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March 20, 2017 | Author: NoridaAnis | Category: N/A
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Jabat anPel aj ar anKel ant an Jal anDokt or,15000Kot aBharu, Kel ant an.
ModulPecut anAkademi k
BI OLOGY SPM
KATA PENGANTAR
Puji dan syukur kita panjatkan kehadrat Allah Yang maha Esa atas segala rahmat dan kurnia-Nya sehingga Modul Pecutan Akademik, Jabatan Pelajaran Kelantan dapat dihasilkan pada tahun ini. Modul Pecutan Akademik ini diharapkan dapat menjadi panduan dan bimbingan kepada para guru dan para pelajar dalam membuat persediaan bagi menghadapi peperiksaan UPSR, PMR dan SPM. Setinggi-tinggi penghargaan dan terima kasih dirakamkan kepada semua guru dan semua pihak yang terlibat dalam menjayakan penghasilan Modul Pecutan Akademik ini. Semoga usaha murni ini dapat diteruskan pada masa hadapan bagi menghasilkan pelajar yang cemerlang, gemilang dan terbilang dan seterusnya menjayakan visi Jabatan Pelajaran Kelantan, “Cakna Pendidikan Kelantan Terbilang 2013”.
HJ MOHD ADNAN BIN MOHD NOOR Ketua Sektor Pengurusan Akademik Jabatan Pelajaran Kelantan
CONTENT 2.0
Cell Structure and Cell Organisation
1-3
3.0
Movement of Substances Across Plasma Membrane
4-6
4.0
Chemical Composition of the Cell
7-9
5.0
Cell Division
10 - 15
6.0
Nutrition
16 - 22
7.0
Respiration
23 - 29
8.0
Dynamic Ecosystem
30 - 34
9.0
Endangered Ecosystem
35 - 37
10.0
Transport
38 - 43
11.0
Locomotion and Support
44 - 48
12.0
Coordination and Response
49 - 51
13.0
Reproduction and Growth
52 - 55
14.0
Inheritance
56 - 58
15.0
Variation
59 - 60
CHAPTER 2: CELL STRUCTURE AND CELL ORGANISATION
1.(a)
Draw and label a green plant cell which is under light microscope. Lukis satu gambar rajah berlabel bagi suatu sel tumbuhan hijau di bawah mikroskop cahaya. [6 marks]
(b)
State the functions of three organelles in the plant cell that you have drawn. Nyatakan fungsi bagi tiga organel yang telah anda lukis pada sel tumbuhan hijau dalam soalan 1(a) [6 marks]
(c)
Compare and contrast a plant cell with an animal cell. Bandingkan dan bezakan antara sel tumbuhan dengan sel haiwan [8 marks]
2.(a)
Diagram 2.1 shows a mature red blood cell, an efferent neurone and a sperm. Rajah 2.1 menunjukkan satu sel darah merah matang, neuron eferen dan satu sel sperma.
Diagram 2.1 / Rajah 2.1 Explain how the structures are adapted to their functions. Terangkan bagaimana struktur sel tersebut disesuaikan untuk menjalankan fungsinya. [10 marks]
1
(b)
Amoeba is a unicellular organism but it can function as a complete unit of life. Amoeba adalah organism unisel tetapi ia dapat menjalankan fungsi sebagai satu kehidupan. (i)
Based on the statement above, describe how Amoeba can carry out the following life processes : nutrition and reproduction. Berdasarkan pernyataan di atas, huraikan bagaimana Amoeba dapat menjalankan proses kehidupan seperti pemakanan dan pembiakan. [6 marks]
(ii)
Explain why Amoeba does not burst when it is immersed in the distilled water. Terangkan mengapa Amoeba tidak meletus apabila ia di masukkan ke dalam air suling. [4 marks]
3.
Diagram 2.2 shows five levels of cell organisation in man. Rajah 2.2 menunjukkan lima aras dalam organisasi pada seorang manusia. Cell sel
Tissue tisu
Organ organ
System (digestive) System (pencernaan)
Diagram 2.2 / Rajah 2.2 2
Organism organism a
(a) Describe the cell organisation in the formation of tissues, organs and systems based on the example of the digestive system of man shown in Diagram 2.2. Huraikan organisasi sel dalam pebentukkan tisu, organ dan sistem berdasarkan contoh sistem pencernaan pada seorang lelaki yang ditunjukkan dalam Rajah 2.2. [10 marks] (b) Diagram 2.3 shows the structure of a dicotyledonous leaf.
Diagram 2.3 / Rajah 2.3
How is the leaf modified to carry out its various function? Terangkan bagaimana daun diubahsuai untuk menjalankan beberapa fungsinya. [10 marks]
3
CHAPTER 3: MOVEMENT OF SUBSTANCES ACROSS THE PLASMA MEMBRANE
1. Diagram 3.1.1 shows the structure of a unicellular organism. Rajah 3.1.1 menunjukkan struktur organisma unisel.
Diagram 3.1.1 / Rajah 3.1.1
(a) Explain the function of organelle which involve in osmoregulation. Terangkan fungsi organel yang terlibat dalam pengosmokawalaturan. [ 4 marks ]
(b) State the similarities and differences between passive transport (facilitated diffusion) and active transport in the movement of molecules across the cell membrane. Nyatakan persamaan dan perbezaan antara pengangkutan pasif
(resapan
berbantu)dan pengangkutan aktif dalam pergerakan molekul-molekul merentas membrane sel. [ 8 marks ]
4
(c) Diagram 3.1.2 and 3.1.3 show two types of food which can be preserved. Rajah 3.1.2 dan 3.1.3 menunjukkan dua jenis makanan yang boleh diawet.
Diagram 3.1.2 / Rajah 3.1.2
Diagram 3.1.3 / Rajah 3.1.3
Explain how vinegar and concentrated salt solution can be used in the food preservation. Terangkan bagaimana cuka dan larutan garam yang pekat boleh digunakan dalam pengawetan makanan tersebut. [ 8 marks ]
2. The following information is about plasma membrane. Maklumat berikut adalah berkaitan dengan membran plasma.
The plasma membrane is semi-permeable and allows certain substances to move across. Membran plasma adalah separa telap dan membenarkan sesetengah bahan melaluinya.
(a) Based on the above statement, describe how an amino acid molecule is transported across the plasma membrane into the cell. Berdasarkan pernyataaan di atas, huraikan bagaimana asid amino diangkut merentasi membran plasma ke dalam sel. [5 marks]
(b) Discuss why the uses of excessive fertilizers can cause wilting in plants. 5
Bincangkan kenapa penggunaan baja berlebihan boleh menyebabkan kelayuan kepada tumbuhan. [5 marks]
(c) Diagram 3.2 shows human erythrocytes after being immersed in distilled water and 20% sodium chloride solution. Rajah 3.2 menunjukkan eritrosit manusia selepas direndam di dalam air suling dan larutan natrium klorida 20%.
Diagram 3.2 / Rajah 3.2
Explain what had happened to the erythrocytes. Terangkan apa yang telah terjadi kepada eritrosit tersebut. [ 6 marks ] (d) Explain the situation occur when a potato strip is transferred from 30% sucrose solution into distilled water. Terangkan situasi yang berlaku apabila jalur kentang dipindahkan dari larutan sukrosa 30% ke dalam air suling. [4 marks]
6
CHAPTER 4: CHEMICAL COMPOSITION OF THE CELL
1. (a) Diagram 4.1.1 shows the organelles involved in the synthesis and secretion of extracellular enzymes in an animal cell. Rajah 4.1.1 menunjukkan organel-organel yang terlibat dalam sintesis dan rembesan enzim-enzim luar sel di dalam sel haiwan.
Diagram 4.1.1 / Rajah 4.1.1
(i). Define intracellular enzymes and extracellular enzymes. Beri definesi enzim intrasel dan enzim luar sel. [ 4 marks ]
(ii) Describe the involvement of the organelles in the production of extracellular enzymes. Huraikan penglibatan organel-organel di dalam penghasilan enzim-enzim luar sel. [ 8 marks ]
7
(b) Diagram 4.1.2 shows three stages in the enzyme reaction. Rajah 4.1.2 menunjukkan tiga peringkat di dalam tindakan enzim.
R Diagram 4.1.2 / Rajah 4.1.2
Identify R . Explain the lock and key hypothesis in the mechanism of enzyme reaction. Kenalpasti R. Terangkan hipotesis kunci dan mangga di dalam mekanisma tindak balas enzim. [ 8 marks ] 2. Diagram 4.2 shows the different structure of protein. Rajah 4.2 menunjukkan struktur protein yang berbeza.
P
Q
R
Diagram 4.2 / Rajah 4.2 (a)
(i) State two examples of protein with structure R that present in our body. Nyatakan dua contoh protein dengan struktur R yang didapati di dalam badan kita. [ 2 marks ] 8
(ii) Name the protein structure P, Q and R. Describe the structures of each protein structure. Namakan struktur protein P, Q dan R. Huraikan struktur setiap struktur protein tersebut. [ 8 marks ]
(b)
Detergent is normally used in washing. Describe three biological enzymes that present in the detergent which help to clean up the stains on clothes. Serbuk pencuci biasanya digunakan dalam pencucian. Huraikan tiga enzim biologi yang hadir dalam serbuk pencuci yang membantu menanggalkan kotoran pada pakaian. [ 6 marks ]
(c)
Using two examples of enzymes, describe the role of the enzymes in the industry field. Dengan menggunakan dua contoh enzim yang dinamakan, huraikan peranan enzim tersebut dalam bidang industri. [ 4 marks ]
9
CHAPTER 5: CELL DIVISION
1.
(a) List the importance of mitosis process. Senaraikan kepentingan proses mitosis. [3 marks] (b) Explain the differences between process of mitosis and meiosis using appropriate diagram. Terangkan perbezaan antara proses mitosis dan meiosis menggunakan rajah yang sesuai. [10 marks] (c) Explain how mitosis is apply in one example of animal to produce large number of animal much faster than through sexual reproduction. Terangkan bagaimana mitosis diaplikasikan dalam satu contoh pada haiwan untuk menghasilkan bilangan haiwan lebih cepat dari melalui kaedah pembiakan seks. [7 marks]
2. (a)(i) Diagram 5.1 shows a cell cycle of an organism. Phase X consist of three sub-phases P, Q and R. Rajah 5.1 menunjukkan satu kitar sel pada satu organism. Fasa X terdiri dari 3 sub-fasa P, Q dan R.
Diagram 5.1 / Rajah 5.1 10
Describe what happens in the cell during phase X. Huraikan apakah yang berlaku semasa fasa X. [2 marks]
(a)(ii) A farmer has succeeded in producing a high quality seedling of oil palm tree as a result of cross-breeding between two species of oil palm. Suggest one method which can be used to help the farmer to produce more of this seedling in the shortest period of time. Seorang petani telah menjalankan kacukan antara dua pokok kelapa sawit dan berjaya menghasilkan anak pokok kelapa sawit yang bermutu tinggi. Cadangkan satu kaedah yang boleh digunakan untuk menolong petani itu menghasilkan anak pokok kelapa sawit yang bermutu tinggi ini dengan banyak dalam masa yang singkat. [6 marks]
(b)
The hormone insulin used by present day diabetics is the result of genetic engineering technology. This hormone which was used to treat diabetics since 1982 is the first technological product approved for the market. Hormone insulin yang digunakan oleh pesakit diabetis adalah hasil dari teknologi kejuruteraan genetik. Hormon yang telah digunakan sejak 1982 adalah produk teknologi pertama yang dibenarkan yang dibenarkan untuk pasaran.
Based on above information, discuss the benefits of genetic engineering method in producing products for the society. Berdasarkan maklumat di atas, bincangkan kebaikan teknik kejuruteraan genetik dalam menghasilkan keperluan masyarakat. [6 marks]
11
(c)
Diagram 5.2 shows a group of cells that is exposed to ultraviolet ray. Rajah 5.2 menunjukkan sekumpulan sel yang terdedah kepada sinar ultraviolet.
Normal cells
Mutated Cells
Sel normal
Sel yang mengalami mutasi
Diagram 5.2 / Rajah 5.2 Diagram 5.2 / Rajah 5.2
The exposure drives the cell cycle malfunctions. Based on the Diagram 5.2 describe effect of cell cycle malfunctions to the body. Pendedahan kepada sinar radioaktif menyebabkan kitar sel tidak berfungsi. Berdasarkan Rajah 1, huraikan kesan kitar sel yang tidak berfungsi ke atas badan. [ 6 marks] 3 (a) “Dolly” the sheep was the first vertebrate cloned from the cell of an adult animal in 1997. That was a remarkable scientific breakthrough and it gained interest and concern from around the world on the of cloning and how it would affect humans. “Dolly” si biri-biri merupakan haiwan vertebrata pertama yang diklon daripada sel haiwan dewasa pada tahun 1997. Ia merupakan satu penemuan saintifik yang amat bermakna dan telah menimbulkan minat dan perhatian daripada pelbagai pihak di seluruh dunia terhadap pengklonan dan bagaimana ia memberi kesan ke atas manusia.
12
i)
Based on the statement above, explain the principles used in the cloning technique. Berdasarkan pernyataan di atas, terangkan prinsip-prinsip yang digunakan dalam teknik pengklonan.
ii)
[3 marks]
State the advantages and disadvantages of this technique. Nyatakan kebaikan dan keburukan teknik tersebut [7 marks]
(b) Describe how cytokinesis occurs in plant and animal cells to produce two daughter cells. Terangkan bagaimana sitokinesis berlaku dalam sel tumbuhan dan sel haiwan untuk menghasilkan dua sel ana [10 marks]
13
4.
Diagram 5.4 shows different stages in a meiosis I. Rajah 5.4 menunjukkan peringkat yang berbeza dalam meiosis 1.
P
Q
S R
Diagram 5.4/ Rajah5. 4 (a)
Based on Diagram 5.4, explain the process that occurs in stage P, Q, R dan S. Berdasarkan Rajah 5.4, terangkan proses yang berlaku dalam peringkat P, Q, R dan S. [6 marks]
(b)
State the different between meiosis I and meiosis II based on stage P, Q, R dan S. Nyatakan perbezaan antara meiosis I dan meiosis II berdasarkan kepada peringkat P, Q, R dan S. [4 marks] 14
(c)
Explain briefly how meiosis involved in genetic variation. Terangkan secara ringkas bagaimana meiosis terlibat dalam variasi genetik. [10 marks ]
15
CHAPTER 6: NUTRITION
1.
Diagram 6.1.1, Diagram 6.1.2 and Diagram 6.1.3 shows the physical condition of three children who suffer from malnutrition. Rajah 6.1.1, Rajah 6.1.2 dan rajah 6.1.3 menunjukkan keadaan fizikal tiga orang kanakkanak yang mengalami malnutrisi.
Diagram 6.1.1 Rajah 6.1.1
(a)
Diagram 6.1.2 Rajah 6.1.2
Diagram 6.1.3 Rajah 6.1.3
State the meaning of malnutrition. Nyatakan maksud malnutrisi. [2 marks]
(b)
Name and explain the disease in Diagram 6.1.1, Diagram 6.1.2 and Diagram 6.1.3 related to malnutrition. Namakan dan terangkan penyakit dalam Rajah 6.1.1, Rajah 6.1.2 dan Rajah 6.1.3 yang berkaitan dengan malnutrisi. [8 marks]
16
(c)
Diagram 6.2 shows a food pyramid. The correct proportion for the various classes of food is shown in a food pyramid. Rajah 6.2 menunjukkan piramid makanan. Nisbah yang betul bagi pelbagai kelas makanan ditunjukkan dalam piramid makanan.
Diagram 6.2 Rajah 6.2
Boy Y does not follow this food pyramid in his diet. He likes to eat hamburger which consist of bread, meat and cheese as his daily diet for a long period of time. Budak lelaki Y tidak mengikut piramid makanan ini dalam amalan pemakanannya. Dia suka makan burger yang terdiri daripada roti, daging dan keju sebagai makanan hariannya dan pengambilannya jangkamasa yang lama.
Explain the long term effects of his diet to his health. Terangkan kesan jangka panjang amalan pemakanannya ini terhadap kesihatannya. [10 marks]
17
2.
Diagram 6.3.1 shows organism P, Diagram 6.3.2 shows digestive system of organism Q and Diagram 6.3.3 shows digestive system of organism R. Rajah 6.3.1 menunjukkan organisma P, Rajah 6.3.2 menunjukkan sistem pencernaan organisma Q dan Rajah 6.3.3 menunjukkan sistem pencernaan organisma R.
Diagram 6.3.1 Rajah 6.3.1
Diagram 6.3.2 Rajah 6.3.2
Diagram 6.3.3 Rajah 6.3.3
(a) (i) Describe the type of nutrition in organism P and organism R. Huraikan jenis nutrisi bagi organisma P dan organisma R. [4 marks] (ii) Compare the process of cellulose digestion in organism Q and organism R. Bandingkan proses pencernaan selulose dalam organisma Q dan organisma R. [6 marks]
18
(b)
Calcium, ferum and iodine are minerals which must be obtained through the diet. Kalsium, ferum dan iodin adalah garam mineral yang mesti diambil melalui pemakanan.
State the function and symptom of deficiency of these minerals. State one source for each mineral. Nyatakan fungsi dan kesan kekurangan garam mineral tersebut. Nyatakan satu sumber bagi setiap garam mineral tersebut. [10 marks]
3.
Diagram 6.4 shows some organs of human digestive system. Rajah 6.4 menunjukkan beberapa organ sistem pencernaan manusia.
X
Y
Z
Diagram 6.4/ Rajah 6.4
(a)
Syarifah eats too many mangoes with vinegar. Explain the effect of eating too much of this kind of mangoes on the digestion of food in Y. Syarifah makan terlalu banyak mangga berserta cuka. Terangkan kesan memakan terlalu banyak mangga jenis ini terhadap pencernaan makanan di Y. [5 marks]
19
(b)
A man is suffering from cancer, his organ Z need to be removed. What should he do to handle health problems that may arise from the removal of organ Z? Seorang lelaki mengidap kanser, organ Z beliau perlu dibuang. Apakah yang patut dilakukan untuk mengawal masalah kesihatannya yang mungkin timbul kesan pembuangan organ Z? [5 marks]
(c)
(i) State the functions of X. Nyatakan fungsi X.
(ii) Explain the process of assimilation of amino acids and glucose in X. Terangkan proses asimilasi asid amino dan glukosa di X. [10 marks] 4.
(a)
Table 1 shows the menu for Adnan’s daily diet. Jadual 1 menunjukkan menu pemakanan harian Adnan.
Fried chicken Ayam goreng
- 2 pieces - 2 ketul
Hamburger Burger
- 2 pieces - 2 biji
Soft drink Minuman ringan
- 2 glasses - 2 gelas
Ice cream Ais krim
- 1 cup - 2 cawan Table 1 Jadual 1
Adnan takes this menu for his breakfast and lunch for a long period of time. Explain the consequences to his health. Adnan mengambil menu ini sebagai sarapan dan makan tengahari dalam jangkamasa yang lama. Terangkan akibat terhadap kesihatannya. [ 10 marks ] 20
(b)
Diagram 6.5 shows organelle P found in a plant cell. Rajah 6.5 menunjukkan organel P yang didapati pada sel tumbuhan.
Diagram 6.5 Rajah 6.5
Organelle P is importance for the process of photosynthesis in plant. Organel P adalah penting bagi proses fotosintesis dalam tumbuhan.
(i)
State the meaning of photosynthesis. Nyatakan maksud fotosintesis. [2 marks]
(ii)
Explain the main stages in photosynthesis. Terangkan peringkat-peringkat utama dalam fotosintesis. [8 marks]
21
5. Food preservation involves methods of preparing food to extend the lifespan and to avoid wastage of food. Pengawetan makanan melibatkan kaedah peyediaan untuk memanjangkan tempoh hayat dan mengelakkan pembaziran makanan.
(a)
Based on the above statement, explain the necessity for food processing. Berdasarkan pernyataan di atas, terangkan keperluan pemprosesan makanan. [10 marks]
(b) Table 6.1 shows several methods of food preservation that being used in food processing. Jadual 6.1 menunjukkan beberapa kaedah pengawetan yang digunakan dalam pemprosesan makanan.
Type of food
Food preservation method
Jenis makanan
Kaedah pengawetan makanan
Milk
Pasteurisation
Susu
Pempasteuran
Fruits
Canning
Buah-buahan
Pengetinan
Meat and fish
Refrigeration
Daging dan ikan
Penyejukbekuan Table 6.1/ Jadual 6.1
Describe how the method can preserve food for a long period of time. Jelaskan bagaimana kaedah itu boleh mengawet makanan untuk satu jangka masa yang panjang. [10 marks] 22
CHAPTER 7: RESPIRATION
1. Diagram 7.1 shows the respiratory system of a human and a fish. Rajah 7.1 menunjukkan sistem respirasi pada manusia dan ikan.
Diagram 7.1/ Rajah 7.1 (a)
Name organ X and organ Y. Namakan organ X dan organ Y. [2 marks]
(b)
Draw and label the respiratory surfaces of organ X and Y. Lukis dan label permukaan respirasi organ X dan Y. [4 marks]
(c) Respiratory system in human and fish are adapted to function in their respective habitats. Compare and complain the adaptations. Sistem respirasi manusia dan ikan disesuaikan untuk befungsi dalam habitat masing masing. Banding dan terangkan penyesuaian tersebut . [10 marks]
(d)
Explain the effects of smoking on the human respiratory system. Terangkan kesan merokok terhadap sistem respirasi manusia. [4 marks]
23
2. Diagram 7.2 shows one stage in human breathing. Rajah 7.2 menunjukkan satu peringkat dalam pernafasan manusia.
R Diagram 7.2/ Rajah 7.2 (a)
Explain the state of breathing shown in Diagram 7.2. Terangkan peringkat pernafasan yang ditunjukkan dalam Rajah 7.2.
[6 marks]
(b)
Explain the effects of the breathing mechanism if structure R is unable to function. Terangkan kesan kepada mekanisma pernafasan jika struktur R tidak boleh berfungsi.
[4 marks]
24
(c)
Diagram 7.3 shows the various structures involved in the regulation of the carbon dioxide concentration in the body fluid. Rajah 7.3 menunjukkan pelbagai struktur yang terlibat dalam pengawalaturan kepekatan karbon dioksida dalam cecair badan.
Central Chemoreceptor Drop / rise in carbon dioxide concentration
Diagram 7.3/ Rajah 7.3 Diagram 7.3 / Rajah 7.3
Based on Diagram 7.3, describe the regulation of the carbon dioxide concentration in body fluid. Berdasarkan Rajah 7.3, terangkan pengawalaturan kepekatan karbon dioksida dalam cecair badan. [10 marks]
25
3 (a)
Diagrams 7.4 shows cell X and tissue Y. Cell X is a microorganism and tissue Y is found
in human body. Respiration in cell X without using oxygen whereas respiration in tissue Y using oxygen. Rajah 7.4 menunjukkan sel X dan tisu Y. Sel X adalah sejenis mikroorganisma dan tisu Y dijumpai di dalam badan manusia. Respirasi dalam sel X tanpa menggunakan oksigen manakala respirasi dalam tisu Y menggunakan oksigen.
Cell X/ Sel X
Tissue Y/ Tisu Y Diagram 7.4/ Rajah 7.4
Based on Diagram7.4, differentiate the cellular respiration process that occurs in cell X and tissue Y. Berdasarkan Rajah 7.4, bezakan respirasi sel yang berlaku di dalam sel X dan tisuY. [8 marks] (b)
Diagram 7.5 shows energy releasing process in a human muscle cell. Rajah 7.5 menunjukkan proses pembebasan tenaga dalam satu sel otot manusia.
Glucose
Molecule X + 2ATP Diagram 7.5/ Rajah 7.5 26
Based on Diagram 7.5, state what is process Q and molecule X. Explain how molecule X can be remove from muscle cell. Berdasarkan Rajah 7.5, nyatakan proses Q dan molekul X. Terangkan bagaimana molekul X dapat disingkirkan dari sel otot. [6 marks] (c)
Diagram 7.6 shows respiratory surfaces of a human. Rajah 7.6 menunjukkan permukaan respirasi pada manusia. CO2 O2
R
S
Diagram 7.6 / Rajah 7.6
Based on the Diagram 7.6, explain the exchange of respiratory gases. Berdasarkan Rajah 7.6, terangkan proses pertukaran gas respirasi. [6 marks]
27
4.(a) Small molecules such as carbon dioxide and oxygen can pass through the cell membrane easily. Molekul kecil seperti karbon dioksida dan oksigen boleh melalui membran sel dengan mudah.
Based on the statement explain how gaseous exchange occurs in the alveoli and blood capillaries? Berdasarkan pernyataan di atas terangkan bagaimana pertukaran gas berlaku di alveolus dan di kapilari darah. [ 10 marks ]
28
(b)
Diagram 7.7 shows the oxygen uptake of an athlete before, during and after a 100 m race. Rajah 7.7 menunjukkan pengambilan oksigen seorang atlet sebelum, semasa dan selepas
larian 100m.
Diagram 7.7/ Rajah 7.7 Based on the above diagram, explain the process of energy production by the athlete during and after the race. Berdasarkan rajah di atas, terangkan proses penghasilan tenaga oleh atlet tersebut semasa dan selepas perlumbaan. [10 marks]
29
CHAPTER 8: DYNAMIC ECOSYSTEM 1. a) Diagram 8.1.1 shows the process of colonisation and succession in a habitat. Rajah 8.1.1 menunjukkan proses pengkolonian dan penyesaran dalam suatu habitat.
Diagram 8.1.1 / Rajah 8.1.1
What is the meant by “colonisation and succestion in a habitat”? Based on Diagram 8.1.1, explain how colonisation and succestion bring about the formation of the primary forest in a habitat Apakah yang dimaksudkan dengan “pengkolonian dan penyesaran dalam suatu habitat”? Berdasarkan Rajah 8. 1.1, terangkan bagaimana pengkolonian dan penyesaran membawa kepada pembentukan hutan primer dalam suatu habitat.
. [10 marks]
30
1. b) Diagram 8.1.2 shows an ecosystem in Malaysia. Rajah 8.1.2 menunjukkan suatu ekosistem di Malaysia.
Diagram 8.1.2/ Rajah 8.1.2
Based on the Diagram 8.1.2, discuss why ecosystem has to be maintained. Berdasarkan Rajah 8.1.2, bincangkan kenapa ekosistem ini perlu dikekalkan. [10 marks]
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2. a) Diagram 8. 2.1 shows a mangrove swamp. Rajah 8. 2.1 menunjukkan kawasan paya bakau.
Diagram 8. 2.1 / Rajah 8. 2.1
i) Explain why most plants cannot colonise and grow in the swamps. Terangkan mengapa kebanyakan tumbuhan tidak boleh hidup dan tumbuh di kawasan paya bakau.
[5 marks]
ii) Explain how the mangrove trees adapt themselves to the harsh living conditi Terangkan bagaimana pokok bakau ini menyesuaikan diri dengan keadaan hidup yang sukar. [5 marks ]
32
3. Diagram 8. 3.1 shows a type of fungi. Rajah 8. 3.1 menunjukkan sejenis fungi.
Diagram 8.3.1 / Rajah 8. 3.1
a) Explain how this organism obtained the nutrient. Terangkan bagaimana organisma ini memperolehi makanan. [4 marks]
b) Diagram 8. 3.2 is a graph shows the population size of rats and owls in an oil palm estate change throughout the year. Rajah 8. 3.2 adalah graf yang menunjukkan perubahan saiz populasi tikus dan burung
hantu dalam sebuah ladang kelapa sawit sepanjang tahun.
Jan
Apr
Aug
Dec
Diagram 8. 3.2 / Rajah 8. 3.2 33
Month
Based on Diagram 8. 3.2, explain the changes in the size of population of the owls and the rats throughout the year. Berdasarkan kepada Rajah 8.3.2, huraikan perubahan saiz populasi bagi burung hantu dan tikus sepanjang tahun.
4.
[ 6 marks ]
Diagram 8. 4.1 shows the nitrogen cycle which plays an important role in the formation of protein. Plants and animals need nitrate to form protein. Explain the role of plants, animals and microorganism in this cycle. Rajah 8. 4.1 menunjukkan kitar nitrogen yang memainkan peranan penting dalam pembentukan protein. Tumbuhan dan haiwan memerlukan nitrat untuk membentuk protein. Huraikan peranan tumbuhan, haiwan dan mikroorganisma dalam kitar ini. [10 marks ]
Diagram 8. 4. 1 / Rajah 8. 4. 1
-
34
CHAPTER 9: ENDANGERED ECOSYSTEM
1
Diagram 9. 1.1 shows evidences of development, urbanization and industrial practices. It is needed to develop the rises of human population. Diagram 9.1.1 menunjukkan bukti-bukti pembangunan, urbanisasi dan aktiviti-aktiviti perkilangan. Ia merupakan keperluan apabila populasi manusia meningkat.
Diagram 9.1.1 / Rajah 9.1. 1
Justify the effects of unplanned development into the ecosystem. Justifikasikan kesan pembangunan yang tidak terancang kepada ekosistem. [10 marks]
35
2.
Diagram 9.2.1 shows one phenomenon in the ecosystem due to an unplanned development. Rajah 9.2.1 menunjukkan satu fenomena dalam ekosistem yang disebabkan oleh pembangunan yang tidak terancang.
Diagram 9.2.1 Rajah 9.2. 1 Based on diagram 9. 2.1, explain the phenomenon and its effects to the
environment.
Berdasarkan rajah 9.2.1, terangkan fenomena ini dan kesan-kesannya terhadap alam sekitar.
[10 marks]
36
3.
Ozone layer is located in the stratosphere, about 15-30 kilometers above the Earth's surface. The ozone hole over Antartica was first discovered by Farman, Gardiner and Shanklin in 1985. Lapisan ozon terletak dalam lapisan stratosfera, yang terletak kira-kira 15-30 kilometer dari permukaan bumi. Lubang pada lapisan ozon di Antartika
ditemukan oleh
Farman, Gardiner dan Shanklin pada tahun 1985.
Describe how the ozone layer becomes thinner. Discuss its effects on humans and the environment and suggest the ways to solve these problems. Huraikan bagaimana lapisan ozon menjadi semakin nipis. Bincangkan
kesan
kesannya kepada manusia serta alam sekitar. Cadangkan langkah-langkah untuk mengatasi masalah ini. [10 marks]
37
CHAPTER 10: TRANSPORT
1. (a)
Diagram 10.1.1 shows a sequence in a process carried out by a leucocyte . Rajah 10.1.1 menunjukkan susunan proses yang dijalankan oleh suatu leukosit.
Diagram 10.1.1 / Rajah 10.1.1 Describe briefly how a leucocyte plays its role in body’s defences. Bincangkan secara ringkas bagaimana leukosit memainkan peranannya dalam pertahanan badan. [ 4 marks ] (b) Diagram 10.1.2 is a graph showing the concentration of antibody in the blood of an individual after receiving two injections. Rajah 10.1.2 adalah graf yang menunjukkan kepekatan antibodi dalam darah individu
Concentration of antibodies in the blood (arbitrary unit)
selepas menerima dua suntikan.
Immunity level
0
1
2
1st vaccination
3
4
5
6
2nd vaccination
7
8
9 Time (weeks)
Diagram 10.1.2 / Rajah 10.1.2 38
Explain how the individual could achieve the immunity level. Terangkan bagaimana individu tersebut boleh mencapai aras keimunan. [ 6 marks ]
2.
Diagram 10.2 shows blood capillary and lymph capillary in a human. Rajah 10.2 menunjukkan kapilari darah dan kapilari limfa dalam manusia.
Blood capillary Kapilari darah Fluid R Bendalir R
Fluid S in lymph capillary Bendalir S dalam kapilari limfa Diagram 10.2 / Rajah 10.2
(a) Explain the differences between the composition of fluid R and fluid S Terangkan perbezaan antara komposisi bendalir R dan bendalir S. [3 marks]
(b)
Based on Diagram 10.2, describe how fluid S is formed from blood plasma until it is brought back into the blood circulatory system. Berdasarkan Rajah10.2, bincangkan bagaimana bendalir S terbentuk dari plasma darah sehingga ia dikembalikan semula ke sistem peredaran darah. [7marks]
39
3.
Diagram 10.3 shows a human blood component. Rajah 10.3 menunjukkan komponen darah manusia
Erythrocyte Sel darah merah Leucocyte Sel darah putih
Platlet Platlet
Diagram 10.3 / Rajah 10.3 (a) Explain how platelets help to stop bleeding when a wound occurs. Terangkan bagaimana platlet membantu menghentikan pendarahan apabila luka. [ 5 marks] (b) A blood test shows that a man’s erythrocytes count is below normal. Explain the possible consequences of this condition on his health. Satu ujian darah manusia menunjukkan bilangan eritrositnya rendah daripada normal. Terangkan kesan keadaan ini ke atas kesihatannya. [ 5 marks]
40
4.
(a) Diagram 10.4.1 shows the part of a stem of a tree where the ring of bark has been removed. The tree is watered everyday. Rajah 10.4.1 menunjukkan bahagian batang satu pokok dimana kulit yang digelang telah dibuang. Pokok itu disiram setiap hari.
Diagram 10.4.1 / Rajah 10.4.1
Based on Diagram 10.4.1, explain what happens to the tree after one month. Berdasarkan Rajah 10.4.1, terangkan apa berlaku kepada pokok itu selepas satu bulan. [ 4 marks] (b) Diagram 10.4.2 shows the blood circulatory system in organism A and organism B. Rajah 10.4.2 menunjukkan sistem peredaran darah dalam organisma A dan organisma B.
Organism A
Organism B Diagram 10.4.2 /Rajah 10.4.2
41
Based on Diagram 10.4.2, Describe the similarities and differences between
the blood
circulatory system in organism A and organism B.. Berdasarkan Rajah 10.4.2, bincangkan persamaan dan perbezaan di antara sistem peredaran darah organism A dan organism B‟ [6 marks]
5.
Diagram 10.5 shows a human heart . Rajah 10.5 menunjukkan jantung manusia. Pulmonary vein Vena pulmonary
Location of Pacemaker Lokasi perentak
Left atrium Atrium kiri
Diagram 10.5/Rajah 10.5 (a)
(i) Explain the pathway of the systemic circulatory system Terangkan laluan sistem peredaran sistemik [ 2 marks] (ii) The pacemaker of a patient’s heart fails to function. An electronic pacemaker is used to replace the original pacemaker. Explain how the electronic pacemaker works. Perentak seorang pesakit jantung gagal berungsi. Satu perentak elektronik digunakan untuk menggantikan perentak asal. Terangkan bagaimana perentak elektronik berkerja. [3 marks]
42
(b)
Explain the effect of taking excess fatty food in the long term to our blood circulatory system / human healthy life. Terangkan kesan pengambilan makanan berlemak secara berlebihan
dalam
jangka masa yang lama kepada sistem peredarah darah manusia dan kesihatan manusia. [ 5 marks ] 6.
Diagram 10.6 shows the structure of Human Immunodeficiency Virus (HIV). The virus causes Acquired Immune Deficiency Syndrome (AIDS) that infects the human immune system. Rajah 10.6 menunjukkan struktur „Human Immunodeficiency Virus‟ (HIV). Virus ini menyebabkan „Acquired Immune Deficiency Syndrome‟ (AIDS) yang menyerang sistem keimunan manusia.
Diagram 10.6/Rajah 10.6 (a)
Descibe the effect of HIV on the body’s defence mechanism Bincangkan kesan HIV ke atas mekanisma pertahanan badan. [ 4 marks ]
(b)
Explain how first line defence can prevent the entry pathogens into the body. Terangkan bagaimana barisan pertahanan pertama boleh mencegah kemasukan pathogen ke dalam badan. [ 6 marks ] 43
CHAPTER 11: LOCOMOTION AND SUPPORT 1. (a) Diagram 11.1.1 shows the earthworm. Rajah 11.1.1 menunjukkan seekor cacing tanah.
Diagram 11.1.1 / Rajah 11.1.1 How this organism support itself? Explain. Bagaimanakah organisma ini menyokong sokong tubuhnya? Terangkan. [ 4 marks ] (b) Diagram 11.1.2 shows the locomotion of earthworm in habitats. Rajah 11.1.2 menunjukkan pergerakan seekor cacing tanah dalam habitatnya.
s Diagram 11.1.2 / Rajah 11.1.2 Explain how locomotion in earthworm occurs. Terangkan bagaimana pergerakan cacing tanah berlaku. [6 marks] 44
(c) Diagram 11.1.3 shows the movement of a human forearm. Rajah 11.1.3 menunjukkan pergerakan lengan manusia.
Diagram 11.1.3(i) / Rajah 11.1.3(i)
Diagram 11.1.3(ii) / Rajah 11.1.3(ii)
Based on Diagram 11.1.3(i) and Diagram 11.1.3(ii), explain the roles of the muscle, tendons, bones and ligaments in the movement of the forearm. Berdasarkan Rajah 11.1.3(i) dan Rajah 11.1.3(ii), terangkan peranan otot, tendon dan tulang dalam pergerakan lengan. [10 marks]
45
2. (a) Diagram 11.2.1 shows Water hyacinth In its habitat. Rajah 11.2.1 menunjukkan pokok keladi bunting di dalam habitatnya.
Diagram 11.2.1 / Rajah 11.2.1 Explain the adaptation of the plant which enable it to float on water surface. Terangkan penyesuaian pada tumbuhan tersebut yang membolehkannya terapung di permukaan air. [ 4 marks ] (b) Diagram 11.2.2 shows a cross-section of a woody plant stem. Rajah 11.2.2 menunjukkan keratan rentas batang pokok berkayu.
Diagram 11.2.2 / Rajah 11.2.2 Explain what modifications are found in the tissues of woody plants which giving support to the plants. Terangkan penyesuaian yang terdapat pada tisu-tisu kayu tumbuhan ini yang memberikan sokongan kepada tumbuhan. [6 marks]
46
(c) Diagram11.2.3 shows two vertebrae. Rajah 11.2.3 menunjukkan dua ruas vetebra.
Diagram 11.2.3(i) Lumbar vertebra Rajah 11.2.3(i) Vetebra lumbar
Diagram 11.2.3(ii) Cervical vertebra Rajah 11.2.3(ii) Vetebra serviks
Compare and contrast between the two vertebrae. Banding bezakan antara keduanya. [10 marks] 3
(a) A bird can fly with its wings. Burung boleh sayapnya.
terbang
dengan
Describe the mechanism of locomotion of a bird flying in the air. Huraikan mekanisme pergerakan seekor burung yang terbang di udara. [10 marks] (b) Impaired musculoskeletal system could cause diseases such as osteoporosis, muscular dystrophy and arthritis. Sistem rangka otot yang tidak sempurna boleh menyebabkan penyakit-penyakit seperti osteoporosis, distrofi otot dan artritis. Discuss how someone could have a healthy musculoskeletal system. Bincangkan bagaimana seseorang itu boleh mendapatkan kesihatan sistem rangka otot yang sihat. [ 10 marks ] 47
4. (a) Diagram 11.4.1 shows a body structure of a fish. Rajah 11.4.1 menunjukkan struktur badan sekor ikan.
Explain how coordination of a myotome and fins of a fish contributes to swimming activities in the aquatic habitat. Terangkan bagaimana koordinasi di antara otot miotom dan sirip-sirip dapat menyumbang kepada aktiviti berenang seekor ikan di dalam habitat akuatik. [10 marks] (b) Unbalanced diet, an unhealthy lifestyle and the process of ageing may cause diseases such as osteoporosis and arthritis. Diet tidak seimbang, gaya hidup tidak sihat dan proses penuaan mungkin menyebabkan penyakit-penyakit seperti osteoporosis dan arthritis. Justify the above statement and explain how such diseases can be avoided. Justifikasikan kenyataan di atas dan terangkan bagaimana penyakit-penyakit tersebut boleh dielakkan. [10 marks]
48
CHAPTER 12: COORDINATION AND RESPONSE
1.
Diagram 12.1 shows the nerve transmission from neurone P to neurone R. Rajah 12.1 menunjukkan pemindahan impuls dari neuron P ke neuron R. Nerve impulse Impulse saraf
Axon of neurone P Akson neuron P
Synaptic knob Bonggol sinaps Dendrite of neurone R Dendrit neuron R
Q Diagram 12.1/Rajah 12.1
(a)
(b)
Explain the transmission of a nerve impulse from neurone P to neurone R across Q. Terangkan pemindahan impuls dari neuron P ke neuron R merentasi Q. [8 marks] Alzheimer’s and Parkinson’s diseases are related to nervous system. Explain the causes and the effects of the diseases on victims. Alzheimer dan Parkinson adalah penyakit-penyakit berkait saraf.Terangkan penyebab dan kesan penyakit ini ke atas pengidap. [4 marks]
(c)
In an accident, a motorist was seriously injured and was sent to a nearby hospital. A doctor is going to do a knee jerk test to examine the patient’s nerve system. Draw the reflect arch and describe the pathway involved in the transmission of nerve impulses which result in the reflect action. Dalam satu kemalangan, seorang penunggang motosikal mengalami kecederaan serius. Doktor menjalankan ujian sentakan lutut untuk menguji sistem saraf ke atas mangsa tersebut. 49
Lukis arka refleks dan huraikan laluan pemindahan impulse yang terlibat bagi menghasilkan tindakan refleks.
2.
(a)
The pituitary gland is regarded as the “master “ endocrine gland.
[8 marks]
Kelenjar pituitary dikenali sebagai “kelenjar induk”.
Explain the statement above. Terangkan pernyataan tersebut. [2 marks] (b)
After watching a horror movie at a cinema, Salim went back to his home. On the way home, he saw a monster exactly the same as he watched in the movie. He was very shocked and frightened. So he ran away as fast as he could.
Selepas menonton filem seram di pawagam, Salim pulang ke rumah. Dalam perjalanannya, dia terserempak dengan satu lembaga yang sama seperti yang dilihatnya dalam filem. Disebabkan sangat terkejut dan takut, dia berlari dengan sekuat hati.
Explain the involvement of both the nervous system and the endocrine system in that critical situation. Terangkan penglibatan sistem saraf dan system endokrin dalam situasi tersebut. [8 marks]
50
3.
(a)
Explain how the body of a healthy person restores the blood sugar level to normal if the level drops too low. Terangkan bagaimana seseorang yang sihat tubuh badannya mengekalkan aras gula dalam darahnya apabila aras gula menjadi rendah . [6 marks]
(a)
Diagram 12.3 shows a part of a nephron structure. Rajah 12.3 menunjukkan sebahagian daripada struktur nefron.
Diagram 12.3/ Rajah 12.3 (i)
Label and complete Diagram 12.3 above. State the role of a nephron. Label dan lengkapkan Rajah 12.3 di atas. Nyatakan fungsi nefron. [4 marks]
(ii)
Describe the urine formation. Huraikan proses pembentukan urin. [10 marks]
4.
(a)
Explain the terms phototropism and geotropism. Terangkan istilah fototropisma dan geotropism. [4 marks]
(b)
Explain the role of auxins in tropism. Terangkan peranan auksin dalam gerakbalas tropisma. [10 marks]
(c)
State three types of hormones and its uses in agriculture. Namakan tiga jenis fitohormon berserta fungsinya dalam pertanian. [6 marks]
51
CHAPTER 13: REPRODUCTION AND GROWTH 1.
Diagram 13.1 shows the development of human zygote. Rajah 13.1 menunjukkan perkembangan zigot manusia.
Fallopian Tube two cells stage
Q P
Diagram 13.1 / Rajah 13.1
(a)
Describe the process which occurred from P to Q. Huraikan proses yang berlaku dari P ke Q. [4 marks]
(b) Pregnant women are advised not to smoke and avoid from drugs and alcohol. Wanita hamil dinasihatkan supaya tidak merokok dan mengelakkan diri dari najis dadah dan alkohol.
Explain the above statement. Terang kenyataan di atas. [6 marks] 52
2
Mrs. Lee is a married woman, she faces problem to get pregnant. Diagram 13.2 shows two laboratory methods which may help Mr. Lee and Mrs. Lee to have their own child. Pn. Lee telah lama berkahwin, beliau menghadapi masalah untuk hamil. Rajah 13.2 menunjukkan dua kaedah makmal yang mungkin boleh membantu En. Lee dan Pn. Lee untuk mendapatkan anak sendiri.
Mr. Lee
Mrs. Lee
Diagram 13.2 / Rajah 13.2 (a)
Explain under what type of condition Method A can be used to help Mr. and Mrs. Lee. Terangkan dalam keadaan yang bagaimanakah Kaedah A dapat membantu En. Lee dan isterinya. [5 marks]
(b) Discuss the role of Madam X in Method B and the issue may arise. Bincangkan peranan Puan X dalam Kaedah B dan isu yang mungkin timbul. [5 marks]
53
3
Diagram 13.3(a) shows the formation of the embryo sac in the ovule, while Diagram 13.3(b) shows the formation of pollen grain in the anther in a flowering plant (angiosperm) Rajah 13.3(a) menunjukkan pembentukan pundi embrio di ovul, manakala Rajah 13.3(b) menunjukkan proses pembentukan butir debunga di anter pada tumbuhan berbunga (angiosperma).
Anther/anter Anter Embryo sac/ Pundi embrio
Mitosis Pollen grain/ Butir debunga Mature embryo sac Pundi embrio matang Diagram 13.3 (a)/ Rajah 13.3 (a)
Diagram 13.3 (b)/ Rajah 113.3 (b)
Based on the above diagrams, explain how the formation of the embryo sac and pollen grain process occurs. Berdasarkan rajah di atas, jelaskan bagaimana proses pembentukan pundi embrio dan butir debunga berlaku. [10 marks]
54
4
Diagram 13.4 shows the fertilization in a flowering plant (angiosperms). Rajah 13.4 menunjukkan proses persenyawaan yang berlaku pada tumbuhan berbunga (angiosperma). Pollen grain/ Butir debunga Butir debunga Pollen tube/Tiub debunga
Ovule/Ovul
Diagram 13.4/ Rajah 13.4
Based on diagram 13.4, describe how pollination leads to the formation of fruit and seed in a flowering plant (angiosperms). Berdasarkan rajah 13.4, jelaskan bagaimana perdebungaan membentuk buah dan biji pada tumbuhan berbunga (angiosperma). [10 marks]
55
CHAPTER 14: INHERITANCE 1. The variation of ABO blood group determined by three different alleles, but an individual can carry only two of the three alleles. Variasi dalam kumpulan darah ABO ditentukan oleh tiga alel yang berbeza, tetapi setiap individu hanya membawa dua daripada tiga alel tersebut.
With schematic diagram, explain the possibilities of the blood group and the genotypes of the offspring if the father’s blood group is A and the mother’s blood group is B. Dengan gambarajah skema, terangkan kebarangkalian kumpulan darah dan genotip pada anak jika ayahnya mempunyai kumpulan darah A dan ibunya kumpulan darah B. [10 marks] 2.
Colour blindness is a genetic disease that could be inherited and occurs within a specific gender. Inheritance of colour blindness can be prevented in a family. Buta warna adalah penyakit genetik yang boleh diwarisi dan barlaku dikalangan jantina tertentu. Pewarisan buta warna dapat dicegah daripada berlaku dalam sesebuah keluarga.
Schematic Diagram 14.2 shows a genetic pedigree of colour blindness in three generations of a family. Rajah skema 14.2 berikut menunjukkan salasilah buta warna bagi tiga generasi dalam sesebuah keluarga. b represents allele for colour blindness and , B represents allele for normal vision. b mewakili alel untuk buta warna dan, B mewakili alel untuk penglihatan yang normal.
56
XBY
XBXb
Generation l
Generation lI
Generation lII
keys:
Normal male
Normal female
Colour blind male
Carrier female
Colour blind female
Diagram 14.2/ Rajah 14.2
Based on Diagram 14.2, give your opinion about the above statements. Berdasarkan Rajah 14.2, berikan pendapat anda tentang pernyataan di atas. [10 marks]
57
3.
The Law of Independent Assortment states that two or more pairs of alleles segregate independently of one another during the formation of gametes. a.Hukum Segregasi menyatakan bahawa dua atau lebih pasangan alel terpisah secara bebas antara satau sama lain semasa pembentukan gamet.
Illustrate the above statements by using the following cross between a pure bred watermelon with green and short fruit with a pure bred watermelon with striped and long fruit. Show the second filial (F2) generation phenotype and its ratio. Gambarkisahkan pernyataan di atas dengan menggunakan kacukan diantara tembikai baka tulen untuk warna hijau dan buah yang bulat dengan tembikai baka tulen untuk warna berjalur dan buah yang bujur.Tunjukkan fenotip dan nisbah fenotip bagi generasi kedua. [10 marks]
58
CHAPTER 15: VARIATION 1.
Diagram 15.1(a) and Diagram 15.1(b) shows the histogram about distribution of genetic variation in human. Rajah 15.1(a) dan Rajah 15.1(b) menunjukkan histogram mengenai taburan variasi genetik dalam manusia.
Diagram 15.1(a)/Rajah 15.1 (a)
Diagram 15.1(b)/ Rajah 15.1 (b)
With a suitable example, explain the comparison of two kinds of variation. Dengan menggunakan contoh yang sesuai, terangkan perbandingan di antara kedua-dua variasi tersebut. [10 marks] 2.
Variation is seen in almost all living organisms. Discuss the importance of variation and give an example.
Variasi dapat dilihat pada hampir semua kehidupan. Bincangkan kepentingan variasi dan berikan satu contoh. [6 marks]
59
3. An ex-champion runner in 800m has two sons. The first son has a well developed body muscles very much like his father. He is also a good athlete, who practices every day. The second son is small in build, choosy in his diet, reluctant to exercise and prefers to spend his time indoors sleeping and reading. Seorang bekas juara pelari 800m mempunyai dua orang anak lelaki. Anak yang pertama mempunyai otot badan yang berkembang baik seperti ayahnya. Dia juga seorang atlit yang baik yang berlatih setiap hari. Anak yang kedua pula berbadan kecil, amat memilih dalam pemakanan, tidak suka bersenam serta lebih suka mengisi masanya di dalam rumah tidur dan membaca.
Discuss the factors affecting the variation in the two sons. Bincangkan faktor-faktor yang memberi kesan variasi terhadap kedua-dua anak lelaki bekas juara tersebut. [10 marks]
60
1 CHAPTER 2: CELL STRUCTURE AND CELL ORGANISATION QUESTION 1: No 1(a)
Marking Criteria
Marks
Able to draw and label a green plant cell
6
Cell wall Chloropla st Cytoplasm vacuole nucleus 2 Diagram – 2 marks [function, neat] Labels : 5 correct – 4 marks
4
4 correct – 3 marks 3 correct – 2 marks 2 correct – 1 mark. (b)
Able to state the functions of three organelles in the plant cell Organelles
Functions 3
1. Nucleus
- controls all activities of the cell // - determine characteristics / cell functions.
2. Vacuole
- store chemicals such as water/ amino acid/ sugar/ mineral / waste products // - regulates water balance // supports
3. Mitochondrion
- the site of energy production // cellular respiration
4. Lysosome
-breaks down complex organic molecules / protein/ lipid / polysaccharides /nucleic acids
5. Chloroplast
-traps sunlight (energy) during photosynthesis
6. Ribosome
- the site of protein / enzyme.synthesis
+ 3
6
2
No
Marking Criteria
7. RER
Marks
- Transports protein (made by ribosomes) to Golgi apparatus/ other parts of the cell.
8. SER
- synthesis of lipid / phospholipids / steroids // - detoxification
of drugs /
poisons. 9. Golgi Body/ Apparatus
-
Processing/
packaging of
/
transporting
centre
the
synthesised
proteins (such as
enzyme / hormone / antibody, phospholipids
and
carbohydrates & glycoproteins (such as mucus). (c)
Compare and contrast a plant cell with an animal cell. Similarities :
8
Both have a nucleus/ a plasma membrane/ mitochondria/
1
ribosomes/ endoplasmic reticulums/ cytoplasm/ Golgi apparatus
1
Any 2 Differences : Animal Cell
Structure/ Characteristic Does not have a fixed Shape shape Absent Cell wall
Has a fixed shape Present
1
Absent
Chloroplast
Present
1
Vacuole
Present
Absent // very small (if any) Food (carbohydrate) is stored in the form of glycogen Present
Food storage
Centriole
Plant Cell
1
Food (carbohydrate) is stored in the form of starch Absent
1
1 1
3
No
Marking Criteria
Marks
Marking Criteria
Marks
QUESTION 2: No 2(a)
Able to state how the structures are adapted to their functions.
10
Red blood cell : Fact
Explanation
F1 : Biconcave E1 : to increase the TSA/V ratio // E2 : allows rapid diffusion of oxygen into or out of disc shaped the rbc F 2: no
E1 : thereby having more space to contain the respiratory pigment / haemoglobin // E2 : the haemoglobin pigment has a high affinity for oxygen
nucleus
F3 : the plasma membrane is elastic
E1 : allowing it to squeeze through the narrow blood capillaries // E2 : to transport oxygen to the whole body
1 1
1 1
1 1 F1/F2/F3 – 1m
Efferent neurone Fact
Explanation
F1: has a long axon
E1 : to conduct /transmit nerve impulses// E2 : to the effector
1 1
F2 : has dendrons / dendrite
E1 : to receive impulse
F3 : Has the node of Ranvier
E1 : speed up impulse transmission
1
F4 : has myelin sheath
E1 : acts as an electrical insulator
1
F5 : contain a lot of mitochondria
E1 : to generate energy / ATP // E2 : that is required for the transmission of nerve impulses
Sperm cell Fact
1 1 1
Explanation
F1: has a tail
E1 : to swim towards the ovum in the fallopian tube
F2 : contain a lot of
E1 : to generate energy / ATP //
1
4 mitochondria
E2 : needed for locomotion
F3 : the head contains hydrolytic enzymes
E1 : can break down the plasma membrane of the ovum E2 : to allow fertilisation
1
1 b(i)
Able to describe how Amoeba can carry out the following life processes: nutrition and reproduction. Answer : Nutrition P1 : holozoic nutrition // feeds on bacteria / small green algae / diatoms found in water
1 1
P2 : pseudopodia are formed to surround food,
1
P3 : forming a food vacuole
1
P4 : food is digested by hydrolytic enzymes
1
P5 : the soluble products are absorbed and assimilated
1
P6 : the undigested food is egested Any 4 Reproduction P1 : Ameoba reproduce asexually through binary fission P2 : a mature amoeba will divide at the nucleus, followed by the a
1
division of the cytoplasm before forming two daughter cells (in
1
favorable condition) P3 : spore with thick protective cyst walls are produced. Any 2 b(ii)
1
Able to explain why Amoeba does not burst when its immerse in the
4
distilled water ; P 1 : Have contractile vacuole to regulates amount of water/ osmotic
1
pressure P2 : Water diffuse into amoeba cause by distilled water is hypotonic
1
than amoeba. P3 : Then excess of water in the cell diffuse into// enters the
1
contractile vacuole by osmosis P4 : When the contractile vacuole enlarges / increase in size to
1
5 maximum size. P5 : its contract to release/ expel / throw out excess water to the
1
external environment
QUESTION 3: No 3(a)
Marking Criteria P1 : Cell is basic units in the organism / man example muscle cell/ epithelium cell P2 : Tissues are groups of similar cells that have a similar structure and performing similar function. P3 : Epithelial tissue consists of one or a few layers of epithelial cell, and is found covering the outside of the body or lining organs and glands.
Marks 1
1 1
P4 : Epithelial tissue acts as a barrier or protection. It helps to protect organs from injury and fluid loss
1
P5 : It is found lining the oesophagus, stomach, intestines, villi and most parts of the digestive system
1
P6 : The epithelial cells are adapted for absorption or secretion
1
P7 : The wall of the upper part of the oesophagus (an organ) contain voluntary muscles
1
P8 : Peristalsis is assisted by the circular muscles and longitudinal muscles (smooth muscle tissue)
1
P9 : All the muscle tissues are formed from groups of muscle cells P10 : The wall of the stomach (an organ) has three layers of smooth muscle tissue
1 1
P11 :There are connective tissues in the digestive system. The connective tissues include all the blood cells
1
P12 : The digestive system is made up of several organs that work together as one unit. They are the stomach, liver, gall bladder, pancreas, small intestine, large intestine (colon) and rectum
1
10
6
3(b)
Able to explain how the leaf modified to carry out its various function.
characteristics
10
Function
1. Flat and broad with a large surface area to volume ratio
Allow efficient absorption of sunlight and carbon dioxide for photosynthesis, absorption of oxygen for transpiration and loss of water through transpiration.
2
2. Thick waterproof cuticle on its upper surface 3. The epidermis has stomatal pores
To reduce excessive loss of water during transpiration
2
Allow gaseous exchange between air space and enviroment Allow loss of water vapour through transpiration To reduce excessive loss of water vapour
2
4. Less stomatal pores on the upper surface than lower surface 5. Palisade mesophyll cells are packed closely together under the leaf‟s upper epidermis 6. Palisade mesophyll cells are also packed with chloroplasts 7. Spongy mesophyll cells are loosely packed.
To absorb as much light as possible
To increase the rate of photosynthesis
Allow efficient diffusion of gases for photosynthesis which uses carbon dioxide and releases oxygen // respiration which uses oxygen and releases carbon dioxide transpiration which results in loss of water vapour.
8. Xylem vessels in the leaf‟s veins
Supply the leaf cells with water for photosynthesis and transpiration
9. Phloem tissue / sieve tubes in the leaf‟s veins
Transport the organic products of photosynthesis to other parts of the plant.
2
2
2
2
2
2
7
CHAPTER 3: MOVEMENT OF SUBSTANCES ACROSS THE PLASMA MEMBRANE Question 1 No 1(a)
Marking Criteria
Marks
Organelle involve in osmoregulation is contractile
Total Marks
1
vacuole Water diffuses into Amoeba sp. by osmosis and fills
1
the contractile vacuole When the vacuole is filled with water to its maximum
1
size, It will contract to expel the excess water to its
1
4
surroundings (b)
Similarities : Both occurs in living cells
1
Both occurs through a semi-permeable membrane
1
Both require carrier protein to bind with the
1
substances
ANY 2
Differences : Facilitated diffusion
Active transport
Molecules move down the
Molecules move against
concentration gradient
the concentration gradient
Molecules move through
Molecules move through
pore proteins or carrier
carrier proteins only
proteins
1
1
Occurs until a dynamic
Results in accumulation of
equilibrium is achieved
substances in the cell or removal of substances from the cell
ATP or energy is not
ATP or energy is required
1
required Not dependent on cellular
Dependent on cellular
respiration
respiration
1
1
2
8 Not affected by inhibitors
Inhibited by inhibitors such as respiratory poisons
1
6
Vinegar is acidic and has low pH
( c)
This prevent the growth of microorganisms in
1
mangoes
1
The mangoes can be preserved to last longer Concentrated salt solutions hypertonic to the tissue
1
of fish
1
Water diffuse out of fish cell by osmosis Water also diffuse out of bacteria cell to the
1
surroundings
1
The bacteria cells become plasmolysed This prevent the growth of bacteria which cause
1 1
food spoilage
8 20
Question 2 No
Marking Criteria
Marks
2(a)
Amino acid is a large water soluble molecule
1
It requires carrier protein to move across the
1
Total Marks
membrane Amino acid will bind to the carrier protein which is
1
specific to it Carrier protein will change its shape to bring the
1
amino acid molecule across the membrane Lastly the carrier protein will release the amino acid
1
and returns to its original shape (b) Fertilizers which are added to the soil dissolve in soil
1
water The soil water become hypertonic to the cell sap of
1
5
9 plant roots Water diffuse from the cell sap into the soil by
1
osmosis
(c)
The cells become plasmolysed
1
Wilting occurs and the plants eventually die
1
20% salt solution is hypertonic to red blood cell
1
Water diffuse out of cell by osmosis
1
5
1
Cell become shrink
1
Crenation occur
3
ANY 3 1 1 Distilled water is hypotonic to red blood cell Water diffuse into the cell by osmosis
1 1 3
Cell swell up and eventually burst * Haemolysis occur
1
ANY 3
1
(d)
1 30% sucrose solution is hypertonic to potato strip
1
Potato cell become plasmolysed
1
Distilled water is hypotonic to potato strip
1
Water diffuse into cell sap of potato strip by osmosis Cell becomes turgid again
4
The cell is said to have undergone deplasmolysis
20
ANY 4
10
CHAPTER 4 – CHEMICAL COMPOSITION OF THE CELL Question 1 No 1(a)(i)
Marking Criteria Intracellular enzymes are produced and retained in
Marks
Total Marks
1
the cell For the use of the cell itself
1
Extracellular enzymes are produced in the cell but
1
secreted from the cell
(a)(ii)
To function externally
1
The nucleus contain DNA which carries the
1
information for the synthesis of enzymes The genetic information is transcribed from DNA to
1
RNA in the nucleus The RNA leaves the nucleus and attaches to
1
ribosomes on the rough endoplasmic reticulum Proteins that are synthesized at the ribosomes are
1
transported through the space within RER Proteins depart from RER wrapped in vesicles that
1
bud off from the membrane of the RER The transport vesicles fuse with the Golgi
1
Apparatus and empty their content into the membranous space The protein are further modified during their
1
transport in Golgi Apparatus Secretory vesicles containing enzymes bud off from Golgi Apparatus and travel to plasma membrane
1
4
11 These vesicles fuse with plasma membrane and release the extracellular enzymes
1
8
ANY 8
(b)
R is enzyme-substrate complex
1
The substrate molecule is represented by the „key‟
1
while the enzyme molecule is represented by the „lock‟ The substrate has a shape that is complementary
1
to the active site of the enzyme The substrate binds to the active site to form an
1
enzyme-substrate complex , like a key fits into a lock Reaction occurs that is the enzyme catalyses the
1
conversion of substrate to products The products are released from the enzymes
1
The enzyme remain unchanged at the end of the
1
reaction The enzyme is now free to bind with another
1
molecule of substrate 8 20
Question 2 ITEM
MARKING SCHEME
NUMBER 2(a)(i)
SUB
TOTAL
MARK
MARKS
Enzyme
1
Antibody
1
Hormone
1
Any
1
2
Plasma protein
12 1 (a)(ii)
P is primary structure
1
It refers to the linear sequence of amino acids in a polypeptide chain
2 1
Q is secondary structure
1
It refers to the polypeptide chain
1
That is coiled to form alpha-helix chain or folded into beta pleated sheet R is tertiary structure
1
3
1
It refers to the way the helix chains or betapleated
1
sheets 3
are coiled or folded into three dimensional shape o polypeptide chain (b)
1 1 1
Amylase
1
Dissolve and remove starch stains
1
Protease
1
Dissolve and remove protein stains and blood 6
Lipase Dissolve and remove fats, oils and grease stains (c)
1
1 Lipase is used in the ripening of cheese in dairy products industry
1
Cellulase is used to extracts agar from seaweed in seaweed products Cellulase is used to breakdown cellulose and
1
removes seed coats from cereal grains in cereal grain products
1
13 Trypsin is used to remove hair from animal hides in leather industry
1
Amylase is used to convert starch flour into sugar in 1
the making of bread in baking industry Zymase is used to convert sugar into ethanol to make alcoholic drink in beer/wine making industry Protease is used to remove the skin of fish in fish product
ANY 4
4 20
CHAPTER 5: CELL DIVISION QUESTION 1 No (a)
Marking Criteria
Marks
Able to state the important of mitosis Sample Answer : P1 : in growth process // to increase the number of cells ( during the
1
growth process) P2 : Cell replacement // To replace dead and damaged
1
P3 : Regeneration // Production of new cells
1
P4 : Asexual reproduction // the production of new individuals ( from
1
parent organism )
Any 3
3
14
No (b)
Marking Criteria
Marks
Able to explain the differences between process of mitosis and
10
meiosis using appropriate diagram mitosis Prophase
meiosis Prophase I
1
location of chromosome is at random
homologous chromosome synapsis
no crossing over/ no chiasmata
crossing over/ chiasmata
Metaphase
Metaphase I
1 1
1
chromosome are arrange at the middle of cell
homologous chromosome are arrange at the middle of cell
Anaphase
Anaphase I
1
1
chromatid move to the opposite
homologous chromosome will
pole
move to the opposite
Telophase
Telophase I
Each daughter cell has the same
Each daughter cell has half number
number of chromosome
1
15
No
Marking Criteria
Marks 1
1 1 1 1 Diagram: 4
(c)
Exp: 6
= Max: 10
Able to explain a cloning in animal to produce large number of animal much faster than through sexual reproduction : P1 : examples: cow/ sheep P2 : remove a somatic cell donor of animal and growth in a culture
1
P4 : the nucleus ovum is removed
1 1 1
P5 : an electric pulse stimulates the fusion between the somatic cell
1
P3 : unfertilised ovum from donor is obtained
and the ovum P6 : the cell divide by mitosis P7 : embryo is form P8 : then, the embryo is planted into surrogate mother
1 1 max 1 7
QUESTION 2:
16 No (a)(i)
Marking Criteria
Marks
Able to explain what happens in the cell during phase X. Sample answer : P 1 : Protein are synthesized // Mitochondria / chloroplast are synthesized
1 1
P2 : DNA are synthesized
1
P3 : The cell accumulate energy.
2
Any 2 P (a)(ii)
Able to describe tissue culture technique. Sample answer: P1 : Tissue culture technique
1
P2 : Tissue culture technique is used to produce (high quality of seedling)oil palm seedlings in vitro/any suitable example.
1
P3 : The leaves/shoot/stem/root tissues are cut out.(These cut out plant tissues are called explants).
1
P4 :The pieces of meristematic tissue (explants) are cultured in sterile nutrient medium, in suitable pH and with addition of plant growth substances.( at least 2 factors)
1
P5 :The flasks containing the tissue are stored in an incubator at 37°C for 2/3 weeks. P6 : The cell divide by mitosis to produce callus. P7 : The callus is then cut into small pieces.
1
1 1
P8 : The small pieces of callus tissues are then cultured in sterile nutrient medium.
1
P9 : When it has grown to a suitable size, the clone is transferred to the nursery. Any 6 P
1 6
(b)
Able to discuss Advantages / strength: P1 : Genetic engineering involves genes manipulation / transfer / modification in organisms to produce certain products.
1
17 No
Marking Criteria
Marks
P2 : Example; products in pharmacy such as insulin / antibiotics; food products based on plants / animals; agricultural / agrochemical products.
1
P3 : The products produced are very similar to the original / natural materials because the same genes are used / particularly chosen genes are transfered.
1
P4 : The production of products is faster especially with the use of microorganisms/bacteria.
1
P5 : Microorganisms such as bacteria are suitable to be used as gene vectors / they have free DNA in the form of ring / plasmid.
1
P6 : High reproduction rate of bacteria/microorgansm in optimal culture mediums able to produce a large amount of chosen genes / products / insulin / antibiotics.
1
P7 : Can be used by thousands of people who need them / widespread usage.
1
P8 : Able to produce a variety of proteins / recombinant proteins / enzymes used in food industries / medicine / agriculture. 1
(c)
P9 : Genetic engineering technique is used to solve criminal cases through DNA finger printing / DNA fragments analysis.
1
P10 : Other uses /examples; metal extraction from oxide/any suitable examples Any 6 P
1
Able to describe the effect of cycle malfunctions to the body. Sample answer P1: The exposure damage the DNA of the cell P2: A cell divides through mitosis repeatedly. P3: Produces cancerous cell
1 1
P4: Due to (severe ) distruption to the mechanism that controls the cell cycle
1 1
P5: Cancerous cells divide freely / uncontrollably heeding the cell cycle control P6: (these cells ) compete with surrounding normal
1
6
18 No
Marking Criteria cells to obtain nutrient / energy (for growth)
Marks
P7: Invade / destroy neighbouring cells
1
P8: (they can spread to other organ and) initiate
1
6
cancers there . Any 6 P TOTAL
1 20
QUESTION 3 No 3(a)i
Marking Criteria
Marks
Able to explain the principles used in the cloning technique P1 : Cloning is an asexual reproductive process of producing
1
clones//does not involve gamete P2 : A clone is a group of cells//organism//a population of organisms produced from a single ancestral cell.
1
P3 : A clones genetically identical
1
3
P4 : The technique can be used to produce high quality of organism /
1
marks
orchids/ oil palm / cocoa plants. 3(a)ii Able to state the advantages and disadvantages of this technique. Advantages
-
multiply copies of useful genes or clones
Any 4 advantages and 3 disadvantages
1
or Any 3
can be produced in a shorter time and in larger numbers
1
advantages and 4 disadvantages
-
produce clones through asexual reproduction
1
-
produce clones that are resistant to diseases
1
Disadvantages
3(b)
-
clones have the same level of resistance towards certain diseases
1
7
-
may undergo mutations which can disrupt natural equilibrium
1
marks
-
do not show any genetic variations
1
-
unable to adapt to the changes of the environment
1
Able to describe how cytokinesis occurs in plant and animal cells to produce two daughter cells.
-
process of cytoplasmic division
-
begins before nucleus division is complete / during telophase to form
1 1
19 No
Marking Criteria
Marks
two daughter cells
-
in animal cells, actin filaments in the cytoplasm contracts
-
to pull a ring of the plasma membrane inwards
-
forming a groove called a cleavage furrow
1
-
the cleavage furrow pinches at the equator of the cell and deepens
1
1 1
progressively until two daughter cells are separated
-
in plants cells, the membranous vesicles are formed along the equator
1
between the two nuclei 1
-
the vesicles fuse to form a cell plate
-
the cell plate grows outwards until its edges fuse with the plasma membrane of the parents cell
-
at the end of cytokinesis , cellulose fibres are produced by the cells to
1 1
strengthen the new cell walls
10 marks
QUESTION 4 : No 4(a)
Marking Criteria
Marks
Able to explain the process that occurs in stage P, Q, R and R. Stage P : Prophase I
-
chromosome becomes shorter/thicker
-
homologous chromosome come together form bivalent through synapsis
1 1
-
non sister chromatids exchange segments of DNA // crossing over
1
-
nucleus membrane disappears
1
-
spindle fibres form
1
Stage Q : Metaphase I
-
pairs of homologous chromosome arrange at metaphase plate
Stage R: Anaphase I
-
spindle fibre pull the homologous chromosome away from one
1
20 No
Marking Criteria another and move to the opposite poles
Marks 1
Stage S: Telophase 1
(b)
-
chromosome arrive at the poles
-
spindle fibre disappears 1
6
1
marks
Able to state the different between meiosis I and meiosis based on stage
4
P, Q, R and R. Stage
P
Meiosis I
Meiosis II
Homologous chromosome come
chromosome locate at
together form bivalent through
random no crossing over
1
synapsis.There is crossing over Q
R
S
(c)
Pairs of homologous
chromosome arrange at
chromosome arrange at
metaphase plate in a
metaphase plate
straight line
Homologous chromosome move
Chromatid move to
towards opposite poles
opposite poles
Chromosome arrive at the poles
Chromatid arrive at poles
1
1 1
10
Explain briefly how meiosis involved in genetic variation. F1 - crossing over during prophase I of meiosis
1
E1 – non sister chromatids of homologous chromosome break at the
1
chiasma E2 – segments of the chromatids exchange places
1
E3 – segment of the marternal chromatids become attached to the
1
paternal chromatids E4 – new combinations of genes are produced on these chromatids
1
F2 – independent assortment of chromosome
1
21 No
Marking Criteria E1 – at metaphase I the homologous pairs of chromosomes arranged on
Marks 1
the metaphase plate at random E2 – each homologous pair of chromosomes is positioned relative to the
1
poles of the cell independent of other pairs E3 – there is independent assortment of maternal and paternal
1
chromosomes into daughter cells E4 – result in a variety of gametes each with different combinations of maternal and paternal chromosome
:
1
22
CHAPTER 6: NUTRITION No Marking Criteria 1(a) Able to state the meaning of malnutrition.
Marks
Sample answers
1- Malnutrition is a condition due to taking an unbalanced
1+1
2
Max 8
8
diet in which certain nutrients are lacking,
2-
(b)
in excess or in the wrong proportions
Able to name and explain the disease in Diagram 6.1.1, Diagram 6.1.2 and Diagram 6.1.3 related to malnutrition. Sample answers
Diagram 6.1.1
1- Kwashiorkor 2- A child does not receive sufficient protein in his diet. 3- has the characteristic sign of scaly skin, thin muscles , thin hair and a swell of the body
Diagram 6.1.2 4 - Rickets 5 - Vitamin D deficiencies 6 - poor teeth and bone formation in children 7 - leads to softening and weakening of the bones.
Diagram 6.1.3 8 - Obesity 9 - excessive intake of food rich in fat 10 - body weight exceed by 20% of ideal/ normal weight
23
No Marking Criteria 1(c) Able to explain the long term effects of his diet to his
Marks Max 10 10
health. Sample answers Bread
1- Bread is rich in carbohydrates. 2- cause high glucose content in blood. 3- leads to Diabetes mellitus. 4- Excess glucose/glycogen is converted to fats. 5- leads to obesity. Meat
6- Meat is rich in protein. 7- Excess protein is converted to urea 8- Causing liver failure/ kidney failure. 9 - Excess uric acid due to consumption of excess protein cause gout
Cheese
9- Cheese is rich in fats/lipids. 10- Results in high cholesterol level in blood 11- Lipid/cholesterol deposited in inner wall of arteries 12- Lumen of arteries become smaller 13- Cause cardiovascular disease. 14- Excess fat also leads to obesity
TOTAL
20
24 No
Marking Criteria
Marks
2(a)(i)
Able to describe the type of nutrition in organism P and organism
2+2
Total Mark 4
6
6
R. Sample answers Organism P 1 - Autotrophic nutrition 2 - Synthesize its own glucose / starch from carbon dioxide and water with the help of light energy through the process of photosynthesis Organism Q 3 - Heterotrophic nutrition/ holozoic 4 - Obtain its food source/organic substances from the surroundings (eat plant/ producer) Able to compare the process of cellulose digestion in organism Q and organism R. (ii)
Sample answers Similarities 1 - Both have alimentary canal which are made up of the oesophagus, stomach, small intestine and large intestine. 2 - Both are unable to produce cellulase to digest cellulose.
Differences Organism Q
Organism R
1 The type of diet is omnivores
The type of diet is herbivores
2 Stomach has one chamber
Stomach has four chambers
3 Microorganisms in the digestive
Bacteria and protozoa in rumen
tract do not play an important
and reticulum secrete the
role in digestion of cellulose/ do
enzyme cellulase to digest
not have enzyme cellulase to
cellulose
25 digest cellulose. 4 The food from the mouth is
The food from the mouth is
swallowed to the stomach
swallowed to the rumen and
without regurgitation.
reticulum, then it is regurgitated into the mouth to be chewed again before being swallowed into the omasum.
Able to state the function and symptom of deficiency of these minerals. Able to state one source for each mineral. Sample answers (b) Calcium 1 - Needed for the formation of bones and teeth. 2 - deficiency in calcium will cause rickets in children 3 - and osteoporosis most often in women who have gone through menopause 4 - the source of calcium is milk / cheese.
Ferum 5 - required in the production of haemoglobin.
6 - Insufficient ferum leads to anemia 7 - Ferum can be found in meat
Iodine 8 - Important component of the hormone (thyroxine) produced by the thyroid
gland.
9 - symptom of deficiency is goiter 10 - the source of iodine is sea food/seaweed/ iodine salt
10
10
26 TOTAL
20
No
Marking Criteria
Marks
3(a)
Able to explain the effect of eating too much of this kind of mangoes
5
Total Mark 5
5
5
Max 5
5
on the digestion of food in Y. Sample answers 1 - mangoes with vinegar contain much acid, so its reduces the pH value/ increases acidity in the duodenum 2 - Acid medium is less suitable for the action of enzyme lipase, amylase and trypsin 3 - less/ no lipid is digested/hydrolysed to fatty acid and glycerol by lipase 4 - less/ no starch is digested/hydrolysed to maltose by amylase 5 - less/ no polypeptide is digested/hydrolysed to peptides by trypsin Able to state what he should do to handle health problems that may arise from the removal of organ Z. (b)
Sample answers 1 - Reduce the intake of high carbohydrate food / protein /fatty
food.
2 - Get insuline injection when needed / if glucose level too high. 3 - Get glucagons injection when needed / if glucose level too low. 4 - Pancreas transplant. 5 - Eat more vegetables / fruits.
Able to state the functions of X. Sample answers (c)(i)
Functions of Y/liver 1 - Maintenance of blood glucose level under the influence of insulin and glucagons.
27 2 - Synthesis plasma protein such as fibrinogen / prothrombin from amino acids. 3 - Synthesis bile. 4 - Storage of nutrients such as fat-soluble vitamins (A & D)/ B12/ ferum/ copper/ potassium. 5 - Detoxification of poisonous substances such as
alcohols/drugs/
toxins/ pesticides/carcinogens /poisons. 6 - Deamination of amino acids. 7 - Produce heat.
Able to explain the process of assimilation of amino acids and glucose in X. Sample answers Max 5 (c)(ii) The process of assimilation amino acids and glucose in Y
Amino acids 1 - Excess amino acids cannot be stored and are broken down into urea by a process called deamination before being excreted by the kidneys Excess amino acid
urea
2 - When there is a short supply of glucose and glycogen, the liver converts amino acids into glucose. 3 - Plasma protein can be synthesized from amino acids and is used for blood clotting and osmoregulation.
Glucose 4 -
Excess glucose is converted into glycogen and stored in the liver. Excess glucose
glycogen
5 - When the blood sugar level falls, glycogen is converted back to glucose. 6-
Glucose in the liver is used for cellular respiration. Glucose + Oxygen
Energy + Carbon dioxide + water
5
28 TOTAL No 4(a)
Marking Criteria Able to explain the consequences to his health. Sample answers
1
- Fried chicken and meat in the hamburger are high in fat and cholesterol
2
- Excessive intake of fats can be lead to obesity
3
- Excessive cholesterol and fats are deposited in the lumen of arterial wall.
4
- This cause the lumen becomes narrow and leads to arteriosclerosis and hypertension
5
- this will cause the decreasing of blood supply to tissues, leads to heart attack and stroke
6
- Bread in hamburger, soft drink and ice cream are sweet and high in sugar
7
- Sweetened food can cause tooth decay
8
- and obesity
9
- Excessive intake of sugars may eventually lead to diabetes mellitus
10 - Vegetables and fruits are good source of vitamins, minerals and fibres 11 - Insufficient intake may lead to deficiency diseases, dry and scaly skin ,scurvy (other suitable examples) 12 - Insufficient in fibres causes constipation
Able to state the meaning of photosynthesis. Sample answers 4(b)(i) 1 - Photoynthesis is a process whereby a green plant produces glucose / starch from carbon dioxide and water
20 Marks Max 10
Total Mark 10
29 2 - in the presence of chlorophyll and sunlight.
No
Marking Criteria
4(b)(ii) Able to explain the main stages in photosynthesis.
2
2
Marks
Total Mark 8
Max 8
Sample answers
Light reaction 1 - Chlorophyll absorbs/ traps light energy to produce ATP and electrons/ chemical energy 2 - Photolysis of water produces H+ and OH- ions 3 - H+ ion combines with electron to form hydrogen atom 4 - Hydrogen/ ATP/ NADPH will be used in the dark reaction 5 - Light reaction occurs in grana
Dark reaction 6 - 6 - The process takes place in the absence of light/ does not need light 7 - CO2combines with hydrogen/ hydrogen and is reduced to form glucose and water 8 - Glucose molecules undergo condensation/ converted/ stored to starch 9 - Formation of glucose and starch occurs in chemical reaction chain requires ATP/ chemical energy 10- Chemical reactions need ATP energy 11 - Dark reaction occurs in stroma
TOTAL
20
30
No
Marking Criteria
5(a) Able to explain the necessity for food processing. Sample answers 1 – prevent food spoilage 2 – (food spoilage) causes by the action of microorganism 3 – decomposing bacteria/fungi on carbohydrate/protein 4 – produced carbon dioxide, water, ammonia hydrogen 5 – make food become toxic 6 – Oxidation of food when cut/expose to air 7 – oxygen react with enzymes/chemicals released by cell 8 – Increase it commercial value 9 – food additives is added in preserving the freshness of food 10 – Improve the taste/appearance/texture 11 – Intention of diversifying the uses of food 12 – increased the variety of products
Marks Max 10
Total Mark 10
31
No
Marking Criteria
Marks
5(b) Able to describe how each method can preserve food for a
Max 10
Total Mark 10
long period of time Sample answers
Pasteurisation 1-
milk is treated to 63oC for 30 minutes//72oC for 15 seconds
2-
followed by rapid cooling to below 10oC
3-
destroy bacteria but not the spores
4-
retains the natural flavour of milk//nutrients//vitamin B
5-
must refrigerated to avoid the growth of sperms
Canning 6-
use heat sterilisation
7-
kill microorganisms and spores
8-
steamed at high temperature and pressure to drive out air
9-
sealed while the food is being cooled
10 - vacuum in the can prevent growth of microorganism
Refrigeration 11 - stored at temperature below 0oC 12 - prevent the growth of microorganisms/the germination of spores
TOTAL
20
32
CHAPTER 7 No 1(a)
Marking Criteria
Marks
Able to name organ X and organ Y
Total
1
X: Lung
1
2
Y: Gills (b)
Able to draw and label the respiratory surfaces of organ X and Y D-2 L-2
Organ X
(c)
Organ Y
Able to compare and complain the respiratory system in human and fish.
adaptations
of
Similarities and explanation 1. both lungs and gills have large surface area for gas
1
exchange. 2. Human lungs have numerous alveoli while the gills have
1
numerous filaments. 3. Both respiratory systems are thin / one cell thick. 4. This ensures more efficient respiratory gases diffusion. 5. Both respiratory systems have many network of blood
1 1 1
capillaries 6. dissolve more respiratory gases. 7. Both systems use muscles to change the pressure of respiratory cavity / thoracic and mouth cavities. 8. Human has diaphragm and intercostals muscles while fish
1 1
4
33 has muscles in the mouth and operculum.
1
9. Enable/ allowing oxygen to be transported to the body cells through blood vessels.
1
10. Both human and fish have a closed blood circulatory system.
1
Differences and explanation 1. To ensure the supply of oxygen is continuous and sufficient 2. The trachea are reinforced with chitin to prevent them from
1
collapsing whereas the density/water pressure prevent the
1
gills from sticking together 3. The position of respiratory organs ensures the respiratory surface is always moist.
1
4.The surface of gas exchange/the alveoli is located inside the body to prevent from dehydrating whereas the gills are located outside of the body because fish lives underwater.
1 Max10
(d)
Able to explain the effects of smoking on the human respiratory system
1. Carbon monoxide competes with oxygen to bind with
1
haemoglobin and forms carboxyhaemoglobin. It reduces the supply of oxygen to the cells.
2. Nitrogen dioxide can dissolve in mucus to form an acidic
1
medium which erodes lung tissue.
3. 3,4 BENZO-(α)-PYRENE is carcinogenic chemical that
1
can cause cancer.
4. Nicotine can stimulate the production of cancer cell in
1
trachea and lung.
5. Heat and dryness irritation the lungs and can lead to laryngitis
Max4 1
34
No 2(a)
Marking Criteria
Marks
Total
1. Inhalation/ breathing in
1
6
2. External intercostal muscles contract, internal intercostal
1
Able to explain the state of breathing shown in Diagram 7.2.
mucles relax.
3. The rib cage move upwards and outwards.
1
4. The diaphragm contracts and flattens.
1
5. Volume of the thoracic cavity increase resulting in
1
reduced air pressure in alveoli.
6. Higher atmospheric pressure outside causes the air to rush in. (b)
1
Able to explain the effects of the breathing mechanism if structure R is unable to function.
1. Structure R is diaphragm.
1
2. Less/no change in volume in the thoracic cavity/ lung
1
3. Less/ no change in air pressure in the thoracic cavity/
1
lung
4. Less/ no air exchange/ less/no intake of O2/ less/no CO2
Max4
1
expelled
5. Resulting difficulty in breathing in and out (c)
1
Able to describe the regulation of the carbon dioxide concentration in body fluid.
1. During vigorous exercise, the partial pressure of carbon dioxide increase.
2. Carbon dioxide dissolve in blood to form carbonic acid
1
10
35 3. and decrease the blood pH value
1
4. The drop in pH / increases in H+ ions detected by central
1
chemoreceptors (in medulla oblongata) and peripheral
1
chemoreceptors at the aortic bodies and carotid bodies.
5. Nerve impuls is then send to respiration centre in m-Ob 6. The respiratory centre send nerve impulse to the diaphragm and the intercostals muscles
1 1
7. Causing the respiratory muscles to contract and relax 1
faster.
8. results the breathing and ventilation rates increase. 1
9. So excess co2 is eliminated from the body, 10.the carbon dioxide
concentration and pH of the blood
1
return to normal
3(a)
1
Able to differentiate the cellular respiration process that occurs in cell X and tissue Y.
Cell X
Tissue Y
Anaerobic respiration
Aerobic respiration
(Because oxygen is absent)
(Because oxygen is present_
Glucose is not completely Glucose oxidized// Glucose
Fermented/ oxidized/ convert
to
quantity
of
completely breakdown//
2
CO2, Glucose is oxidize to CO2,
ethanol and energy. The
is
2
water and energy. energy The
quantity
of
energy
produced is lower/ 2 ATP/ 210 produced is higher/ 38 ATP/ kJ (per molecule of glucose)
2
2898 kJ (per molecule of glucose)
Respiration equation:
Respiration equation:
Glucose → Carbon dioxide +
Glucose + Oxygen → Carbon
Ethanol + Energy
dioxide + Water + Energy
2
8
36
(b)
Able to state what is process Q and molecule X. Process Q
- Anaerobic respiration
Molecule X - Lactic acid
Max 6 1 1
Able to explain how molecule X can be remove from muscle cell. 1. Inhale more oxygen by doing fast and deep breathing.
1
2. Excess oxygen taken in during inhalation is used to oxidise
1
lactic acid to carbon dioxide and water.
1
3. This oxidation process takes place in the liver. 4. Thus the oxygen debt is the amount of oxygen needed to
1
remove the lactic acid from the muscle cells. Lactic acid + oxygen
carbon dioxide + water +
1
energy (c)
Able to explain the exchange of respiratory gases.
1. Respiratory surfaces in human is alveoli.
1
2. The concentration of oxygen in the alveoli is higher than its
1
concentration in the blood capillaries. 3. Oxygen in the alveoli diffuses into the blood capillaries.
1
4. The concentration of carbon dioxide in the blood capillaries is
1
higher than its concentration in the alveoli. 5. Carbon dioxide diffuses from the blood capillaries of the lungs
1
into the alveoli. 6. Blood leaving the blood capillary of the lungs has higher concentration of oxygen and lower concentration of carbon dioxide.
1
6
37
4(a)
Able to explain how gaseous exchange occurs in the alveoli and blood capillaries
P1: Gas exchange is driven by diffusion //
1
max 10
Diffusion of a gas depends on differences in partial pressure between the two regions P2: thus does not require energy (for exchange).
1
P3: The molecules move down a concentration
1
gradient. P4: Oxygen moves from the alveoli which is high oxygen concentration P5: to the blood which has lower oxygen concentration
1 1 1
P6: due to the continuous consumption of oxygen in the body. P7: Conversely, carbon dioxide is produced by
1
metabolism P8: has a higher concentration in the blood than in the air of alveoli P9: carbon dioxide diffuses out of the blood capillaries
1 1
into the alveoli P10: Oxygen in the lungs first diffuses through the alveolar wall and dissolves in the blood plasma. P11: then diffuse into red blood cells
1 1
P12 (Oxygen) bind to hemoglobin. P13: allows a greater amount of oxygen to be transported in the blood
(b)
1 1
Able to explain the process of energy production by the athlete during and after the race.
1. The muscle cells of the athlete undergoes anaerobic respiration to produce energy
1
max 10
38 2. During intensive physical activity / running / sprinting// when the athlete start running (t = 0), oxygen requirement increase
1
immediately to produce large amount of energy 3. The athlete holds his breath for a short period of time // the athlete breath is shallow during running 4. The oxygen supplied by breathing between t = 0 minute to 6
1 1
minute is insufficient for cellular respiration 5. The muscle cells are now in the state of oxygen debt // undergo oxygen deficit
1
6. Glucose is broken down incompletely without the presence of oxygen 7. Small amount of energy is released to continue the activity
1
8. Lactic acids produced accumulate in the muscle causing the
1
muscular pain and fatigue 9. The anaerobic respiration occurs in the cytoplasm
1
10. (after the activity is over), the athlete breathes faster and
1
deeper to supply more oxygen
1
11. Oxygen is used to oxidize the lactic acid into carbon dioxide, water and energy // converted into glucose and stored as glycogen
1
39
CHAPTER 8: DYNAMIC ECOSYSTEM Question 1 No
Marking Criteria
Marks
1(a)
- Colonisation is a process in which organism starts to inhibit an
1
unhabitat area such as bare ground and forms a colony. - Succession is the gradual process where one community changes its
1
environment so that it is replaced by another community. - Diused pond, pioneer species are Phytoplankton (microscopic algae)
1
and submerged plant (Hydrilla sp., Cabomba sp., elodea sp.) begin to grow and carry out photosynthesis. - When they die and decompose, organic matter converted into humus
1
at the pond base, the pond become shallow. -The condition becomes favourable for floating plants such as water
1
hyacinths (Eichornia sp.) and duckweeds (Lemna sp.) -They spread covering water surface and prevent sunlight from reaching the submerged plant causing these plants to die since they
1
cannot photosynthesise. - The decomposed plants add more organic matter and the pond becomes more shallow.
1
- The emergent plants (sedges, cattails) replace the floating plants. - They grow from the edge of the pond towards the middle of the pond
1
as the pond becomes more shallow.
1
- When these plants die, their decomposed remains are added as
Total
40 sediments to the base of the pond thus reduces the depth of the pond.
1
- The condition becomes suitable for land plants like small herbaceous weeds. - Gradually, the land becomes much drier and more land plants
1
(shrubs, herbs and large woody plant) start to grow. ANY 10
1
- A jungle emerges and turns into a tropical rainforest which form a climax community.
1
– It is a balanced ecosystem which involves the interaction 1. b)
between the abiotic and biotic factors.
1
- Ensures conversation of biodiversity, preservation of flora, fauna and organism. 1
- Preventing the extinction of flora and fauna. - Maintaining major sources of human food such as ulam, meat, honey
1
and sources of traditional medicinal herbs.
1
- Sustains the food webs in the ecosystem. - Preventing disruption of the natural cycles of water/carbon and also balancing photosynthesis and respiration.
1
- Provides natural water catchment areas.
1
- Preserves natural resources for recreational activities and eco- tourism.
1
- Reduces stress and promotes a healthy lifestyle.
1
- A balanced ecosystem prevents the loss of plants which will cause a reduction in food resources and food chains.
1
- It will also cause soil erosion and flash floods.
1
- And the extinction of some animal and plant species. ANY 10
1 1
41
Question 2
No
Marking Criteria
Marks
2. a) - The ground is too soft and muddy soil, unable to support plants. i)
1
- Waterlogged conditions of the soil decrease the amount of oxygen 1 for root respiration. - The swamp water has a high concentration of salt and is hypertonic
1
compared to the cell sap of the root cells. - The plants growing in swamp will have the problem of dehydration.
1
- Seeds that fall into the muddy swamp will die of dehydration/
1
insufficiency of oxygen. - The swamp is exposed to the sun leads to a high rate of
1
transpiration. - As a result, the plants growing there will lose water very fast by transpiration.
1
ANY 5
- Root system which is highly branched and widespreads underground 2.
cable roots to give good support to the plants.
a) ii)
- Pneumatophores (breathing root) which grow protruding upwards
1
Total
42 above the ground to allow gaseous exchange.
1
- The cell sap of the roots cells has a higher osmotic pressure than the soil water that surrounds them to enable water enter the root by 1 osmosis. - Hence, the cells are able to withstand the high salt content of the swamp.
1
- Excess salt is eliminated through hydatodes found at the lower epidermis of leaves.
1
- Viviparous seeds which germinate while still attached to the mother plants.
1
- The long radical produced will let the seedling stick into the ground, not submerged into the soft and waterlogged soil or drift away. - Thick layer of cuticle covers the leaves and succulent which help to
1
reduce the rate of transpiration and store water. ANY 5
1
10
Question 3 No 3.a)
Marking Criteria
Marks
-The diagram shows a saprophytic fungus.
Total
1
- It obtain its nutrient by secreting digestive enzymes onto the 1 substrate.
3. b)
- The enzymes digest the complex substances into simple forms.
1
- The simple forms are then absorbed by hypha.
1
- Owls are predators and rats are preys.
1
- In the month of January until April, the increase in the prey‟s is
1
followed by an increase in predator population. - Due to abundance of food.
1
- However from April to August, when the number of predator 1 increases, the number of preys will then decrease. - This is because the high number of predators will easily
consume
1
4
43 the prey. - When the prey reduces, the predators will have less to eat.
1
- There is intraspecific competition.
1
- The number of predators also reduces in the following months from 1 August to December. -The prey-predator relationship takes place in cycle.
1
- This keep the population of both organisms in a dynamic equilibrium.
1
ANY 6 10
Question 4: No 4.
Marking Criteria
Marks
Total
- Rhizobium bacteria inside the root nodules of legumes and Nostoc 1 sp. found freely in the soil fixed the nitrogen in air. - Decaying bacteria/ fungi decompose plant/ animal/ dead organism/
1
waste product - To form ammonium compound.
1
- Nitrosomonas sp. / nitrifying bacteria converted ammonium 1 compound to nitrite. - Nitrobacter sp. /nitrifying bacteria convert nitrite to nitrate.
1
- Nitrate is absorbed by plant to form plant protein.
1
- ( Plant protein ) eaten by an animal to form animal protein.
1
- Denitrifying bacteria reduce the nitrate content in the soil.
1
- by converting the nitrate into the nitric oxide and nitrogen gas.
1
- Nitrogen gas goes back into the atmospheric to complete the
1
nitrogen cycle. 10
44
CHAPTER 9: ENDANGERED ECOSYSTEM Question 1 No 1
Marking Criteria
Marks
Total
- Industries/ factories/ vehicle contribute to air pollution. - Smoke/ fine solid particles can cause respiratory problem.
1
- oxides of nitrogen/ sulphur dioxide dissolve in rain to produce acid
1
rain. - (acid rain) causing the soil become acidic/ unsuitable for cultivation
1
of crops/ leaching of mineral/ corrosion of metal. - Increase Carbon dioxide in atmosphere causes the greenhouse / global warming. - Industrial/ domestic/ agricultural activities produce waste to contribute water pollution.
1
4
45 - Agrochemical / pesticides/ insecticides used by farmer flow into the
1
river/ lead to the poisoning of aquatic organism. - Agricultural run-offs contain excess nitrates/ phosphates lead to 1 eutrophication. - (eutrophication) causes the BOD value will increase thus may harm
1
the aquatic organisms. - Effluents from electronics factories contain heavy metals/ mercury/ cadmium kill the aquatic organism/ disturb food chain.
1
- Discharged of hot water from industries / glass building cause 4
thermal pollution. - Increase the water temperature in the river causing died aquatic 1
organisms - Increase the atmosphere temperature.
1 2
Question 2: No 2
Marking Criteria - The phenomenon is acid rain.
Marks 1
- Factories / motor vehicles released large amount of oxides of nitrogen to the atmosphere.
1
- and sulphur dioxide.
1
- oxides of nitrogen combines with water vapour (in the atmosphere) to 1 form nitric acid. - Sulphur dioxide combines with water vapour (in the atmosphere) to form sulphuric acid.
1
Total
46 - The rain falls as acid rain.
1
Effects:
6
- May corrode buildings.
1
- Aquatic lives may die due to acidic water/ low pH
1
- Minerals in soil will be dissolved/ washed into rivers.
1
- Soil becomes infertile / not suitable for agriculture.
1
- Plants may die due to infertile soil/ acidic soil.
1
- The ecological balance of ecosystem disrupted.
1 ANY 4 10
Question 3: No 3
Marking Criteria
Marks
- The destruction of ozone layer is due to the increasing levels of
Total
1
chlorofluorocarbons (CFCs) in the atmosphere. - CFCs are a group of chemical compounds that contain chlorine, 1 fluorine and carbon. - CFCs are widely used as coolants in air conditioners and refrigerators, as propellants in aerosol cans and as foaming agents in 1 the making of polystyrene packaging, pillow, cushions etc. - The CFCs in the atmosphere are struck by UV light forming chlorine
1
atoms. - Chlorine atom then breaks the ozone molecule into chlorine
1
monoxide and oxygen gas. - Chlorine monoxide then reacts with the free oxygen atom in the 1
atmosphere to form chlorine atom and oxygen molecule. - The chlorine atoms repeat the breaking of the ozone molecules
1
causing the depletion of the ozone layer continuously. ANY 4 The effects of excessive ultraviolet radiation on human:
4
- reduction of the body‟s immune system. - Skin cancer - Cataract of the eyes.
1
47 ANY 2
1
The effect on plants:
2
- reduction of the growth therefore reducing crop yields. Effect on aquatic organism
1
- Death of plankton, reduce food supply to aquatic organism, fisherman‟s catch is reduced.
1 1
Steps to overcome this problem:
1
- Reduce or stop using CFC or chlorine based products. - Replace CFC with HCFC.
1
- Use wrapping papers instead of polystyrene boxes.
1
- Patch up the holes in the ozone layer by firing frozen ozone balls into the atmosphere.
1 1
ANY 2 2
10
CHAPTER 10: Transport
No
Mark Scheme
1(a) Able to describe the role of leucocyte in body defence.
Sub Mark Any 4
Total Mark 4
48
P1 - The process is called phagocytosis P2 - Phagocytes move towards pathogens using its pseudopodium (Approaching) P3 - Phagocytes surround the pathogens ( Engulfing) P4 - Pathogen is hydrolysed by lysosome (Digestion) and reabsorbed P5 - Destroyed pathogen is removed from the phagocyte 1(b) Able to explain how individual could achieve immunity level.
Any 6
6
P1 - The graph shows Artificial Active Immunity P2 - The person has been injected with a vaccine P3 -
The vaccine contain
killed or weakened
antigens/
bacteria/viruses P4 - antigens/bacteria stimulates lymphocyte / WBC to produce antibodies P5 - 1st dose usually induces a slow production of antibody (and shorter lasting) P6 - Booster dose (2nd and 3rd ) are needed to stimulates more antibody to achieve immunity level ( and larger lasting response). P7 -
any invasion of the pathogenic microorganisms, the body is able to destroy them immediately
P8 - Eg of vaccination: BCG / Hepatitis / Polio / HPV (cervix cancer) TOTAL
No
Mark Scheme
10
Sub Mark
Tota l Mar k
49 2(a)
Able to differenciate the composition between fluid R and fluid Fluid R (blood plasma) 1. Has less lymphocyte
Fluid S (lymph) 1. Has a larger numbers of
plasma protein / eg:
1
lymphocyte
Explain : Lymphocyte is produced by the lymph node. 2. Contain erythrocyte &
3
1
2. No erythrocyte, no plasma protein / eg: fibrinogen
fibrinogen
1
Explain: RBC & plasma protein are too big molecule to pass
2(b)
through
Any
3. Has high content of oxygen
3. Has lower contents of oxygen
Explain : oxygen has been used up by the cell S
Able to explain how fluids is formed 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 / S P7 - lymph / S passes through lymphatic vessels into the thoracic
7
7
50 duct P8 - lymph/ S eventually drains into the right subclavian vein. (Hence, lymph drains back into the blood) TOTAL
No
10
Sub Total Mark Mark
Mark Scheme
3(a) Able to explain how platlet stop bleeding P1 - Platelets clump together/ expose to air and produce
5
5
Any
5
thrombokinase// thromboplastine P2 - Thrombokinase/ thromboplastine converts prothrombin to thrombin (calcium ions must be present P3 - Thrombin converts fibrinogen (a soluble protein plasma) to fibrin (Vitamin K is needed in the formation of prothrombin) P4 - Fibrin forms a network to trap the erythrocytes P5 - to form a clot// scab
3(b) Able to explain possible consequence due to low count of his RBC P1 - Less red blood cells(rbc)/ haemoglobin to combine with oxygen
5
P2 - to form oxyhaemoglobin P3 - less oxygen transported to body cells/ tissues// less oxygen diffuses into the body tissue P4 - for cellular respiration P5 - less energy is produced P6 - resulting in tiredness/ breathlessness/ weakness// fatigue P7 - pale looking appearance // anaemia TOTAL
10
51 No 4 (a)
Sub Total Mark Mark
Mark Scheme Able to explain what happen to the tree after one month P1 - Phloem is removed
4
4
S1 - Both have a closed circulation
Any
2
S2 - Blood flows in blood vessels
2
P2 - glucose / organic substance cannot be transported to the root/ below the ring / downwards P3 - so, glucose accumulate at this part (upper part of the ring) P4 - Upper part of the ring swells / become bigger 4(b) Able to describe the similarities between A and B
S3 - Both have hearts/ atrium and ventricle S4 - Hearts pump blood/ act as a pumping organ
Able to describe the differences between the circulatory system of
4
organisms A and B
Any 4
Organism A
Organism B
Single circulation// Blood flow Double circulation// Blood flow only once/ one time through twice/ two time through the heart the heart Heart have 2 chambers/ heart Heart have 4 chambers/ heart consists of 1 atrium and 1 consists of 2 atriums and 2 ventricle
ventricles
Absence of septum
Presence of septum
Oxygenated blood flow from Oxygenated blood flow from lungs gill to body cell/ tissues body
to the heart
Deoxygenated blood flow from Deoxygenated blood flow from the the heart to the gill
heart to the lungs TOTAL
10
52
No 5(a)(i)
Sub Total Mark Mark
Mark Scheme Able to explain the pathway of systemic circulatory system P1 - Oxygenated blood is carried from left ventricle (heart) to
2
2
3
3
Any
5
the body cells(body), P2 - deoxygenated blood is carried from body cells (body) to the right atrium &right ventricle (heart). 5(a)(ii) Able to explain how the electronic pacemaker works. P1 - Electronic pacemaker work as sino-atrial node (SAN) P2 - It generates a wave of excitatory impulses P3 - spread to atria, causing the heart to contract simultaneously 5(b)
Able to explain the effect of taking excess fatty acid
P1 - Fatty food contain high cholesterol level in blood
5
P2 - Our body has better ability to store fats rather than use it ( release energy) P3 - Cholesterol deposited (in inner wall) of artery cause arteriosclerosis P4 - Lumen of artery become smaller / narrow cause high blood pressure P5 - If coronary artery is blocked, cause angina / heart attack (no oxygen and nutrient to the heart is supplied) P6 - If artery to the brain is blocked, cause stroke (no oxygen and nutrients to the brain is supplied) TOTAL
10
53
No 6(a)
Mark Scheme
Sub Mark
Total Mark
4
4
Any 6
6
Able to describe effect of HIV on body’s defence mechanism F
- HIV reproduce inside the lymphocyte
P1 - less antibody is produced // destruction of immune system P2 - it take 8 -10 years for the symptoms of disease to appear. P3 - the body prone to various infections P4 - immune system collapses and victim dies. (1F + any 3 P) 6(b)
Able to explain how first line defence prevent pathogen entering the body The skin P1 - Skin consist of keratin that make difficult enough to penetrate. P2 - Sweat contain acid / pH 3-pH 5/ is not conducive for the growth of
microorganism.
P3 - Sweat and sebum contain lysozyme that kill the microorganisms.
The repiratory tract //nasal cavity and trachea P4 - mucus on the respiratory tract traps microbs/ dust/ particles P5 - the cilia which lines the respiratory tract
The stomach P6 - HCl can kill the micorganisms ( that present in the foods and drinks) Tears and saliva P7 - contain lysozyme protect eyes and mouth from invasion of pathogen. TOTAL
10
54 Chapter 11: Locomotion and Support
No
Mark Scheme
Sub mark
Total Mark
Any 4
4
Any 6
6
1(a) Able to explain how earth worm support itself. P1 – Has hydrostatic skeleton P2 – Body wall consist of outer circular muscle P3 – (and) inner longitudinal muscle. P4 – Body cavity is filled with a fluid which is held in compartments. P5 – The muscles act antagonistically. 1(b) Able to explain the mechanism of locomotion in earthworm. P1 – (When earthworm is crawling over a surface), the chaetae in posterior end of the body pushed into the ground to anchor it. P2 – The muscle in the anterior end of the body contracts, while the longitudinal muscle relaxes. P3 – (Hence) the anterior end of the body elongates P4 – The hydrostatic pressure builds up in the body P5 – The body fluid is pushed backward. P6 – The chaetae in the posterior end of the body are withdrawn while the chaetae in the anterior end of the body are push into the ground. P7 – The longitudinal muscle in the anterior end of the body contract, while the circular muscle relax. P8 – causes the anterior end of the body become short and thick. P9 – The body fluid flows into the anterior end of the body P10 – causing the posterior end of the body pulled forward.
55
P11-
The earthworm moves on the ground by alternately lengthening and shortening its body, assisted by chaetae. 1(c) Able to explain the role of muscles, tendons and ligaments in the movement of forearms. P1 – Forearm has two sets of muscles; biceps and triceps
Any
P2 – acts antagonistically
10
10
P3 – muscles connected to bone by tendons. P4 – Bones are held together by ligaments. P5 – When the biceps contracts, the triceps relaxes. P6 – Biceps becomes shorter (and thicker), triceps becomes longer (and thinner). P7 – This exerts a pulling force which transmitted to the radius through the tendons. P8 – The radius is pulled upward and the fore arm is bent. P9 – When the triceps contracts, the biceps relaxes. P10 – The triceps becomes shorter and thicker while the biceps becomes longer and thinner. P11 – This exerts a pulling force on ulna through tendons. P12 – The ulna and radius pulled downward, causing the forearm to straighten. 2(a) Able to explain the adaptation of the plant which enable it to float on water surface. P1 – Stem and enlarged petiole with many air sacs E1 – Provide buoyancy P2 – Many fibrous roots can trap air E2 – allow the plant to float
Any 4
4
56 P3 – Stem and roots have aerenchyma tissues E3 – makes the plants light and enable plants to float. E dependent on P
2(b) Able to explain the modification found in the woody plant tissues which giving support to the plant. P1 – Xylem tissues
Any 6
6
E2 – Xylem vessels and tracheids are strengthened with lignin P2 – Parenchyma tissues E3 – Store starch and sugar and water. E4 – Turgid cells give support to the plant. P2 – Collenchyma tissues E5 – Walls thickened with cellulose and pectin P3 – Schlerenchyma tissues E5 – Wall thickened with lignin to provide support
2(c) Able to compare and contrast between the two vertebrae. Similarities: S1 – Both have centrum E1 – Gives support and able to withstand compression force S2 – Both have neural canal E2 – to contain spinal nerve S3 – Both have neural spine E3 – For muscle attachment S4 – Both have transverse process E4 – For muscle attachment S5 – Both have neural arch E5 – Form neural canal which protect the spinal cord
10
10
57 E6 – Both have zygapofisis E6 – To articulate with another vertebra Differences: Vertebra cervical
Vertebra lumbar
D1: Flat (small) centrum
Large and thick centrum
E1:
Give more support
D2: Short neural spine
Long neural spine
E2:
Attachment more muscles
D3: Broad transverse
Well develop transverse
prosess
process
E3:
For attachment more muscles
D4: Has to vertebrarterial
No vertebrarterial canal
Canals E4: Enable blood supply to Head D5: Bigger neural canal
Small neural canal
E5: Contain bigger spinal cord/ brain trunk
E dependent on S/D ANY 5 Correct S & E similarities : 5 marks ANY 5 Correct D & E differences: 5 marks 3(a) Able to describe the mechanism of locomotion of a bird flying in the air. P1 – Bird fly by flapping their wings / gliding P2 – The wings of bird is in the shape of aerofoil During flying: P3 - (To fly) the pectoralis major contract P4 – The pectoralis minor relax P5 – The pectoralis muscles are antagonistic muscles
10
10
58 P6 – The wings moving downward and backward P7 – The air resistance produced as a result of moving wing downward P8 - provide an upthrust on the wings P9 – The thrust is transmitted from wings to the coracoids until sternum P10 - (As result) the whole body is lifted up. P11 – (then) the pectoralis minor contract P12 – The wings are pulled up P13 – The air resistance is very low P14 – The wings are ready to move downward.
During gliding: P15 – The wings spread (to act as aerofoil) P16 – The air move faster on the upper of the wings compared to the
lower of the wings
P17 – The air pressure is lower in the upper surface than below the wings P18 – Upward thrust produced enable the birds to glide. ANY 10 3(b) Able to discuss how someone could has a healthy musculoskeletal system. P1: Having a well-balanced diet E1: contain sufficient calcium and phosphorus E2: Contain sufficient vitamin D E3: To build strong bones / prevent osteoporosis P2: Having a good posture E4: While standing, our body should be erect straight, so that the weight of our body is supported bu both our feet. E5: While sitting, the thorax is vertical/the thigh is comfortable/ almost all muscle relaxed
10
10
59 E6: While walking, our body should be upright and straight E7; While lying down, use a mattress that is firm so that the body is evenly supported E8: Bend both knees when lifting heavy object from the floor. P3: Using proper attire for daily activities E9: Wearing tight could restrict the movement E10: Wearing high-heeled shoes could injure the back bone. P4: Taking appropriate precautions during vigorous activities E15: Consistent and moderate exercise can increase the bone mass and prevent osteoporosis E16: Very vigorous activity could results in pain/strain/ dislocation/fractures. P5: practicing correct and safe exercise techniques E17: Warming up before exercise can raise the temperature of our muscle to enabling them to make more efficient use of energy/ preventing injuries. ANY 10 4(a) Able to explain how coordination of myotomes and fins of a fish contributes to swimming activities in the aquatic habitat. P1: Myotomes are muscle block P2: arranged in segments on both sides of the body / vertebral column. P3: The muscles acts antagonistically / contraction of myotomes on one side of vertebral column and relaxation of the myotomes on the other side. P4: the contraction of myotomes on the right side of the body will bend the tail to the right // the contraction of myotomes on the left side of the body will bend the tail to the left. P5: Alternate contraction of the right and left myotome block causes the body to bend side to side. P6: This produces the forward thrust which propel the fish
10
10
60 forward P7: The paired fins and unpaired fin used to maintain the balanced of body during swimming. P8: The pectoral fins used to steering and brake. P9: The pelvic fin are used to prevent diving and rolling movements P10: Dorsal and ventral fins used to stay on course without yawing. P11: Tail/caudal fin used to propel the fish.
ANY 10 4(b) Justify that unbalanced diet, an unhealthy lifestyle and the process of aging may cause diseases such as osteoporosis and arthritis and explain how such diseases can be avoided .P1: (unbalanced diet) such as diet less in calcium / P2: less in phosphorus could lead P3: less in vitamin D P4: unhealthy lifestyle such as consume liquor P5: Process of ageing such as life after menopause P6: (could) cause osteoporosis / bone becomes porous/ soft and brittle P7: (could) cause arthritis / inflammation of the joints.
10
10
61 Way to overcome osteoporosis: P8: Optimize calcium intake to increase the bone mass P9: Optimize vitamin D intake to enhance calcium absorption P10: Exercise regularly to help strengthened the muscle and bone P11: undergo hormone replacement therapy during menopause to prevent osteoporosis Way to overcome gouts: P12: Reduction of offal and protein in diet P13: Taking medication to lessen the joint inflammation and to reduce the level of uric acid in the body. P14: Less/stop consumption of liquor. P15: Massage on the surrounding muscles using heat therapy. ANY 10
62 Chapter 12 Coordination and Support
No
Mark Scheme
1(a)
Able to explain the transmission of a nerve impulse
Sub Mark
Total Mark
Max 8
8
from neurone P to neurone R across Q
P1 P2
Q is a synapse/ synaptic cleft. - The transmission of information across a synapse involves
the conversion of electrical signal into
chemical signal in the form of neurotransmitter. P3
- Neurotransmitter is produced in vesicles in the axon terminal called synaptic knob.
P4
- Synaptic knob contains abundant mitochondrion to generate energy for the nerve transmission.
P5
- When an impulse arrived at the synaptic knob, the vesicles release the neurotransmitters into the synapse.
P6
- The neurotransmitters molecules diffuse across the synapse
P7
- to the dendrite of another neurons.
P8
- Reaching R, impulse is converted back into electrical
P9
signal.
- The transmission of impulse in one way direction
P10 -
since the vesicle containing neurotransmitter is only found in pre-synaptic membrane.
(b)
Able to explain the causes and the effects of the Alzheimer’s and Parkinson’s diseases on victims. Alzheimer‟s
Parkinson‟s
Caused by
Caused by
4
63 -the shrinkage of brain -the tissues
and
lack
reduced
level
of
of neurotransmitter in the
neurotransmitter.
brain caused tremors and
Usually affects the
weakness of the muscles
elderly
-The hardening of the
1+1
cerebral arteries 1+1
Effect:
Effect:
-Loss of intelligent
-The
-Loss of memory
function
-Poor concentration
become stiff and jerky in
muscle
cannot
smoothly
and
their action (c)
Able to draw the reflect arch and describe the pathway involved in the transmission of nerve impulses which result in the reflect action.
D -1
L -1
P1
- The knee jerk action involves two types of neurons
64 named afferent and efferent neurons. P2
6
- As the hammer strike, the force stretches the quadriceps muscle and stimulates the stretch receptors in the muscles triggering a nerve impulses
P3
- Afferent neurons transmit the information to the efferent neuron in the spinal cord
P4
- The efferent neurons transmit the information to the quadriceps muscle as an effector and the muscle contracts thus swing the leg forward
P5
- If the patient is able to swing the leg forward, it indicates that the patient‟s nerve system is still functioning
P6
- If there is no response, it shows that the patient‟s nervous system fails to function properly
2(a) Able to explain why the pituitary gland is known as the “ master gland” 2 P1
2
- because it secretes several hormones that control other endocrine glands
P2
- for example, TSH is secreted to stimulate thyroid gland to release thyroxine //accept any correct hormones and their function
(b)
Able to explain the involvement of both the nervous
8
system and the endocrine system in this situation 1 P1-The situation is called “fight or flight” situation P2-Nerve impulses from the eyes (receptors) travel to the 1 brain P3-The information is interpreted and the brain sends nerve 1
65 impulses to the adrenal glands P4-The adrenal glands are stimulated to release adrenaline
1
P5-The hormone increases the heartbeat rate, blood pressure and blood flow to the muscle
1
P6-The breathe become faster and deeper P7-metabolic activity and glucose level in blood increase
1
P8-The skeletal muscles become more energized and enable a person to fight off an attacker or flee immediately
3(a)
Able to explain how the body of a healthy person restores the blood sugar level to normal if the level drops too low 6 P1 - The islet cells in the pancreas are stimulated to release glucagon P2 - Glucagon stimulates the liver to break down glycogen to glucose P3 - This restores the blood sugar level to normal P4 - Glucagon also promotes lipid breakdown P5 - This releases fatty acids that can be metabolized to generate energy P6 - This restores the blood sugar level to the normal range
(b)(i)
Able to complete the diagram and state the role of the
6
66 nephron
4 D–2
Bowman‟s capsule Proximal convoluted tubule Collecting duct Loop of Henle Distal convoluted tubule
P1 - A nephron is a basic structural and the functional unit
1
of the kidney P2
- It is responsible for the actual purification of blood in producing urine
1
Able to describe the urine formation (b)(ii) -The formation of urine involves three main processes :ultrafiltration, reabsorpton and secretion
1
-Ultrafiltration o The high hydrostsatic pressure in the glomerulus o causes many constituents of the blood such as
1
water,glucose, amino acid, vitamin & salt and ures to be filtered out from the glomerulus into the Bowman capsule. 1 The filtrate is called glomerular filtrate o and the presence of pores in capillary of glomerulus increase its permeability o The glomerular filtrate then passes along the nephron to let the reabsorption to occur
1
67
-Reabsorption
1
o When the glomerular filtrate which reaches the proximal convoluted tubul, 65% of the water is re reabsorb to the
1
capillary by osmosis o All glucose, amino acid, vitamin and some salt are reabsorbed by active transport. Urea is not reabsorbed
1
o The filtrate now contains only water, some salt and urea o It continues to pass along the loop of Henle where 20% of the water and some salt are reabsorbed into the blood capillaries
1
o The filtrate continues to pass through the distal convoluted tubule and collecting duct whereby some water and salt
1
are reabsorbed into the blood capillaries
1
o The amount of water and salts reabsorbed is actually regulated by the amount of ADH and aldosterone
1
-Secretion o The process involves the pumping out of the waste
1
product such as urea /uric acid/ ammonia / creatinine from the blood capillaries into the tubule o by active transport o The glomerular filtrate that reaches the collecting duct is
1
called urine. It composes of 69% water, 2.5% nitrogenous product and 1.5 % salts and other trace elements o Urine is channeled out into the bladder by the ureter and finally secreted out of the body through the urethra 4 (a)
Able to explain the terms: Phototropism, geotropism phototropism
4 1
- growth movement/ growth of plants towards light
1
- growth towards a source of light is called positive
1
68 phototropism - growth away from the source is termed negative phototropism geotropism - The response of plants towards gravity
Max2 1 1 1 Max2
- The root shows positive geotropism. - The shoot shows negative geotropism (b)
Able to explain the role of auxins in tropism
10
- Phototropism is controlled by auxins
1
- When a plant is given light from all directions/uniform light,
1
the distribution of auxins in the shoot meristem is uniform -hence, elongation of cells takes place evenly in all direction,
1
-resulting in the plant/ coleoptile grow upwards.
1
- when the plant is given unilateral light, auxins will be
1
accumulated to the shaded side -As a result, cells on the shaded side of the stem elongate
1
more than those on the exposed side. -Thus, the shoot bends in the direction of light
1
- this explains why the shoot is positively phototropic.
1
-As for the root, a high concentration of auxins inhibits growth
1
- Cells on the exposedt side elongates more than those on the 1 shaded side.
( c)
-hence, the root grows away from light.
1
-this explains why the root is negatively phototropic.
1
Able to state 3 types of hormones and its uses in agriculture
6 1
-Auxins - is used to promote the growth of crop plants, induce 1 parthenocarpy and as herbicides. -Gibberellins -are used to promote growth and parthenocarpy.
1
69 -Cytokinins
1
-are used together with Auxins to promote growth. -Ethylene
1
-(a gaseous plant hormone) is used to promote the 1 ripening of fruits. 1 max 3 name of hormone with relevant function
1
70 Chapter 13 Reproduction and Growth No 1
Mark Sub Mark Scheme (a) – Ovulation releases a secondary oocyte , which enters the oviduct. - The secondary oocyte starts meiosis II which progresses until metaphase II. - The nuclei of a sperm cell (n) and the ovum (n) fuse and form a diploid zygote (2n). // A sperm fertilize the ovum to form a zygote. - Zygote begins to divide repeatedly by mitosis as it travels along the fallopian tube towards the uterus. - Morula is form followed by blastula. - Implantation occur / The blastocyst attaches itself to the endometrium. (b) Sample Answer : - Cigarette contain nicotine / DDT / lead particles.
Total Mark 1 1 1
1
1 1
Max 4
1
- The wall of maternal blood capillaries and the wall of foetal blood capillaries are semi-permeable.
1
- Nicotine, drugs and alcohol are small in size.
1 1
- Nicotine, drugs and alcohol can diffuse from maternal blood capillaries to foetal blood capillaries
No
1
Max 6
- through the placenta - The substances carried by umbilical vein to the foetus. - Nicotine, drugs or alcohol can affect the development of foetus
1 1 1
- (example) cause disable / miscarriage . birth defect/ illness in the resulting baby. 2
(a)
- Method A is use if the fallopian tubes of Mrs. Ali are blocked. - sperm cannot reach the ovum, fertilization fail to occur. - fertilization has to be done outside the body. - developed zygote/embryo then retransfer and implant in the uterus of Mrs. Ali. - the embryo then undergo normal development in the uterus of Mrs. Ali as normal pregnancy.
1 1 1 1 1
5
71
(b)
- Method B is used if the uterus of Mrs. Ali fail to carry the implanted embryo because of damaged or abnormal uterus. - Madam X is the woman who is willing / hired to carry the implanted embryo to full term. - Madam X is known as surrogated mother. - Genetically the baby belongs to Mr. and Mrs. Ali. - Who is the real biological mother of the baby, Mrs. Ali or Madam X? - There are cases that the surrogated mother refuse to return the baby to the couple after giving birth.
1
1 1 1 Max 5
1 1
No 3
Mark Scheme
Sub Mark
Able to explain formation of the embryo sac F1-The ovule develops from the ovarian tissue. It has a diploid embryo sac mother cell(2n) F2- Embryo sac mother cell undergoes meiosis to form a row of four haploid cells called megaspores
1
1
F3- Three of the four megaspores degenerated, leaving one in the ovule
1
F4- (The megaspore continues to grow and enlarges, filling up most of the ovule.) The nucleus of the megaspore then undergoes mitosis three times to form eight haploid nuclei
1
F5- Three of the eight nuclei (migrate to one end of the cell) to form antipodal cells, another two nuclei to form polar nuclei and one of the three nuclei develops into an egg cell/female gamete/ovum and flanked two synergid cells
1
Able to explain formation of pollen grain F1- Pollen grain are formed in the anther, an anther
1
Total Mark
72 has four pollen sacs. F2- Each pollen sac contains hundred of cells called pollen mother cells (2n)
1
F3- Each pollen mother cell undergoes meiosis to produce four haploid microspores(n)
1
F4- The nucleus of each microspores then divided by mitosis to form a tube nucleus and generative nucleus.
1
1 No. Item 4
F5- The microspores develop into pollen grains Explanation
Mark
Able to describe pollination leads to the formation of fruit and seed in a flowering plant (angiosperms). F1- Pollen grains have been released from the anther to the stigma for pollination by insects or wind
1
1 F3- The pollen tube grows down the style towards the ovule 1 1 F5- The male gamete nuclei move down the pollen tube led by the tube nucleus 1 F6- When the pollen tube reaches the ovary, it penetrates the ovule through the micropyle 1 F7- The tube nucleus degenerates, leaving a clear passage for the male nuclei to enter the embryo sac 1 F8- Double fertilization occurs in the ovule. One male nucleus fuses with the egg nucleus to form a diploid zygote(2n) 1 F9- The other male nucleus fuses with the two polar nuclei to form a triploid nucleus(3n) 1 F10- (After fertilization), the triploid nucleus divides rapidly by mitosis to forms an endosperm and zygote divides by mitosis
Total mark Max 10 marks
F2- The sugar solution (sucrose) secreted by the stigma stimulates the pollen grain to germinate and form a pollen tube
F4- The generative nucleus divides by mitosis to form two male gamete nuclei
Max 10 marks
73 develops into suspensor and embryo. 1 F11- The ovule develops into a seed while the ovary enlarges and develops into a fruit
74 Chapter 14 Inheritance No
Sub Mark
Mark Scheme 1
Able to explain the possibilities of the blood group and the genotypes of the offspring when the father‟s blood group is A and the mother‟s blood group is B. Sample answer: 1. The ABO blood group in humans is controlled by three alleles I A, , IB and Io. 2 Alleles IA and IB are codominant but allele Io is recessive.
1
3. There are four possibilities; Blood group AB, A, B, O
1
Father
Mother IA IA
(a) Parental genotypes:
IB IB
X
1
meiosis
Gametes
IA
IB 1 Fertilisation
IA IB
Genotype F1
1 Phenotype F1
All offspring have Blood group AB 1 Father
(b) Parental genotypes:
Mother IA IA
IB IO
X
1
meiosis
Gametes
IA
IB
IO Fertilisation
1
Total Mark
75 Genotypes F1
IA IB
IA IO
Phenotypes F1
AB
A
Phenotipic ratio
1
:
1 1
@ Phenotype F1
50% of offspring have blood group AB
1
and 50% have blood group A Father
Mother IA IO
(c) Parental genotypes:
1
IB IB
X
meiosis
1 Gametes
IA
IO
IB Fertilisation
1 A B
B O
Genotype F1
I I
I I
Phenotypes F1
AB
B
Phenotipic ratio
1
:
1
1
@ Phenotype F1
50% of offspring have blood group AB
1
and 50% have blood group B
Father
Mother IA IO
(d) Parental genotypes:
IB IO
X
meiosis
1 Gametes
I
A
I
O
B
I
I
O
Fertilisation
1 Genotypes F1
IA IB
IA IO
IB IO
IO IO
76 Phenotypes F1 Phenotipic ratio
AB 1
A :
1
B : 1
O :
1 1
1
@ 25% chance that offspring has blood group AB, A, B, O
@ Able to explain : example (d) 1. Genotype of Fother is IA IO , produces two types of sperms, one containing allele IA and the other containing allele IO . 2. Genotype of Mother is IB IO, produces ovum containing allele IO or allele IB 3. When the sperm containing allele IA fertilizes with the ovum containing allele IB , the offspring produced will have the genotype IA IB and the phenotype is blood group AB 4. When the sperm containing allele IA fertilizes with the ovum containing allele IO , the offspring produced will have the genotype IA IO and the phenotype is blood group A 5. When the sperm containing allele IO fertilizes with the ovum containing allele IB , the offspring produced will have the genotype IO IB and the phenotype is blood group B 6.
No
2
Mark Scheme Able to explain how the inheritance can be prevented based on the schematic diagram given. Sample Answer 1. Colour blindness are sex-linked inheritance disease and is carried by recessive gene.
Sub Mark
Total Mark
77 2. Males are homozygous, receiving only X chromosome from their mother. 3. Females are heterozygous, inheriting X chromosomes from both parents. 4. Daughters, as long as one parent is genotypically normal, can only be carriers. 5. The normal gene on the second X chromosome counteracts the defect and the daughters do not suffer from trait. 6. When a son receives the defective gene from his mother he will be colour blindness because the Y chromosome cannot counteract the defective gene located on his X chromosome. 7. Thus, colour blindness is more common in males than females. Able to state opinion for statement II 8. Hemophilia can be prevented by removing the hemophilic gene
1
1
1
1
1
in the gamete stage and insert the normal gene // the hemophilic gene is replaced with a healthy gene through
1
genetic engineering 9. Hemophilia can be prevented by artificial insemination. 10. Hemophilia can be prevented by choosing sperms free of hemophilic trait// difficult to choose sperms which are free from hemophilic trait 11. Choose partner which is free from hemophilic trait . 12. Inheritance of colour blindness can be prevented by avoid marriage of the grandchildren which consists of carriers or colour blindness through a few generation
1
78
No
3
Mark Scheme
Able to explain the type of cross breed that involve two pairs of characteristics -this type of cross breed is the result of breeding between two opposite pairs of characteristics Able to state the type of inheritance -known as dyhibrid inheritance and follows Mendel‟s Second Law G represents the allele for green colour g represents the allele for striped colour L represents the allele for short shape l represents the allele for long shape Able to show the schematic diagram of how to get the second filial generation phenotype and the ratio. Parental : GGLL x ggll genotypes Parental : (short green) Phenotypes
Sub Mark
1
1
(long striped) 1
Gametes:
GL
gl 1
First filial generation F1
GgLl (all short green)
F1 Self cross
x
GgLl
GgLl
GL Gl gL gl GL Gl gL gl Punnett square is prepared to determine the phenotypic ratio in F2 generation. gametes GL Gl gL gl GL GGLL GGLl GgLL GgLl Gl GGLl GGll GgLl Ggll gL GgLL GgLl ggLL ggLl gl GgLl Ggll ggLl ggll F2 generation Phenotypes:
1
Round : Long : Green Green
Short : Long Striped Striped
1 1
1
Tota l Mar k
79
F2 generation Phenotypic Ratio:
9
:
3
:
3
:
1
1
max 10
80 Chapter 15: Variation No
Mark Scheme Able to (i) 1 State the example of continuous variation and discontinuous variation (ii) Explain the similarity and the contrast of continuous variation and discontinuous variation Sample answer: Example of continuous variation: Height or weight Example of discontinuous variation: ABO blood group Similarity: - both create varieties in the population of species - both type of variations are caused by environment factor or genetic factors or both Differences Continuous variation - Graf distribution shows a normal distribution - The characters are quantitative / can be measured and graded (from one extreme to the other) - Exhibits a spectrum of phenotypes with intermediate values between the highest and the shortest in the population. - Influenced by environmental factors
-
Two or more genes control the same character
Sub Mark
Total Mark
1 1
1 1 Discontinuous variation -
Graf distribution shows a discrete distribution
-
The characters are qualitative / cannot be measured and graded (from one extreme to the other)
1
1 -
Exhibits a few distinctive phenotypes with no intermediate values in between them.
-
Is not Influenced by environmental factors/ Is caused by genetic factors and also by the mutation of genes and chromosomes.
-
A single genes determines the differences in the traits of the character
81 -
The phenotype is usually controlled by many pair of alleles
No
-
The phenotype is controlled by a pair of alleles
Mark Scheme
Sub Mark
Total Mark
Able to discuss genetic and environment factor affecting variation Sample answer Genetic factors F1: crossing over during prophase 1/meiosis 1 P1:occur between chromatid from a pair of homologous chromosomes
1
P2:the exchange of parts between chromatid results in new genetic combination.
1
P3:produced a large number of gametes with different genetic
1
composition. F2:independent assortment
1
P4:homologous part of chromosome are arranged randomly on 2 metaphase plate/during metaphase 1
1
P5:during anaphase 1,each homologous pair of chromosomes
1
separate. P6:resulting in an independent assortment of maternal and paternal
1
chromosomes into daughter cells
1
F: Random fertilization P7; sperms and ovum with a variety of combinations of chromosomes/genetically different are randomly fertilized.
1
P8:Thus,variation exists between individuals from the same species//zygote produces wll have a variety of diploid combination. F:Mutation P10:mutation causes permanent change in the genetic
1 1
82 composition/genotype of an organism Environmental factor F1: (can cause variation among individuals at same species)by interacting with genetic factor.
P: examples of factor at least 2 type of food/exercise/skill/experience/education/sunlight/climatic Any 9 from genetic factor And any 1 from environment factors
No
Mark Scheme
Sub Mark
Total Mark
Able to state The importance of variation Sample answer -
Variation provides new genetic material for the survival of the
fittest, 3 -
e.g. the mutated genes of the black peppered moth Variation prepares a species to survive when there are changes
in the external environment, like after a volcanic eruption, or in global warming; -
e.g. the black peppered moths survive well in a soot-covered
1
1 1 1
83 environment. -
A population with a varied genotypes or genotypes is useful in
spreading the particular species over a wider range of habitats; -
e.g. the house sparrow.
-
Produced phenotype/physical differences among individuals.
-
E.g. No human are alike eventhough they are identitical
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