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

Ujian Akhir Sesi AkademikEnd of Academic Session Test

BiologiBiology

Tingkatan 4, 2 jam 30 minit / 2 hours 30 minutes, 100 marksmarkah

Nama / NameNama

Kelas / ClassKelas

Tarikh / DateTarikh

Markah / marksmarkah: / 100

InstructionsArahan

Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C.This question paper consists of three sections: Section A, Section B and Section C.
Jawab semua soalan.Answer all questions.
Bahagian A: pilih satu jawapan bagi setiap soalan.Section A: choose one answer for each question.
Bahagian B dan C: tulis jawapan anda di ruang yang disediakan. Kerja mengira mesti ditunjukkan.Sections B and C: write your answers in the spaces provided. Working must be shown for calculations.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.The diagrams in this paper are not drawn to scale unless stated.
Kalkulator saintifik yang tidak boleh diprogramkan dibenarkan.A non-programmable scientific calculator is allowed.

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.

Sebelum menjalankan eksperimen, seorang pelajar menulis, "Semakin banyak cahaya matahari yang diterima oleh sebatang pokok, semakin banyak kanji yang dihasilkan." Apakah pernyataan ini?Before an experiment, a student writes, "The more sunlight a plant receives, the more starch it makes." What is this statement?

HipotesisHypothesis
KesimpulanConclusion
PemerhatianObservation
Pemboleh ubahVariable

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A hypothesis is a statement that predicts the relationship between the manipulated and responding variables and is tested by the experiment. A conclusion is written after the results are analysed; an observation is something seen or measured.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.4

TextbookBuku teks p.ms. 15

Struktur manakah memberikan bentuk tetap kepada sel tumbuhan dan menghalangnya daripada pecah dalam larutan hipotonik?Which structure gives a plant cell a fixed shape and stops it from bursting in a hypotonic solution?

Membran plasmaPlasma membrane
Dinding selCell wall
NukleusNucleus
RibosomRibosome

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The cell wall is made of cellulose and is strong and rigid. It resists the pressure of the water that enters the cell, so the cell becomes turgid and does not burst. The plasma membrane is thin and elastic.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 2, standard 2.1

TextbookBuku teks p.ms. 25

Sel P ialah kubus bersisi 2 µm dan sel Q ialah kubus bersisi 4 µm. Bagaimanakah nisbah luas permukaan kepada isi padu sel P berbanding dengan sel Q?Cell P is a cube with sides of 2 µm and cell Q is a cube with sides of 4 µm. How does the surface area to volume ratio of cell P compare with that of cell Q?

Separuh daripada sel QHalf that of cell Q
Sama dengan sel QThe same as cell Q
Dua kali ganda sel QTwice that of cell Q
Empat kali ganda sel QFour times that of cell Q

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Cell P: surface area = 6 × 2 × 2 = 24 µm², volume = 8 µm³, ratio = 3. Cell Q: surface area = 6 × 4 × 4 = 96 µm², volume = 64 µm³, ratio = 1.5. P has twice the ratio of Q, so a small cell exchanges substances faster per unit volume.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 2, standard 2.1

TextbookBuku teks p.ms. 24

Pernyataan yang manakah menerangkan pengangkutan aktif dengan betul?Which statement describes active transport correctly?

Zarah bergerak menuruni kecerunan kepekatan.Particles move down the concentration gradient.
Zarah bergerak tanpa menggunakan protein pembawa.Particles move without using a carrier protein.
Hanya berlaku dalam sel tumbuhan.It occurs only in plant cells.
Zarah bergerak menentang kecerunan kepekatan dengan menggunakan tenaga ATP.Particles move against the concentration gradient using ATP energy.

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Active transport uses a carrier protein and energy from ATP to move particles from a region of lower concentration to a region of higher concentration. Diffusion and facilitated diffusion move particles down the gradient without ATP.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 3, standard 3.2

TextbookBuku teks p.ms. 52

Empat beg tiub Visking berisi larutan sukrosa 0.2 M, 0.4 M, 0.6 M dan 0.8 M direndam dalam bikar larutan sukrosa 0.5 M selama 30 minit. Tiub Visking telap kepada air tetapi tidak telap kepada sukrosa. Beg yang manakah mengalami penurunan jisim yang paling besar?Four Visking tubing bags containing 0.2 M, 0.4 M, 0.6 M and 0.8 M sucrose solution are placed in a beaker of 0.5 M sucrose solution for 30 minutes. Visking tubing is permeable to water but not to sucrose. Which bag shows the greatest decrease in mass?

Beg 0.8 MThe 0.8 M bag
Beg 0.2 MThe 0.2 M bag
Beg 0.4 MThe 0.4 M bag
Beg 0.6 MThe 0.6 M bag

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Water leaves a bag only if the solution inside is weaker than the 0.5 M outside, so the 0.2 M and 0.4 M bags lose mass. The 0.2 M bag has the largest difference in concentration (0.3 M), so it loses the most water. The 0.6 M and 0.8 M bags gain water.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 3, standard 3.3

TextbookBuku teks p.ms. 57

Pasangan yang manakah menunjukkan monomer dan polimer dengan betul?Which pair shows the monomer and the polymer correctly?

Asid lemak dan glikogenFatty acid and glycogen
Asid amino dan selulosaAmino acid and cellulose
Glukosa dan glikogenGlucose and glycogen
Gliserol dan proteinGlycerol and protein

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Glycogen, starch and cellulose are polysaccharides made of glucose units. Proteins are polymers of amino acids. Lipids are not true polymers; a triglyceride is made of glycerol and fatty acids.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 4, standard 4.2

TextbookBuku teks p.ms. 73

Bagaimanakah enzim mempercepat tindak balas biokimia?How does an enzyme speed up a biochemical reaction?

Dengan merendahkan tenaga pengaktifan tindak balasBy lowering the activation energy of the reaction
Dengan membekalkan tenaga haba kepada substratBy supplying heat energy to the substrate
Dengan bergabung dengan hasil tindak balas secara kekalBy combining permanently with the products
Dengan meninggikan suhu tindak balasBy raising the temperature of the reaction

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An enzyme is a biological catalyst. It lowers the activation energy by forming an enzyme-substrate complex, so the reaction takes place faster at body temperature. It does not supply energy and is not changed permanently.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 5, standard 5.1

TextbookBuku teks p.ms. 86

Dalam satu eksperimen, substrat berlebihan, suhu 37 °C dan pH optimum. Kepekatan enzim digandakan. Apakah kesan kepada kadar tindak balas awal?In an experiment the substrate is in excess, the temperature is 37 °C and the pH is optimum. The concentration of the enzyme is doubled. What is the effect on the initial rate of reaction?

Tidak berubah kerana substrat sudah berlebihanNo change because the substrate is already in excess
Menjadi separuh kerana enzim bersaing antara satu sama lainIt halves because the enzymes compete with each other
Menurun kerana enzim ternyahasliIt decreases because the enzymes are denatured
Menjadi dua kali ganda kerana bilangan tapak aktif bertambah dua kali gandaIt doubles because the number of active sites is doubled

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When substrate is in excess, the number of active sites is the limiting factor. Doubling the enzyme concentration doubles the number of active sites, so the initial rate of reaction doubles.

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

TextbookBuku teks p.ms. 90

Sel soma seekor haiwan mempunyai 8 kromosom (2n = 8). Berapakah bilangan bivalen yang terbentuk semasa profasa I meiosis?A somatic cell of an animal has 8 chromosomes (2n = 8). How many bivalents are formed during prophase I of meiosis?

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In prophase I the homologous chromosomes pair up, each pair forming one bivalent. Eight chromosomes make 4 homologous pairs, so 4 bivalents are formed.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 6, standard 6.3

TextbookBuku teks p.ms. 107

Apakah yang berlaku kepada asid piruvik dalam sel otot yang kekurangan oksigen semasa senaman lasak?What happens to pyruvic acid in muscle cells that lack oxygen during vigorous exercise?

Ia ditukarkan kepada etanol dan karbon dioksida.It is changed into ethanol and carbon dioxide.
Ia ditukarkan kepada asid laktik.It is changed into lactic acid.
Ia dioksidakan kepada karbon dioksida dan air.It is oxidised to carbon dioxide and water.
Ia ditukarkan semula kepada glukosa.It is changed back into glucose.

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Without oxygen, muscle cells carry out anaerobic respiration and change pyruvic acid into lactic acid. Ethanol and carbon dioxide are the products of fermentation in yeast.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 7, standard 7.3

TextbookBuku teks p.ms. 119

Tiga penyakit sistem respirasi diterangkan dalam jadual.

Penyakit / Disease	Penerangan / Description
P	Bronkiol menjadi sempit kerana keradangan dan otot licin mengecut, biasanya dicetuskan oleh alergen / Bronchioles narrow because of inflammation and contraction of smooth muscle, usually triggered by allergens
Q	Dinding alveolus musnah dan bergabung menjadi kantung udara yang besar / Alveolar walls break down and join into larger air sacs
R	Bronkus meradang dan menghasilkan banyak mukus, menyebabkan batuk berpanjangan / Bronchi are inflamed and produce much mucus, causing a long-lasting cough

Baris yang manakah menamakan P, Q dan R dengan betul?Three diseases of the respiratory system are described in the table. Which row names P, Q and R correctly?

P: asma; Q: emfisema; R: bronkitis kronikP: asthma; Q: emphysema; R: chronic bronchitis
P: emfisema; Q: asma; R: bronkitis kronikP: emphysema; Q: asthma; R: chronic bronchitis
P: bronkitis kronik; Q: emfisema; R: asmaP: chronic bronchitis; Q: emphysema; R: asthma
P: asma; Q: bronkitis kronik; R: emfisemaP: asthma; Q: chronic bronchitis; R: emphysema

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Asthma is narrowing of the bronchioles, often triggered by allergens. Emphysema is destruction of the alveolar walls, which reduces the surface area for gaseous exchange. Chronic bronchitis is inflammation of the bronchi with excess mucus and a persistent cough.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, Chapter 8, standard 8.4

TextbookBuku teks p.ms. 140

Mineral manakah diperlukan untuk membentuk hemoglobin dalam sel darah merah?Which mineral is needed to make haemoglobin in red blood cells?

KalsiumCalcium
IodinIodine
NatriumSodium
FerumIron

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Haemoglobin contains iron, which binds oxygen. A lack of iron causes anaemia. Calcium is for bones and teeth, iodine is for thyroxine and sodium is for nerve impulses and water balance.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 9, standard 9.1

TextbookBuku teks p.ms. 146

Sebungkus makanan ringan mengandungi 10 g karbohidrat, 5 g protein dan 6 g lemak. Nilai tenaga karbohidrat dan protein ialah 17 kJ g⁻¹ dan nilai tenaga lemak ialah 37 kJ g⁻¹. Berapakah jumlah nilai tenaga makanan ringan itu?A packet of snack contains 10 g of carbohydrate, 5 g of protein and 6 g of fat. The energy value of carbohydrate and of protein is 17 kJ g⁻¹ and the energy value of fat is 37 kJ g⁻¹. What is the total energy value of the snack?

357 kJ
477 kJ
557 kJ
577 kJ

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(10 × 17) + (5 × 17) + (6 × 37) = 170 + 85 + 222 = 477 kJ. 357 kJ uses 17 kJ g⁻¹ for fat as well.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 9, standard 9.1

TextbookBuku teks p.ms. 148

Kebuk jantung yang manakah mengepam darah beroksigen ke seluruh badan?Which chamber of the heart pumps oxygenated blood to the whole body?

Atrium kananRight atrium
Ventrikel kananRight ventricle
Ventrikel kiriLeft ventricle
Atrium kiriLeft atrium

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The left ventricle has the thickest wall and pumps oxygenated blood into the aorta, which sends it to the whole body. The right ventricle pumps deoxygenated blood to the lungs.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 10, standard 10.2

TextbookBuku teks p.ms. 176

Tekanan darah seorang lelaki ialah 160/100 mm Hg. Pernyataan yang manakah betul tentang keadaannya?A man has a blood pressure of 160/100 mm Hg. Which statement about his condition is correct?

Tekanan darahnya normal.His blood pressure is normal.
Dia mengalami hipotensi, iaitu tekanan darah yang terlalu rendah.He has hypotension, which is blood pressure that is too low.
Tekanan sistolnya normal tetapi tekanan diastolnya tinggi.His systolic pressure is normal but his diastolic pressure is high.
Dia mengalami hipertensi, yang meningkatkan risiko strok dan serangan jantung.He has hypertension, which increases the risk of stroke and heart attack.

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Normal blood pressure is about 120/80 mm Hg. Readings persistently above 140/90 mm Hg are hypertension. Both his systolic (160) and diastolic (100) values are high, and hypertension damages blood vessels and the heart.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 10, standard 10.4

TextbookBuku teks p.ms. 190

Mengapakah vaksin yang disuntik ke dalam badan seorang kanak-kanak merangsang imuniti tanpa menyebabkan penyakit?Why does a vaccine injected into a child stimulate immunity without causing the disease?

Vaksin mengandungi patogen yang dilemahkan atau dimatikan atau antigennya, yang merangsang penghasilan antibodi dan sel memori.The vaccine contains weakened or dead pathogens, or their antigens, which stimulate production of antibodies and memory cells.
Vaksin mengandungi antibodi siap yang memusnahkan patogen.The vaccine contains ready-made antibodies that destroy the pathogen.
Vaksin mengandungi fagosit yang menelan patogen.The vaccine contains phagocytes that engulf the pathogen.
Vaksin membunuh semua patogen dalam badan.The vaccine kills all the pathogens in the body.

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A vaccine contains weakened or killed pathogens (or their antigens). They are antigens that cause the body to make its own antibodies and memory cells, giving artificial active immunity, but they cannot cause the disease. Ready-made antibodies are given in artificial passive immunity.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, Chapter 11, standard 11.3

TextbookBuku teks p.ms. 207

Sejurus sebelum perlumbaan, jantung Aisha berdegup kencang dan dia bernafas dengan cepat. Hormon daripada kelenjar manakah yang bertanggungjawab?Just before a race, Aisha's heart beats fast and she breathes quickly. A hormone from which gland is responsible?

PankreasPancreas
Kelenjar tiroidThyroid gland
Kelenjar adrenalAdrenal gland
Kelenjar paratiroidParathyroid gland

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In a stressful or exciting situation the adrenal glands secrete adrenaline, which raises the heartbeat rate, the breathing rate and the blood glucose level to prepare the body for action.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, Chapter 12, standard 12.5

TextbookBuku teks p.ms. 236

Glukosa dikesan dalam air kencing seorang pesakit. Apakah penjelasan yang paling mungkin?Glucose is found in a patient's urine. What is the most likely explanation?

Glomerulus tidak menapis glukosa.The glomerulus does not filter glucose.
Kepekatan glukosa darah terlalu tinggi sehingga tidak semua glukosa dapat diserap semula di tubul berlingkar proksimal.The blood glucose concentration is so high that not all of the glucose can be reabsorbed in the proximal convoluted tubule.
Terlalu banyak ADH dirembeskan.Too much ADH is secreted.
Sel tubul distal merembeskan glukosa ke dalam air kencing.Cells of the distal tubule secrete glucose into the urine.

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Glucose is filtered at the glomerulus and normally all of it is reabsorbed in the proximal convoluted tubule. When the blood glucose level is very high, as in untreated diabetes mellitus, the carriers are saturated and some glucose remains in the filtrate and urine.

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

TextbookBuku teks p.ms. 255

Tulang yang manakah ialah sebahagian daripada rangka paksi?Which bone is part of the axial skeleton?

Humerus
Femur
SkapulaScapula
TengkorakSkull

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The axial skeleton consists of the skull, vertebral column, ribs and sternum. The humerus, femur and scapula belong to the appendicular skeleton (limbs and girdles).

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 14, standard 14.2

TextbookBuku teks p.ms. 267

Di manakah persenyawaan biasanya berlaku dalam sistem pembiakan wanita?Where does fertilisation normally take place in the female reproductive system?

Oviduk (tiub Fallopio)Oviduct (Fallopian tube)
Uterus
OvariOvary
Vagina

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The sperm swim up through the uterus to meet the secondary oocyte in the upper part of the oviduct, where fertilisation takes place. The zygote then travels to the uterus and implants in the endometrium.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 15, standard 15.1

TextbookBuku teks p.ms. 290

Bahagian BSection B

50 marksmarkah

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

Seorang pelajar menjalankan ujian makanan ke atas tiga sampel tidak diketahui, K, L dan M. Untuk ujian hidrolisis, sampel dipanaskan dengan asid hidroklorik cair, kemudian dineutralkan dengan natrium hidrogen karbonat sebelum diuji dengan larutan Benedict. Jadual menunjukkan keputusan.

Sampel / Sample	Iodin / Iodine	Benedict (langsung / direct)	Benedict selepas hidrolisis / after hydrolysis	Biuret
K	Perang kekuningan / Yellowish-brown	Biru / Blue	Merah bata / Brick-red	Biru / Blue
L	Biru tua / Blue-black	Biru / Blue	Merah bata / Brick-red	Biru / Blue
M	Perang kekuningan / Yellowish-brown	Merah bata / Brick-red	Merah bata / Brick-red	Ungu / Violet

A student carries out food tests on three unknown samples, K, L and M. For the hydrolysis test, the sample is heated with dilute hydrochloric acid, then neutralised with sodium hydrogen carbonate before it is tested with Benedict's solution. The table shows the results.

[1 markmarkah] (a)Namakan karbohidrat yang terdapat dalam sampel L.Name the carbohydrate present in sample L.

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Starch (blue-black with iodine; it is hydrolysed to glucose, which gives brick-red with Benedict's)1

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[2 marksmarkah] (b)Sampel K tidak memberikan perubahan dengan larutan Benedict secara langsung tetapi menghasilkan mendakan merah bata selepas hidrolisis. Terangkan keputusan ini.Sample K gives no change with Benedict's solution directly but gives a brick-red precipitate after hydrolysis. Explain this result.

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K contains a non-reducing sugar (sucrose), which does not react with Benedict's solution directly1
Acid hydrolysis breaks it down into reducing monosaccharides (glucose and fructose), which give a brick-red precipitate1

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[1 markmarkah] (c)Mengapakah hasil hidrolisis perlu dineutralkan sebelum larutan Benedict ditambah?Why must the hydrolysed mixture be neutralised before Benedict's solution is added?

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Benedict's test works only in alkaline conditions; the acid would prevent the reaction1

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[1 markmarkah] (d)Namakan dua kelas makanan dalam sampel M.Name two food classes in sample M.

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Reducing sugar and protein (both needed)1

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[1 markmarkah] (e)Nyatakan ikatan yang dikesan oleh ujian Biuret.State the bond that is detected by the Biuret test.

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Peptide bond1

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For the teacherUntuk guru: AnalysingMenganalisis, lowrendah difficultyaras lowrendah, Chapter 4, standard 4.2

TextbookBuku teks p.ms. 74

Seorang pelajar mengkaji kesan suhu air ke atas kadar denyutan jantung kutu air (Daphnia). Satu ekor kutu air diletakkan dalam air pada setiap suhu selama lima minit dan bilangan denyutan jantung dalam satu minit dikira di bawah mikroskop. Jadual menunjukkan keputusannya.

Suhu air / Water temperature (°C)	10	15	20	25	30	35
Denyutan jantung seminit / Heartbeats per minute	180	226	275	330	385	310

A student investigates the effect of water temperature on the heartbeat rate of a water flea (Daphnia). One water flea is placed in water at each temperature for five minutes and the number of heartbeats in one minute is counted under a microscope. The table shows the results.

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

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Manipulated variable: water temperature1
Responding variable: heartbeat rate (number of heartbeats per minute)1

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[1 markmarkah] (b)Pada suhu manakah kadar denyutan jantung paling tinggi?At which temperature is the heartbeat rate the highest?

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30 °C1

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[1 markmarkah] (c)Nyatakan jenis graf yang paling sesuai untuk mempersembahkan data ini dan beri satu sebab.State the most suitable type of graph to present these data and give one reason.

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Line graph, because both variables are numerical and continuous (the graph shows the trend as the temperature changes)1

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[1 markmarkah] (d)Cadangkan bagaimana kebolehpercayaan keputusan dapat ditingkatkan.Suggest how the reliability of the results can be improved.

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Repeat the experiment with several water fleas at each temperature and calculate the mean heartbeat rate1

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[1 markmarkah] (e)Daphnia ialah haiwan hidup. Nyatakan satu langkah untuk melayannya dengan baik semasa eksperimen.Daphnia is a living animal. State one step to treat it humanely during the experiment.

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Do not keep it at the high temperature for long / use only a little water and a short observation time / return it to its container of pond water afterwards (any one)1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 1, standard 1.4

TextbookBuku teks p.ms. 17

Rajah 1 menunjukkan empat peringkat meiosis, P, Q, R dan S, dalam satu sel haiwan (2n = 4). Peringkat itu tidak disusun mengikut urutan.Diagram 1 shows four stages of meiosis, P, Q, R and S, in an animal cell (2n = 4). The stages are not arranged in order.

Rajah 1Diagram 1

[1 markmarkah] (a)Namakan peringkat dalam P.Name the stage in P.

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Metaphase I1

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[1 markmarkah] (b)Susun P, Q, R dan S mengikut urutan peristiwa yang betul.Arrange P, Q, R and S in the correct order of events.

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S → P → Q → R1

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[1 markmarkah] (c)Berapakah bilangan kromosom dalam setiap sel pada R? Nyatakan istilah bagi bilangan kromosom ini.How many chromosomes are in each cell at R? State the term for this chromosome number.

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2 chromosomes; haploid (n = 2)1

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[1 markmarkah] (d)Nyatakan satu peristiwa dalam S yang menyebabkan variasi genetik.State one event in S that causes genetic variation.

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Crossing over between non-sister chromatids of homologous chromosomes (exchange of genetic material at the chiasma)1

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[2 marksmarkah] (e)Nyatakan dua perbezaan tentang kelakuan kromosom dalam meiosis I dengan dalam mitosis, dari metafasa hingga anafasa.State two differences between how chromosomes behave in meiosis I and in mitosis, from metaphase to anaphase.

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Meiosis I: homologous chromosomes are paired as bivalents at the equator; mitosis: chromosomes are not paired and line up singly1
Meiosis I: homologous chromosomes separate to opposite poles (centromeres do not divide); mitosis: sister chromatids separate (centromeres divide)1

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[1 markmarkah] (f)Seorang bayi dilahirkan dengan sindrom Down. Berapakah bilangan kromosom dalam sel somanya?A baby is born with Down syndrome. How many chromosomes are in its somatic cells?

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47 (an extra chromosome 21)1

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For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, Chapter 6, standard 6.3

TextbookBuku teks p.ms. 106

Rajah 3 menunjukkan struktur sendi engsel pada lutut manusia. P, Q, R dan S ialah struktur pada sendi itu.Diagram 3 shows the structure of a hinge joint at the human knee. P, Q, R and S are structures of the joint.

Diagram of a knee joint with the femur above and the tibia below. P points to the blue layer on the end of the bone, Q to the yellow space between the bones, R to a brown band joining the two bones outside the joint and S to the pink membrane lining the joint capsule — Rajah 3Diagram 3

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

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P: articular cartilage1
Q: synovial fluid (in the synovial cavity)1
R: ligament1
S: synovial membrane1

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[1 markmarkah] (b)Nyatakan satu fungsi P.State one function of P.

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Reduces friction between the bones / absorbs shock (any one)1

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[1 markmarkah] (c)Nyatakan fungsi R.State the function of R.

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Joins one bone to another bone and holds them in position (prevents dislocation)1

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[1 markmarkah] (d)Seorang warga emas mengadu sendi lututnya sakit dan kaku. Nyatakan satu amalan yang membantu menjaga kesihatan sendi dan tulang.An elderly woman complains that her knee joints are painful and stiff. State one practice that helps to keep the joints and bones healthy.

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Regular exercise (such as walking) / a diet with enough calcium and vitamin D / maintain a healthy body weight / correct posture (any one)1

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[1 markmarkah] (e)Kebanyakan mamalia yang berjalan menggunakan sendi engsel pada kaki. Terangkan satu kelebihan sendi engsel berbanding dengan sendi lesung dan bebola untuk pergerakan itu.Walking in most mammals uses hinge joints in the legs. Explain one advantage of a hinge joint over a ball-and-socket joint for this movement.

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A hinge joint moves in one plane only (forward and backward), so it is stable and bears the weight of the body without twisting / dislocating1

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For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, Chapter 14, standard 14.2

TextbookBuku teks p.ms. 272

Jadual menunjukkan tiga haiwan, P, Q dan R, dan ciri sistem respirasi masing-masing.

Haiwan / Animal	Ciri / Feature
P: belalang / grasshopper	Udara masuk melalui lubang di sepanjang abdomen ke dalam tiub bercabang yang berakhir dengan tiub halus berisi cecair yang menyentuh sel otot / Air enters through openings along the abdomen into branching tubes that end in tiny fluid-filled tubes touching the muscle cells
Q: ikan / fish	Air mengalir ke dalam mulut dan melalui struktur seperti sikat merah di belakang kepala; darah mengalir dalam arah bertentangan dengan air / Water flows in through the mouth and over red comb-like structures behind the head; blood flows in the opposite direction to the water
R: katak yang berehat dalam air kolam / frog resting in pond water	Pertukaran gas melalui permukaan badan yang nipis, lembap dan mempunyai banyak kapilari darah / Gas exchange takes place through a thin, moist body surface with many blood capillaries

The table shows three animals, P, Q and R, and the features of their respiratory systems.

[3 marksmarkah] (a)Namakan struktur respirasi bagi P, Q dan R.Name the respiratory structure of P, Q and R.

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P: tracheae (spiracles and tracheoles)1
Q: gills1
R: skin1

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[1 markmarkah] (b)Namakan bahagian insang Q yang menjadi tapak pertukaran gas.Name the part of the gills of Q where gaseous exchange takes place.

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Gill lamellae1

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[2 marksmarkah] (c)Terangkan kelebihan aliran darah dalam arah bertentangan dengan aliran air pada lamela insang.Explain the advantage of blood flowing in the opposite direction to the water over the gill lamellae.

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A concentration gradient of oxygen between the water and the blood is maintained along the whole length of the lamella1
So more oxygen diffuses into the blood (almost all the oxygen in the water is absorbed)1

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[1 markmarkah] (d)Mengapakah kulit R mesti sentiasa lembap?Why must the skin of R always be moist?

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Oxygen dissolves in the moisture before it diffuses through the skin into the blood capillaries1

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[1 markmarkah] (e)Seorang perokok tegar didiagnosis menghidap emfisema. Terangkan mengapa dia sesak nafas.A heavy smoker is diagnosed with emphysema. Explain why he is short of breath.

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The alveolar walls are destroyed, so the surface area for gaseous exchange is reduced and less oxygen enters the blood1

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For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, Chapter 8, standard 8.1

TextbookBuku teks p.ms. 130

Rajah 4 menunjukkan kesan pH ke atas kadar tindak balas dua enzim pencernaan manusia, P dan Q. Satu enzim ialah pepsin yang bertindak dalam perut dan satu lagi ialah tripsin yang bertindak dalam duodenum.Diagram 4 shows the effect of pH on the rate of reaction of two human digestive enzymes, P and Q. One is pepsin, which acts in the stomach, and the other is trypsin, which acts in the duodenum.

Graph of rate of reaction against pH. P peaks at 100 at pH 2 and falls to 0 by pH 6. Q is 0 at pH 5, peaks at 100 at pH 8 and falls to 15 at pH 10 and 0 at pH 11 — Rajah 4Diagram 4

[1 markmarkah] (a)Nyatakan pH optimum bagi enzim Q.State the optimum pH of enzyme Q.

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pH 81

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[2 marksmarkah] (b)Enzim yang manakah ialah pepsin? Beri satu sebab.Which enzyme is pepsin? Give a reason.

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P1
Its optimum pH is 2 (acidic), which matches the acidic conditions of the stomach1

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[2 marksmarkah] (c)Terangkan apa yang berlaku kepada enzim P pada pH 7.Explain what happens to enzyme P at pH 7.

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The pH is far from the optimum, so the bonds that hold the enzyme in shape are broken and the enzyme is denatured1
The shape of the active site changes, so the substrate cannot fit and the rate falls to zero1

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[1 markmarkah] (d)Hitung peratus penurunan kadar tindak balas enzim Q apabila pH berubah daripada pH optimum kepada pH 10.Calculate the percentage decrease in the rate of reaction of enzyme Q when the pH changes from the optimum pH to pH 10.

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(100 − 15) ÷ 100 × 100 = 85%1

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[1 markmarkah] (e)Penguraian protein oleh pepsin ialah tindak balas katabolisme atau anabolisme? Beri satu sebab.Is the digestion of protein by pepsin a catabolic or an anabolic reaction? Give a reason.

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Catabolic, because a large molecule (protein) is broken down into smaller molecules (polypeptides)1

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[1 markmarkah] (f)Nyatakan satu kegunaan enzim dalam industri makanan.State one use of enzymes in the food industry.

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Lactase to make lactose-free milk / pectinase to clear fruit juice / amylase to make glucose syrup from starch / protease to tenderise meat (any one)1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, Chapter 5, standard 5.2

TextbookBuku teks p.ms. 91

Bahagian CSection C

30 marksmarkah

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

Seorang pelajar mengkaji kesan suhu ke atas aktiviti enzim amilase. Dia mencampurkan 2 cm³ larutan kanji 1% dengan 1 cm³ larutan amilase dalam setiap tabung uji dan merendam setiap tabung dalam kukus air pada suhu yang berlainan. Setiap 30 saat, setitik campuran diuji dengan larutan iodin pada jubin. Jadual menunjukkan masa untuk campuran tidak lagi memberikan warna biru tua dengan iodin.

Suhu / Temperature (°C)	10	20	30	40	50	60
Masa / Time taken (min)	16.0	8.0	4.0	2.0	5.0	Tiada perubahan selepas 20 min / No change after 20 min

A student investigates the effect of temperature on the activity of the enzyme amylase. She mixes 2 cm³ of 1% starch solution with 1 cm³ of amylase solution in each test tube and places each tube in a water bath at a different temperature. Every 30 seconds, a drop of the mixture is tested with iodine solution on a tile. The table shows the time taken for the mixture to stop giving a blue-black colour with iodine.

[3 marksmarkah] (a)Nyatakan pemboleh ubah dimanipulasikan, pemboleh ubah bergerak balas dan satu pemboleh ubah dimalarkan.State the manipulated variable, the responding variable and one constant variable.

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Manipulated: temperature of the water bath1
Responding: time taken for the starch to be completely digested (iodine no longer blue-black)1
Constant: volume or concentration of starch / volume or concentration of amylase / pH (any one)1

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[1 markmarkah] (b)Nyatakan hipotesis bagi eksperimen ini.State a hypothesis for this experiment.

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As the temperature increases up to the optimum temperature, the rate of digestion of starch by amylase increases, then it decreases at higher temperatures1

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[3 marksmarkah] (c)Huraikan bagaimana pelajar itu memastikan campuran dalam setiap tabung mencapai suhu yang dikehendaki sebelum enzim bertindak.Describe how the student makes sure that the mixture in each tube reaches the required temperature before the enzyme acts.

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Place the tube containing starch and the tube containing amylase separately in the water bath for a few minutes (equilibrate)1
Then pour the amylase into the starch and mix, starting the stopwatch at once1
Keep the water bath at the same temperature using a thermometer, and keep the tubes in the water bath while testing1

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[2 marksmarkah] (d)Hitung kadar tindak balas pada 20 °C dan 40 °C dengan menggunakan kadar = 1 ÷ masa. Berapa kali lebih cepat kadar pada 40 °C berbanding dengan 20 °C?Calculate the rate of reaction at 20 °C and at 40 °C using rate = 1 ÷ time. How many times faster is the rate at 40 °C than at 20 °C?

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20 °C: 1 ÷ 8.0 = 0.125 min⁻¹; 40 °C: 1 ÷ 2.0 = 0.5 min⁻¹ [1 mark]1
0.5 ÷ 0.125 = 4 times [1 mark]1

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[2 marksmarkah] (e)Terangkan keputusan pada 10 °C hingga 40 °C dan pada 50 °C hingga 60 °C.Explain the results from 10 °C to 40 °C and from 50 °C to 60 °C.

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10 °C to 40 °C: kinetic energy of enzyme and substrate molecules increases, so collisions and enzyme-substrate complexes form more often, and the rate increases1
50 °C to 60 °C: the enzyme is denatured, the shape of the active site changes and the substrate no longer fits, so the rate falls to zero1

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[2 marksmarkah] (f)Satu sampel daripada tabung 40 °C (selepas 2 minit) dan satu sampel daripada tabung 60 °C (selepas 20 minit) dididihkan dengan larutan Benedict. Nyatakan keputusan dan terangkan.A sample from the 40 °C tube (after 2 minutes) and a sample from the 60 °C tube (after 20 minutes) are each boiled with Benedict's solution. State the results and explain.

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40 °C: brick-red (orange or green) precipitate, because amylase has hydrolysed starch to maltose, a reducing sugar1
60 °C: remains blue, because the enzyme is denatured so no maltose was formed and starch is not a reducing sugar1

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[2 marksmarkah] (g)Pelajar itu membuat kesimpulan bahawa suhu optimum amilase ialah tepat 40 °C. Nilaikan kesimpulan itu dan cadangkan satu penambahbaikan.The student concludes that the optimum temperature of amylase is exactly 40 °C. Evaluate this conclusion and suggest one improvement.

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The conclusion is not fully supported: only temperatures at 10 °C intervals were tested, so the optimum could lie between 30 °C and 50 °C (e.g. 45 °C)1
Improvement: test more temperatures at smaller intervals, such as 35 °C, 40 °C, 45 °C and 50 °C, and repeat each to find the mean1

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For the teacherUntuk guru: CreatingMencipta, hightinggi difficultyaras hightinggi, Chapter 5, standard 5.2

TextbookBuku teks p.ms. 91

Test specification (JSU)Jadual Spesifikasi Ujian (JSU)

Chapter	RememberingMengingat	UnderstandingMemahami	ApplyingMengaplikasi	AnalysingMenganalisis	EvaluatingMenilai	CreatingMencipta	Items	marksmarkah
1. Introduction to Biology and Laboratory Rules	–	1 (1)	1 (6)	–	–	–	2	7
2. Cell Biology and Organisation	1 (1)	–	–	1 (1)	–	–	2	2
3. Movement of Substances across a Plasma Membrane	1 (1)	–	–	1 (1)	–	–	2	2
4. Chemical Compositions in a Cell	1 (1)	–	–	1 (6)	–	–	2	7
5. Metabolism and Enzymes	1 (1)	–	2 (9)	–	–	1 (15)	4	25
6. Cell Division	–	1 (7)	1 (1)	–	1 (15)	–	3	23
7. Cellular Respiration	–	1 (1)	–	–	–	–	1	1
8. Respiratory Systems in Humans and Animals	–	1 (8)	–	1 (1)	–	–	2	9
9. Nutrition and the Human Digestive System	1 (1)	–	1 (1)	–	–	–	2	2
10. Transport in Humans and Animals	1 (1)	–	1 (1)	–	–	–	2	2
11. Immunity in Humans	–	1 (1)	–	1 (7)	–	–	2	8
12. Coordination and Response in Humans	–	–	1 (1)	–	–	–	1	1
13. Homeostasis and the Human Urinary System	–	–	–	1 (1)	–	–	1	1
14. Support and Movement in Humans and Animals	2 (9)	–	–	–	–	–	2	9
15. Sexual Reproduction, Development and Growth in Humans and Animals	1 (1)	–	–	–	–	–	1	1
TotalJumlah	9 (16)	5 (18)	7 (19)	6 (17)	1 (15)	1 (15)	29	100

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

BiologiBiology

Bahagian ASection A

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Bahagian BSection B

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Bahagian CSection C

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