Biologi f4 sbp akhir tahun 2008
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Transcript of Biologi f4 sbp akhir tahun 2008
SULIT 4551
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SULIT 45514551BiologiOktober20082½ jam
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUHDAN SEKOLAH KLUSTER
KEMENTERIAN PELAJARAN MALAYSIA
________________________________________________________________
PEPERIKSAAN DIAGNOSTIK TAHUN 2008
TINGKATAN EMPAT
BIOLOGI
Kertas 1
Satu jam lima belas minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. Kertas soalan ini adalah dalam Bahasa Inggeris.
2. Calon dikehendaki membaca maklumat di bawah
4551
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INFORMATION FOR CANDIDATES
1. This question paper consists of 50 questions.2. Answer all questions.3. Answer each question by blackening the correct space on the answer
sheet.4. Blacken only one space for each question.5. If you wish to change your answer, erase the blackened mark that you
have made. Then blacken the space for the new answer.6. The diagrams in the questions provided are not drawn to scale unless
stated.
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Kertas soalan ini mengandungi 26 halaman bercetak.
[Lihat sebelah
7. You may use a non-programmable scientific calculator.
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Answer all questions
1. The following information refers to an organelle of a plant cell.
Whi
2. The
Whi
A
B
C
D
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A membranous sac which occupies a large part of a mature plant cell.
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ch of the labelled parts, A, B, C and D is the organelle?
diagram shows an organelle of a cell.
ch process only occurs in this organelle?
Osmosis
Diffusion
Respiration
Photosynthesis
Contains water, stored food, salt and waste materials.
A B C
D
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3. The diagram shows a type of animal tissue.
What is the tissue?
A Epithelial tissue
B Connective tissue
C Smooth muscle tissue
D Skeletal muscle tissue
4. The diagram shows the process of feeding by Amoeba sp.
Which organelle is not involved in the process shown in the diagram ?
A Lysosome
B Food vacuole
C Pseudopodia
D Contractile vacuole
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5. The table shows the characteristics of a plant tissue X.
Consists of companion cells
Contains sieve tubes
Which of the processes is related to the tissue X?
A Absorption
B Reproduction
C Transpiration
D Translocation
6. The diagram shows the structure of a plasma membrane.
The proteins labelled X and Y are
X Y
A Pore protein Pore protein
B Carrier protein Carrier protein
C Carrier protein Pore protein
D Pore protein Carrier protein
XY
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7. The diagram shows a sodium-potassium pump.
What process involves in the mechanism?
A Osmosis
B Diffusion
C Plasmolysis
D Active transport
8. The information refers to a condition of a red blood cell after being immersed in a
solution.
What process experienced by the red blood cell?
A Crenation
B Haemolysis
C Plasmolysis
D Deplasmolysis
The cell shrinks
The plasma membrane crinkles up
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9. Which process does not use energy released by respiration?
A Passage of nerve impulses
B Active transport of glucose into the villi
C Formation of gametes in the gonads
D Diffusion of oxygen across the alveolar surface
10. Which process takes place in a root hair cell when oxygen is not available?
A Active transport and osmosis
B Active transport and diffusion
C Active transport
D Diffusion
11 . Onion cells have been placed in four solution of different concentration.
Which cells is immersed in hypotonic solution.
A C
B D
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12. The graph shows the changes of the length of the potato slices in different
solutions.
Which curves labelled A, B, C and D shows the changes of the length of the slices
which were placed in distilled water?
13. Which of the following shows the structure of a polysaccharide?
A B
C D
D
C
B
A
Time / min
Length of potatoslices /cm
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14. How many basic units are required to form the DNA molecule as shown in the
diagram .
A. 3
B. 6
C. 9
D. 18
15. The diagram shows an action of an enzyme and a sucrose.
Q
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What are P, Q and R?
P Q R
A Lactase Glucose Lactose
B Sucrase Glucose Fructose
C Sucrase Lactose Fructose
D Lactase Galactose Lactose
P
R
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16. The diagram shows one of the structures of a protein.
Which compound is made up of th
A. Enzymes
B. Hormones
C. Antibodies
D. Haemoglobins
17. Which graph represents the effec
A.
C.
Rate ofreaction
Rate ofreaction
0
0 7
7
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is structure?
t of pH on the rate of reaction of pepsin.
B.
D.
Rate ofreaction
Rate ofreaction
pH
pH pH
pH
0
014
14 14
14
7
7
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18. The figure shows a cell cycle of an organism.
Which of the following is the correct sequence of process X?
A. Telophase Anaphase Metaphase Prophase
B. Anaphase Metaphase Prophase Telophase
C. Prophase Metaphase Anaphase Telophase
D. Prophase Anaphase Metaphase Telophase
19. The number of chromosomes in a skin cell of a cat has a dipl
chromosomes 24.
What is the chromosomal number in its gametes?
i. 12
ii. 24
iii. 36
iv. 48
X
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oid number of
Interphase
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20. The diagram shows a cell of an organism.
What is the stage of cell division shown by the diagram.
A. Prophase I
B. Prophase II
B. Metaphase I
D. Metaphase II
21. The diagram shows several stages in mitosis.
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What is the behaviour of chromosome in stage Y.
A. Chromosomes condense and thicken
B. Chromosomes arrange themselves at the equatorial plane
C. Homologous chromosomes pair together and cross-over occurs
D. Homologous chromosomes separate and move the opposite poles.
Y
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22. Which of the followings shows a stage in meiosis?
A. B.
C.
23. Which is
A. Para
B. Holo
C. Sapr
D. Chem
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D.
an autotrophic organism?
sites
phytes
ophytes
osynthetic organism
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24. The information shows the characteristic of a compound.
Which compound has the characteristics?
A. Carbohydrate
B. Protein
C. Water
D. Lipid
25. The diagram shows a longitudinal section of a human villus.
Which of these compounds can be found in S?
A. Vitamin A
B. Vitamin D
C. Amino Acids
D. Tiny droplets of lipids
o Provides protection to organs against physical injuries
o Provides energy for physical activities
o Insulates body against heat loss
o Forms sex hormones
S
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26. Which of the following is correct about the light and dark reaction
Light reaction Dark reaction
A. Occurs in stroma Occurs in grana
B. ATP and hydrogen are used ATP and hydrogen are produced
C. No reduction of carbon dioxide There is reduction of carbon dioxide
D. No oxygen is produced Oxygen is produced
27. The diagram shows the graph of the rate of photosynthesis against light intensity.
● ●
●
The rate of photosyntehsis increases from X to Y. Which limiting factor has been
overcomed?
A. Light intensity
B. Carbon dioxide concentration
C. Temperature
D. Water
Rate ofphotosynthesis 0∙06 % CO2
0∙02 % CO2
Light intensity
X
Y
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28. When 0.4 g of groundnut is completely burnt, the temperature of 20 ml of water
rises up from 30°C to 70°C. ( Secific heat capacity of water is 4.2 Jg °C )
What is the energy value of the groundnut?
A 1.4 kJg−¹
B 3.4 kJg−¹
C. 8.4 kJg−¹
D. 76.2 kJg−¹
29. A food sample is tested to determine its contents.
The table below shows the results obtained.
Food test Result
Benedict’s solution is added and the mixture is boiled Brick red precipitate
Iodine solution is added Brownish yellow colour
Millon’s reagent is added and is heated in a water bath Red precipitate
What are the food classes found in the food sample?
A Starch and protein
B Lipids and protein
C Only reducing sugar
D Protein and reducing sugars
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30. Organ P is a part of the human digestive system.
When a baby drinks milk, what would happen to the milk protein in organ P?
A. Protein is digested by pepsin into amino acids
B. Polypeptides is digested by trypsin into dipeptide
C. Caseinogen is coagulated into casein by rennin
D. Caseinogen is coagulated into casein by hydrochloric acid
31. The diagramshows arespiratorystructure ofan organism.
Which organism has this structure?.
A Frog
B Fish
C Crocodile
D Grasshopper
Organ P
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32. The diagram shows the structure of alveoli .
What process occurs between an alveolus and structure P during gaseousexchange?
A. Osmosis
B. Simple diffusion
C. Active transport
D. Facilitated diffusion
33. Which structure involves in the gaseous exchange in Ameoba sp?.
A. Cytoplasma
B. Food Vacuole
C. Contractile vacuole
D. Plasma membrane
P
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34. The equation shows the process that takes place in yeast.
What is S?.
A. Water
B. Ethanol
C. Oxygen
D. Lactic acid
35. The figure shows a model of a rib cage
Which of the following is represented by P, Q and
P Q
A Sternum Rib
B Intercostal muscle Sternum
C Sternum Intercostal mu
D Intercostal muscle Rib
Glucose S + Energy + Carbon dioxide
R
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R?
R
Intercostal muscle
Rib
scle Rib
Sternum
Backbone
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36. The diagram shows a human respiratory system.
What happen to structure R during exhalation?.
A. Relaxes and becomes flatten
B. Contracts and becomes flatten
C. Relaxes and becomes dome shape
D. Contracts and becomes dome shape
37. The information shows the results of an experiment to determine the oxygen
content in exhaled air using J-tube.
Th
A.
B.
C.
D.
R
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Length of exhaled air column = 10.0 cm
Length of exhaled air column after
treatment with potassium hydroxide = 9.6 cm
Length oh of exhaled air column after
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e percentage of oxygen content in the exhaled air is
4.0 %
11.0 %
16.0 %
21.0 %
treatment with potassium pyrogallol = 8.5 cm
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38. The diagram shows gaseous exchange in human lungs.
Which of the follow
R
A Contains bloo
high concentroxygen
B Contains bloo
High concentoxygen
C Contains blooconcentrationdioxide
D Contains blooconcentrationdioxide
T
S
Bloodcapillary
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Alveolus
ing is true about blood ves
S
d with
ation of
Allows oxygdiffuse into talveolus
d with
ration of
Allows carboto diffuse intalveolus
d with highof carbon
Allows carboto diffuse intcapillary
d with highof carbon
Allows oxygdiffuse into tcapillary
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sels R, S and T?
T
en tohe
Contains blood withhigh concentrationof carbon dioxide
n dioxideo the
Contains blood withhigh concentrationof carbon dioxide
n dioxideo the
Contains blood withhigh concentrationof oxygen
en tohe
Contains blood withhigh concentrationof oxygen
R
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39. The diagram shows a food web and the pyramid of number in an ecosystem.
Which of the following is K?
A. Snake
B. Bees
C. Eagle
D. Green plant
40. The diagram shows the changes in size of the populations of two organism in a
palm oil plantation. The organism are interdependent in a food chain.
What is the interaction between organism X and organism Y.
A. Mutualism
B. Parasitism
C. Prey – predator
D. Commensalism
K
J
M
L
Time / year
Populationof animal Owl
Rats
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41. The diagram shows an interaction between two species.
Which of the follow
Species
A Benefit
B Harm
C Not affect
D Benefit
42. An officer in the
Cengal Trees in
Which is the mo
A. Direct Count
B. Quadrat 1m
C. Quadrat 10m
D. Capture, ma
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ing describe the effect of the interaction between species P and Q?
P Species Q
Benefit
Harm
ed Harm
No affected
Forrestry Depatment wants to determine the distribution of
Belum National Park.
st suitable method to be used ?
ing
X 1m
X 10 m
rk, release and recapture
P
Q
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43. Figure (a) shows the profile of a mangrove swamp while figure (b) shows three
types of root of the trees that are found in the mangrove swamp.
Figure (a)
44.
P
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Figure (b)
Which of the mangrove root(s) can be found in P?
A. X
B. Y
C. Z
D. X and Y
Penicillin was the first antibiotic found.
What is the microorganism that produces penicillin?
A. Virus
B. Fungi
C. Bacteria
D. Protozoa
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45. The diagram shows the root of a leguminous plant.
What is the role of
cycle.
A. Nitrogen fixatio
B. Decomposition
C. Denitrification
C. Nitrification
46. The following inform
The process is
A. colonization
B. competition
C. eutrophication
D. phosphorilation
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the bacteria found in the root nodule of the plant in nitrogen
n
ation is related to a process occurs in an ecosystem
Phosphate runs off into lakes
Rapid growth of autotrophic organisms
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47. The diagram shows a human activity.
Which is the effect of the activity ?
A. Decrease in BOD level
B. Increase the habitat of the fauna
C. Decrease the temperature in north pole
D. Increase the carbon dioxide level in the atmosphere
48. What is the effect of dust and smog on leaves of plants ?
A. Decrease the water transport
B. Damages the chlorophyll in the leaf
C. Decreases the amount of light received by the leaf
D. Damages the mesophyll cells in leaf
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49. Which of the following are non renewable resources?
I Iron ore
II Timber
III Petroleum
IV Tin ore
A I and III only
B II and IV only
C I, III and IV only
D I, II, III and IV
50. Which diseases become worse when air pollution occurs?
I Ashma
II Skurvi
III Pellagra
IV Bronchitis
A. I and III
B. I and IV
C. II and III
D. I, III and IV
END OF THE QUESTIONS
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SULIT Nama ...................................................... Tingkatan .................... 45514551BiologiOktober20082½ jam
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUHDAN SEKOLAH KLUSTER
KEMENTERIAN PELAJARAN MALAYSIA
________________________________________________________________
PEPERIKSAAN DIAGNOSTIK TAHUN 2008
TINGKATAN EMPAT
BIOLOGIKertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. Kertas soalan ini adalah dalam Bahasa Inggeris.2. Calon dikehendaki membaca maklumat di bawah.
4
Untuk Kegunaan Pemeriksa
Bahagian Soalan MarkahPenuh
Markah
A
1 12
2 12
3 12
4 12
5 12
B
6 20
7 20
8 20
9 20
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INFORMATION FOR CANDIDATES
1. This question paper consists of two sections :Section A and Section B.
2. Answer all questions in Section A. Write youranswers for Section A clearly in the spaces providedin the question paper.
3. Answer any two questions from Section B. Write youranswer for Section B on the lined paper in detail. Youmay use equations, diagrams, tables, graphs andother suitable methods to explain your answer.
4. Show your working, it may help you to get marks.5. If you wish to cancel any answer, neatly cross out the
answer.6. The diagrams in the questions are not drawn to scale
unless stated.7. The mark allocated for each question or part of
question is shown in brackets.8. The time suggested to complete Section A is 90
minutes, and Section B is 60 minutes.9. You may use a non-programmable scientific calculator10. Hand in this question paper at the end of the
Kertas soalan ini mengandungi 15 halaman bercetak.
551 [Lihat sebelah
Jumlahexamination.
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2
1.
Section A
[ 60 marks ]
Answer all questions
Diagram 1 shows cell organisation in plant. Cells J undergo differentiation and
specialisation to form several tissues in a leaf of a green plant.
Cell Specialisation
DIAGRAM 1
(a) Name tissue K and tissue L.
K : ..………………………………………………………………………………
L: …………………………………………………………………………………
[2 marks]
(b) State the function of cells K and M in a leaf.
K : ..………………………………………………………………………………
M: ……………...…………………………………………………………………
[2 marks]
ForExaminer’s
Use
1(a)
1(b)
Cells J
Cross-sectionif aleaf
K
L
Xylemtissue
M
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(c) (i) Explain the differentiation of cells J to form the xylem tissue.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(ii) During the formation of the xylem tissue, the plant was unable to
synthesise lignin.
Explain the effect on the function of the leaf.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(d) Based on diagram 1, state the meaning of cell specialization.
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(e) Leaf is the main photosynthetic organ of a plant.
Explain the adaptation of tissue L to enable the leaf to carry out its
function.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[3 marks]
1(c)(i)
1(c)(ii)
1(d)
1(e)
TOTAL
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2. Diagram 2.1 shows the synthesis and secretion of an enzyme in an animal cell.
(a) Name the
P : ..………
R: ………
(b) State the f
……………
……………
(c) Explain the
……………
……………
……………
Diagram 2.2 sho
Y.
En
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DIAGRAM 2.1
parts labelled P and R.
………………………………………………………………………
……...…………………………………………………………………
[2 marks]
unction of organelle S.
………………………………………………………………………
………………………………………………………………………
[1 mark]
role of organelle Q in the synthesis of an enzyme.
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
[2 marks]
ws the structure of an enzyme and three substrates W, X and
DIAGRAM 2.2
2(a)
2(b)
2(c)
P
Q
R
S
zyme W X Y
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(d) (i) Based on Diagram 2.2, choose the correct substrate and complete the
schematic diagram to show the mechanism of enzyme action on the
substrate.
[2 marks]
(ii) State two characteristics of the enzyme, which are illustrated in (d)(i).
1 : ..………………………………………………………………………………
2: ……………...…………………………………………………………………
[2 marks]
(e) A child with high fever has body temperature of 41ºC.
Explain the effect on the process of digestion in his digestive system.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[3 marks]
2(d)(i)
2(d)(ii)
2(e)
TOTAL
+
Enzyme + Substrate
Enzyme–Substrate complex
+
Enzyme + Products
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3. Diagram 3 shows the phases in a cell division in cell X. The phases are not in
the correct sequence.
DIAGRAM 3
(a) (i) Name the type of cell division.
……………………………………………………………………………………
[1 mark]
(ii) State an organ where this cell division takes place.
……………………………………………………………………………………
[1 mark]
(b) State the number of chromosomes of cell X during phase P and phase U.
P: …….……………………… U: …….………………………
[2 marks]
(c) (i) State the name of each phase of the cell division in Table 1
Phase Name of the phase
P
Q
R
S
T
U
TABLE 1
[3 marks]
3(a)(i)
3(a)(ii)
3(b)
3(c)(i)
P Q R
UTS
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(ii) State the phase that brings about variation in organism.
Explain your answer.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[3 marks]
(d)) Cell X is treated with a type of chemical that retards the function of
centrioles.
Explain what will happen to the chromosomal behavior in stage P.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
3(c)(ii)
3(d)
TOTAL
4. Diagram 4 shows human digestive system.
(a) (i) Name organ
………………
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DIAGRAM 4
Q.
………………………………………
P
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……………………………
[1 mark]
4(a)(i)
R
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(ii) Explain the process of food digestion in Q.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[3 marks]
(b) State two effects if the gastric glands in Q are unable to produce
hydrochloric acid.
1 : ..………………………………………………………………………………
2: ……………...…………………………………………………………………
[2 marks]
(c) Explain one difference between the food that enters P and R.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(d) (i) Explain the absorption of proteins in R.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(ii) Explain the importance of food digestion which enables nutrients to be
absorbed by the villi.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
4(a)(ii)
4(b)
4(c)
4(d)(i)
4(d)(ii)
TOTAL
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5. Diagram 5 shows part of the nitrogen cycle.
DIAGRAM 5
(a) Name process X and compound Y.
Process X: ………………………………………………………………………
Compound Y: …………………………………………………………………..
[2 marks]
5(a)
Nitrogen in theatmosphere
Process X
Ammonium
Nitrites NO2
Compound Y
Fixation bynitrogen
fixingbacteria in
the rootnodules ofleguminous
plants
Deadorganisms
and animalswaste
Absorption
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(b) (i) Plants will convert compound Y into an organic compound.
Name the organic compound.
……………………………………………………………………………………
[1 mark]
(ii) Describe the functions of the organic compound named in (b)(i) in plants.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(c) (i) Name the bacteria involve in the fixation of nitrogen in nodules of a
leguminous plant.
……………………………………………………………………………………
[1 mark]
(ii) The microorganism name in d (i) interacts with the leguminous plant.
Name and explain the type of relationship between the two species.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[2 marks]
(d) Relate the importance of nitrogen cycle in the growth of rabbits.
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………
[3 marks]
5(b)(i)
5(b)(ii)
5(c)(i)
5(c)(ii)
5(d)
TOTAL
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
11
Section B
[ 40 marks ]
Answer any two questions from this section.
6. (a) State two differences between passive transport and active transport. [2 marks]
(b)
DIAGRAM 6.1
Diagram 6.1 shows a red blood cell immersed in different salt solutions.
Explain what happen to the red blood cell after being immersed in 3% of sodium
chloride solution and 0.1% of sodium chloride solution for half an hour.
[8 marks]
(c) Diagram 6.2 shows the appearance of the plants cell which is immersed in different
concentration of sucrose solution one after another.
DIAGRAM 6.2
Based on Diagram 6.2, describe what happen to the cell in each concentration of
sucrose solution. [10 marks]
Plant cell in 17% ofsucrose solutionfor 30 minutes
Plant cell in 0.1%of sucrose solutionfor 30 minutes
Plant cell in 30% ofsucrose solutionfor 30 minutes
Plant cell in 0.1%of sucrose solutionfor 30 minutes
Red blood cellsin 3% sodiumchloride solutionafter 30 minutes
Red blood cellsin 0.1% sodiumchloride solutionafter 30 minutes
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
12
7. Diagram 7.1 shows a respiratory structure of an insect.
P
tracheol
body cell
DIAGRAM 7.1
(a) (i) Explain the gases exchange between tracheol and body cell. [4 marks]
(ii) Chitin is a polysaccharide on the outer surface of structure P. Due to the change
in the environment, the insect is unable to form the polysaccharide.
Explain how the absence of chitin affects inhalation and the energy production.
[6 marks]
(b) Diagram 7.2 shows the rate of oxygen intake before, during and after a vigorous
exercise of an athlete.
Time (min)
DIAGRAM 7.2
(i) Based on the graph, compare the respiration before and during the vigorous
exercise.
(ii) Explain how the oxygen intake by the athlete returns to the normal level at the
25th minute.
[10 marks]
Vigorous exercise
Oxygen intake(litre/minute)
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
13
8. (a) Diagram 8.1 shows the daily menu of a pregnant woman.
Breakfast
A plate of fried rice
A can of carbonated drink
An apple
Lunch
A bowl of chicken rice complete with a
piece of roasted drumstick and a bowl
of chicken soup
A plate of fried prawn
A glass of sweetened fruit juice
Dinner
A plate of fried noodle
2 slices of cucumber
A cup of coffee
DIAGRAM 8.1
Does the menu provide a balanced diet for the pregnant woman?
Evaluate the nutrients content and the effects of consuming the foods. .
[10 marks]
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
14
(b) Diagram 8.2 shows a schematic diagram of the photosynthesis process.
DIAGRAM 8.2
Based on Diagram 8.2,
(i) write the chemical equation of photosynthesis.
(ii) define photosynthesis.
[4 marks]
(c) Diagram 8.3 shows organisms in a pond ecosystem
DIAGRAM 8.3
Based on Diagram 8.3, describe how Hydrilla sp is able to obtain all the requirement
for photosynthesis.
[6 marks]
chlorophyll
CO2
H2O
C6H12O6
O2
Hydrilla sp Elodea sp
lotus
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551
15
9. (a)
Based on the statement, explain how the use of excessive inorganic fertilizers in a
long period of time reduces the population of aquatic organisms.
[10 marks]
(b) Diagram 9 shows a tree grown wildly in an agricultural area.
A farmer pla
Discuss the
Bird`s nest
Insects
Roots
entering the water system.
MOZ@C
The use of fertiliser increases the amount of phosphate and nitrate in the soil.
This enhances the growth of plants. There are two main types of fertilizers,
which are organic and inorganic fertilisers. Farmers are advised to use organic
fertilizers rather than the inorganic fertilizers. The organic fertilizers release the
nutrients gradually into the soil and this will reduce the amount of nutrients
SULIT
DIAGRAM 9
ns to cut down the tree to grow vegetables.
good and bad effects of this action on human and the ecosystem.
[10 marks]
END OF QUESTION PAPER
Soil
SMS MUZAFFAR SYAH , MELAKA
SULIT Nama ...................................................... Tingkatan .................... 45514551BiologiOktober20082½ jam
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUHDAN SEKOLAH KLUSTER
KEMENTERIAN PELAJARAN MALAYSIA
________________________________________________________________
PEPERIKSAAN DIAGNOSTIK TAHUN 2008
TINGKATAN EMPAT
BIOLOGIKertas 3
Satu jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
Untuk KegunaanPemeriksa
SoalanMarkahpenuh
Markah
1 33
2 17
JUMLAH
1.2. Calon dikehendaki membaca maklumat di bawah.
MOZ@C
INFORMATION FOR CANDIDATES1. This question paper consists of two questions. Answer all the
questions.2. Write your answers for Question 1 in the spaces provided in
the question paper3. Write your answers for Question 2 on the lined pages at the
end of the question paper in detail. You may use equations,diagrams, tables, graph and other suitable methods to explainyour answer.
4. Show your working, it may help you to get marks.5. If you wish to cancel any answer, neatly cross out the answer.6. The diagrams in the questions are not drawn to scale unless
stated.7. Marks allocated for each question or part question are shown
in brackets8. The time suggested to complete Question 1 is 45 minutes
and Question 2 is 45 minutes9. You may use a non-programmable scientific calculator10. Hand in this question paper at the end of the examination.
Marks awarded:Score Description
3 Excellent: The best response2 Satisfactory: An average response1 Week: An inaccurate response0 No response or wrong response
45
Kertas soalan ini adalah dalam Bahasa Inggeris.
Kertas soalan ini mengandungi 6 halaman bercetak.
51 [Lihat sebelah
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
2
Answer all questions.
Question 1
Hydrogen peroxide is a chemical that bubbles when it reacts with a catalase enzyme.
Catalase enzyme is an antioxidant enzyme in living cells. Hydrogen peroxide is converted
into water and oxygen by the catalase enzyme
2H2O2 2H2O + O2
An experiment is carried out to investigate the effect of pH on catalase enzyme in potatoes.
Bubbling of gases is used to indicate that a reaction is occurring. The rate of reaction is
determined by measuring the volume of bubbles produced in a unit time. The experiment is
conducted as followed:
1. Three measuring cylinders, P, Q and R are filled with 3.0 cm3 of hydrogen peroxide.
2. 6.0 cm3 of 0.1% hydrochloric acid is added to P, 6.0 cm3 of distilled water is added to Q
and 6.0 cm3 of 0.1% sodium hydroxide is added to R.
3. pH paper is used to measure the pH value of each tube.
4. The potato is cut into three cubes, with the size of 1.0 cm3 each.
5. One potato cube is added into each measuring cylinder.
6. The volume of bubbles produced in each measuring cylinder is observed after
5 minutes and recorded.
7. The results are shown in Diagram 1.
25
20
15
10
5
PpH 2
25
20
15
10
5
QpH 7
DIAGRAM 1
25
20
15
10
5
RpH 10
3.0 cm3
hydrogenperoxide+6.0 cm3
0.1%hydrochloricacid+1.0 cm3
potato
3.0 cm3
hydrogenperoxide+6.0 cm3
distilledwater+1.0 cm3
potato
3.0 cm3
hydrogenperoxide+6.0 cm3
0.1%sodiumhydroxide+1.0 cm3
potato
Volumeofbubbles
=
Volumeofbubbles
=
Volumeofbubbles
=
cm3cm3 cm3
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
3
(b) Record the volume of bubbles formed in each measuring cylinder after5 minutes in Diagram 1.
[3 marks]
(a) List all materials and apparatus used in this experiment.
Materials Apparatus
[3 marks]
ForExaminer’s
Use
(c) (i) State two observations made on Diagram 1.
Observation 1
………………………………………………………………………….…………
……………………………………………………………………………….……
Observation 2
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
(ii) State the inference for each observation made in (b) (i).
Inference for observation 1
………………………………………………………………………………….…
…………………………………………………………………………………….
Inference for observation 2
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
1(a)
1(b)
1(c)(i)
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
4
(d) (i) Construct a table and record all the data collected in the experiment
based on the following criteria:
pH value Volume of bubbles formed Rate of reaction (cm3 minute-1)
[6 marks]
(ii) Explain the relationship between the test tube content and the volume of
bubbles formed in Q.
………………………………………………………………………………….…
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
(e) (i) State the variables and explain how the variables are operated.
Variables How the variables are operated
Manipulated variable
……………………………
……………………………
………………………………………………
………………………………………………
Responding variable
……………………………
……………………………
………………………………………………
………………………………………………
Fixed variables
……………………………
……………………………
………………………………………………
………………………………………………
[3 marks]
1(d)(ii)
1(d)(i)
1(e)(i)
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
5
(ii) State the hypothesis of the experiment.
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
(f) State the relationship between volume of bubbles formed and time in a
medium of pH 7.
………………………………………………………………………………….…
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
(g) Based on the experiment, what is enzyme?
...……………………………………………………………………………….…
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
(h) The experiment is repeated by using 2 potato cubes sized 0.5 cm3 each.
Predict the observation in measuring cylinder R. Explain your answer.
………………………………………………………………………………….…
………………………………………………………………………………….…
…………………………………………………………………………………….
[3 marks]
1(f)
1(g)
1(e)(ii)
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
SULIT 4551
4551 SULIT
6
Question 2
Acorbic acid, or vitamin C, is found in fruits and green vegetables. Ascorbic acid is a
reducing agent which decolourises the blue colour of DCPIP solution. The vitamin C in
solutions will deteriorate when exposed to oxygen.
Plan an experiment to determine the vitamin C content in orange, papaya and watermelon
juices.
Your experimental planning need to include the following aspects:
Statement of identified problem
Objective of study
Variables
Statement of hypothesis
List of materials and apparatus
Technique used
Experimental procedures
Presentation of data
Conclusion
[17 marks]
END OF QUESTION PAPER
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME
1
SULIT 45514551BiologiOktober20081¼ jam
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUHDAN SEKOLAH KLUSTER
KEMENTERIAN PELAJARAN MALAYSIA
________________________________________________________________
PEPERIKSAAN DIAGNOSTIK TAHUN 2008
TINGKATAN EMPAT
PERATURAN PEMARKAHAN
BIOLOGIKertas 1, 2 dan 3
Peraturan pemarkahan ini men
UNTUK KEGUNAAN
MOZ@C
SMS MUZAFFAR
4551gandungi 21 halaman bercetak.
PEMERIKSAN SAHAJA
SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
2
PAPER 1
No Answer No Answer No Answer No Answer No Answer1 D 11 A 21 B 31 D 41 D2 D 12 A 22 C 32 B 42 C3 A 13 C 23 D 33 D 43 B4 D 14 B 24 D 34 B 44 B5 D 15 B 25 C 35 C 45 A6 C 16 D 26 C 36 C 46 C7 D 17 A 27 B 37 B 47 D8 A 18 C 28 C 38 D 48 C9 D 19 A 29 D 39 B 49 C10 D 20 A 30 C 40 C 50 B
PAPER 2
Question 1
No Criteria Marks
(a) Able to name tissue K and tissue L.Answer: K: Upper epidermis (cells / tissue) L: Palisade mesophyll (cells / tissue)
11 2
(b) Able to state the function of cells K and M in a leaf.Sample answer: K: Protect the inner tissues. // Allows light to penetrate. M: Controls the size of stoma / transpiration / gaseous exchange
// Allows gaseous exchange through the stoma.
11
2
(c) (i) Able to explain the differentiation of cells J to form the xylem tissue.Sample answer: Cells J join end to end, / the wall of cells J at the joints dissolved, to form a hollow tube / continuous tube (from root to leaves). The wall of xylem vessel is thickened by lignin. (Any 2)
111 2
(ii) Able to explain the effect on the function of the leaf when the plantunable to synthesise lignin during the formation of the xylem tissue.Sample answer: Xylem cannot be strengthened / cannot uphold leaf. Less sunlight received / absorbed. Slow down the rate of photosynthesis / less glucose produced
Or (Any 2)
Xylem vessels collapsed. Less water supplied to leaves. Slow down the rate of photosynthesis / less glucose produced
(Any 2)
111
111 2
(d) Able to state the meaning of cell specialisation.Sample answer: Cells grow, change shape / differentiate. To carry out / perform specific function.
11 2
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SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
3
(e) Able to explain the adaptation of palisade mesophyll tissue to enablethe leaf to carry out its function.Sample answer: Upright and closely packed. Contains large number of chloroplast. All cells receive maximum amount of sunlight.
// Absorb maximum amount of sunlight // energy.
111
3
TOTAL 13
Question 2
No Criteria Marks
(a) Able to name the parts labelled P and R.Answer : P: Golgi apparatus R: vesicle // lysosomes
11 2
(b) Able to state the function of mitochondrion.Sample answer : Site of (cellular) respiration // To produce ATP. 1 1
(c) Able to explain the role of nucleus in the synthesis of an enzyme.Sample answer : (Nucleus / DNA) carries genetic information (for protein synyhesis). Ribosome synthesis the protein.
11 2
(d) (i) Able to complete the schematic diagram to show the mechanism ofenzyme action.Sample answer : Suitable substances : Y Complete drawing
11 2
(ii) Able to name two characteristic of enzyme based on answer (d)(i).Sample Answer: The reaction of enzyme are highly specific // One enzyme only for
one substrate. Enzyme reactions are reversible. Enzymes are not destroyed in the reaction.
(Any 2)
1
11
2
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
4
(e) Able to explain the effect of temperature on the digestion in themouth.Sample Answer: Reaction of enzyme decreases. The active sites change. Enzymes could not bind with substrates. Digestion becomes slow.
(Any 3)
1111
3
TOTAL 12
Question 3
No Criteria Marks
(a) (i) Able to name the type of cell division.Answer: meiosis 1 1
(ii) Able to state an organ where meiosis takes place.Answer: Testis // Ovary 1 1
(b) State the number of chromosomes of cell X during phase P andphase U.Answer: P = 4; U = 2
11 2
(c) (i) Able to state the name of each phase of the cell division.Answer:
Phase Name of the phaseP Anaphase IQ Metaphase IR Telophase IS Anaphase IIT Prophase IU Telophase II
(6 correct=3m; 4-5 correct=2m; 2-3 correct=1m) 3 3
(ii) Able to state and explain the phase that brings about variation inorganism.Answer: Phase T Crossing-over occurs. Exchange of genetic material / segment of chromatid between
members of homologous chromosomes. Daughter cells have different gene combination. (Any 2)
111
1 3
(d) Able to explain the chromosomal behavior in stage P when Cell X istreated with a type of chemical that retards the function of centrioles.Sample answer: The centrioles form spindle fibers to separate chromosomes
(during anaphase). so the retarded centrioles will cause spindle fibres cannot be
formed. (As a result) the chromosomes do not line up at equator //
metaphase cannot occur. (Any 2)
1
1
12
TOTAL 12
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
5
Question 4
No Criteria Marks
(a) (i) Able to name organ Q.Answer: Stomach 1 1
(ii) Able to explain the process of food digestion in the stomach.Sample answer: The food digested in stomach is protein. Stomach secrete gastric juices contain pepsin. Pepsin hydrolyses protein to polypeptide.
111 3
(b) Able to state two effects if the gastric glands in the stomach areunable to produce hydrochloric acid.Sample answer: Bacteria in the food cannot be killed. Cannot provide acidic medium for enzyme reaction Pepsin inactive, so proteins are unable to be hydrolysed. Renin is inactive, so protein in milk cannot be coagulated
// caseinogen unable to be transform into insoluble casein.(Any 2)
1111
2
(c) Able to explain one difference between the foods that enters P(oesophagus) and R (ileum).Sample answer: The food in R has less content of starch / protein / lipid (than in P).
// The food in R has higher content of glucose / amino acid / fattyacid and glycerol (than in P). Starch is broken down / hydrolysed into maltose by amylase / into
glucose by maltase / sucrase / lactase.// Protein is broken down / hydrolysed into polypeptides / peptonesby pepsin / into peptides by trypsin / into amino acid by protease /erepsin. // Lipid is broken down / hydrolysed into fatty acids andglycerols by lipase.
1
1
2
(d) (i) Able to explain the absorption of proteins in the ileum.Sample answer: Amino acids are absorbed (by the villi). by facilitated diffusion into blood capillaries.
// The remaining amino acids is absorbed by active transport.
11
2
(ii) Able to explain the importance of food digestion which enablesnutrients to be absorbed by the villi.Sample answer: Complex food material is digested into simpler form. // Protein is
hydrolysed / broken down into amino acids // Other examples. Simpler molecules / amino acids / other examples are able to pass
through the plasma membrane of the villi.
1
12
TOTAL 12
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
6
Question 5
No Criteria Marks
(a) Able to name process X and compound Y.Answer: Process X : Decomposition / decaying Process Y : Nitrates
11 2
(b) (i) Able to Name the organic compound in plants.Answer: Amino acids / protein 1 1
(ii) Able to describe the functions of the amino acids / protein in plants.Sample answer: Growth / build new cells / tissue repairs in plants. Forms nucleic acid / chlorophyll / photosynthetic and respiration
enzyme / other examples.
11
2
(c) (i) Able to name the bacteria involved in the fixation of nitrogen innodules of a leguminous plant.Answer: Rhizobium sp. 1 1
(ii) Able to name and explain the relationship between Rhizobium sp.and the leguminous plant.Sample answer: Mutualism Relationship / interaction between two spesies of organism which
live closely together and give benefit to both.
11
2
(d) Able to relate the importance of nitrogen cycle in the growth ofrabbits.Sample answer: Provides nitrate / nitrogen (elements / compound) Absorb by the plant to synthesise protein. (When the rabbit eats the plant) the protein is transferred to the
rabbit to be used for producing plasma membrane / enzyme /hormones / growth / build new cells / tissue repairs.
111
3
TOTAL 11
Question 6
No Criteria Marks
(a) Able to state two differences between passive transport and activetransport.Answer:
Passive transport Active transport
Does not require energy Require energy
Occurs down theconcentration gradient
Occurs against theconcentration gradient
1
1 2
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
7
(b) Able to explain what happen to the red blood cell after beingimmersed in 3% of sodium chloride solution and 0.1% of sodiumchloride solution for half an hour.Sample answer:
Red blood cell in 3% of sodium chloride solution: 3% sodium chloride solution is a hypertonic solution compare to
the cell. Water diffuses out from the cell / red blood cell by osmosis. The cell becomes flaccid / shrunken. This is known as crenation.
Red blood cell in 0.1% of sodium chloride solution: 0.1% sodium chloride solution is a hypotonic solution compare to
the cell. Water diffuses into the cell/ red blood cell by osmosis. The cell becomes swollen and burst. This is known as haemolysis
1
111
1
111 8
(c) Able to describe what happen to the cell in each concentration ofsucrose solution.Sample answer:
Plant cell in 17% of sucrose solution for 30 minutes: The plant cell maintain its shape size 17% sucrose solution is isotonic to the concentration of the cell sap
in the plant cell The rates of movement of water molecule in and out of the cell sap
in the plant cell The size of the plant cell is maintained. (Any 3)
Plant cell in 0.1% of sucrose solution for 30 minutes: 0.1% sucrose solution is hypotonic to the concentration of cell sap
of the plant cell As a result, water molecules diffuse into the cell by osmosis The vacuoles will expand, causing a pressure to be exerted on the
cell wall The pressure causes the plant cell expands and become turgid.
(Any 3)
Plant cell in 30% of sucrose solution for 30 minutes: 30% sucrose solution is hypertonic to the concentration of cell sap
in the plant cell As a result, water molecules diffuse out of the plant cell by osmosis The vacuole becomes smaller, plasma membrane is pulled away
from the cell wall The plant cell is plasmolysed and becomes flaccid. (Any 3)
Plant cell in 0.1% of sucrose solution for 30 minutes: 0.1% sucrose solution is hypotonic to the concentration of cell sap As a result, water molecules diffuse in the plant cell by osmosis The cell undergoes deplasmolysis The cell return to its normal condition. (Any 3)
11
1
1
1
11
1
1
11
1
1111
Max10
TOTAL 20
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
8
Question 7
No Criteria Marks
(a) (i) Able to explain the exchange of gases between tracheole and bodycell.Sample answer: Partial pressure/concentration of oxygen in the tracheole is higher
than partial pressure/concentration of oxygen in body cell . Oxygen diffuse from tracheole to body cell Partial pressure/concentration of carbon dioxide in the body cell is
higher than partial pressure/concentration of carbon dioxide intracheole . Carbon dioxide diffuse from tracheole to body cell
1
1
1
1 4
(ii) Able to explain how the absent of chitin affect the process ofinhalation and energy production of the insect.Sample answer: The function of chitin is to prevent trachea from collapsing/sustain
the air pressure During inhalation high pressure air moves into the trachea. The absent of chitin will cause the trachea / P to collapse / burst /
rupture. Air with oxygen cannot reach tracheal. Body cell cannot get enough oxygen for cellular respiration The insect does not produce enough energy and respire
anaerobically. Less energy produced. (Any 6)
1
11
111
1 6
b (i) Able to compare and explain the respiration before and duringvigorous exercise.Sample answer:
Before (A) During (B) Explanation (E)1.
AerobicRespiration
AnaerobicRespiration
Before - oxygen intake islow/the same as oxygenrequired/enough oxygenis supplied to the cell
During – oxygen requiredis more than oxygenintake
2.
Themuscles arein normalcondition
The musclesare in the stateof oxygen debt
Before – oxygen issufficient
During – oxygen isinsufficient / oxygensupplied is less thanoxygen supplied.
3.
Energyproduced ismore/38ATP
Energyproduced isless / 2 ATP
Before – complete breakdown of glucose (producemore energy)
During – incomplete breakdown of glucose (produceless energy)
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
9
4.
No/lessaccumulation of lacticacid in themuscles
Highaccumulationof lactic acid inthe muscles
Before – complete breakdown of glucose producecarbon dioxide and water
During – Incompletebreakdown of glucoseproduce lactic acid
A + B = 1m
E=1m (Any one E)
Max = 8m 8
(b) (ii) Able to explain how the oxygen intake by the athlete returns to thenormal level at the 25th minute.Sample answer: Lactic acid has been removed from the muscles The lactic acid has been converted to energy/ convert to glucose
1
1 2
TOTAL 20
Question 8
No Criteria Marks
(a) Able to state that the daily menu for the pregnant woman does notprovide a balanced diet.
Able to evaluate the nutrients content and the effects of consumingthe foods in the menu correctly.Criteria: Food class / nutrients Effects
Sample answer:
Pregnant woman: Carbohydrate in fried rice / chicken rice / (fried) noodle Protein in chicken drumstick / fried prawn Vitamin C in apple / fruit juice / cucumber Lipids in fried prawn / fried rice Caffein in coffee Water in apple / fruit juice / chicken soup Excess sugar in carbonated drink / fruit juice with sugar Lack of / No calcium Lack of fibre (Any 5)
Effects of nutrients intake: Carbohydrate provides energy Protein for the foetus growth / repairing the damaged tissue Lipid provides energy Vitamin C for a good skin / preventing scurvy Caffein may increase the blood pressure Excess sugar may lead to diabetis / hyperglysaemia Lack of / no calcium will lead to osteoporosis / teeth problem Lack of fibre may lead to constipation / defecaetion problem Water for replacing the water loss during daily activities / any
suitable functions of water. (Any 4)
1
111111111
111111111
1
5
4
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
10
(b) (i) Able to write chemical equation of photosynthesis correctly.Answer:
6 H2O + 6CO2
Or equation is not balance.
1+1
1
1 2
(ii) Able to define photosynthesis.Sample answer: Photosynthesis is a process where a leaf / green plant absorbs
carbon dioxide and water to produce glucose and oxygen in the presence of chlorophyll and
light.
1
1 2
(c) Able to describe how Hydrilla sp is able to obtain all the requirementfor photosynthesis.Sample answer:
F ENo stomata Enables easy diffusion of
photosynthetic gases.Many small green leaves Contain a lot of chloroplast /
chlorophyllMany fishes/ and aquaticorganism in the pond
Supply carbon dioxide throughrespiration
Penetration of sunlight Supply lightPenetration of sunlight Increase temperature in water
(Any 6) 6
TOTAL 20
Question 9
No Criteria Marks
(a) Able to explain how the use of inorganic fertilizer reduces thepopulation of aquatic organism.Criteria P: The effect of excessive inorganic fertilizer. Explanation of europhication.
Sample answers : Inorganic fertilisers / phosphates / nitrates from agricultural area
enter the river water. Excess inorganic fertilizer will encourage the high rate of growth of
algae / blue green bacteria. This causes an algal bloom The algae increase in number and form a thick scum on the
surface of river. Prevent the penetration of sunlight reaching the bottom of the
water.
1
1
11
1
Light
chlorophyll
C6H12O6 + 6O2
6 H20 + 6CO2 C6H12O6 + 6O2
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
11
Alga and other aquatic organism in the bottom of the water cannotdo photosynthesis The decomposition by aerobic bacteria of dead organic matter will
use up the dissolved oxygen This will raise the BOD level. Aquatic organism with low oxygen content will die. The river water will be polluted.
1
1
111 10
(b) Able to discuss the good and bad effects of this action on human andthe ecosystem.Sample answers :
Good effect: It can provide more space for another crop It can increase the economic use of the land It allow the crop to receives enough sunlight More product will be produces. It can reduces the possibility for the crop to be infected by insect /
pest. So the crop can growth much healthier. It can increase the consuming of water and nutrient The crop will get enough nutrient
(Any 7)Bad effect: No windbreaker / shelter /niche for bird/small animal No protection for the crop Soil erosion Roots of the plant will improve soil stability
(Any 3)
11111
111
1111
10
TOTAL 20
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
12
PAPER 3
Question 1
1 (a) [KB0602 - Classifying]
Score Criteria
3 Able to categorise all the materials and apparatus used in the experiment correctly.
Sample answer:Material Apparatus
Potato* Hydrogen peroxide* Hydrochloric acid Sodium hydroxide Distilled water pH paper
Knife Measuring cylinder* Stopwatch
2 Able to categorise any 4 materials and any 2 apparatus into correctly.(*compulsory)
1 Able to categorise 2 materials (*) and 1 apparatus (*) correctly.
1 (b) [KB0603 - Measuring Using Number]
Score Criteria
3 Able to record all the volume of bubbles formed in each measuring cylinderaccurately with correct unit.Answer:
P Q QVolume of bubbles= 0 cm3
Volume of bubbles= 11.5 cm3
Volume of bubbles= 0 cm3
2 Able to record two readings (which include Q) accurately with correct unit.
1 Able to record any one reading accurately with correct unit.
1 (c) (i) [KB0601 - Observation]
Score Criteria
3 Able to state any two observations correctly according to the criteria: pH value test tube bubbles produce
Sample answers:1. At pH 7 in test tube R, volume of bubbles produced in 5 minutes time is 11.5 cm3.2. At pH 2 in test tube P, volume of bubbles produced in 5 minutes time is 0 cm3.3. At pH 10 in test tubes Q, volume of bubbles produced in 5 minutes time is 0 cm3.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
13
2 Able to state any one observation correctly.or
Able to state any two incomplete observations ( any 2 criteria)
Sample answers:1. pH affect the volume of bubbles formed.2. The volume of bubbles formed depends on pH.
1 Able to state any one idea of observation (any 1 criterion)
Sample answers:1. Bubbles formed in neutral pH.2. No bubbles in test tube P / Q.3. Rate of enzyme reaction is higher at pH 7.4. Bubbles are produced when potatoes react with hydrogen peroxide.
1 (c) (ii) [KB0604 - Making inferences]
Score Criteria
3 Able to make one logical inference for each observation based on the criteria: pH enzyme in potato react with H2O2 producing oxygen
Sample answers:1. Neutral medium is suitable for enzyme catalase in potato to react with hydrogen
peroxide and produce oxygen.2. Acidic / alkali medium is not suitable for enzyme catalase in potato to react and
no oxygen is produced.
2 Able to make one logical inference for any one observation.orAble to make one logical and incomplete inference base on 2 criteria for eachobservation.
Sample answers:1. pH of the medium affects enzyme catalase in potato to react with hydrogen
peroxide and produce oxygen.2. Acid / alkali medium is not suitable for enzyme reaction.3. Neutral medium is suitable for enzyme reaction.
1 Able to make an idea of inference with one criterion.
Sample answers:1. Reaction of enzyme is affected by pH.2. Gas produced when the medium is suitable for enzyme reaction.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
14
1 (d) (i) [KB0606 - Communicating]
Score Criteria
3 Able to tabulate a table and fill in data accurately base on three criteria: pH value Volume of bubbles formed (cm3) Rate of reaction (cm3 minute-1)
Sample answers:pH value Volume of bubbles formed
(cm3)Rate of reaction(cm3 minute-1)
2 0 07 11.5 2.3
10 0 0
2 Able to tabulate a table base on two criteria.
1 Able to Able to tabulate a table base on one criterion.
1 (d) (ii) [KB0607 - Interpreting Data]
Score Criteria
3 Able to explain clearly and accurately the relationship between the test tubecontent and the volume of bubbles formed in Q, base on 3 criteria. Test tube content Volume of bubbles formed Explanation: Relationship between pH value / medium and the rate of reaction.
Sample answer:1. The content of test tube Q is hydrogen peroxide, distilled water and potato cube,
at pH 7 / neutral the rate of reaction is the highest / optimum pH for enzymereaction which produces highest volume of bubbles / 11.5 cm3.
2 Able to state / explain clearly but less accurate the relationship base on 2 criteria.
Sample answers:1. The content of test tube Q is hydrogen peroxide, distilled water and potato cube,
which produces highest volume of bubbles / 11.5 cm3.2. The content of test tube Q is hydrogen peroxide, distilled water and potato cube,
at pH 7 / neutral the rate of reaction is the highest / optimum pH for enzymereaction.
3. At pH 7 / neutral the rate of reaction is the highest / optimum pH for enzymereaction which produces highest volume of bubbles / 11.5 cm3.
1 Able to state the idea of the relationship base on 2 criterion
Sample answer:1. At pH 7 / neutral the rate of reaction is the highest / optimum pH for enzyme
reaction.2. Test tube with hydrogen peroxide, distilled water and potato cube, produced
bubbles.3. The volume of bubbles in pH 7 / neutral is 11.5 cm3.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
15
1 (e) (i) [KB061001 - Controling Variables]
Score Criteria
3 Able to state all the variables and the method to handle the variables correctly.
Sample answers:Variables Method to handle the variables
Manipulated variable:pH (value of mediums) //acidic, neutral andalkaline mediums
Three mediums of different pH are used //Hydrochloric acid, distilled water andsodium hydroxide solutions are used.
Responding variable:Volume of bubblesformed //The rate of reaction /(catalase) enzyme action
Observe / measure and record the volumeof bubble formed (in 5 minutes) by using ameasuring cylinder (and a stopwatch) //Calculate the rate of reaction by dividing thevolume of bubble formed with time in cm3
minute-1 / using formula (show the formula)Fixed variable:Volume / concentration ofsolutions / H2O2 / acid /alkali / distilled water /potato cube
Use 3.0 cm3 of H2O2 for all experiment //Use 6.0 cm3 of acid, alkali and distilledwater //Put 1 potato cube with size of 1.0 cm3 eachfor all experiment
2 Able to state 4 - 5 of the variables and the method to handle the variables correctly.
1 Able to state 1 - 3 of the variables and the method to handle the variables correctly.
1 (e) (ii) [KB0611 - Making Hypothesis]
Score Criteria
3 Able to state a hypothesis to show a relationship between the manipulatedvariable and responding variable and the hypothesis can be validated, base on 3criteria: Manipulated variable Responding variable Relationship
Sample answers:1. In neutral medium / pH 7 the reaction between the potato enzyme and H2O2
produces the highest volume of bubbles / the highest rate of reaction.2. In acidic / alkali medium / pH 2 / pH 10 the potato enzyme will not react with H2O2
producing no bubbles.
2 Able to state less accurate hypothesis to show a relationship between manipulatedvariable and responding variable base on 2 criteria.
Sample answers:1. The pH affects the reaction of potato enzyme with H2O2 producing bubbles.2. The bubbles produce by potato’s enzyme reaction depends on different pH.
1 Able to state idea of hypothesis to show a relationship between manipulatedvariable and responding variable base on 1 criterion.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
16
Sample answers:1. pH medium affect the reaction.2. Suitable pH causes the production of bubbles.
1 (f) [KB0608 - Space and Time Relationship]
Score Criteria
3 Able to state clearly and accurately the relationship between the volume of bubblesformed and the time in a medium of pH 7 base on criteria: Volume of bubbles formed Time Relationship
Sample answers:1. In a medium of pH 7, the volume of bubbles formed in 5 minutes is 11.5 cm3.
2 Able to state clearly but less accurate the relationship between the volume ofbubbles formed and the time in a medium of pH 7 base on 2 criteria.
Sample answers:1. In a medium of pH 7, the rate of reaction is 2.3 cm3 minute-1.
1 Able to state the idea of the relationship base on 2 criteria.
Sample answer:1. Bubbles are formed in 5 minutes.
1 (g) [KB0609 - Define Operationally]
Score Criteria
3 Able to state what an enzyme is base on experiment correctly according to thecriteria: Chemical / substance / molecule / enzyme in potato React with H2O2 which produce bubbles Affected by pH
Sample answer:1. Enzyme is a chemical in potato that able to react with H2O2 producing bubbles
and the reaction is affected by pH of the medium.2. Enzyme in potato react with H2O2 producing bubbles and affected by pH.
2 Able to state what an enzyme is base on experiment less accurately according to 2criteria.
Sample answers:1. Enzyme react with H2O2 producing bubbles.
1 Able to state the idea of an enzyme or the theoretical definition of enzyme.
Sample answers:1. Enzyme reacts with H2O2.2. Enzyme in potato reacts to produce bubbles.3. Enzyme is a biological catalyses that accelerate the reaction.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
17
1 (h) [KB0605 - Predicting]
Score Criteria
3 Able to predict and explain the observation accurately when 2 potato cubes 0.5 cm3
is used in measuring cylinder P. Expected observation Compare to which Reason
Sample answer:1. The volume of bubbles produce is 23 cm3 (any value between 11,5 - 23 cm3),
more than the experiment when using 1 potato cube (sized 1 cm3), because moreenzyme for the reaction / larger surface area of potato (in contact with H2O2).
2 Able to predict the result less accurately (2 criteria).
Sample answers:1. The volume of bubbles produce is 23 cm3 (any value between 11,5 - 23 cm3),
more than the experiment when using 1 potato cube (sized 1 cm3).2. The volume of bubbles produce is more than the experiment when using 1 potato
cube (sized 1 cm3), because more enzyme for the reaction / rate of reaction ishigher.
1 Able to give idea of the result.
Sample answers:1. The volume of bubbles produce is 23 cm3.2. The rate of reaction is higher.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
18
Question 2
Problem Statement
Score Criteria
3
Able to state the problem statement of the experiment correctly that include criteria:
Manipulate variables Responding variables Relation in question form and question symbol [?]
Sample answers:
1. Does orange juice contain higher amount / more vitamin C than papaya andwatermelon juices.
2. Which fruit juice has the highest amount of vitamin C?3. What is the amount of vitamin C in orange, papaya and watermelon juices?
2
Able to state the problem statement of the experiment with two criteria.
Sample answers:
1. Do different fruit juices have different amount of vitamin C?2. Does the content of vitamin C in fruit juices differ?
1
Able to state the of problem statement with one criteria.
Sample answers:
1. What is the amount of vitamin C in orange juice?
Aim
Score Criteria
To determine the concentration / percentage of vitamin C in orange, papaya andwatermelon juices.
Hypothesis
Score Criteria
3
Able to state the hypothesis correctly according to the criteria:
Manipulate variables Responding variables Relationship of the variables
Sample answers:
1. Orange juice has the highest concentration / percentage of vitamin C thanpapaya and watermelon juices.
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
19
2
Able to state the hypothesis with two criteria
Sample answers:
1. The vitamin C content depends on types of fruit juice.2. Different fruit juices have different amount of vitamin C
1
Able to state the idea of the hypothesis.
Sample answers:
1. Fruit juices have vitamin C.
Variables
Score Criteria
Able to state the three variables correctly
Sample answers:Manipulated variable: Fruit juices // orange, papaya and watermelon juicesResponding variable: Concentration / percentage of vitamin CFixed variable: Volume DCPIP solution // Concentration of ascorbic acid.
Materials and Apparatus
Score Criteria
3
Able to state all functional materials and apparatus / 2*materials + 1 other materialand 1*apparatus + 2 other apparatus for the experiment.
Materials: *Orange, papaya and watermelon juices, 0.1% ascorbic acid solution,*DCPIP solution.
Apparatus: *Syringes with needles, beakers, test tubes / specimen tubes, gauze /muslin cloth.
2
Able to state 2*materials and 1*apparatus + 1 other apparatus for the experiment.
1
Able to state 2*material and 1*apparatus for the experiment.
Technique
Score Criteria
Bonus
1m
Able to state the action on responding variable with an apparatus / formula.
Sample answer:Calculating and recording the concentration / percentage of vitamin C in fruit juicesby using the following formula.
Concentration of Vitamin C = Volume of 0.1% ascorbic acid mg cm3
Volume of fruit juiceOr / Percentage of Vitamin C = Volume of 0.1% ascorbic acid x 0.1 %
Volume of fruit juice
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
20
Procedure
Score Criteria
3
Able to state five procedures P1, P2, P3, P4 and P5 correctly.P1 : How to Set Up The Apparatus (3P1)P2 : How to Keep Constant The Control Variable (2P2)P3 : How to Manipulate The Manipulated Variable (1P3)P4: How to Record The Responding Variable (1P4)P5 : Precaution (1P5)
2
Able to state three of any procedures P1, P2, P3, P4 and P5 correctly
1
Able to state two of any procedures P1, P2, P3, P4 and P5 correctly
Example of Procedure:
1. Measure (1 cm3) of DCPIP by using a syringe and place in a test tube.(1 cm3)
P1P2
2. Fill a 5 cm3 syringe with (0.1%) ascorbic acid.(0.1%)
P1P2
3. Place the needle of syringe into the DCPIP solution. P54. Add the acid ascorbic solution,
drop by drop into the DCPIP solution.P1P5
5. Stir the mixture gently with the needle of the syringe. P56. Add the acid ascorbic solution continuously until the DCPIP solution decolourise. P17. Record the volume of acid ascorbic used. P48. Repeat steps 1 to 7 by using orange juice, papaya juice and watermelon juice. P39. Record the results in a table. P4
Data
Score Criteria
Bonus
1m
Able to tabulate the correct table with observations.
Sample answers:Solution / Fruit juice Volume of solution / fruit
juice needed to decolourise1 cm3 DCPIP solution (cm3)
Concentration / Percentageof vitamin C in fruit juice
(mg cm3) / %0.1% ascorbic acid
Orange juicePapaya juice
Watermelon Juice
Conclusion
Score Criteria
Able to rewrite the hypothesis correctly.
Sample answers:
Orange juice has the highest concentration / percentage of vitamin C than papayaand watermelon juices. (Hypothesis is accepted).
MOZ@C
SMS MUZAFFAR SYAH , MELAKA
F4 BIOLOGY DIAGNOSTIC 2008 MARKING SCHEME
MARKING SCHEME 4551
21
Planning the Experiment
Score Criteria
3 Able to plan the experiment based on 7 – 9 ( ) of the following criteria: Statement of identified problem Objective of study Variables Statement of hypothesis List of materials and apparatus Technique used Experimental procedures Presentation of data Conclusion
2 Able to plan the experiment based on 4 – 6 ( ) of the criteria.
1 Able to plan the experiment based on 1 – 3 ( ) of the criteria.
END OF MARKING SCHEME
MOZ@C
SMS MUZAFFAR SYAH , MELAKA