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PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 1 
SECTION A [60 marks] Answer all the questions Jawab semua soalan dalam bahagian ini 
1. Diagram 1.1 shows the structure of animal cell 
Rajah 1.1 menunjukkan struktur sel haiwan 
Diagram 1.1 Rajah 1.1 
(a) On Diagram 1.1, label X and Y. 
Pada Rajah 1.1, labelkan X dan Y [1marks] 
(b) Explain one characteristic of Y related to cell division. 
Terangkan satu ciri Y berkaitan dengan pembahagian sel Sample answer: F: Composed of a complex arrangement of microtubules E: Form spindle fibers during cell division (in animal cell) 
[2marks] 
Y: centriole 
X: nucleus
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 2 
(c) Explain the function of X in cell division 
Terangkan fungsi X dalam pembahagian sel Sample answer: F: nucleus containchromosomes which carry genetic information E: behavior of chromosome in stage of mitosis/meiosis/cell division transfer genetic information to daughter cell [2marks] 
(d) Diagram 1.2 below shows stage of cell division in somatic cell of human. 
In the box given, draw another two stage in that cell division. Rajah 1.2 di bawah menunjukkan peringkat pembahagian sel dalam sel soma manusia. Dalam kotak yang diberi, lukis dua peringkat dalam pembahagian tersebut 
[2marks] 
Diagram 1.2 
Rajah 1.2 
(a) 
(b) 
(c) 
(d)
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 3 
(e) If spindle fibre is not form in Diagram 1.2 (d), explain the effect on the number of chromosome in the daughter cell. 
Jika gentian gelendong tidak terbentuk dalam Rajah 1.2 (d), terangkan kesan keatas bilangan kromosom dalam sel anak Sample answer: F:the number of chromosome in the daughter cell less/ extra E: no contraction of spindle fibre to pull chromosome toward the pole [2marks] 
(f) Explain how the cell division above can be used to increase in a short time the number of given example of the plant in the farm. 
Terangkan bagaimanakah pembahagian sel di atas dapat digunakan untuk meningkatkan bilangan dalam masa yang singkat tumbuhan yang dinamakan dalam ladang. Sample answer: Example of the plant: banana/ palm oil/ any other example F:tissue culture E1: cut off explants / part of plant / young shoot, leaves, roots, seeds, embryos E2: explants are sterilized and placed in cultured medium containing nutrient such as glucose, amino acids, minerals and growth hormone/ auxin E3:the culture medium need to be maintained at optimum pH and temperature 25 -35°C E4: explants divide by mitosis form callus// undifferentiated mass of tissues E5: callus develop into embryoid/ somatic embryos and later into plantlets E6: plantlets are transferred to the soil where they grow into adult plant [3marks]
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 4 
2.(a) Diagram 2.1 shows the shape of red blood cells after being immersed for 30 minutes in three solutions with different concentration. Rajah 2.1 menunjukkan bentuk sel darah merah selepas direndam selama 30 minit dalam tiga larutan yang berbeza kepekatannya. 
Based on the Diagram 2.1 Berdasarkan Rajah 2.1 (i) State the condition of the red blood cells after being immersed in Nyatakan keadaan sel darah merah selepas direndam di dalam Sample answer: Solution P: Crenation / shrink / shrivel Solution Q: Haemolysis / swell and burst [2 marks] 
Diagram 2.1 
Red blood cells in Q solution 
Sel darah merah dalam larutan Q 
Red blood cells in P solution 
Sel darah merah dalam larutan P 
Red blood cells in R solution 
Sel darah merah dalam larutan R
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 5 
(ii) Name the type of solution R in which the red blood cells are immersed. Namakan jenis larutan R yang mana sel darah merah direndam. Sample answer: Solution R is isotonic solution. [1 mark] (iii) Explain your answers given in a(ii) Terangkan jawapan yang anda berikan di a(ii) Sample answer: P1:The cell retains its normal shape/ biconcave disc shape. P2:The water diffuses in and out of the cells at equal rate by osmosis P3:Solution R has the same osmotic concentration as the cytoplasmic fluid in the red blood cells [3 marks] 
(b) Based on the statement, explain why vinegar is suitable to be used as the natural preservative for the preservation of garlic. Berdasarkan pernyataan di atas ,terangkan mengapa cuka adalah sesuai digunakan sebagai pengawet semulajadi untuk bawang putih. Sample answer: F1:Vinegar has a low pH/acidic E1: Vinegar diffuses into the tissues of the garlic E2: The tissues of the garlic becomes acidic E3: The low pH prevents the growth of microorganisms in garlics E4: The garlic can be preserved to last longer 
[3 marks] 
Food such as mushrooms, fruits, vegetables and fish can be preserved longer by using natural preservatives such as salt, sugar and vinegar. 
Makanan seperti cendawan, buah-buahan, sayur-sayuran dan ikan boleh diawet untuk tahan lama menggunakan bahan-bahan pengawet semulajadi seperti garam,gula dan cuka .
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 6 
(c) Diagram 2.2 shows the condition of herbaceous plant due to water shortage in soil. Rajah 2.2 menunjukkan keadaan pokok herba disebabkan oleh kekurangan air dalam tanah. 
Explain the condition of the plant in Diagram 2.2 after one week. Terangkan keadaan pokok dalam Rajah 2.2 selepas satu minggu. Sample answer: F:The plant wilt E1: The cells become flaccid / plasmolysed // both the vacuole the vacuole and cytoplasm shrink // the plasma membrane of the root cells pull away from the cell wall. E2: Water molecules diffuse out from the cell sap of the root hair cell by osmosis E3: (the remaining) soil water becomes hypertonic to the cell sap of the root hair cell as the soil dries out. [3 marks] 
Water shortage after one week 
Kekurangan air selepas satu minggu 
Diagram 2.2
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 7 
4. Diagram 4.1 shows the process of phagocytosis as second line of defence to destroy the bacteria 
Rajah 4.1 menunjukkan proses fagositosis sebagai barisan pertahanan kedua untuk memusnahkan bakteria. 
Stage 1 Stage 2 
Peringkat 1 Peringkat 2 
Stage 4 Stage 3 Peringkat 4 Peringkat 3 Diagram 4.1/ Rajah 4.1 
(a) (i) Name cell M involved in mechanism above. 
Namakan sel M yang terlibat dalam mekanisma di atas. Neutrophil / eosinophil / basophil / granulocyte / phagocyte [1 mark] 
(ii) Draw a diagram in stage 3 
Lukis rajah dalam peringkat 3. [1 mark] 
(b) Explain the function of lisosome in mechanism above 
Terangkan peranan lisosom dalam mekanisma di atas F : digest bacteria E : because lisosome contain hydrolytic enzymes / cellulase which digest cellulose / by breaking down the bacteria cell wall. [2 mark] 
M
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 8 
(c) Explain what happen in stage 2. 
Terangkan apa yang berlaku dalam peringkat 2. F : phagocytes surrounds / engulfs the bacteria using pseudopodia E : forming phagocytic vacuole / fagosome /food vacuole [2 mark] 
(d) How the action of M to pathogen is different to lymphocyte? 
Bagaimanakah tindakan M terhadap patogen adalah berbeza dengan limfosit? M kill the pathogen by engulf and digest the pathogen but lymphocyte produce antibody then antibody kill the pathogen / neutralize the toxin from pathogen [1 marks] 
(e) Diagram below show a type of immunity occurs in human. 
Rajah di bawah menunjukkan jenis immuniti yang berlaku dalam manusia. 
(i) Name the type of immunity shows in the diagram. 
Namakan jenis immuniti yang ditunjukkan dalam rajah. Answer: Artificial active immunity [1 mark]
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 9 
(ii) Name the second injection and why the person should take second injection? 
Namakan suntikan kedua dan mengapa individu ini perlu mengambil suntikan kedua? Sample answer: N: booster dose // an additional administration of a vaccine E: to stimulate lymphocyte produce more antibody until achieve immunity level. [2 mark] 
(f) The above immunity is example of third line of defence. What make it different to the second line of defence? 
Sample answer: P1: Third line of defence – specific response to pathogen infection but second line of defence – non-specific response/generalized responses to pathogen infection P2: Third line of defence involved production of antibody(active immunity)/ used supply antibody (passive immunity) from leucocyte but second line of defence involve the physical structure of leucocyte [2 marks]
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 10 
Q 
5. Diagram 5.1 shows the stages of the ovarian cycle in human ovary Diagram 5.2 shows the thickness of the endometrium of uterus before the fertilisation in the second menstrual cycle. Rajah 5.1 menunjukkan peringkat kitaran ovary dalam ovary manusia. Rajah 5.2 menunjukkan ketebalan endometrium dalam uterus sebelum berlaku persenyawaan dalam kitarhaid yang kedua. 
Diagram 5.1 
Diagram 5.2 
Y 
P 
Secondary follicle 
Secondary oocyte 
X
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 11 
a) (i) Describe the change in the structure of follicle P into a secondary follicle. 
Terangkan perubahan struktur folikel P dalam pembentukan folikel sekunder. [2 marks] Sample answer: P1 : FSH concentration increases // is released (by the pituitary gland). P2 : Stimulates the development of follicle cells. (1m) P3 : Primary follicle developed into secondary / Graafian follicle // Primary oocyte developed into secondary oocyte (ii) Relate the change in (a)(i)to the thickness of the endometrium Hubungkan perubahan dalam (a) (i) dengan ketebalan endometrium [1 mark] Sample answer: P1 : The thickness of the endometrial wall increases 
b) Explain the process that occurs at X. 
Terangkan proses yang berlaku pada X. [2 marks] Sample answer: P1 : Ovulation P2 : The release of secondary oocyte from the (matured) secondary follicle / Graafian follicle to the oviduct / Fallopian duct 
c) Explain the effect of the change of structure Q to the thickness of the endometrium. 
Terangkan kesan perubahan struktur Q keatas ketebalan dinding endometrium. [2 marks] Sample answer: P1 : The thickness of the endometrial wall / uterine lining decreases. P2 : The level of progesterone decreases. 
d) (i) Fertilisation takes place in the second menstrual cycle. 
Complete the graph in Diagram 5.2 to show the changes in the thickness of the endometrium after point Y Persenyawaan berlaku dalam kitar haid yang kedua 
Lengkapkan graf dalam Rajah 5.2 untuk menunjukkan perubahan ketebalan endometrium selepas titik Y
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 12 
[1 mark] Answer: 
(ii) Explain your answer in (d)(i). Terangkan jawapan di dalam (d)(i). [2 marks] Sample answer: P1 : The corpus luteum / placenta developed. P2 : The corpus luteum / placenta released progesterone (and oestrogen). 
e) (i) State the changes in the thickness of the endometrium after point Y relating to the secretion of hormones secreted by the ovary. 
Terangkan hubungan perubahan ketebalan endometrium selepas titik Y dengan perembesan hormon oleh ovari. [1 mark] Sample answer: P1 : The thickness of the endometrial wall increases / is maintained. (ii) State the importance of thickened endometrium to the continuity of life Nyatakan kepentingan ketebalan endometrium dalam kesinambungan hidupan [1 mark] Sample answer: P1 : Increase the chance of implantation // development of embryo / blastocyst.
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 13
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 14 
SECTION B [40 marks] Answer any two questions from this section Jawab mana-mana duasoalandaripadabahagianini 6. Diagram 6.1 shows the growth and development process at the shoot tip. Rajah 6.1 menunjukkan proses pertumbuhan dan perkembangan pada hujung pucuk. 
Diagram 6.1 
a) Explain the process of primary growth that shown in Diagram 6.1 
Terangkan proses pertumbuhan primer yang ditunjukkandalam Rajah 6.1 [8 marks] Sample answer: Cell division P1: Cell division take place by mitosis P2: Each cell divides to become two cells which are identical to the parent cell P3: This process repeats itself until a mass of cells consisting of many identical cells are formed
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 15 
Cell elongation P4: Cell elongation cause by intake of water and nutrient into the cell from the environment P5: Water accumulates in the vacuoles of plant cells to form large central vacuole, causing the primary wall to stretch P6: The nutrients are used in the building up of the protoplasm// more organelles leading to an increase in the cell size and volume Cell differentiation P7: Cells begin to differ from each other to form groups of specialised cells P8: to perform new and specialised functions // Example: cell differentiation in the epidermis of roots to form root hair to enable the cell to have a large total surface area for absorption of water from the soil P9: Cells differentiation causing the changes of shape and complexity of organism 
b) Explain the importance of primary growth to plant. 
Terangkan kepentingan pertumbuhan primer kepada tumbuhan [4 marks] Sample answer: P1: During this time the stem and roots of plant increase in length. This allows a plant to achieve its maximum height P2: Its bring about the formation of primary xylem that helps in the transport of water and mineral P3: Its bring about the formation of primary phloem that helps in the transports organic substances P4: Its provides support because the walls of xylem tissue are thickened with lignin
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 16 
c) Diagram 6.2 shown the tropism respond at shoot tip and root tip. 
Rajah 6.2 menunjukkan gerakbalas tropisme pada hujung pucuk dan hujung akar. 
Diagram 6.2 Base on diagram 6.2 explain how the tropism response occurred. Berdasarkan rajah 6.2 terangkan bagaimana gerakbalas tropisme berlaku. [8 marks] Sample answer: P1: auxins produced by shoot and root P2: auxins diffuse into zone of elongation P3: (Owing to gravity) auxins move to lower side of shoot and root P4: The lower side of shoot and root has a higher concentration of auxins than the upper side P5: height concentration of auxins in the shoot promotes elongation of cells. P6: the lower side of the shoot will faster than the upper side P7: the shoot curves and grows upward (negative geotropism) P8: height concentration of auxins in the root inhibits elongation of cells. P9: So the upper side of the root will grow faster than the lower side P10: the root curves and grows downwards (positive geotropism)
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 17 
7. (a) Explain the interaction based on Diagram 7.1 Terangkan interaksi berdasarkan Rajah 7.1 
Diagram 7.1 Rajah 7.1 [4marks] Sample answer: F: Commensalism E1:relationship between two species that benefits one species/ commensal but neither benefits nor harms the other species/ host E2: fern has a sponge-like root mass that soaks up rain water and absorbs nutrients released from the decaying litter. E3: fern leaves has mesophyll cell contain chloroplast do photosynthesis (b) Diagram 7.2 shows mechanism of photosynthesis in plant. Explain why the product from light reaction need for dark reaction. Rajah 7.2 menunjukkan mekanisma fotosintesis dalam tumbuhan.Terangkan Mengapa produk dari tindakbalas cahaya diperlukan untuk tindakbalas gelap.
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 18 
[6marks] Sample answer: F: Mechanism of photosynthesis in plant consist of light reaction anddark reaction E1: Chlorophyll absorbs light energy released electrons andproduce ATP E2: Light energy is also split the water molecules intohydrogen ion(H+)and hydroxyl ions(OH-)/Photolysis of water E3: Hydrogen ions combine with electron from chlorophyll to form hydrogen atoms E4: Hydrogen atom and ATP will be used in dark reaction E5: Each hydroxyl ion loses an electron to form hydroxyl group (the electron is received by the chlorophyll) E6: The hydroxyl groups then combine to form water and gaseous oxygen E7: Hydrogen atomfit/ reduce carbon dioxide in dark reaction to form glucose E8: Reduction of carbon dioxide need ATP from light reaction E9: The reaction catalysed by photosynthetic enzyme in stroma E10: Produced glucose molecules, then glucose undergo condensation/ converted to starch for storage 
Diagram 7.2 
Rajah 7.2
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 19 
(c) Diagram 7.3 shows the condition of the town. Rajah 7.3 menunjukkan sebuah bandar. 
Diagram 7.3 Rajah 7.3 
(i) Discuss the effect of air pollution may occur in the town. 
Bincangkan kesan pencemaran udaya yang mungkin berlaku dalam bandar. [6 marks] Sample answer: F1: formation of haze / smog E1: cause by fine particle matter/smoke/ soot E2: prevents vision /reduce light intensity photosynthesis/ reduce oxygen content F2: acid rain E1: cause by SO2 / NO2 E2: estroy building F3: increasetemperature E1: cause by CO2 E2: green house effect/ global warming F4: depletion of ozonelayer E1: cause by CFC gases E4: more uv penetration F5: respiratory problems/allergies/risk for cancer E1: cause by CO/ SO2 / NO2 E2: cause health hazards
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 20 
(ii) If you are an environmental activist, suggest how you would explain to the government about the measures needed to overcome the type of pollution. 
Jika anda seorang aktivis alam sekitar, cadangkan bagaimana anda akan menjelaskan kepada kerajaan mengenai langkah-langkah yang diperlukan untuk mengatasi jenis pencemaran. [4 marks] Sample answer: F1: implementation of laws E1: control and prevent pollution using the environmental act F2: use of technology E1: using unleaded petrol for cars/ fit catalytic converter in factory F3: education E3: media massa/ internate/ scholl 8. (a) Diagram 8.1 shows three types of neurons. Rajah 8.1 menunjukkantigajenis neuron. 
Diagram 8.1 Rajah 8.1 Name types of neurone P, Q and R and state two differences between the structure of neurone P and neurone Q Namakan jenis neuron P, Q dan R dan nyatakan dua perbezaan di antara struktur bagi neuron P dan neuron Q 
[4 marks] 
Neurone P 
Neuron P 
Neurone Q 
Neuron Q 
Neurone R 
Neuron R
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 21 
Sample answer: Neurone P is afferent neurone Neurone Q is efferent neurone Neurone R isinterneurone 3√ = 2 marks 2√ = 1 marks 
Neurone P 
Neurone Q 
P1: has a long dendron but a short axon 
Has a short dendron but a long axon 
P2: cell body is located in the middle of the cell 
Cell body is located at the terminal / end of the cell 
(b) Diagram 8.2 shows the transmission of a nerve impulse from neurone P to neurone R. Rajah 8.2 menunjukkan pemindahan impuls saraf dari neuron P ke neuron R. 
Diagram 8.2 Rajah 8.2 Explain the transmission of a nerve impulse from neurone P to neurone R Terangkan pemindahan impuls saraf dari neuron P ke neuron R [6 marks]
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 22 
Sample answer: P1: when impulse reaches the synaptic knob / terminal / terminal axon / presynaptic membrane P2: it stimulates the synaptic vesicles P3: to release neurotransmitter P4: mitochondrion (in the synaptic terminal) produces energy / ATP P5: for active transport / transmission of the impulse P6: (neurotransmitter) diffuse across / into synaptic cleft / synapse to the next dendrite / neurone R / postsynaptic membrane P7: transmission of impulse from neurone P to neurone R is in the form of chemicals (c) Diagram 8.3a shows example of voluntary action and Diagram 8.3b shows example of involuntary action. By using pathway of transmission of information from receptors to effectors, explain similarities and difference between voluntary action and involuntary action. Rajah 8.3a menunjukkan contoh tindakan terkawal dan Rajah 8.3b menunjukkan contoh tindakan luar kawal. Dengan menggunakan laluan pemindahanmaklumat dari reseptor kepada efektor, terangkan persamaan dan perbezaan tindakan terkawal dan tindakan luar kawal. 
Diagram 8.3a Diagram 8.3b Rajah 8.3a Rajah 8.3b [10 mark]
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 23 
Sample answer: Pathway of transmission of information from receptors to effectors for voluntary action 
Pathway of transmission of information from receptors to effectorsfor involuntary action 
Similarities S1: both voluntary action and involuntary action involved receptor E1: to detect the stimulus and trigger impulse S2: both voluntary action and involuntary action involved 3 neuron E2: carry impulse from receptor to efector S3: both voluntary action and involuntary action involved efector 
receptor 
efector
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 24 
E3: contract to show the response Differences D1: Voluntary action is controlled by conscious thoughts but Involuntary action occurs automatically without any conscious control E1: because voluntary action Involves the integration and interpretationof information in the cerebrum but involuntary action involve spinal cord only D2: Voluntary action are under the control of the will of the Individual but involuntaryaction are not control by the will E2: because voluntary action involve the action of doing thing for activity but involuntary action involve the action to protect the person from danger
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 25 
9. (a) Diagram 9.1 and 9.2 shows the histogram about distribution of genetic variation in human. 
Rajah 9.1 dan 9.2 menunjukkan histogram mengenai taburan variasi genetik dalam manusia. 
Diagram 9.1 Diagram 9.2 
(i) With a suitable example, explain the differences of two kinds of variation. 
Dengan menggunakan contoh yang sesuai, terangkan perbezaan di antara kedua- dua variasi tersebut. [7 marks] Sample answer: Example of continuous variation: Height or weight Example of discontinuous variation: ABO blood group Differences: 
Continuous variation 
Discontinuous variation 
Graf distribution shows a normal distribution 
Graf distribution shows a discrete distribution 
The characters are quantitative / can be measured and graded (from one extreme to the other) 
The characters are qualitative / cannot be measured and graded (from one extreme to the other) 
Exhibits a spectrum of phenotypes with intermediate character 
Exhibits a few distinctive phenotypes with no intermediate character 
Influenced by environmental factors 
Is not Influenced by environmental factors 
Two or more genes control the same character 
A single genes determines the differences in the traits of the character 
The phenotype is usually controlled by many pair of alleles 
The phenotype is controlled by a pair of alleles
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 26 
(ii) What is the importance of variation to organism? Apakah kepentingan variasi kepada organisma? [3 marks] Sample answer: P1: variation provided better adaptation for organism to survive in the changing environment P2: variation are essential to the survival of species / to survive more successfully P3: variation be able to organism explore a new habitat P4: to ensure organism survival from predator
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 27 
(b) Diagram 9.3a, 9.3b and 9.3c shows the genetic factors that affected on the variation of organism. Rajah 9.3a, 9.3b dan 9.3c menunjukkan faktor-faktor genetik yang member kesan ke atas variasi pada organisma. 
Diagram 9.3a 
Diagram 9.3b Diagram 9.3c Explain how these factors in the diagram above will cause the variation among the organism. Terangkan bagaimana faktor-faktor dalam rajah di atas akan menyebabkan variasi dikalangan organisma. [10 marks] Sample answer: F1: meiosis P1: produce varies gamete with different genetic content P2: through homologous chromosomes random assortment during metaphase I 
Combination 4 
Gabungan4 
Combination 1 
Gabungan 1 
Combination 2 
Gabungan2 
Combination 3 
Gabungan3
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 28 
F2: crossing over P3: two homologous chromosomes are paired up / synapsis during prophase I P3: crossing over occurs between non-sister chromatids at the chiasma P4: chromatids break and rejoin in such a way that segments of chromatids are exchange // causing a genetic recombination P5: genes in the chromosomes is altered and gametes with various combinations of chromosomes are produced F3: Fertilization P6: random fertilization between sperm and ovum P7: produce zygote with varies genetic material End…
PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 1 
SECTION A [60 marks] Answer all questions in this section. Jawab semua soalan di bahagian ini 
1. 
Diagram 1 shows a microscopic structure of a part of pancreatic cell. Rajah 1 menunjukan struktur mikroskopik sebahagian sel pancreas. 
(a) 
(i) 
Name the organelle K and organelle L Namakan organel K dan organel L. K: Rough Endoplasmic Reticulum L : Golgi apparatus / body [2 marks] 
(ii) 
Explain how the organelle K and organelle L are interrelated in their function Terangkan bagaimana organel K dan organel L adalah saling berkaitan dari segi fungsi mereka. P1 : Rough endoplasmic reticulum /K transports protein synthesized in the ribosomes P2 : then forms a transport vesicle which carries the protein to Golgi body / L P3 : Golgi body processes, modifies the protein into a functional one / 
Diagram 1 
Rajah 1 
Note 
Catatan
PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 2 
enzyme / hormone (before forming a secretory vesicle) [2 marks] 
(b) 
(i) 
Name one chemical substance in the structure R which is involved in the synthesis of protien in a cell. Namakan satu bahan kimia dalam structur R yang terlibat dalam sintesis protien dalam suatu sel. The chemical substance in the chromosome - DNA / Deoxyribonucleic acid [1 marks] 
(ii) 
Draw the structure of the chemical compound in (b)(i) in the blank space below. Lukis struktur bahan kimia dalam (b)(i) dalam ruang kosong di bawah. DNA structure: 
Nitrogenous base DNA strand comprises phosphate and pentose sugar [2 marks] 
(c) 
(i) 
Based on the diagram, describe the synthesis of a specific pancreatic hormone in the cell. Berdasarkan gambarajah itu, huraikan bagaimana suatu hormon tertentu disintesiskan dalam sel itu The synthesis of hormone in the pancreas cell: P1 : The genetic information for the synthesis of the protein/ hormone (eg. Insulin contained in the DNA) is copied to RNA / messenger – RNA P2: (RNA) carries the information to ribosome P3: (Ribosome) synthesize the protein and 
Drawing = 1 mark 
2 correct labels = 1 mark
PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 
BIOLOGY 4551 | MODUL PERFECT SCORE SBP 3 
P4 : (Protein) is transfer to the Rough Endoplasmic Reticulum P5: then protein is transported to Golgi Apparatus P6: (In Golgi Apparatus) protein is modified to hormone // processed /packed and sorted / transport to plasma membrane [3 marks] 
(ii) 
The structure R in Diagram 1 undergoes some changes due to exposure to radioactive rays. Explain the possible effect to the synthesis of the hormone. Struktur R dalam Rajah 1 mengalami perubahan akibat pendedahan kepada sinaran radioaktif. Terangkan kemungkinan kesanya ke atas sintesis hormon itu. Effect of changes of the structure R / chromosome on hormone synthesis: P1: there will be some changes in the gene / base sequence / gene mutation responsible for the synthesis of the hormone P2: protein synthesis changes , a different protein / not the original hormone is synthesized or no hormone is being synthesized [2 marks] 
2. 
Diagram 2.1 shows two individual, P and Q in two different situations. P is in a vigorous activity while Q is at rest. Processes of R and S occurs in a human muscle cell. Rajah 2.1 menunjukkan dua individu, P dan Q dalam dua situasi yang berbeza. P sedang melakukan satu aktiviti cergas manakala Q berada dalam keadaan rehat. Proses R dan S berlaku dalam satu sel otot manusia.
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(a) 
Based on Diagram 2.1, name the processes R and S. Berdasarkan Rajah 2.1, namakan proses R dan S. Process R/ Proses R : Anaerobic respiration Process S/ Proses S : Aerobic respiration [2 marks] 
(b) 
Write the equation of process S. Tuliskan persamaan bagi proses S. 
[1 marks] 
(c) 
State two differences between process R and process S. Nyatakan perbezaan diantara proses R dan proses S. 
Process R/ Anaerobic respiration 
Process S/ Aerobic respiration 
Glucose + Water Carbon dioxide + Water + Energy 
Diagram 2.1 
Rajah 2.1
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Glucose is broken down partially 
Glucose is broken down completely 
ATP produced – 2 ATP 
ATP produces – 38 ATP 
Product – Lactic acid and energy 
Product – Carbon dioxide , water and energy 
Occurs in the absence of oxygen 
Occurs in the presence of oxygen 
[2 marks] 
(d) 
Diagram 2.2(a) shows fish respiratory structure and Diagram 2.2(b) shows human respiratory structure. 
Rajah 2.2(a) menunjukkan struktur respirasi ikan dan Rajah 2.2(b) menunjukkan struktur respirasi manusia. 
(i) 
What is X? Apakah X? 
Answer : Gills [1 mark] 
(ii) 
Structure X has adaptation for good gases exchange in fish. Explain one adaptation of X. 
Answer: 1. A :(Gills ) have lamellae E :to increase total surface area (for gases exchange) 
Diagram 2.2(a) 
Rajah 2.2(a) 
Diagram 2.2(b) 
Rajah 2.2(b) 
X
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2. A : (Gills) have network of blood capillary E : to transport gases rapidly 3. A : Counter current of blood and water E: increase diffusion of gases in and out of the gills 4: A :Numerous in number E: to increase total surface area (for gases exchange) [2 marks] 
(e) A man is a heavy smoker. Explain how this habit affect the efficiency of gases exchange on the respiratory structure in Diagram 2.2(b). Seorang lelaki adalah seorang perokok tegar. Terangkan bagaimana tabiat ini mempengaruhi kecekapan pertukaran gas pada struktur respirasi dalam Diagram 2.2(b). 
Answer: P1: Tobacco smoke contain tar P2: (Tar) deposit on the surface of alveolus P3 : Tobacco smoke contain heat P4 : reduce moisture on the surface of alveolus P5 : Tobacco smoke contain NO2 P6 : increase acidity / corrode the surface of alveolus [4 marks]
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3. 
Diagram 3.1 shows the sequence of hydrolysis of starch to molecules P and molecule Q by enzymes. Rajah 3.1 menunjukkan urutan hidrolisis kanji kepada molekul P dan molekul Q oleh enzim. 
(a) 
(i) 
Complete Table 3.1. Lengkapkan Jadual 3.1 
Molecule Molekul 
Name of molecule Nama molekul 
Name the enzyme involved in hydrolysis Namakan enzim yang terlibat dalam hidrolisiss 
P 
Maltose 
Amylase 
Q 
Glucose 
Maltase 
[4 marks] 
(ii) 
Based on your biological knowledge, explain the effect of consuming food that 
Enzyme 
Enzim 
Starch 
Kanji 
+ water 
air 
+ water 
air 
Molecule P 
Molekul P 
Enzyme 
Enzim 
Molecule Q 
Molekul Q 
Diagram 3 
Rajah 3 
Table 3.1 / Jadual 3.1
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contain excessive of starch on health. Berdasarkan pengetahuan biologi anda, terangkan kesan pengambilan makanan yang mengandungi kanji berlebihan keatas kesihatan. 
Answer : P1 : Glucose level in blood increase/ hyperglycemia P2 : Starch (finally) is digested into glucose P3 : Glucose is absorbed into blood (capillary) P4 : causes diabetes mellitus / obesity [2 marks] 
(b) 
Table 3.2 shows the energy value and nutrient content in a few types of food taken by student. Jadual 3.2 menunjukkan nilai tenaga dan kandungan nutrient di dalam beberapa jenis makanan yang diambil oleh seorang pelajar. 
Food Makanan (/100g) 
Energy Tenaga (kJ) 
Carbohydrate Karbohidrat (g) 
Fats Lemak (g) 
Protein Protein (g) 
Vitamin C Vitamin C (μg) 
Rice Nasi 
1530 
86.8 
1.0 
6.4 
0.0 
Fish Ikan 
320 
0.0 
0.5 
17.5 
0.0 
Egg Telur 
612 
0.0 
10.9 
12.4 
0.0 
Orange Oren 
150 
8.5 
0.0 
0.8 
50 
(i) 
Based on Table 3.2, which type of food supplies the most energy? 
Table 3.2/ Jadual 3.2
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Berdasarkan Jadual 3.2, jenis makanan manakah yang membekalkan paling banyak tenaga? 
Answer: Rice [1 mark] 
(ii) 
Which type of food should be taken regularly to prevent scurvy? Jenis makanan manakah yang perlu kerap diambil untuk mengelakkan penyakit skurvi? 
Answer: Orange [1 mark] 
(iii) 
Calculate the amount of energy obtained by the student if he eats a meal which contain 200 g rice and 150g fish. Kirakan jumlah tenaga yang diperolehi oleh pelajar tersebut jika dia mengambil 200 g nasi dan 150 g ikan. 
Answer: Rice - 1530 kJ x 2 = 3060 kJ Fish - 320 kJ x 1.5 = 480 kJ Total : 3060 + 480 = 3540 kJ [3 marks] 
(c) 
Why does an egg produces double amount of energy compared to a fish? Mengapakah sebiji telur menghasilkan jumlah tenaga dua kali ganda berbanding seekor ikan? 
Answer: Egg contain more fat/lipid than fish [1 mark]
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4. 
Diagram 4.1 shows a longitudinal section of the reproductive parts of a flower during fertilization. 
Rajah 4.1 menunjukkan keratan memanjang bahagian pembiakan bunga semasa persenyawaan. 
(a) 
(i) 
In Diagram 4.1, label P,Q, R and S Pada Rajah 4.1, labelkan P, Q, R dan S [2 marks] 
(ii) 
In the space below, draw a section through the ovule showing all the cells in S. Label the cells involved in fertilization. Dalam ruang di bawah , lukiskan keratan melalui ovul menunjukkan semua sel- sel dalam S. Labelkan sel-sel yang terlibat dalam persenyawaan 
Pollen tube 
Male gamete nucleus 
Embryo sac 
Ovary 
Diagram 4.1 
Rajah 4.1
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Drawing: clear diagram with 8 nucleus – 1 mark Label : 2 label = 1 mark 
(iii) 
Describe the fertilization process that occurs. Huraikan proses persenyawaan yang berlaku . [2marks] P1 :One of the Q/ male nucleus fertilizes an egg to form the diploid zygote P2:One of the Q/ male nucleus fertilizes 2 polar nuclei to form the triploid endosperm 
(b) 
(i) 
In Diagram 4.1, the structure Y has to be kept dormant for future research purposes. Explain how Y can be prevented from germinating. Dalam Rajah 4.1, struktur Y perlu disimpan tidak aktif untuk tujuan penyelidikan pada masa hadapan. Terangkan bagaimana Y boleh dihalang daripada bercambah . [2 marks] P1 : Keep Y in dry place/ low temperature P2 : Because moisture initiate germination// enzyme is in inactive state 
(ii) 
Suggest one method to stimulate the germination of Y Cadangkan satu kaedah untuk merangsang percambahan Y. [1mark] Dropping/ spraying sucrose / sugary solution on Y 
(e) 
Diagram 4.2 shows a watermelon with seed and watermelon without seed.. Rajah 4.2 menunjukkan buah tembikai dengan biji dan buah tembikai tanpa biji. 
Polar cell 
Egg cell
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Nowadays, it is more easier to find seedless watermelon in market. Shoppers also can find varieties of seedless oranges, grapes, and cucumbers. Explain how to produce varieties of fruits without seed. Pada masa kini, adalah lebih lebih untuk mencari tembikai tanpa biji di pasaran. Pembeli juga boleh mendapatkan buah limau , anggur dan timun tanpa biji . Terangkan bagaimana untuk menghasilkan buah tanpa biji . [3marks] P1 : Parthenocarpy P2 : Spraying flower with auxin, P3 : stigma and anther becomes degenerate P4 : auxin diffuse into ovary and stimulate ovary to develop. 
5 
Diagram 5.1 shows a uriniferous tubule and its associated blood vessels. Diagram 5.2 shows cells from structure P as seen through an electron microscope. Gambarajah 5.1 menunjukkan tubul uriniferus dan salurdarah yang berkaitan. Gambarajah 5.2 menunjukkan struktur sel P seperti yang dilihat di bawah mikroskop elektron. 
Diagram 4.2 / Rajah 4.2 
Diagram 5.1 
Rajah 5.1 
Y 
P 
X 
Q 
Blood 
Darah 
Blood 
Darah 
Blood vessel 
Salur darah
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(a) 
State the difference in the urea composition between the blood vessel X and Y. Nyatakan perbezaan kandungan urea antara salur darah X dan Y. Urea concentration is lowest in Y but higher in X. [1 mark] 
(b) 
Based on the Diagram 5.2 explain how the cells are structured for reabsorption of substances. Berdasarkan Gambarajah 5.2, terangkan bagaimana sel distrukturkan untuk penyerapan semula bahan. P1 :They have many/abundant mitochondria P2 : Produce a lot of energy needed for active transport or P1 :Numerous/many microvilli P2 : Increase total surface area for reabsorption [2 marks] 
(c) 
Table 5.1 shows the concentration of certain substances found in structure Q. 
Diagram 5.1 
Rajah 5.1 
Diagram 5.2 
Rajah 5.2
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Jadual 5.1 menunjukkan kepekatan beberapa bahan yang terdapat di dalam struktur Q. 
Substances Bahan 
Water Air 
Protein Protein 
Glucose Glukosa 
Urea Urea 
Salts Garam 
Concentration Kepekatan (%) 
95.0 
0.00 
0.00 
2.00 
1.50 
Explain how the concentration of the substances present in Q would change after eating meat and eggs. Terangkan bagaimana kepekatan bahan-bahan yang terdapat dalam Q akan berubah selepas memakan daging dan telur. P1 :meat and eggs contains high protein/ main source of amino acid P2:(Excess) amino acids are deaminated / converted into ammonia / urea in the liver P3 :The urea is transported to the kidneys and removed as urine P4 :The concentration of urea in the urine would increase then 2.00 [3 marks] 
(d) 
Diagram 5.2 shows the flow of blood and dialysis fluid through a dialysis machine. Rajah 5.2 menunjukkan aliran darah dan bendalir dialisis melalui suatu mesin dialisis. 
Table 5.1 / Jadual 5.1
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DIAGRAM 5.2/ RAJAH 5.2 Tube P contain more nitrogenous waste compared to tube Q. Explain why. Tiub P mengandungi banyak bahan kumuh bernitrogen berbanding tiub Q. Terangkan mengapa. P1 - The concentration of dialysis fluid is maintained at a concentration similar to the blood plasma of healthy person P2 - the concentration of nitrogenous waste / urea / salt in P higher than dialysis fluid P3 - urea / salt diffused out into dialysis fluid P4 - through semi-permeable tubing (Any 3) [3 marks] 
(e) 
Explain the importance of kidney in maintaining human health. Terangkan kepentingan ginjal dalam mengekalkan kesihatan manusia. 
 To eliminate waste materials / urea / toxics / excess water / salts 
from the blood. 
 Maintaining normal osmotic pressure in the blood / constant 
internal environment. 
 Ensure an optimal physical / chemical condition (in the internal 
environment). [3 marks] 
Section B Bahagian B [40 markah]
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Answer all question from this section Jawab semua soalan dalam bahagian ini. 
6(a) 
Diagram 6.1 shows a forearm of humans. Rajah 6.1 menunjukkan lengan atas manusia. 
SKEMA JAWAPAN 
6(a) 
Similarities: S1 : Both joint has cavity filled with synovial fluid// lines with synovial membrane S2 : (synovial fluid) act as lubricant to reduce friction between two bones. S3 :The end surface of bone are covered with cartilage 
A joint is the location at which bones connect. They are constructed to allow movement and provide mechanical support. Explain the similarity and difference between joint S and T? Sendi adalah tempat di mana tulang-tulang bertemu. Sendi dibina untuk membolehkan pergerakan dan member sokongan mekanikal berlaku. Terangkan persamaan dan perbezaan antara sendi S dan T? [6 marks/6 markah] 
Joint S 
Sendi S 
Diagram 6.1 
Rajah 6.1 
Joint T 
Sendi T
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S4 : (Cartilage) to protect the bone /reduce friction between the bones S5 : Both joint are connected with ligament S6 : (Ligament) allow movement/ avoid dislocation of bone during movement Differences: 
Joint S 
Joint T 
D1 
Hinge joint /Elbow joint 
Ball and socket joint/Shoulder joint 
D2 
Allow the movement in one plane 
Allow rotation movement // all direction movement 
D3 
Articulation between humerus, ulna and radius 
Articulation between humerus,scapula and clavicle. 
(b) 
Based on your biological knowledge, discuss the statement above. Dengan menggunakan pengetahuan biologi anda, bincangkan pernyataan di atas. [6 marks/6 markah] 
SKEMA JAWAPAN 
(b) 
P1 : problem / disease : arthritis/gout P2 : (diet high protein intake) cause accumulation of uric acid in the joint P3: inflammation at joint // joint become stiff and pain P4 : Lack of exercise P5 : Diet lack of calcium / vitamin D P6 : reduce the mass of bone //bone become lighter P7: practice wrong posture during activity 
A man has swollen ankle and is painful during movement after having a habit of taking high protein diet and practicing unhealthy lifestyle. 
Seorang lelaki mengalami bengkak pada buku lali dan berasa sakit ketika bergerak setelah mengamalkan pengambilan diet yang tinggi kandungan yang protein dan tidak mengamalkan gaya hidup sihat.
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(c) 
Diagram 6.2 show an earthworm and structure in its body. 
Rajah 6.2 menunjukkan seekor cacing tanah dan struktur pada badannya. 
Explain how the structure in the earthworm involve in their movement . Terangkan bagaimana struktur pada cacing tanah terlibat dengan pergerakannya. [4 marks/4 markah] 
SKEMA PEMARKAHAN 
(c) 
P1 : Hydrostatic skeleton P2 : fluid in the body cavity helps the earthworm to move / give support P3 : muscle at the body wall are longitudinal and circular muscle //antagonistic muscle P4 : contraction of circular muscle( and relaxation of longitudinal muscle) cause segment of body extended/longer/thinner P5 : contraction of longitudinal muscle (and relaxation of circular muscle) cause segment of body shorten/ thicken P6 : (The presence of) chaetae P7 : secure/anchor the shorted segment on the ground [4 marks] 
P8 : give pressure to skeleton system [6 marks] 
Diagram 6.2 /Rajah 6.2
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(d) 
Diagram 6.3 shows flight muscle of a bird. 
Rajah 6.3 menunjukkan otot penerbangan seekor burung. 
Explain the effect to locomotion of bird if structure W is torn. Terangkan kesan terhadap pergerakan burung kastuktur W terkoyak. [4 marks/4 mark] 
SKEMA PEMARKAHAN: 
(d) 
P1: W is tendon P2 : Tendon is inelastic /strong/ tough P3 : Function of tendon is to connect (pectoralis minor) muscle to bone (/humerus) P4 : Contarction of (pectoralis minor) muscle produces (pulling) force P5 : (If tendon is torn), (pulling) force (that produced by contraction of muscle) 
W 
Pectoralis minor muscle 
Ototpektoralis minor 
Humerus 
Humerus
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cannot be transferred to the bone P5 : the bone/humerus is not pulled upward P6 : no movement of wing [4 marks] 
7 Diagram 7.1 shows part of the blood circulatory system and the lymphatic system in the human body. Rajah 7.1 menunjukkan system peredaran darah dan sistem limfa dalam badan manusia. Diagram 7.1 Rajah 7.1 
(i) Explain the differences between the composition of fluid P and fluid Q
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Terangkan perbezaan antara komposisi bendalir P dan bendalir Q. [4 marks] [4 markah] 
(a) (i) 
Able to explain the diffrences of composition fluid P and fluid Q Sample answer: F1: Fluid Q/lymph has a larger numbers of lymphocyte compare to fluid P/blood P1: lymphocyte is produced by the lymph nodes in lymph system F2: Fluid Q/lymph has lower contents of oxygen compare to fluid P/blood P2: oxygen has been used up by the cell 
1 1 1 1 
4 
(ii) Describe how the fluid Q is formed from blood until it is incorporated back into the blood circulatory system. Huraikan bagaimana bendalir Q terbentuk daripada darah sehingga bendalir tersebut masuk semula ke dalam sistem peredaran darah. 
(a) (ii) 
Able to describe how lymph is formed from blood until it is brought back into the blood circulatory system. Sample Answer : P1: (When the blood flows from arteries into capillaries) there is higher hydrostatic pressure at the arterial end of the capillaries P2: (This high pressure) forces some plasma to pass through the capillary walls into the intercellular spaces (between the cells) P3: Once the fluid leaves the capillary walls, it is called interstitial/tissue fluid // The interstitial fluid fills the spaces between the cells and constantly bathes the cells P4: 90% of the interstitial fluid diffuses back into blood capillary P5: 10% of the interstitial fluid that has not been reabsorbed into the bloodstream goes into the lymph capillaries.(Once inside the lymph capillaries) the fluid is known as lymph. 
1 1 1 1 1 1 
6
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[6 marks][6 markah] 
(c) (i) Describe how are lacteals in the villi related with the lymphatic system? 
Huraikan bagaimana lakteal di dalam vilus dapat dikaitkan dengan sistem limfa? 
(c) (i) 
Able to describe the how are lacteals related with the lymphatic 
P6: The lymph capillaries unite to form larger lymphatic vessels. P7: From the lymphatic vessels, lymph eventually passes into the thoracic duct/the right lymphatic duct. P8:The thoracic duct empties its lymph into the right subclavian vein. (Hence, lymph drains back into the blood). Any 6 P 
1 1 Max 6
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system. Sample answer: P1: A lacteal is a lymphatic capillary P2: absorbs fatty acids and glycerol in the villi of the small intestine P3: lacteals merge to form larger lymphatic vessels P4: that transport the fats to the thoracic duct which empties into the left subclavian vein. 
1 1 1 1 
4 
(ii) Helmie takes fried chicken at lunch. Explain the absorption and assimilation process of lipid content in the fried chicken. Helmie mengambil ayam goreng semasa makan tengah hari. Huraikan proses penyerapan dan asimilasi lemak yang terkandung dalam ayam goreng tersebut. [6 marks] 
(c) (i) 
Able to explain the absorption and assimilation of lipid Absorption P1: Digestion of lipid produce fatty acid and glyserol P2: Absorption of lipid occur at ileum P3: At ileum there are villi which have lacteal P4: Fatty acid and glyserol are absorbed into lacteal P5: In the lacteal condensation of fatty acid and glyserol forms lipid P6: The lipids then transported via the subclavian vein into the blood steam Assimilation P7: In the cells lipid is use as a main component of plasma membrane P8: lipid also is use as a main component of some hormone and vitamin P9: Excess lipid will be stored underneath the skin as adipose tissue 
1 1 1 1 1 1 1 1 1 1 
max 6
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8(a) Diagram 8.1 shows the process of colonisation and succession in a habitat. Rajah 8.1 menunjukkan proses pengkolonian dan penyesaran dalarn suatu habitat. 
Diagram 8.1 What is meant by "colonisation and succession in a habitat”? Based on Diagram 8.1 .explain how colonisation and succession bring about the formation of the primary forest. [10 marks] Apakah yang dimaksudkan dengan "pengkolonian dan penyesaran dalam suatu habitat"? Berdasarkan Rajah 8.1, terangkan bagaimana pengkolonian dan penyesaran membawa kepada pembentukan hutan primer dalam suatu habitat. [ 10 markah] 
8.0 
F1: COLONISATION The process whereby living organisms move into this newly formed area which is completely devoid of life. 
1 
Rhizophora sp 
Sonneratia sp 
Bruguiera sp 
Low 
tidal 
High tidal
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F2: SUCCESSION The gradual process where one community changes its environment so that it is replaced by another community. Zone 1 / Avicennia sp. and Sonneratia sp. P1: The pioneer species in a mangrove swamp are the Avicennia sp. and Sonneratia sp. P2: The Avicennia sp. grows in the part of the mangrove swamp that faces the sea while Sonneratia sp. grows at the mouth of the river which is sheltered. P3: A root system that spreads out widely to give support to the trees in the soft muddy soil P4: The Avicennia sp. and Sonneratia sp. have asparagus-shaped pneumatophores that grow vertically upwards from the main roots through the mud into the air. P5: The widely spread roots of the Avicennia sp. and Sonneratia sp. trap mud. P6: As more and more mud accumulate, the bank is slowly raised and would then contain less water. P7: The mangrove swamp is now more suitable for another mangrove tree which is the Rhizophora sp. Hence the Rhizophora sp. as the successor will slowly replace the pioneer species. 
Zone 2 / Rhizophora sp. zone P8: This zone is higher and less waterlogged. P9: The Rhizophora sp. has prop roots to support and anchor the tree in the soft muddy soil. P10: The Rhizophora sp. has viviparity seeds to ensure that the seedlings can grow and are not carried away by the seawater. P11: The prop roots of the Rhizophora sp. are able to trap mud. The pioneer species and the Rhizophora sp. that are old, will die and decay, adding humus to the soil. P12: The banks are raised up even higher. The soil becomes more solid/compact, more fertile and less saline. 
1 1 1 1 1 1 1 1 1 1 1 1 1 1
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(b) Diagram 8.2 shows sources of greenhouse gases arising from human activities and natural processes. Rajah 8.2, menunjukan sumber-sumber gas rumah hijau yang dihasilkan daripada aktiviti manusia dan proses-proses semulajadi. P13: The soil that is harder and drier now is not suitable for the Rhizophora sp. Hence, the Rhizophora sp. is replaced by the Bruguiera sp. Zone 3 / Bruguiera sp. zone P14: Trees of Bruguiera sp. grow well in hard clay soil that subjects to flooding during high tide. P15: Trees of Bruguiera sp. have buttress roots for support and knee- shaped pneumatophores for gaseous exchange (Figure 8.24(c)). P16: As more sedimentation of decayed substances occur, new mud banks are being built up seawards while the old banks move further inland, away from the sea. P17 : The soil becomes harder and dry land is formed. P18: Bruguiera sp. are replaced by other types of plants such as coconut trees, palm trees and Pandanus sp. P19: These are later replaced by other land plants. P20: Finally, after a few hundred years, the process of succession stops and a tropical rain forest, which is the climax community, is formed. At least one point in each zone. [10 marks] 
1 1 1 1 1 1 Max 10
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Diagram 8.2 Rajah 8.2 Based on Diagram 8.2, Explain the green house effectand global warming as a result of human activities. Berdasarkan Rajah 8.2, terangkan kesan rumah hijau dan pemanasan global akibat aktiviti- aktiviti manusia [5 marks]. 
b) The formation of greenhouse effect caused by; P1: Solar radiation , containing uv rays penetrate earth atmosphere and reaches the earth surface P2: Part of uv rays is reflected back by earth's surface to atmosphere in the form of infrared radiation / light which contains heat P3: Heat (energy) is trapped by greenhouse gases (such as carbon dioxide, oxides of nitrogen, methane) P4: Human activities such as combustion of fossil fuel by factories and vehicles increase the amount of greenhouse gases P5: Higher concentration of greenhouse gases in the atmosphere results in more heat being absorbed / trapped P6: Extensive forest burning, burning of fossil fuel and higher rate of evaporation worldwide causes accumulation of great amount of water vapour in the air P7: which also contribute to the increase in the earth's temperature / causes global warming 
1 1 1 1 1 1 1 Max 5 
8 (c) 
Diagram 8.3 shows the emission of gases from factories. Rajah 8.3 menunjukkan pelepasan gas daripada kilang. 
Nitrogen oxide (NO) ,sulphur dioxide (SO2) 
Nitrogen oksida(NO), sulfur dioksida (SO2) 
Pond 
Kolam 
Forest 
Hutan
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(i) 
Explain the effects of the emission of the gases to the ecosystem. Terangkan kesan pembebasan gas tersebut kepada ekosistem. [5 marks] 
SKEMA PEMARKAHAN 
8(c) 
P1: (The release of nitrogen oxides / sulphur dioxides) leads to the formation of acid rain P2 : the gases dissolve in the rain water P3 : Acid rain causes damage on the leaves / chloroplast P3 : Lower rate of photosynthesis P4 : Leads to stunted growth / death of plants//population reduced P5 : Acid rain lowers pH of the pond// more acidic P6 : causes death to aquatic organisms /fishes P7 : pH of soil lower//more acidic P8 : crop yield decrease [max : 5 marks] 
9.(a) Diabetics do not correctly produce or use their insulin hormone. The insulin hormone helps control how much sugar is in your bloodstream. Millions of diabetics need to take insulin. Insulin from cows and pigs has been used since the early 1900s to treat diabetes. Now human insulin hormone can be mass-produced through genetic engineering processes. Pesakit kencing manis tidak dapat menghasilkan atau menggunakan insulin dengan betul. Hormone insulin membantu mengawal kandungan gula dalam aliran darah . Berjuta-juta pesakit kencing manis perlu mengambil insulin. Insulin daripada lembu dan babi telah digunakan seja kawal 1900-an untuk merawat kencing manis. Sekarang hormon insulin manusia boleh dihasilkan secarabesar-besaran melalui proses kejuruteraan genetik. 
Diagram 9.1shows a few stage that involves in the production of insulin hormone through genetic engineering technology. Rajah 9.1 menunjukkan sebahagian daripada peringkat yang terlibat dalam proses penghasilan hormon insulin melalui teknologi kejuruteraan genetik. 
Diagram 8.3/Rajah 8.3
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By using your knowledge , explain how this technology can be used in insulin hormone production. Berdasarkan pengetahuan biologi anda huraikan bagaimana teknologi ini dijalankan bagi menghasilkan hormone insulin. [6 marks/ 6 markah] 
SKEMA JAWAPAN 
9(a) 
P1 -The gene for the insulin is isolated from human pancreas cell P2 - The bacterial plasmid is isolated (DNA found in bacteria) P3 - The bacterial plasmid is cut by using enzyme P4 - The enzyme used to incorporate gene for insulin production into the plasmid P5 - the bacteria are cultured in bioreactor P6 - the plasmid replicate as a bacteria divide asexually . P7 - the bacteria can produce insulin in large quantity, purified and isolate. 
(b) 
Genetic engineering (GE) is the manipulation of genetic material (DNA or genes) in a cell or an organism in order to produce desired characteristics and to eliminate unwanted ones. GE includes a range of different techniques with many different uses, 
Diagram 9.1 
Rajah 9.1
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and can be applied to plants, animals and humans. For example, the genetic modification of food is a form of GE that involves manipulating the cells of plants. Kejuruteraan genetik (GE) adalah manipulasi bahan genetik (DNA atau gen) di dalam selatau organism untuk menghasilkan ciri-ciri yang dikehendaki dan untuk menghapuskan organisma yang tidak diingini. GE termasuk pelbagai teknik yang berbeza dengan kegunaan yang berbeza, dan boleh digunakan untuk tumbuh- tumbuhan, haiwan dan manusia. Sebagai contoh, pengubahsuaian genetic makanan adalah satubentuk GE yang melibatkan memanipulasi sel-sel tumbuh-tumbuhan. 
Diagram 9.2 shows twotoma to leaves which have been exposed to a bacterial pathogen, Pseudomonas syringae. Leaf A is the normal leaf show disease when infected with the bacteria while Leaf B, the genetically engineered leaf shows practically no signs of damage. Rajah 9.2 menunjukkan dua helai daun tomato yang telah didedahkan kepada sejenis pathogen bacteria ,Pseudomonas syringae. Daun A adalah daun bias yang menunjukkan tanda-tanda penyakit setelah dijangkiti oleh bacteria tersebut, manakala daun B yang telah mengalami pengubahsuaian kandungan genetiknya tidak menunjukkan tanda kerosakan. 
Discuss the benefits and the risks of using the genetically engineered organisms in agriculture and food production. Bincangkan faedah dan risiko menggunakan organism yang terubah suai kandungan genetiknya dalam pertanian dan penghasilan makanan. [10 marks/10 markah] 
Diagram 9.2/ Rajah 9.2 
Leaf A /Daun A 
Leaf B /Daun B
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SKEMA PEMARKAHAN 
9(b) 
The benefit: P1: Genetic engineering used to produce disease resistant /pest resistant plant. P2 : Less pesticide are used P3 : Less pollution to the environment //better health for consumer P4: Increase crop yield P5 : help to solve problem of insufficient food P6 :better livehood for farmer P7 : Increase resistance to herbicide P8 : which allow weeds to be killed without affecting the crop plant P9 : Able to survive on poorer quality grassland P10 : can resist drought // climatic changes P11 : create crops with better nutrition value P12 : with high vitamin A / protein content P13 : help to solve problems of malnutrition P14: create crops with longer shelf live P15: less food wastage P16: genetically modified livestock (eg :cow) P17: produce meat with less fat / more milk [Max : 6 marks] The risk: K1 : Pest resistant genes may be transferred to weed K2 : may be difficult to control the growth of weed K3: some transgenic crops may have animal genes K4: this may not be acceptable to certain groups for religious reasons K5: genetically modified food may be harmful to health K6: may activate human genes to cause cancer K7: transgenic organism may affected the survival of other organism in the ecosystem K8: may cause the imbalance of nature / ecosystem K9 : decrease biodiversity K10: certain cultivar are being planted to the exclusion of others K11: this will less the genetic variation in environment. [Max: 4 marks] 
(c) 
a. 
b. 
c. 
d. 
e. 
f. 
g. 
h. 
i. 
j. Diagram 9.3 shows a cloning process of a plant.
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k. Rajah 9.3 menunjukkan proses pengklonan satu tumbuhan. 
Explain the characteristic of cloned plant. Terangkan ciri-ciri tumbuhan yang diklon. [4 marks] 
SKEMA PEMARKAHAN 
9(C) 
P1: Clones are genetically identical to the parent cell P2: no exchange of genetic materials P3: Clones have the same chromosomal number as the parent cell P4: no reduction in the chromosomal number P5: Clones easily get disease // shorter life span P6: Clones have the same body resistance against disease 
End .. 
Leaf cells form calluses in culture medium 
Sel daun membentuk kalus di dalam medium kultur 
Calluses develop into tiny plantlets 
Kalus berkembang menjadi anak pokok kecil 
Cloned plants 
Tumbuhan klon 
Leaf cells are taken from the parent plant 
Sel daun diambil dari tumbuhan induk 
Diagram 9.3 
Rajah 9.3
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 1 
QUESTION 1: 
1. A group of students carried out an experiment to study the effect of intraspecific competition on the growth of maize seedlings. 
Sekumpulan pelajar menjalankan satu eksperimen untuk mengkaji kesan persaingan intraspesifik ke atas pertumbuhan biji benih jagung. Diagram 1 shows the apparatus set-up of the experiment Rajah 1 menunjukkan susunan radas untuk eksperimen tersebut 
Step 1 : Three seedlinging trays are filled with 4 kg of garden soil. Langkah 1 : Tiga kotak semaian diisikan dengan 4 kg tanah kebun. Step 2 : The trays are labeled as P, Q and R. Langkah 2 : Kotak-kotak semaian dilabelkan P, Q dan R. Step 3 : In tray P, 50 maize seedlings are seedlinged at a distance of 15 cm intervals. In tray Q, 50 maize seedlings are seedlinged at a distance of 10 cm intervals. In tray R, 50 maize seedlings are seedlinged at a distance of 5 cm intervals. Langkah 3 : Dalam kotak P, 50 anak benih jagung ditanam pada jarak 15 cm berselang seli. Dalam kotak Q, 50 anak benih jagung ditanam pada jarak 10 cm berselang seli. Dalam kotak R, 50 anak benih jagung ditanam pada jarak 5 cm berselang seli. Step 4 : Each tray is watered daily with the same amount of water for 30 days. Langkah 4 : Setiap kotak semaian disiram tiap-tiap hari dengan jumlah air yang sama banyak untuk 30 hari. Step 5 : After 30 days, remove 30 maize seedlings randomly from tray P, tray Q and tray R. The root of seedlings are washed and wipedry. Langkah 5 : Selepas 30 hari, 30 anak benih jagung secara rawak dari kotak P, kotak Q dan kotak R. Akar anak benih dibersihkan dan dilapkan sehingga kering. Step 6 : The dry weight of the maize seedlings is recorded in Table 1. Langkah 6 : Berat kering anak benih jagung dicatatkan dalam Jadual 1. 
O 0 0 
O 0 0 
0 0 0 
0 0 0 
0 0 0 
0 0 0 0 
0 0 0 0 
0 0 0 0 
0 0 0 0 
0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
Tray P/kotak P 
Tray Q /kotak Q 
Tray R/kotak R 
Diagram 1/rajah 1
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Distance between maize seedlings / jarak antara anak benih jagung (cm) 
Dry weight of 30 maize seedlings (g) / berat kering 30 anak benih jagung (g) 
15 
10 
200 
150
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BIOLOGY 4551 MODUL PERFECT SCORE SBP 3 
5 
Table 1/Jadual 1 (a) Record the dry weight of the maize seedlings in the boxes provided in Table 1 Rekodkan berat kering anak benih jagung di dalam kotak yang disediakan dalam jadual 1. (b) (i) Based on the results in Table 1 , state two observations that can be made from this experiment. Berdasarkan keputusan di dalam Jadual 1, nyatakan dua pemerhatian yang dapat dibuat daripada eksperimen ini Observation 1/pemerhatian 1: At distance 15 cm, the dry weight of 30 paddy seedlings is 200g Observation 2/pemerhatian 2: At distance 5 cm, the dry weight of 30 paddy seedlings is 100g [3 marks] (ii) State the inference from the observations in (b) (i). Nyatakan inferens berdasarkan pemerhatian di (b) (i) Inference from observation 1/inferen dari pemerhatian 1: (At distance 15 cm), there is low intraspecific competition so the growth rate of maize plant is high Inference from observation 2/inferen dari pemerhatian 2: (At distance 5 cm), there is highintraspecific competition so the growth rate of maize plant is low [3 marks] 
100
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(c) Complete table 2 based on the experiment. Berdasarkan eksperimen, lengkapkan jadual 2 di bawah 
[3 marks] (d) State the hypothesis for this experiment. Nyatakan hipotesis bagi eksperimen ini The further the distance between the maize seedlings, the higher the growth rate of/dry weight of maize plant//vice versa [3 marks] (e) Construct a table and record all your data collected in the experiment which include the following aspects : Bina satu jadual untuk merekodkan semua keputusan eksperimen meliputi aspek berikut : 
 Distance between maize seedlings/jarak antara anak benih jagung 
 Dry weight of 30 maize seedlings/berat kering 30 anak benih jagung 
 Growth rate of maize seedling/kadar pertumbuhan anak benih jagung 
Growth rate = Dry weight of maize seedlings Number of days Kadar pertumbuhan = Berat kering anak benih jagung Bilangan hari 
Variable Pembolehubah 
Particulars to be implemented Cara mengendalikan pembolehubah 
Manipulated/ manipulasi: Distance between maize seedlings 
Used different distance between maize seedling//used the distances at 15cm, 10cm and 5cm 
Responding / bergerakbalas: Dry weight of maize seedlings// growth rate 
Record dry weight of maize seedlings by using a weight balance // calculate the growth rate using formula Growth = Dry weight of 30 maize plant 30 days 
Controlled/ dimalarkan Volume garden soil // type of maize plant // size of tray 
Fix the volume of garden soil at 4kg // fix the same type of maize plant // fix the same size of tray 
Table 2// Jadual 2
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Distance between maize seedlings (cm) 
Dry weght of 30 maize seedlings (g) 
Growth rate (g/days) 
15 
200 
6.67 
10 
150 
5.00 
5 
100 
3.33 
[3 marks] (f) Use the graph paper provided on page 55 to answer this question. Using the data in 1 (e)draw a graph of the growth rate of maize seedlings against the distance between the maize seedlings. Dengan menggunakan kertas graf yang dibekalkan pada muka surat 55 untuk menjawab soalan ini. Dengan menggunakan data di dalam 1 (e), lukiskan graf kadar pertumbuhan anak benih jagung melawan jarak di antara anak benih jagung [3 marks] (g) Based on graph in 1 (f), explain the relationship between the growth rate of maize seedlings and distance between seedling Berdasarkan graf di 1 (f), terangkan hubungan di antara kadar pertumbuhan anak benih jagung dan jarak antara anak benih. P1: As the distance between maize increases, the growth rate of maize seedlings increases. P2: This is because there is lower intraspecific competition P3: causes the dry weight of maize seedlings increase [3 marks] (h) If the experiment is repeated by increasing the distance between the maize seedlings to 20 cm, predict the observation . Explain your prediction.. Jika eksperimen diulang dengan meningkatkan jarak di antara anak benih jagung pada 20 cm,ramalkanpemerhatian. Terangkan ramalan anda. P1: The dry weight more than 200g P2: because longer distance give more water/ nutrient / space to the maize seedling, P3: so the growth rate of maize seedlings increases [3 marks] (i) Based on this experiment, what can you deduce about intraspecific competition? Berdasarkan eksperimen ini, apakah yang dapat anda rumuskan tentang persaingan intraspesifik? P1: Intraspecific competition is the growth maize plants when it compete between themselves P2: and shown by the dry weight of maize seedling. P3: The growth rate of maize is affected by the distance between the seedlings [3 marks]
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(j) When resources are limited supply, organisms living in the same habitat will compete for the same resources. The following is a list of the resources. Apabila sumber-sumber menjadi terhad, organisma hidup di habitat yang sama akan bersaing untuk sumber yang sama. Berikut ialah senarai sumber-sumber tersebut. 
In Table 3, classify the resources given according to what are the resources competed by animals and resources compete by plants. Dalam Jadual 3, klasifikasikan sumber-sumber yang diberi mengikut apakah sumber- sumber yang dsaingi oleh haiwan dan sumber-sumber yang disaingi oleh tumbuhan. 
The resources competed by animals/ Sumber-sumber yang dsaingi oleh haiwan 
Resources compete by plants/ Sumber-sumber yang disaingi oleh tumbuhan 
Food Space Breedingmate 
Nutrien Water Light 
[3 marks] 
Food/makananSpace/ruang Breedingmate/pasangan mengawan 
Nutrient/nutrien Water/air Light/cahaya 
Table 3/Jadual 3
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Distance between 
maize seedlings (cm) 
0 5 10 15 
6 
5 
Growth rate 
Of maize seedlings 
(g/days)
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QUESTION 2 
Score 
Explanation 
01 √ 
Identified the problem 
3 
Able to state problem statement correctly P1 – light intensity P2 – rate of transpiration Sample answer: Is the light intensity increase the rate of transpiration of plant? 
2 
Able to state problem statement but slightly incorrect 
1 
Able to state idea only (not in question)//Hypothesis form. 
0 
No response or wrong response. 
√ 
Objective of study/Aim Able to state the objective of study correctly Sample answer: To investigate the effects of light intensity on the rate of transpiration of a balsam plant. 
√ 
Variables Able to state any one item for each variable given. Manipulated Variable : distance light sources// ligh intensity Responding Variable : Time taken for the air bubble move// rate of transpiration Fixed / Controlled Variable: temperature//type of plant 
02 √ 
Statement of hypothesis P1 – light intensity P2 – rate of transpiration P3 – The rate transpiration / air bubble movement / is influence by light intensity 
3 
Able to state the hypothesis correctly by relating two variable correctly. Sample answer: The higher the light intensity, the rate of transpiration of a balsam plant increase. 
2 
Able to state hypothesis but slightly incorrect. 
1 
Able to state idea only. 
0 
No response or wrong response. 
05 √ 
List of apparatus Photometer, stopwatch, cutter (knife), beaker, fluorescent lamp, meter ruler List of materials Balsam plant, Vaseline, water, tissue 
3 
Able to list down 4 apparatus and 3 material. 
2 
Able to list down 2 apparatus and 2 material. 
1 
Able to list down 1 apparatus and 1 material. 
0 
No response or wrong response. 
B1 – 1 √ 
Technique used Measure and record the time taken for the air bubble to move in a distance for 10 cm by (B1-1). 
04 √ 
Experimental procedure 
1. A suitable balsam plant is selected (K1) and is cut using a sharp knife (K1). The cut end is immediately immersed in a beaker filled
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with distilled water. (K1) 
2. The cut plant is then fixed onto a photometer (K1) and the joints between the plant and the photometer are sealed using Vaseline to make them airtight (K5). 
3. The laboratory curtains and doors are pulled and closed so that outside lightning will not affect the outcome of the experiment (K1). 
4. A 40W(K2) fluorescent lamp is set 30 cm (K3) away from the edge of the (K3) photometer with a meter rule placed to measure the distance. 
5. The air bubble in the photometer is set to 0 cm (K4). The lamp is switched on and the stopwatch is started (K4) when the air bubble cross the X mark . 
6. The movement of air bubble is observed and the stopwatch is stopped when the bubble reaches Y mark, that is 10 cm (K2). 
7. Record the time taken into a table(K4) . 
8. Steps 4 to 7 are repeated, with the distance of the lamp are put at 40 cm(K3), 50 cm(K3), 60 cm (K3) away from the photometer. 
9. All the findings are recorded into the table(K4). 
3 
All 5Kcriteria correct K1 – any three criteria K2 – any one criteria K3 – any three criteria K4 – any two criteria K5 – any one criteria 
2 
3K – 4Kcriteria correct. 
1 
At least 2Kcriteria correct. 
0 
No response or wrong response. 
B2 – 1 √ 
Presentation of data Data is present in a table with right unit for rate of transpiraton (for B2 – 1 cm/second or cm second-1) 
If without the unit for the rate of transpiration, give no an idea (x) and B2 - 0. 
√ 
Conclusion Write the hypothesis or another hypothesis. Sample answer: The higher the light intensity the higher the rate of transpiration. Hypothesis is accepted. 
03 
Report writing 
3 
Score 3 = 7-9 
2 
Score 2 = 4-6 √ 
Distance of lamp from the edge of the photometer (cm) 
Time taken for the air bubble to travel for X to Y (s) 
Rate of Transpiration 
(cm/second)
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1 
Score 1 = 1-3 √ 
0 
No response or wrong response. 
Question 1: 33 Marks Question 2: 17 Marks (Total = 50 marks) 
END OF THE SCHEME MARKING
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
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SKEMA BIOLOGI KERTAS 3 
Question 
Criteria 
Score 
1 (a) 
Able to record all the data for volume of the urine collected and average of urine produce correctly. Sample answers 
Type of Vegetables 
Amount of Filtrate needed to decolorize DCPIP solution (ml) 
Cauliflower 
4.2 
Broccoli 
2.5 
Lime 
3.6 
Ascorbic Acid 
1.0 
3 
Able to record 3 data correctly 
2 
Able to record 2 data correctly 
1 
Able to record only 1 data / wrong response 
0 
(b) (i) 
Able to state two different observation correctly base on the criteria: C1 – Type of vegetables C2 – Amount of Filtrate needed to decolorize DCPIP solution (ml) Sample answers 
1. The amount of Cauliflower/ Broccoli/ Lime/ Ascorbic Acid filtrate needed to decolorize DCPIP solution is 4.2 ml/ 2.5 ml/ 3.6ml/ 1.0ml 
2. The amount of Cauliflower filtrate needed to decolorize DCPIP solution is higher than the amount of Broccoli filtrate. 
3 
Able to state one correct observation and one inaccurate observation. Sample answer (inaccurate) 
1. The amount of cauliflower filtrate needed to decolorize DCPIP solution the highest 
2 
Able to state only one correct observation or two observations at idea level. Sample answer (idea) 
1. The amount of vegetables filtrate needed to decolorize DCPIP solution is different 
1 
No response or incorrect response or inaccurate observation or one idea only. 
0
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Question 
Criteria 
Score 
(b) (ii) 
Able to make two accurate inferences base on two criteria: C1 – amount of vegetable filtrate C2 – content of vitamin C Sample answer 
1. Amount of vegetable filtrate needed to decolorize DCPIP solution is high, the content of vitamin C in the vegetables filtrate is low. 
2. Amount of vegetable filtrate needed to decolorize DCPIP solution is low, the content of vitamin C in the vegetables filtrate is high. 
3 
Able to make one correct inference and one inaccurate inference or able to state two inaccurate inferences. Sample answer (inaccurate) 
1. More / high / much (amount) of vegetable filtrate. 
2. Low/ high content of Vitamin C 
2 
Able to sate one correct inference or two inferences at idea level. Sample answer (idea level) 
1. The content of vitamin C is different 
1 
No response or inaccurate response 
0 
Summary of scoring for 1(b)(i) and 1(b)(ii) 
Score 
Correct 
Inaccurate 
Idea 
Wrong 
3 
2 
- 
- 
- 
2 
1 
1 
- 
- 
- 
2 
- 
- 
1 
1 
- 
1 
- 
- 
- 
2 
- 
- 
1 
1 
- 
0 
- 
1 
- 
1 
- 
- 
1 
1
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Question 
Criteria 
Score 
(c) 
Able to state all variables and methods to handle each variable correctly. Sample answer 
Variable 
Method 
Manipulated variable: Types of Filtrate 
Use different type of filtrate 
Responding variable: 
 Volume of filtrate needed to decolorize DCPIP solution 
 Measure and record the volume of vegetable filtrate needed to decolorize DCPIP solution using syringe 
 Calculate and record the percentage of Vitamin C using the formula : 
Amount of Ascorbic Acid needed to decolorize 
DCPIP solution(ml) x 0.1% Amount of Filtrate needed to decolorize DCPIP solution(ml) 
Initial volume of filtrate/ volume of DCPIP solution 
Fix the initial volume of filtrate which is 5ml / volume of DCPIP solution which is 1ml 
Able to state 4 – 5 ticks 
2 
Able to state 2 – 3 ticks 
1 
No response or incorrect response or 1 tick only 
0 
(d) 
Able to state the hypothesis relating the manipulated variable and the responding variable correctly based on three criteria: P1 : Manipulated variable (volume of water intake) P2 : Responding variable ( volume of urine collected) H : Relationship of the variables. Sample answer 
1. Broccoli has the highest amount of Vitamin C compared to 
3
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Cauliflower, and Spinach. 
P1 + P2 + H 
Able to state the hypothesis based on any two criteria. Sample answer 
1. The amount of filtrate needed to decolorize DCPIP solution affects the content of Vitamin C. 
2. Different amount of filtrate needed to decolorize DCPIP solution, different content of Vitamin C. 
P1 + P2 // P1/P2 + H 
2 
Able to state the hypothesis based on any one criteria or idea level. Sample answer 
1. The content of Vitamin C is different. 
1 
Question 
Criteria 
Score 
(e) (i) 
Able to construct a table correctly with the following aspects: T : Title with correct unit - 1 mark D : Data - 1 mark C : Percentage of Vitamin C - 1 mark Sample answer 
Type of filtrate 
The amount of filtrate needed to decolorize DCPIP solution (ml) 
Percentage of Vitamin C (%) 
Cauliflower 
4.2 
0.02//0.022 
Broccoli 
2.5 
0.04 
Lime 
3.6 
0.03//0.027 
Ascorbic Acid 
1.0 
0.1 
3 
Any two aspect correct 
2 
Any one aspect correct 
1 
Incorrect response 
0 
(ii) 
Able to draw the graph of volume of water reabsorb against volume of water intake based on the following aspects: P (Paksi): Title of x-axis and y-axis with unit - 1 mark Title (Title) : Four points plotted correctly - 1 mark B (Bentuk) : All points connected smoothly - 1 mark All three aspect correct 
3
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Any two aspect correct 
2 
Any one aspect correct 
1 
No response or Incorrect response 
0 
(f) 
Able to explain the relationship between the amount of vegetable filtrate needed to decolorize DCPIP solution and type of filtrate base on the following criteria. R1 : Relationship - R2 : the amount of vegetable filtrate to decolorize DCPIP R3 : the concentration of vitamin C Sample answer Broccoli has the highest amount of vitamin C compared to cauliflower and Lime. This is because the amount of vegetable filtrate needed to decolorize DCPIP is low, thus the concentration of vitamin C is high enough which needed only small amount of flitrate to reduce DCPIP solution. 
3 
Able to explain the relationship using any two aspects 
2 
Able to explain the relationship using any one aspect only 
1 
No response or Incorrect response 
0 
Question 
Criteria 
Score 
(g) 
Able to predict and explain the volume of urine produced based on the following criteria : P1 : Prediction - More than 4.5 ml P2 : Explanation - denaturation of vitamin C P3 : Concentration of filtrate needed to decolorize DCPIP solution Sample answer Volume of filtrate needed to decolorize DCPIP solution is more/ higher than 4.5ml. This is because boiling causes Vitamin C to denature. Thus there is less/ no vitamin C concentration in filtrate to decolorize DCPIP solution 
Able to predict and explain the volume of urine produced based on any two criteria 
2 
Able to predict and explain the volume of urine produced based on any one criteria 
1 
No response or Incorrect response 
0 
(h) 
Able to define Vitamin C operationally based on the following criteria. D1 : D2 : D3 :
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 6 
Sample answer 
3 
Any two criteria stated 
2 
Any one criteria stated 
1 
No response or Incorrect response 
0 
Question 
Criteria 
Score 
(i) 
Able to classify apparatus and materials into their respective variables. Sample answer 
Materials 
Apparatus 
Ascorbic Acid Fruit Juices DCPIP Solution 
Syringe with needle Specimen Tube Measuring Cylinder 
All 6 corrects in its appropriate correct. 
3 
5 – 4 correct 
2 
3- 2correct 
1 
1 – 0 No response or Incorrect response 
0
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 7 
Paper 3 – No 2 (Scheme) 
Item 
Criteria 
Marks 
Problem statement Able to state the problem statement of the experiment correctly based on three criteria: 
 Manipulated variables – (different) water sources/ samples (P1) 
 Responding variables – time taken to decolourise methylene blue solution 
// level of polution (P2) 
 Relationship in question form and question symbol [?] (H) 
Sample Answer 1. What is the time taken to decolourise methylene blue solution// level of pollution of different water sources/samples ? 2. Does different water sources/samples has different time taken to decolourise methylene blue solution// level of pollution ? 
3 marks 
Able to state the problem statement of the experiment correctly based on any two criteria. Sample Answer 1. What is the time taken to decolourise methylene blue solution// level of pollution of different water sources/ samples. 
2 marks 
Able to state the problem statement of the experiment correctly based on only one criteria. Sample Answer 
1. What is the time taken to decolourise methylene blue solution / level of pollution of water sources/ sample. 
2. Water sources/ samples have different level of pollution. 
1 mark 
Wrong or no response 
0 mark
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 8 
Item 
Criteria 
Marks 
Hypothesis Able to state the hypothesis of the experiment correctly based on three criteria: 
 Manipulated variables – different water sources/ samples (P1) 
 Responding variables – time taken to decolourise methylene blue solution 
/level of pollution (P2) 
 Relationship – more/less than//higher/lower than // shortest//most (H) 
Sample Answer 1. Water sample A/ drain water took the shortest time to decolourise methylene blue solution compared to water sample B, C and D. 2. Drain water /water sample A is the most polluted samples of water collected. 
3 marks 
Able to state the hypothesis of the experiment correctly based on two criteria: Sample Answer 1. Different sources of water samples affect the time taken for the methylene blue solution to decolourise . 
2 marks 
Able to state the hypothesis of the experiment correctly based on one criteria: Sample Answer 1. The drain water is polluted 
1 mark 
Wrong or no response 
0 mark
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 9 
Item 
Criteria 
Marks 
Variables Able to state the variables of the experiment correctly that include three criteria: Manipulated : different water sources/ samples Responding : time taken to decolourise methylene blue solution // level of polution Constant : volume of water samples// volume of methylene blue solution 
3 marks 
Able to state the variables of the experiment correctly that include two criteria. 
2 marks 
Able to state the variables of the experiment correctly that include one criteria 
1 mark 
Wrong or no response 
0 mark 
Item 
Criteria 
Marks 
Materials Methylene blue solution, water samples from A, B, C and D Apparatus Syringe with needle, stop watch, Reagent bottle, stopper and beaker 
3 marks 2M + 5A 
Materials
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 10 
Two materials Apparatus Any three apparatus (syringe and stop watch are compulsory ) 
2 marks 2M + 3A 
Materials Two materials Apparatus Two apparatus ( syringe and stop watch ) 
1 mark 2M + 2A 
Cannot state the functional materials and apparatus 
0 mark 
Item 
Criteria 
Marks 
Procedure Able to state five Ks correctly. K1 : Preparation of materials & apparatus (3K1) K2 : Operating fixed variable (1K2) K3 : Operating responding variable (1K3) K4: Operating manipulated variable (1K4) K5 : Precaution (1K5) 
3 marks 
Able to state any three Ks correctly 
2 marks
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 11 
Able to state any two Ks correctly 
1 marks 
Able to state only one K correctly or no response 
0 marks 
Example for procedure K1: Preparation of materials & apparatus 
1. Water samples are collected from four different sources A,B,C and D 
2. The reagent bottles are labelled 1, 2, 3 and 4 
3. The reagent bottles are closed with the stopper immediately 
4. The stopwatch is activated 
5. The bottles are examined from time to time 
6. The results are tabulate. 
K2: Operating fixes variable 
1. Measure 250 ml of water sample from A, B, C and D separately and pour into the reagent bottle labelled 1, 2, 3 and 4 respectively 
2. 1 ml of methylene blue solution is added to each water sample using a syringe 
K3: Operating responding variable 
1. The time taken for the methylene blue solution to decolourise is measured and recorded in the table using stop watch. 
K4: Operating manipulated variable 
1. Measure 250 ml of water samples from A, B, C and D separately and pour into the reagent bottle labeled 1, 2, 3 and
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 12 
4 respectively 
Item 
Criteria 
Marks 
K5: Precaution 
1. 1 ml of methylene blue solution is added to the base of each water sample using a syringe with needle 
2. The reagent bottles are closed with stopper immediately 
3. The contents of the bottles cannot be shaken. 
4. All the reagent bottles are kept in a dark cupboard 
Sample answer: Method / procedure : 
1. Water samples are collected (K1) from 4 different water resources (K3) 
*(A,B,C and D). 
2. The reagent bottles are labelled 1, 2, 3 and 4 
3. 250 ml of water samples (K2) is measured (K1) from 4 different water 
Resources (A, B, C and D) separately and pour (K1) into the reagent bottle labeled 1, 2, 3 and 4 respectively. 
4. 1 ml (K2) of methylene blue solution is added (K1) to the base (K5) of each water sample using a syringe with needle. 
5. The reagent bottles are closed with stopper immediately. (K5) 
6. The content of the bottles cannot be shaken. (K5) 
7. All the reagent bottles are kept in a dark cupboard. (K5) 
8. The stop watch is activated. (K1) 
9. The bottles are examined (K1) at one hour interval. 
10. The time taken for the methylene blue solution to decolourise is measured and recorded for all the water samples using stop watch. (K3) 
11. The result is tabulated. (K1) 
*Tap, drain, river, pond.
PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 
BIOLOGY 4551 MODUL PERFECT SCORE SBP 13 
Presentation of data 1. Able to state all the titles correctly with units. 2. Able to state 4 water samples. 
Water samples 
Time taken for methylene blue solution to decolourise (hour ) 
Tap water 
Drain water 
River water 
Pond water 
2 marks 
Able to state any one criteria only 
1 mark 
END OF MARKING SCHEME

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Modul perfect score sbp biology spm 2014 skema

  • 1. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 1 SECTION A [60 marks] Answer all the questions Jawab semua soalan dalam bahagian ini 1. Diagram 1.1 shows the structure of animal cell Rajah 1.1 menunjukkan struktur sel haiwan Diagram 1.1 Rajah 1.1 (a) On Diagram 1.1, label X and Y. Pada Rajah 1.1, labelkan X dan Y [1marks] (b) Explain one characteristic of Y related to cell division. Terangkan satu ciri Y berkaitan dengan pembahagian sel Sample answer: F: Composed of a complex arrangement of microtubules E: Form spindle fibers during cell division (in animal cell) [2marks] Y: centriole X: nucleus
  • 2. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 2 (c) Explain the function of X in cell division Terangkan fungsi X dalam pembahagian sel Sample answer: F: nucleus containchromosomes which carry genetic information E: behavior of chromosome in stage of mitosis/meiosis/cell division transfer genetic information to daughter cell [2marks] (d) Diagram 1.2 below shows stage of cell division in somatic cell of human. In the box given, draw another two stage in that cell division. Rajah 1.2 di bawah menunjukkan peringkat pembahagian sel dalam sel soma manusia. Dalam kotak yang diberi, lukis dua peringkat dalam pembahagian tersebut [2marks] Diagram 1.2 Rajah 1.2 (a) (b) (c) (d)
  • 3. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 3 (e) If spindle fibre is not form in Diagram 1.2 (d), explain the effect on the number of chromosome in the daughter cell. Jika gentian gelendong tidak terbentuk dalam Rajah 1.2 (d), terangkan kesan keatas bilangan kromosom dalam sel anak Sample answer: F:the number of chromosome in the daughter cell less/ extra E: no contraction of spindle fibre to pull chromosome toward the pole [2marks] (f) Explain how the cell division above can be used to increase in a short time the number of given example of the plant in the farm. Terangkan bagaimanakah pembahagian sel di atas dapat digunakan untuk meningkatkan bilangan dalam masa yang singkat tumbuhan yang dinamakan dalam ladang. Sample answer: Example of the plant: banana/ palm oil/ any other example F:tissue culture E1: cut off explants / part of plant / young shoot, leaves, roots, seeds, embryos E2: explants are sterilized and placed in cultured medium containing nutrient such as glucose, amino acids, minerals and growth hormone/ auxin E3:the culture medium need to be maintained at optimum pH and temperature 25 -35°C E4: explants divide by mitosis form callus// undifferentiated mass of tissues E5: callus develop into embryoid/ somatic embryos and later into plantlets E6: plantlets are transferred to the soil where they grow into adult plant [3marks]
  • 4. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 4 2.(a) Diagram 2.1 shows the shape of red blood cells after being immersed for 30 minutes in three solutions with different concentration. Rajah 2.1 menunjukkan bentuk sel darah merah selepas direndam selama 30 minit dalam tiga larutan yang berbeza kepekatannya. Based on the Diagram 2.1 Berdasarkan Rajah 2.1 (i) State the condition of the red blood cells after being immersed in Nyatakan keadaan sel darah merah selepas direndam di dalam Sample answer: Solution P: Crenation / shrink / shrivel Solution Q: Haemolysis / swell and burst [2 marks] Diagram 2.1 Red blood cells in Q solution Sel darah merah dalam larutan Q Red blood cells in P solution Sel darah merah dalam larutan P Red blood cells in R solution Sel darah merah dalam larutan R
  • 5. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 5 (ii) Name the type of solution R in which the red blood cells are immersed. Namakan jenis larutan R yang mana sel darah merah direndam. Sample answer: Solution R is isotonic solution. [1 mark] (iii) Explain your answers given in a(ii) Terangkan jawapan yang anda berikan di a(ii) Sample answer: P1:The cell retains its normal shape/ biconcave disc shape. P2:The water diffuses in and out of the cells at equal rate by osmosis P3:Solution R has the same osmotic concentration as the cytoplasmic fluid in the red blood cells [3 marks] (b) Based on the statement, explain why vinegar is suitable to be used as the natural preservative for the preservation of garlic. Berdasarkan pernyataan di atas ,terangkan mengapa cuka adalah sesuai digunakan sebagai pengawet semulajadi untuk bawang putih. Sample answer: F1:Vinegar has a low pH/acidic E1: Vinegar diffuses into the tissues of the garlic E2: The tissues of the garlic becomes acidic E3: The low pH prevents the growth of microorganisms in garlics E4: The garlic can be preserved to last longer [3 marks] Food such as mushrooms, fruits, vegetables and fish can be preserved longer by using natural preservatives such as salt, sugar and vinegar. Makanan seperti cendawan, buah-buahan, sayur-sayuran dan ikan boleh diawet untuk tahan lama menggunakan bahan-bahan pengawet semulajadi seperti garam,gula dan cuka .
  • 6. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 6 (c) Diagram 2.2 shows the condition of herbaceous plant due to water shortage in soil. Rajah 2.2 menunjukkan keadaan pokok herba disebabkan oleh kekurangan air dalam tanah. Explain the condition of the plant in Diagram 2.2 after one week. Terangkan keadaan pokok dalam Rajah 2.2 selepas satu minggu. Sample answer: F:The plant wilt E1: The cells become flaccid / plasmolysed // both the vacuole the vacuole and cytoplasm shrink // the plasma membrane of the root cells pull away from the cell wall. E2: Water molecules diffuse out from the cell sap of the root hair cell by osmosis E3: (the remaining) soil water becomes hypertonic to the cell sap of the root hair cell as the soil dries out. [3 marks] Water shortage after one week Kekurangan air selepas satu minggu Diagram 2.2
  • 7. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 7 4. Diagram 4.1 shows the process of phagocytosis as second line of defence to destroy the bacteria Rajah 4.1 menunjukkan proses fagositosis sebagai barisan pertahanan kedua untuk memusnahkan bakteria. Stage 1 Stage 2 Peringkat 1 Peringkat 2 Stage 4 Stage 3 Peringkat 4 Peringkat 3 Diagram 4.1/ Rajah 4.1 (a) (i) Name cell M involved in mechanism above. Namakan sel M yang terlibat dalam mekanisma di atas. Neutrophil / eosinophil / basophil / granulocyte / phagocyte [1 mark] (ii) Draw a diagram in stage 3 Lukis rajah dalam peringkat 3. [1 mark] (b) Explain the function of lisosome in mechanism above Terangkan peranan lisosom dalam mekanisma di atas F : digest bacteria E : because lisosome contain hydrolytic enzymes / cellulase which digest cellulose / by breaking down the bacteria cell wall. [2 mark] M
  • 8. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 8 (c) Explain what happen in stage 2. Terangkan apa yang berlaku dalam peringkat 2. F : phagocytes surrounds / engulfs the bacteria using pseudopodia E : forming phagocytic vacuole / fagosome /food vacuole [2 mark] (d) How the action of M to pathogen is different to lymphocyte? Bagaimanakah tindakan M terhadap patogen adalah berbeza dengan limfosit? M kill the pathogen by engulf and digest the pathogen but lymphocyte produce antibody then antibody kill the pathogen / neutralize the toxin from pathogen [1 marks] (e) Diagram below show a type of immunity occurs in human. Rajah di bawah menunjukkan jenis immuniti yang berlaku dalam manusia. (i) Name the type of immunity shows in the diagram. Namakan jenis immuniti yang ditunjukkan dalam rajah. Answer: Artificial active immunity [1 mark]
  • 9. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 9 (ii) Name the second injection and why the person should take second injection? Namakan suntikan kedua dan mengapa individu ini perlu mengambil suntikan kedua? Sample answer: N: booster dose // an additional administration of a vaccine E: to stimulate lymphocyte produce more antibody until achieve immunity level. [2 mark] (f) The above immunity is example of third line of defence. What make it different to the second line of defence? Sample answer: P1: Third line of defence – specific response to pathogen infection but second line of defence – non-specific response/generalized responses to pathogen infection P2: Third line of defence involved production of antibody(active immunity)/ used supply antibody (passive immunity) from leucocyte but second line of defence involve the physical structure of leucocyte [2 marks]
  • 10. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 10 Q 5. Diagram 5.1 shows the stages of the ovarian cycle in human ovary Diagram 5.2 shows the thickness of the endometrium of uterus before the fertilisation in the second menstrual cycle. Rajah 5.1 menunjukkan peringkat kitaran ovary dalam ovary manusia. Rajah 5.2 menunjukkan ketebalan endometrium dalam uterus sebelum berlaku persenyawaan dalam kitarhaid yang kedua. Diagram 5.1 Diagram 5.2 Y P Secondary follicle Secondary oocyte X
  • 11. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 11 a) (i) Describe the change in the structure of follicle P into a secondary follicle. Terangkan perubahan struktur folikel P dalam pembentukan folikel sekunder. [2 marks] Sample answer: P1 : FSH concentration increases // is released (by the pituitary gland). P2 : Stimulates the development of follicle cells. (1m) P3 : Primary follicle developed into secondary / Graafian follicle // Primary oocyte developed into secondary oocyte (ii) Relate the change in (a)(i)to the thickness of the endometrium Hubungkan perubahan dalam (a) (i) dengan ketebalan endometrium [1 mark] Sample answer: P1 : The thickness of the endometrial wall increases b) Explain the process that occurs at X. Terangkan proses yang berlaku pada X. [2 marks] Sample answer: P1 : Ovulation P2 : The release of secondary oocyte from the (matured) secondary follicle / Graafian follicle to the oviduct / Fallopian duct c) Explain the effect of the change of structure Q to the thickness of the endometrium. Terangkan kesan perubahan struktur Q keatas ketebalan dinding endometrium. [2 marks] Sample answer: P1 : The thickness of the endometrial wall / uterine lining decreases. P2 : The level of progesterone decreases. d) (i) Fertilisation takes place in the second menstrual cycle. Complete the graph in Diagram 5.2 to show the changes in the thickness of the endometrium after point Y Persenyawaan berlaku dalam kitar haid yang kedua Lengkapkan graf dalam Rajah 5.2 untuk menunjukkan perubahan ketebalan endometrium selepas titik Y
  • 12. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 12 [1 mark] Answer: (ii) Explain your answer in (d)(i). Terangkan jawapan di dalam (d)(i). [2 marks] Sample answer: P1 : The corpus luteum / placenta developed. P2 : The corpus luteum / placenta released progesterone (and oestrogen). e) (i) State the changes in the thickness of the endometrium after point Y relating to the secretion of hormones secreted by the ovary. Terangkan hubungan perubahan ketebalan endometrium selepas titik Y dengan perembesan hormon oleh ovari. [1 mark] Sample answer: P1 : The thickness of the endometrial wall increases / is maintained. (ii) State the importance of thickened endometrium to the continuity of life Nyatakan kepentingan ketebalan endometrium dalam kesinambungan hidupan [1 mark] Sample answer: P1 : Increase the chance of implantation // development of embryo / blastocyst.
  • 13. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 13
  • 14. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 14 SECTION B [40 marks] Answer any two questions from this section Jawab mana-mana duasoalandaripadabahagianini 6. Diagram 6.1 shows the growth and development process at the shoot tip. Rajah 6.1 menunjukkan proses pertumbuhan dan perkembangan pada hujung pucuk. Diagram 6.1 a) Explain the process of primary growth that shown in Diagram 6.1 Terangkan proses pertumbuhan primer yang ditunjukkandalam Rajah 6.1 [8 marks] Sample answer: Cell division P1: Cell division take place by mitosis P2: Each cell divides to become two cells which are identical to the parent cell P3: This process repeats itself until a mass of cells consisting of many identical cells are formed
  • 15. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 15 Cell elongation P4: Cell elongation cause by intake of water and nutrient into the cell from the environment P5: Water accumulates in the vacuoles of plant cells to form large central vacuole, causing the primary wall to stretch P6: The nutrients are used in the building up of the protoplasm// more organelles leading to an increase in the cell size and volume Cell differentiation P7: Cells begin to differ from each other to form groups of specialised cells P8: to perform new and specialised functions // Example: cell differentiation in the epidermis of roots to form root hair to enable the cell to have a large total surface area for absorption of water from the soil P9: Cells differentiation causing the changes of shape and complexity of organism b) Explain the importance of primary growth to plant. Terangkan kepentingan pertumbuhan primer kepada tumbuhan [4 marks] Sample answer: P1: During this time the stem and roots of plant increase in length. This allows a plant to achieve its maximum height P2: Its bring about the formation of primary xylem that helps in the transport of water and mineral P3: Its bring about the formation of primary phloem that helps in the transports organic substances P4: Its provides support because the walls of xylem tissue are thickened with lignin
  • 16. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 16 c) Diagram 6.2 shown the tropism respond at shoot tip and root tip. Rajah 6.2 menunjukkan gerakbalas tropisme pada hujung pucuk dan hujung akar. Diagram 6.2 Base on diagram 6.2 explain how the tropism response occurred. Berdasarkan rajah 6.2 terangkan bagaimana gerakbalas tropisme berlaku. [8 marks] Sample answer: P1: auxins produced by shoot and root P2: auxins diffuse into zone of elongation P3: (Owing to gravity) auxins move to lower side of shoot and root P4: The lower side of shoot and root has a higher concentration of auxins than the upper side P5: height concentration of auxins in the shoot promotes elongation of cells. P6: the lower side of the shoot will faster than the upper side P7: the shoot curves and grows upward (negative geotropism) P8: height concentration of auxins in the root inhibits elongation of cells. P9: So the upper side of the root will grow faster than the lower side P10: the root curves and grows downwards (positive geotropism)
  • 17. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 17 7. (a) Explain the interaction based on Diagram 7.1 Terangkan interaksi berdasarkan Rajah 7.1 Diagram 7.1 Rajah 7.1 [4marks] Sample answer: F: Commensalism E1:relationship between two species that benefits one species/ commensal but neither benefits nor harms the other species/ host E2: fern has a sponge-like root mass that soaks up rain water and absorbs nutrients released from the decaying litter. E3: fern leaves has mesophyll cell contain chloroplast do photosynthesis (b) Diagram 7.2 shows mechanism of photosynthesis in plant. Explain why the product from light reaction need for dark reaction. Rajah 7.2 menunjukkan mekanisma fotosintesis dalam tumbuhan.Terangkan Mengapa produk dari tindakbalas cahaya diperlukan untuk tindakbalas gelap.
  • 18. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 18 [6marks] Sample answer: F: Mechanism of photosynthesis in plant consist of light reaction anddark reaction E1: Chlorophyll absorbs light energy released electrons andproduce ATP E2: Light energy is also split the water molecules intohydrogen ion(H+)and hydroxyl ions(OH-)/Photolysis of water E3: Hydrogen ions combine with electron from chlorophyll to form hydrogen atoms E4: Hydrogen atom and ATP will be used in dark reaction E5: Each hydroxyl ion loses an electron to form hydroxyl group (the electron is received by the chlorophyll) E6: The hydroxyl groups then combine to form water and gaseous oxygen E7: Hydrogen atomfit/ reduce carbon dioxide in dark reaction to form glucose E8: Reduction of carbon dioxide need ATP from light reaction E9: The reaction catalysed by photosynthetic enzyme in stroma E10: Produced glucose molecules, then glucose undergo condensation/ converted to starch for storage Diagram 7.2 Rajah 7.2
  • 19. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 19 (c) Diagram 7.3 shows the condition of the town. Rajah 7.3 menunjukkan sebuah bandar. Diagram 7.3 Rajah 7.3 (i) Discuss the effect of air pollution may occur in the town. Bincangkan kesan pencemaran udaya yang mungkin berlaku dalam bandar. [6 marks] Sample answer: F1: formation of haze / smog E1: cause by fine particle matter/smoke/ soot E2: prevents vision /reduce light intensity photosynthesis/ reduce oxygen content F2: acid rain E1: cause by SO2 / NO2 E2: estroy building F3: increasetemperature E1: cause by CO2 E2: green house effect/ global warming F4: depletion of ozonelayer E1: cause by CFC gases E4: more uv penetration F5: respiratory problems/allergies/risk for cancer E1: cause by CO/ SO2 / NO2 E2: cause health hazards
  • 20. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 20 (ii) If you are an environmental activist, suggest how you would explain to the government about the measures needed to overcome the type of pollution. Jika anda seorang aktivis alam sekitar, cadangkan bagaimana anda akan menjelaskan kepada kerajaan mengenai langkah-langkah yang diperlukan untuk mengatasi jenis pencemaran. [4 marks] Sample answer: F1: implementation of laws E1: control and prevent pollution using the environmental act F2: use of technology E1: using unleaded petrol for cars/ fit catalytic converter in factory F3: education E3: media massa/ internate/ scholl 8. (a) Diagram 8.1 shows three types of neurons. Rajah 8.1 menunjukkantigajenis neuron. Diagram 8.1 Rajah 8.1 Name types of neurone P, Q and R and state two differences between the structure of neurone P and neurone Q Namakan jenis neuron P, Q dan R dan nyatakan dua perbezaan di antara struktur bagi neuron P dan neuron Q [4 marks] Neurone P Neuron P Neurone Q Neuron Q Neurone R Neuron R
  • 21. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 21 Sample answer: Neurone P is afferent neurone Neurone Q is efferent neurone Neurone R isinterneurone 3√ = 2 marks 2√ = 1 marks Neurone P Neurone Q P1: has a long dendron but a short axon Has a short dendron but a long axon P2: cell body is located in the middle of the cell Cell body is located at the terminal / end of the cell (b) Diagram 8.2 shows the transmission of a nerve impulse from neurone P to neurone R. Rajah 8.2 menunjukkan pemindahan impuls saraf dari neuron P ke neuron R. Diagram 8.2 Rajah 8.2 Explain the transmission of a nerve impulse from neurone P to neurone R Terangkan pemindahan impuls saraf dari neuron P ke neuron R [6 marks]
  • 22. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 22 Sample answer: P1: when impulse reaches the synaptic knob / terminal / terminal axon / presynaptic membrane P2: it stimulates the synaptic vesicles P3: to release neurotransmitter P4: mitochondrion (in the synaptic terminal) produces energy / ATP P5: for active transport / transmission of the impulse P6: (neurotransmitter) diffuse across / into synaptic cleft / synapse to the next dendrite / neurone R / postsynaptic membrane P7: transmission of impulse from neurone P to neurone R is in the form of chemicals (c) Diagram 8.3a shows example of voluntary action and Diagram 8.3b shows example of involuntary action. By using pathway of transmission of information from receptors to effectors, explain similarities and difference between voluntary action and involuntary action. Rajah 8.3a menunjukkan contoh tindakan terkawal dan Rajah 8.3b menunjukkan contoh tindakan luar kawal. Dengan menggunakan laluan pemindahanmaklumat dari reseptor kepada efektor, terangkan persamaan dan perbezaan tindakan terkawal dan tindakan luar kawal. Diagram 8.3a Diagram 8.3b Rajah 8.3a Rajah 8.3b [10 mark]
  • 23. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 23 Sample answer: Pathway of transmission of information from receptors to effectors for voluntary action Pathway of transmission of information from receptors to effectorsfor involuntary action Similarities S1: both voluntary action and involuntary action involved receptor E1: to detect the stimulus and trigger impulse S2: both voluntary action and involuntary action involved 3 neuron E2: carry impulse from receptor to efector S3: both voluntary action and involuntary action involved efector receptor efector
  • 24. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 24 E3: contract to show the response Differences D1: Voluntary action is controlled by conscious thoughts but Involuntary action occurs automatically without any conscious control E1: because voluntary action Involves the integration and interpretationof information in the cerebrum but involuntary action involve spinal cord only D2: Voluntary action are under the control of the will of the Individual but involuntaryaction are not control by the will E2: because voluntary action involve the action of doing thing for activity but involuntary action involve the action to protect the person from danger
  • 25. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 25 9. (a) Diagram 9.1 and 9.2 shows the histogram about distribution of genetic variation in human. Rajah 9.1 dan 9.2 menunjukkan histogram mengenai taburan variasi genetik dalam manusia. Diagram 9.1 Diagram 9.2 (i) With a suitable example, explain the differences of two kinds of variation. Dengan menggunakan contoh yang sesuai, terangkan perbezaan di antara kedua- dua variasi tersebut. [7 marks] Sample answer: Example of continuous variation: Height or weight Example of discontinuous variation: ABO blood group Differences: Continuous variation Discontinuous variation Graf distribution shows a normal distribution Graf distribution shows a discrete distribution The characters are quantitative / can be measured and graded (from one extreme to the other) The characters are qualitative / cannot be measured and graded (from one extreme to the other) Exhibits a spectrum of phenotypes with intermediate character Exhibits a few distinctive phenotypes with no intermediate character Influenced by environmental factors Is not Influenced by environmental factors Two or more genes control the same character A single genes determines the differences in the traits of the character The phenotype is usually controlled by many pair of alleles The phenotype is controlled by a pair of alleles
  • 26. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 26 (ii) What is the importance of variation to organism? Apakah kepentingan variasi kepada organisma? [3 marks] Sample answer: P1: variation provided better adaptation for organism to survive in the changing environment P2: variation are essential to the survival of species / to survive more successfully P3: variation be able to organism explore a new habitat P4: to ensure organism survival from predator
  • 27. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 27 (b) Diagram 9.3a, 9.3b and 9.3c shows the genetic factors that affected on the variation of organism. Rajah 9.3a, 9.3b dan 9.3c menunjukkan faktor-faktor genetik yang member kesan ke atas variasi pada organisma. Diagram 9.3a Diagram 9.3b Diagram 9.3c Explain how these factors in the diagram above will cause the variation among the organism. Terangkan bagaimana faktor-faktor dalam rajah di atas akan menyebabkan variasi dikalangan organisma. [10 marks] Sample answer: F1: meiosis P1: produce varies gamete with different genetic content P2: through homologous chromosomes random assortment during metaphase I Combination 4 Gabungan4 Combination 1 Gabungan 1 Combination 2 Gabungan2 Combination 3 Gabungan3
  • 28. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 28 F2: crossing over P3: two homologous chromosomes are paired up / synapsis during prophase I P3: crossing over occurs between non-sister chromatids at the chiasma P4: chromatids break and rejoin in such a way that segments of chromatids are exchange // causing a genetic recombination P5: genes in the chromosomes is altered and gametes with various combinations of chromosomes are produced F3: Fertilization P6: random fertilization between sperm and ovum P7: produce zygote with varies genetic material End…
  • 29. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 1 SECTION A [60 marks] Answer all questions in this section. Jawab semua soalan di bahagian ini 1. Diagram 1 shows a microscopic structure of a part of pancreatic cell. Rajah 1 menunjukan struktur mikroskopik sebahagian sel pancreas. (a) (i) Name the organelle K and organelle L Namakan organel K dan organel L. K: Rough Endoplasmic Reticulum L : Golgi apparatus / body [2 marks] (ii) Explain how the organelle K and organelle L are interrelated in their function Terangkan bagaimana organel K dan organel L adalah saling berkaitan dari segi fungsi mereka. P1 : Rough endoplasmic reticulum /K transports protein synthesized in the ribosomes P2 : then forms a transport vesicle which carries the protein to Golgi body / L P3 : Golgi body processes, modifies the protein into a functional one / Diagram 1 Rajah 1 Note Catatan
  • 30. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 2 enzyme / hormone (before forming a secretory vesicle) [2 marks] (b) (i) Name one chemical substance in the structure R which is involved in the synthesis of protien in a cell. Namakan satu bahan kimia dalam structur R yang terlibat dalam sintesis protien dalam suatu sel. The chemical substance in the chromosome - DNA / Deoxyribonucleic acid [1 marks] (ii) Draw the structure of the chemical compound in (b)(i) in the blank space below. Lukis struktur bahan kimia dalam (b)(i) dalam ruang kosong di bawah. DNA structure: Nitrogenous base DNA strand comprises phosphate and pentose sugar [2 marks] (c) (i) Based on the diagram, describe the synthesis of a specific pancreatic hormone in the cell. Berdasarkan gambarajah itu, huraikan bagaimana suatu hormon tertentu disintesiskan dalam sel itu The synthesis of hormone in the pancreas cell: P1 : The genetic information for the synthesis of the protein/ hormone (eg. Insulin contained in the DNA) is copied to RNA / messenger – RNA P2: (RNA) carries the information to ribosome P3: (Ribosome) synthesize the protein and Drawing = 1 mark 2 correct labels = 1 mark
  • 31. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 3 P4 : (Protein) is transfer to the Rough Endoplasmic Reticulum P5: then protein is transported to Golgi Apparatus P6: (In Golgi Apparatus) protein is modified to hormone // processed /packed and sorted / transport to plasma membrane [3 marks] (ii) The structure R in Diagram 1 undergoes some changes due to exposure to radioactive rays. Explain the possible effect to the synthesis of the hormone. Struktur R dalam Rajah 1 mengalami perubahan akibat pendedahan kepada sinaran radioaktif. Terangkan kemungkinan kesanya ke atas sintesis hormon itu. Effect of changes of the structure R / chromosome on hormone synthesis: P1: there will be some changes in the gene / base sequence / gene mutation responsible for the synthesis of the hormone P2: protein synthesis changes , a different protein / not the original hormone is synthesized or no hormone is being synthesized [2 marks] 2. Diagram 2.1 shows two individual, P and Q in two different situations. P is in a vigorous activity while Q is at rest. Processes of R and S occurs in a human muscle cell. Rajah 2.1 menunjukkan dua individu, P dan Q dalam dua situasi yang berbeza. P sedang melakukan satu aktiviti cergas manakala Q berada dalam keadaan rehat. Proses R dan S berlaku dalam satu sel otot manusia.
  • 32. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 4 (a) Based on Diagram 2.1, name the processes R and S. Berdasarkan Rajah 2.1, namakan proses R dan S. Process R/ Proses R : Anaerobic respiration Process S/ Proses S : Aerobic respiration [2 marks] (b) Write the equation of process S. Tuliskan persamaan bagi proses S. [1 marks] (c) State two differences between process R and process S. Nyatakan perbezaan diantara proses R dan proses S. Process R/ Anaerobic respiration Process S/ Aerobic respiration Glucose + Water Carbon dioxide + Water + Energy Diagram 2.1 Rajah 2.1
  • 33. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 5 Glucose is broken down partially Glucose is broken down completely ATP produced – 2 ATP ATP produces – 38 ATP Product – Lactic acid and energy Product – Carbon dioxide , water and energy Occurs in the absence of oxygen Occurs in the presence of oxygen [2 marks] (d) Diagram 2.2(a) shows fish respiratory structure and Diagram 2.2(b) shows human respiratory structure. Rajah 2.2(a) menunjukkan struktur respirasi ikan dan Rajah 2.2(b) menunjukkan struktur respirasi manusia. (i) What is X? Apakah X? Answer : Gills [1 mark] (ii) Structure X has adaptation for good gases exchange in fish. Explain one adaptation of X. Answer: 1. A :(Gills ) have lamellae E :to increase total surface area (for gases exchange) Diagram 2.2(a) Rajah 2.2(a) Diagram 2.2(b) Rajah 2.2(b) X
  • 34. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 6 2. A : (Gills) have network of blood capillary E : to transport gases rapidly 3. A : Counter current of blood and water E: increase diffusion of gases in and out of the gills 4: A :Numerous in number E: to increase total surface area (for gases exchange) [2 marks] (e) A man is a heavy smoker. Explain how this habit affect the efficiency of gases exchange on the respiratory structure in Diagram 2.2(b). Seorang lelaki adalah seorang perokok tegar. Terangkan bagaimana tabiat ini mempengaruhi kecekapan pertukaran gas pada struktur respirasi dalam Diagram 2.2(b). Answer: P1: Tobacco smoke contain tar P2: (Tar) deposit on the surface of alveolus P3 : Tobacco smoke contain heat P4 : reduce moisture on the surface of alveolus P5 : Tobacco smoke contain NO2 P6 : increase acidity / corrode the surface of alveolus [4 marks]
  • 35. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 7 3. Diagram 3.1 shows the sequence of hydrolysis of starch to molecules P and molecule Q by enzymes. Rajah 3.1 menunjukkan urutan hidrolisis kanji kepada molekul P dan molekul Q oleh enzim. (a) (i) Complete Table 3.1. Lengkapkan Jadual 3.1 Molecule Molekul Name of molecule Nama molekul Name the enzyme involved in hydrolysis Namakan enzim yang terlibat dalam hidrolisiss P Maltose Amylase Q Glucose Maltase [4 marks] (ii) Based on your biological knowledge, explain the effect of consuming food that Enzyme Enzim Starch Kanji + water air + water air Molecule P Molekul P Enzyme Enzim Molecule Q Molekul Q Diagram 3 Rajah 3 Table 3.1 / Jadual 3.1
  • 36. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 8 contain excessive of starch on health. Berdasarkan pengetahuan biologi anda, terangkan kesan pengambilan makanan yang mengandungi kanji berlebihan keatas kesihatan. Answer : P1 : Glucose level in blood increase/ hyperglycemia P2 : Starch (finally) is digested into glucose P3 : Glucose is absorbed into blood (capillary) P4 : causes diabetes mellitus / obesity [2 marks] (b) Table 3.2 shows the energy value and nutrient content in a few types of food taken by student. Jadual 3.2 menunjukkan nilai tenaga dan kandungan nutrient di dalam beberapa jenis makanan yang diambil oleh seorang pelajar. Food Makanan (/100g) Energy Tenaga (kJ) Carbohydrate Karbohidrat (g) Fats Lemak (g) Protein Protein (g) Vitamin C Vitamin C (μg) Rice Nasi 1530 86.8 1.0 6.4 0.0 Fish Ikan 320 0.0 0.5 17.5 0.0 Egg Telur 612 0.0 10.9 12.4 0.0 Orange Oren 150 8.5 0.0 0.8 50 (i) Based on Table 3.2, which type of food supplies the most energy? Table 3.2/ Jadual 3.2
  • 37. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 9 Berdasarkan Jadual 3.2, jenis makanan manakah yang membekalkan paling banyak tenaga? Answer: Rice [1 mark] (ii) Which type of food should be taken regularly to prevent scurvy? Jenis makanan manakah yang perlu kerap diambil untuk mengelakkan penyakit skurvi? Answer: Orange [1 mark] (iii) Calculate the amount of energy obtained by the student if he eats a meal which contain 200 g rice and 150g fish. Kirakan jumlah tenaga yang diperolehi oleh pelajar tersebut jika dia mengambil 200 g nasi dan 150 g ikan. Answer: Rice - 1530 kJ x 2 = 3060 kJ Fish - 320 kJ x 1.5 = 480 kJ Total : 3060 + 480 = 3540 kJ [3 marks] (c) Why does an egg produces double amount of energy compared to a fish? Mengapakah sebiji telur menghasilkan jumlah tenaga dua kali ganda berbanding seekor ikan? Answer: Egg contain more fat/lipid than fish [1 mark]
  • 38. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 10 4. Diagram 4.1 shows a longitudinal section of the reproductive parts of a flower during fertilization. Rajah 4.1 menunjukkan keratan memanjang bahagian pembiakan bunga semasa persenyawaan. (a) (i) In Diagram 4.1, label P,Q, R and S Pada Rajah 4.1, labelkan P, Q, R dan S [2 marks] (ii) In the space below, draw a section through the ovule showing all the cells in S. Label the cells involved in fertilization. Dalam ruang di bawah , lukiskan keratan melalui ovul menunjukkan semua sel- sel dalam S. Labelkan sel-sel yang terlibat dalam persenyawaan Pollen tube Male gamete nucleus Embryo sac Ovary Diagram 4.1 Rajah 4.1
  • 39. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 11 Drawing: clear diagram with 8 nucleus – 1 mark Label : 2 label = 1 mark (iii) Describe the fertilization process that occurs. Huraikan proses persenyawaan yang berlaku . [2marks] P1 :One of the Q/ male nucleus fertilizes an egg to form the diploid zygote P2:One of the Q/ male nucleus fertilizes 2 polar nuclei to form the triploid endosperm (b) (i) In Diagram 4.1, the structure Y has to be kept dormant for future research purposes. Explain how Y can be prevented from germinating. Dalam Rajah 4.1, struktur Y perlu disimpan tidak aktif untuk tujuan penyelidikan pada masa hadapan. Terangkan bagaimana Y boleh dihalang daripada bercambah . [2 marks] P1 : Keep Y in dry place/ low temperature P2 : Because moisture initiate germination// enzyme is in inactive state (ii) Suggest one method to stimulate the germination of Y Cadangkan satu kaedah untuk merangsang percambahan Y. [1mark] Dropping/ spraying sucrose / sugary solution on Y (e) Diagram 4.2 shows a watermelon with seed and watermelon without seed.. Rajah 4.2 menunjukkan buah tembikai dengan biji dan buah tembikai tanpa biji. Polar cell Egg cell
  • 40. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 12 Nowadays, it is more easier to find seedless watermelon in market. Shoppers also can find varieties of seedless oranges, grapes, and cucumbers. Explain how to produce varieties of fruits without seed. Pada masa kini, adalah lebih lebih untuk mencari tembikai tanpa biji di pasaran. Pembeli juga boleh mendapatkan buah limau , anggur dan timun tanpa biji . Terangkan bagaimana untuk menghasilkan buah tanpa biji . [3marks] P1 : Parthenocarpy P2 : Spraying flower with auxin, P3 : stigma and anther becomes degenerate P4 : auxin diffuse into ovary and stimulate ovary to develop. 5 Diagram 5.1 shows a uriniferous tubule and its associated blood vessels. Diagram 5.2 shows cells from structure P as seen through an electron microscope. Gambarajah 5.1 menunjukkan tubul uriniferus dan salurdarah yang berkaitan. Gambarajah 5.2 menunjukkan struktur sel P seperti yang dilihat di bawah mikroskop elektron. Diagram 4.2 / Rajah 4.2 Diagram 5.1 Rajah 5.1 Y P X Q Blood Darah Blood Darah Blood vessel Salur darah
  • 41. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 13 (a) State the difference in the urea composition between the blood vessel X and Y. Nyatakan perbezaan kandungan urea antara salur darah X dan Y. Urea concentration is lowest in Y but higher in X. [1 mark] (b) Based on the Diagram 5.2 explain how the cells are structured for reabsorption of substances. Berdasarkan Gambarajah 5.2, terangkan bagaimana sel distrukturkan untuk penyerapan semula bahan. P1 :They have many/abundant mitochondria P2 : Produce a lot of energy needed for active transport or P1 :Numerous/many microvilli P2 : Increase total surface area for reabsorption [2 marks] (c) Table 5.1 shows the concentration of certain substances found in structure Q. Diagram 5.1 Rajah 5.1 Diagram 5.2 Rajah 5.2
  • 42. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 14 Jadual 5.1 menunjukkan kepekatan beberapa bahan yang terdapat di dalam struktur Q. Substances Bahan Water Air Protein Protein Glucose Glukosa Urea Urea Salts Garam Concentration Kepekatan (%) 95.0 0.00 0.00 2.00 1.50 Explain how the concentration of the substances present in Q would change after eating meat and eggs. Terangkan bagaimana kepekatan bahan-bahan yang terdapat dalam Q akan berubah selepas memakan daging dan telur. P1 :meat and eggs contains high protein/ main source of amino acid P2:(Excess) amino acids are deaminated / converted into ammonia / urea in the liver P3 :The urea is transported to the kidneys and removed as urine P4 :The concentration of urea in the urine would increase then 2.00 [3 marks] (d) Diagram 5.2 shows the flow of blood and dialysis fluid through a dialysis machine. Rajah 5.2 menunjukkan aliran darah dan bendalir dialisis melalui suatu mesin dialisis. Table 5.1 / Jadual 5.1
  • 43. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 15 DIAGRAM 5.2/ RAJAH 5.2 Tube P contain more nitrogenous waste compared to tube Q. Explain why. Tiub P mengandungi banyak bahan kumuh bernitrogen berbanding tiub Q. Terangkan mengapa. P1 - The concentration of dialysis fluid is maintained at a concentration similar to the blood plasma of healthy person P2 - the concentration of nitrogenous waste / urea / salt in P higher than dialysis fluid P3 - urea / salt diffused out into dialysis fluid P4 - through semi-permeable tubing (Any 3) [3 marks] (e) Explain the importance of kidney in maintaining human health. Terangkan kepentingan ginjal dalam mengekalkan kesihatan manusia.  To eliminate waste materials / urea / toxics / excess water / salts from the blood.  Maintaining normal osmotic pressure in the blood / constant internal environment.  Ensure an optimal physical / chemical condition (in the internal environment). [3 marks] Section B Bahagian B [40 markah]
  • 44. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 16 Answer all question from this section Jawab semua soalan dalam bahagian ini. 6(a) Diagram 6.1 shows a forearm of humans. Rajah 6.1 menunjukkan lengan atas manusia. SKEMA JAWAPAN 6(a) Similarities: S1 : Both joint has cavity filled with synovial fluid// lines with synovial membrane S2 : (synovial fluid) act as lubricant to reduce friction between two bones. S3 :The end surface of bone are covered with cartilage A joint is the location at which bones connect. They are constructed to allow movement and provide mechanical support. Explain the similarity and difference between joint S and T? Sendi adalah tempat di mana tulang-tulang bertemu. Sendi dibina untuk membolehkan pergerakan dan member sokongan mekanikal berlaku. Terangkan persamaan dan perbezaan antara sendi S dan T? [6 marks/6 markah] Joint S Sendi S Diagram 6.1 Rajah 6.1 Joint T Sendi T
  • 45. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 17 S4 : (Cartilage) to protect the bone /reduce friction between the bones S5 : Both joint are connected with ligament S6 : (Ligament) allow movement/ avoid dislocation of bone during movement Differences: Joint S Joint T D1 Hinge joint /Elbow joint Ball and socket joint/Shoulder joint D2 Allow the movement in one plane Allow rotation movement // all direction movement D3 Articulation between humerus, ulna and radius Articulation between humerus,scapula and clavicle. (b) Based on your biological knowledge, discuss the statement above. Dengan menggunakan pengetahuan biologi anda, bincangkan pernyataan di atas. [6 marks/6 markah] SKEMA JAWAPAN (b) P1 : problem / disease : arthritis/gout P2 : (diet high protein intake) cause accumulation of uric acid in the joint P3: inflammation at joint // joint become stiff and pain P4 : Lack of exercise P5 : Diet lack of calcium / vitamin D P6 : reduce the mass of bone //bone become lighter P7: practice wrong posture during activity A man has swollen ankle and is painful during movement after having a habit of taking high protein diet and practicing unhealthy lifestyle. Seorang lelaki mengalami bengkak pada buku lali dan berasa sakit ketika bergerak setelah mengamalkan pengambilan diet yang tinggi kandungan yang protein dan tidak mengamalkan gaya hidup sihat.
  • 46. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 18 (c) Diagram 6.2 show an earthworm and structure in its body. Rajah 6.2 menunjukkan seekor cacing tanah dan struktur pada badannya. Explain how the structure in the earthworm involve in their movement . Terangkan bagaimana struktur pada cacing tanah terlibat dengan pergerakannya. [4 marks/4 markah] SKEMA PEMARKAHAN (c) P1 : Hydrostatic skeleton P2 : fluid in the body cavity helps the earthworm to move / give support P3 : muscle at the body wall are longitudinal and circular muscle //antagonistic muscle P4 : contraction of circular muscle( and relaxation of longitudinal muscle) cause segment of body extended/longer/thinner P5 : contraction of longitudinal muscle (and relaxation of circular muscle) cause segment of body shorten/ thicken P6 : (The presence of) chaetae P7 : secure/anchor the shorted segment on the ground [4 marks] P8 : give pressure to skeleton system [6 marks] Diagram 6.2 /Rajah 6.2
  • 47. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 19 (d) Diagram 6.3 shows flight muscle of a bird. Rajah 6.3 menunjukkan otot penerbangan seekor burung. Explain the effect to locomotion of bird if structure W is torn. Terangkan kesan terhadap pergerakan burung kastuktur W terkoyak. [4 marks/4 mark] SKEMA PEMARKAHAN: (d) P1: W is tendon P2 : Tendon is inelastic /strong/ tough P3 : Function of tendon is to connect (pectoralis minor) muscle to bone (/humerus) P4 : Contarction of (pectoralis minor) muscle produces (pulling) force P5 : (If tendon is torn), (pulling) force (that produced by contraction of muscle) W Pectoralis minor muscle Ototpektoralis minor Humerus Humerus
  • 48. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 20 cannot be transferred to the bone P5 : the bone/humerus is not pulled upward P6 : no movement of wing [4 marks] 7 Diagram 7.1 shows part of the blood circulatory system and the lymphatic system in the human body. Rajah 7.1 menunjukkan system peredaran darah dan sistem limfa dalam badan manusia. Diagram 7.1 Rajah 7.1 (i) Explain the differences between the composition of fluid P and fluid Q
  • 49. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 21 Terangkan perbezaan antara komposisi bendalir P dan bendalir Q. [4 marks] [4 markah] (a) (i) Able to explain the diffrences of composition fluid P and fluid Q Sample answer: F1: Fluid Q/lymph has a larger numbers of lymphocyte compare to fluid P/blood P1: lymphocyte is produced by the lymph nodes in lymph system F2: Fluid Q/lymph has lower contents of oxygen compare to fluid P/blood P2: oxygen has been used up by the cell 1 1 1 1 4 (ii) Describe how the fluid Q is formed from blood until it is incorporated back into the blood circulatory system. Huraikan bagaimana bendalir Q terbentuk daripada darah sehingga bendalir tersebut masuk semula ke dalam sistem peredaran darah. (a) (ii) Able to describe how lymph is formed from blood until it is brought back into the blood circulatory system. Sample Answer : P1: (When the blood flows from arteries into capillaries) there is higher hydrostatic pressure at the arterial end of the capillaries P2: (This high pressure) forces some plasma to pass through the capillary walls into the intercellular spaces (between the cells) P3: Once the fluid leaves the capillary walls, it is called interstitial/tissue fluid // The interstitial fluid fills the spaces between the cells and constantly bathes the cells P4: 90% of the interstitial fluid diffuses back into blood capillary P5: 10% of the interstitial fluid that has not been reabsorbed into the bloodstream goes into the lymph capillaries.(Once inside the lymph capillaries) the fluid is known as lymph. 1 1 1 1 1 1 6
  • 50. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 22 [6 marks][6 markah] (c) (i) Describe how are lacteals in the villi related with the lymphatic system? Huraikan bagaimana lakteal di dalam vilus dapat dikaitkan dengan sistem limfa? (c) (i) Able to describe the how are lacteals related with the lymphatic P6: The lymph capillaries unite to form larger lymphatic vessels. P7: From the lymphatic vessels, lymph eventually passes into the thoracic duct/the right lymphatic duct. P8:The thoracic duct empties its lymph into the right subclavian vein. (Hence, lymph drains back into the blood). Any 6 P 1 1 Max 6
  • 51. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 23 system. Sample answer: P1: A lacteal is a lymphatic capillary P2: absorbs fatty acids and glycerol in the villi of the small intestine P3: lacteals merge to form larger lymphatic vessels P4: that transport the fats to the thoracic duct which empties into the left subclavian vein. 1 1 1 1 4 (ii) Helmie takes fried chicken at lunch. Explain the absorption and assimilation process of lipid content in the fried chicken. Helmie mengambil ayam goreng semasa makan tengah hari. Huraikan proses penyerapan dan asimilasi lemak yang terkandung dalam ayam goreng tersebut. [6 marks] (c) (i) Able to explain the absorption and assimilation of lipid Absorption P1: Digestion of lipid produce fatty acid and glyserol P2: Absorption of lipid occur at ileum P3: At ileum there are villi which have lacteal P4: Fatty acid and glyserol are absorbed into lacteal P5: In the lacteal condensation of fatty acid and glyserol forms lipid P6: The lipids then transported via the subclavian vein into the blood steam Assimilation P7: In the cells lipid is use as a main component of plasma membrane P8: lipid also is use as a main component of some hormone and vitamin P9: Excess lipid will be stored underneath the skin as adipose tissue 1 1 1 1 1 1 1 1 1 1 max 6
  • 52. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 24 8(a) Diagram 8.1 shows the process of colonisation and succession in a habitat. Rajah 8.1 menunjukkan proses pengkolonian dan penyesaran dalarn suatu habitat. Diagram 8.1 What is meant by "colonisation and succession in a habitat”? Based on Diagram 8.1 .explain how colonisation and succession bring about the formation of the primary forest. [10 marks] Apakah yang dimaksudkan dengan "pengkolonian dan penyesaran dalam suatu habitat"? Berdasarkan Rajah 8.1, terangkan bagaimana pengkolonian dan penyesaran membawa kepada pembentukan hutan primer dalam suatu habitat. [ 10 markah] 8.0 F1: COLONISATION The process whereby living organisms move into this newly formed area which is completely devoid of life. 1 Rhizophora sp Sonneratia sp Bruguiera sp Low tidal High tidal
  • 53. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 25 F2: SUCCESSION The gradual process where one community changes its environment so that it is replaced by another community. Zone 1 / Avicennia sp. and Sonneratia sp. P1: The pioneer species in a mangrove swamp are the Avicennia sp. and Sonneratia sp. P2: The Avicennia sp. grows in the part of the mangrove swamp that faces the sea while Sonneratia sp. grows at the mouth of the river which is sheltered. P3: A root system that spreads out widely to give support to the trees in the soft muddy soil P4: The Avicennia sp. and Sonneratia sp. have asparagus-shaped pneumatophores that grow vertically upwards from the main roots through the mud into the air. P5: The widely spread roots of the Avicennia sp. and Sonneratia sp. trap mud. P6: As more and more mud accumulate, the bank is slowly raised and would then contain less water. P7: The mangrove swamp is now more suitable for another mangrove tree which is the Rhizophora sp. Hence the Rhizophora sp. as the successor will slowly replace the pioneer species. Zone 2 / Rhizophora sp. zone P8: This zone is higher and less waterlogged. P9: The Rhizophora sp. has prop roots to support and anchor the tree in the soft muddy soil. P10: The Rhizophora sp. has viviparity seeds to ensure that the seedlings can grow and are not carried away by the seawater. P11: The prop roots of the Rhizophora sp. are able to trap mud. The pioneer species and the Rhizophora sp. that are old, will die and decay, adding humus to the soil. P12: The banks are raised up even higher. The soil becomes more solid/compact, more fertile and less saline. 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  • 54. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 26 (b) Diagram 8.2 shows sources of greenhouse gases arising from human activities and natural processes. Rajah 8.2, menunjukan sumber-sumber gas rumah hijau yang dihasilkan daripada aktiviti manusia dan proses-proses semulajadi. P13: The soil that is harder and drier now is not suitable for the Rhizophora sp. Hence, the Rhizophora sp. is replaced by the Bruguiera sp. Zone 3 / Bruguiera sp. zone P14: Trees of Bruguiera sp. grow well in hard clay soil that subjects to flooding during high tide. P15: Trees of Bruguiera sp. have buttress roots for support and knee- shaped pneumatophores for gaseous exchange (Figure 8.24(c)). P16: As more sedimentation of decayed substances occur, new mud banks are being built up seawards while the old banks move further inland, away from the sea. P17 : The soil becomes harder and dry land is formed. P18: Bruguiera sp. are replaced by other types of plants such as coconut trees, palm trees and Pandanus sp. P19: These are later replaced by other land plants. P20: Finally, after a few hundred years, the process of succession stops and a tropical rain forest, which is the climax community, is formed. At least one point in each zone. [10 marks] 1 1 1 1 1 1 Max 10
  • 55. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 27 Diagram 8.2 Rajah 8.2 Based on Diagram 8.2, Explain the green house effectand global warming as a result of human activities. Berdasarkan Rajah 8.2, terangkan kesan rumah hijau dan pemanasan global akibat aktiviti- aktiviti manusia [5 marks]. b) The formation of greenhouse effect caused by; P1: Solar radiation , containing uv rays penetrate earth atmosphere and reaches the earth surface P2: Part of uv rays is reflected back by earth's surface to atmosphere in the form of infrared radiation / light which contains heat P3: Heat (energy) is trapped by greenhouse gases (such as carbon dioxide, oxides of nitrogen, methane) P4: Human activities such as combustion of fossil fuel by factories and vehicles increase the amount of greenhouse gases P5: Higher concentration of greenhouse gases in the atmosphere results in more heat being absorbed / trapped P6: Extensive forest burning, burning of fossil fuel and higher rate of evaporation worldwide causes accumulation of great amount of water vapour in the air P7: which also contribute to the increase in the earth's temperature / causes global warming 1 1 1 1 1 1 1 Max 5 8 (c) Diagram 8.3 shows the emission of gases from factories. Rajah 8.3 menunjukkan pelepasan gas daripada kilang. Nitrogen oxide (NO) ,sulphur dioxide (SO2) Nitrogen oksida(NO), sulfur dioksida (SO2) Pond Kolam Forest Hutan
  • 56. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 28 (i) Explain the effects of the emission of the gases to the ecosystem. Terangkan kesan pembebasan gas tersebut kepada ekosistem. [5 marks] SKEMA PEMARKAHAN 8(c) P1: (The release of nitrogen oxides / sulphur dioxides) leads to the formation of acid rain P2 : the gases dissolve in the rain water P3 : Acid rain causes damage on the leaves / chloroplast P3 : Lower rate of photosynthesis P4 : Leads to stunted growth / death of plants//population reduced P5 : Acid rain lowers pH of the pond// more acidic P6 : causes death to aquatic organisms /fishes P7 : pH of soil lower//more acidic P8 : crop yield decrease [max : 5 marks] 9.(a) Diabetics do not correctly produce or use their insulin hormone. The insulin hormone helps control how much sugar is in your bloodstream. Millions of diabetics need to take insulin. Insulin from cows and pigs has been used since the early 1900s to treat diabetes. Now human insulin hormone can be mass-produced through genetic engineering processes. Pesakit kencing manis tidak dapat menghasilkan atau menggunakan insulin dengan betul. Hormone insulin membantu mengawal kandungan gula dalam aliran darah . Berjuta-juta pesakit kencing manis perlu mengambil insulin. Insulin daripada lembu dan babi telah digunakan seja kawal 1900-an untuk merawat kencing manis. Sekarang hormon insulin manusia boleh dihasilkan secarabesar-besaran melalui proses kejuruteraan genetik. Diagram 9.1shows a few stage that involves in the production of insulin hormone through genetic engineering technology. Rajah 9.1 menunjukkan sebahagian daripada peringkat yang terlibat dalam proses penghasilan hormon insulin melalui teknologi kejuruteraan genetik. Diagram 8.3/Rajah 8.3
  • 57. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 29 By using your knowledge , explain how this technology can be used in insulin hormone production. Berdasarkan pengetahuan biologi anda huraikan bagaimana teknologi ini dijalankan bagi menghasilkan hormone insulin. [6 marks/ 6 markah] SKEMA JAWAPAN 9(a) P1 -The gene for the insulin is isolated from human pancreas cell P2 - The bacterial plasmid is isolated (DNA found in bacteria) P3 - The bacterial plasmid is cut by using enzyme P4 - The enzyme used to incorporate gene for insulin production into the plasmid P5 - the bacteria are cultured in bioreactor P6 - the plasmid replicate as a bacteria divide asexually . P7 - the bacteria can produce insulin in large quantity, purified and isolate. (b) Genetic engineering (GE) is the manipulation of genetic material (DNA or genes) in a cell or an organism in order to produce desired characteristics and to eliminate unwanted ones. GE includes a range of different techniques with many different uses, Diagram 9.1 Rajah 9.1
  • 58. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 30 and can be applied to plants, animals and humans. For example, the genetic modification of food is a form of GE that involves manipulating the cells of plants. Kejuruteraan genetik (GE) adalah manipulasi bahan genetik (DNA atau gen) di dalam selatau organism untuk menghasilkan ciri-ciri yang dikehendaki dan untuk menghapuskan organisma yang tidak diingini. GE termasuk pelbagai teknik yang berbeza dengan kegunaan yang berbeza, dan boleh digunakan untuk tumbuh- tumbuhan, haiwan dan manusia. Sebagai contoh, pengubahsuaian genetic makanan adalah satubentuk GE yang melibatkan memanipulasi sel-sel tumbuh-tumbuhan. Diagram 9.2 shows twotoma to leaves which have been exposed to a bacterial pathogen, Pseudomonas syringae. Leaf A is the normal leaf show disease when infected with the bacteria while Leaf B, the genetically engineered leaf shows practically no signs of damage. Rajah 9.2 menunjukkan dua helai daun tomato yang telah didedahkan kepada sejenis pathogen bacteria ,Pseudomonas syringae. Daun A adalah daun bias yang menunjukkan tanda-tanda penyakit setelah dijangkiti oleh bacteria tersebut, manakala daun B yang telah mengalami pengubahsuaian kandungan genetiknya tidak menunjukkan tanda kerosakan. Discuss the benefits and the risks of using the genetically engineered organisms in agriculture and food production. Bincangkan faedah dan risiko menggunakan organism yang terubah suai kandungan genetiknya dalam pertanian dan penghasilan makanan. [10 marks/10 markah] Diagram 9.2/ Rajah 9.2 Leaf A /Daun A Leaf B /Daun B
  • 59. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 31 SKEMA PEMARKAHAN 9(b) The benefit: P1: Genetic engineering used to produce disease resistant /pest resistant plant. P2 : Less pesticide are used P3 : Less pollution to the environment //better health for consumer P4: Increase crop yield P5 : help to solve problem of insufficient food P6 :better livehood for farmer P7 : Increase resistance to herbicide P8 : which allow weeds to be killed without affecting the crop plant P9 : Able to survive on poorer quality grassland P10 : can resist drought // climatic changes P11 : create crops with better nutrition value P12 : with high vitamin A / protein content P13 : help to solve problems of malnutrition P14: create crops with longer shelf live P15: less food wastage P16: genetically modified livestock (eg :cow) P17: produce meat with less fat / more milk [Max : 6 marks] The risk: K1 : Pest resistant genes may be transferred to weed K2 : may be difficult to control the growth of weed K3: some transgenic crops may have animal genes K4: this may not be acceptable to certain groups for religious reasons K5: genetically modified food may be harmful to health K6: may activate human genes to cause cancer K7: transgenic organism may affected the survival of other organism in the ecosystem K8: may cause the imbalance of nature / ecosystem K9 : decrease biodiversity K10: certain cultivar are being planted to the exclusion of others K11: this will less the genetic variation in environment. [Max: 4 marks] (c) a. b. c. d. e. f. g. h. i. j. Diagram 9.3 shows a cloning process of a plant.
  • 60. PROGRAM KECEMERLANGAN AKADEMIK SEKOLAH BERASRAMA PENUH 2014 BIOLOGY 4551 | MODUL PERFECT SCORE SBP 32 k. Rajah 9.3 menunjukkan proses pengklonan satu tumbuhan. Explain the characteristic of cloned plant. Terangkan ciri-ciri tumbuhan yang diklon. [4 marks] SKEMA PEMARKAHAN 9(C) P1: Clones are genetically identical to the parent cell P2: no exchange of genetic materials P3: Clones have the same chromosomal number as the parent cell P4: no reduction in the chromosomal number P5: Clones easily get disease // shorter life span P6: Clones have the same body resistance against disease End .. Leaf cells form calluses in culture medium Sel daun membentuk kalus di dalam medium kultur Calluses develop into tiny plantlets Kalus berkembang menjadi anak pokok kecil Cloned plants Tumbuhan klon Leaf cells are taken from the parent plant Sel daun diambil dari tumbuhan induk Diagram 9.3 Rajah 9.3
  • 61. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 1 QUESTION 1: 1. A group of students carried out an experiment to study the effect of intraspecific competition on the growth of maize seedlings. Sekumpulan pelajar menjalankan satu eksperimen untuk mengkaji kesan persaingan intraspesifik ke atas pertumbuhan biji benih jagung. Diagram 1 shows the apparatus set-up of the experiment Rajah 1 menunjukkan susunan radas untuk eksperimen tersebut Step 1 : Three seedlinging trays are filled with 4 kg of garden soil. Langkah 1 : Tiga kotak semaian diisikan dengan 4 kg tanah kebun. Step 2 : The trays are labeled as P, Q and R. Langkah 2 : Kotak-kotak semaian dilabelkan P, Q dan R. Step 3 : In tray P, 50 maize seedlings are seedlinged at a distance of 15 cm intervals. In tray Q, 50 maize seedlings are seedlinged at a distance of 10 cm intervals. In tray R, 50 maize seedlings are seedlinged at a distance of 5 cm intervals. Langkah 3 : Dalam kotak P, 50 anak benih jagung ditanam pada jarak 15 cm berselang seli. Dalam kotak Q, 50 anak benih jagung ditanam pada jarak 10 cm berselang seli. Dalam kotak R, 50 anak benih jagung ditanam pada jarak 5 cm berselang seli. Step 4 : Each tray is watered daily with the same amount of water for 30 days. Langkah 4 : Setiap kotak semaian disiram tiap-tiap hari dengan jumlah air yang sama banyak untuk 30 hari. Step 5 : After 30 days, remove 30 maize seedlings randomly from tray P, tray Q and tray R. The root of seedlings are washed and wipedry. Langkah 5 : Selepas 30 hari, 30 anak benih jagung secara rawak dari kotak P, kotak Q dan kotak R. Akar anak benih dibersihkan dan dilapkan sehingga kering. Step 6 : The dry weight of the maize seedlings is recorded in Table 1. Langkah 6 : Berat kering anak benih jagung dicatatkan dalam Jadual 1. O 0 0 O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Tray P/kotak P Tray Q /kotak Q Tray R/kotak R Diagram 1/rajah 1
  • 62. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 2 Distance between maize seedlings / jarak antara anak benih jagung (cm) Dry weight of 30 maize seedlings (g) / berat kering 30 anak benih jagung (g) 15 10 200 150
  • 63. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 3 5 Table 1/Jadual 1 (a) Record the dry weight of the maize seedlings in the boxes provided in Table 1 Rekodkan berat kering anak benih jagung di dalam kotak yang disediakan dalam jadual 1. (b) (i) Based on the results in Table 1 , state two observations that can be made from this experiment. Berdasarkan keputusan di dalam Jadual 1, nyatakan dua pemerhatian yang dapat dibuat daripada eksperimen ini Observation 1/pemerhatian 1: At distance 15 cm, the dry weight of 30 paddy seedlings is 200g Observation 2/pemerhatian 2: At distance 5 cm, the dry weight of 30 paddy seedlings is 100g [3 marks] (ii) State the inference from the observations in (b) (i). Nyatakan inferens berdasarkan pemerhatian di (b) (i) Inference from observation 1/inferen dari pemerhatian 1: (At distance 15 cm), there is low intraspecific competition so the growth rate of maize plant is high Inference from observation 2/inferen dari pemerhatian 2: (At distance 5 cm), there is highintraspecific competition so the growth rate of maize plant is low [3 marks] 100
  • 64. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 4 (c) Complete table 2 based on the experiment. Berdasarkan eksperimen, lengkapkan jadual 2 di bawah [3 marks] (d) State the hypothesis for this experiment. Nyatakan hipotesis bagi eksperimen ini The further the distance between the maize seedlings, the higher the growth rate of/dry weight of maize plant//vice versa [3 marks] (e) Construct a table and record all your data collected in the experiment which include the following aspects : Bina satu jadual untuk merekodkan semua keputusan eksperimen meliputi aspek berikut :  Distance between maize seedlings/jarak antara anak benih jagung  Dry weight of 30 maize seedlings/berat kering 30 anak benih jagung  Growth rate of maize seedling/kadar pertumbuhan anak benih jagung Growth rate = Dry weight of maize seedlings Number of days Kadar pertumbuhan = Berat kering anak benih jagung Bilangan hari Variable Pembolehubah Particulars to be implemented Cara mengendalikan pembolehubah Manipulated/ manipulasi: Distance between maize seedlings Used different distance between maize seedling//used the distances at 15cm, 10cm and 5cm Responding / bergerakbalas: Dry weight of maize seedlings// growth rate Record dry weight of maize seedlings by using a weight balance // calculate the growth rate using formula Growth = Dry weight of 30 maize plant 30 days Controlled/ dimalarkan Volume garden soil // type of maize plant // size of tray Fix the volume of garden soil at 4kg // fix the same type of maize plant // fix the same size of tray Table 2// Jadual 2
  • 65. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 5 Distance between maize seedlings (cm) Dry weght of 30 maize seedlings (g) Growth rate (g/days) 15 200 6.67 10 150 5.00 5 100 3.33 [3 marks] (f) Use the graph paper provided on page 55 to answer this question. Using the data in 1 (e)draw a graph of the growth rate of maize seedlings against the distance between the maize seedlings. Dengan menggunakan kertas graf yang dibekalkan pada muka surat 55 untuk menjawab soalan ini. Dengan menggunakan data di dalam 1 (e), lukiskan graf kadar pertumbuhan anak benih jagung melawan jarak di antara anak benih jagung [3 marks] (g) Based on graph in 1 (f), explain the relationship between the growth rate of maize seedlings and distance between seedling Berdasarkan graf di 1 (f), terangkan hubungan di antara kadar pertumbuhan anak benih jagung dan jarak antara anak benih. P1: As the distance between maize increases, the growth rate of maize seedlings increases. P2: This is because there is lower intraspecific competition P3: causes the dry weight of maize seedlings increase [3 marks] (h) If the experiment is repeated by increasing the distance between the maize seedlings to 20 cm, predict the observation . Explain your prediction.. Jika eksperimen diulang dengan meningkatkan jarak di antara anak benih jagung pada 20 cm,ramalkanpemerhatian. Terangkan ramalan anda. P1: The dry weight more than 200g P2: because longer distance give more water/ nutrient / space to the maize seedling, P3: so the growth rate of maize seedlings increases [3 marks] (i) Based on this experiment, what can you deduce about intraspecific competition? Berdasarkan eksperimen ini, apakah yang dapat anda rumuskan tentang persaingan intraspesifik? P1: Intraspecific competition is the growth maize plants when it compete between themselves P2: and shown by the dry weight of maize seedling. P3: The growth rate of maize is affected by the distance between the seedlings [3 marks]
  • 66. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 6 (j) When resources are limited supply, organisms living in the same habitat will compete for the same resources. The following is a list of the resources. Apabila sumber-sumber menjadi terhad, organisma hidup di habitat yang sama akan bersaing untuk sumber yang sama. Berikut ialah senarai sumber-sumber tersebut. In Table 3, classify the resources given according to what are the resources competed by animals and resources compete by plants. Dalam Jadual 3, klasifikasikan sumber-sumber yang diberi mengikut apakah sumber- sumber yang dsaingi oleh haiwan dan sumber-sumber yang disaingi oleh tumbuhan. The resources competed by animals/ Sumber-sumber yang dsaingi oleh haiwan Resources compete by plants/ Sumber-sumber yang disaingi oleh tumbuhan Food Space Breedingmate Nutrien Water Light [3 marks] Food/makananSpace/ruang Breedingmate/pasangan mengawan Nutrient/nutrien Water/air Light/cahaya Table 3/Jadual 3
  • 67. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 7 Distance between maize seedlings (cm) 0 5 10 15 6 5 Growth rate Of maize seedlings (g/days)
  • 68. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 8 QUESTION 2 Score Explanation 01 √ Identified the problem 3 Able to state problem statement correctly P1 – light intensity P2 – rate of transpiration Sample answer: Is the light intensity increase the rate of transpiration of plant? 2 Able to state problem statement but slightly incorrect 1 Able to state idea only (not in question)//Hypothesis form. 0 No response or wrong response. √ Objective of study/Aim Able to state the objective of study correctly Sample answer: To investigate the effects of light intensity on the rate of transpiration of a balsam plant. √ Variables Able to state any one item for each variable given. Manipulated Variable : distance light sources// ligh intensity Responding Variable : Time taken for the air bubble move// rate of transpiration Fixed / Controlled Variable: temperature//type of plant 02 √ Statement of hypothesis P1 – light intensity P2 – rate of transpiration P3 – The rate transpiration / air bubble movement / is influence by light intensity 3 Able to state the hypothesis correctly by relating two variable correctly. Sample answer: The higher the light intensity, the rate of transpiration of a balsam plant increase. 2 Able to state hypothesis but slightly incorrect. 1 Able to state idea only. 0 No response or wrong response. 05 √ List of apparatus Photometer, stopwatch, cutter (knife), beaker, fluorescent lamp, meter ruler List of materials Balsam plant, Vaseline, water, tissue 3 Able to list down 4 apparatus and 3 material. 2 Able to list down 2 apparatus and 2 material. 1 Able to list down 1 apparatus and 1 material. 0 No response or wrong response. B1 – 1 √ Technique used Measure and record the time taken for the air bubble to move in a distance for 10 cm by (B1-1). 04 √ Experimental procedure 1. A suitable balsam plant is selected (K1) and is cut using a sharp knife (K1). The cut end is immediately immersed in a beaker filled
  • 69. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 9 with distilled water. (K1) 2. The cut plant is then fixed onto a photometer (K1) and the joints between the plant and the photometer are sealed using Vaseline to make them airtight (K5). 3. The laboratory curtains and doors are pulled and closed so that outside lightning will not affect the outcome of the experiment (K1). 4. A 40W(K2) fluorescent lamp is set 30 cm (K3) away from the edge of the (K3) photometer with a meter rule placed to measure the distance. 5. The air bubble in the photometer is set to 0 cm (K4). The lamp is switched on and the stopwatch is started (K4) when the air bubble cross the X mark . 6. The movement of air bubble is observed and the stopwatch is stopped when the bubble reaches Y mark, that is 10 cm (K2). 7. Record the time taken into a table(K4) . 8. Steps 4 to 7 are repeated, with the distance of the lamp are put at 40 cm(K3), 50 cm(K3), 60 cm (K3) away from the photometer. 9. All the findings are recorded into the table(K4). 3 All 5Kcriteria correct K1 – any three criteria K2 – any one criteria K3 – any three criteria K4 – any two criteria K5 – any one criteria 2 3K – 4Kcriteria correct. 1 At least 2Kcriteria correct. 0 No response or wrong response. B2 – 1 √ Presentation of data Data is present in a table with right unit for rate of transpiraton (for B2 – 1 cm/second or cm second-1) If without the unit for the rate of transpiration, give no an idea (x) and B2 - 0. √ Conclusion Write the hypothesis or another hypothesis. Sample answer: The higher the light intensity the higher the rate of transpiration. Hypothesis is accepted. 03 Report writing 3 Score 3 = 7-9 2 Score 2 = 4-6 √ Distance of lamp from the edge of the photometer (cm) Time taken for the air bubble to travel for X to Y (s) Rate of Transpiration (cm/second)
  • 70. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 10 1 Score 1 = 1-3 √ 0 No response or wrong response. Question 1: 33 Marks Question 2: 17 Marks (Total = 50 marks) END OF THE SCHEME MARKING
  • 71. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 1 SKEMA BIOLOGI KERTAS 3 Question Criteria Score 1 (a) Able to record all the data for volume of the urine collected and average of urine produce correctly. Sample answers Type of Vegetables Amount of Filtrate needed to decolorize DCPIP solution (ml) Cauliflower 4.2 Broccoli 2.5 Lime 3.6 Ascorbic Acid 1.0 3 Able to record 3 data correctly 2 Able to record 2 data correctly 1 Able to record only 1 data / wrong response 0 (b) (i) Able to state two different observation correctly base on the criteria: C1 – Type of vegetables C2 – Amount of Filtrate needed to decolorize DCPIP solution (ml) Sample answers 1. The amount of Cauliflower/ Broccoli/ Lime/ Ascorbic Acid filtrate needed to decolorize DCPIP solution is 4.2 ml/ 2.5 ml/ 3.6ml/ 1.0ml 2. The amount of Cauliflower filtrate needed to decolorize DCPIP solution is higher than the amount of Broccoli filtrate. 3 Able to state one correct observation and one inaccurate observation. Sample answer (inaccurate) 1. The amount of cauliflower filtrate needed to decolorize DCPIP solution the highest 2 Able to state only one correct observation or two observations at idea level. Sample answer (idea) 1. The amount of vegetables filtrate needed to decolorize DCPIP solution is different 1 No response or incorrect response or inaccurate observation or one idea only. 0
  • 72. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 2 Question Criteria Score (b) (ii) Able to make two accurate inferences base on two criteria: C1 – amount of vegetable filtrate C2 – content of vitamin C Sample answer 1. Amount of vegetable filtrate needed to decolorize DCPIP solution is high, the content of vitamin C in the vegetables filtrate is low. 2. Amount of vegetable filtrate needed to decolorize DCPIP solution is low, the content of vitamin C in the vegetables filtrate is high. 3 Able to make one correct inference and one inaccurate inference or able to state two inaccurate inferences. Sample answer (inaccurate) 1. More / high / much (amount) of vegetable filtrate. 2. Low/ high content of Vitamin C 2 Able to sate one correct inference or two inferences at idea level. Sample answer (idea level) 1. The content of vitamin C is different 1 No response or inaccurate response 0 Summary of scoring for 1(b)(i) and 1(b)(ii) Score Correct Inaccurate Idea Wrong 3 2 - - - 2 1 1 - - - 2 - - 1 1 - 1 - - - 2 - - 1 1 - 0 - 1 - 1 - - 1 1
  • 73. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 3 Question Criteria Score (c) Able to state all variables and methods to handle each variable correctly. Sample answer Variable Method Manipulated variable: Types of Filtrate Use different type of filtrate Responding variable:  Volume of filtrate needed to decolorize DCPIP solution  Measure and record the volume of vegetable filtrate needed to decolorize DCPIP solution using syringe  Calculate and record the percentage of Vitamin C using the formula : Amount of Ascorbic Acid needed to decolorize DCPIP solution(ml) x 0.1% Amount of Filtrate needed to decolorize DCPIP solution(ml) Initial volume of filtrate/ volume of DCPIP solution Fix the initial volume of filtrate which is 5ml / volume of DCPIP solution which is 1ml Able to state 4 – 5 ticks 2 Able to state 2 – 3 ticks 1 No response or incorrect response or 1 tick only 0 (d) Able to state the hypothesis relating the manipulated variable and the responding variable correctly based on three criteria: P1 : Manipulated variable (volume of water intake) P2 : Responding variable ( volume of urine collected) H : Relationship of the variables. Sample answer 1. Broccoli has the highest amount of Vitamin C compared to 3
  • 74. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 4 Cauliflower, and Spinach. P1 + P2 + H Able to state the hypothesis based on any two criteria. Sample answer 1. The amount of filtrate needed to decolorize DCPIP solution affects the content of Vitamin C. 2. Different amount of filtrate needed to decolorize DCPIP solution, different content of Vitamin C. P1 + P2 // P1/P2 + H 2 Able to state the hypothesis based on any one criteria or idea level. Sample answer 1. The content of Vitamin C is different. 1 Question Criteria Score (e) (i) Able to construct a table correctly with the following aspects: T : Title with correct unit - 1 mark D : Data - 1 mark C : Percentage of Vitamin C - 1 mark Sample answer Type of filtrate The amount of filtrate needed to decolorize DCPIP solution (ml) Percentage of Vitamin C (%) Cauliflower 4.2 0.02//0.022 Broccoli 2.5 0.04 Lime 3.6 0.03//0.027 Ascorbic Acid 1.0 0.1 3 Any two aspect correct 2 Any one aspect correct 1 Incorrect response 0 (ii) Able to draw the graph of volume of water reabsorb against volume of water intake based on the following aspects: P (Paksi): Title of x-axis and y-axis with unit - 1 mark Title (Title) : Four points plotted correctly - 1 mark B (Bentuk) : All points connected smoothly - 1 mark All three aspect correct 3
  • 75. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 5 Any two aspect correct 2 Any one aspect correct 1 No response or Incorrect response 0 (f) Able to explain the relationship between the amount of vegetable filtrate needed to decolorize DCPIP solution and type of filtrate base on the following criteria. R1 : Relationship - R2 : the amount of vegetable filtrate to decolorize DCPIP R3 : the concentration of vitamin C Sample answer Broccoli has the highest amount of vitamin C compared to cauliflower and Lime. This is because the amount of vegetable filtrate needed to decolorize DCPIP is low, thus the concentration of vitamin C is high enough which needed only small amount of flitrate to reduce DCPIP solution. 3 Able to explain the relationship using any two aspects 2 Able to explain the relationship using any one aspect only 1 No response or Incorrect response 0 Question Criteria Score (g) Able to predict and explain the volume of urine produced based on the following criteria : P1 : Prediction - More than 4.5 ml P2 : Explanation - denaturation of vitamin C P3 : Concentration of filtrate needed to decolorize DCPIP solution Sample answer Volume of filtrate needed to decolorize DCPIP solution is more/ higher than 4.5ml. This is because boiling causes Vitamin C to denature. Thus there is less/ no vitamin C concentration in filtrate to decolorize DCPIP solution Able to predict and explain the volume of urine produced based on any two criteria 2 Able to predict and explain the volume of urine produced based on any one criteria 1 No response or Incorrect response 0 (h) Able to define Vitamin C operationally based on the following criteria. D1 : D2 : D3 :
  • 76. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 6 Sample answer 3 Any two criteria stated 2 Any one criteria stated 1 No response or Incorrect response 0 Question Criteria Score (i) Able to classify apparatus and materials into their respective variables. Sample answer Materials Apparatus Ascorbic Acid Fruit Juices DCPIP Solution Syringe with needle Specimen Tube Measuring Cylinder All 6 corrects in its appropriate correct. 3 5 – 4 correct 2 3- 2correct 1 1 – 0 No response or Incorrect response 0
  • 77. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 7 Paper 3 – No 2 (Scheme) Item Criteria Marks Problem statement Able to state the problem statement of the experiment correctly based on three criteria:  Manipulated variables – (different) water sources/ samples (P1)  Responding variables – time taken to decolourise methylene blue solution // level of polution (P2)  Relationship in question form and question symbol [?] (H) Sample Answer 1. What is the time taken to decolourise methylene blue solution// level of pollution of different water sources/samples ? 2. Does different water sources/samples has different time taken to decolourise methylene blue solution// level of pollution ? 3 marks Able to state the problem statement of the experiment correctly based on any two criteria. Sample Answer 1. What is the time taken to decolourise methylene blue solution// level of pollution of different water sources/ samples. 2 marks Able to state the problem statement of the experiment correctly based on only one criteria. Sample Answer 1. What is the time taken to decolourise methylene blue solution / level of pollution of water sources/ sample. 2. Water sources/ samples have different level of pollution. 1 mark Wrong or no response 0 mark
  • 78. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 8 Item Criteria Marks Hypothesis Able to state the hypothesis of the experiment correctly based on three criteria:  Manipulated variables – different water sources/ samples (P1)  Responding variables – time taken to decolourise methylene blue solution /level of pollution (P2)  Relationship – more/less than//higher/lower than // shortest//most (H) Sample Answer 1. Water sample A/ drain water took the shortest time to decolourise methylene blue solution compared to water sample B, C and D. 2. Drain water /water sample A is the most polluted samples of water collected. 3 marks Able to state the hypothesis of the experiment correctly based on two criteria: Sample Answer 1. Different sources of water samples affect the time taken for the methylene blue solution to decolourise . 2 marks Able to state the hypothesis of the experiment correctly based on one criteria: Sample Answer 1. The drain water is polluted 1 mark Wrong or no response 0 mark
  • 79. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 9 Item Criteria Marks Variables Able to state the variables of the experiment correctly that include three criteria: Manipulated : different water sources/ samples Responding : time taken to decolourise methylene blue solution // level of polution Constant : volume of water samples// volume of methylene blue solution 3 marks Able to state the variables of the experiment correctly that include two criteria. 2 marks Able to state the variables of the experiment correctly that include one criteria 1 mark Wrong or no response 0 mark Item Criteria Marks Materials Methylene blue solution, water samples from A, B, C and D Apparatus Syringe with needle, stop watch, Reagent bottle, stopper and beaker 3 marks 2M + 5A Materials
  • 80. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 10 Two materials Apparatus Any three apparatus (syringe and stop watch are compulsory ) 2 marks 2M + 3A Materials Two materials Apparatus Two apparatus ( syringe and stop watch ) 1 mark 2M + 2A Cannot state the functional materials and apparatus 0 mark Item Criteria Marks Procedure Able to state five Ks correctly. K1 : Preparation of materials & apparatus (3K1) K2 : Operating fixed variable (1K2) K3 : Operating responding variable (1K3) K4: Operating manipulated variable (1K4) K5 : Precaution (1K5) 3 marks Able to state any three Ks correctly 2 marks
  • 81. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 11 Able to state any two Ks correctly 1 marks Able to state only one K correctly or no response 0 marks Example for procedure K1: Preparation of materials & apparatus 1. Water samples are collected from four different sources A,B,C and D 2. The reagent bottles are labelled 1, 2, 3 and 4 3. The reagent bottles are closed with the stopper immediately 4. The stopwatch is activated 5. The bottles are examined from time to time 6. The results are tabulate. K2: Operating fixes variable 1. Measure 250 ml of water sample from A, B, C and D separately and pour into the reagent bottle labelled 1, 2, 3 and 4 respectively 2. 1 ml of methylene blue solution is added to each water sample using a syringe K3: Operating responding variable 1. The time taken for the methylene blue solution to decolourise is measured and recorded in the table using stop watch. K4: Operating manipulated variable 1. Measure 250 ml of water samples from A, B, C and D separately and pour into the reagent bottle labeled 1, 2, 3 and
  • 82. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 12 4 respectively Item Criteria Marks K5: Precaution 1. 1 ml of methylene blue solution is added to the base of each water sample using a syringe with needle 2. The reagent bottles are closed with stopper immediately 3. The contents of the bottles cannot be shaken. 4. All the reagent bottles are kept in a dark cupboard Sample answer: Method / procedure : 1. Water samples are collected (K1) from 4 different water resources (K3) *(A,B,C and D). 2. The reagent bottles are labelled 1, 2, 3 and 4 3. 250 ml of water samples (K2) is measured (K1) from 4 different water Resources (A, B, C and D) separately and pour (K1) into the reagent bottle labeled 1, 2, 3 and 4 respectively. 4. 1 ml (K2) of methylene blue solution is added (K1) to the base (K5) of each water sample using a syringe with needle. 5. The reagent bottles are closed with stopper immediately. (K5) 6. The content of the bottles cannot be shaken. (K5) 7. All the reagent bottles are kept in a dark cupboard. (K5) 8. The stop watch is activated. (K1) 9. The bottles are examined (K1) at one hour interval. 10. The time taken for the methylene blue solution to decolourise is measured and recorded for all the water samples using stop watch. (K3) 11. The result is tabulated. (K1) *Tap, drain, river, pond.
  • 83. PROGRAM KECEMERLANGAN SPM SEKOLAH BERASRAMA PENUH, KPM 2014 BIOLOGY 4551 MODUL PERFECT SCORE SBP 13 Presentation of data 1. Able to state all the titles correctly with units. 2. Able to state 4 water samples. Water samples Time taken for methylene blue solution to decolourise (hour ) Tap water Drain water River water Pond water 2 marks Able to state any one criteria only 1 mark END OF MARKING SCHEME