kelantan schema answer paper 1 & 2
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SULIT 4551
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4551 Biologi SEPT. 2008
PKPSM
PEPERIKSAAN PERCUBAAN SPM TAHUN 2008
Kertas soalan ini mengandungi 13 halaman bercetak.
BIOLOGY
PERATURAN PERMARKAHAN
KERTAS 1,2, 3
UNTUK KEGUNAAN PEMERIKSA SAHAJA
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ANSWER PAPER 1 ;
1 C 16 D 31 C 46 D
2 B 17 D 32 A 47 A
3 C 18 D 33 D 48 A
4 A 19 B 34 D 49 B
5 C 20 D 35 C 50 B
6 B 21 C 36 A
7 A 22 B 37 B BIO
8 C 23 C 38 D PAPER 1
9 C 24 A 39 C
10 B 25 D 40 A
11 A 26 A 41 D
12 D 27 C 42 B
13 B 28 D 43 C
14 A 29 B 44 C
15 A 30 B 45 B
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1 Diagram 1 shows structures of protein. The protein structures can be
classified into four levels J, K, L and M based on organization of their
structures.
Rajah 1 menunjukkan struktur protein. Struktur protein boleh dikelaskan
kepada empat aras J,K L dan M berdasarkan organisasi bentuk strukturnya.
Diagram 1
(a) ( i) Label amino acid and peptide bond in J Labelkan asid amino dan ikatan peptide pada J X : Amino acid Y : Peptide bond [2 marks] (ii) Name the protein structures of L and M Namakan struktur protein L dan M. L: Secondary structure
M: Quaternary structure [2 marks]
(b) (i) Name the organelle in the cell where the protein is synthesized? Namakan organel dalam sel di mana protein disintesis? Ribosome [1 mark]
(ii) Name the process P in the following reaction? Namakan proses P dalam tindakbalas berikut?
P + H2O Hydrolysis
1 (a)(i)
1 (a)(ii)
1 (b)(i)
1 (b)(ii)
J K L M
X
Y
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All enzymes are protein. Enzymes are sensitive to temperature.
Semua enzim adalah protein. Enzim peka kepada suhu.
[ 1 mark ] (iii) By using the letters J,K L and M, which protein structure is represented Dengan menggunakan huruf J, K, L dan M, struktur protein manakah mewakili Enzyme / enzim : K Haemoglobin /hemoglobin : M [ 2 marks ]
(c) Explain why food is kept in refrigerator?
Terangkan kenapa makanan disimpan dalam peti sejuk?
F : Temperature in refrigerator is low
P1 : Enzyme (in bacteria) is inactive
P2 : food cannot be decomposed
[3 marks ]
d) A branded washing machine is provided with temperature regulator.
A housewife uses the detergent containing enzyme at 40oC to wash
the clothes. Using the information given, explain why?
Mesin basuh berjenama dibekalkan dengan pengawalatur suhu.
Seorang surirumah menggunakan pencuci yang mengandungi enzim
pada suhu 40oC untuk mencuci pakaian. Dengan mengunakan
maklumat yang diberi, terangkan mengapa?
F: 40oC is an optimum temperature
P1: enzyme activity is maximum
P2: the cleaning is more effective.
[ 2 marks]
1 (b)(iii)
1 (c)
1 (d)
TOTAL
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2 Diagram 2 shows a food pyramid.
Rajah 2 menunjukkan piramid makanan.
DIAGRAM 2
Rajah 2
(a) Label A, B, C and D are the different classes of food that make up the
balanced diet.
Label A, B, C and D adalah kelas-kelas makanan yang berbeza
dalam gizi seimbang.
(i) Name the classes of food labeled C and D.
Namakan kelas makanan dilabelkan C dan D.
C : Fibre / roughage /dietary fibre
D: Carbohydrate
[2 marks]
(ii) State two functions of C ?
Nyatakan dua fungsi bagi C
To prevent constipation //
Source of vitamins // mineral (any example accepted)
[2 marks]
2 (a)(i)
2 (a)(ii)
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(b) Explain the statement above.
Terangkan pernyataan di atas..
• Malnutrition due to the lacking , the excessive or the wrong proportion of nutrients intake for a long term. (1 mark)
• Example of protein: bean/ meat/fish (1 mark)
• Lack of protein intake cause kwashiorkor (1mark)
• Health problem/ Symptom: scaly skin/ thin muscles/thin hair /swell of the body. (1 mark)
OR
• Excessive protein intake cause gout/ kidney stone/ kidney damage (1 mark)
• Health problem/ symptom :inflammation of joint /urination trouble
[4 marks]
(c) Explain how the methods above extend the life span of the foods.
Terangkan bagaimana kaedah-kaedah di atas dapat memanjangkan jangka hayat makanan.
Method Explanation /Biological concept
1 Fermentation
Food substances are added with yeast(1 mark)
-Fermentation yields ethanol which, at high concentration, will stop the activity of bacteria that cause food spoilage.(1 mark)
2 UHT treatment
Fresh milk is heated to 1320C for 1-5 seconds (under high pressure).(1 mark)
The high temperature kills microorganisms/ bacteria /fungus and microorganisms/ bacteria /fungus spores. (1 mark)
[4 marks]
2 (b)
2 (c)
TOTAL
Malnutrition caused by the unbalanced diet. Malnutrition of B
for a long term will affects certain health problems.
Malnutrisi berpunca dari gizi yang tidak seimbang. Malnutrisi
bagi B dalam tempoh masa yang lama akan menyebabkan
beberapa masalah kesihatan.
Fermentation and UHT treatment are food processing methods.
Penapaian dan UHT adalah kaedah pemprosesan makanan
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3 Diagram 3.1 shows agricultural activities which give effect P along the river at Zone R, Zone S and Zone T. Diagram 3.2 shows the changes of dissolved oxygen and the changes of certain bacteria population in the river. Rajah 3.1 shows aktiviti pertanian yang memberikan kesan P di sepanjang aliran sungai di Zon R, Zon S dan Zon T. Rajah 3.2 menunjukkan perubahan kepekatan oksigen terlarut dan perubahan populasi bakteria tertentu dalam sungai itu.
Diagram 3.1
Diagram 3.2
R T S
Co
nce
ntr
ation
o
f d
isso
lve
d
oxyge
n. (
arb
itra
ry u
nit) P
opu
latio
n o
f ba
cte
ria
(arb
itrary
un
it)
Oxygen Bacteria
X
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(a) (i) What is the distance from X where the concentration of dissolved
oxygen and the bacterial population begin to change?.
Berapakah jarak dari X dimana kepekatan oksigen terlarut dan
populasi bakteria mulai berubah?.
15 km
[1 mark]
(ii) What is the ecological term for the effect P? Give one example of
substance which cause the effect P.
Apakah istilah ekologi bagi kesan P? Berikan satu contoh bahan yang
menyumbangkan kesan P.
Water pollution
Pesticide/ herbicide / fungicide/ insecticide
OR
Eutrophication
Faeces / fertilizer’s
[2 marks]
(b) The fish population in the river was also affected by P. Draw in the
Diagram 3.2 to show the change of fish population at Zone R, S and T.
Give a reason.
Populasi ikan dalam sungai tersebut juga mengalami kesan dari P. Lakar
graf dalam Diagram 3.2 untuk menunjukkan populasi ikan pada Zon R, S
dan T. Berikan alasan anda.
(On the Diagram 3.2)
Oxygen level in the water drops, they die / their population
decrease.
[2 marks]
3 (a)(i)
3 (a)(ii)
3 (b)
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(c) Describe the phenomenon based on above statement.
Berdasarkan pernyataan di atas, huraikan fenomena tersebut.
o Agriculture activities releases fertilizer / nitrates / phosphates
o Decomposition of the waste product/ sewage release mineral / ions
o Leaching / Washing down the fertilizers and ions/minerals into
the river, accelerate / promotes alga and aquatic plant growth/algal
blooming
o Prevent penetration of sunlight into the river and inhibit
photosynthesis process
o Oxygen content decrease, aquatic organisms die.
[4 marks]
(d) Based on Diagram 3.2, explain the relationship among the agriculture
sewage, bacterial population and the oxygen content in the river.
o High agriculture sewage provides food / organic substances for
microorganisms
o Encourages the population of bacterial in the river.
o There is increase in oxygen consumption by the bacteria / High
BOD
o Therefore the oxygen content in the river decrease.
[3 marks]
3 (c)
3 (d)
TOTAL
After 3 years, there is abundant of algal population at Zone S which
result the death of aquatic organisms.
Selepas 3 tahun didapati Zon S mengalami pertumbuhan populasi alga
yang sangat ketara dan menyebabkan kematian organisma akuatik.
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4 Diagram 4 shows the pathway of water movement from the soil to the upper
plant of the plant.
Rajah 4 menunjukkan laluan pergerakan air dari tanah ke bahagian atas
tumbuhan.
DIAGRAM 4
RAJAH 4
(a) State one characteristic of R and its importance.
Nyatakan satu ciri R dan kepentingannya
o Characteristics of R : small in size // have a large total surface
area to volume (TSA/V) // Increases the surface area
o Important of R: for water (and mineral) absorption //
o Characteristics of R : Have thin cell wall // have no cuticle
o Important of R: for water (and mineral) ions absorption //
o Charateristics of R : The cell sap in R is usually hypertonic to
the surrounding soil water.
o Important of R: for water absorption.
(any 2) [ 2 marks ]
4 (a)
Soil
R S
T
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(b) (i) Based on Diagram 4, explain how water from soil move to structure T.
Berdasarkan Rajah 4, terangkan bagaimana air dari tanah bergerak
ke struktur T
• The cell sap of R is hypertonic to the soil water.
• So, water diffuses into R by osmosis
• The entry of water dilutes cell sap of R// cell sap of R becomes
hypotonic compared to cell sap of S/ the next cells.
• Therefore water diffuses into S/to these adjacent cells which
become more diluted themselves, so osmosis continues across
the S
• The continuous flow of water in S creates a force known as root
pressure to push water into xylem.
(any 4) [ 4 marks ]
(ii) The flow of water along the T structure is carried out by capillary
action. Name the forces involved to ensure the continuous flow of
water.
Pergerakan air sepanjang struktur T disebabkan oleh tindakan
kapilari. Namakan daya yang terlibat untuk memastikan pergerakan air
yang berterusan
Adhesion & Cohesion (force)
[ 1 mark ]
(iii) Explain how these forces in b(ii) enables the movement of water to the
top of the plant.
Terangkan bagaimana daya dalam b(ii) membolehkan pergerakan air
ke bahagian atas tumbuhan.
o Adhesion – attraction force between water molecules and the xylem
wall enable water to move up along the narrow xylem vessels
o Cohesion – attraction force between water molecules form a
continuous water column in the xylem vessels
[ 2 marks ]
4(b)(i)
4(b)(ii)
4(b)(iii)
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(d) Describe how the plant control their water loss.
Terangkan bagaimana tumbuhan mengawal kehilangan air.
o water is lost from the plant in the form of water vapour (into the
surrounding air) by transpiration,
o mainly through the stomata in the leaves.
o The amount of water lost depends on the size of the stomatal
pore
o (the size of the stomatal pore) is controlled by the guard cell//
through the opening and closing of stomata
OR
Examples: in daylight /light intensity is high, the guard cell
become turgid// stomata open, thus the water loss increases
OR
Examples: in dark /light intensity is low, the guard cell become
flaccid// stomata closed, thus the water loss decreases
(any 3) [ 3 mark
4(d)
TOTAL
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5 Diagram 5 shows the changes of four types of hormone which control the
menstrual cycle and follicle development in the ovaries.
Rajah 5 menunjukkan perubahan empat jenis hormon yang mengawal kitar
haid dan perkembangan folikel dalam ovari.
Diagram 5
(a) Based on Diagram 5 name the hormone labeled P and R.
Berdasarkan Rajah 5, namakan hormon yang berlabel P and R:
P : LH / Luteinising hormone
R : Oestrogen
[2 marks]
(b) Complete the follicle development in boxes M and N in the Diagram 5.
Lengkapkan perkembangan folikel dalam petak M dan K pada Rajah 5.
[1 marks ]
5 (a)
5(b)
Q P
R S
M N
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(c) Based on the Diagram 5, explain the relationship between the structure M
and the level of hormone S.
Terangkan hubungan di antara aras hormon S dengan struktur M.
• After ovulation, M / corpus luteum secretes S / progesterone
• the level of S/ progesterone increases to maintain the thickness
of the endometrium.
• When the M / corpus luteum degenerates, the level of S/
progesterone decreases, the endometrium begin to disintegrates.
[3 marks]
(d) If fertilization occurred, the level of hormones S is maintained and the
pregnancy is proceed.
Explain the importance of hormone S.
Jika persenyawaan telah berlaku, aras hormone S dikekalkan dan
kehamilan terus berlaku.
• To inhibit the secretion of FSH and LH from pituitary gland
• No development of follicle / secondary oocyte
• Then the secretion of oestrogen is reduced
• Repair/ rejuvenation of endometrium is not happened
• Hence no new early embryo develops.
[3 marks]
(e) If the sperm counts of a husband is too low, artificial insemination can be
carried out to overcome this infertility problem. Discuss the appropriate
technique should be used.
Jika jumlah sperma suami terlalu rendah, teknik permanian beradas boleh
digunakan untuk mengatasi masalah ketidaksuburan ini. Bincangkan
teknik yang sesuai digunakan.
• The sperms are collected from the husband / taken from sperm
banks.
• And inserted directly into the Fallopian tube of the wife during
ovulation phase.
[2 marks]
5(c)
5(d)
5(e)
TOTAL
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Section B
[ 40 marks ]
Answer any two question. The time suggested to complete this section is 60 minutes
6 (a) According to the stages metaphase, anaphase and telophase in cell division,
differentiate the events happening during mitosis and meiosis.
Berdasarkan kepada peringkat metafasa, anafasa dan telofasa dalam
pembahagian sel, bezakan peristiwa yang berlaku semasa mitosis dan meiosis.
[4 marks]
(b) Diagram 6.1 Diagram 6.2 Diagram 6.1 is a new variety of a vegetable which has a great commercial value.
Diagram 6.2 is the original parent of the plant.
Based on above Diagram and with your biological knowledge, explain how a farmer
can propagate this variety to give a large scale of yield and at the same time maintains
its quality.
Rajah 6.1 adalah satu variati sayuran di mana mempunyai nilai komersial.
Rajah 6.2 adalah induk tempatan pokok tersebut.
Berdasarkan Rajah di atas dan pengetahuan biologi anda, terangkan bagaimana
seorang petani dapat memperbanyakkan variati ini supaya hasil ladang dapat
meningkat dan pada masa yang sama, kualiti dapat dipelihara.
[6 marks] c) Discuss how mutation can lead to the formation of tumour Bincangkan bagaimana mutasi dapat menyebabkan pembentukan tumor.
[10 marks]
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7 Diagram 7 shows the colour variation in a species of moth, Biston betularia in polluted
environment. Due to natural selection, the white coloured moth’s become extinct as well
as the time passed.
Rajah 7 menunjukkan variasi warna bagi satu spesis kupu-kupu di suatu kawasan
persekitaran tercemar. Akibat dari pemilihan semulajadi, lama-kelamaan kupu-kupu
putih mengalami kepupusan.
Diagram 7 / Rajah 7
(a) Based on the Diagram 7 , explain the meaning of ‘natural selection’.
Berdasarkan Rajah 7, terangkan pengertian ‘pemilihan semulajadi’
[4 marks] (b) The variation of human blood group is determined by three different alleles A, B and
O. By using a schematic diagram, show the possibilities of phenotypes and
genotypes of the offsprings if mother’s blood group is AB and father’s blood group is
A.
Variasi kumpulan darah manusia ditentukan oleh tiga alel yang berbeza A, B dan O.
Dengan mengunakan rajah skema, tunjukkan kemungkinan fenotip dan genotip
anak-anak jika kumpulan darah ibu AB dan kumpulan darah bapa adalah A.
[6 marks ]
(c) Table 7.1 and 7.2 below show the blood group and height variations of the
students in a secondary school.
Jadual 7.1 dan 7.2 di bawah menunjukkan variasi kumpulan darah dan ketinggian
pelajar di sebuah sekolah menengah.
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i) Blood Group / Kumpulan darah
Blood group
Kumpulan darah
A B AB 0
Number of students
Bilangan pelajar
13
9
3
20
TABLE 7.1
JADUAL 7.1
ii) Height / Ketinggian
TABLE 7.1
JADUAL 7.1
Based on Table 7.1 and 7.2, construct two different histograms on a graph paper to
show the number of students against blood group and the number of students against
height. Explain the similarities and differences between these two types of variations.
Berdasarkan Jadual 7.1 dan 7.2, bina dua histogram yang berbeza pada satu kertas
graf untuk menunjukkan bilangan pelajar melawan kumpulan darah dan bilangan pelajar
melawan ketinggian. Terangkan persamaan dan perbezaan antara dua jenis variasi
tersebut.
[10 marks]
Height /
Tinggi
( cm )
135-
139
140-
144
145-
149
150-
154
155-
159
160-
164
165-
169
170-
174
Number of students
Bilangan pelajar
2
5
7
9
10
7
4
1
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SUGGESTED ANSWER :
No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
Stages Mitosis Meiosis
1. Metaphase
- homologous chromosome are
arranged in linear sequence
/randomly at the metaphase
plate
- homologous chromosome line up
side by side at the metaphase plate
2. Anaphase
- separation of sister chromatids
to the opposite pole// the
centromere of each chromosome
divides into two and allows sister
chromatid to move to opposite
pole.
- separation of the homologous
chromosome to the opposite pole //
sister chromatids still remain
attached to each other during
movement to the opposite pole
3. Telophase - two daughter nuclei are formed
- diploid (2n) number of
chromosome is remained
- daughter cells are genetically
identical to each other and to
the parent cell.
- four daughter nuclei are formed
- diploid (2n) number of chromosome
is reduced to haploid (n)
- daughter cells are differ
from the parent and from
each other// variation
occurs among daughter
cells.
1m each = max 4 marks
6(a) o Able to differentiate the events happening during mitosis
and meiosis. 4
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No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
6(b) Able to explain how a farmer can propagate this variety to
give a large scale of yield and at the same time maintains its
quality.
• The technique used is tissue culture technique
• A piece of tissue/explant is taken from the young part of the
parent plant eg. Shoot/ root and cut into smaller pieces
• The tissues are sterilized (with dilute sodium hypochlorite
solution) to prevent the growth of pathogens / bacteria
/fungus.
• Each pieces of sterilised tissue is placed onto a growth
medium/ gel containing nutrients (eg. Glucose, amino
acid, minerals etc.) and hormone/auxin with optimum pH
level
• The apparatus and culture medium used must be in sterile
conditions and kept under the suitable temperature/ 30-
35°C.
• The tissue cells then divide repeatedly by mitosis to
produce a mass of undifferentiated cells/ callus
• After several weeks, callus differentiated to produce shoots
and roots /organogenesis.
• Once the roots grow, the plantlets/little plant are removed
and transferred to the soil for growth into the adult plant.
• All the plantlets produced this way are genetically identical
and known as clones.
• Therefore, all adults plants that develop from them share
the same traits, for example has large fruits.
1
1
1
1
1
1
1
1
1
1
Max
6
1 point - 1 mark
10 point - 10 marks Max: 6 marks
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No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
6(b) Discuss how mutation can lead to the formation of
• certain substance/carcinogen such as benzo - A -
pyrene etc…
• can cause the change in DNA structure (that control
the cell cycle)
• an abnormal cell is formed/ cancer cell / mutant cell
• this change disrupts the coded DNA genetic instruction
for mitosis control
• this leads to uncontrolled mitosis (which is non-stop
division of the cell) producing a mass of new daughter
cells called tumour
• tumour cells have no function, but instead compete
with surrounding normal cells to obtain nutrients and
energy for their own growth
• some tumours remain inactive and are relatively
harmless (not cancerous) and called benign tumour
• Benign tumour cells remain at its original site and do
not spread to other part of the body. It can be removed
by surgery.
• Other tumour, called malignant tumours are very
active (cancerous), spread locally and some cancer
cell migrates through bloodstream to invade other
organ.
• when this happens, secondary tumous develop in
other body tissue, then lead to the malfunction of the
tissue and ultimately death
• An individual with a malignant tumour is said to have
cancer.
1
1
1
1
1
1
1
1
1
1
1
Max 10
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QUESTION 7
No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
7a
Able to explain the meaning of natural selection.
• Over the time, one species better adapted to the environmental
changing.
• Example : camouflage of colour protect themselves from predator
• That particular ( white/ black) species increase their population
while the others cannot.
• White species / black species becomes dominant in their
community
1
1
1
1
[ 4 marks]
7b
Enable to draw a schematic diagram
Max 6
Parents Mother Father Group AB X Group A …………….. 1m Genotypes AB AA …………...1m Gamets A B A A ………1m Children’s genotypes AA AA AB AB ………..1m Children’s phenotypes A A AB AB …………1m # 50% of the children have group A and 50% of the children have group AB .1m
[Max 6/ 8 ] OR
Meiosis 1m
Fertilization 1m
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No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
Parents Mother Father Group AB Group A Genotypes AB AO Gamets A B A O Children’s genotypes AA AO AB BO Children’s phenotypes A A AB B # 50% of the children have group A , 25% of the children have group AB and 25% of the children have group B.
[ max 6 / 8 mark] 7(c)
Graph: Height variation – continuous
Blood group Variation – discontinuous
Able to explain the similarity and contrast of height variation and blood group variation. # Similarity – both height variation and blood group
variation create varieties in population.
# differences
4 marks
1m
5m
Height Blood group
1. Continuous variation Discontinuous variation
2. Graph shows normal distribution Graph shows discrete distribution
3. influence by environmental factors Influence by genetic
4. Traits are controlled by two or more
genes
Traits are controlled by a single gene
5 The phenotype is usually controlled
by many pairs of gene.
The phenotypes is controlled by a pair of alleles.
6. the characters can be measured and graded // quantitative
The characters cannot be measured and graded // qualitative.
Absorption
Serapan
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Section B
[ 40 marks ]
Answer any two question. The time suggested to complete this section is 60 minutes
8 (a) Diagram 8 shows the schematic diagram the regulatory mechanism of carbon dioxide
content in the human body.
Rajah 8 menunjukkan rajah skema mekanisme kawalatur kandungan karbon dioksida
dalam badan manusia.
DIAGRAM 8
RAJAH 8
(a) Based on Diagram 8, explain how the regulatory mechanism of carbon dioxide assists a
person during vigorous activity to maintain the carbon dioxide content in his blood.
[10 marks]
Berdasarkan Rajah 8 terangkan bagaimana mekanisme kawal atur kandungan
karbon dioksida membantu seseorang yang melakukan aktiviti cergas menstabilkan
kandungan karbon dioksida dalam darahnya.
[10 markah]
Carbon dioxide level in the blood increases Aras karbon dioksida dalam darah meningkat
Detected by Dikesan oleh
Nerves Impuls Impuls
Detected by Dikesan oleh
Nerves Impuls Impuls
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(b)
You are asked to prepare a talk on” Good eating habits” for a group of parents.
Discuss the good eating habits that you may want to educate them.
Anda dikehendaki menyediakan suatu ceramah yang bertajuk “ Amalan tabiat
pemakanan yang baik” untuk sekumpulan ibu bapa. Bincangkan amalan
pemakanan yang baik untuk disampaikan kepada mereka.
[ 10 marks]
Some families often eat fast food because working parents do not
have time to prepare home cooked food. These eating habits lead
to many health problems.
Sesetengah keluarga kerap mengambil makanan segera kerana
ibubapa mereka yang bekerja tidak mempunyai masa untuk
menyediakan sarapan di rumah. Tabiat makan ini menyebabkan
berlakunya pelbagai masalah kesihatan.
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9 (a) Diagram 9 shows the Nitrogen Cycle. Rajah 9 menunjukkan kitar Nitrogen.
Diagram 9.1 Based on Diagram 9.1, discuss the role of microorganisms in the Nitrogen Cycle.
Berdasarkan Rajah 9.1 , bincangkan peranan mikroorganisma di dalam Kitar Nitrogen [10 marks]
Urine Air kencing
Organic materials Bahan organik
Root nodules of legumes Nodul akar
kekacang
Ammonium compound Sebatian ammonium
Nitrite Nitrit
Nitrate Nitrat
Faeces Najis
Nitrogen in the atmosphere
Nitrogen dalam atmosfera
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(b) Diagram 9.2 shows the ozone layer in atmosphere that protects earth from ultraviolet rays from the sun.
Rajah 9.2 menunjukkan lapisan ozon dalam atmosfera yang melindungi bumi daripada sinar ultraviolet daripada matahari.
DIAGRAM 9.2 Describe how the ozone layer becomes thinner. Discuss its effects on humans and
the environment and suggest the ways to solve these problems.
Huraikan bagaimana lapisan ozon menjadi semakin nipis. Bincangkan kesannya
kepada manusia serta alam sekitar dan cadangkan langkah-langkah untuk
mengatasi masalah ini.
[10 marks]
END OF QUESTION PAPER
Solar radiation
Sinaran suria
Stratosphere Stratosfera
Absorption
Serapan Ozone layer
Lapisan ozon
Harmful ultraviolet radiation Sinaran ultra ungu
berbahaya
Troposphere
Trofosfera
Earth
Bumi
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SUGGESTED ANSWER :
No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
Stages Mitosis Meiosis
1. Metaphase
- homologous chromosome are
arranged in linear sequence
/randomly at the metaphase
plate
- homologous chromosome line up
side by side at the metaphase plate
2. Anaphase
- separation of sister chromatids
to the opposite pole// the
centromere of each chromosome
divides into two and allows sister
chromatid to move to opposite
pole.
- separation of the homologous
chromosome to the opposite pole //
sister chromatids still remain
attached to each other during
movement to the opposite pole
3. Telophase - two daughter nuclei are formed
- diploid (2n) number of
chromosome is remained
- daughter cells are genetically
identical to each other and to
the parent cell.
- four daughter nuclei are formed
- diploid (2n) number of chromosome
is reduced to haploid (n)
- daughter cells are differ
from the parent and from
each other// variation
occurs among daughter
cells.
1m each = max 4 marks
6(a) o Able to differentiate the events happening during mitosis
and meiosis. 4
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No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
8a
Able to explain how the regulatory mechanism of carbon
dioxide content in the human body.
1. Vigorous exercise produces more carbon dioxide/ the
carbon dioxide content increase as a result of active cellular respiration
2. Carbon dioxide dissolved in the plasma to form carbonic
acids // CO2 + H2O H+ + HCO3
-
3. Carbonic acids converted into bicarbonate ion and hidrogen ion.
4. Concentration hidrogen ion increased// drop in pH value of
blood (and cerebrospinal fluid)
5. The drop in pH is detected by Central Chemoreceptor in Medulla Oblongata
6. (also) detected by peripheral chemoreceptor / carotid bodies / aortic bodies.
7. Chemoreceptor (Central and peripheral ) triggers nerve
impulse and sends to medulla oblongata 8. Then the Central Chemoreceptor send the nerve impulses
to the diaphragm and intercostal muscle in the lung 9. Causing (respiratory muscles) to contract and relax, Finally
increases the breathing and ventilation rate
10 As excess CO2 eliminated from the the body, carbon dioxide concentration and pH value of blood return to normal 11. (also) the Central Chemoreceptor send the nerve impulses
to cardiac muscle
12. Cardiac muscles contract and relax faster// Heart beat increased
13. Blood transport more O2 to tissue/ body cell for cell
respiration ( to produced energy)// more CO2 can be
1
1
1
1
1
1
1
1
1
1
1
1
1
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No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
transported to the lung (to eliminated from the body)
14. Carbon dioxide concentration and pH value of blood return to normal levels
Maks: 9 markah
KA – P5,P7,P9 & P11 – 1m Jumlah: 10 markah
1
1
Max 9
10
8b
Able to discuss the good eating habits
o Good eating habits mean taking food in the correct quantity
at the correct time
o Also should refrain from overeating or eating too little during
a meal
o Improper eating habits can lead health problems
Example Health problems related to eating habits
o obesity is caused by consumption of excess energy rich
food such as liked carbohydrates and fats
o and by lack of exercise
o Diabetes mellitus is caused by overeating of sweet foods.
o Gastritis is caused by inflammation and erosion of the
stomach/ by drugs such as aspirin or by drinking high
concentration of alcohol/ due to poor diet/ irregular food
intake
some of the good eating habits that we need to inculcate
are
o always eat a balanced diet that include all the different
classes of food / protein, carbohydrate, fat, vitamins,
minerals, water and fibre
o in right quantities and in correct proportions
o always refer to the Food Guide Pyramid. Choose from four
level of food
- Level 1 – bread, cereals, rice, pasta group (eat most
/ 50 % from the energy value)
1
1
1
1
1
1
1
1
1
1
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No. ACCEPTED POINTS / DESCRIPTION / EXPLANATION M Sum
- Level 2 – fruits and vegetables
- Leve 3 – meat, fish, eggs, nuts, poultry, milk, yogurt
and cheese
- Level 4- Lipids, oils and sweets group (eat less)
o Take proper meals a regular times of the day / in three meals a
day
o involve choosing the types of food we eat wisely/aware of the
contents of the food that consume every day
o check the food labels for information regarding the nutrient
contents/ the total calories of the food
o also check the freshness and various safety aspects
o for example whether the food contains additives which effects
on health
o avoid consuming unhealthy food / junk foods, salty snack foods,
because junk food includes food that is high in salt/sugar/fat but
low in nutritional value
o avoid taking excessive fatty food and food rich in sugar.
o avoid overeating, it will lead to obesity
o avoid under eating, it cause tiredness, malnourishment
o and may lead to health problem like anorexia nervosa and
bulimia
o avoid eating too fast and eat slowly.
o Take time to chew the food to avoid indigestion
o avoid smoking, drinking too much alcohol and coffee
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Max: 10
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Item
number Criteria Marks Notes
9 (a)
F1 - Rhizobium bacteria inside the root nodules of legumes and Nostoc sp. found freely in the soil fixed the nitrogen in air F2 - decaying bacteria / fungi decompose plant /animal / dead organism / waste product P1 - to form ammonium compound F3 - Nitrosomonas sp./ nitrifying bacteria converted ammonium compound to nitrite F4 - Nitrobacter sp /nitrifying bacteria convert nitrite to nitrate P2 - nitrate is absorbed by plant to form plant protein P3 - (plant protein) eaten by an animal to form animal protein F4 - Denitrifying bacteria reduce the nitrate content in the soil P4 - by converting the nitrate into nitric oxide and nitrogen gas P5 - nitrogen gas goes back into the atmospheric to complete the nitrogen cycle
1 1 1 1 1 1 1 1 1 1
10 mark
s
9 (b)
Able to explain how the ozone layer becomes
thinner / ozone depletion occur
• Thinning of the ozone layer is due to the
widespread use of CFC
• It is used in aerosol, industrial solvents,
electronics and Freon in air conditioners
• Ultraviolet radiation strikes a CFC molecule cause
the chlorine atom to break away
• Then the chlorine atom collides with an ozone
molecule and combines with an oxygen atom to
form chlorine monoxide and oxygen
1
1
1
1
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Item number
Criteria Marks Notes
• Then the free atom of oxygen collides with the
chlorine monoxide, the two oxygen atoms form a
molecule of oxygen
• The chlorine atom is released and free to destroy
more ozone molecules
• The chlorine produced re-enters the cycle
• When the ozone layer becomes thinner, more
ultraviolet radiation reaches the Earth
The effect of excessive ultraviolet radiation on
human
• reduction of the body’s immune system
• skin cancer
• cataract of the eye
Effect on plants
• reduction of the rate of growth therefore reducing
crop yields
Effect on aquatic organism
• death of plankton, reduce food supply to aquatic
organism, fisherman’s catch is reduced.
Steps to overcome this problem
• Reduce or stop using CFC or chlorine-based
products
• Replace CFC with HCFC
• Use wrapping papers instead of polystyrene
boxes
• Patch up the holes in the ozone layer by firing
frozen ozone balls into the atmosphere
1
1
1
1
1
1
1
1
1
1
1
1
1
Max 5 Max 3 Max 2
TOTAL 10marks
END OF MARKING SCHEME
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