Practice paper, not an official KPM paperKertas latihan, bukan kertas rasmi KPM

Ujian Akhir Sesi AkademikEnd of Academic Session Test

BiologiBiology

Tingkatan 5, 2 jam / 2 hours, 100 marksmarkah

Nama / NameNama

Kelas / ClassKelas

Tarikh / DateTarikh

Markah / marksmarkah: / 100

InstructionsArahan

Kertas soalan ini mengandungi empat bahagian: Bahagian A, Bahagian B, Bahagian C dan Bahagian D.This question paper consists of four sections: Section A, Section B, Section C and Section D.
Bahagian A: jawab semua soalan. Pilih satu jawapan bagi setiap soalan.Section A: answer all questions. Choose one answer for each question.
Bahagian B: jawab semua soalan.Section B: answer all questions.
Bahagian C: jawab satu soalan sahaja daripada dua soalan. Bahagian D: jawab satu soalan sahaja daripada dua soalan.Section C: answer one question only out of two. Section D: answer one question only out of two.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.The diagrams in this paper are not drawn to scale unless stated.
Anda dibenarkan menggunakan kalkulator saintifik.You may use a scientific calculator.

Answer schemes are in English (DLP).Skema jawapan dalam Bahasa Inggeris (DLP).

Practice shows answers as you go. Exam times you and hides answers until you submit.Latihan menunjukkan jawapan semasa menjawab. Peperiksaan menetapkan masa dan menyembunyikan jawapan sehingga dihantar.

Bahagian ASection A

20 marksmarkah

Jawab semua soalan. Setiap soalan diikuti oleh empat pilihan jawapan, A, B, C dan D. Pilih satu jawapan sahaja.Answer all questions. Each question is followed by four options, A, B, C and D. Choose one answer only.

Ciri manakah terdapat pada batang monokot tetapi tidak terdapat pada batang eudikot?Which feature is found in the stem of a monocot but not in the stem of a eudicot?

Berkas vaskular tersusun dalam satu cincinVascular bundles arranged in a ring
Berkas vaskular bertaburan di seluruh tisu asasVascular bundles scattered throughout the ground tissue
Kambium vaskular di antara xilem dan floemVascular cambium between the xylem and the phloem
Korteks dan empulur yang dapat dibezakan dengan jelasA cortex and a pith that can be clearly told apart

Show answerTunjuk jawapanHide answerSembunyi jawapan

In a monocot stem the vascular bundles are scattered through the ground tissue and there is no cambium, so there is no secondary growth. A, C and D describe a eudicot stem.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.1

TextbookBuku teks p.ms. 8

Seorang murid memerhati epidermis bawah sehelai daun di bawah mikroskop. Luas medan pandangan ialah 0.25 mm² dan terdapat 9 stoma di dalam medan pandangan itu. Berapakah ketumpatan stoma pada daun itu?A student observes the lower epidermis of a leaf under a microscope. The area of the field of view is 0.25 mm² and there are 9 stomata in the field of view. What is the stomatal density of the leaf?

2.25 stoma per mm²2.25 stomata per mm²
9 stoma per mm²9 stomata per mm²
18 stoma per mm²18 stomata per mm²
36 stoma per mm²36 stomata per mm²

Show answerTunjuk jawapanHide answerSembunyi jawapan

Density = number ÷ area = 9 ÷ 0.25 = 36 stomata per mm². 2.25 comes from multiplying instead of dividing; 9 ignores the area.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.2

TextbookBuku teks p.ms. 31

Jadual 1 menunjukkan kadar fotosintesis suatu tumbuhan akuatik pada lima keamatan cahaya dan dua kepekatan karbon dioksida, pada 30 °C.

Keamatan cahaya (unit)	1	2	3	4	5
Kadar pada 0.03% CO₂ (unit)	2	4	6	6	6
Kadar pada 0.10% CO₂ (unit)	2	4	6	8	10

Pernyataan yang manakah betul?Table 1 shows the rate of photosynthesis of an aquatic plant at five light intensities and two concentrations of carbon dioxide, at 30 °C.

Light intensity (units)	1	2	3	4	5
Rate at 0.03% CO₂ (units)	2	4	6	6	6
Rate at 0.10% CO₂ (units)	2	4	6	8	10

Which statement is correct?

Pada 0.03% CO₂, kepekatan karbon dioksida menghadkan kadar apabila keamatan cahaya melebihi 3 unit.At 0.03% CO₂, the concentration of carbon dioxide limits the rate once the light intensity is above 3 units.
Pada 0.03% CO₂, keamatan cahaya menghadkan kadar pada 5 unit cahaya.At 0.03% CO₂, light intensity limits the rate at 5 units of light.
Pada 2 unit cahaya, kepekatan karbon dioksida menghadkan kadar.At 2 units of light, the concentration of carbon dioxide limits the rate.
Pada 0.10% CO₂, kadar tidak meningkat lagi selepas 3 unit cahaya.At 0.10% CO₂, the rate does not increase any more after 3 units of light.

Show answerTunjuk jawapanHide answerSembunyi jawapan

At 0.03% the rate stops rising at 6 units even though more light is given, but at 0.10% it keeps rising. So above 3 units of light it is the carbon dioxide that limits the rate. At 1 to 3 units the two curves are the same, so light is the limiting factor there.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 2, standard 2.5

TextbookBuku teks p.ms. 53

Pasangan nutrien dan fungsinya yang manakah betul?Which pair of nutrient and its function is correct?

Nitrogen - membentuk lamela tengahNitrogen - makes the middle lamella
Kalium - bahagian tengah molekul klorofilPotassium - the central part of the chlorophyll molecule
Fosforus - membentuk ATP dan asid nukleikPhosphorus - makes ATP and nucleic acids
Kalsium - membentuk asid aminoCalcium - makes amino acids

Show answerTunjuk jawapanHide answerSembunyi jawapan

Phosphorus makes nucleic acids, ATP and the phospholipids of the plasma membrane. Calcium forms the middle lamella, magnesium is the centre of chlorophyll, and nitrogen (not calcium) is in amino acids and proteins.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.1

TextbookBuku teks p.ms. 61

Tumbuhan X tumbuh pada batang pokok getah. Tumbuhan X mempunyai daun hijau yang berfotosintesis dan akarnya menyerap nutrien yang terkumpul dalam celah kulit kayu. Pokok getah itu tidak mengalami kerosakan. Apakah tumbuhan X?Plant X grows on the trunk of a rubber tree. Plant X has green leaves that photosynthesise and its roots absorb the nutrients collected in the gaps of the bark. The rubber tree is not harmed. What is plant X?

Tumbuhan parasitA parasitic plant
Tumbuhan epifitAn epiphytic plant
Tumbuhan karnivorA carnivorous plant
Tumbuhan mesofitA mesophyte

Show answerTunjuk jawapanHide answerSembunyi jawapan

An epiphyte lives on a taller plant for support and light, makes its own food and does not harm the host. A parasite would take food from the host and harm it; a carnivorous plant traps insects; a mesophyte is described by its habitat, not by growing on another plant.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.3

TextbookBuku teks p.ms. 70

Pernyataan manakah betul tentang gutasi?Which statement about guttation is correct?

Ia berlaku pada waktu siang melalui stoma yang terbuka.It occurs in the day through open stomata.
Ia membebaskan wap air tulen ke atmosfera.It releases pure water vapour into the atmosphere.
Ia berlaku pada waktu malam apabila tekanan akar tinggi dan kelembapan udara tinggi.It occurs at night when the root pressure is high and the air humidity is high.
Ia berlaku pada semua tumbuhan, termasuk pokok berkayu yang tinggi.It occurs in all plants, including tall woody trees.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Guttation is the release of droplets of water with minerals at the ends of leaf veins, caused by high root pressure when humidity is high and transpiration is low (night, early morning). It happens only in herbaceous plants, and the droplets are liquid, not vapour.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.2

TextbookBuku teks p.ms. 82

Sekumpulan murid meletakkan keladi bunting (Eichhornia crassipes) dalam bekas air yang dicemari kuprum dan plumbum. Selepas dua minggu, kepekatan logam berat dalam air berkurang. Apakah penjelasan yang paling sesuai?A group of pupils place water hyacinths (Eichhornia crassipes) in a container of water polluted with copper and lead. After two weeks the concentration of heavy metals in the water has decreased. What is the most suitable explanation?

Akar panjang keladi bunting mengumpulkan logam berat daripada air.The long roots of the water hyacinth accumulate the heavy metals from the water.
Stoma pada daun menyerap logam berat daripada air.The stomata on the leaves absorb the heavy metals from the water.
Kloroplas menukarkan logam berat kepada karbohidrat.The chloroplasts change the heavy metals into carbohydrates.
Floem membebaskan logam berat ke udara sebagai gas.The phloem releases the heavy metals into the air as gases.

Show answerTunjuk jawapanHide answerSembunyi jawapan

In phytoremediation the long roots of the water hyacinth take up and accumulate heavy metals such as copper and lead from the water. Stomata exchange gases, chloroplasts make food and phloem carries organic substances; none of them removes metals.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 4, standard 4.4

TextbookBuku teks p.ms. 91

Seorang penjual bunga mahu melambatkan penuaan dan pengguguran daun pada bunga ros yang dipotong. Fitohormon manakah yang patut digunakan?A florist wants to delay the ageing and yellowing of the leaves of cut roses. Which phytohormone should be used?

AuksinAuxin
GiberelinGibberellin
Asid absisikAbscisic acid
SitokininCytokinin

Show answerTunjuk jawapanHide answerSembunyi jawapan

Cytokinin delays leaf senescence, so it keeps cut flowers fresh. Abscisic acid stimulates abscission and leaf fall, gibberellin lengthens stems and auxin is used for rooting and parthenocarpy.

For the teacherUntuk guru: RememberingMengingat, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.3

TextbookBuku teks p.ms. 108

Seorang tukang kebun memotong pucuk apeks pokok bunga raya. Selepas dua minggu, banyak cabang sisi tumbuh. Apakah penjelasan yang paling tepat?A gardener cuts off the apical bud of a hibiscus shrub. After two weeks many lateral branches grow. What is the most accurate explanation?

Pucuk apeks menghasilkan giberelin yang merencat pertumbuhan tunas sisi.The apical bud makes gibberellin, which inhibits the growth of the lateral buds.
Pucuk apeks menyimpan makanan yang memanjangkan tunas sisi apabila dipotong.The apical bud stores food that elongates the lateral buds when it is cut.
Pucuk apeks menghasilkan auksin yang merencat tunas sisi; apabila pucuk dibuang, tunas sisi tidak direncat lagi dan tumbuh.The apical bud makes auxin, which inhibits the lateral buds; when it is removed the lateral buds are no longer inhibited and they grow.
Pucuk apeks menghasilkan etilena yang menggalakkan tunas sisi tumbuh.The apical bud makes ethylene, which stimulates the lateral buds to grow.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Auxin from the apical bud inhibits the lateral buds (apical dominance). Cutting the bud removes the source of this auxin, so the lateral buds, helped by cytokinin, grow into branches. Gibberellin and ethylene do not control lateral buds this way.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 5, standard 5.2

TextbookBuku teks p.ms. 105

Sebuah anter mempunyai 120 sel ibu mikrospora diploid. Setiap sel membahagi secara meiosis dan semua mikrospora berkembang menjadi butir debunga. Berapakah bilangan butir debunga yang dihasilkan?An anther has 120 diploid microspore mother cells. Each cell divides by meiosis and all the microspores develop into pollen grains. How many pollen grains are produced?

Show answerTunjuk jawapanHide answerSembunyi jawapan

One microspore mother cell divides by meiosis into four haploid microspores (a tetrad) and each becomes a pollen grain. 120 × 4 = 480 pollen grains.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.2

TextbookBuku teks p.ms. 118

Pasangan ciri tumbuhan dan fungsinya yang manakah betul?Which pair of plant feature and function is correct?

Kutikel tebal pada daun teratai - mengurangkan kehilangan airThick cuticle on the leaf of a lotus - reduces loss of water
Aerenkima dalam batang teratai - menyimpan airAerenchyma in the stem of a lotus - stores water
Stoma terbenam pada daun xerofit - meningkatkan transpirasiSunken stomata on the leaf of a xerophyte - increase transpiration
Pneumatofor pokok api-api - pertukaran gas dalam lumpur yang tergenang airPneumatophores of the api-api tree - gaseous exchange in waterlogged mud

Show answerTunjuk jawapanHide answerSembunyi jawapan

Pneumatophores grow upwards out of the mud and have lenticels, so the roots can exchange gases. Aerenchyma holds air (buoyancy), not water; sunken stomata reduce transpiration; a lotus leaf has stomata on the upper surface, not a thick cuticle for saving water.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 7, standard 7.1.2

TextbookBuku teks p.ms. 136

Organisma Y ialah organisma eukariot yang berbilang sel. Dindingnya daripada kitin, badannya terdiri daripada hifa dan ia menyerap nutrien daripada kayu mati. Dalam kingdom manakah organisma Y diletakkan?Organism Y is a multicellular eukaryote. It has a chitin cell wall, its body is made of hyphae and it absorbs nutrients from dead wood. In which kingdom is organism Y placed?

Protista
Fungi
Plantae
Animalia

Show answerTunjuk jawapanHide answerSembunyi jawapan

A chitin cell wall, a body of hyphae (mycelium) and feeding as a saprophyte are characteristics of kingdom Fungi. Plants have photosynthesis and cellulose walls; animals have no cell walls; protists have no specialised tissue.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 8, standard 8.1.1

TextbookBuku teks p.ms. 146

Apakah yang dimaksudkan dengan niche sesuatu organisma?What is meant by the niche of an organism?

Peranan organisma dalam ekosistem, termasuk tingkah laku dan interaksinya dengan komponen biotik dan abiotikThe role of the organism in the ecosystem, including its behaviour and its interactions with biotic and abiotic components
Tempat semula jadi tempat organisma itu hidupThe natural place where the organism lives
Semua organisma spesies yang sama dalam satu habitatAll the organisms of the same species in one habitat
Semua populasi yang hidup dan berinteraksi dalam satu habitatAll the populations that live and interact in one habitat

Show answerTunjuk jawapanHide answerSembunyi jawapan

The niche is the role of the organism in its ecosystem. B is the habitat, C is a population and D is a community.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 9, standard 9.1.1

TextbookBuku teks p.ms. 170

Seorang murid menandakan 45 ekor kumbang di sebuah padang seluas 150 m² dan melepaskannya. Sehari kemudian dia menangkap 40 ekor kumbang, dan 8 ekor daripadanya mempunyai tanda. Berapakah anggaran ketumpatan populasi kumbang itu?A student marks 45 beetles in a field of area 150 m² and releases them. The next day she captures 40 beetles, 8 of which are marked. What is the estimated population density of the beetles?

0.8 ekor per m²0.8 beetles per m²
1.2 ekor per m²1.2 beetles per m²
1.5 ekor per m²1.5 beetles per m²
225 ekor per m²225 beetles per m²

Show answerTunjuk jawapanHide answerSembunyi jawapan

Population size = (45 × 40) ÷ 8 = 225 beetles. Density = 225 ÷ 150 = 1.5 beetles per m². 225 is the population size, not the density.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.2.2

TextbookBuku teks p.ms. 187

Tiga sampel air S, T dan U diuji dengan metilena biru. Masa untuk warna biru hilang ialah 8 jam bagi S, 2 jam bagi T dan 5 jam bagi U. Sampel manakah yang mempunyai BOD paling tinggi dan berapakah kadar pencemarannya?Three water samples S, T and U are tested with methylene blue. The times for the blue colour to disappear are 8 hours for S, 2 hours for T and 5 hours for U. Which sample has the highest BOD and what is its rate of pollution?

S, 0.125 j^-1S, 0.125 h^-1
T, 0.50 j^-1T, 0.50 h^-1
T, 2.00 j^-1T, 2.00 h^-1
U, 0.20 j^-1U, 0.20 h^-1

Show answerTunjuk jawapanHide answerSembunyi jawapan

The faster the blue colour disappears, the more microorganisms and the higher the BOD, so T is the most polluted. Rate of pollution = 1 ÷ time = 1 ÷ 2 = 0.50 h^-1. 2.00 is the time written as a rate.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 10, standard 10.1

TextbookBuku teks p.ms. 208

Amalan manakah dapat menjimatkan air di sebuah rumah?Which practice saves water in a household?

Membasuh kereta dengan hos yang dibiarkan mengalirWashing the car with a hose that is left running
Membiarkan paip terbuka semasa memberus gigiLeaving the tap open while brushing teeth
Menyiram pokok pada waktu tengah hari yang panasWatering plants at midday when it is hot
Menadah air hujan dan menggunakan semula air basuhan sayur untuk menyiram pokokCollecting rain water and reusing the water used to wash vegetables to water plants

Show answerTunjuk jawapanHide answerSembunyi jawapan

Collecting rain water and reusing water are water saving practices (together with water-efficient products with the SPAN label). The other three waste water or lose it by evaporation.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 10, standard 10.3

TextbookBuku teks p.ms. 214

Sebatang pokok kacang pea berbunga ungu (B ialah alel dominan) disilangkan dengan sebatang pokok berbunga putih. Anaknya terdiri daripada 48 pokok berbunga ungu dan 52 pokok berbunga putih. Apakah genotip kedua-dua induk?A pea plant with purple flowers (B is the dominant allele) is crossed with a plant with white flowers. The offspring are 48 plants with purple flowers and 52 plants with white flowers. What are the genotypes of the two parents?

BB × bb
Bb × Bb
Bb × bb
BB × Bb

Show answerTunjuk jawapanHide answerSembunyi jawapan

The white parent must be bb. A ratio of about 1 purple : 1 white shows that the purple parent is heterozygous, Bb. BB × bb gives all purple; Bb × Bb gives 3 : 1; BB × Bb gives all purple.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 11, standard 11.1

TextbookBuku teks p.ms. 235

Sebatang pokok ros dibiak secara keratan batang. Sepuluh keratan ditanam dalam petak P yang subur dan sepuluh keratan dalam petak Q yang kurang subur. Selepas tiga bulan, tinggi purata anak pokok dalam petak P ialah 42 cm dan dalam petak Q ialah 25 cm. Apakah kesimpulan yang paling sesuai?A rose plant is propagated by stem cuttings. Ten cuttings are planted in the fertile plot P and ten cuttings in the poor plot Q. After three months the mean height of the plants in plot P is 42 cm and in plot Q is 25 cm. What is the most suitable conclusion?

Perbezaan ketinggian disebabkan oleh genotip yang berbeza dan boleh diwarisi.The difference in height is caused by different genotypes and can be inherited.
Perbezaan ketinggian disebabkan oleh faktor persekitaran dan tidak diwarisi.The difference in height is caused by environmental factors and is not inherited.
Perbezaan ketinggian disebabkan oleh mutasi dalam keratan di petak Q.The difference in height is caused by a mutation in the cuttings in plot Q.
Perbezaan ketinggian disebabkan oleh pindah silang semasa meiosis.The difference in height is caused by crossing over during meiosis.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Cuttings from one plant are genetically identical (asexual reproduction, no meiosis), so the difference must come from the environment (soil fertility). Variation caused by the environment is not inherited.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 12, standard 12.1

TextbookBuku teks p.ms. 260

Semasa meiosis I pada seorang wanita, sepasang kromosom 21 gagal berpisah. Ovum yang mengandungi dua kromosom 21 (24 kromosom kesemuanya) disenyawakan oleh sperma yang normal. Berapakah bilangan kromosom dalam zigot dan apakah sindrom yang terhasil?During meiosis I in a woman, the pair of chromosome 21 fails to separate. An egg that contains two chromosomes 21 (24 chromosomes in all) is fertilised by a normal sperm. What is the number of chromosomes in the zygote and what syndrome results?

46, sindrom Down46, Down syndrome
47, sindrom Klinefelter47, Klinefelter syndrome
45, sindrom Turner45, Turner syndrome
47, sindrom Down47, Down syndrome

Show answerTunjuk jawapanHide answerSembunyi jawapan

24 chromosomes from the egg plus 23 from the sperm = 47, with three copies of chromosome 21 (trisomy 21), which is Down syndrome. Klinefelter (44 + XXY) and Turner (44 + XO) involve the sex chromosomes.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 12, standard 12.3.2

TextbookBuku teks p.ms. 272

Apakah satu kelemahan makanan ubah suai genetik?What is one disadvantage of genetically modified food?

Gen asing seperti gen rintang antibiotik mungkin berpindah kepada manusia.A foreign gene such as an antibiotic resistance gene may be transferred to humans.
Kos pengeluaran makanan menjadi lebih tinggi.The cost of producing food becomes higher.
Lebih banyak racun perosak perlu digunakan.More pesticides have to be used.
Nilai pemakanan makanan berkurang.The nutritional value of the food decreases.

Show answerTunjuk jawapanHide answerSembunyi jawapan

A possible risk of genetically modified food is that a foreign gene, for example an antibiotic resistance gene, might transfer to humans, and its effect on health is not fully known. B, C and D are the opposite of the advantages of GMF (lower cost, less pesticide, higher nutritional value).

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 13, standard 13.1.1

TextbookBuku teks p.ms. 282

Bahagian BSection B

40 marksmarkah

Jawab semua soalan dalam bahagian ini.Answer all questions in this section.

Sebuah tasik bersebelahan dengan sebuah estet kelapa sawit dipantau selama lima bulan. Baja nitrat ditabur di estet itu pada bulan 2 dan hujan lebat turun pada bulan 3. Jadual 2 menunjukkan keputusannya.

Jadual 2
Bulan	Nitrat (mg dm^-3)	Bilangan alga (sel per cm³ × 1 000)	Oksigen terlarut (mg dm^-3)
1	2	8	7.8
2	3	10	7.6
3	9	40	6.9
4	12	90	4.1
5	6	25	2.0

A lake next to an oil palm estate is monitored for five months. Nitrate fertiliser is spread in the estate in month 2 and heavy rain falls in month 3. Table 2 shows the results.

Table 2
Month	Nitrate (mg dm^-3)	Number of algae (cells per cm³ × 1 000)	Dissolved oxygen (mg dm^-3)
1	2	8	7.8
2	3	10	7.6
3	9	40	6.9
4	12	90	4.1
5	6	25	2.0

[1 markmarkah] (a)Pilih proses yang berlaku di dalam tasik itu.Choose the process taking place in the lake.

EutrofikasiEutrophicationPencemaran termaThermal pollutionHujan asidAcid rain

Show answerTunjuk jawapanHide answerSembunyi jawapan

Eutrophication

Nitrate from fertiliser flows into the lake with the rain, and the algae multiply (algal bloom). This is eutrophication.

[2 marksmarkah] (b)Hitung peratus pertambahan bilangan alga dari bulan 2 hingga bulan 4.Calculate the percentage increase in the number of algae from month 2 to month 4.

Show answerTunjuk jawapanHide answerSembunyi jawapan

(90 – 10) ÷ 10 × 1001
= 800%1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (c)Terangkan mengapa oksigen terlarut paling rendah pada bulan 5 walaupun bilangan alga telah berkurang.Explain why the dissolved oxygen is lowest in month 5 although the number of algae has decreased.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Many algae die after the bloom1
Decomposer bacteria (microorganisms) multiply and decompose the dead algae1
The bacteria use up a large amount of dissolved oxygen (high BOD), so the dissolved oxygen is lowest1

Tick each point you got.Tandakan setiap isi yang betul.

[1 markmarkah] (d)Nyatakan satu kesan bahan cemar ini kepada hidupan akuatik dalam tasik itu.State one effect of this pollution on the aquatic life in the lake.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Any 1 pointsMana-mana 1 isi

Fish and other aquatic organisms die (suffocate) because there is not enough dissolved oxygen1
Sunlight cannot reach submerged plants, so they die1

Tick each point you got.Tandakan setiap isi yang betul.

[1 markmarkah] (e)Sampel air daripada tasik itu diuji dengan metilena biru pada bulan 1 dan bulan 5. Pilih perbandingan masa untuk warna biru hilang pada bulan 5 berbanding bulan 1.Samples of the lake water are tested with methylene blue in month 1 and in month 5. Choose how the time for the blue colour to disappear in month 5 compares with month 1.

Lebih pendekShorterLebih panjangLongerSamaThe same

Show answerTunjuk jawapanHide answerSembunyi jawapan

Shorter

In month 5 there are many more decomposing microorganisms, so the BOD is higher and the blue colour disappears faster.

[2 marksmarkah] (f)Cadangkan dua cara pengurus estet itu boleh mengurangkan masalah ini.Suggest two ways in which the manager of the estate can reduce this problem.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Any 2 pointsMana-mana 2 isi

Use the correct amount of fertiliser, or organic fertiliser made from compost, so less nitrate is washed away1
Leave a strip of vegetation between the estate and the lake to trap the fertiliser1
Do not apply fertiliser before heavy rain1
Treat the run-off water before it enters the lake1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 10, standard 10.1

TextbookBuku teks p.ms. 207

Bahagian CSection C

20 marksmarkah

Jawab satu soalan sahaja daripada dua soalan dalam bahagian ini.Answer one question only out of the two questions in this section.

Bahagian DSection D

20 marksmarkah

Jawab satu soalan sahaja daripada dua soalan dalam bahagian ini. Jawapan hendaklah jelas dan berasas.Answer one question only out of the two questions in this section. Your answers must be clear and reasoned.

27Choose 1Pilih 1

Seorang murid mengkaji hipotesis bahawa semakin banyak daun pada sebatang pucuk, semakin banyak air yang hilang melalui transpirasi. Dia menggunakan pucuk pokok balsam yang sama saiz dengan 0, 2, 4 dan 6 helai daun. Setiap pucuk diletakkan dalam tabung yang mengandungi 50 cm³ air dengan satu lapisan minyak di atas air. Isi padu air yang hilang dalam satu jam direkodkan dalam Jadual 3.

Jadual 3
Bilangan daun	0	2	4	6
Air yang hilang dalam 1 jam (cm³)	0.4	2.2	4.0	5.8

A student investigates the hypothesis that the more leaves a shoot has, the more water it loses through transpiration. He uses shoots of balsam plant of the same size with 0, 2, 4 and 6 leaves. Each shoot is placed in a tube containing 50 cm³ of water with a layer of oil on top of the water. The volume of water lost in one hour is recorded in Table 3.

Table 3
Number of leaves	0	2	4	6
Water lost in 1 hour (cm³)	0.4	2.2	4.0	5.8

[3 marksmarkah] (a)Nyatakan pemboleh ubah dimanipulasikan, pemboleh ubah bergerak balas dan dua pemboleh ubah dimalarkan dalam eksperimen ini.State the manipulated variable, the responding variable and two constant variables in this experiment.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Manipulated: number of leaves1
Responding: volume of water lost in 1 hour1
Constant (any two): size of the shoot, volume of water at the start, light intensity, temperature, humidity, wind (air movement)1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (b)Mengapakah lapisan minyak diletakkan di atas air dan mengapakah satu pucuk tanpa daun disertakan? Hitung air yang hilang bagi setiap helai daun dalam pucuk dengan 6 helai daun selepas kehilangan air daripada batang diambil kira.Why is a layer of oil put on top of the water and why is a shoot with no leaves included? Calculate the water lost per leaf in the shoot with 6 leaves after the loss from the stem is taken into account.

Show answerTunjuk jawapanHide answerSembunyi jawapan

The oil stops water evaporating from the surface of the water, so all the loss is from the shoot1
The shoot with no leaves shows the loss from the stem (control) to be subtracted1
(5.8 – 0.4) ÷ 6 = 0.9 cm³ per leaf1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (c)Adakah hipotesis itu diterima? Terangkan hubungan antara bilangan daun dengan kehilangan air dengan menggunakan konsep transpirasi dan daya tarikan transpirasi.Is the hypothesis accepted? Explain the relationship between the number of leaves and the loss of water using the ideas of transpiration and transpirational pull.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Yes, the more leaves, the more water is lost (0.9 cm³ per leaf in every shoot, so the loss increases evenly)1
More leaves have more stomata, so more water vapour leaves by transpiration1
This increases the transpirational pull, so more water is drawn up the xylem to the leaves1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (d)Terangkan bagaimana struktur xilem dan trakeid sesuai untuk mengangkut air dan menyokong tumbuhan.Explain how the structure of xylem vessels and tracheids suits them for transporting water and supporting the plant.

Show answerTunjuk jawapanHide answerSembunyi jawapan

The cells are dead and empty with no cytoplasm (vessel elements also have no end walls), joined end to end into a continuous tube, so water flows without obstruction1
The walls are thickened with lignin, which strengthens the vessel so it does not collapse and which supports the plant1
The lumen is narrow, so adhesion and cohesion (capillary action) can raise the water column1
Tracheids have pits, so water can pass sideways to neighbouring cells1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (e)Seorang murid lain ingin membuktikan bahawa air bergerak dalam xilem dan bukan dalam floem. Cadangkan satu eksperimen, termasuk pemerhatian yang dijangka.Another student wants to prove that water moves in the xylem and not in the phloem. Suggest an experiment, including the expected observation.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Place a freshly cut shoot (or a plant with roots) in water coloured with red ink (eosin) and leave it in the light for some hours1
Cut a transverse section (and a longitudinal section) of the stem and observe under a microscope or magnifying glass1
Observation: the red colour is found only in the xylem tissue, so water moves up in the xylem1
Suggestion: a control shoot in plain water shows no colour; the phloem stays uncoloured1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (f)Nyatakan tiga faktor persekitaran yang meningkatkan kadar transpirasi.State three environmental factors that increase the rate of transpiration.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Any 3 pointsMana-mana 3 isi

High light intensity (stomata open wide)1
High temperature1
Low humidity of the air1
Strong wind (air movement)1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.2

TextbookBuku teks p.ms. 80

28Choose 1Pilih 1

Pulau Karang mempunyai tiga zon terumbu karang. Zon P ialah zon tanpa masuk, zon Q ialah zon yang dibuka untuk menyelam rekreasi dan zon R terletak berhampiran sangkar ikan komersial. Seorang penyelidik mencatatkan bilangan spesies yang ditemui dalam kawasan sampel yang sama luas dalam Jadual 4.

Jadual 4
Zon	Spesies karang	Spesies ikan	Jumlah spesies
P	20	40	60
Q	14	27	41
R	5	13	18

Kerajaan negeri sedang mempertimbangkan sama ada untuk mewartakan seluruh pulau itu sebagai taman laut atau membenarkan lebih banyak sangkar ikan.Pulau Karang has three zones of coral reef. Zone P is a no-entry zone, zone Q is open to recreational diving and zone R is next to commercial fish cages. A researcher records the number of species found in sample areas of the same size in Table 4.

Table 4
Zone	Coral species	Fish species	Total species
P	20	40	60
Q	14	27	41
R	5	13	18

The state government is considering whether to gazette the whole island as a marine park or to allow more fish cages.

[3 marksmarkah] (a)Nyatakan jenis biodiversiti yang diukur dalam Jadual 4 dan nyatakan dua jenis biodiversiti yang lain.State the type of biodiversity measured in Table 4 and name the two other types of biodiversity.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Species diversity (species richness)1
Genetic diversity1
Ecosystem diversity1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (b)Hitung peratus penurunan jumlah spesies di zon Q dan zon R berbanding zon P, betul kepada satu tempat perpuluhan.Calculate the percentage decrease in the total number of species in zone Q and in zone R compared with zone P, correct to one decimal place.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Zone Q: (60 – 41) ÷ 60 × 100 = 31.7%1
Zone R: (60 – 18) ÷ 60 × 1001
= 70.0%1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (c)Cadangkan dua sebab mengapa zon R mempunyai bilangan spesies yang paling rendah.Suggest two reasons why zone R has the lowest number of species.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Waste and uneaten feed from the fish cages causes water pollution (eutrophication) that lowers the dissolved oxygen and kills corals and fish1
Antibiotics or chemicals used in the cages harm the organisms1
Boats and anchors damage the corals, and fishing reduces the fish1

Tick each point you got.Tandakan setiap isi yang betul.

[3 marksmarkah] (d)Pasangkan setiap tindakan dengan jenis usaha yang betul: mewartakan zon P sebagai taman laut, menanam semula karang yang rosak dan membiakkan ikan karang dalam akuarium.Match each action with the correct type of effort: gazetting zone P as a marine park, replanting the damaged corals and breeding reef fish in an aquarium.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Gazetting zone P as a marine park: preservation or in situ conservation1
Replanting damaged corals: restoration1
Breeding reef fish in an aquarium: ex situ conservation1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (e)Nilaikan dua pilihan kerajaan negeri itu dan beri keputusan anda dengan sebab-sebab yang kukuh.Evaluate the two options of the state government and give your decision with sound reasons.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Gazetting the whole island: advantage is that biodiversity (about 60 species per sample area as in zone P) is kept and the reef is a source of food, medicine and tourism for future generations1
Gazetting the whole island: disadvantage is that fishermen and the fish cage operators lose their income, so some compensation or an alternative job is needed1
More fish cages: advantage is more income and food, but the pollution and loss of species in zone R (a 70.0% decrease) will spread1
Decision: gazette the whole island (the reef is lost for good if destroyed), and move the cages to a place far from the reef so that development stays sustainable1

Tick each point you got.Tandakan setiap isi yang betul.

[4 marksmarkah] (f)Terangkan dua kepentingan biodiversiti terumbu karang kepada manusia dan satu amalan lestari yang boleh dilakukan oleh pelancong.Explain two importances of the biodiversity of the coral reef to humans and one sustainable practice that tourists can follow.

Show answerTunjuk jawapanHide answerSembunyi jawapan

Source of food: fish and seafood for the local people and for income1
Source of medicine or recreation, education and research: tourism and ecotourism give income and the reef is used for scientific research1
Keeps nature in balance: the reef is a breeding and feeding place for many organisms and protects the coast from waves1
Practice: do not touch or collect corals, do not throw rubbish, use environment-friendly boats or follow the 5R (any one)1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 10, standard 10.2

TextbookBuku teks p.ms. 211

Test specification (JSU)Jadual Spesifikasi Ujian (JSU)

ChapterBab	RememberingMengingat	UnderstandingMemahami	ApplyingMengaplikasi	AnalysingMenganalisis	EvaluatingMenilai	CreatingMencipta	Items	marksmarkah
1. Organisation of Plant Tissues and Growth	1 (1)	–	–	–	–	–	1	1
2. Leaf Structure and Function	–	–	1 (1)	1 (1)	–	–	2	2
3. Nutrition in Plants	1 (1)	1 (20)	1 (1)	–	–	–	3	22
4. Transport in Plants	–	1 (1)	1 (1)	–	1 (20)	–	3	22
5. Response in Plants	1 (1)	–	–	2 (11)	–	–	3	12
6. Sexual Reproduction in Flowering Plants	–	1 (10)	1 (1)	–	–	–	2	11
7. Adaptations of Plants in Different Habitats	1 (1)	–	–	–	–	–	1	1
8. Biodiversity	–	–	1 (1)	–	–	–	1	1
9. Ecosystem	1 (1)	–	1 (1)	–	–	–	2	2
10. Environmental Sustainability	–	1 (1)	2 (11)	–	1 (20)	–	4	32
11. Inheritance	1 (20)	–	1 (1)	–	–	–	2	21
12. Variation	–	1 (10)	–	2 (2)	–	–	3	12
13. Genetic Technology	–	1 (1)	–	–	–	–	1	1
TotalJumlah	6 (25)	6 (43)	9 (18)	5 (14)	2 (40)	0 (0)	28	140

Each cell shows items (marks).Setiap petak menunjukkan item (markah).

BiologiBiology

Bahagian ASection A

QuestionSoalan 1

QuestionSoalan 2

QuestionSoalan 3

QuestionSoalan 4

QuestionSoalan 5

QuestionSoalan 6

QuestionSoalan 7

QuestionSoalan 8

QuestionSoalan 9

QuestionSoalan 10

QuestionSoalan 11

QuestionSoalan 12

QuestionSoalan 13

QuestionSoalan 14

QuestionSoalan 15

QuestionSoalan 16

QuestionSoalan 17

QuestionSoalan 18

QuestionSoalan 19

QuestionSoalan 20

Bahagian BSection B

QuestionSoalan 21

QuestionSoalan 22

QuestionSoalan 23

QuestionSoalan 24

Bahagian CSection C

QuestionSoalan 25

QuestionSoalan 26

Bahagian DSection D

QuestionSoalan 27

QuestionSoalan 28