BIOLOGI Kertas 2 Dua jam tiga puluh minit

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1 455/2 Biologi kertas 2 Ogos/ Sept ½ jam Nama: PEPERIKSAAN PERCUBAAN BERSAMA SIJIL PELAJARAN MALAYSIA 2008 ANJURAN BERSAMA PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH CAWANGAN NEGERI PERLIS DAN JABATAN PELAJARAN PERLIS BIOLOGI Kertas 2 Dua jam tiga puluh minit JANGAN BUKA SOALAN SEHINGGA DIBERITAHU. Tuliskan nama anda pada ruang yang disediakan. Bahagian Soalan Markah penuh Markah diperolehi 2. Kertas soalan ini adalah dalam dwibahasa. 3. Soalan dalam Bahasa Inggeris mendahului soalan yang sepadan dalam Bahasa Melayu. 4. Calon dibenarkan menjawab keseluruhan atau sebahagian soalan sama ada dalam Bahasa Melayu atau Bahasa Inggeris. 5. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini. A B Jumlah Kertas soalan ini mengandungi 23 halaman bercetak termasuk kulit. 455/2 [Lihat sebelah

2 SECTION A BAHAGIAN A [60 marks] Diagram shows the structure of a leaf cell as seen under electron microscope. Rajah menunjukkan struktur sel daun dilihat di bawah mikroskop elektron. Diagram Rajah (a) On Diagram, label the parts P, Q, R and S. Pada rajah, labelkan bahagian P, Q, R dan S. (a) [2 marks] (b)(i) On Diagram, use letter T to label the part of the cell where glucose is synthesised. Pada Rajah, gunakan huruf T untuk melabelkan bahagian pada sel yang mensintesis makanan. [ mark] (b)(i) (ii) Name the part of the leaf where T is most abundantly found. Namakan bahagian pada daun di mana T terdapat dengan banyak. [ mark] (b)(ii) (iii) Explain how T is adapted to synthesise glucose. Terangkan bagaimana T diadaptasikan untuk mensintesis glucose..... [2 marks] (b)(iii) 455/2 [Lihat sebelah

3 (c) (i) Water enters the plant through the root hair cells. In the space below, draw and label root hair cell as seen under a microscope. Air memasuki tumbuhan melalui sel rerambut akar. Pada ruang yang disediakan di bawah, Lukis dan label sel rerambut akar seperti yang dilihat di bawah mikroskop (c)(i) [2 marks] (ii) Give two differences between the root hair cell and the cell shown in diagram. Berikan dua perbezaan di antara sel rerambut akar dan sel pada rajah.... (c)(ii) [2 marks] (d) Describe how water is absorbed by the root hair. Terangkan bagaimana air diserap oleh rerambut akar. (d) [2 marks] JUMLAH 455/2 [Lihat sebelah

4 2 X : C 6 H 2 O 6 + 6O 2 Y : C 6 H 2 O 6 6CO 2 + 6H 2 O + energy 2C 2 H 5 OH + 2CO 2 + energy The equation X and Y above shows reaction that accurs in the respiratory process. Persamaan X dan Y di atas menunjukkan tindakbalas yang berlaku semasa proses pernafasan (a)(i) State the type of respiration represented by the above equation. Nyatakan jenis respirasi yang diwakili oleh persamaan di atas. X :. Y :. 2(a)(i) [2 marks] (ii) Define process X. Definisikan proses X 2(a)(ii) [ mark ] (iii) Define process Y Definisikan proses Y 2(a)(iii) [ mark ] (b) Besides energy, what are the end products of anaerobic respiration in plants and animals? Selain tenaga, apakah hasil akhir respirasi anaerobik dalam tumbuhan dan haiwan. [2 marks] 2(b) 455/2 [Lihat sebelah

5 (c) Anaerobic respiration produces less energy compared to aerobic respiration. State the reason for this. Respirasi anaerobik menghasilkan tenaga yang kurang berbanding dengan respirasi aerobik. Berikan alasan anda. [ 2 marks] 2(c) (d) Yeast carries out fermentation. Write word equation for the process mentioned. Yis menjalankan fermentasi. Tuliskan persamaan perkataan bagi proses tersebut.... [ mark ] 2(d) (e) A person who does strenuous exercise experiences oxygen debt ". Describe "oxygen debt ". Seseorang yang melakukan latihan aktif yang berterusan mengalami keadaan "hutang oksigen". Terangkan hutang oksigen".... [ 3 marks ] 2(e) JUMLAH 455/2 [Lihat sebelah

6 3 A study on rat population in an oil palm plantation was carried out. On the first day, a total of 00 rats were caught with rat traps. The rats were marked and then released. After week the traps were used again. There were 40 marked rats out of a total of 40 caught. Satu kajian ke atas populasi tikus di sebuah ladang kelapa sawit telah dijalankan. Pada hari pertama, sejumlah 00 ekor tikus telah ditangkap menggunakan perangkap tikus. Tikus-tikus itu ditanda dan dilepaskan. Selepas satu minggu perangkap yang sama telah digunakan semula. Terdapat 40 ekor tikus yang bertanda daripada 40 yang ditangkap. (a) Name the technique used in this study. Namakan teknik yang digunakan dalam kajian ini. [ mark] 3(a) (b)(i) Suggest a suitable material which can be used to mark the specimens. Cadangkan satu bahan yang sesuai yang boleh digunakan untuk menandakan spesimen.... [ mark ] 3(b)(i) (ii) Give two reasons for your answer in (b) (i). Beri dua sebab bagi jawapan anda di (b) (i) [ 2 marks ] 3(b)(ii) 455/2 [Lihat sebelah

7 (c) Calculate the rat population in the oil palm plantation. Hitung populasi tikus di dalam ladang kelapa sawit itu. 3(c) [2 marks ] (d) Why the second capture was only carried out after one week? Kenapa tangkapan kedua hanya dibuat selepas satu minggu?. [ mark ] 3(d) (e) (f) If another survey is carried out after three months, name the factors which would affect the size of the rat population in the oil palm plantation. Explain briefly. Jika satu lagi kajian dijalankan selepas tiga bulan, namakan faktor-faktor yang akan mempengaruhi saiz populasi tikus di ladang kelapa sawit itu. Terangkan secara ringkas. [ 2 marks ] Suggest what would happen if excessive nitrate fertilizer from the oil palm plantation flows into a nearby river. Cadangkan apa yang akan berlaku jika baja nitrat berlebihan dari ladang kelapa sawit mengalir ke dalam sungai berhampiran. 3(e) 3(f) [3 marks ] 455/2 [Lihat sebelah

8 JUMLAH 4 An experiment was carried out to determine how the level of glucose in blood varies in a normal person. Blood samples were taken at hourly intervals for a period of 6 hours. During that period three meals were taken as the following: Satu eksperimen telah dijalankan untuk menentukan bagaimana aras glukosa berbeza dalam seorang manusia yang normal. Sampel darah diambil setiap satu jam selama 6 jam. Dalam waktu itu tiga hidangan berikut telah diambil. Breakfast Sarapan Lunch Makan tengahari Tea Minum petang The results are tabulated in table Keputusan direkodkan dalam jadual Time Masa Blood glucose level (mmoldm - ) Aras glukosa darah (mmoldm - ) (a) Plot a graph to show the changes in blood glucose level during the 6-hour period. Plot satu graf untuk menunjukkan aras glukosa dalam darah dalam masa 6 jam. 455/2 [Lihat sebelah

9 [3 marks] 4(a) 455/2 [Lihat sebelah

10 455/2 [Lihat sebelah

11 (b) (i) Based on the graph, state the changes in the blood glucose level: Berdasarkan graf, nyatakan perubahan aras glukosa dalam darah: between 0700 and 0800 hours antara jam 0700 dan (b)(i) [mark] (ii) between 0800 and 200 hours antara jam 0800 dan 200. [ mark] 4(b)(ii) (c) Explain how the changes in (b)(i) and (b)(ii) are brought about. Terangkan bagaimana perubahan di (b)(i) dan (b)(ii) berlaku. 4(c) [3 marks] 455/2 [Lihat sebelah

12 (d) Explain the variations in blood glucose level during the period between 800 and 2200 hours. Terangkan perubahan aras glukosa dalam darah dalam jangkamasa antara 800 dan 2200 jam (d). [3 marks] (e) State one reason why it is important to maintain blood glucose within the normal range. Nyatakan satu sebab mengapa ia sangat penting untuk mengekalkan aras glukosa darah dalam julat yang normal. [ marks] 4(e) JUMLAH 455/2 [Lihat sebelah

13 5 Diagram 5 (i) shows the fusion between two varieties of homozygote corn, one coloured and smooth seed and another one colourless and shrink seed. The relationship between alleles Y, y, L, l are as the following: Rajah 5 (i) menggambarkisahkan kacukan antara dua variasi jagung yang homozigot, satu mempunyai biji berwarna dan licin, manakala satu lagi mempunyai biji tak berwarna dan lisut. Hubungan antara alel-alel Y, y, L, l adalah seperti berikut: Diagram 5 (i) Allel alel Physical of seed Sifat biji Dominant / Recessive kedominanan Y Colored/ Berwarna Dominant/Dominan y Colorless/ Tidak berwarna Recessive/Resesif L Smooth/ Licin Dominant/Dominan l Shrink/ Lisut Recessive/Resesif 455/2 [Lihat sebelah

14 (a)(i) On diagram 5 (i) label alleles that are found in the F generation of corn. Pada Rajah 5(i), labelkan alel-alel yang terdapat dalam jagung generasi F. 5(a)(i) [ mark] (ii) What is the phenotype shown by F generation of corn? Apakah fenotip yang akan ditunjukkan oleh jagung generasi F?. 5(a)(ii) [ mark] Diagram 5 (ii) show two probabilities that undergo by F generation of corn in its gamete formation. Rajah 5 (ii) menunjukkan dua kemungkinan yang dialami oleh jagung generasi F dalam pembentukan gamet-gametnya st probability kemungkinan pertama F generation Generasi F 2 nd probability Kemungkinan kedua Diagram 5 (ii) Rajah 5 (ii) 455/2 [Lihat sebelah

15 (b)(i) Name the process that happen during meiosis to produce a different gamete in 2 nd probability. Namakan proses yang berlaku dalam meiosis yang menghasilkan gamet berlainan jenis untuk kemungkinan kedua.... 5(b)(i) [ mark] (ii) State at which stage does the process happen in b (i)? Nyatakan pada peringkat manakah berlaku proses di b (i)?. [ mark] 5(b)(ii) (c) On Diagram 5 (ii), draw the different gametes produced for the 2 nd probabilities. Pada Rajah 5(ii), lukiskan gamet-gamet berlainan yang terhasil bagi kemungkinan kedua. 5(c) [ mark] (d) Which probability will give more variation to the corn in producing the young? Explain the reason for your answer. Kemungkinan yang manakah akan memberikan lebih variasi kepada jagung ini dalam penghasilan zuriatnya? Terangkan sebab bagi jawapan anda (d) [2 marks] (e) Why must the meiosis process happen? Mengapakah proses meiosis mesti berlaku? 5(d) [2 marks] 455/2 [Lihat sebelah

16 (f) F generation of corn that undergoes 2 nd probability process is fused again with the colourless and shrink seed. Complete Table 2 by filling in the F generation gamete of 2 nd probability drawn in diagram 5 (ii), F2 genotype generation and F2 phenotype generation produced. Jagung generasi F yang telah mengalami proses kemungkinan kedua, dikacukkan semula dengan induk yang mempunyai biji tak berwarna dan lisut. Lengkapkan Jadual 2 dengan mengisikan gamet generasi F yang dilukis di (c), genotip generasi F2 dan fenotip generasi F2 yang akan terhasil, F generation gamete (2 nd probability) Gamet Generasi F Parent s Gamete Gamet Induk F2 generation (genotype) Genotip Generasi F2 generation (phenotype) Fenotip Generasi (kemungkinan kedua) F2 F2 Table 2 5(f) Jadual 2 [3 marks] JUMLAH 455/2 [Lihat sebelah

17 SECTION B BAHAGIAN B 6. (a) P, Q and R show the mechanisms of enzyme reaction that occur in living cells. P, Q dan R menunjukkan mekanisme tindak balas enzim yang berlaku dalam selsel hidup. Diagram 6 (a) Rajah 6 (a) Enzymes which are isolated from cells can function outside the cells. Enzyme can be used as catalysts in industries. The use of enzymes in industrial processes is known as Enzyme Technology. Enzim yang dipencilkan daripada sel boleh berfungsi dengan baik di luar sel. Enzim boleh digunakan sebagai pemangkin dalam industri. Kegunaan enzim di dalam proses industri dikenali sebagai Teknologi Enzim. Based on diagram 6 (a) and the above statement, Berdasarkan rajah 6 (a) dan pernyataan di atas, (i). List the general characteristics of enzymes. Senaraikan ciri-ciri am enzim. [ 4 marks ] (ii). Using suitable examples, discuss the uses of enzymes in industrial processess and our daily life. Dengan menggunakan contoh yang sesuai, bincangkan kegunaan enzim dalam proses industri dan kegunaan harian. [ 6 marks ] 455/2 [Lihat sebelah

18 (b) Diagram 6 (b) shows the organelles involved during the synthesis and secretion of an enzyme in animal cell. Rajah 6(b) menunjukkan organel-organel yang terlibat dalam sintesis dan rembesan enzim pada sel haiwan. Diagram 6 (b) Rajah 6 (b) Based on diagram 6(b), explain how extracellular enzymes are produced by emphasizing on the role of P, Q, R and S. Berdasarkan rajah 6(b), terangkan bagaimana enzim ekstrasel dihasilkan dengan menekankan peranan P, Q, R dan S. [0 marks] 455/2 [Lihat sebelah

19 7. Diagram 7 (a) shows an animal stomach. Rajah 7 (a) menunjukkan perut sejenis haiwan. Diagram 7(a) Rajah 7 (a) (a) (i) Identify the type of this animal and give two examples of this type of animal. (i) Kenalpasti jenis haiwan ini dan berikan dua contoh haiwan jenis ini. [2 marks] (ii) How does this animal s digestive system differ from human s digestive system? (ii) Bagaimanakah sistem pencernaan haiwan ini berbeza dengan sistem pencernaan manusia? [2 marks] 455/2 [Lihat sebelah

20 Diagram 7 (b) shows human digestive system. Diagram 7 (b) Rajah 7(b) (b)(i) Describe the functions of T in diagram 7(b). (i) Huraikan fungsi-fungsi bahagian yang berlabel T pada rajah 7(b). [8 marks] (ii) Digestion of food occurs actively in the part labelled R which receives juices secreted by the parts labelled T and Q. Explain the digestion of food in the part labelled R. Equations in words should be given to show the reactions which occur during digestion. (ii) Pencernaan makanan berlaku secara aktif pada bahagian yang berlabel R yang menerima rembesan enzim dari bahagian berlabel T dan Q. Terangkan pencernaan makanan yang berlaku di bahagian R. Persamaan dalam bentuk perkataan perlu dinyatakan untuk menunjukkan tindak balas yang berlaku semasa pencernaan. [8 marks] 455/2 [Lihat sebelah

21 8. Diagram 8 (a) shows a few types of food which teenagers like to eat. Rajah 8 (a) menunjukkan beberapa jenis makanan yang digemari oleh remaja. Salted plum Potato chips Prawn crackers Plum jeruk kerepek kentang keropok udang Diagram 8 (a) Rajah (a) (a) (i) State the main preservative containing in the food shown in Figure 8 (a). Nyatakan kandungan bahan awet utama di dalam makanan yang ditunjukkan dalam Rajah 8 (a). [ mark] (ii) Discuss the effect of eating large quantity of the above food on the internal environment. Bincangkan kesan pengambilan makanan diatas dalam kuantiti yang banyak terhadap persekitaran dalam. [9 marks] 455/2 [Lihat sebelah

22 Diagram 8 (b) shows the human organs that involved in regulating internal environment. Rajah 8 (b) menunjukkan organ-organ manusia yang terlibat dalam pengawalaturan persekitaran dalam. Diagram 8(b) Rajah 8(b) (b) Based on diagram 8 (b) explain how these organs involve in monitoring changes in internal environment through negative feedback mechanism. Berdasarkan rajah 8(b) terangkan bagaimana organ-organ di atas dapat mengawal atur perubahan persekitaran dalam melalui mekanisme suap balik negatif. [0 marks] 455/2 [Lihat sebelah

23 9. Variation appears among the members of the same species. Variasi wujud di antara ahli-ahli dalam satu spesies (a) Based on the above statement and your biological knowledge, describe how genetic factor becomes the cause of variation. Berdasarkan penyataan di atas dan pengetahuan biologi anda, huraikan bagaimana faktor genetik menyebabkan berlakunya variasi. (0 marks] (b) A study about haemophilia disease was done to two couples of teenagers who plan to get married. Table 9 shows the result of the study. Satu penyiasatan tentang penyakit hemofilia telah dijalankan ke atas dua pasangan remaja yang ingin berkahwin. Jadual 9 menunjukkan keputusan kajian tersebut. Couple / Pasangan A B Genotype / Genotip Man Lelaki Woman Perempuan Man Lelaki Woman Perempuan X h Y X H X H X H Y X h X h Table 9: The result of the experiment on haemophilia disease. Jadual 9 : Keputusan kajian tentang penyakit hemofilia Based on your biological knowledge, give your advice and justification to both couples. Berdasarkan pengetahuan biologi anda, berikan nasihat dan penerangan anda kepada kedua-dua pasangan. [0 marks] END OF QUESTION PAPER 455/2 [Lihat sebelah

24 INFORMATION FOR CANDIDATE MAKLUMAT UNTUK CALON. This question paper consists of two sections: Section A and Section B. Kertas soalan ini mengandungi dua bahagian: Bahagian A dan Bahagian B. 2. Answer all questions in Section A. Write your answers for Section A in the spaces provided in the question paper. Jawab semua soalan dalam Bahagian A. Tulis jawapan bagi Bahagian A dalam ruang yang disediakan dalam kertas soalan. 3. Answer two questions from Section B. Write your answers for Section B in detail. You can use equation, diagram, table, graph and any other suitable ways to clarify your answer. Jawab two soalan daripada Bahagian B. Tulis jawapan bagi Bahagian B dengan terperinci. Anda boleh menggunakan persamaan, rajah, jadual, graf dan cara lain yang sesuai untuk menjelaskan jawapan anda. 4. Show your working. It may help you to get marks. Tunjukkan kerja mengira, ini membantu anda mendapatkan markah. 5. If you wish to change your answer, neatly cross out your answer that you have done. Then write down the new answer. Sekiranya anda hendak menukar sesuatu jawapan, batalkan dengan kemas jawapan yang telah dibuat. Kemudian tulis jawapan yang baru. 6. The diagrams in the questions are not drawn to scale unless stated. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. 7. Marks allocated for each question or part question are shown in brackets. Markah yang diperuntukkan bagi setiap soalan atau ceraian soalan ditunjukan dalam kurungan. 8. The time suggested to answer Section A is 90 minutes and Section B is 30 minutes. Masa yang dicadangkan untuk menjawab Bahagian A ialah 90 minit,dan Bahagian B ialah 30 minit. 9. You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogramkan. 0. Hand in all your answer sheets at the end of the examination. Serahkan kertas soalan dan jawapan anda diakhir peperiksaan. 455/2 [Lihat sebelah

25 Structured question: No. Item No. Suggested Answer: Mark (a) P: Cell wall Q: Cytoplasm R: Vacuole S: Nucleus Award mark for two correct answers: maximum: 2 (b) (i) correct label of chloroplast (T) (ii) (iii) Palisade mesophyll Chloroplast contains chlorophyll for absorption of light energy for photosynthesis (c)(i) 2 (ii) *Root hair cell does not have chloroplast while the cell in Diagram 3 has chloroplast * Root hair cell has a protruding structure but the cell in Diagram 3 does not have a protruding structure 2 (d) * Water is absorbed by the root hair through osmosis * Cell sap of the root hair has a higher solute concentration than the soil water. 2 TOTAL 2 Skema 455/2 [Lihat sebelah

26 No. 2 Question no. 2 (a)(i) (a)(ii) (a)(iii) (b)(i) (b)(ii) (b)(iii) Answer X : Aerobic respiration Y : Anaerobic respiration in plants Aerobic respiration is the oxidation of glucose to release energy in the presence of oxygen Anaerobic respiration is the process whereby glucose is broken down to release energy in the absence of ( without using ) oxygen. Fermentation Glucose ethanol + carbon dioxide + energy Plants Alcohol (ethanol), carbon dioxide, energy. Animals Lactic acid, energy Mark 2 2 (c)(i) (c)(ii) Glucose is broken down completely in aerobic respiration. In anaerobic respiration glucose is not broken down completely but partially to give lactic acid or ethanol and carbon dioxide. Some of the energy is still stored in the lactic acid molecule and ethanol molecule. [Any two correct answers, award 2 marks ] The condition known as oxygen debt arises when the lactic acid accumulates in the muscle tissue as a result of anaerobic respiration. This causes the person to fell tired because of the muscle fatigue. Inhalation of oxygen when resting oxidises the lactic acid to release energy, water and carbon dioxide. When this occurs the oxygen debt is said to have been paid. [Any two correct answers, award 2 marks ] TOTAL No /2 [Lihat sebelah

27 Item No. Suggested answer 3 (a) The capture and recapture technique. Mark (b)(i) Mark the specimens using a non-toxic permanent ink marker. (ii) The mark must not be lost and must not inhibit normal body activities. The mark does not prevent the rat from randomly mixing with the other unmarked rats. (c) Population = (00 x 40) / 40 = 350 rats 2 2 (d) To give sufficient time for the random dispersal and mixing among the rats in the population. (e) Changes in the size of population after three months can be caused by: increase in number of the rats due to increase in birth rate. decrease in number of the rats due to death of old rats, diseases or eaten by predators. migration (immigration or emigration) of the rats. 2 (f) The nitrate fertilizer in the river water is absorbed by the algal cells. Eutrophication occur The algae grow and reproduce rapidly that they completely cover the water. They block out the light for plants growing beneath them, which causes death. Decomposing bacteria acting on the dead plants and algae compete for the oxygen in the water. As a result, fish and other organisms in the river die due to the lack of oxygen. Maximum: 4 marks TOTAL /2 [Lihat sebelah

28 No. 4 Item No. Suggested answers Marks 4(a) 3 (b)(i) The blood glucose level rises sharply between 0700 and 0800 hours. (ii) The blood glucose level decreases gradually between 0800 and 200 hours. (c) The blood glucose rises sharply because the carbohydrates in the meal are digested. High glucose level in blood stimulates the secretion of insulin. Insulin promotes the conversion of glucose to glycogen in the liver (and the uptake of glucose for cellular respiration). Thus the blood glucose level decreases gradually between 0800 and 300 hours. MAXIMUM (d) The blood glucose level drops during the period between 800 to 900 hours because the last meal was at 600 hour. Between 900 hours to 2200 hours the blood glucose level maintains at about 4.6 mmoldm -3. This occurs because of the secretion of glucagon from the pancreas. Glucagon promotes the conversion of glycogen to glucose. As a result the blood glucose level rises again at 2200 hours. MAXIMUM (e) Blood glucose level needs to be maintained within the normal range to provide a constant environment for the optimal functioning of cells /2 [Lihat sebelah

29 (f) To avoid the adverse effects of hypoglycaemia or hyperglycaemia. TOTAL 2 No. 5 Item No Suggested Answer Markah 5 (a)(i) (ii) Coloured and smooth seed/ Biji berwarna dan licin (b)(i) (c) (ii) Crossing over/ Pindah silang Prophase I meiosis/ Profasa I Meiosis 455/2 [Lihat sebelah

30 (d) (e) Second Probablity/ Kemungkinan kedua. Berlaku proses pindah silang yang akan menghasilkan lebih banyak gamet yang membawa ciri genetik yang berbeza. Percantuman gamet-gamet ini akan menghasilkan individu yang bervariasi. Untuk menghasilkan gamet yang haploid, Percantuman antara gamet jantan yang haploid dan gamet betina yang haploid akan menghasilkan zigot yang diploid, yang sama bilangan kromosom induk. Oleh itu bilangan kromosom dalam generasi spesis organisma dapat dikekalkan (f) Gamet generasi F Gamet induk Genotip generasi F2 Fenotip generasi F2 Y l y l Yyll Biji berwarna dan lisut y L y l yyll Biji tak berwarna dan licin /2 [Lihat sebelah

31 No. 6 ITEM NO 6(a)(i) SCORING CRITERIA Able to list the general characteristic of enzymes Sample answers: P Enzymes are proteins which are synthesized by living Organisms P2 Enzymes bind to their substrates and convert them into product in the enzymatic reaction P3 Enzymes have specific site called active sites to bind to specific substrates // enzymes are highly specific in their reaction. P4 Enzymes speed up the rate of bilchemical reaction but remain unchanged // are not destroyed at the end of the reaction. P5 Enzymes are needed in small quantities because they are not used up P6 - Most enzyme- catalysed reaction are reversible P7 The activity of an enzyme can be slowed down or completely stop by inhibitors. MARK MAXIMUM 4 6(a)(ii) Able to discuss the uses of enzymes in industrial processes and our daily life, using suitable examples Sample answers: Type of industry/ Application (A) Enzyme used (E) Uses (U). Food Rennin solidifies milk processing Lipase Ripening of the cheese industry (a) dairy product Lactase Hydrolysed lactose to glucose in the making of ice cream (b) bakery products Amylase Convert starch flour into sugar in making bread Protease Convert protein in the making of biscuits (c) Alchoholic drinks(beer/wine Amylase Convert malt into glucose for the fermentation of yeast industry) Zymase Convert sugar into ethanol during fermentation of yeast (d) Fish products Protease Removes the skin of fish (e) Meat Protease Tenderises meat products (f) cereal grain products Cellulose Break down cellulose and removes seed coats from cereal grain 455/2 [Lihat sebelah

32 (g) Seaweed products (h) Starch products 2 Leather products 3. Madical products 4. Biological washing powder/detergent Cellulose Amylase Glucose isomerase Trypsin/protease Pancreatic trypsin Protease and amylase Digest cell wall and extract agar from seaweed Change starch into sugar in making syrups Convert glucose into fructose/production of high fructose syrup Removal of hair from animal hides Treats inflammation Dissolve protein and starch stains in clothes MAXIMUM 6 6(b) Able to explain how extracellular enzyme is produced by emphasizing on the role of P,Q, R and S. P Q is nucleus, contain DNA which store genetic information for enzyme synthesis in chromosome. P2 The genetic information is transferred to ribosome by RNA P3 R is mitochondrion, carry out cellular respiration and produces energy which is needed in enzyme synthesis. P4 Protein that are synthesized at ribosome are transported in the space within the rough endoplasmic reticulum P4 Protein depart from the rough endoplasmic reticulum wrapped in vesicles that bud off from the membrane of rough endoplasmic reticulum P5 The transport vesicles then fuse with the membrane of the golgi apparatus. P6 The proteins are further modified during their transport in the golgi apparatus P7 For example carbohydrates are added to proteins to make glycoprotein P8 Secretory vesicles, S containing these modified proteins bud off from the golgi apparatus and travel to the plasma membrane P9 These vesicles fuse with the plasma membrane before releasing the proteins as enzyme outside the cell Grant marks: Student mention the role of P,Q,R and S before or after the explanation. 455/2 [Lihat sebelah

33 No. 7 Item No. Suggested answer 7(a) Mark 7(b) Functions of the liver The liver secretes bile which is stored in the gall bladder and released into the duodenum when required. The liver regulates the blood sugar level by converting excess glucose to glycogen and storing it. The liver removes excess amino acids by breaking them down first into ammonia and then into urea which is excreted; and this process is known as deamination. The liver stores and metabolises fats. The liver synthesises fibrinogen and prothrombin which are bloodclotting substances and heparin which is an anticoagulant. The liver detoxifies many harmful chemicals. The liver breaks down worn out red blood cells and removes circulating hormones from the blood when they are no longer required and breaks them down. The liver synthesises vitamin A and stores it together with vitamins D, K and B 2 MAXIMUM [8 marks] 4 8 (c) The type of animal is ruminant Examples cow and buffalo/ goats (any two correct answers) The differences between ruminant digestive system and human digestive system are: A ruminant digestive system has four-chamber stomach namely rumen, reticulum, omasum and abomasum whereas a human 2 455/2 [Lihat sebelah

34 digestive system has only one stomach. A ruminant is aided by bacteria and protozoa (that produce cellulose enzyme) to digest cellulose (whereas ) human digestive system cannot digest cellulose/ is not aided in digestion of cellulose, thus cellulose is egested without digested. Cellulose is digested in ruminant but cellulose is not digested in human. A ruminant can vomit out the content of its rumen to regurgitate and rechew food (rumination) but in human the food is chewed only once/ pass through the digestive tract only once. Maximum: 8 455/2 [Lihat sebelah

35 No. 8 8(a)(i) Salt Maximum: (a)(ii) (b). a large intake of salty food causes the osmotic pressure of the blood to increase. 2. sensory cells in the hypothalamus / osmoreceptor cells detect changes in the osmotic pressure of the blood. 3. this tringgers the pitutary gland to secrete ADH. 4. ADH carried by the blood to the distal convoluted tubule and the collecting ducts 5. ADH increases the permeability of the walls of these ducts to water. 6. more water is reabsorbed, causing the volume of urine to decrease 7. no aldosterone is secreted from the adrenal glands because of the high salt content in the blood. 8. less salt is reabsorbed in the kidneys. 9. as a result, the urine produced is concentrated and the volume is low until osmotic pressure of the blood returns to normal MAXIMUM:. when the athlete is running, the hypothalamus sends nerve impulses to the adrenal gland. 2. the nerve impulses stimulate the cells of adrenal gland. 3. to secrete adrenaline 4. adrenaline stimulates the heart to beat faster, and increases the breathing rate, blood pressure, blood glucose level and metabolic rate. 5. the heart contracts more vigorously to pump larger amounts of oxygen and glucose to the sceletal muscles 6. this is to increase cellular respiration and the production of ATP for muscle contractions. 7. an increase in the metabolic rate causes the temperature of the body to increase. 8. the sweat glands in the skin secrete more sweat. The evaporation of sweat brings about a cooling effect which helps to lower the body temperature. 9. at the same time, vasolidation of the arterioles occurs to increase the amount of heat lose through the skin. 0. the erector muscles in the skin relax, lowering the skin hairs so that there is no warm air trapped against the skin. MAXIMUM /2 [Lihat sebelah

36 No. 9 9(b) P penyakit hemofilia di kawal oleh gen resesif yang terdapat pada kromosom seks // X h. P2 kehadiran satu gen hemofilia pada anak lelaki menyebabkan ia menjadi pengidap // X H Y. P3 kehadiran satu gen hemofilia pada perempuan menjadikannya pembawa // X H X h. P4 kehadiran sepasang gen hemofilia / 2 gen pada kedua-dua kromosom seks menjadikannya pengidap // X h X h. X h Y X H X H P5 X h X h Y X H XH P6 X H X h X H X h X H Y X H Y P7 Perempuan perempuan lelaki lelaki Pembawa pembawa normal normal X H Y X h X h P8 X H Y X h X h P9 X H X h X H X h X h Y X h Y Perempuan perempuan lelaki lelaki Pembawa pembawa hemofilia hemofilia P0 Pasangan A boleh meneruskan perkahwinan kerana anak yang dihasilkan adalah normal dan bebas dari penyakit hemofilia. P pasangan B tidak meneruskan perkahwinan kerana anak lelaki mendapat hemofilia. MAXIMUM: 0 455/2 [Lihat sebelah