BiologiBiology

Print questions onlyCetak soalan sahaja Print with answersCetak dengan skema Print answer key onlyCetak kunci jawapan sahaja 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.

1

QuestionSoalan 1

Sel kolenkima terdapat di bawah epidermis batang muda dan tangkai daun. Apakah fungsi utama sel-sel ini?Collenchyma cells are found beneath the epidermis of young stems and leaf stalks. What is the main function of these cells?

• AMengangkut air dan garam mineral ke daunTo transport water and mineral salts to the leaves
• BMenyokong bahagian muda sambil membenarkannya melenturTo support young parts while allowing them to bend
• CMemberi kekuatan kepada pokok kerana sel-selnya mati dan berdinding tebalTo give the plant strength because its cells are dead with thick walls
• DMenyimpan kanji dan menjalankan fotosintesisTo store starch and carry out photosynthesis

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B

Collenchyma cells are living and have thickened but flexible walls of pectin and hemicellulose, so they support growing stems and stalks while letting them bend. A is xylem, C is sclerenchyma and D is parenchyma.

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

TextbookBuku teks p.ms. 5

2

QuestionSoalan 2

Biji benih kacang hijau dicambahkan dalam gelap sepenuhnya selama tujuh hari. Jisim kering anak benih diukur setiap hari. Apakah yang akan diperhatikan?Mung bean seeds are germinated in complete darkness for seven days. The dry mass of the seedlings is measured each day. What will be observed?

• AJisim kering bertambah dengan cepat kerana daun menjalankan fotosintesisThe dry mass increases quickly because the leaves photosynthesise
• BJisim kering tidak berubah kerana anak benih tidak mendapat apa-apa daripada persekitaranThe dry mass stays constant because the seedling gets nothing from its surroundings
• CJisim kering bertambah perlahan kerana anak benih menyerap airThe dry mass increases slowly because the seedling absorbs water
• DJisim kering berkurang kerana makanan simpanan digunakan dalam respirasi dan tiada fotosintesisThe dry mass decreases because the stored food is used in respiration and there is no photosynthesis

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D

In the dark the seedling cannot photosynthesise, so it lives on the food stored in the cotyledons, which respiration uses up. The dry mass falls (stage A of the growth curve). Water raises the fresh mass, not the dry mass, so C is wrong.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 1, standard 1.3

TextbookBuku teks p.ms. 19

3

QuestionSoalan 3

Jadual 1 menunjukkan kadar fotosintesis daun Hydrilla pada suhu 28 °C dalam bilik yang sama.

Jadual 1

Keamatan cahaya (klux)	Kepekatan CO2 (%)	Gelembung oksigen seminit
8	0.04	20
12	0.04	20
12	0.10	36

Apakah faktor penghad kadar fotosintesis pada 8 klux dan 0.04% CO₂?Table 1 shows the rate of photosynthesis of a Hydrilla leaf at 28 °C in the same room.

Table 1

Light intensity (klux)	CO2 concentration (%)	Oxygen bubbles per minute
8	0.04	20
12	0.04	20
12	0.10	36

What is the limiting factor of the rate of photosynthesis at 8 klux and 0.04% CO₂?

• AKepekatan karbon dioksidaCarbon dioxide concentration
• BKeamatan cahayaLight intensity
• CSuhuTemperature
• DJumlah klorofilAmount of chlorophyll

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A

Raising the light from 8 to 12 klux did not change the rate (20), so light is not limiting. Raising CO2 from 0.04% to 0.10% raised it to 36, so CO2 concentration is the limiting factor.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.4

TextbookBuku teks p.ms. 44

4

QuestionSoalan 4

Mineral manakah menjadi komponen sesetengah asid amino dan vitamin B dalam tumbuhan?Which mineral is a component of some amino acids and vitamin B in plants?

• AKaliumPotassium
• BFerumIron
• CSulfurSulphur
• DFosforusPhosphorus

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C

Sulphur is part of some amino acids, vitamin B and some coenzymes; its deficiency yellows the leaves or the whole plant. Potassium keeps cells turgid, iron helps to make chlorophyll and phosphorus makes ATP and nucleic acids.

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

TextbookBuku teks p.ms. 61

5

QuestionSoalan 5

Dodder (Cuscuta) mempunyai batang kuning yang nipis, tiada daun dan membelit pada pokok lain. Bagaimanakah tumbuhan ini mendapat makanan?Dodder (Cuscuta) has thin yellow stems with no leaves and twines around another plant. How does this plant obtain its food?

• AMenjalankan fotosintesis dan hanya menumpang untuk mendapat cahayaIt photosynthesises and only uses the host to reach light
• BAkarnya menembusi berkas vaskular perumah untuk menyerap bahan organik, air dan mineralIts roots penetrate the vascular bundles of the host to absorb organic substances, water and minerals
• CMenangkap dan mencerna serangga dengan enzim untuk mendapat nitrogenIt traps and digests insects with enzymes to obtain nitrogen
• DMenyerap bahan organik daripada bahan reput dalam tanahIt absorbs organic matter from decaying material in the soil

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B

Dodder lacks chlorophyll and is a parasite: it takes organic substances, water and minerals from the vascular bundles of its host, which becomes malnourished. A describes an epiphyte, C a carnivorous plant and D a saprophyte.

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

TextbookBuku teks p.ms. 70

6

QuestionSoalan 6

Sebahagian kulit batang pokok, termasuk floem, dibuang dalam bentuk cincin. Selepas beberapa minggu, bahagian batang di atas cincin itu membengkak. Apakah sebabnya?A ring of bark, including the phloem, is removed from a tree stem. After a few weeks the part of the stem above the ring swells. What is the reason?

• AAir tertumpu di atas cincin kerana xilem tersumbatWater collects above the ring because the xylem is blocked
• BGaram mineral tidak dapat naik melepasi cincinMineral salts cannot move up past the ring
• CSel kambium di bawah cincin berhenti membahagiThe cambium cells below the ring stop dividing
• DSukrosa daripada daun tidak dapat bergerak ke bawah melepasi cincin dan terkumpul di atasnyaSucrose from the leaves cannot move down past the ring and collects above it

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D

The phloem translocates sucrose from the leaves downwards. Where the phloem is removed, sucrose is stopped and collects above the ring, so the stem swells. The xylem is not cut, so water still moves up.

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

TextbookBuku teks p.ms. 85

7

QuestionSoalan 7

Sebatang pokok mengalami kemarau dan stomata daunnya tertutup. Fitohormon manakah yang meningkat dalam daun untuk menyebabkan keadaan ini?A plant suffers a drought and the stomata of its leaves close. Which phytohormone increases in the leaves to cause this?

• AAsid absisikAbscisic acid
• BGiberelinGibberellin
• CSitokininCytokinin
• DEtilenaEthylene

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A

Abscisic acid inhibits growth and closes the stomata in drought, so less water is lost. Gibberellin and cytokinin promote growth and cell division, and ethylene ripens fruits.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.2

TextbookBuku teks p.ms. 103

8

QuestionSoalan 8

Sebuah ovari mengandungi 25 ovul. Setiap ovul mempunyai satu sel induk megaspora yang membahagi secara meiosis dan satu megaspora yang hidup berkembang menjadi pundi embrio. Berapakah jumlah nukleus dalam semua pundi embrio yang matang?An ovary contains 25 ovules. Each ovule has one megaspore mother cell that divides by meiosis, and the one surviving megaspore develops into an embryo sac. How many nuclei are there in all the mature embryo sacs?

• A25
• B100
• C200
• D400

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C

Only one of the four megaspores survives. Its nucleus divides by mitosis three times to give 8 nuclei per embryo sac. 25 × 8 = 200. B counts 4 nuclei per sac and D counts 16.

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

TextbookBuku teks p.ms. 117

9

QuestionSoalan 9

Bahagian ovul manakah yang berkembang menjadi kulit biji (testa)?Which part of the ovule develops into the seed coat (testa)?

• ANuselusNucellus
• BIntegumenIntegument
• CFunikelFunicle
• DMikropilMicropyle

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B

The integument that covers the nucellus becomes the seed coat. The ovule becomes the seed, the ovary becomes the fruit and the ovary wall becomes the pericarp.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 6, standard 6.4

TextbookBuku teks p.ms. 125

10

QuestionSoalan 10

Elodea ialah tumbuhan akuatik yang tenggelam sepenuhnya, tanpa stomata dan kutikel berlilin. Bagaimanakah tumbuhan ini bertukar gas dengan air?Elodea is a fully submerged aquatic plant with no stomata and no waxy cuticle. How does it exchange gases with the water?

• AGas meresap melalui kutikel yang tebal pada daunGases diffuse through the thick cuticle of the leaves
• BGas masuk melalui stomata pada epidermis atasGases enter through stomata on the upper epidermis
• CGas diserap oleh akar daripada lumpurGases are absorbed by the roots from the mud
• DGas meresap terus melalui epidermis daun yang nipis dan kecil dengan nisbah luas permukaan kepada isi padu yang besarGases diffuse directly through the epidermis of the thin, small leaves, which have a large surface area to volume ratio

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D

A submerged plant has thin, small leaves with a large surface area for their volume and no cuticle, so water, mineral salts and gases diffuse directly through the epidermis. Only floating plants have stomata, on the upper surface.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.1.2

TextbookBuku teks p.ms. 137

11

QuestionSoalan 11

Daun tua pokok bakau gugur secara berkala. Apakah kebaikan keadaan ini kepada pokok itu?The old leaves of a mangrove tree fall off from time to time. How does this benefit the tree?

• ADaun tua menyimpan garam berlebihan, jadi pengguguran membuang garam daripada pokokThe old leaves store excess salt, so shedding them removes salt from the tree
• BDaun tua menyimpan air untuk digunakan pada musim kemarauThe old leaves store water for use in the dry season
• CPengguguran membekalkan oksigen kepada akar dalam lumpurShedding supplies oxygen to the roots in the mud
• DPengguguran menambah bilangan stoma pada pokok ituShedding increases the number of stomata on the tree

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A

Besides excreting salt through hydathodes, a mangrove stores excess salt in its old leaves, which then fall off, removing the salt. The pneumatophores, not the leaves, supply oxygen to the roots.

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

TextbookBuku teks p.ms. 135

12

QuestionSoalan 12

Nama saintifik bagi pokok mangga ialah genus Mangifera dan spesies indica. Manakah cara penulisan yang betul menurut sistem tatanama binomial?The scientific name of the mango tree is genus Mangifera and species indica. Which is the correct way to write it in the binomial nomenclature system?

• AMangifera Indica
• Bmangifera indica
• CMangifera indica
• DMANGIFERA INDICA

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C

The scientific name has two words in italics: the genus starts with a capital letter and the species starts with a small letter.

For the teacherUntuk guru: RememberingMengingat, mediumsederhana difficultyaras mediumsederhana, ChapterBab 8, standard 8.1

TextbookBuku teks p.ms. 150

13

QuestionSoalan 13

Daun-daun mati reput di atas tanah. Pokok berdekatan kemudian menyerap nitrat daripada tanah. Antara urutan berikut, yang manakah menunjukkan mikroorganisma yang terlibat mengikut tertib yang betul?Dead leaves decay on the ground. Nearby plants later absorb nitrate from the soil. Which sequence shows the microorganisms involved in the correct order?

• ARhizobium → Nitrobacter → Nitrosomonas
• BPengurai → Nitrobacter → NitrosomonasDecomposers → Nitrobacter → Nitrosomonas
• CNitrosomonas → bakteria denitrifikasi → NitrobacterNitrosomonas → denitrifying bacteria → Nitrobacter
• DPengurai → Nitrosomonas → NitrobacterDecomposers → Nitrosomonas → Nitrobacter

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D

Decomposers change the protein in dead leaves to ammonium ions (ammonification). Nitrosomonas changes ammonium to nitrite, then Nitrobacter changes nitrite to nitrate (nitrification). Denitrifying bacteria remove nitrate, and Rhizobium fixes nitrogen gas.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 8, standard 8.3

TextbookBuku teks p.ms. 159

14

QuestionSoalan 14

Sebuah spesies rumput ditemui dalam 12 daripada 20 petak kuadrat berukuran 0.5 m × 0.5 m. Jumlah rumput yang dikira ialah 30. Apakah kekerapan dan ketumpatan populasi rumput itu?A grass species is found in 12 of 20 quadrats measuring 0.5 m × 0.5 m. The total number of grass plants counted is 30. What are the frequency and the population density of the grass?

• A60% dan 1.5 per m²60% and 1.5 per m²
• B60% dan 6.0 per m²60% and 6.0 per m²
• C40% dan 6.0 per m²40% and 6.0 per m²
• D60% dan 3.0 per m²60% and 3.0 per m²

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B

Frequency = 12 ÷ 20 × 100 = 60%. Area sampled = 20 × 0.25 = 5 m2, so density = 30 ÷ 5 = 6.0 per m2. A divides by the 20 quadrats without their area; C uses the wrong frequency; D takes the area of one quadrat as 0.5 m2 instead of 0.25 m2.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.2

TextbookBuku teks p.ms. 193

15

QuestionSoalan 15

Tiga sampel air parit X, Y dan Z diuji dengan metilena biru. Warna biru hilang dalam 1.5 jam bagi X, 3 jam bagi Y dan 6 jam bagi Z. Manakah menunjukkan sampel daripada yang paling kurang tercemar kepada yang paling tercemar?Three drain water samples X, Y and Z are tested with methylene blue. The blue colour disappears in 1.5 hours for X, 3 hours for Y and 6 hours for Z. Which shows the samples from the least polluted to the most polluted?

• AZ, Y, X
• BX, Y, Z
• CY, X, Z
• DZ, X, Y

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A

The faster the blue colour disappears, the more microorganisms there are, so the higher the BOD and the more polluted the water. Rates (1 ÷ time): X 0.67 h-1, Y 0.33 h-1, Z 0.17 h-1, so Z is the least polluted and X the most.

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

TextbookBuku teks p.ms. 207

16

QuestionSoalan 16

Sebatang pokok mempunyai genotip TtRr. Antara berikut, yang manakah menunjukkan semua jenis gamet yang boleh dihasilkan oleh pokok ini?A plant has the genotype TtRr. Which of the following shows all the types of gametes that this plant can produce?

• ATT, Tt, tt, RR, Rr, rr
• BTtRr sahajaTtRr only
• CTR, Tr, tR, tr
• DTR dan tr sahajaTR and tr only

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C

By independent assortment each allele of one pair can combine with either allele of the other pair, so four gametes form in equal numbers: TR, Tr, tR and tr. A gamete has only one allele of each gene.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 11, standard 11.2

TextbookBuku teks p.ms. 238

17

QuestionSoalan 17

Albinisme dikawal oleh alel resesif (a). Sepasang suami isteri normal mempunyai seorang anak albino. Apakah kebarangkalian anak mereka yang seterusnya ialah anak perempuan yang albino?Albinism is controlled by a recessive allele (a). A normal couple has an albino child. What is the probability that their next child is an albino girl?

• A1/4
• B1/2
• C3/8
• D1/8

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D

Both parents must be Aa. P(albino) = 1/4 and P(girl) = 1/2. The two events are independent, so P(albino girl) = 1/4 × 1/2 = 1/8.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 11, standard 11.3

TextbookBuku teks p.ms. 243

18

QuestionSoalan 18

Antara berikut, yang manakah ialah variasi yang tidak diwarisi oleh anak?Which of the following is a variation that is not inherited by the children?

• AOtot lengan yang membesar akibat latihan angkat beratArm muscles that enlarge because of weight-lifting training
• BAlbinismeAlbinism
• CKumpulan darahBlood group
• DCuping telinga bebasFree earlobes

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A

Bigger muscles from training are an environmental variation that changes the phenotype but not the genes in the gametes, so it cannot be inherited. Albinism, blood group and earlobe type are controlled by genes.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 12, standard 12.1

TextbookBuku teks p.ms. 261

19

QuestionSoalan 19

Rajah 1 menunjukkan profil DNA seorang bapa, seorang ibu dan tiga orang anak. Salah seorang kanak-kanak itu bukan anak kandung pasangan ini. Kanak-kanak yang manakah?Diagram 1 shows the DNA profiles of a father, a mother and three children. One of the children is not the biological child of this couple. Which child is it?

• AAnak 1Child 1
• BAnak 2Child 2
• CAnak 3Child 3
• DKetiga-tiga anak adalah anak kandung merekaAll three are their biological children

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B

Every band of a child must come from the mother or the father. Child 2 has a band at row 4, which neither parent has. Children 1 and 3 have bands matching one or both parents.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 13, standard 13.2

TextbookBuku teks p.ms. 287

20

QuestionSoalan 20

Dalam terapi gen, gen normal dimasukkan ke dalam sel pesakit untuk menggantikan gen yang tidak berfungsi. Apakah yang membawa gen normal itu ke dalam sel?In gene therapy a normal gene is put into a patient's cells to replace the faulty gene. What carries the normal gene into the cells?

• AEnzim penyekatA restriction enzyme
• BElektroforesis gelGel electrophoresis
• CVirusA virus
• DProb radioaktifA radioactive probe

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C

In gene therapy a virus is used as the vector that carries the normal gene into the cell, which then works normally. A restriction enzyme cuts DNA, electrophoresis separates fragments and a probe marks bands in DNA profiling.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 13, standard 13.2

TextbookBuku teks p.ms. 285

21

QuestionSoalan 21

Rajah 2 menunjukkan keratan rentas batang sebatang pokok getah yang berumur lima tahun. Jadual 2 menunjukkan lilitan batang pokok itu pada ketinggian 1.5 m selama lima tahun. Tinggi pokok bertambah daripada 3.0 m kepada 9.5 m dalam tempoh yang sama.

Jadual 2

Tahun	0	1	2	3	4	5
Lilitan batang (cm)	20	31	42	54	64	75

Diagram 2 shows the cross-section of the stem of a five-year-old rubber tree. Table 2 shows the girth of the stem at a height of 1.5 m over five years. The height of the tree increased from 3.0 m to 9.5 m in the same period.

Table 2

Year	0	1	2	3	4	5
Girth of stem (cm)	20	31	42	54	64	75

[1 markmarkah] (a)Pilih nama tisu R.Choose the name of tissue R.

AKambium vaskularVascular cambiumBKambium gabusCork cambiumCMeristem apeksApical meristem

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Vascular cambium

R is the ring between the secondary phloem (Q) and the secondary xylem (S). That ring is the vascular cambium.

[2 marksmarkah] (b)Nyatakan dua fungsi lapisan P.State two functions of layer P.

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• Protects the stem from insects and pathogens1
• Reduces water loss / protects against injury1

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

[1 markmarkah] (c)Pilih pertumbuhan yang menyebabkan pertambahan lilitan batang.Choose the growth that causes the increase in the girth of the stem.

APertumbuhan primerPrimary growthBPertumbuhan sekunderSecondary growth

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Secondary growth

An increase in the thickness (girth) of a stem is secondary growth. Primary growth lengthens the stem.

[2 marksmarkah] (d)Hitung purata pertambahan lilitan batang setiap tahun dan peratus pertambahan lilitan dari tahun 0 hingga tahun 5.Calculate the mean increase in girth per year and the percentage increase in girth from year 0 to year 5.

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• (75 – 20) ÷ 5 = 11 cm per year1
• (55 ÷ 20) × 100 = 275%1

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

[2 marksmarkah] (e)Terangkan bagaimana tisu R menyebabkan lilitan batang bertambah.Explain how tissue R causes the girth of the stem to increase.

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• The vascular cambium divides actively (by mitosis) to form secondary xylem inwards and secondary phloem outwards1
• The new tissues add to the thickness of the stem, so the stem widens1

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

[2 marksmarkah] (f)Nyatakan jenis meristem yang menyebabkan pertambahan tinggi pokok itu dan nyatakan kedudukannya.State the type of meristem that causes the increase in the height of the tree and state its position.

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• Apical meristem1
• At the tip of the shoot (and of the root)1

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

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2

TextbookBuku teks p.ms. 11

22

QuestionSoalan 22

Rajah 3 menunjukkan keratan rentas daun dua jenis tumbuhan. Daun P terapung di atas permukaan kolam manakala daun Q tumbuh di padang pasir yang kering.Diagram 3 shows cross-sections of the leaves of two types of plants. Leaf P floats on the surface of a pond while leaf Q grows in a dry desert.

[1 markmarkah] (a)Pilih jenis tumbuhan P berdasarkan habitatnya.Choose the type of plant P according to its habitat.

AHidrofitHydrophyteBXerofitXerophyteCHalofitHalophyte

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Hydrophyte

P floats on a pond, an aquatic habitat, so it is a hydrophyte.

[1 markmarkah] (b)Pilih jenis tumbuhan Q berdasarkan habitatnya.Choose the type of plant Q according to its habitat.

AHidrofitHydrophyteBXerofitXerophyteCHalofitHalophyte

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Xerophyte

Q grows in the desert, a hot, dry habitat, so it is a xerophyte.

[2 marksmarkah] (c)Namakan tisu pada ruang 1 dan nyatakan fungsinya pada tumbuhan P.Name the tissue at space 1 and state its function in plant P.

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• Aerenchyma (a tissue with large air spaces)1
• Gives buoyancy so the leaf floats / stores air for gaseous exchange1

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

[2 marksmarkah] (d)Terangkan mengapa stoma pada daun P terdapat terutamanya pada epidermis atas.Explain why the stomata of leaf P are mostly on the upper epidermis.

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• The lower surface is in contact with water, so gases cannot pass through it1
• The upper surface is exposed to the air, so the stomata stay open for gaseous exchange1

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

[2 marksmarkah] (e)Terangkan bagaimana struktur 3 dan struktur 4 pada daun Q membantu tumbuhan itu hidup di padang pasir.Explain how structure 3 and structure 4 of leaf Q help the plant to live in the desert.

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• Structure 3 is a thick waxy cuticle that reduces evaporation of water from the leaf1
• Structure 4 is a sunken stoma, in which humid air collects, which reduces the rate of transpiration1

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

[2 marksmarkah] (f)Nyatakan satu ciri akar tumbuhan Q dan terangkan kelebihannya.State one feature of the roots of plant Q and explain its advantage.

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• The roots spread widely (or go deep)1
• To absorb as much water and mineral salts as possible from the dry soil (or to reach the water deep underground)1

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

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.1.2

TextbookBuku teks p.ms. 137

23

QuestionSoalan 23

Sebuah syarikat bioteknologi menghasilkan kacang soya yang tahan racun rumpai. Gen rintangan daripada sejenis bakteria tanah dipotong dengan enzim P. Plasmid bakteria dipotong dengan enzim P yang sama, kemudian gen itu disambungkan ke dalam plasmid dengan enzim Q. Plasmid rekombinan itu dimasukkan ke dalam sel kacang soya. Dalam ujian ladang, anak benih kacang soya transgenik dan anak benih biasa disembur racun rumpai. Jadual 3 menunjukkan keputusan.

Jadual 3

Kumpulan	Anak benih yang disembur	Anak benih yang hidup
Kacang soya transgenik	400	368
Kacang soya biasa	400	28

A biotechnology company produces soya beans that resist a weedkiller. A resistance gene from a soil bacterium is cut out with enzyme P. A bacterial plasmid is cut with the same enzyme P, then the gene is joined into the plasmid with enzyme Q. The recombinant plasmid is put into soya bean cells. In a field trial, transgenic and ordinary soya bean seedlings are sprayed with the weedkiller. Table 3 shows the results.

Table 3

Group	Seedlings sprayed	Seedlings that survived
Transgenic soya bean	400	368
Ordinary soya bean	400	28

[1 markmarkah] (a)Pilih istilah bagi kacang soya yang mengandungi gen daripada organisma lain.Choose the term for a soya bean that contains a gene from another organism.

AKlonCloneBOrganisma transgenikTransgenic organismCBaka tulenPurebreed

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Transgenic organism

An organism that carries recombinant DNA, with a gene from another species, is a transgenic organism (GMO).

[2 marksmarkah] (b)Namakan enzim P dan enzim Q.Name enzyme P and enzyme Q.

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• P: restriction enzyme1
• Q: DNA ligase1

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

[1 markmarkah] (c)Mengapakah gen dan plasmid dipotong dengan enzim P yang sama?Why are the gene and the plasmid cut with the same enzyme P?

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• So that the cut ends match / are complementary and the gene can be joined into the plasmid1

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[1 markmarkah] (d)Apakah peranan plasmid dalam proses ini?What is the role of the plasmid in this process?

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• It acts as a vector (cloning vector) that carries the gene into the cell1

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

[3 marksmarkah] (e)Hitung peratus kemandirian anak benih transgenik dan anak benih biasa, kemudian buat satu kesimpulan.Calculate the percentage survival of the transgenic seedlings and the ordinary seedlings, then make one conclusion.

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• Transgenic: 368 ÷ 400 × 100 = 92%1
• Ordinary: 28 ÷ 400 × 100 = 7%1
• The gene gives resistance to the weedkiller, so more transgenic seedlings survive1

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

[2 marksmarkah] (f)Nyatakan satu kelebihan dan satu keburukan penanaman kacang soya ini.State one advantage and one disadvantage of growing this soya bean.

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• Advantage: weeds can be killed without harming the crop, so yield is higher / less labour and cost1
• Disadvantage: the gene may spread to weeds or natural species (or its long-term effect on human health is unknown)1

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

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 13, standard 13.1

TextbookBuku teks p.ms. 281

24

QuestionSoalan 24

Pada jagung manis, biji ungu (P) dominan terhadap biji kuning (p), dan biji licin (S) dominan terhadap biji berkedut (s). Seorang pembiak menyilangkan jagung baka tulen berbiji ungu dan licin dengan jagung baka tulen berbiji kuning dan berkedut. Pokok F₁ disilang sesama sendiri dan menghasilkan 320 biji F₂.In sweet corn, purple kernels (P) are dominant to yellow kernels (p), and smooth kernels (S) are dominant to wrinkled kernels (s). A breeder crosses a purebreed corn with purple, smooth kernels with a purebreed corn with yellow, wrinkled kernels. The F₁ plants are self-crossed and produce 320 F₂ kernels.

[1 markmarkah] (a)Pilih fenotip biji F₁.Choose the phenotype of the F₁ kernels.

AUngu dan licinPurple and smoothBUngu dan berkedutPurple and wrinkledCKuning dan licinYellow and smoothDKuning dan berkedutYellow and wrinkled

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Purple and smooth

The F1 genotype is PpSs and both dominant alleles are present, so the kernels are purple and smooth.

[2 marksmarkah] (b)Nyatakan genotip F₁ dan semua jenis gamet yang dihasilkan oleh pokok F₁.State the genotype of the F₁ and all the types of gametes produced by an F₁ plant.

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• F1 genotype: PpSs1
• Gametes: PS, Ps, pS and ps1

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

[3 marksmarkah] (c)Dengan menggunakan petak Punnett, tunjukkan genotip dan nisbah fenotip F₂.Using a Punnett square, show the genotypes and the phenotypic ratio of the F₂.

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• Gametes of each parent correctly placed: PS, Ps, pS, ps along both sides1
• 16 boxes with the correct genotypes (for example PpSs where PS meets ps)1
• Phenotypic ratio 9 purple smooth : 3 purple wrinkled : 3 yellow smooth : 1 yellow wrinkled1

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

[2 marksmarkah] (d)Hitung bilangan biji F₂ yang dijangka berwarna ungu dan berkedut.Calculate the number of F₂ kernels that are expected to be purple and wrinkled.

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• 3/16 × 3201
• = 60 kernels1

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

[1 markmarkah] (e)Nyatakan Hukum Mendel yang menerangkan mengapa biji ungu berkedut dan biji kuning licin muncul dalam F₂.State the Mendelian law that explains why purple wrinkled and yellow smooth kernels appear in the F₂.

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• Mendel's Second Law (Law of Independent Assortment): each allele of one pair can combine randomly with any allele of the other pair during gamete formation1

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

[1 markmarkah] (f)Pokok F₁ disilang dengan pokok berbiji kuning dan berkedut (ppss). Nyatakan nisbah fenotip yang dijangka.An F₁ plant is crossed with a plant with yellow, wrinkled kernels (ppss). State the expected phenotypic ratio.

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• 1 purple smooth : 1 purple wrinkled : 1 yellow smooth : 1 yellow wrinkled1

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

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 11, standard 11.2

TextbookBuku teks p.ms. 238

25Choose 1Pilih 1

QuestionSoalan 25

Rajah 4 menunjukkan keratan rentas sebahagian daun sebuah tumbuhan eudikot. Seorang murid mengkaji kesan keamatan cahaya terhadap fotosintesis Hydrilla dengan mengira gelembung oksigen yang dibebaskan dalam 5 minit. Jadual 4 menunjukkan keputusannya.

Jadual 4

Keamatan cahaya (klux)	0	2	4	6	8	10
Bilangan gelembung dalam 5 minit	0	8	17	26	30	30

Diagram 4 shows the cross-section of part of the leaf of a eudicot plant. A pupil studies the effect of light intensity on the photosynthesis of Hydrilla by counting the oxygen bubbles released in 5 minutes. Table 4 shows the results.

Table 4

Light intensity (klux)	0	2	4	6	8	10
Number of bubbles in 5 minutes	0	8	17	26	30	30

[4 marksmarkah] (a)Namakan struktur P, Q, R dan U.Name structures P, Q, R and U.

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• P: upper epidermis1
• Q: palisade mesophyll1
• R: spongy mesophyll1
• U: guard cells1

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

[3 marksmarkah] (b)Terangkan bagaimana struktur Q sesuai untuk fotosintesis.Explain how structure Q is adapted for photosynthesis.

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• The cells are tall and packed closely together in a vertical arrangement just below the upper epidermis1
• They contain many chloroplasts with chlorophyll1
• So they absorb the maximum amount of light for photosynthesis1

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

[4 marksmarkah] (c)Terangkan bagaimana sel pengawal membuka stoma pada waktu siang.Explain how the guard cells open the stoma in the daytime.

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• Potassium ions enter the guard cells and sucrose is made by photosynthesis, so the solute concentration rises1
• The water potential in the guard cells falls1
• Water enters the guard cells from the epidermal cells by osmosis1
• The guard cells become turgid; the thin outer wall stretches more than the thick inner wall, so the cells curve and the stoma opens1

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

[2 marksmarkah] (d)Nyatakan pemboleh ubah dimanipulasikan dan pemboleh ubah bergerak balas dalam eksperimen ini.State the manipulated variable and the responding variable in this experiment.

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• Manipulated variable: light intensity1
• Responding variable: number of oxygen bubbles released in 5 minutes (rate of photosynthesis)1

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

[2 marksmarkah] (e)Hitung bilangan gelembung oksigen seminit pada 6 klux.Calculate the number of oxygen bubbles per minute at 6 klux.

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• 26 ÷ 51
• = 5.2 bubbles per minute1

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

[5 marksmarkah] (f)Huraikan dua peringkat fotosintesis dengan menyatakan tapak dan hasil bagi setiap peringkat.Describe the two stages of photosynthesis, stating the site and the products of each stage.

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• Light-dependent reaction takes place in the thylakoids (grana)1
• Chlorophyll absorbs light energy; photolysis splits water into H+ and OH–1
• Products: ATP, NADPH and oxygen1
• Light-independent reaction takes place in the stroma1
• Carbon dioxide is fixed and reduced by ATP and NADPH to glucose, which joins to form starch1

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

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.4

TextbookBuku teks p.ms. 42

26Choose 1Pilih 1

QuestionSoalan 26

Jadual 5 menunjukkan kunci dikotomi untuk mengenal pasti lima jenis tumbuhan. Jadual 6 menunjukkan ciri lima spesimen tumbuhan P, Q, R, S dan T.

Jadual 5

Pasangan	Pernyataan	Pergi ke / jawapan
1a	Mempunyai tisu vaskular	2
1b	Tiada tisu vaskular	Lumut
2a	Membiak dengan spora; tiada biji	Paku-pakis
2b	Membiak dengan biji	3
3a	Biji pada kon, tidak terbungkus dalam buah	Pokok pine
3b	Biji terbungkus dalam buah	4
4a	Urat daun selari	Padi
4b	Urat daun berjaring	Bunga raya

Jadual 6

Spesimen	Ciri
P	Mempunyai xilem dan floem; daun berbentuk fron dengan spora di bawahnya; tiada biji
Q	Tiada xilem dan floem; hanya mempunyai rizoid; tumbuh pada dinding lembap
R	Daun berbentuk jarum; biji pada kon; tiada buah
S	Berbunga dan berbuah; urat daun selari
T	Berbunga dan berbuah; urat daun berjaring

Table 5 shows a dichotomous key to identify five types of plants. Table 6 shows the features of five plant specimens P, Q, R, S and T.

Table 5

Couplet	Statement	Go to / answer
1a	Has vascular tissue	2
1b	No vascular tissue	Moss
2a	Reproduces by spores; no seeds	Fern
2b	Reproduces by seeds	3
3a	Seeds on cones, not enclosed in a fruit	Pine tree
3b	Seeds enclosed in a fruit	4
4a	Parallel leaf veins	Paddy
4b	Net-like leaf veins	Hibiscus

Table 6

Specimen	Features
P	Has xylem and phloem; leaves are fronds with spores underneath; no seeds
Q	No xylem and phloem; has only rhizoids; grows on a damp wall
R	Needle-like leaves; seeds on cones; no fruit
S	Has flowers and fruits; parallel leaf veins
T	Has flowers and fruits; net-like leaf veins

[2 marksmarkah] (a)Padi dikenali dengan nama saintifik Oryza sativa. Nyatakan dua peraturan sistem tatanama binomial berdasarkan nama ini.Paddy has the scientific name Oryza sativa. State two rules of the binomial nomenclature system, based on this name.

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Any 2 pointsMana-mana 2 isi

• The name has two words: the genus first, then the species1
• The genus starts with a capital letter and the species with a small letter1
• The name is printed in italics (when handwritten, each word is underlined separately)1

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

[5 marksmarkah] (b)Dengan menggunakan Jadual 5 dan Jadual 6, kenal pasti spesimen P, Q, R, S dan T.Using Table 5 and Table 6, identify specimens P, Q, R, S and T.

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• P: fern1
• Q: moss1
• R: pine tree1
• S: paddy1
• T: hibiscus1

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

[3 marksmarkah] (c)Padi liar, Oryza rufipogon, dan jagung, Zea mays, dibandingkan dengan padi, Oryza sativa. Namakan taksa yang sama bagi kedua-dua Oryza, nyatakan yang mana lebih rapat hubungannya dengan padi dan beri sebabnya.Wild rice, Oryza rufipogon, and maize, Zea mays, are compared with paddy, Oryza sativa. Name the taxon that the two Oryza share, state which is more closely related to paddy and give the reason.

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• Genus (Oryza)1
• Oryza rufipogon is more closely related to paddy1
• They are in the same genus, so they share more characteristics and are more closely related genetically than maize, which is in a different genus1

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

[4 marksmarkah] (d)Jadual 7 menunjukkan ciri empat organisma.

Jadual 7

Organisma	Ciri
Amoeba	Eukariot, unisel, tiada dinding sel, bergerak dengan pseudopodia
Yis	Eukariot, unisel, dinding sel berkitin, tiada kloroplas
Spirogyra	Eukariot, berbentuk filamen, kloroplas berbentuk lingkaran, tiada akar, batang atau daun sebenar
Cendawan	Eukariot, multisel, hifa, dinding sel berkitin, tiada klorofil

Namakan kerajaan bagi setiap organisma.Table 7 shows the features of four organisms.

Table 7

Organism	Features
Amoeba	Eukaryotic, unicellular, no cell wall, moves by pseudopodia
Yeast	Eukaryotic, unicellular, cell wall of chitin, no chloroplasts
Spirogyra	Eukaryotic, filament, spiral chloroplast, no true roots, stems or leaves
Mushroom	Eukaryotic, multicellular, hyphae, cell wall of chitin, no chlorophyll

Name the kingdom of each organism.

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• Amoeba: Protista1
• Yeast: Fungi1
• Spirogyra: Protista1
• Mushroom: Fungi1

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

[5 marksmarkah] (e)Bina kunci dikotomi untuk mengenal pasti empat organisma dalam Jadual 7 (Amoeba, yis, Spirogyra dan cendawan).Construct a dichotomous key to identify the four organisms in Table 7 (Amoeba, yeast, Spirogyra and mushroom).

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• Each step has two contrasting statements (a couplet) using clear features that do not overlap1
• A first couplet divides the organisms correctly, for example unicellular / multicellular (or has chloroplasts / has no chloroplasts)1
• A second couplet separates the two organisms of one group correctly, for example Amoeba (no cell wall) from yeast (cell wall)1
• A third couplet separates the two organisms of the other group correctly, for example Spirogyra (has chloroplasts) from mushroom (no chloroplasts)1
• All four organisms are named at the end of the key, each at a different end point, and the couplets are numbered or lead on correctly1

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

[1 markmarkah] (f)Siapakah yang dikenali sebagai bapa taksonomi?Who is known as the father of taxonomy?

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• Carolus Linnaeus1

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

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.1

TextbookBuku teks p.ms. 150

27Choose 1Pilih 1

QuestionSoalan 27

Tikus memakan buah kelapa sawit di sebuah blok ladang seluas 10 hektar. Pengurus ladang memasang enam kotak sarang burung hantu tuan (Tyto alba) sebagai kawalan biologi, sebagai ganti racun tikus. Setiap kotak berharga RM150. Setiap ekor tikus memusnahkan buah bernilai RM24 setahun. Populasi tikus dianggarkan dengan kaedah tangkap, tanda, lepas dan tangkap semula sebelum dan setahun selepas kotak dipasang. Tenaga dalam buah kelapa sawit yang dimakan oleh tikus ialah 50 000 kJ. Jadual 8 menunjukkan keputusan.

Jadual 8

	Tikus ditanda dan dilepaskan	Saiz sampel kedua	Tikus bertanda dalam sampel kedua
Sebelum kotak dipasang	60	45	9
Setahun selepas	60	40	16

Rats eat the fruits of oil palm in a 10-hectare block of an estate. The estate manager installs six nest boxes for barn owls (Tyto alba) as biological control, instead of rat poison. Each box costs RM150. Each rat destroys fruits worth RM24 a year. The rat population is estimated by the capture, mark, release and recapture method before and one year after the boxes are installed. The energy in the oil palm fruits eaten by the rats is 50 000 kJ. Table 8 shows the results.

Table 8

	Rats marked and released	Size of second sample	Marked rats in second sample
Before the boxes	60	45	9
One year after	60	40	16

[2 marksmarkah] (a)Tulis satu rantai makanan yang melibatkan buah kelapa sawit, tikus dan burung hantu tuan. Nyatakan aras trofik burung hantu.Write a food chain that involves the oil palm fruit, the rat and the barn owl. State the trophic level of the owl.

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• Oil palm fruit → rat → barn owl1
• Secondary consumer (third trophic level)1

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

[3 marksmarkah] (b)Hitung anggaran populasi tikus sebelum dan setahun selepas kotak dipasang.Calculate the estimated rat population before and one year after the boxes are installed.

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• Population = (marked × second sample) ÷ marked recaptured1
• Before: 60 × 45 ÷ 9 = 300 rats1
• After: 60 × 40 ÷ 16 = 150 rats1

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

[2 marksmarkah] (c)Nyatakan dua syarat yang perlu dipatuhi supaya anggaran populasi itu tepat.State two conditions that must be met so that the population estimate is accurate.

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Any 2 pointsMana-mana 2 isi

• The marks do not harm the rats or make them easier for predators to see1
• The marked rats mix fully with the population before recapture1
• No migration, births or deaths during the study (or equal chance of capture)1

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

[3 marksmarkah] (d)Hitung tenaga yang sampai kepada burung hantu melalui tikus dan terangkan mengapa bilangan burung hantu jauh lebih kecil daripada bilangan tikus.Calculate the energy that reaches the owls through the rats and explain why the number of owls is much smaller than the number of rats.

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• Rats: 10% of 50 000 = 5 000 kJ; owls: 10% of 5 000 = 500 kJ1
• Only about 10% of the energy is passed to the next trophic level; about 90% is lost as heat, in life processes and in excretion1
• So the energy at the top is small and supports only a small number of owls1

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

[4 marksmarkah] (e)Bandingkan penggunaan burung hantu dengan racun tikus untuk mengawal tikus dari segi kesannya kepada alam sekitar dan kemampanan.Compare using owls with using rat poison to control the rats in terms of the effect on the environment and sustainability.

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• Owls: a natural enemy that controls the rats and is not a pollutant, so it is environment-friendly1
• Owls: self-sustaining because they breed and keep the rat population low for a long time1
• Poison: may kill other organisms (such as the owls that eat poisoned rats) and pollute the soil and water1
• Poison: has to be bought and applied repeatedly, and the rats may become resistant by natural selection, so it is not sustainable1

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

[3 marksmarkah] (f)Hitung kerugian yang dapat dielakkan dalam setahun dan kos kotak sarang. Nilaikan sama ada pemasangan kotak itu berbaloi pada tahun pertama.Calculate the yearly loss that is avoided and the cost of the nest boxes. Evaluate whether installing the boxes is worth it in the first year.

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• Loss avoided: (300 – 150) × RM24 = RM3 6001
• Cost: 6 × RM150 = RM9001
• Worth it: the saving (RM3 600) is more than the cost (RM900), a net gain of RM2 700 in the first year, and no pesticide pollution1

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

[3 marksmarkah] (g)Cadangkan satu langkah lain untuk memastikan burung hantu terus tinggal dan membiak di ladang itu, dan beri satu sebab.Suggest one other step to make sure the owls stay and breed in the estate, and give one reason.

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• Do not use rat poison near the nest boxes (or use no poison at all)1
• Keep trees or nesting sites and leave some areas undisturbed for roosting1
• Reason: poisoned rats would kill the owls; owls need undisturbed sites to breed, so the biological control continues1

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

For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 10, standard 10.3

TextbookBuku teks p.ms. 215

28Choose 1Pilih 1

QuestionSoalan 28

Sebuah institut penyelidikan padi menyinari 1 000 biji benih padi dengan sinar gama pada setiap satu daripada lima dos (serta kawalan 0 Gy) untuk menghasilkan mutan yang lebih tinggi hasilnya. Rajah 5 menunjukkan peratus biji benih yang bercambah (X) dan peratus anak benih mutan di kalangan anak benih yang bercambah (Y). Jadual 9 memberikan nilai tersebut.

Jadual 9

Dos (Gy)	0	100	200	300	400	500
Biji benih bercambah (%)	98	92	80	61	38	15
Anak benih mutan (%)	0	2	5	8	9	6

A rice research institute irradiates 1 000 rice seeds with gamma rays at each of five doses (and a 0 Gy control) to produce mutants with a higher yield. Diagram 5 shows the percentage of seeds that germinate (X) and the percentage of mutant seedlings among the seeds that germinate (Y). Table 9 gives the values.

Table 9

Dose (Gy)	0	100	200	300	400	500
Seeds germinated (%)	98	92	80	61	38	15
Mutant seedlings (%)	0	2	5	8	9	6

[2 marksmarkah] (a)Sinar gama ialah suatu mutagen. Nyatakan jenis mutagen ini dan beri satu contoh lain bagi jenis yang sama.Gamma rays are a mutagen. State the type of this mutagen and give one other example of the same type.

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• Physical mutagen1
• Ultraviolet rays or X-rays (alpha or beta rays)1

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

[3 marksmarkah] (b)Huraikan kesan peningkatan dos terhadap peratus biji benih yang bercambah dan peratus anak benih mutan berdasarkan Rajah 5.Describe the effect of increasing the dose on the percentage of seeds that germinate and on the percentage of mutant seedlings, based on Diagram 5.

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• Germination falls steadily as the dose increases (98% to 15%)1
• Mutant seedlings increase from 0% to a maximum of 9% at 400 Gy1
• Above 400 Gy the percentage of mutants decreases (to 6% at 500 Gy)1

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

[4 marksmarkah] (c)Hitung bilangan anak benih mutan yang diperoleh daripada 1 000 biji benih pada 200 Gy, 300 Gy dan 400 Gy. Tentukan dos yang terbaik.Calculate the number of mutant seedlings obtained from 1 000 seeds at 200 Gy, 300 Gy and 400 Gy. Decide which dose is the best.

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• 200 Gy: 800 seedlings × 5% = 401
• 300 Gy: 610 seedlings × 8% = 48.8 (about 49)1
• 400 Gy: 380 seedlings × 9% = 34.2 (about 34)1
• Best dose: 300 Gy, because it gives the most mutant seedlings1

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[3 marksmarkah] (d)Kebanyakan mutasi gen adalah memudaratkan. Terangkan bagaimana mutasi gen boleh menghasilkan protein yang tidak berfungsi.Most gene mutations are harmful. Explain how a gene mutation can produce a protein that does not function.

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• A base is substituted, inserted or deleted, which changes the base sequence of the gene1
• The genetic code changes, so a different amino acid sequence is made1
• The protein (enzyme) has a different shape and cannot work1

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[3 marksmarkah] (e)Satu pokok mutan menghasilkan hasil yang tinggi. Terangkan bagaimana kultur tisu boleh menghasilkan banyak pokok yang sama dengannya.One mutant plant has a high yield. Explain how tissue culture can produce many plants identical to it.

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• A small piece of tissue from the mutant plant is grown on a sterile nutrient medium1
• Auxin and cytokinin make the cells divide by mitosis and differentiate to form roots and shoots1
• The plantlets are genetically identical (clones), so all keep the high yield1

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[5 marksmarkah] (f)Selain pembiakan mutasi, pakar boleh menggunakan kejuruteraan genetik untuk menghasilkan padi tahan serangga. Nilaikan kaedah ini berbanding pembiakan mutasi.Besides mutation breeding, experts can use genetic engineering to produce insect-resistant rice. Evaluate this method compared with mutation breeding.

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• Mutation breeding is random: most mutations are harmful and only a few useful mutants are obtained1
• Genetic engineering transfers a chosen gene (for example the Bt gene), so the result is specific and faster1
• Advantage of genetic engineering: less pesticide is used, yield is higher and the cost is lower1
• Disadvantage: the gene may spread to natural species, and the long-term effect on human health or the environment is not known1
• Decision with a reason, for example genetic engineering is better for a specific trait if the biosafety is tested and regulated1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 12, standard 12.3

TextbookBuku teks p.ms. 268

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