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Biology STPM Trials 2010 Gombak District with answers. this is actually the 2010 paper but the front page had a typo resulting to the year being 2008, but this is actually the 2010 trial paper
Citation preview
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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1
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,
1
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