Mark scheme P1 and P2 Kedah
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Transcript of Mark scheme P1 and P2 Kedah
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PEPERIKSAAN PERCUBAAN STPM 2010
Anjuran Bersama
JABATAN PELAJARAN NEGERI KEDAH DARULAMAN
DAN
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH MALAYSIA (PKPSM) KEDAH
MARKING SCHEME
BIOLOGY (964)PAPER 1 & PAPER 2
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Answer
Paper 1
1 A 26 B
2 D 27 D
3 D 28 B
4 C 29 C
5 C 30 B
6 C 31 B
7 B 32 A
8 A 33 B
9 C 34 C
10 D 35 B
11 A 36 A
12 C 37 C
13 C 38 C
14 D 39 B
15 A 40 B
16 D 41 A
17 D 42 D
18 A 43 D
19 C 44 B
20 C 45 A
21 A 46 C
22 C 47 D
23 C 48 B
24 A 49 B
25 D 50 A
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Answer
Paper 2 Section A
1 (a) A: Golgi apparatus/Golgi body - 1
D: rough endoplasmic reticulum/rough ER - 1
E : mitochondrion - 1
(b) (i) Fluid mosaic model//Singers model//Singer and Nicholson model - 1
(ii) Phospholipid (bilayer) - 1
(c) 1 Regulates the fluidity of the plasma membrane - 1
2 Regulates the movement of phospholipid in different temperature//Helps to stabilise the membrane structure - 1
3 Regulates the movement of hydrophobic molecules or/ polar molecules
across the plasma membrane - 1
Max 2m
(d) (i) Centrioles/Centriole - 1
(ii) 1 Organise the formation of spindle fibres during cell division - 1
2 To produce basal bodies in which the flagella and cilli develop - 1
Total 10 marks
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2 (a) (i) glycolysis - 1
(ii) Stage A : cytoplasm/cytosol - 1
Stage B : matrix of mitochondrion/mitochondrial matrix - 1
Stage C : inner membrane of mitochondrion/mitochondrial inner membrane - 1
(b) oxaloacetate - 1
4C - 1
(c) (i) Electron transport chain/ETC/Electron transport system/Stage C - 1
(ii) Oxygen acts as the last/final acceptor of hydrogen atom in electrontransport chain - 1
(d) Stage A : 4 molecules of ATP - 1
Stage B : 2 molecules of ATP - 1
Total 10 marks
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3 (a) (i) A: hormone receptor - 1B: G protein - 1
C: enzyme adenylyl cyclase (adenylyl cyclase) - 1
(ii) 1 Cyclic AMP activates the protein kinase (an enzyme that is
previously present in an inactive form) - 1
2 which then activates the enzyme phophorylase which catalyses
the hydrolysis of glycogen into glucose - 1
(b) (i) The cascade effect - 1
(ii) 1 reaction N/Cascade effect causes the signal amplification - 1
2 a small amount of hormones stimulates the production of manycAMP molecules that activate many enzymes along the endocrine
pathway
(c) 1 acts as first messenger/initiates the Cascade effect - 1
2 stimulates the conversion of glycogen to glucose phosphate (in order
to increase the blood sugar level.) - 1
Total 10 marks
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4 (a) (i) 1 the entire process by which holozoic obtain their energy andnutrient - 1
2 through feeding on complex, solid organic food material - 1
(ii) 1 Earthworm/ centipedes/ millipedes/ woodlice - 1
2 Detritivores specialise in feeding on decaying organic matter and
digesting it internally - 1
(b) Columnar epithelium - 1
(c) (i) Z : goblet cell - 1
X: mucus - 1A: microvilli - 1
(ii) 1 Mucus /X acts as a lubricant (facilitate the movement of foodalong the alimentary canal)//
Mucus /X forms a protective lining, (protecting the cells of
alimentary canal e.g. stomach cells against digestion by
protease and acid) - 1
2 Microvilli/ A increase the surface area for absorption
of nutrients. - 1
Total 10 marks
Answer
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Paper 2 Section B
5 1. - HIV binds to a receptor at the surface of helper T cells//CD4
lymphocytes 1
2. - lipoprotein membrane of HIV fuses with lipoprotein
membrane of helper T cells 1
3. - viral RNA and reverse transcriptase enzyme enters the helper
T cells//once inside the host cell, the viral enzyme uses the
RNA as a template 1
4. - DNA produced then inserted into a host
chromosome//incorporates with the DNA of the host cells 1
5. - transcription process yields copies of viral RNA 1
6. - some transcripts are translated into viral proteins 1
7. - others get enclosed as hereditary material in the proteins whennew virus particles are put together 1
8. - the particles bud off from the hosts cells plasmamembrane//exocytosis and are released to start a new round of
infection 1
9. - with each round, more macrophages, antigen-presenting cell,
and helper T-cells are killed 1
10. - the host immune system produces antibodies in response to
HIV antigenic proteins 1
11. - helper T and cytotoxic T cells also are produced 1
12. - gradually, the immune system destroys about half of these
virus cells and replaces half of the helper T cells lost in the
battle
13. - huge reservoir of HIV and masses of infected T cells
accumulate in lymph nodes 1
14. - as the battle proceeds, the number of virus particles in the
general circulation rises 1
15. - the body produces fewer and fewer helper T cells to replace
the ones it lost 1
16. - by time, the erosion of helper T cell count cause to lose its
capacity to mount effective immune responses//eventually theimmune system collapses 1
Max 15 marks
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6 (a) Functions of placenta :
1. - allows exchange of substances or materials between the
foetus and the mother 1
2. - nourish the development of foetus by transferring essential
materials from mothers blood 1
3. - excretes metabolic waste from the foetal blood 1
4. - transfers certain maternal antibodies to provide the foetus
with passive natural immunity 1
5. - acts as barrier for the foetus against certain pathogen and
their toxins except the HIV virus 1
6. - shields the foetus from harmful substances of the mothers
blood
7. - permits the maternal and foetal blood systems to function
at different pressures 1
8. - produces hormones to prevent mothers ovulation and
menstruation 1
Max : 6 marks
Functions of the amnion :
9. - secretes amniotic fluid to fill the amniotic cavity that actsas a water cushion to help maintain the suitable
temperature for the foetus 1
10. - encloses the amniotic fluid to protect the foetus from
mechanical shock 1
11. - expands as the embryo increases in size, allows space for
foetal growth 1
12. - protects against fluid loss from the foetus due to
dehydration and against tissue adhesion 1
13. - amnion is suspended and allows the movement of foetus
without hindrance 1
Max : 4 marks
(10 marks)
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(b) 1. - pathogens and their toxins can reach the still developingfoetus 1
2. - drugs taken by mother can reach the foetus causing damageto the unborn child 1
3. - hormones present in the maternal blood can adverselyaffect the development of foetus 1
4. - drugs like nicotine and alcohol may lead to mental and
physical retardation in the baby 1
5. - maternal circulation has a blood pressure that is too high
for the foetus to handle 1
6. - since the foetus carries paternal genes, it will produces
foreign antigens to mother if their blood group is
incompatible 1
7. - if incompatible blood types mix, they will clot 1
8. - this leads to blockage of vital organs, possibly resulting in
death 1
9. - if the maternal antibodies leak to the foetus, it will lead to abreakdown in the foetal antigens that should be a source of
protection to the foetus 1
Max: 5 marks
Total : 15 marks
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7 (a) Difficulties and problems encountered when measuring growth:
1. - method//parameter : length//height 1*
2. - does not take into account growth in other directions,which may be considerable 1
3. - does not consider allometric growth// different growth rates
of body parts 1
4. - method//parameter : volume 1*
5. - difficult to measure if the organism is irregular shape
(dependent on p4) 1
6. - method//parameter : fresh mass 1*
7. - does not measure true growth//provide inconsistent reading
due to fluctuations in water content (dependent on p6) 1
8. - method//parameter : dry mass 1*
9. - organism will be killed during drying, 1
10. - thus the method cannot be used to monitor growth of an
organism over a period of time 1
11. - the sample must be large enough to obtain representativereflection of growth 1
12. - therefore many organisms must be killed 1
13. - fat accumulation//increase in fat content is not considered
as growth because it is reversible 1
14. - problem of irregular growth due to fluctuations in diet or
environment 1
Max : 8 marks(4* + 4)
dependent
on p1- any 1
dependent
on p8
- any 1
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(b) Absolute growth curve in plants :
correct axes and the unit 1/0
correct shape 1/0
1. - annual plants complete their entire life cycles, from seed to
reproduction to death in one year//limited growth 1
the curve consists of three part :
part I
2. - slight decreases in growth parameter in the initial stage of
germination 1
3. - because food reserves in the seed are used for (aerobic)
respiration to provide energy for germination 1
4. - glucose is oxidised during respiration to carbon dioxide andwater 1
5. - the loss of carbon dioxide in the form of gas does notcontribute to growth parameter 1
part II
6. - there is a large increase in dry mass after (the first week of)
germination 1
7. - a net increase in dry mass due to the higher rate of photosynthesis than respiration 1
any 2
Time (arbitrary unit)
Growth parameter
(example : drymass (g))
II IIII
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part III
8. - decrease in dry mass before the death of the plant. 1
9. - negative growth due to senescence, fall of leaves anddispersal of seeds 1
10. - due to senescence, the rate of cell death is greater than that
of cell division 1
Max: 7 marks
Total: 15 marks
any 2
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8 (a) 1. - chromosonal mutation is a random, spontaneous changes 1
2. - involving changes to the structure of the chromosome 1
3. - or number of chromosome 1
(b) 4. - Translocation 1
5. - is the movement of a portion of a chromosomes to another
portion of chromosome or the genome 1
6. - Internal translocation is the transfer that occurred within the
same chromosome 1
7. - Cross translocation is the transfer between the two non-
homologous chromosomes 1
8. - Alternating /reciprocal cross translocation involves thetransfer of interchangeable portion between the two non
homologous chromosomes 1
9. - Duplication : 1
10.- occurs when one or several chromosome segment undergoes
replication so that a set of gene is repeated 1
11.- also occurs when a chromosomes received a segment from
its homologue 1
12.- Deletion 1
13.- is the lost of chromosomes segment either from the end orinternally 1
14.- Inversion 1
15.- occurs when a chromosome segment breaks and the segment
breaks and the segment is reinserted in the opposite
orientation will respect to the rest of the chromosome. 1
Max: 12 marks
Total: 15 marks
dependent
on p4
dependent
on p9
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9 (a) 1. - Genetic screening is prenatal diagnostic procedure usuallyundertaken by woman whose pregnancies are considered
high-risk 1
2. - High-risk pregnancies are those in which the mothers are
over 35 years old or when the expectant parents are carriers
of genetic disorders 1
3. - May also refer to identification of carriers of defective
genes 1
4. - Two commonly used procedures are amniocentesis and
chorionic villus sampling 1
Max : 3 marks
(b) (i) amniocentesis
1. - Is usually done in the fourth month of pregnancies 1
2. - A sterile hypodemic needle is inserted through the
abdominal wall and into the uterus of the pregnant
woman 1
3. - During the procedure, the position of the needle andfoetus is monitored by ultra-sound 1
4. - A small sample of amniotic fluid is removed 1
5. - The amniotic fluid contains living cells derived from
the foetus 1
6. - The cells are cultured and then analysed for
chromosomal abnormalities, metabolic disorders, aswell as to determine the sex of the foetus 1
(ii) chorionic villus sampling (CVS)
7. - This procedure can be done much earlier (by the 8th
week) in pregnancy 1
8. - Produces results faster compared to amniocentesis 1
9. - Cells from the chorion (the foetal part of placenta) areremoved 1
10. - Chorionic villi cells are actively dividing, so karyotype
analysis can be performed without having to grow the
cell in tissue culture first 1
Max : 9 marks
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(c) 1. - Both amniocentesis and chorionic villus sampling are not100% accurate 1
2. - Many disorders cannot be diagnosed 1
3. - Procedure may harm foetus 1
4. - Since the conditions that are detected are incurable, it
leaves the expectant parents with the difficult choice of
terminating the pregnancy 1
5. - Or the prospect of ensuring lifelong care for the
child//parents have difficult choice when faced with the
prospect of having a child with an incurable genetic disease 1
Max : 3 marks
Total: 15 marks
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10 (a) 1. - It does not have any cellular structure and they are notmembrane- bound 1
2. - Not capable of reproducing outside of the host cell 1
3. - Other characteristics of the cell like movement, feeding,
respiration, excretion, growth, and sensitivity do not exist 1
4. - Virus are small (20 nm to 300 nm) being about 50 times
smaller than bacteria 1
5. - Visible only under the electron microscope 1
6. - Do not have cellular structure, no organelle or ability to
synthesis protein 1
7. - Have a simple structure consisting of a length of genetic
material forming a core surrounded and protected by a coat
of protein called a capsid 1
8. - Often made up of identical repeating subunits called
capsomeres 1
9. - Can be crystallized and inactive outside the host 1
10. - They are highly specific to their host 1
Max : 8 marks
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(b) Bryophytes are adapted to live on the land although they aremainly found in damp and shady areas , they are being able to
survive periods of dryness in the following ways:
1. - Depend on water to obtain mineral salts close to the surface
of the soil which assisted by the rhizoids(thin filaments
growing from the stem) 1
2. - Water as a medium of transport of the sperm to the
archegonia during fertilization 1
3. - The gametophyte thallus has a large surface area to absorb
water and salts 1
4. - They are able to absorb water and salts through the whole
surface of the plant, including the rhizoids 1
5. - The presence of rhizoids serves as anchorage to the
substratum 1
6. - Has no cuticle to prevent water lost 1
7. - The gametes develop in protective structures such as
antheridia and the archegonia 1
8. - Lack of strengthening and supporting tissues, they are
confined to an upward growth pattern to increase thesurface area for exposure to a water absorption. 1
Max : 7 marks
Total : 15 marks