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37 SULIT 2 455/2 Section A Bahagian A [60 marks/ 60 markah] Answer all questions this section Jawab semua soalan dalam bahagian ini. Figure shows human s cheek cell observed through an electron microscope. Rajah menunjukkan sel pipi manusia di bawah mikroskop elektron. (a)(i) Name the structure labeled R and. Namakan struktur berlabel P dan Q. Figure / Rajah P :. (a)(i) Q : [2 marks/2 markah] 2 (a)(ii) P and Q involve in the process of enzyme synthesis. State the roles of P and Q in this process. P dan Q terlibat di dalam proses sintesis enzim. Nyatakan peranan P dan Q di dalam proses ini.... [2 marks/ 2 markah] (a)(ii) 2

38 SULIT 3 455/2 (b)(i) Cheek cell and onion epidermal cells are observed under light microscope. Draw a labeled diagram of an onion epidermal cell observed under the light microscope. Sel pipi dan sel epidermis bawang dilihat di bawah mikroskop cahaya. Lukiskan gambarajah berlabel untuk sel epidermis bawang yang dilihat di bawah mikroskop cahaya. (b)(i) [2 marks/2 markah] 2 (ii) State two differences in structure between a cheek cell and an onion epidermal cell that is observed under the microscope. Nyatakan dua perbezaan antara struktur sel pipi dan sel epidermis bawang yang dilihat di bawah mikroskop... [2 mark/ 2 markah] (b)(ii) 2 (c) If structure R is removed from a cell. Explain what would happen to the growth of the cell. Jika struktur R dibuang daripada sel. Terangkan apakah yang akan berlaku kepada tumbesaran sel... [2 marks/2 markah] (c) 2

39 SULIT 4 455/2 (d) Explain why meristematic cells have more organelle S compared to cheek cells. Terangkan mengapa sel meristem mempunyai lebih banyak organel S berbanding sel pipi..... [2 marks/2 markah] (d) 2 TOTAL 2 2 Figure 2 shows two cell division processes P and Q that occur in two different types of cell. Rajah 2 menunjukkan dua proses pembahagian sel iaitu P dan Q yang berlaku dalam dua jenis sel yang berbeza. Figure 2/ Rajah 2

40 SULIT 5 455/2 (a)(i) State the importance of processes P and Q. Nyatakan kepentingan bagi proses P dan Q P:. Q: [2 marks/2 markah] (ii) Where do processes P and Q occur in an animal? Di manakah proses-proses P dan Q berlaku dalam haiwan? P: Q: [2marks/ 2 markah] 2(a)(i) 2 2(a) (ii) 2 (b) (i) X and Y are stages in processes P and Q respectively. Describe the behavior of chromosomes in X and Y. X dan Y adalah peringkat-peringkat dalam proses P dan proses Q. Huraikan perlakuan kromosom dalam X dan Y. X: Y: [2 marks/ 2 markah] 2(b)(i) 2 (ii) Draw the cell behavior in stages X and Y in the space provided. Lukiskan perlakuan sel di peringkat X dan Y dalam ruang yang disediakan. X Y 2(b) (ii) 2 [2marks/ 2markah]

41 SULIT 6 455/2 (c) State one differences between process P and Q. Nyatakan satu perbezaan antara proses P dan Q.... [ marks/ markah] 2(c) (d)(i). Which process related to the formation of cancerous cells? Proses yang manakah berkaitan dengan pembentukan sel-sel kanser? [mark/ markah] 2(d) (i) (ii) Explain how cancerous cells are formed in human body. Terangkan bagaimana sel-sel kanser terbentuk dalam badan manusia.. [2marks/ 2 markah] 2(d)(ii) 2 TOTAL 2

42 SULIT 7 455/2 3. Figure 3 shows a part of paddy field ecosystem. Rajah 3 menunjukkan sebahagian daripada ekosistem dalam sawah padi. Figure 3/ Rajah 3 3(a) Based on Figure 3, name the producers in the paddy field ecosystem. Berdasarkan Rajah 3, namakan pengeluar di dalam ekosistem sawah padi. 3(a) [ mark/ mark] (b) Give two ways in which energy may be lost in the food web. Nyatakan dua cara bagaimana tenaga boleh hilang di dalam siratan makanan. [ marks/ markah] 3(b) 2

43 SULIT 8 455/2 (c) Calculate how much energy is received by the secondary consumer in the food web if the energy in producer is 0 000kJ. Kirakan jumlah tenaga yang diterima oleh pengguna kedua di dalam siratan makanan jika jumlah tenaga di dalam pengeluar adalah 0 000KJ. 3(c) 2 [2 marks/ 2 markah] 3(d)(i) Based on the food web, draw and name the organism in each trophic level in the pyramid of numbers. Berdasarkan siratan makanan, lukis dan namakan organisma pada setiap aras tropik dalam pyramid nombor itu. 3(d)(i) [2 marks/ 2 markah] 2

44 SULIT 9 455/2 3(d)(ii) (e) (f) State two differences that you can see from the base to the top of the pyramid number. Nyatakan dua perbezaan yang dapat dilihat dari tapak ke puncak piramid nombor. :.. 2 :.. [2 marks/ 2 markah] Explain the interaction between frog and snake in controlling the population size in the ecosystem. Terangkan hubungan di antara katak dan ular dalam mengawal saiz populasi dalam ekosistem.. [3 marks/ 3 markah] Decomposers play an important role in an ecosystem. Name one example of decomposer and explain its role in an ecosystem. Pengurai memainkan peranan penting di dalam ekosistem. Namakan satu jenis pengurai dan terangkan peranannya di dalam ekosistem. 3(d)(ii) 2 3(e) 3 [2 marks/ 2 markah] 3(f) 2 TOTAL 4

45 SULIT 0 455/2 4. Figure 4. shows a human vertebra. Rajah 4. menunjukkan vertebra manusia. P Q R Figure 4. / Rajah 4. (a) Name the part labeled P and R. Namakan bahagian yang berlabel R. [ mark/ markah] 4(a) (b) State the function of Q. Nyatakan fungsi Q.... [ mark/ markah] 4(b) (c)(i) State one type of mineral that is essential for the strength of this vertebra. Nyatakan satu jenis mineral yang diperlukan untuk menguatkan vertebra ini.. [ mark/ markah] 4(c)(i) (c)(ii) Explain how the lack of mineral stated in (c)(i) leads to osteoporosis. Huraikan bagaimana kekurangan mineral yang dinyatakan di (c)(i) membawa kepada osteoporosis... [2 marks/ 2 markah] 4(c)(ii) 2

46 SULIT 455/2 (c)(iii) Suggest one way on how to reduce the risk of this disease. Cadangkan satu cara bagaimana untuk mengurangkan risiko mendapat penyakit ini... [ mark/ markah] 4(c)(iii) Figure 4.2(a) shows the cross section of a water hyacinth stem and figure 4.2 (b) shows the cross section of a woody plant. Rajah 4.2(a) menunjukkan keratan rentas batang pokok keladi bunting dan rajah 4.2 (b) menunjukkan keratin rentas bagi pokok berkayu. Figure 4.2(a) Rajah 4.2(a) Figure 4.2 (b) Rajah 4.2 (b) (e)(i) What is structure P? Apakah structure P? [ mark/ markah] 4(e)(i) (e)(ii) Explain how structure P helps the plant in support. Terangkan bagaimana struktur P membantu tumbuhan dalam sokongan. [3 marks/ 3 markah] 4(e)(ii) 3

47 SULIT 2 455/2 (f) Describe how woody plant is supported by structure Q in figure 4.2 (b). Terangkan bagaimana pokok berkayu disokong oleh struktur Q dalam rajah 4.2 (b). 4(f) 2 [2 marks/ 2 markah] TOTAL 2 5. Figure 5. shows the structure of a type of nucleic acid. Rajah 5. menunjukkan struktur sejenis asid nukleik. 5(a) Figure5./ Rajah 5. Based on the Figure 5, name the type of nucleic acid as shown. Berdasarkan Rajah 5, namakan jenis asid nukleik yang ditunjukkan... [ mark / markah] 5(a)

48 SULIT 3 455/2 5(b) Draw and label a basic unit of structure show in figure 5.. Lukis dan labelkan unit asas bagi struktur yang ditunjukkan dalam rajah 5.. 5(b) [2 marks / 2 markah] 2 5(c) 5(d) Explain the role of structure in figure 5. in determining the characteristics of organisms. Terangkan peranan struktur dalam rajah 5. dalam menentukan ciri-ciri organism.. [3 marks / 3 markah] Name another type of nucleic acid. Namakan satu lagi jenis asid nukleik. 5(d)(i) 5(e) 3... [ mark / markah]

49 SULIT 4 455/2 5(e)(i) (e) (i) 2 Figure 5.3/ Rajah5.3 Complete the structure in figure 5.3. Lengkapkan struktur dalam rajah 5.3. [2 marks / 2 markah] (e)(ii) State the component that determines the characteristics of an organism. Nyatakan komponen yang menentukan ciri-ciri sesuatu organisma. [ marks / markah] TOTAL 2

50 SULIT 5 455/2 PAPER 2 SECTION B Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini. 6 The photographs in Figure 6. show three individuals with different needs for energy. Fotograf dalam Rajah 6. menunjukkan tiga individu dengan keperluan tenaga yang berbeza. A lady athlete/ Olahragawati ( 000kJ) A pregnant lady/ Wanita mengandung (0 000kJ) An old lady/ Perempuan tua (6500kJ) Figure 6./ Rajah 6. (a)(i) What is a balanced diet? Why do we need a balanced diet? Apakah gizi yang seimbang? Kenapa kita memerlukan gizi yang seimbang? [2 marks / 2 markah]

51 SULIT 6 455/2 (a)(ii) Based on your biological knowledge about balanced diet, explain the factors that determine the energy requirement for the three individuals in figure 6.. Berdasarkan pengetahuan biologi anda tentang gizi seimbang, terangkan faktor-faktor yang menentukan keperluan tenaga bagi tiga individu dalam rajah 6.. [8 marks / 8 markah] (b) Figure 6.2 shows human s digestive system. X is part of the cross section of structure Y. Rajah 6.2 menunjukkan sistem pencernaan manusia. X ialah sebahagian daripada keratan rentas struktur Y. X Y Figure 6.2/ Rajah 6.2

52 SULIT 7 455/2 (i) What are the processes that occur in structure Y? Describe the processes by giving examples. Apakah proses-proses yang berlaku dalam struktur Y? Dengan memberikan contoh huraikan proses-proses itu. [5 marks/ 5 markah] (b)(ii) Describe the adaptations of X that allow structure Y to carry out its function efficiently. Huraikan penyesuaian-penyesuaian X yang membolehkan struktur menjalankan fungsinya dengan efisien. [5 marks/ 5 markah] 7. Figure 7 shows the events leading to the greenhouse effect Gambarajah 7 menunjukkan fenomena kesan rumah hijau. Figure 7/ Rajah7 (a) With reference to figure 7, describe the steps involved in the formation of the greenhouse effect Dengan merujuk kepada rajah 7, huraikan langkah-langkah yang terlibat dalam pembentukan kesan rumah hijau. [5 marks /5 markah]

53 SULIT 8 455/2 (b) In many parts of the world, forests are being cut down and burned. Explain why this may be contributing to the greenhouse effect. Di kebanyakan negara, hutan ditebang dan dibakar. Terangkan mengapa keadaan ini menyumbang kepada kesan rumah hijau. [5 marks/5 markah] (c) Describe the other causes besides deforestation and the overall effects of the enhanced greenhouse effect. Huraikan penyebab lain selain penebangan dan kesannya ke atas kesan rumah hijau. [0 marks/0 markah] 8(a) Graphs 8. (a) and 8. (b) show the growth curve of human and insect. Based on the graphs, compare the growth process in human and insect. Graf 8.(a) dan 8.(b) menunjukkan lengkung pertumbuhan bagi manusia dan serangga. Berdasarkan graf, bandingkan proses pertumbuhan manusia dan serangga. Height (cm) Time (year) Graph 8..(a) : Growth curve for human : Lengkung Pertumbuhan manusia.

54 SULIT 9 455/2 Length (cm) Time (day) Graph 8. (b) : Growth curve for insect : Lengkung pertumbuhan serangga. [0 marks/ 0 markah] (b) A couple, Mr Zafrie and Mrs Munirah had married almost ten years but still do not have any child. After undergone an inspection, they found that the husband does not have any problem but the Fallopian tube of Mrs Munirah is blocked. Pasangan suami isteri, Encik Zafrie dan Puan Munirah telah berkahwin hampir sepuluh tahun tetapi masih tidak mempunyai anak. Setelah pemeriksaan dilakukan, didapati suaminya tidak mempunyai sebarang masalah tetapi tiub Fallopio Puan Munirah telah tersumbat. Based on your biological knowledge in reproduction technology, explain how the couple can have children. Berdasarkan pengetahuan biologi anda dalam teknologi pembiakan, terangkan bagaimanakah pasangan ini untuk memperolehi anak. [0 marks/0 markah]

55 SULIT /2 9 (a) Figure 9 shows various types of thumbprints. Rajah 9 menunjukkan pelbagai jenis cap jari. Figure 9/ Rajah 9 (i) Explain the type of variation shown in Figure 9. Terangkan jenis variasi yang ditunjukkan dalam Rajah 9. [4 marks/ 4 markah] (ii) Compare the variation shown in figure 9 with the type of variation shown by body mass. Bandingkan variasi yang ditunjukkan dalam rajah 9 dengan jenis variasi yang ditunjukkan oleh jisim badan. [6 marks/ 6 markah]

56 SULIT 2 455/2 (b) Two parents who are both thin and are able to roll their tongues have a son who is fat and is able to roll their tongue. Explain how this happens. Ibubapa yang mana kedua-duanya adalah kurus dan boleh menggulung lidah mempunyai seorang anak yang gemuk dan boleh menggulung lidah. Terangkan bagaimana ia berlaku. [0 marks/ 0 markah] END OF QUESTION PAPER

57 SULIT 2 455/3 Answer all questions. Jawab semua soalan.. A group of students conducted an experiment to investigate how different amount of water intake influence the urine production. The volumes of water intake and urine produced after half an hour is shown in Table.. This experiment was repeated twice. Sekumpulan pelajar telah menjalankan eksperimen untuk mengkaji bagaimana pengambilan amaun air yang berbeza mempengaruhi penghasilan air kencing. Isipadu air yang diambil dan air kencing yang dihasilkan selepas setengah jam ditunjukkan di dalam Jadual.. Eksperimen ini diulangi sebanyak 2 kali. Table. shows the volume of water intake and the urine produced after half an hour. Jadual. menunjukkan isipadu air yang diambil dan air kencing yang dihasilkan selepas setengah jam. 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

58 SULIT 3 455/3 Table. Jadual. 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

59 SULIT 4 455/3 (a) Complete Table. to show the volumes of urine production after half an hour. Lengkapkan Jadual. untuk menunjukkan isipadu air kencing yang terhasil selepas setengah jam. [ 3 marks / 3 markah ] (a) 3 (b)(i) State two observations that can be made from this experiment based on Table. Nyatakan dua pemerhatian yang boleh dibuat daripada eksperimen berdasarkan Jadual. Observation / Pemerhatian : Observation 2 / Pemerhatian 2: [ 3 marks / 3 markah ] (b)(i) 3 (ii) State the inferences from the observations in (c) (i) Nyatakan inferens daripada pemerhatian di (c) (i) Inference from observation / Inferens daripada pemerhatian : Inference from observation 2 / Inferens daripada pemerhatian 2: [ 3 marks / 3 markah ] (b)(ii) 3 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

60 SULIT 5 455/3 (c) Complete Table.2 based on this experiment. Lengkapkan Jadual.2 berdasarkan eksperimen ini. Variable Method to handle variable Pembolehubah Cara mengendali pembolehubah Manipulated variable: Pembolehubah dimanipulasi.. Responding variable: Pembolehubah bergerakbalas Constant variable: Pembolehubah dimalarkan Table.2 Jadual.2 (c) [ 3 marks / 3 markah ] 3 (d) State the hypothesis for this experiment. Nyatakan hipotesis bagi eksperimen ini. [ 3 marks / 3 markah ] (d) 3 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

61 SULIT 6 455/3 (e) (i) Construct a table and record all the data collected in this experiment. Bina satu jadual dan rekodkan semua data yang dikumpul dalam eksperimen ini. Your table should have the following aspects. Jadual anda hendaklah mengandungi aspek- aspek berikut: - Volume of water intake / Isipadu air yang diambil - Average of volume of urine produced / Purata isipadu air kencing yang terhasil - Percentage of urine produced / Peratus air kencing yang terhasil Percentage of urine produced Peratus air kencing yang dihasilkan = = Volume of urine produced X 00% Volume of water intake Isipadu air kencing terhasil X 00% Isipadu air yang diambil (e)(i) [ 3 marks / 3 markah ] 3 (ii) Use the graph paper provided on page 8 to answer this question. Using the data in (e) (i) draw a bar chart to show the relationships between volume of water intake with the percentage of urine produced. Guna kertas graf yang disediakan di halaman 8 untuk menjawab soalan ini. Dengan menggunakan data di (e) (i) lukis satu carta bar untuk menunjukkan hubungan antara isipadu pengambilan air dengan peratusan air kencing yang dihasilkan. [ 3 marks / 3 markah ] (e)(ii) 3 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

62 SULIT 7 455/3 (f) Based on the bar chart in (e) (ii), explain the relationships between the volume of water intake to the percentage of urine produced. Berdasarkan carta bar dalam (e) (ii), terangkan perkaitan isipadu pengambilan air dengan peratusan air kencing yang dihasilkan.... [ 3 marks / 3 markah ] (f) 3 (g) State the operational definition of urine production based on this experiment. Nyatakan definisi secara operasi bagi penghasilan air kencing berdasarkan eksperimen ini..... [ 3 marks / 3 markah ] (g) 3 (h) If the students in this experiment is asked to stay in a room with 6ºC temperature, predict the volume of urine produced after half an hour. Explain your prediction. Jika pelajar yang terlibat di suruh berada di dalam bilik bersuhu 6ºC, ramalkan isipadu air kencing yang dihasilkan selepas setengah jam. Terangkan ramalan anda. [ 3 marks / 3 markah ] (h) 3 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

63 SULIT 8 455/3 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

64 SULIT 9 455/3 (i) (i) Based on Table., classify the apparatus and materials used in this experiment. Berdasarkan Jadual., kelaskan radas dan bahan yang digunakan di dalam eksperimen ini. Apparatus / Radas Materials / Bahan (i) [ 3 marks / 3 markah ] 3 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

65 SULIT 0 455/3 2. Diagram 2 / Rajah 2 Eutrofication occurs usually due to the inflow of organic matter into the river. The increase in nutrients may cause producers such as algae to grow in abundance. When a large number of algae die, decomposition by bacteria and fungi utilizes a large amount of dissolved oxygen in the water causing the BOD value and level of pollution to increase. Eutrofikasi biasanya berlaku disebabkan pengaliran bahan organik ke dalam sungai. Peningkatan bahan-bahan nutrien menyebabkan pertumbuhan pengeluar seperti alga berlaku dengan pesat. Apabila sebahagian besar alga ini mati, penguraian oleh bakteria dan kulat menggunakan banyak oksigen terlarut yang menyebabkan nilai BOD dan aras pencemaran meningkat. Based on the above statement and Diagram 2, design a laboratory experiment to study the level of water pollution in three river water samples taken from Village X, Y and Z. Berdasarkan maklumat dan Rajah 2 di atas, rancangkan satu eksperimen makmal untuk mengkaji aras pencemaran air pada tiga sampel air sungai yang diambil daripada Kampung X, Y dan Z. 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

66 SULIT 455/3 The planning of your experiment must include the following aspects: Perancangan eksperimen anda hendaklah meliputi aspek-aspek berikut: Problem statement / Pernyataan masalah Objective of investigation / Objektif penyiasatan Hypothesis / Hipotesis Variables / Pembolehubah List of apparatus and material / Senarai radas dan bahan Technique used / Teknik yang digunakan Experimental procedure / Kaedah eksperimen Conclusion / Kesimpulan [7 marks / 7 markah] END OF QUESTION PAPER 455/3 200 Hak cipta Panitia Biologi Negeri Perlis [Lihat sebelah

67 SKEMA JAWAPAN Biologi Kertas 200 No Answers No Answers. C 26. C 2. C 27. D 3. C 28. B 4. A 29. D 5. C 30. A 6. B 3. B 7. C 32. A 8. A 33. C 9. C 34. A 0. B 35. C. D 36. C 2. D 37. A 3. C 38. D 4. B 39. D 5. C 40. C 6. A 4. D 7. C 42. C 8. B 43. C 9. A 44. A 20. C 45. D 2. D 46. A 22. A 47. A 23. C 48. B 24. D 49. C 25. C 50. A

68 SULIT Skema Trial Biology 200 SCHEME PAPER 2 SECTION A: structured questions Item No. Suggested Answers: Marks (a) (i) P : Rough endoplasmic reticulum Q : Golgi body (2marks) (ii) - Ribosome at P synthesizes protein - Protein that is synthesized is transported by P to Q - Q modifies the protein to enzymes/q packages the enzyme/protein (3 marks) (b) (i) Cell wall vacuole nucleus Shape and structure: cytoplasm 3-4 labels correct: (2 marks) (ii) The onion epidermal cell has cell wall while cheek cell do not have cell wall ( mark) (c) - The cell does not divide/differentiate - because there are no genes (that control mitosis/differentiation) (2 marks) (d) - S (mitocondria) : it generates energy - Meristematic cells require more energy to undergo mitosis/cell Division (2 marks) TOTAL: 2 PANITIA BIOLOGI NEGERI PERLIS 200

69 SULIT 2 Skema Trial Biology (a)(i) P : To produce gamete cell Q : Repair damage tissue / for growth (ii) P: In reproductive organ// ovary// testis Q: in all somatic cells (2 marks) (2 marks) (b)(i) X: metaphase Y: anaphase Both answers correct: ( mark) (ii) X : metaphase Y : anaphase Correct diagram award mark (2 marks) (c). crossing over occur in P but not in Q 2. Process P produces 4 daughter cells whereas process Q produces 2 daughter cells. 3. Daughter cells produced in P have variation whereas daughter cells produced in Q are identical to parent cell. Any two (2 marks) (d)(i) Process Q ( mark) (ii) F : The cancer cells occur due to severe disruption of mechanism that controls the cell cycle. E : The cells divide without control and regulation to form cancerous cells// uncontrolled division of cells forms cancerous cells. F & E = 2, F=, E = 0 TOTAL: 2 PANITIA BIOLOGI NEGERI PERLIS

70 SULIT 3 Skema Trial Biology (a) Paddy plant ( mark) 3(b) Sunlight ( mark) 3(c) (i) respiration (ii) excretion (iii) defaecation any two (2 marks) 3(d) Kj 0% 000 Kj 0% 00Kj 3(e)(i) Producer Primary consumer Secondary consumer 0 / 00 X kj = 000 kj 0 / 00 X 000 kj = 00 kj Snake Bird / frog Grasshopper / caterpillar / bird Paddy plant (2 marks) Correct Shape m Label: 2-4 correct m correct 0m (2 marks) (e)(ii) (f). Dari tapak ke atas bilangan organism semakin bertambah 2. Saiz organism dari tapak ke atas semakin bertambah Bacteria and fungi. They break down waste products and dead bodies of other organisms into simpler substances to be used again by plants. (2 marks) 2 (2 marks) Total : 3 PANITIA BIOLOGI NEGERI PERLIS

71 SULIT 4 Skema Trial Biology (a) R : centrum ( mark) (b) P : muscle attachment Q : the placing of spinal cord (2 marks) (c) Lumbar vertebra ( mark) (d)(i) calcium/ phosphorus ( mark) (ii) osteoporosis // Porous / brittle bone ( mark) (iii) taking a diet rich in calcium/phosphorus and vitamin D // drink milk, regular exercise (any one) ( mark) (e) (i) P: air sac ( mark) (ii) P: to reduce the density of aquatic plant // to keep aquatic plant light ( mark) (iii) -able to float -get enough sun light - carry out photosynthesis (any two) (2 marks) TOTAL: 5(a) DNA ( mark) 5(b) Carbon, hydrogen, oxygen, nitrogen and phosphorus ( mark) 5(c ) - Carry genetic information - Direct protein synthesis (2 marks) 5(d)(i) Nucleotide ( mark) (ii) X : Phosphate group Y : Pentose sugar / Sugar Z : Nitrogenous base (3 marks) (iii) Z ( mark) 5(e) RNA ( mark) 5(f) P : Adenine / Q:Thymine // P : Cytosine / Q: Guanine // P: Thymine / Q: Adenine // P : Guanine / Q: Cytosine (any two correct pairs) (2 marks) TOTAL: 2 PANITIA BIOLOGI NEGERI PERLIS

72 SULIT 5 Skema Trial Biology 200 Section B: ESSAY Item No. Suggested Answers: Marks 6 (a)(i) A balanced diet is the foods that contain correct proportion of nutrients which include carbohydrates, proteins, lipids, vitamins, minerals, water and dietary fibre / roughage// A balanced diet is one which contains the correct proportions of all the different food requirement of the body. We need a balanced diet to supply enough energy for each day s activities (2 marks) (ii) A lady athlete: F: An athlete is a very active person and has high rate of metabolism to produce energy. E: The diet should include more carbohydrates to supply enough energy to carry out the vigorous activity in sports.// She needs to contract and relax her muscles frequently for her vigorous activities. //Energy is needed to contract the muscles. E2: The diet should include more protein to build new tissues to replace tissues that are dead or damaged. E3: She also needs calcium, sodium and potassium to strengthen the bones and to prevent muscular cramp. A pregnant lady: F2: A pregnant lady has a high rate of metabolism to provide energy for herself and the baby. E4: The pregnant lady also needs more iron and calcium to build red blood cells to avoid anemia. E5: She needs a high quantity of calcium and phosphate to form strong teeth and bones for the baby. An old lady: F3: An old lady has low rate of metabolism as she does not need energy to grow. (age) PANITIA BIOLOGI NEGERI PERLIS

73 SULIT 6 Skema Trial Biology 200 E6: An old lady needs less carbohydrates and fats because she is less active and thus do not need much energy. E7: she needs more proteins, vitamins and minerals to replace dead tissues and maintain her daily activities E8:She needs calcium and phosphorus to prevent osteoporosis E9: She should avoid food that contains a lot of fats, sugar and salt because excess fat can lead to heart diseases, excess sugar can cause diabetes mellitus and excess salt can cause high blood pressure. F, F2 and F3 and any five E: (a) (i) F: Digestion // Intestinal glands of the wall of ileum secrete a few enzymes to complete the digestion process. Digestion process is completed in ileum to produce simple sugars (glucose, fructose and galactose), amino acids, fatty acids and glycerol. Example: (any correct enzymes/ intestinal juices and substrate reaction). Enzyme erepsin(peptidase) --- peptide to amino acids. 2. Enzyme sucrose -----sucrose to glucose and fructose F2: Absorption // The wall of ileum has many projections called villus to absorb the products of digestion. Blood capillaries in the villus absorb simple sugars, amino acids, minerals, vitamins B and C Lacteal of the villus absorb fatty acids, glycerol, fat soluble vitamins (A,D,E,K) Must have F and F2 and other 3 (8 marks) (5 marks) (ii) F: X is villus - has very thin epithelium that is only one cell thick. E: The thin epithelium facilitate the diffusion of digested food// enable digested food to move across easily. F2: X is greatly folded structure E2: Provides a large surface area for efficient absorption of digested food. F3: It has a mass network of blood capillaries PANITIA BIOLOGI NEGERI PERLIS

74 SULIT 7 Skema Trial Biology 200 E3: to transport the digested food such as glucose, amino acids, minerals and vitamins B and C that has been absorbed. F4: Each villus has a lacteal E4: to transport lipid soluble nutrient// fatty acids, glycerol and lipids soluble vitamins. F, F2, F3 and any 2 suitable E (5 marks) 7(a) - solar radiation warms the earth s surface - heat energy is reradiated back into the atmosphere - some heat energy escapes into space - heat energy is reradiated back to the earth - by greenhouse gases such as carbon dioxide and methane (5 marks) (b) - in slash-and-burn agriculture, the burning trees increases the concentration of carbon dioxide in the atmosphere - trees photosynthesis using carbon dioxide thus deforestation leads to less carbon dioxide being removed from atmosphere - less carbon dioxide is able to be locked up in trees. - As trees die and tree stumps left to rot, microbial activity during decomposition also releases carbon dioxide - carbon dioxide is greenhouse gas. It absorbs energy radiated by earth, some of these energy reradiated back thus increasing the surface temperature (5 marks) (c) - the greenhouse effect is further worsened by the combustion of fossil fuel in motor vehicles in power stations to generate electricity and in industries - these activities produce large quantities of carbon dioxide into the atmosphere - the setting up of large cattle ranches to cater for the growing human PANITIA BIOLOGI NEGERI PERLIS

75 SULIT 8 Skema Trial Biology 200 population leads to massive emissions of methane gas from the belching of these animals and the decomposition of their wastes. - As the world population strives for higher standards of living, there is an increased use of CFCs in products like air-conditioning and during the production of packaging materials - The rapid decomposition of humus, combustion of fossil fuels use of nitrogen or nitrate fertilizer contributes to the production of nitrogen oxides which is another greenhouse gas - The increase in the concentration of these greenhouse gases would then lead to increasing surface temperatures on earth as they trap and reflect heat back to the earth - This may lead to an increase in photosynthesis which is about the only positive effect of an enhanced greenhouse effect - More alarmingly however the greenhouse effect may change the local and global climate causing a shift in vegetation and vectors - This may also lead to extinction of certain species of organisms which are unable to adapt to the changing environment - Glaciers may start to melt thus increasing the sea levels. This leads to flooding in low lying areas around the globe (0 marks) 8(a) Similarities - height of man / length of instar increases by time - both show horizontal line / constant growth during adult *2 marks for similarities, Difference - Form of graph Sigmoid form for human and like series of steps in insect - Age of organism the height measured yearly, but in insect used day for measuring the length - Caused of different human have endoskeleton but insect have exoskeleton - Stages involve in human, the curve has three different phases, but there are five steps in insect // nymphal stages - Vertical and horizontal line : curve for human did not shows different line (only the curve from continuous points), but there are five different horizontal and vertical lines each - Zero growth no point to show zero growth in human, but there are 5 8 marks for differences PANITIA BIOLOGI NEGERI PERLIS

76 SULIT 9 Skema Trial Biology 200 time of zero growth (at horizontal line) - Sudden growth : no sudden growth for human, but there are sudden growth in insect (at vertical line) - Ecdysis : no ecdysis in human but ecdysis occurred in insect - Mitosis : the cells in human undergo mitosis all the time, but in insect, mitosis only occurred at certain time (during ecdysis) - Absorption of air : in human, there are no absorption of air, but in insect, during ecdysis Max 0 marks (b) F : In vitro technique P : means fertilization occurs outside of the body P2 : the wife injected with hormone to fasten the development of the secondary oocyte / ovum in ovary P3 : secondary oocyte / ovum release out by using before ovulation) laparoscope (from ovary Max 0 marks P4 : sperms from husband fertilize with the secondary oocyte in a Petri dish (contains culture medium) P5 : (after fertilization) zygote will divide by mitosis P6 : (after 2 days) formed embryo until eight cells stage P7 : embryo transferred to the endometrium of the uterus wall through cervix (using pippet) P8 : embryo implant at the endometrium of uterus wall and develop P9 : baby that is delivered is called test tube baby P0 : This technique is complicated / expensive and the probability to success is low PANITIA BIOLOGI NEGERI PERLIS

77 SULIT 0 Skema Trial Biology 200 9(a)(i) - Discontinuous variation - Contrasting features - No intermediate values - Caused by genetic factors - Can be inherited Max : 4 (ii) Body mass Thumbprints - Continous variation - Discontinous variation - Features change gradually - Contrasting features - Intermediate values - No intermediate values Caused by genetic factor and affected by environmental factors - Cannot be inherited if characteristic affected by environmental factors - Caused by genetic factor - Can be inherited Graph shows normal distribution - Graph shows discrete distribution 2 Max : 0 (b) - Thin parents can have a fat son even though the son inherits genes for thinness from his parents - Body size is affected by environmental factors - Continuous variation PANITIA BIOLOGI NEGERI PERLIS

78 SULIT Skema Trial Biology Due to diet, the size of body changes - The ability to roll the tongue is determined by genetic factors - Alleles for the ability to roll the tongue can be inherited - Genes for this characteristic is dominant - Hence, the son can roll his tongue Max : 6 PANITIA BIOLOGI NEGERI PERLIS 200

79 Skema jawapan biologi MARKING SCHEME : PAPER THREE TRIAL BIOLOGY 200 Question : (a) Score Explanation Able to record all readings of volume of urine produced 3 Student Volume of urine in the first experiment/ml Volume of urine in the second experiment/ml A B C D Able to record any 4-5 volume Able to record any 2-3 volume 0 No response or wrong response Score 3 (b)(i) Explanation Able to state two correct observations based on following criteria. C volume of water intake C2 volume of urine produced Sample Answer:(either 2):. When the volume of water intake is 00 ml, the volume of urine produced in first experiment is 80ml and second experiment is 82 ml. 2. When the volume of water intake is 400 ml, the volume of urine produced in first experiment is 360 ml and second experiment is 370 ml. 2 Able to state one correct observation and one inaccurate response. Able to state one correct observation or two inaccurate response or idea. 0 No response or wrong response (response like hypothesis) Score 3 (b) (ii) Explanation Able to state two reasonable inferences for the observation. Sample answer:. When the volume of water intake is less, more water is reabsorbed, less urine is produced 2. When the volume of water intake is more, less water is reabsorbed, more urine is produced 2 Able to state one correct inference and one inaccurate inference. Able to state one correct inference or two inaccurate inference or idea. 0 No response or wrong response (inference like hypothesis) Peperiksaan Percubaan Biologi SPM 200 Panitia Biologi Negeri Perlis

80 Skema jawapan biologi 2 Score 3 (c) Explanation Able to state all the variables and the method to handle variable correctly ( ) for each variable and method Manipulated Variable: Volume of water intake ( ) Method to handle: Repeat the experiment by drinking different volume of water ( ) Responding Variable: Volume of urine produced ( ) Method to handle: Measure and record the volume of urine produced after half an hour by using measuring cylinder.( ) Controlled variable : type of water intake/ duration to collect urine ( ) Method to handle:. drink same type of water/ fix the time to collect urine ( ) Able to get 6 (with the correct key words) 2 Able to get 4 5 Able to get No response or wrong response Score 3 (d) Explanation Able to state the hypothesis correctly based on the following criteria: V State the volume of water intake V2 State the volume of urine produced R - State the relationship between V and V2. The more the volume of water intake, the more the volume of urine produced. 2 Able to state the hypothesis but less accurate. Able to state the idea of the hypothesis 0 No response or wrong response Score (e)(i) Explanation Able to construct a table and record the result of the experiment with the following criteria: volume of water intake(ml) ( ) average of volume of urine produced (ml) ( ) - percentage of urine produced (%) ( ) If without unit (x) 3 Students Volume of Volume of urine Average of Percentage of water produced/ml volume of volume of urine intake/ml First Second urine produced / % experiment experiment produced/ml A B C D Able to construct a table and record any two criteria Able to construct a table and record any one criteria 0 No response or wrong response Peperiksaan Percubaan Biologi SPM 200 Panitia Biologi Negeri Perlis

81 Skema jawapan biologi 3 (e)(ii) Score Explanation Able to draw a bar chart of percentage of urine produced against the volume l of water intake. Axes (A) both axis are labeled with units, uniform scales, independent variable on horizontal axis. ( ) Point (P) All points are correctly plotted. ( ) Shape (S) All bars are correctly drawn ( ). 3 2 Graph with any two criteria. Graph with any one criteria. 0 No response or wrong response. Score 3 (f) Explanation Able to explain the relationship between the volume of water intake and the volume of urine produced correctly. When the volume of water intake is more, the volume of urine produce also more because less water is reabsorbed Able to explain briefly the relationship between the volume of water intake and volume 2 of urine produced Able to explain the idea the relationship between the volume of water intake and volume of urine produced 0 No response or wrong response Peperiksaan Percubaan Biologi SPM 200 Panitia Biologi Negeri Perlis

82 Skema jawapan biologi 4 Score 3 (g) Explanation Able to state the definition of urine correctly, based on the following criteria. C waste product (in the form of liquid) C2 excreted by human C3 influence by volume of water intake Urine is a waste product in the form of liquid excreted by human and influent by the volume of water intake 2 Able to state the definition of urine based one of the two criteria. Able to state the idea of urine 0 No response or wrong response Score 3 (h) Explanation Able to predict correctly and explain the prediction based on the following item: C the volume of urine produced C2 blood osmotic pressure C3 - reabsorption of water The volume of urine produced is less than 78 ml because after drinking 5% sodium chloride solution, the blood osmotic pressure increases, therefore more water is reabsorbed thus the volume of urine produced is less. 2 Able to predict based on any two criteria. Able to predict based on any one criteria. 0 No response or wrong response Score (i) Explanation Able to classify the apparatus and material used in the experiment 3 Apparatus Measuring cylinder Drinking bottle Stopwatch Materials Plain water Able to classify all the apparatus and material correctly. Able to classify two apparatus and one material correctly. 2 Able to classify one apparatus and one material correctly. 0 No response or wrong response Peperiksaan Percubaan Biologi SPM 200 Panitia Biologi Negeri Perlis

83 Skema jawapan biologi 5 Question 2 : Aspect Sample Answer Remarks Problem statement Which water sample is most polluted? / Which area contains water that is most polluted? 3 marks Aim/objective Hypothesis Variables To study the level of water pollution in three different water samples. The nearer the area of water sample to the factory/farm, the higher the level of water pollution. Manipulative variable: river water samples Responding variable: time taken for methylene blue solution to decolourise Fixed variable: volume of water sample/concerntration of methylene blue solution 3 marks Apparatus and materials Apparatus: measuring cylinder, stopwatch, reagent bottle, collecting cup Materials : 0.% methylene blue solution, water samples All present 3 marks 2 materials and 3-4 app 2 marks 2 materials and 2 app mark Technique used Procedure Observe and record the time taken by methylene blue solution to decolourise. Water sample are collected from three different area by using collecting cup KP,KMV 2. Three reagent bottles are labelled X,Y and Z and filled with 00 ml of water sample from village X, Y and Z respectively.-kfv & KMV 3. The tests are run for all the water samples on the same day-kp, KFV 4. A syringe is used to add ml of 0.% methylene blue solution to the base of each the water sample.kp, KFV 5. Make sure the reagent bottles are closed quickly and placed in the dark cupboard.-kc 6. The stopwatch is started.-kp 7. The bottles are examined at hour intervals- KP, KFV 8. The time taken for the methylene blue to decolourise is recorded for all the water samples- KRV 9. The results are recorded in the table.kp B 8-9P 3 m 6-7P 2 m 3-5P m 3 marks KP Step,3,4,6,7,9 (any 5) KMV Step,2 KRV Step 8, KFV- Step2,3,4,7 KC Step 5 Peperiksaan Percubaan Biologi SPM 200 Panitia Biologi Negeri Perlis

84 Skema jawapan biologi 6 Presentation of data Able to draw a complete table to record the relevant data base on the 3 criteria: Water samples Time taken for methylene blue to decolourise Level of water pollution Sample Answer Water sample X Y Z Time taken for methylene blue solution to decolourise/min B Level of water pollution Conclusion Planning KB06203 Hypothesis is accepted. The nearer the water sample to the farm/factory, the higher the level of water pollution / The nearer the water sample to the farm/factory, the shorter the time taken for methylene blue to decolourise Able to state correctly 8 9 aspects (correct) - 3 marks 6 7 aspects (correct) - 2 marks 3 5 aspects (corect) - mark < 3-0 mark 3 marks Report Able to state correctly: presentation of data and technique. 2 correct - 2 marks correct - mark TOTAL 2 marks 7 Marks END OF MARKING SCHEME Peperiksaan Percubaan Biologi SPM 200 Panitia Biologi Negeri Perlis