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Section A / Bahagian A [60 marks] / [60 markah]

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

1/KELANTAN/2014/Cell structure and cell organisation

Diagram 1 .1 shows the structure of a plant cell as seen under an electron microscope.

Rajah 1.1 menunjukkan struktur satu sel tumbuhan yang dilihat di bawah mikroskop electron.

Diagram 1.1 / Rajah 1.1

(a) (i) Name organelles P and X / Namakan organel P dan X.

P: Mitochondrion X: Chromosome

[2 marks]

(ii) Explain the function of organelle R in the plant cell.

Terangkan fungsi organel R dalam sel tumbuhan.

-R is a chloroplast

- chloroplast contain chlorophyll to absorb light energy -to carry out photosynthesis

[2 marks

(b) The ribosomes in Q synthesize proteins to form hormone.

Explain the effect to the plant growth if the cells in the shoot tip are lacking in ribosome.

Ribosome dalam Q mensintesis protein untuk membina hormone. Terangkan kesan ke atas

pertumbuhan suatu tumbuhan jika sel pada bahagian hujung pucuk tumbuhan kekurangan

ribosom.

-Less protein is being synthesized -Less auxin hormone will be produced -the rate of cell division will be slowed/ -Less growth occur

[2 marks]

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(c)

Diagram 1 .2 shows X behaviour in meiosis of an animal cell.

Rajah 1.2 menunjukkan perlakuan X semasa meiosis satu sel haiwan.


(i)

Stage T, U, V and W are arranged in order. Complete the Diagram 1.2 in stage U to

show the chromosome behaviour.

Peringkat T, U, V dan W disusun mengikut tertib. Lengkapkan Rajah 1.2 di peringkat

U untuk menunjukkan perlakuan kromosom.

[1 mark]

(ii) Explain the chromosomal behaviour in stage T.

Terangkan perlakuan kromosom pada peringkat T.

-Homologous chromosomes come together to form bivalent -Synapsis occur -Crossing over occur between non- sister chromatid

[2 marks]

(iii) A male has sex chromosome of 45 + XY. This genetic disorder is due to failure of

chromosome to separate during Stage V. Explain how this condition happens.

Seorang lelaki mempunyal kromosom seks 45 + XY. Kecacatan genetic ini berlaku

disebabkan kegagalan kromosom berpisah pada peringkat V. Terangkan bagaimana

keadaan ini boleh berlaku.

- the man is with Down Syndrome - caused by chromosome mutation - caused by chromosome 21 failed to separate during anaphase - produced sperm with 23 +Y// ovum with 23+ X -After fertilization the Zygote with 45+XY is formed.

[3 marks]

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2/KELANTAN/2014/Movement of substances across the plasma membrane

Diagram 2.1 shows the structure of a plasma membrane.

Rajah 2.1 menunjukkan struktur membran plasma.


(a) Plasma membrane is a semi permeable membrane and has a fluidity characteristic.

Membran plasma bersifat separa telap dan mempunyai sifat bendalir.

(i)

State the meaning of semi permeable membrane.

Nyatakan maksud membran separa telap.

- Semipermeable membrane is a membrane that allows certain substances to pass through it.

[1 mark]

(ii)

Explain the cause that makes the plasma membrane has the fluidity characteristic.

Terangkan apakah yang menyebabkan sifat bendalir pada membran plasma.

- phospholipid, protein, and other membrane component are freely move and not static(dynamic) -protein molecules float about freely in the phospholipid bilayer.

[2 marks]

(b) Diagram 2.2 shows two strips of mustard green stem after 20 minutes immersed into two

different solutions P and Q.

Rajah 2.2 menunjukkan dua jalur batang sawi selepas 20 minit direndam di dalam dua

larutan yang berbeza, iaitu P dan Q.


(i) Based on the diagram 2.2, state the type of solution Q.

Berdasarkan rajah 2.2, nyatakan jenis larutan Q.

Hypertonic solution [1 mark]

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(ii) Explain how the solution Q affects the condition of the cells in the strip that have been

immersed.

Terangkan bagaimana larutan Q mempengaruhi keadaan sel di dalam jalur sawi yang

telah direndamkan.

P1 –30% sucrose solution in beaker Q is hypertonic to the cell sap of the mustard green cell P2 - Water molecule will diffuse out from vacuole through osmosis P3 – Causing both vacuole and cytoplasm shrink / plasma member is pulled away from cell wall. P4: Cell become flaccid and plasmolysis occur.

[2 marks]

(iii) If strip from the solution Q is transferred into solution P, explain what will happen to

the cells in the strip after 20 minutes.

Jika jalur dan larutan Q dipindahkan ke dalam larutan P, terangkan apa yang akan

berlaku kepada sel dalam jalur tersebut selepas 20 minit.

P1 –Distilled water in beaker P is hypotonic to the to the cell sap of the mustard green cell P2 - Water molecule will diffuse into vacuole through osmosis P3 - Vacuole expand and swells out// the vacuole and the cytoplasm press outward P4 – The plant cells undergo deplasmolysis

[3 marks]

(c)

Diagram 2.3 shows the red blood cells in different concentrations of solutions.

Rajah 22 menunjukkan sel darah merah di dalam larutan yang berlainan kepekatan.

Red blood cells X in 3% sodium

Chloride solution after 30 minutes.

Sel darah merah X dalam larutan

Natrium klorida 3% selepas 30 minit

Red blood cells Y in 0.1% sodium chloride

solution after 30 minutes.

Sel darah merah Y dalam larutan natrium

klorida 0.1% selepas 30 minit

Explain the differences between the process experienced by the red blood cell

X and Y after being immersed for half an hour.

Terangkan perbezaan di antara proses yang dialami oleh sel darah merah X dan Y setelah

direndam 30 minit.

Red blood cell X Red Blood cell Y

3% sodium Chloride solution is hypertonic to the red blood cell

0.1% sodium Chloride solution is Hypotonic to the red blood cell

Water will diffuse out of the cell by osmosis

Water will diffuse into of the cell by osmosis

Red blood cell become shrink Red blood cell become swell and burst

[3 marks]

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3/KELANTAN/2014/Chemical Composition In The Cell

Diagram 3.1 shows types of carbohydrate found in the plant cell wall.

Rajah 3.1 menunjukkan jenis karbohidrat yang didapati pada dinding set tumbuhan.

Diagram 3.1/ Rajah 3.1

(a) (i) X is a basic unit of carbohydrate. Name X and Y.

X ialah unit asas karbohidrat. Namakan X dan Y.

X - Monosaccaride Y – Polisaccaride [2 marks]

(ii) Name the type of Y in Diagram 3.1 / Namakan jenis Y pada Rajah 3.1.

Cellulose [1 mark]

(iii) Name the process where Y can be converted to X.

Namakan proses di mana Y boleh ditukarkan kepada X.

Hydrolysis [1 mark]

(b)

A man’s diet consists of white bread, sweet food, meat and high fat foods. This leads to

constipation. The doctor advises him to eat more fruits, vegetables and cereals.

Gizi seorang lelaki terdiri daripada roti putih, makanan manis, daging dan makanan

berlemak tinggi. Ini telah menyebabkan sembelit. Doktor menasihatkan beliau untuk

memakan lebih banyak buah, sayuran dan bijirin.

(i) What class of food found in fruits, vegetables and cereals that will help him to prevent

constipation?

Apakah kelas makanan yang didapati dalam makanan, sayuran dan bijirin yang

membantu dia mengelakkan sembelit?

Roughage/ fibre [1 mark]

(ii) Explain why this class of food prevents constipation.

Terangkan mengapa kelas makanan tersebut dapat mengelakkan sembelit.

- Roughage have high water absorbance - this will increase its bulk - help peristalsis - which stimulate the muscles to push out the residue, preventing constipation[2 marks]

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(iii) Give one example of disease (other than constipation ) that can be prevented with

more intake this class of food.

Berikan satu contoh penyakit (selain daripada sembelit) yang boleh dielakkan jika

mengambil lebih banyak kelas makanan tersebut.

Colon cancer/ haemorroids [1 mark]

(c) Diagram 3.2 and Diagram 3.3 shows the physical condition of two children who suffer from

malnutrition.

Rajah 3.2 dan Rajah 3.3 menunjukkan keadaan fizikal dua orang kanak-kanak yang

mengalami malnutrisi.



Name and explain the disease in Diagram 3.2 and Diagram 3.3 related to malnutrition.

Namakan dan terangkan penyakit dalam Rajah 3.2 dan Rajah 3.3 yang berkaitan dengan

malnutrisi.

(i) Diagram 3.2 / Rajah 3.2

-Obesity - the excessive intake of fats

- increase 20% of ideal/normal body weight [2 marks]

(ii) Diagram 3.3 / Rajah 3.3

-Ricket - Lack of Vitamin D - Formation of weak bone and teeth Pembinaan tulang dan gigi yang lemah

- the bone become soft causing bowed legs and knock knee [2 marks]

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4/KELANTAN/2014/Support and movement

Diagram 4.1 shows the arrangement of skeleton and flight muscles in a bird.

Rajah 4.1 menunjukkan susunan rangka dan otot penerbangan pada seekor burung.


(a) Label two types of muscles in Diagram 4.1 that enable the bird to fly.

Labelkan dua jenis otot pada Rajah 4.1 yang membolehkan burung terbang.

[2 marks]

Pectoralis major n minor

(b) State skeleton and body characteristics that enable a bird to fly in air.

Nyatakan ciri rangka dan badan yang membolehkan burung terbang di udara.

- The bone is hollow and light to reduce body weight

- The wing is in the shape of aerofoil - have a streamline shape body

[3 marks]

(c) Explain why the bones of a female bird that lays eggs are more brittle as its age increases.

Terangkan mengapa tulang burung betina yang bertelur adalah Iebih rapuh apabila

umurnya semakin meningkat.

- reduced bone weight jisim - because more cesium / phosphorus/ phosphate is used to form egg shells - causing lack of calcium/ phosphorus/ phosphate to form strong bone

[3 marks]

(d) Diagram 4.2 shows the movement of an earthworm.

Rajah 4.2 menunjukkan pergerakan seekor cacing tanah.


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(i) Explain the muscles action for the forward movement of an earthworm.

Terangkan tindakan otot untuk pergerakan ke hadapan cacing tanah.

-Circular muscles and longitudinal muscles works antagonistically -Contractions of the circular muscles and relaxation of the longitudinal muscles cause the segments to extend. -The segment becomes thinner and longer. -Conversely, contractions of the longitudinal muscles and relaxation of circular muscles -cause the segments to shorten and thicker.

[3 marks]

(ii) State one differences between the musculoskeletal system of earthworm and the

musculoskeletal system of birds.

Nyatakan satu perbezaan antara sistem otot rangka cacing tanah dan sistem otot

rangka burung.

- earthworm has a hydrostatic muscle where as bird has a skeletal muscle that consist of bone and cartilage

[1 mark]

4/KEDAH/2014/Transport

Diagram 4 .1 shows the blood circulatory system in organism P and organism Q.

Rajah 4.1 menunjukkan sistem peredaran darah bagi organisma P dan organism Q.

(a) State the type of blood circulatory system of organism P and organism Q.

Nyatakan jenis sistem peredaran darah bagi organisma P dan organisma Q.

Organism P: Double closed circulatory system Organism Q: Single closed circulatory system

[2 marks]

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(b) State one difference between the heart of organism P and organism Q.

Nyatakan satu perbezaan di antara jantung organisma P dan organisma Q.

Organism P Organism Q

P1 Heart has 4 chambers Heart has 2 chambers

P2 has two atriums and two ventricles // has right atrium, left atrium, right ventricle and left ventricle

has one ventricle and one atrium // has ventricle and an atrium

P3 heart receives both oxygenated blood and deoxygenated separately.

heart receives the mixed oxygenated blood and deoxygenated blood

[1 mark]

(c) Explain why blood vessel X has higher pressure than vessel Y.

Terangkan mengapa salur darah X mempunyai tekanan Iebih tinggi dan salur darah Y

F : contraction of ventricle/ heart E1: generates a (high) pressure E2: (to) propel/ force/ pump the blood flow from the heart/ ventricle to vessel A. Any 2 [2 marks]

(d) The blood circulatory system is also involved in the production of antibodies in the body

defense mechanism. Diagram 4.2 shows the concentration of antibody in the blood of

individuals A and individuals B for a period of 10 weeks to acquire immunity.

Both of them were given two injections respectively.

Sistem peredaran darah terlibat dalam penghasilan antibodi untuk mekanisme pertahanan

badan. Rajah 4.2 menunjukkan kepekatan antibodi di dalam darah kedua-dua individu A

dan B dalam jangka masa IQ minggu untuk mendapatkan keimunan. Kedua - dua mereka

masing- masing telah diberikan dua suntikan.


(i) What type of immunity is obtained by A and B respectively?

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Apakah jenis keimunan yang diperolehi oleh individu A dan B masing-masing?

Individual A: Artificial/ (Acquired) active immunity Individual B: Artificial/ (Acquired) passive immunity[2 markS]

(ii) Explain why individual A has to be given the second injection which contained the

same vaccine.

Terangkan mengapa individu A harus diberi suntikan kedua bagi vaccine yang sama.

P1 : The first dose results the production of low level of Antibody // the concentration of antibody still not reach the immunity level P2: Second dose is needed to stimulate lymphocyte to produce more antibodies P3: until it reaches the immunity level. P3: that protects the person against the disease. [2 marks]

(iii) AIDS is caused by HIV (Human Immunodeficiency Virus). AIDS patients are easily

infected with various types of diseases and will eventually die, even from harmless

disease. Explain how HIV affects the body’s immune system.

Penyakit AIDS disebabkan oleh virus HIV (Human Immunodeficiency Virus).

Pesakit AIDS mudah dijangkiti oleh pelbagai jenis penyakit dan akhirnya akan mati

walaupun oleh penyakit yang tidak berbahaya. Terangkan bagaimana HIV memberi

kesan kepada sistem pertahanan badan.

P1 : HIV weakens the immune system P2 : By attacking helper T cell (which coordinates the immune system.) P3 : Helper T cells are essential to activate other lymphocytes in the body immunity. P4 : HIV also attack the central nervous system P5 : Decrease in the function of nervous system. P6 : The patient can expose himself to secondary infections. [3 marks]

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5/SBP/2014/Reproduction And Growth

Diagram 5.1 shows the formation of cells P.

Diagram 5.2 shows the pollination of cells P on the stigma of a matured pistil.

Rajah 5.1 menunjukkan pembentukkan sel-sel P.

Rajah 5.2 menunjukkan pendebungaan sel-sel P di atas stigma satu pistil yang matang.



(a) Describe how cells P are formed from the pollen mother cell.

Huraikan bagaimana sel-sel P terbentuk daripada sel induk debunga.

P1 Pollen mother cell (2n/diploid) undergoes meiosis P2 produces 4 haploid microspores (n) P3 The nucleus of each microspore divide by mitosis P4 to form a tube nucleus and a generative nucleus (Any 2)

[2 marks]

(b) Compare the number of nuclei in cell P and in the embryo sac. What makes it different?

Bandingkan bilangan nukleus di dalam sel P dan nukleus di dalam pundi embrio,

Apakah yang menyebabkan perbezaan ini?

P1 Cell P contains 2 nuclei, but the embryo sac contains 8 nuclei. P2 The nucleus in each microspore divide by mitosis one time only, but the nucleus of megaspore/embryo sac divides (by mitosis) 3 times.

[2 marks]

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(c) (i) Cell P germinates in response to a sugary fluid secreted by the matured stigma

forming a pollen tube. The pollen tube carries two male gametes towards the

ovary. Inside the ovary, an embryo sac developed producing haploid nuclei.

Three of the female nuclei are involved in double fertilisation. Suggest what

will happen to both nuclei R and nucleus S if the pollen tube fails to develop.

Sel P bercambah setelah dirangsang oleh cecair bergula yang dirembeskan

oleh stigma matang membentuk satu tiub debunga. Tiub debunga mi membawa

dua garnet jantan menghala ke ovari. Di dalam ovari, satu pundi embrio

berkembang menghasilkan nukleus-nukleus haploid. Hanya tiga daripada

nukleus - nukleus betina mi terlibat dalam persenyawaan ganda dua.

Cadangkan apa yang akan berlaku kepada kedua-dua nukleus R dan nukleus S

jika tiub debunga gagal untuk berkembang.

P1 the male gametes will not reach the ovary P2 will not penetrate the ovule through micropyle// not enter the embryo sac P3 one of the male gamete will not fertilise/fuse with the egg cell//no diploid zygote form P4 another male gamete will not fused with the two polar body//the triploid nucleus/endosperm will not form [3 marks]

(ii) Double fertilisation in plants ensures the survival of plant species. What is your

opinion about this statement?

Persenyawaan ganda dua dalam tumbuhan memastikan kemandirian spesies

tumbuhan. Apakah pendapat anda berkenaan pernyataan ini?

P1 The formation of diploid zygote involves meiosis P2 This produces variation in plants P3 That increase the ability to survive // Avoid extinction P3 Triploid nucleus/endosperm tissues provide nutrients/nourishment to the developing zygote/embryo.

[2 marks]

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(d) The ovary is able to develop into a fruit without undergoing fertilisation, It is the

phenomenon of fruiting without the union of male and female gametes and

artificially induced by applying hormone X on the stigma.

Ovari boleh berkembang menjadi buah tanpa melalui persenyawaan. Ini adalah satu

fenomena pembuahan tanpa melibatkan paduan garnet betina dan garnet jantan dan

dirangsang secara buatan dengan menyemburkan hormone X ke atas stigma.


(i) Name the phenomenon in producing seedless fruits and give one example of

hormone X.

Namakan fenomena dalam menghasilkan buah tanpa biji dan berikan satu contoh

hormone X.

The phenomenon : Parthenocarpy Hormone X: Auxin [2 marks]

(ii) Suggest one benefit of producing seedless fruits.

Cadangkan satu kebaikan menghasilkan buah tanpa biji.

P1 useful when pollination is poor / for instance, during freezing temperatures// difficult for some flower to pollinate or fertilize P2 can increase the texture//shelf life of fruits//increase sugar content P3 seedless fruits are very desirable because of their convenience to eat without the hard testa. [1 mark]

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6/KELANTAN/2014/ Coordination and response

Diagram 6 shows the effects of light on auxin.

Rajah 6 menunjukkan kesan cahaya ke atas auksin.

Diagram 6 / Rajah 6

6 (a) (i)

Based on the diagram, explain the response of auxin towards:

Berdasarkan rajah, terangkan gerak balas auksin terhadap:

• Light from above / Cahaya dan atas

• Light from one side / Cahaya dan satu sisi

[5 marks]

i) The response of auxin towards light from above:

- When the light from above, auxin is synthesized uniformly in the shoot. -the cell growth and elongate uniformly from all sides - causing the plant to grow up straight. Any two

ii) The response of auxin towards light from one side:

- When the light from one side, auxin will move away from light source and accumulates on the shaded side. -cell in the shaded side will grow quickly than the cells from the light side. -the shoots will bend and grow towards light.

(ii) Explain how auxins used to stimulate parthenocarphy

Terangkan bagaimana auksin digunakan untuk merangsang partenokarpi.

[3 marks]

–Parthenocarphy is the process of fruit formation without being fertilised. - when the flower is sprayed by the auxin, the ovary will develop to form fruit without fertilization. - producing the seedless fruits

(b) Predict the conditions of the coleoptiles kept in the box with the light source from one

direction after three days.

Ramalkan keadaan koleoptil yang disimpan dalam kotak dengan sumber cahaya dan satu

hala selepas tiga hari.

i. Tip of coleoptile cut / Hujung koleoptil dipotong

ii. Tip of coleoptile covered with cloth / Hujung koleoptil ditutup dengan kain

iii. Tip of coleoptile separated by a piece of metal

Hujung koleoptil diasingkan dengan sekeping logam

[5 marks]

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i) Tip of coleoptile cut: - No auxin is produced -No further growth of coleoptile ii) Tip of coleoptile covered with cloth:

-auxin that produced in the coleoptile is not affected by light - auxin is uniformly produced - Coleoptile will grow up straight iii) Tip of coleoptile separated by a piece of metal

- No growth occurred - a piece of metal will stop the diffusion of auxin to the zone of elongation.

(c) Diagram 6.2 shows the agricultural activity / Rajah 6.2 menunjukkan aktiviti pertanian.

Plants hormones are very important in growth and development of plants. Therefore they are

use widely in the agricultural sector.

Hormon tumbuhan sangat penting dalam pertumbuhan dan perkembangan tumbuh-

tumbuhan. OIeh itu, Ia digunakan secara meluas dalam bidang pertanian.

Explain the usage of another three hormones in the agricultural sector.

Terangkan kegunaan tiga hormon lain dalam bidang pentanian.

Ethylene hormone: -to induce ripening of fruits - speed up ripening by producing cellulase -cellulase hydrolyse cellulose and the fruits become soft - farmer used this hormone to speed up ripening of fruits when picking the unripe fruits . Cytokinins hormone -promote cell division and differentiation in tissue culture when used with auxins - to stimulate bud development when using leaf cutting. Gibberelin hormone - used to induce seed germination -induced parthenocarphy when used together with auxin

[6 marks]

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7 (a) Diagram 7.1 shows two neurones. The chemical substances involved in transmission of nerve impulse are known as neurotransmitter. For examples; Acetylcholine, Noradrenaline, Serotonin and Dopamine. Rajah 7.1 menunjukkan dua neuron. Bahan kimia yang terlibat dalam pemindahan impuls

dikenali sebagai neurotransmitter. Sebagai contoh; Asetilkolina, Noradrenalin, Serotonin dan

Dopamin

(i) Explain how the neurotransmitter helps in the transmission of nerve impulse from neurone A to neurone B. Terangkan bagaimana neurotransmitter membantu dalam pemindahan impuls saraf dari neuron A ke neuron B.

P1: transferring information along the neurone is in the form of electrical signals.

P2 : Impulse arrived at the axon terminal of Neurone A

P3 : Electrical signals is change to the chemical signals (diffusion of neurotransmitter)

P4 : Induced synaptic vesicles to release neurotransmitter

P5 : neurotransmitter diffuse across synapse

P6 : to the dendrite of neurone B

P7 : transmission of impulse is need energy

P8 : Synaptic knob contain many mitochondria

P9 : Chemical signals is change back to the electrical signals at the dendrite of neurone B

P10 : Neuron B transmits the electrical signals to the next neurone/ effector.

[6 marks] [6 markah]

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(ii) Based on Diagram 7.1, explain why the impulse transmission through a neuron occurs in one direction only. Berdasarkan Rajah 7.1 , terangkan mengapa pemindahan impuls melalui neuron adalah satu hala .

- Transmission of impulse occur across synapse - Synaptic vesicles are only present at synaptic terminals, hence only pre-synaptic terminal can discharge neurotransmitter. - Receptors are only present on the postsynaptic membrane, ensuring that only the postsynaptic membrane can receive the chemical signals. [3 marks][3 markah]

(b) Diagram 7.2 show impulses transmissions when a boy accidentally stepped on a pin. He yelled as he bounced on one foot while holding the injured foot up Rajah 7.2 menunjukkan pemindahan impuls apabila seorang budak lelaki secara tidak sengaja terpijak pin. Beliau berteriak dan melompat di atas satu kaki sambil memegang kaki yang satu lagi ke atas.

Explain the transmission of nerve impulse in the response. Terangkan pemindahan impuls saraf dalam gerakbalas tersebut.

P1 : Sharp pin causes the pain receptor to detect stimulus P2:Pain receptor will trigger nerve impules P3 : Afferent neurone will transmit nerve impulse to the spinal cord P4 : (At spinal cord), interneurone will transmit impulse to the efferent neurone P6 : Interneurone will transmit impulse to the effector/ muscle at leg P7 : Reflex action occur/ the leg will move up. [5 marks] [5 markah]

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(c) Pn Aminah puts pot of orchid by the window in her house for two months. Diagram 7.3 shows the orchids plant grows towards the sunlight. Pn Aminah meletakkan pasu orkid di tepi tingkap dalam rumahnya selama dua bulan. Rajah 7.3 menunjukkan pokok orkid tersebut bertumbuh ke arah cahaya.

Name and describe tropism response shown in Diagram 7.3. Namakan dan huraikan gerakbalas tropisme yang ditunjukkan di dalam Rajah 7.3. [5 marks] [5 markah]

F : Positive phototropism P1 : Shoot tip/ coleoptile orchid contain auxin P2 : Auxin will stimulate the cell growth and elongation/ shoot growth P3 : Auxin diffuse away from light and accumulate at the shaded side P4 : Accumulation of auxin is higher at the shaded side compared to the light side. P5 : Cells on the shaded side will grow more quickly than the cells on the light side P6 : Orchid shoot will bend towards light

F any 4 P

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