentrated

urine.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[4 marks]

d) Explain why glucose may be found in the urine of diabetics.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[2 marks]

e) Why would the consumption of a large amount of alcohol cause the

production of a high volume of diluted urine and the body to suffer from

dehydration?

_____________________________________________________________

_____________________________________________________________

[1 mark]

25

25

3. The diagram below summarises the hormonal control of ecdysis and

metamorphosis in insects.

Prothoracic

glands Pupa

(a) Name the hormones R, S and T.

R - ____________________________________________________________

________________________________________________________________

S - _____________________________________________________________

________________________________________________________________

T - _____________________________________________________________

________________________________________________________________

[3 marks]

Neurosecretory

cells in the brain

Corpus

allatum

First

instar

Final

instar

26

26

(b) Explain how hormone S is produced.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[2 marks]

(c) What is the role of hormone S in the insect?

____________________________________________________________

_____________________________________________________________

[1 mark]

(d) What is the effect of the presence of the hormone T on the activity of

hormone S?

_____________________________________________________________

_____________________________________________________________

[1 mark]

(e) What are the changes in the levels of the hormones S and T in the:

(i) final instar?

_____________________________________________________________

_____________________________________________________________

[1 mark]

(ii) pupa?

_____________________________________________________________

_____________________________________________________________

[1 mark]

27

27

(f) Some plants are able to synthesise compounds that are similar to that of S and T. What

is the significance of this ability in such plants?

_____________________________________________________________

_____________________________________________________________

[1 mark]

28

28

4. The graphs below show three different types of graphs showing the mode of natural

selection on populations.

(a) State the types of selection shown in X, Y and Z.

X : ____________________________________________________________ Y : ____________________________________________________________

Z : ____________________________________________________________

[3 marks]

(b) Using an example, describe how the type of natural selection in X can occur.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

29

29

[2 marks]

(c) Give one difference between natural selection and artificial selection.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[2 marks]

(d) State three disadvantages of inbreeding.

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

_____________________________________________________________

[3 marks]

30

30

Essay Questions

5. a. With the aid of labelled diagram, explain

i. The structure of a stoma [2 marks]

ii. The mechanism of stomata opening and closing based on the potassium

ions accumulation hypothesis [8 marks]

b. Outline the environmental factors which influence the opening and

closing of a stoma. [5 marks]

6. Human hormones can be classified into two main types based on their

mechanisms of actions.

a. What is meant by endocrine glands?

[1 mark]

b. Name the two main types of human hormones and give an example for

each type.

[4 marks]

c. Describe the differences in mechanism of action between both types of

human hormones [10 marks]

7. a. Name the organs and structures of the lymphatic system and describe their

functions. [8 marks]

b. A malfunctioned kidney may be replaced by a healthy one which is

transplanted from a donor.

I) Explain why the healthy kidney may later be rejected by the recipient’s

body.

II) What are the steps taken to overcome the problem of rejection by the

recipients body [7 marks]

8. (a) Explain briefly the following ecological terms:

(i) Biotic potential

[3 marks]

(ii) Environmental resistance

[3 marks]

(iii) Carrying capacity

[3 marks]

(b) By using a named example and a graph, explain the population growth that

produces a sigmoid growth curve.

[6 marks]

31

31

9. a) Discuss the role of the following in gene cloning:

i) Vector

ii) Restriction enzymes

iii) Transformation

[ 9 marks]

b) Explain how cDNA library for humans is constructed [ 6 marks]

10. The greatest threat to the biodiversity in Malaysia is human activities. Discuss how

human activities pose a threat to the biodiversity. [15 marks]

32

32

Marking Scheme

Paper 2

Structure Question

Questions Sample answers Marks

1.(a) Biological organic catalysts produced by living cells that speed up

the rate of biochemical reactions.

1

1.(b) Only one type of key with specific configuration can fit

into the keyhole of a specific lock to open it, just like only

one type of substrate that

have specific configuration can fit into the active side of

a specific enzyme to form enzyme-substrate complex.

1

1

1

1.(c ) Prosthetic group 1

1.(d) Coenzyme 1

1.(e) Allostearic site 1

1.(f) Non-competitive inhibitor that attaches temporarily to the

allostearic site causes changes to the configuration of the

enzyme and its active site.

The substrate is prevented from attaching to the active

side, therefore preventing the formation of the enzyme-

substrate complex.

The rate of the enzyme catalysed reaction is lowered until

the non-competive inhibitor detaches from the allostearic

site.Vmax value of the enzyme catalysed reaction is

lowered.

1

1

1

Total: 10

33

33

2. (a) Ultra filtration 1

2. (b) - the afferent arteriole has a larger diameter

than that of the efferent arteriole

- A hydrostatic pressure is generated which forces the fluid

through the pores of the capillary wall of glomerulus into the

Bowman capsule.

1

1

2.(c) The loop of Henle is a hairpin-shaped structure which operates as a

countercurrent multiplier.

- The descending limb is more permeable to water. As the

filtrate moves down, water diffuses out of descending limb

by osmosis. The filtrate becomes more concentrated.

- The ascending limb is relatively impermeable to water but

more permeable to salts. As the filtrate moves up, sodium

and chloride ions diffuse out passively at first and then

higher up actively transported out into the surrounding

tissue.

- This creates an area of high solute concentration mainly in

medulla region.

- More water is reabsorbed from the ducts and distal

convoluted tubules into the region of high solute

concentration. The water is carried away by the vasa recta.

A small volume of concentrated urine is produced.

1

1

1

1

2.(d) - Diabetes mellitus is caused by the failure of

beta cells in the pancreas to produce

insulin .

- The liver and other cells are unable to take up and

metabolise glucose. The blood glucose level rises to above

normal level.

- The kidney tubules are unable to reabsorb the high levels of

glucose. Excess glucose is removed in the urine

1

1

1

Any

two

2.(e) The consumption of a large amount of alcohol inhibits the release

34

34

of Antidiuretic hormone (ADH) by the posterior lobe of the

pituitary gland. Less ADH is produced. The permeability of

walls of distal convoluted tubule and collecting duct to water

decreases, less water is reabsorbed from the filtrate. A large

volume of dilute urine is produced.

1

Total: 10

3.(a) R – Prothoracicotrophic hormone

S – Ecdysone/moulting hormone

T – Juvenile hormone/neotonin

1

1

1

3.(b) Neurosecretory cells in the brain secrete prothoracictrophic hormone/

which is transported to the prothoracic glands/

the prothoracic glands are stimulated to produce edysone/moulting

hormone

Any

2

1

mark

each

3.(c ) Promotes growth and controls moulting/

Controls the synthesis of enzymes involved in growth and changes in

epidermal cells.

1

3.(d) Retention of larval/nymphal characteristics 1

3.(e)(i) Juvenile hormone/Hormone T decreases, ecdysone/hormone S increases

1

3.(e)(ii

)

No production of juvenile hormone/hormone T, high levels of

ecdysone/hormone S

1

3.(f) Plant is protected from the insects as these compounds interfere with

the development of the insects.

1

Total: 10

4.(a) X : Stabilising selection

Y : Disruptive selection

Z : Directional selection

1

1

1

4.(b) An example of a stabilising selection can been

seen in the human birth weight

35

35

Babies with low birth weights loose heat easily and get sick

from infectious diseases easily also. Babies with large body

weight are more difficult to deliver passing through the pelvis.

Therefore, babies with average (intermediate) weights have

higher chances of survival and will be favoured by natural

selection.

2

mark

s

4.(c ) Natural Selection

Artificial Selection

Occurs naturally Involves human action

2

mark

s

4.(d) Reduced health // reduced fertility both in litter size and sperm

viability // congenital disorders // facial asymmetry // lower birth rate // higher infant mortality // slower growth rate // smaller adult size // loss of immune system

function.

Any

3

Total: 10

Essay Questions

5.(a)(i)

Label

Diagram

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5.(a)(ii) The mechanism of stomatal opening (during day)

- potassium ion (K+) are pumped from subsidiary cells

into the guard cell, H+ are pumped out of the

subsidiary cells to maintain the electro neutrality

- the increase of ion K+ and sugar(from photosynthesis)

concentration makes the water potential of the guard

cells more negative (lower), therefore

- the water from subsidiary cells moves into the guard

cell

- the resultant increase in hydrostatic pressure causes the

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guard cells to become turgid

- the uptake of water causes increased bowing of the

guard cell (owing to the greater expansion of the outer

walls than the inner wall ) and the stoma open

The mechanism of stomatal closing (during night)

- K+ ion are actively transported out from the guard

cells into the subsidiary cells, H+ ions are transported

into the guard cells

- photosynthesis does not occur and the carbon dioxide

concentration increases and the pH of the guard cell

fall

- sugar is converted into insoluble starch, therefore the

water potential of the guard cell increases

8 marks

5.(b) - light/blue light stimulate guard cells to accumulate

potassium and become turgid, stoma open; or by driving

photosynthesis in guard cells chloroplast making ATP

available for active transport of H+

- temperature – increased temperature stoma opens

- air movement

- dehydration(water stress) –in case of water

deficiency, guard cells lose turgid and stoma closes.

- Mesoplhyll cells produce hormone abscisic acid which

signals the guard cells to close.

- concentration of carbon dioxide – depletion of CO2

within the air spaces of the leaf causes the stoma to opens

- moisture/humidity

5 marks

6. (a) Ductless glands

Secrete chemicals/hormones

Directly into blood capillaries

To reach targeted cells

Any 2:

1 mark

6.(b) Steroid hormones

Example:

Oestrogen, progesterone, testosterone

Non-steroid / peptide hormones

Examples:

Adrenaline, glucagon, insulin,

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Any 1:

1 mark

1

Any 1:

1 mark

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6.(c) Steroid Non-steroid

It can easily pass through

the lipoprotein membrane

of target cells.

It cannot pass through the

lipoprotein membrane of

target cells.

It binds to the receptor in

the cytoplasm or

nucleoplasm.

It binds to the receptor on

the surface of the plasma

membrane.

The complex formed does

not activate adenyl

cyclase.

The complex formed

activates adenyl cyclase.

No cAMP is produced. cAMP is produced.

No cascading effect

occurs.

Cascading effects occurs

and effects take place

immediately

The complex enters the

nucleus.

The complex does not

enter the nucleus.

Certain genes are activated

by the complex

Genes are not activated by

the complex

Transcription takes place

followed by the production

of proteins.

Transcription does not take

place and no protein is

produced.

Effects take place through

the proteins formed.

Effects take place through

the activated enzymes and

their products.

Effects are usually long

term such as muscular

body by testosterone.

Effects are short term such

as increase in blood

glucose by adrenalin.

Any 5

pairs:

5 × 2m =

10m

7.(a)

Lympahatic system consists of lymphatic vessels , bone marrow

, spleen, thymus, tonsils and Peyer’s patches in small intestine.

The network of lymphatic vessels transport lymph derived from

the tissue fluids back to the bloodstream.

When blood flows from arterioles into capillaries, the

hydrostatic pressure forces some fluid through the capillary

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walls into the intercellular space. Most of the tissue fluid is

absorbed at the venule end of the capillary

A small amount of the fluid enters into lymphatic capillaries and

its called lymph

Lymph is colourless ,similar to blood plasma but has no red

blood cell and large protein molecules

The is moved through the vessels by contraction of surrounding

skeletal muscles

The lymph flows in one direction ,backflow is prevented by

valves present in the large vessels

The vessel join to form larger lymphatic vessles.The lymph is

vessel from the left side of the body , the alimentary canal and

the right side of lower side flow into thoracic duct.

The thoracic duct carries the lymph to the left subclavian vein ,

back into the bloodstream

The right lymphatic duct drains the lymph from the right side of

the head and chest into the right subclavian vein

Lymph notes are found at intervals of the lymphatic vessels.

Function

The lymph nodes filter the lymph and phagocytes engulf and

destroy forein particle e.g bacteria

When stimulated , the lynphocytes can rapidly to produce new

cells.The B cells produce antibodies responcible for humoral

immunity. The T cell trigger cell – mediated ommunity

Lactealsin the villi of ileum absorb fats and fat – soluble

vitamins and transport them to the blood circulatory system

The lymph transports fluid containing food and excretory waste

materials back into the blood circulatory system

8 marks

7.(b)(i)

At fetal stage , lymphocytes bearing specific receptors that can

bind to MHC proteins of own body cells destroyed. This leaves

only lymphocytes that can react to foreign molecules, non-self

molecules,the capacity to distinguish ‘self’ feom ‘non –self’

In organ transplant , the MHC molecules on cell membranes of

foreign non-self cell act as antigen .

The bodys immune system is stimulated.

The helper T cell recognize the foregn ‘non-self’ protein (MHC

molecules)and secrete interleulins to stimulate cytoxic T cell to

divide rapidly

The cytoxic Tcells produced perforins which cause lysis of

cells of transplanted organ and loss of function

The T lymphocytes system is mainly responsible for the

rejection of transplanted tissue

However , antibodies produced by specific B cells ,

inflammation and nonspecific phagocytosis by

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7.(b)(ii)

macrophages can cause damage to the blood vessels of the

grafted organs

To overcome the problem og rejection by the recipient’s body

Close tissue matching between donor and recipient is carried

out . The lesser variance are the MHC proteins. The lesser the

chances of organ rejection

The recipient has to take medicines to suppress the immune

response

Newer drugs e.g cyclosporine reduces the number of cytotoxic

T cells but B cells are unaffected thus retaining humoral

immunity

Irradiation with X-ray of bone and lymph tissue reduces

production of lymphocytes that cause tissue rejection.

7 marks

8.(a)(i) -The biotic potential of a population refers to the ability of its

individual organisms to reproduce under optimum conditions.

-The actual capacity of a natural population will only reach biotic

potential if all the individuals in the population survive and

reproduce at the maximum rate.

-Full biotic potential is reached if environmental resistance is

absent.

-Example : Growth of bacteria in a Petri dish with ideal nutrients

under a perfect conditions in laboratory.

-Population reproducing at its biotic potential produces an

exponential growth curve or J-shaped curve.

3 marks

8.(a)(ii) -Environmental resistance refers to the biotic and abiotic factors

that oppose the achievement of biotic potential by a population.

-Environmental resistance limits the size of a population.

-Increase in the population size also increases the usage of limited

resources in the environment, and the depleted limited resources

may become an environmental resistance as it limits the growth of

population.

-Increase in the waste products produced by a larger population

size can act also as environmental resistance as it limits the

population growth.

-The growth rate slows down to zero and the population size tends

to stabilise, producing a S-Shaped or sigmoid growth curve.

3 marks

8.(a)(iii) - Carrying capacity is the total number of organisms that can be

supported by the environmental resources in an ecosystem.

-The area occupied by a population has limited the resources and

this limits the population growth by maintaining an equilibrium

between the natality rate and the mortality rate.

-Population size increases until it reaches saturation within the

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carrying capacity of its ecosystem.

-The carrying capacity of an ecosystem is not constant as it is

affected by environmental conditions.

3 marks

8.(b) -An example of a population with the S-shaped curve is the

Paramecium population grown in the laboratory.

*The S-shaped growth curve can be divided into four phases:

-(i) Initial phase in which the population growth rate is slow and

the organisms are still adapting to the new environment.

-(ii) Logarithmic /exponential phase in which exponential growth

of the population size occurs.

-(iii)Transitional phase in which the growth of the population size

slows down as environmental resistance starts to increase.

-(iv)Equilibrium phase in which the population size is stable and

varies around the fixed average value.

-(v)The growth of the population size is limited by environmental

resistance. An equilibrium between the natality rate and mortality

rate occurs. The population has reached maximum carrying

capacity.

6 marks

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9.(a)(i) Vector

Agent for carrying foreign DNA segment

Able to replicate freely

Able to express cloned gene

Possesses multiple cloning site

Possesses marker genes for screening

Cloned gene can be converted

Examples : plasmid @ phage

Maks: 3

9.(a)(ii) Restriction enzymes

Enzymes isolated from bacteria

Can cut DNA molecules at specific location

Can recognize short nucleotide sequence in DNA

Produces sticky ends

Enable isolation of target DNA and vector DNA

Examples : Eco R1 @ BAM H1

Maks:3

9.(a)(iii) Transformation

Refers to uptake of foreign DNA by bacretia cells

In gene cloning, plasmids that have taken up

foreign DNA fragment are call recombinant

plasmids

Need to be transferred into bacteria by

transformation process

Bacteria that received recombinant plasmid can be

used for cloning process to multiply required gene.

Maks:3

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9.(b) RNA from a specific cell is isolated and used as a template to make DNA by a method called complementary DNA spesies

Reverse transcriptase catalyses the reverse transcription of cDNA from the mRNA

mRNA is degraded with ribonuclease or an alkaline solution

DNA polymerase is used to synthesise the the complementary DNA strand to cDNA

The DNA produced is cut with a specific restriction enzyme

A cloning vector such as the plasmid is cut with the same restriction enzyme

Both the DNA fragments and the cloning vector DNA are mixed together in a test tube

DNA ligase is added to join the two parts to form recombinant DNA

rDNA produced will be introduced into host cells such as the E. coli bacrteria

The transform bacterial cells will be cultured in a medium

containing antibiotics. The collection of colonies containing the

coding region of the genome forms a cDNA Library………..

10. Human activities pose the greatest threat to the biodiversity in

Malaysia in the following ways:

(a) Loss of habitat

* Large areas of forest are cleared for agriculture,

housing and industry as well as undergo excessive

logging.

* The destruction of mangrove forest as a result of

development and logging also leads to habitat loss.

* Deforestation causes changes in climate, soil erosion

and leaching of mineral salts which further degrades

once pristine habitats.

* Silting of water bodies such as lakes and rivers from

development projects and logging changes the aquatic

environment making it not suitable for organisms

anymore.

* The loss of various types of habitats leads to the

extinction of many types of species.

Any 6

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(b) Air and Water Pollution

* The large amount of sulphur dioxide and other

greenhouse gases causes acid rain and increases the

temperature of the environment. Acid rain changes the

pH of water bodies and thus affects aquatic organisms.

The temperature of the water also increases.

* Leaching of fertilizers and nutrients from farms, leach

ate from sewage and landfills, as well as toxic waste

run-off from factories contributes to water and soil

pollution. This can also lead to eutrophication in lakes

and can cause the death of organisms.

* Rubbish from human settlements that find its way to

the sea also cause many sea creatures to die as they eat

the rubbish and choke on materials such as plastic

bags.

(c) Commercial overexploitation of species.

* Certain animals and plant species are commercially

hunted and collected for their medicinal values,

fashion (fur and reptile skin), exotic dishes (eggs,

meat, body parts) and sold as pets. Some have been

drastically reduced due to this.

* Certain fishing practice, such as trawling and fish

bombing in order to increase the catch, has led to large

reduction in numbers of sea animals. Leatherback

turtles, sharks and dolphins die as they are accidently

caught in the trawlers’ nets.

(d) Introduction of foreign species

* When humans settle in a new place, introducing a

foreign species can threaten the original/native species

of that particular area as these new species can cause

new diseases to that area or the newly introduced

species may prey on the native species.

[15 marks]

15 marks