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964/1
PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA NEGERI PAHANG DARUL MAKMUR 2009
BIOLOGY PAPER 1 MULTIPLE-CHOICE ONE HOUR AND FORTY-FIVE MINUTES
Instructions to candidates DO NOT OPEN THIS BOOKLET 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 multiplechoice answer sheet provided. Read the instructions on the multiple-choice answer sheet carefully. Answer all questions. Marks will not be deducted for wrong answers.
CONFIDENTIAL *
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1
2
3
Which one of the following is a characteristic of saturated fats that distinguishes them from unsaturated fats? A
They have no double bonds between their carbon atoms
B
They contain only unbranched fatty acids
C
They contain high proportion by mass of oxygen
D
They do not contain glycerol
Which part of a phospholipid molecule contributes most to the thickness of a cell surface membrane? A
Glycerol
B
Hydrocarbon chain
C
Hydrophilic head
D
Phosphate group
What does a haemoglobin molecule contain? A
Four iron (Fe2+) ions attached to each haem group
B
Four oxygen molecules attached to each haem group
C
Four polypeptide chains each with four attached haem groups
D
Four polypeptide chains each with one attached haem group
4 Potassium cyanide is known to interfere with the formation and use of ATP in cell metabolism. If the use of potassium cyanide resulted in an accelerated entry of a solute into a cell, it may be reasonably assumed that, under normal circumstances, the solute enters by A
Active transport
B
Osmosis
C
Passive diffusion
D
Pinocytosis
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5
6
Which cell component forms pinocytic vesicles? A
Lysosome
B
Nucloelus
C
Cell surface membrane
D
Endoplasmic reticulum
A culture of bacteria had their entire DNA labelled with the heavy isotope of nitrogen, 15
N. The culture was then allowed to reproduce using nucleotides containing normal 14N.
The DNA was examined using a centrifuge after one generation and again after two generations. The diagram shows the position of the DNA band at Z in the centrifuge tube when the DNA was first labelled.
X Y Z In which patterns would the DNA be found after the first and after the second cell generations? After first generation
After second generation
A Half at X and half at Y
Quarter at X and at Z and half at Y
B Half at X and half at Z
Quarter at X and at Z and half at Y
C
All at Y
Half at X and half at Y
D
All at Z
Half at Y and half at Z
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7
Turgid plant tissue is placed in a solution which has the same solute potential as the contents of the cells. The diagram shows a cell after one hour.
Which equation describes the value of the pressure potential for this cell? A Pressure potential = solute potential of the cell B Pressure potential = solute potential of the external solution C Pressure potential = water potential of the cell D Pressure potential = zero
8 A metabolic pathway is shown below. Enzyme 1 Reactant
Enzyme 2 substance X
Enzyme 3 substance Y
end
product
What would be the effect of adding a small amount of a non-competitive inhibitor to enzyme 2? A Enzyme 2 would be partially denatured B Substance X would increase in concentration C Substance Y would no longer be formed D The initial reactant would no longer be metabolised
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9 Which statement is true of photophosphorylation? I
High energy phosphate bond is split
II High energy phosphate bond is formed III It occurs in the thylakoid membrane IV Light is transformed to chemical energy A II, III, and IV B II and III C II and IV D I and III
10 The following scheme shows a pathway in the formation of fatty acids in plants. CO₂ RuBP
W
X
Y
Z Fatty acid
Which of the following compounds are represented by W, X, Y, and Z in the above scheme? A B C D
W Pyruvate Phosphoglyceraldehyde Phosphoglycerate Phophoglycerate
X Phosphoglyceraldehyde Acetyl CoA Phosphoglyceraldehyde Phosphoglyceraldehyde
5
Y Phosphoglycerate Phosphoglycerate Acetyl CoA Pyruvate
Z Acetyl CoA Pyruvate Pyruvate Acetyl CoA
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11 Six tubes containing preparations from animals were set up as shown below: Tube 1 2 3 4 5 6
Contents Glucose + homogenised cells Glucose + mitochondria Glucose + cytoplasm lacking organelles Pyruvic acid + homogenised cells Pyruvic acid + mitochondria Pyruvic acid + cytoplasm lacking organelles
After incubation, in which three tubes would carbon dioxide be produced? A
1, 2 and 3
B
1, 4 and 5
C
2, 4 and 5
D 3, 5 and 6 12
13
When a glucose molecule is completely oxidised to carbon dioxide and water, 38 moles of ATP molecules are formed. How many ATP molecules can be formed if a respiratory poison such as cyanide is added to the oxidation process? A
2 ATP
B
8 ATP
C
9 ATP
D
30 ATP
The following statements are related to the nutrition of living organisms W- Photosynthetic bacteria synthesise organic food from carbon dioxide and water using light energy absorbed by bacteriochlorophyll X- Chemosynthetic bacteria use energy obtained from the oxidation of organic matter Y- Human nutrition is holozoic nutrition Z- Bacteria and fungi involved in decomposition are saprophytic organisms Which of the above statements are true? A W and Y B
Y and Z
C W, Y and Z D X, Y and Z
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14 Graph 1 shows oxygen dissociation curves which illustrate the Bohr effect. Percentage of saturation
X
P O2 mmHg
Which one of the following statements could represent the X factor? I
High pH
II
Less CO2 partial pressure
III
Smaller size of animals
IV
Higher altitude
V
Myoglobin
A I, II, IV and V B I, III, IV and V C II, III, IV and V D I, II, III, IV and V
15 Which of the following is not a function of transpiration? A Cooling of the leaf B Excretion of mineral salts C Transport of mineral salts D Transport of water upwards
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16 Which nerve impulses can initiate normal expiration and prevent over expansion of the lungs? A Impulses from chemoreceptors in the lungs B Impulses from stretch receptors in the diaphragm C Impulses from the external intercostal muscles D Impulses from the stretch receptors in the walls of the alveoli
17 Which one of the following is a function of the lymphatic system? A To assist in the clotting mechanism of the blood B To return tissue fluid to the bloodstream C To transport dissolved food molecules to all cells of the body D To transport hormones from endocrine glands to their target organs
18 What effect would be caused by cutting the sympathetic nerve fibres to the heart? A A decrease in heart rate B A decrease in the length of the diastole phase C A decrease in the length of the systole phase D A decrease in the stroke volume
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19 The passage of some of the contents of the plasma through the capillary wall depends on the effective hydrostatic pressure (HP) of the blood in the capillary and the effective osmotic pressure (OP) of the blood plasma. The diagram below shows the HP of the blood at the arteriole and venule ends of a capillary and the HP of the tissue fluid, together with the OP of the blood plasma protein and the OP of the tissue fluids. All pressures are in mm Hg.
OP = 25 HP = 32
OP = 25
capillary
HP = 12
Tissue fluid OP = 10 HP = 8 venule
arteriole
Which one of the following is the pressure of filtration at the arteriole end of the capillary? A 1mm Hg B 9 mm Hg C 15 mm Hg D 25 mm Hg
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20 The diagram below shows a nephron of the mammalian kidney.
Which one of the regions A, B, C or D is the main region responsible for restoring the metabolite level of the blood plasma to its pre-renal level?
21 The following sequence of events occurs at the neuromuscular junction Nerve Impulses
Release of V
end plate potential
X released from sarcoplasmic reticulum
formation of Y
W produced in muscle fibre Muscle contraction
Which one of the following shows the correct sequence from V to Y? V
W
X
Y
A Acetylcholine
Action potential
Calcium ions
Actomyosin
B Acetylcholine
Action potential
Actomyosin
Calcium ions
C Actomyosin
Acetylcholine
Calcium ions
Action potential
D Calcium ions
Action potential
Actetylcholine
Actomyosin
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22 Which statement is correct? I
Phytochrome is produced in the dark
II
Phytochrome is a simple protein
III Phytochrome is a simple photoreceptor IV
Phytochrome is initially produced in an inactive form before it is converted into the active form.
A
I and II
B
I and III
C
III and IV
D
II and III
23 The female oestrous cycle is controlled by the secretion and sequential release of at least four hormones: FSH-follicle stimulating hormone LH-luteinising hormone Oestrogen Progesterone Which one of the following shows the correct order in which they are released? A
FSH
Oestrogen
LH
Progesterone
B
FSH
LH
Progesterone
Oestrogen
C
Oestrogen
FSH
LH
Progesterone
D
FSH
LH
Oestrogen
Progesterone
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24 Which one of the following chemical substances causes apical dominance in plants? A
Indole acetic acid
B
Giberellic acid
C
Cytokinine
D
Abscissic acid
25 Which one of the following is true of T cells? I
T cells produce complement proteins
II
T cells mature in the bone marrow
III
T cytotoxic cells kill body cells that are infected with viruses
IV
T suppressor cells inhibit the acivities of T cytotoxic cells
A
I and II
B
I and IV
C
II and III
D
III and IV
26 Which is not correct about the concept of self and non-self ? A
MHC protein markers on the surface of cells act as specific antigens
B
Each organism carries a set of antigens which is unique to each individual
C
In the embryo, T lymphocytes which can bind to the receptors (proteins and polysaccharides) of their own body cells are induced to proliferate to form a clone of T cells
D The immune system can distinguish “self” (the body’s own tissue) from “non-self” (foreign tissue)
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27 Which one of the following represents the general pattern of the alternation of generations? A
gametophyte
meiosis
B
gametophyte
C
sporophyte
meiosos
D
sporophyte
meiosis
meiosis
sporophyte
spores
spores
sporophyte
gametophyte spores
spores
gametophyte
28 If the diploid state is represented by (D) and the haploid state by (H), which one of the following sequences correctly describes the chromosome number in the named plant structures? Polar Nucleus Pollen grain Pollen tube nucleus
Cell in the testa Cell in the pericarp
A
(D)
(D)
(D)
(D)
(D)
B
(D)
(H)
(H)
(H)
(H)
C
(H)
(D)
(H)
(H)
(D)
D
(H)
(H)
(H)
(D)
(D)
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29 Some seeds, in which the major food reserve is lipid, were analysed for lipid and sugar content during germination in the dark. During germination, most of the lipid is converted to sugar. The results of this experiment are shown in the graph below.
lipid
sugar
Lipid Dry mass
Sugar dry mass
0
Time from sowing/ days
12
Which one of the following explains why the total sugar content decreases after one week? A The rate of consumption of sugar in respiration suddenly increases B Sugar is no longer being produced by photosynthesis C The rate of consumption of sugar in metabolism exceeds the rate of production of sugar from lipid D The sugar has moved from the endosperm, where it was produced, to the embryo, where it is needed
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30 The diagram below shows the scheme of hormonal control in insects. External stimulation
Neurosecretory cells in brain Larva Neurosecretory hormones
I
I
Pupa Adult
III
II
IV Different hormone Concentrations in blood
Which of the following identifies I, II, III, and IV?
I
II
III
IV
A
Corpus allatum
Prothoracic gland
Ecdysone hormone
Juvenile hormone
B
Corpus allatum
Prothoracic gland
Juvenile hormone
Ecdysone hormone
C
Prothoracic gland
Corpus allatum
Juvenile hormone
Ecdysone hormone
D
Prothoracic gland
Corpus allatum
Ecdysone hormone
Juvenile hormone
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31 Drosophilia flies heterozygous for the size of the wing and body colour were crossed with flies with vestigial wings and ebony bodies. The F1 offspring obtained were as follows: • Normal wing, grey body: 522 • Vestigial wing, grey body: 98 • Normal wing, ebony body: 104 • Vestigial wing, ebony body: 496 What is the crossover value? A 8% B 9% C 17% D 20% 32 Two animals are mated. One is homozygous dominant for one character and homozygous recessive for another. The other animal is heterozygous for both characters. How many phenotypes are expected in the offspring of this cross? A 1 B 2 C 3 D 4
33 Two parents each of blood group A have a daughter of blood group O. What is the probability that their next child will be a boy who has blood group O? A 0.125 B 0.25 C 0.50 D 0.75 E 0.825
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34 Haemophilia is caused by a sex-linked, recessive allele. Two parents have a haemophiliac son and a haemophiliac daughter. What are the most likely genotypes of the parents? mother
father
A
XHX h
X Y
B
XHY h
X Y
C
XhX h
X Y
D
XHY H
X Y
h
H
H
h
35 During the formation of an ovum, non-disjunction of the sex chromosomes occurred. The ovum was then fertilised by a normal Y bearing sperm cell. Which one of the following shows the sex chromosome complement of the resulting zygote? A XO B XY C XXY D XXXY E XXYY
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36 The diagram below shows an intergeneric cross performed in a breeding experiment
X plant
Y plant
(2n=18)
(2n=20)
Sterile hybrid
Fertile hybrid
What is the chromosome number in the sterile hybrid and fertile hybrid? Sterile hybrid
Fertile hybrid
A
18
36
B
19
19
C
19
38
D
20
40
37 In a population of Africans, 1.0% of its individuals suffered from sickle-cell anaemia and died. What is the percentage of the heterozygous individuals in that population that possess abnormal haemoglobin? A 1% B 10% C 18% D 99%
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38 Which one of the following phenotypic features of Man can be affected only by genotype? A height B intelligence C skin colour D number of different blood group antigens
39 The genotype and phenotype for a population of rabbits are as given below:
Genotype RMRM RM RP RPRP
Phenotype Black Black, White White
Number of rabbits 110 150 48
The frequencies of allele RM and RP in this population are RM
RP
A
0.4
0.6
B
0.3
0.7
C
0.2
0.8
D
0.6
0.4
40 Which one of the following statements is not true of the lactose operon? A The lactose operon hypothesis was proposed by Jacob and Monod B This system consists of four parts and they are the regulator gene, the promoter gene, the terminator gene and the three structural genes C This system is activated by the presence of lactose in the medium D A repressor molecule bound to the operator gene will prevent the transcription of the structural genes
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41 Resctriction enzymes EcoRI and BamHI both have the following properties I
They are extracted from bacteria
II Their action produces DNA fragments with sticky ends III They identify a sequence of six bases during restriction IV They carry out incomplete cutting of DNA V Their action is very specific A I, II and IV only B I, II, III and V only C II, III and IV only D I, II, III, IV and V
42 The following dichotomous key was made to identify different groups of organisms A1 A2 B1 B2 C1 C2 D1 D2
Unicellular organism ………… Multicellular organism ………... Enteron present ………... Enteron absent ………… Coleom present ………… Coleom absent ……….. Chitinous exoskeleton present ….. Chitinous exoskeleton absent ……
organism Q go to B organism R go to C go to D organism S go to E organism T
Among organisms Q, R, S and T, which organism belongs to the phylum Annelida A Organism Q B Organism R C Organism S D Organism T
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43 Which one of the following statements describes a feature of a natural system of classification which an artificial system does not have? A A natural system of classification is based on phenotypic characters. B A natural system of classification is based on analogous structures. C A natural system of classification is based on evolutionary relationships between organisms. D A natural system of classification is based on single characteristics rather than several similar features. 44 Human blood when mixed with antibodies to human blood, will give maximum precipitation. If another animal’s blood is mixed with antibodies to human blood, the percentage of precipitation indicates how similar the animal is to a human. The following experimental results were obtained Species: H 100%;
H ON
M 37%;
N 75%;
M P
H O N
A H M N O
O 79%;
M
P
B P
H N M
O
D
C
Which phylogeny would fit these results?
21
P
P 17%
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45 Refer to the following graphs which show the distribution of fang length in two populations A and B of adult female wolf spiders that live in different environments. The dotted line indicates the optimum fang length for capturing the most common prey in each environment. 500
500
Population A
Number of individuals
Population B
Number of individuals
0.5
1.5
2.5
0.5
1 .5
2.5
Fang length (mm)
Fang length (mm)
Which one of the following statements is consistent with the information shown in the graphs? A The average fang length in population A is greater than the average fang length in population B B The environment of population A has a wider range of prey than the environment of population B C The number of individuals in population A is the same as that in population B D Population A would be more likely than population B to become extinct if the average size of the prey in both environments decreased
46 Excess nitrate is a common pollutant in rivers and lakes causing rapid growth of planktonic algae and the death of submerged plants rooted in the mud. Lack of which factor causes the death of the submerged plants? A Carbon dioxide B Light C Organic debris D Oxygen
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47 Competition between species will be greatest if they try to occupy the same A Community B Ecosystem C Habitat D Niche 48 Humans affect the environment in the following ways 1
Felling of tropical rain forests
2
Harvesting of marine algae
3
Re-afforestation
4
Combustion of fossil fuels
5
Over-use of nitrate fertilisres
Which activities lead to an increase in the level of carbon dioxide in the earth’s atmosphere? A 1, 2 and 3 B 1, 2 and 4 C 2, 3 and 4 D 2, 4 and 5
23
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49 The diagram represents a pyramid of energy which shows energy loss from a food chain to decomposers, upward transfer of energy to the next trophic level and energy loss through repiration. All figures are in kJ, m-², y-1.
Loss to decomposers
Total energy of trophic level 40
24
Respiratory loss 16
114
360
206
1188
3600
2052
11 880
36 360 primary producers
20 880
What is illustrated by this diagram? A
A pyramid of energy shows the nutrient transfer less clearly than a pyramid of numbers
B
Energy loss to decomposers is higher than respiratory loss
C
Food chain efficiency is about 10%
D
The energy of the final trophic level is not used
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50 A student analyses the population of species X in an area S by using 1m X 1m quadrats (as shown in the diagram below). The numbers in the quadrats represent the number of species X. Calculate the density and frequency of species X
19
30
41
19
28
11
15
22
26
17
19
27
24
19
20
Density (m-2)
Frequency (%)
A
22.4
100
B
22.4
50
C
44.8
100
D
44.8
50
25
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964/2 PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA NEGERI PAHANG DARUL MAKMUR 2009
BIOLOGY PAPER 2 Two and a half hours
Instruction 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 your own answer sheets. Begin each answer on a fresh sheet of paper. Answers should be illustrated by large and clearly labelled diagrams wherever suitable.
For examiner’s use 1 2 3 4 5
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
6 7 8 9 10
Total
This question paper consists of 11 printed pages
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CONFIDENTIAL*
2 Section A [ 40 marks ] Answer all questions in this section.
1.
The diagram below shows the DNA replication fork during DNA replication
(a)
When does the replication occur in a cell cycle?
[1 mark]
………………………………………………………………………………………………………… (b)
Name enzymes C and D involved in replication and state their function. [4 marks]
C : …………………………………………………………………………………………………….. Function : ……………………………………………………………………………………………. ………………………………………………………………………………………………………… D : ……………………………………………………………………………………………………. Function : ……………………………………………………………………………………………. ………………………………………………………………………………………………………… (c)
Based on the diagram, there are two newly synthesized strands, strand X and Y (i)
Name both strands and state the way they are synthesized.
[2 marks]
X: …………………………………………………………………………………………….. Y: ………………………………………………………………………………………………
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(d)
3
State why DNA replication is said to be semi-conservative.
[1 mark]
……………………………………………………………………………………………………….. ……………………………………………………………………………………………………….. (e)
Explain why strand X is synthesized as fragments
[2 marks]
……………………………………………………………………………………………………….. ………………………………………………………………………………………………………… …………………………………………………………………………………………………………
964/2 This Question paper is CONFIDENTIAL until the examination is over.
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2.
4
The diagram below shows changes in the membrane potential of a neurone during the production of an action potential
(a)
Label phases A,B,C and the period labelled D as shown in the diagram.
[4 marks]
A : …………………………………………………………………………………………………….. B : ……………………………………………………………………………………………………. C : ……………………………………………………………………………………………………. D : …………………………………………………………………………………………………….
(b)
Describe how A is maintained
[2 marks]
………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ……………………………………………………………………………………………………..
(c)
Distinguish between B and C
[2 marks]
………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………..
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5
………………………………………………………………………………………………………. (d)
Why is the potential of D less than A?
[1 mark]
……………………………………………………………………………………………………….. ……………………………………………………………………………………………………….. (e)
State the importance of D
[1 mark]
………………………………………………………………………………………………………. ……………………………………………………………………………………………………….
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6
3. Graphs S and T show the measurements of growth of a same plant, Zea mays.
(a)
Name the type of curve illustrated by graphs S and T.
[2 marks]
S:………………………………………………………………………………………………………… T:………………………………………………………………………………………………………… (b)
Explain what graphs S and T represent.
[2 marks]
S: ……………………………………………………………………………………………………….. T: ……………………………………………………………………………………………………….. (c)
Based on its lifespan, what type of plant is Zea mays known as?
[1 mark]
…………………………………………………………………………………………………………..
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(i)
7
Describe the growth pattern in Zea mays
[2 marks]
………………………………………………………………………………………………………….. …………………………………………………………………………………………………………. (ii)
Explain how this type of growth pattern differs from that of woody plants [3 marks]
………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….
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CONFIDENTIAL* 4.
8
The figure below shows a summary of a procedure designed to clone genes
Foreign DNA
Isolated plasmid
restriction
I
Recombinant DNA
II
Bacterial cell
III
SCREENING
964/2 This Question paper is CONFIDENTIAL until the examination is over.
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(a)
9
What is meant by recombinant DNA?
[1 mark]
…………………………………………………………………………………………………………. (b)
State briefly what happens at step I, II and III
[3 marks]
I : ……………………………………………………………………………………………………… II: ……………………………………………………………………………………………………… III: …………………………………………………………………………………………………….. r (c)
Why are the plasmid and foreign gene cut by using the same restriction enzyme? [1 mark]
………………………………………………………………………………………………………… (d)
State two ways in which bacteria can be screened for the presence of recombinant plasmids [ 2 marks]
………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….. (e) (i) State two benefits of insulin produced by genetic engineering
[2 marks]
………………………………………………………………………………………………………….. ………………………………………………………………………………………………………….. …………………………………………………………………………………………………………. (ii) Unlike insulin, blood cannot be produced by genetic engineering. Explain why ?
[1 mark]
………………………………………………………………………………………………………… …………………………………………………………………………………………………………
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10
Section B [60 marks] Answer any four questions in this section.
5. (a) With reference to cellulose, explain briefly the term polymerization. State briefly the differences between the formation of polysaccharides and polypeptides [7 marks] (b) Describe the mechanism of action of an enzyme based on the induced fit model [8 marks] . 6. (a) State briefly the differences between anaerobic respiration that occur in plants and animals. [6 marks] (b) NAD and NADP are important molecules in plant cells. Describe in detail, the role of these molecules within a palisade mesophyll cell . [9 marks] 7. (a) Drugs affect the nervous system by altering the mechanism of synaptic transmission. By using a suitable example, describe the effects of a stimulant drug on the nervous system. [6 marks] (b) AIDS is an immunodeficiency disease caused by an RNA virus. Describe how the virus replicates in the host cell . [9 marks] 8. (a) Explain why no flowering response is shown by some plants that grow near highways. [5 marks] (b) Explain the changes that occur during the onset of labour in humans. [10marks] 9. (a) With reference to an example, explain what is meant by the following genetic terms (i) multiple alleles (ii) polygenic inheritance [6 marks] (b) In sweet corn plants, the alleles for red and smooth seed are dominant over the alleles for white and wrinkled seeds. Sweet corn plants with red and smooth seeds were crossed with plants with white and wrinkled seeds. A test cross was done using the F1 progeny and the results are as follows: Red, smooth 348 White. wrinkled 335 Red, wrinkled 40 White, smooth 39 By using suitable symbols, draw a genetic cross diagram to explain the results. [9 marks]
964/2 This Question paper is CONFIDENTIAL until the examination is over.
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11
10. (a) Explain why proper management and conservation of ecosystems are needed. [5 marks] (b) (i) Define sustainable development
[2 marks]
(ii) Explain the benefits of sustainable development techniques used in the management of forests. [8 marks]
964/2 This Question paper is CONFIDENTIAL until the examination is over.
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964/1
PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA NEGERI PAHANG DARUL MAKMUR 2009
BIOLOGY PAPER 1 MULTIPLE-CHOICE
ANSWERS
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1
A
11
B
21
A
31
C
41
B
2
B
12
A
22
C
32
B
42
D
3
D
13
C
23
A
33
A
43
C
4
A
14
A
24
A
34
A
44
B
5
C
15
B
25
D
35
C
45
B
6
C
16
D
26
C
36
C
46
B
7
D
17
B
27
D
37
C
47
D
8
B
18
D
28
D
38
D
48
B
9
A
19
B
29
C
39
D
49
C
10
D
20
B
30
B
40
B
50
A
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PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA NEGERI PAHANG DARUL MAKMUR 2009
BIOLOGY PAPER 2
MARKING SCHEME
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1. (a) - during S phase / interphase (b) - enzyme C :
1
1
DNA polymerase
1
- function :
adds complementary nucleotides to the growing strand
1
- enzyme D:
DNA ligase
1
- function :
links Okazaki fragments by forming phosphodiester bonds between the fragments
1
4
lagging strand, which is synthesized discontinuously leading strand, which is synthesized continuously
1 1
2
(d) - because each new DNA molecule contains one old and one new strand
1
1
(e) - the template for strand X is in the 5’ to 3’ end direction
1
(c) - strand X : - strand Y :
- DNA polymerase only adds nucleotides at the 3’ of a growing Strand - therefore, strand X has to be synthesized away from replication fork any 2
1 1 2 ------total 10
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2
(a)
(b)
A: B: C: D:
Resting potential depolarization repolarisation refractory period
1 1 1 1
- The active transport of sodium and potassium ions in and out across the membrane/ 3Na+ out, 2 K + in - The membrane is more permeable to potassium ions compared to sodium ions ; thus more potassium ions diffuse out of the neurone compared to sodium ions diffuse in
4
1 1 2
(c) B / depolarization Sodium-gated channels open
C / Repolarisation Potassium-gated channesl open
Sodium ions enter the neurone
Potassium ions leave the neurone
The inside of a neurone becomes positive
The inside of a neurone becomes negative
(d)
(e)
Any 2 pairs - the potassium-gated channels remain open / slow to close This allows further efflux of potassium ions
1
- Limits the frequency in which impulses may flow - Ensures that impulses flow in one direction along the nerve
2 1
1 1 any 1
1 total
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3 (a) - graph S : absolute growth curve - graph T : absolute growth rate curve
1 1
2
(b) - graph S : actual increase in size / area/ growth over a period of time - graph T : the increase in size / area / growth per unit time over a period of time
1 1
2
(c) - annual plant
1
1
1
2
(d) (i) (ii)
- Limited growth - the growth curve is a sigmoid curve
1
- woody plants are perennial plants - display unlimited growth - the growth curve is a series of sigmoid curve
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1 1 1
3 ----Total 10
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4 (a) - DNA that contains genes from more than one source
1
(b) - step I : Insertion of the DNA fragment into the plasmid
1
1
- step II : Transformation / introducing recombinant DNA into host cell - step III: DNA cloning / amplification making multiple copies of target gene
1 1
3
(c)
- to produce the same sticky / blunt ends which are complementary to each other
1
1
(d)
- cultured in a medium containing antibiotics/ ampicillin and X-gal before a blue-white screening - hybridization by using genetics / radioactive probes
1
1
1
1
(i)
1
(e)
- insulin produced through genetic engineering is similar to human insulin - insulin produced is non –allergic - cheaper / not costly
1 1
Any 2 (ii)
- because DNA / gene only codes for the synthesis of a protein and blood is not a protein
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1
Total
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5. (a)
i.
answer polymerization is formation of long repeated units of monomer
ii.
by condensation with the removal of water molecules
1
iii.
in cellulose the monomer unit is β-glucose
1
iv
β-glucose linked together by β- 1,4 glycosidic bonds
1
[any 3]
i.
Polysaccharide does not involve ribosome
marks 1
maximum
polypeptide Involve ribosome
3
marks 1/0
ii. only involve one type of monomer – hexose molecule
Involve 20 different types of amino acid
1/0
iii. formation of glycosidic bonds between the monomers
formation of peptide bonds between the monomers
1/0
iv. does not involve gene / DNA
Involve gene / DNA to code the amino acid sequences
1/0
v. does not involve mRNA and tRNA
Involve mRNA and tRNA
1/0
vi. no transcription and translation stages
Involves transcription and translation stages
1/0
[ Any 4 pairs]
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4
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(b)
i.
Answer the active site of an enzyme is flexible
marks 1
ii.
and not exactly complementary / fitting to the shape of the substrate
1
iii.
the active site of an enzyme can be modified when its interacts with substrate
1
iv.
the active site changes its shape slightly as the substrate enters
1
v.
the active site continues to change until the substrate is completely bound and final shape is determined
1
vi
making the fit more precise
1
vii
these structural changes may help to stabilized the transition state
1
viii.
when the substrate binds to the enzyme, it forms an enzyme substrate complex
1
ix
the substrate is then converted into products
1
x.
the products are released from the active sites
1
xi,
the enzymes return to its original conformation
1
xi.
Products are released from the active sites
1
xii.
the enzymes return to its original conformation
1
[any 8]
maximum
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6. (a) Plants i. occurs in roots / seed / storage organ /tuber
Animals Occurs in skeletal muscles
ii. occurs when root are submerged // oxygen supply decreases
Occurs when muscles are actively Contracting
1/0
iii. products are ethanol and carbon dioxide
Product is lactic acid
1/0
iv. pyruvate changed / reduced to ethanal first then to its final product, ethanol
Pyruvate directly changed / reduced to its final product , lactic acid
1/0
v. ethanol is poisonous / can kill the plant cells
Lactic acid is still useful / not poisonous
1/0
vi. ethanol cannot be converted back to glucose
Lactic acid can be converted back to glucose
1/0
vii. anaerobic respiration is not needed to sustain the supply of ATP
anaerobic respiration is useful to sustain the supply of ATP
1/0
Any 6 pairs
marks 1/0
maximum
6
6 (b)
i.
Answer NAD involved in cellular respiration
ii.
2 molecules of reduced NAD / NADH produced in glycolysis
1
iii.
1 / 2 molecules of NADH produced in link reaction
1
iv.
3 / 6 molecules of NADH produced in Kreb cycle
1
v.
NAD accepts hydrogen / electron and transfers / carry to the inner mitochondrial membranes / cristae / cytochromes / ETC / matrix
1
vi.
NADP involved in photosynthesis
1
vii
viii.
marks 1
NADP accept electron and hydrogen ion forming NADPH in non-cyclic photophosphorylation / light dependent reaction / light reaction NADPH used in Calvin cycle / dark reaction / light independent reaction
1
1
ix.
NADPH is used to reduce CO2 / in reduction phase
1
x.
NADPH reduces glycerate -1,3-diphosphate into
1
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CONFIDENTIAL* glyceraldehydes- 3-phosphate / [ Any 9]
PGAL
maximum
9
7. (a)
i.
Answer cocaine present in the synaptic cleft
marks 1
ii.
cocaine bind to the transporter molecules
1
iii.
cocaine delays the breakdown of norepinephrine / dopamine / noradrenaline
1
iv.
reuptake/reabsorption of the norepinephrine / dopamine / noradrenalin from synaptic cleft is prevented / blocked
1
v.
norepinephrine / dopamine / noradrenalin remains / accumulates in the synaptic cleft
1
vi.
norepinephrine / dopamine / noradrenalin keep on binding to the receptors on the post-synaptic membrane
1
vii.
impulses will be generated repeatedly
1
viii.
causes continuous euphoria / intense pleasure
1
[ Any 6]
maximum
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6
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(b) i.
Answer HIV glycoprotein (gp 120) binds to CD4 receptor on the surface of the helper T cell / TH cell / host cell
marks 1
ii.
HIV protein envelope / lipoprotein / fuse with / to host cell membrane
1
iii.
Reverse transcriptase and viral RNA enter the host / TH cell by endocytosis
1
iv.
reverse transcriptase makes a DNA copy (cDNA) of the viral DNA // The first DNA strand is formed from RNA viral by using reverse transcriptase
1
v.
DNA copy (cDNA) / the first DNA strand becomes a template for the synthesis of complementary / second strand of viral DNA
1
vi.
the double stranded viral DNA formed then enters the nucleus and incorporated into the DNA of the host cell called provirus
1
vii.
each time the host’s DNA divides / replicates, viral DNA / provirus also replicates
1
viii.
this carries on for six years without any symptoms of disease are exhibited
1
ix.
the activated provirus will direct the host cell to synthesis new viral protein and viral RNA
1
x.
new viral protein and RNA assembled into new retroviruses
1
xi.
new retrovirus particles bud off from host cell membrane by exocytosis
1
xii.
finally, the virus spreads to all the helper T cells and destroys them
1
xiii.
resulting the destruction of the immune system
1
Any 9 Maximum
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9
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8. (a) marks 1
i.
answer these types of plants are short day plants The plants require a period of darkness equal to / longer than critical night length light / spotlight flashes from vehicles interrupt / shorten the period of darkness
1
ii. iii. iv.
causes the conversion of Pr into Pfr
1
v.
Pfr is the (biological) active form
1
vi.
high concentration / level of Pfr inhibit flowering in SDP’s
1 [any 5] Maximum
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1
5
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(b) Explain the changes happens and action of hormone on the onset of labour in human [10marks]
i.
answer estrogen level high in the end / last week of pregnancy
marks 1
ii.
stimulate the development of oxytocin receptor in uterine wall // increase the sensitivity of uterus muscle/myometrium to oxytocin
1
iii.
baby’s head / growing fetus press against the mother cervix
1
vi.
causing the cervix and uterine wall to stretch
1
v.
this stimulates maternal posterior pituitary to release oxytocin
1
vi.
and also stimulates fetal pituitary gland secretes ACTH
1
vii.
ACTH triggers fetal adrenal gland to secrete corticosteroids hormone
1
viii.
corticosteroids hormone triggers placenta to secrete prostaglandin
1
ix.
oxytocin also stimulates placenta / fetal to produce prostaglandin
1
x.
both prostaglandin and oxytocin stimulates a powerful contraction of myometrium / uterus wall
1
xi.
The more frequent the myometrium contracts, the more oxytocin / prostaglandin produced the uterus contraction also stimulates stretch receptors in the uterine wall and the cervix to trigger further release of oxytocin
1
xiii.
the cervix dilates / widen, its tissue softens and become more flexible and amniotic bag burst
1
xiv.
contractions become stronger and stronger / more frequent help to push baby down/out (Any 10) Maximum
1
xii.
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1
10
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9.(a) Multiple alleles:
Answer one particular characteristic / phenotype is controlled by three or more alleles
marks 1
ii.
each allele can occupy the same gene locus on a pair of homologous chromosomes
1
iii.
only two alleles can occupy a locus on a pair of homologous chromosomes at any time e.g. human ABO blood group
1
i.
iv.
1
Maximum
3
Polygenic inheritance:
i.
Answer one particular characteristic is controlled by two or more than two different genes
ii.
genes located at different loci on different chromosomes
1
iii.
each dominant allele has a small quantitative effect individually on the phenotype// phenotype produced is the total effect of all dominant genes present
1
iv.
e.g. height in human
1 Maximum
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Marks 1
3
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(b) [9 marks] Parents (p)
1
Phenotype
Red smooth seed R
1
x
White wrinkled seed
R
Genotypes
r
r
s
s
x S
S
Meiosis
1
R
r
S
s
Gametes
Fertilisation
1
F1 generation genotype
Phenotype
R
r
S
s
All red and smooth seed
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Test cross for F1 progeny
] R 1
r
Genotype
1
r
s
s
x S
1
r
Phenotype
s
Red smooth seed
x
White wrinkled seed
R
r
R
r
r
S
s
s
S
s
Gametes
Recombinant/new combination of genes
1
1
1
Genotype offspring
Phenotypes
R
r
r
r
R
r
r
r
S
s
s
s
s
s
S
s
Red smooth
White wrinkled
Prental combination 346
335
Red wrinkled
White smooth
Recombinants forms as a result of crossing over 40
39
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10. (a)
i.
answer to sustain the biodiversity / flora and fauna of ecosystems
ii.
to maintain / improve the quality of life
1
iii.
to prevent the useful biological resources that have economical /medicinal value from extinct
1
iv.
to prevent the biogeochemical cycles from disrupted by the extinction of some species
1
v.
to prevent an extreme global climatic changes
1
vi.
to keep the ecosystems in their natural state (e.g. ecotourism) which provides aesthetic values for humans
1
vii.
to protect the environment for future generation
1
maximum
marks 1
5
(b)(i)
i.
answer is development that can continue indefinitely
ii.
can be achieved by minimizing the use of non-renewable resources
1
iii.
and controlling the use of renewable resources of the earth
1
Maximum
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marks 1
2
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(b) (ii) techniques i. Replanting trees in areas that have been logged ii. Creating forest reserves
benefits Can decrease the carbon dioxide level in the Atmosphere // can reduced the greenhouse Effect
marks 1+1
- Able to avoid the complete destruction and maintain the biodiversity of the forests. - to keep the ecosystems in their natural state(e.g. ecotourism) which provides aesthetic values for humans
1+1
- to avoid lost of wildlife and potential resources [any 1] iii. Selective logging / only selects tree with certain diameters / species can be felled iv. Enforcement of laws and surveillance iv.
Recycling of paper
- to maintain soil fertility - to prevent flood, soil erosion - to avoid species extinct - to avoid loss of watershed areas - to avoid water pollution To prevent illegal logging
1+1
1+1
To reduce the demand for new raw materials
Any 4 pairs
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Total
1+1
8
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