CHAPTER 2: CELL STRUCTURE AND FUNCTION No 1 (a)

Mark Scheme

Sub Mark

Total Mark

Able to name Q,R and S. Answer :

(b)

Q: Golgi apparatus / jasad golgi

1

R: Rough endoplasmic reticulum / jalinan endoplasma kasar

1

S: Mitochondria / mitokondrion

1

3

Able to state the function of organelle S Sample answer : F : Generates/ release/produces energy /ATP// Menjana/membebaskan/menghasilkan tenaga/ATP E1: require more energy // memerlukan lebih banyak tenaga E2: to swim/move towards the ovum// untuk berenang/ bergerak ke arah ovum

(c)

1

1 1 3

Able to name two extra cellular enzyme that are secreted by the cell Answers : 1. Trypsin 2. Lipase 3. Amylase

1 1 1 Any 2 2

(d)

Able to explain the reason what will happen to the production of extracellular enzyme if Q and R are absent Sample answers : E1 If R absent protein synthesized (by the ribosomes) cannot be transported throughout to Q. Jika R tiada, protein yang disintesis oleh ribosom tidak dapat diangkut keluar ke Q E2 If Q absent protein /carbohydrate/glycoprotein cannot be modified / packaged /transport (to form secretory substances such as extra cellular enzyme out of the cell) Jika Q tiada, protein/karbohidrat/ glikoprotein tidak dapat diubahsuai/ dibungkus/ dihantar

(e)

1

1

2

Able to state and explain the number of P if the cell undergoes mitosis Sample answers : F : 6

1

E1: mitosis maintain the diploid number of the cell Mitosis mengekalkan bilangan diploid bagi sel

1 2

TOTAL

12

CHAPTER 3: MOVEMENT OF SUBSTANCES ACROSS THE PLASMA MEMBRAN No 1 (a)(i)

(a)(ii)

Mark Scheme

Total Mark

Able to name structure labeled P ,Q and R. Answer P: Cell wall Dinding sel

1

Q: Cytoplasma Sitoplasma

1

R: Vacoule /Cell sap Vakuol

1

3

Able to state the condition Y of cell. Answer Turgid Segah

(b)

Sub Mark

1

Able to explain how conditions X and Y occur. Answer: Condition X: Keadaan X: 





Cell is in normal condition / maintain its shape / not change. Sel dalam keadaan normal / kekal dalam bentuknya / tidak berubah. Solution outside the cell is isotonic to the sap cell of plant cell. Larutan di luar sel adalah isotonic terhadap sap sel tumbuhan. Water diffuses in and out of the cell at equal rate. Air meresap masuk dan keluar dalam kadar yang sama. (Any 2)

Condition Y: Keadaan Y:

1

1

1 2

1



 

Solution outside the cell is hypertonic to cell sap of plant cell. Larutan luar sel adalah hipertonik terhadap sap sel tumbuhan. Water diffuse out from the cell by osmosis. Air meresap keluar daripada sel secara osmosis. Plasma membrane is pulled away from the cell wall. Membrane plasma tertarik menjauhi dinding sel. (Any 2)

1

1 1 2

(c )

Able to explain the effect to the transport of mineral ions into the root hair cell. Answer: 

 

The cell unable to produce energy // energy is not generated. Sel tidah dapat menghasilkan tenaga // tidak dapat menjana tenaga Active transport does not occurs . pengangkutan aktif tidak berlaku. Thus, mineral ions cannot be transported into the cell. Oleh itu, ion mineral tidak dapat dihantar ke dalam sel.

1

1 1

2 (d)

4

( any 2) Able to draw a diagram to show the condition of a leaf cell.  Diagram  Label

2

1 1

2

CHAPTER 4: CHEMICAL COMPOSITION OF THE CELL No 1 (a)

Mark Scheme

Total Mark

Able to name P and Q. Answer P: Sucrose Sukrase Q: Glucose // Fructose Glukosa // Fruktose

(b)

Sub Mark

1 2 1

(i) Able to explain the statement; the action of enzyme sucrase on substrate P is specific Answer F : Enzyme sucrase only acts on (substrate) P // One enzyme only acts on one substrate only. Enzim sukrase hanya bertindak balas dengan (substrat) P// Satu enzim hanya bertindak balas dengan satu substrat sahaja E : The active site (of the enzyme) is specific to certain substrate Tapak aktif (enzim) adalah spesifik/khusus kepada substrat tertentu.

1 2

1

(ii) Able to state two characteristics of enzyme sucrase Answer P1: Enzyme is not destroyed by the reaction Enzim tidak boleh dimusnahkan oleh tindak balas P2: Enzyme has a specific active site Enzim mempunyai tapak aktif yang spesifik

1 1 1

P3: Enzyme can catalyse a reversed reaction Tindak balas enzim adalah berbalik [Any two] [Mana-mana dua]

2

(c)

(i) Able to explain the observation based on diagram 1.2 Answer F : The apple part X remains the same but part Y turns brown/black Epal bahagian X kekal sama/tiada mengalami perubahan tetapi bahagian Y bertukar kepada coklat/hitam P1: Alkaline condition is not suitable for the enzyme. Keadaan alkali tidak sesuai untuk enzim P2: Neutral condition is suitable for the enzyme. Keadaan neutral sesuai untuk enzim

1

1 3 1

P3: The alkali neutralises/change the charges on the active sites of the enzyme// The enzyme cannot start the chemical reaction/oxidation process/no oxidation in part X Alkali meneutralkan/mengubah cas pada tapak aktif enzim// Enzim tidak dapat memulakan tindak balas kimia/proses pengoksidaan/tiada pengoksidaan pada bahagian X

1

[Any three] [Mana-mana tiga] (ii) Able to explain another treatment to avoid sliced apples from turning brown Answer 1

F1: Soak apple in warm/hot water Rendam epal dalam air panas/suam P1: Enzymes are denatured Enzim akan ternyahasli P2: No oxidation process occur Tiada proses pengoksidaan berlaku @ F2: Coat the sliced apple in sugar/oil Salut/Lumur hirisan epal dengan gula /minyak P1: Enzymes are not exposed to air/oxygen Enzim tidak terdedah dengan udara/oksigen P2: No oxidation process occur Tiada proses pengoksidaan berlaku

1 1 3 1 1 1

[Any 1F+2P] [Mana-mana 1F+2P ] JUMLAH

12

CHAPTER 5: CELL DIVISION NO 3 (a)(i)

MARK SCHEME

MARKS

Able to state the type of cell division Answer: Mitosis

1

1

Able to give two reasons for the type of the cell division Sample answer: (a)(ii)

P1: each daughter cells has identical number of chromosome to the parent cell / setiap sel anak mempunyai bilangan kromosom yang seiras/serupa dengan induknya

1

P2: chromosomes in daughter cells have same genetic material to the parent cell./ kromosom di dalam sel anak mempunyai maklumat genetik sama dengan induk

1

1 P3: two daughter cells are produced by each parent cell / dua sel anak dihasilkan bagi setiap induk Any 2 (b)

2

Able to explain the importance of mitosis Sample answer: F1: increases the number of cells / menambahkan bilangan sel

1 1

P1. replacing dead cell/repair the damaged tissue/organ / menggantikan tisu mati/ membaik pulih tisu rosak/organ

1

P2. for growth/development in living organisme / untuk pertumbuhan/ perkembangan organisma hidup

1

F2: to produce genetically identical for daugther cells / untuk menghasilkan sel anak yang seiras

1 1

P3. asexual reproduction (for unicellular organisms) / pembiakan asexual P4. maintain the chromosomal number(of daughter cells) /

Any 2

2

mengekalkan nombor bilangan kromosom (c)(i)

Able to draw the stage of Metaphase I correctly.

Criteria: 2 homologus chromosomesare aligned on either side of the metaphase plate= 1m / 2 kromosom homolog tersususn cecara berpasangan di plat metafasa One chromosome of each pair is attached to the spindle fibre from one pole while its homologue is attached to the fibre from the opposite pole = 1m / satu kromosom dari setiap pasangan melekat pada gentian gelendong di satu kutub manakala kromosom homolog melekat pada gentian di kutub bertetangan

1

1

Sample answer

2

(c)(ii)

Able to explain events during cell division Sample Answer: P1: independent assortment of chromosomes, / pengaturan bebas kromosom P2: which are randomly arranged during metaphase I, / yang secara spontan disusun semasa metafasa 1 P3 : to produce different haploid gametes. / untuk menghasilkan gamet haploid yg berbeza.

1 1 1 Any two

2

(d)

Able to name two chemical substances which cause cancer

Sample Answer:

(e)

P1: benzo-alfapyrene

1

P2: nicotine

1

P3: x-rays

1

P4: ultra-violet rays

1

P5: tar ( in tobacco )

1

P6: formaldehyde

1

P7: gamma rays

1 Any two

2

1

1

Able to name the disease cause by uncontrolled meiosis

Sample answer: Down / Turner/Klinefelter syndrom

TOTAL

12

CHAPTER 6: NUTRITION No

Mark Scheme

3(a)

Sub Total Mark Mark

Able to name X, Y and Z X = glucose Y = amino acid Z = fatty acid and glycerol

3(b)

1 1 1

3

Able to draw one vilus that show the parts and following label.

3(c ) (i)

R- Able to show following parts: - blood capillaries, epithelium, lacteal.

1

L- Able to label any of two from the following: Blood capillaries, epithelium, lacteal,

1

2

1

1

1 1

1

Able to state the process that occurs in the projection -

(ii)

absorption / simple diffusion / facilitated diffusion / active transport (reject diffusion)

Able to state one characteristic of the projections - one cell thick wall / very thin wall - a lot of blood capillaries

3(d)

Able to name the correct process. Deamination

3(e)

(i)

1

1

1

Able to name the circulatory system involved in transporting nutrient X and nutrient Y Blood circulatory system

(ii)

1

Able to describe how nutrient Z is drained back into the circulatory system. - nutrient Z / fatty acid and glycerol absorbed into the lacteal - (nutrient Z / lipid / lymph) travel into lymphatic vessels by contraction of skeletal muscles - (the lymph) from the left side drains into the thoracic duct // the lymph from the right side drains into the right lymphatic duct (the thoracic duct) empties its lymph into the left subclavian vein // (the right lymphatic duct) empties its lymph into the right subclavian vein

1 1

1

3

CHAPTER 7: RESPIRATION No 1 (a)(i)

Mark Scheme

Total Mark

Process R : Anaerobic respiration Process S : Aerobic respiration

Sub Mark 1 1

Reactant : Glucose + oxygen Product : Carbon dioxide + water +

1 1

2

2

(b) 2898 kJ

(c) D1 D2 D3

R Absent of oxygen Glucose is partially oxidised Produce lactic acid

S Present of oxygen Glucose is completely oxidised Do not produce lactic acid

1 1

2

1

D4 Produce less energy / 150kJ/2ATP

 (d)

Produce more energy/2898kJ/36ATP

1

ANY 2D

i)gills ii) P1 : have lamella and filament to increase total surface area P2 : numerous blood capillaries for efficient transport of respiratory gasses

iii) P1 : thin membrane / one cell thick for easily diffusion of respiratory gases. P2 : moist surface for respiratory gases easily dissolve P3 : numerous blood capillaries for efficient transport of respiratory gases

1

1

1 1

2

1 3 1 1 TOTAL

12

CHAPTER 8: DYNAMIC ECOSYSTEM ANSWER a. J : ammonium compound

MARKS 1

K : nitrites

1

L : nitrates

1

b. X : Nitrosomonas sp.

1

Y: Nitrobacter sp.

1

c. Nostoc sp and Rhizobium sp.

2

d. F: Nitrogen exists as animal protein in animal bodies

1

E1 :When animals die, their bodies will be decomposed by

1

decomposers (such as fungi) E2: Converted into ammonium compound e. Z : Denitrifying bacteria

1 1

E1: They breakdown nitrates into nitrogen and oxygen

1

E2 : Returned to the atmosphere

1

TOTAL

12

CHAPTER 9: ENDANGERED ECOSYSTEM No 1 (a)

Mark Scheme Able to state two substances X Answer P1- Carbon dioxide// Karbon dioksida

(c)

1

P2-Nitrogen monoxide / nitrogen dioxide// Oksida nitrogen / nitrogen dioksida

1

P3-Sulphur dioxide / oxides of sulphur// Sulfur dioksida / oksida sulfur

1

P4- Carbon monoxide / soot // Karbon monoksida / jelaga

(b)

Sub Total Mark Mark

1

2

1

1

[any two] [mana-mana dua] Able to give the name of phenomenon as a result by substance X Answer Greenhouse effect Kesan rumah hijau Able to explain how greenhouse effect occur Answer F1- sunlight enters the earth’s atmosphere Cahaya matahari memasuki atmosfera bumi P1- some solar radiation is reflected back to the space Sebahagian pancaran matahari dipantul balik ke luar atmosfera

1

1

P2-most of the radiation is absorb by earth sebahagian/kebanyakan pancaran matahari diserap oleh bumi

1

P3- this warm the surface of Earth Menyebabkan permukaan bumi panas.

1

F2- Deforestation/a lot of logging increasing the greenhouse gasses Penyahutanan /pembalakan meningkatkan kepekatan gas-gas rumah hijau P4- a lot of heat wat trapped by greenhouse gasses banyak memerangkap haba oleh gas rumah hijau

1

1

(d)

(e)

P5-cause increasing of Earth’s temperature/ global warming menyebabkan suhu bumi meningkat / pemanasan global

1

[1F + 2P] Able to state two implication of greenhouse effect Answer P1- acid rain hujan asid

1

P2- increasing of Earth’s temperature peningkatan suhu bumi

1

P3- climate change perubahan iklim

1

P4-polar ice caps melt pencairan ais / paras air laut meningkat

1

P5- loss of biodiversity / loss of habitats of many of flora and fauna kepupusan flora dan fauna / kehilangan biodiversity

1

P6- Any correct answers Mana-mana jawapan yang betul

1

[any two] [mana-mana dua] Able to state the steps to reduce greenhouse effect Answer P1- Limit deforestation Hadkan penerbangan hutan

1

P2-replant the plants after logging tanam semula pokok/ tumbuhan hijau selepas pembalakan

1

P3- Enhance laws to industry and deforestation to control and air pollution Menguatkuasakan undang-undang kepada industri dan pembalakan untuk mengawal pencemaran udara P4- Use the high funnel for factory Gunakan cerobong asap yang tinggi untuk kilang P5- Public must be educated about air pollution and deforestation/ importance of protecting and caring for environment Orang ramai mesti mengetahui tentang pencemaran udara dan pembalakan/ kepentingan menjaga dan menyayangi alam sekitar.

1

1

1

3

2

P6- Do campaigns through mass media, school and social media about the importance of protecting and caring for environment Lakukan kempen melalui madia masa, sekolah, media sosial tentang kepentingan menjaga dan menyayangi alam sekitar. P7- any suitable answers Mana-mana jawapan yang sesuai

1

1

2

[any two] [mana-mana dua] JUMLAH 2 (a) (i)

(ii)

(b) (i)

(ii)

(iii)

Able to give meaning B.O.D Answer B.O.D refers to amount/quantity/level of oxygen that is utilized by microorganism to decompose organic matter in the water. B.O.D ialah kuatiti oksigen yang diperlukan oleh mikroorganisma untuk mengurai bahan organik dalam air.

10

1

1

1

1

1

1

1

1

1

3

Able to state the relationship between B.O.D value with level of water pollution Answer B.O.D value increase, the level of water pollution increase Semakin tinggi nilai B.O.D, semakin tercemar air sungai. Able to state which river water in which zone most polluted Answer Zone T / Zon T Able to name two types of pollutant which discharged into the river in zone T Answer Organic matter // nitrates /phosphate (in fertilizer)// herbiside/ pesticide//domestic waste Bahan organik//nitrat/fosfat(dalam baja) // herbisid/pestisid// bahan kumbahan

Able to describe the changes in the population of bacteria in zone T Answer P1- the population of bacteria increase Populasi bacteria meningkat

P2- because Zone T is polluted by organic matter /nitrates /phosphate / herbiside/ pesticide/domestic waste Kerana sungai dicemari oleh baha organik/nitrat/fosfat/ herbisid/pestisid/bahan kumbahan P3- the supply of organic nutrient/ dissolve oxygen in the water is high Bekalan nutrient organik / oksigen terlarut dalam air adalah tinggi

(c)

Able to state two methods to reduce river pollution. Answer P1- Stop all the activities through out the domestic waste into the river Hentikan semua aktiviti pembuangan bahan kumbahan dalam sungai

1

1

2 1

P2- Treat sewage before enter the rivers Rawatan sisa kumbahan 1 P3- give education about the importance of protecting and caring the river. Pendidikan tentang kepentingan menjaga dan menyayangi sungai. P4- Do campaigns through mass media, school and social media about the importance of protecting and caring our river. Lakukan kempen melalui madia masa, sekolah, media sosial tentang kepentingan menjaga dan menyayangi sungai kita. P5- enhance the law to control water pollution Kuatkuasa undang-undang untuk mengawal pencemaran air/sungai P6- any suitable answers Mana-mana jawapan yang sesuai

1

1

1

1

[any two] [mana-mana dua] Able to opinion and explain the river water is not suitable to be use by villagers as their water sources. Answer F- No/ Tidak 1

3

P1- more microorganism/bacteria/protozoa in water Banyak mikroorganisma dalam air 1 P2- can cause water-borne disease / cholera Boleh menyebabkan penyakit taun

1 JUMLAH

12

CHAPTER 3: MOVEMENT OF SUBSTANCES ACROSS THE PLASMA MEMBRANE QUESTION 1 1 (a) [KB0603 - Measuring Using Number]

Criteria

Markah

Dapat merekod kesemua 4 data jisim akhir telu asin dgn betul Cth jwpn: Kepekatan larutan garam %

Jisim akhir, g

5 15 30 45

73 71 68 66

Dapat merekod 3 data betul Dapat merekod 2 data betul Dapat merekod 1 data betul atau respon salah

3

2 1 0

1 (b) (i) [KB0601 - Observation]

Kriteria Dapat menyatakan pemerhatian yang berbeza dengan tepat: P1 : Kepekatan larutan garam P2- Jisim akhir telur asin selepas direndam selama 1 jam Sample answers: ( Horizontal observations ) 1. Jisim akhir telur asin selepas direndam dlam larutan garam 5% ialah 73 g. 2. Jisim akhir telur asin selepas direndam dalam larutan garam 45% ia lah 66g. 3. Jisim akhir telur asin yang direndam dalam larutan garam 5% paling tinggi berbanding larutan garam 15%,30% dan 45%

Markah

3

DApat menyatakan psatu pemerhatian dgn tepat atau 2 pemerhatian yang kurang tepat. Sample answers: 1. Jisim akhir telur asin selepas direndam dalam larutan garam 5% ialah 73g. 2. Jisim akhir telur asin ygn direndam dalam larutan garam 5% lebih tinggi. 3. Jisim akhir telur asin yg direndam dalam larutan garam

2

45% lebih rendah. Dapat menyatakan pemerhatian di peringkat idea sahaja. Contoh : 1 1. Jisim akhir telur asin bertambah / berkurang. 2. Kepekatan larutan garam semakin tinggi. Tiada respon atau repon salah

0 Penskoran

correct

innacurate

idea

Wrong

2

-

-

-

1

1

-

-

-

2

-

-

1

-

1

-

-

--

2

-

1

-

-

1

-

1

1

-

-

1

-

1

-

-

1

1

Score 3

2

1

0

B (ii) Membuat Inferens Dapat membuat inferens dgn betul berdasarkan kriteria di bawah (manamana 2) P1: laruutan hipertonik / hipotonik terhadap kepekatan zat terlarut dalam telur asin. P2: molekul air meresap keluar / masuk secara osmosis P3: jisim berkurang / bertambah Contoh jawapan: 1. Larutan garam 5%/ a5% bersifat hipotonik terhadap kepekatan zat terlarut telur asin menyebabkan air meresap masuk secara osmosis, jisim bertambah. 2. Larutan garam 30%/ 45% bersifat hipertonik terhadap kepekatan zat terlaut telur asin menyebabkan air mersap masuk secara osmosis, jisim berkurang. 3 Dapat membuat inferens kurang tepat berdasarkan 1 kriteria sahaja. Contoh : 1. LArutan garam 5% /15% menyebabkan air meresap masuk secara osmosis. 2. Larutan garam 30% /45% bersifat hipertonik terhadap zat terlarut.

2

Dapat menyatakan inferens di peringkat idea sahaja. Contoh :

1

1. Jisim telur asin berubah. Tiada respon atau respon salah

( c ) pembolehubah Dapat menyatakan semua pembolehubah dan cara mengendalikannya dengan betul Contoh jawapan:

0

Pembolehubah

Cara mengendalikan pembolehubah 3

Dimanipulasi: Kepekatan larutan garam

Gunakan kepekatan larutan garam yg berbeza (5%, 15%, 30%, 40%)

Bergerak balas 1. Jisim akhir telur asin

2. Peratus perubahan jisim Dimalarkan 1. Jisim awal 2. Isipadu larutan garam yg digunakan 3. Suhu 4. Jenis larutan

1. Ukur dan rekod jisim akhir telur asin menggunakan penimbang. 2. Mengira peratus perubahan jisim dengan menggunakan formula :wal Jisim akhir – jisim awal x 100.% Jisim

1. Tetapkan jisim awal yang sama iaitu 70 g. 2. Tetapkan isipadu larutan garam yang sama iaitu 250ml 3. Gunakan suhu yang sama iaitu suhu bilik 4. Gunakan jenis larutan yg sama iaitu larutan garam

4-5 kotak yang betul

2

2-3 kotak yang betul

1

1-0 kotak yang betul

0

( d) Hipotesis Dapat menyatakn hipotesis dengan menghubungkan pembolehubah dimanupilasi dengan bergerak balas dengan betul berdasarkan kriteria berikut: P1 ; Manipulasi : kepekatan larutan garam P2 :bergerakbalas: Jisim akhir telur asin / peratus perubahan jisim telur 3

P3: hubungan : tinggi / rendah Contoh : 1. Semakin tinggi kepekatan larutan garam semakin berkurang jisim akhir telur asin. Dapat menyatakan hipotesis dengan menghubungkan 2 pembolehubah dengan kurang tepat. Contoh:

2

1. Kepekatan larutan garam yg berbeza menghasilkan jisim akhir telur asin yang berbeza. 2. Kepekatan larutan garam mempengaruhi perubahan peratusan jisim akhir telur asin Dapat menyatakan satu idea bg hipotesis

1

Contoh : 1. Jisim akhir telur asin semakin berkurang. Tiada respon atau respon salah

( e ) (i)

0

Membina jadual

Dapat membina jadual yang mengandungi kriteria berikut: (T) : tajuk dan unit yg betul (D) : semua data betul (C ) : mengira peratus perubahan jisim telur asin Kepekatan Jisim awal Jisim akhir Perbezaan larutan telur asin, g telur asin, g jisim garam % 5 70 73 3

3 Peratusan perubahan jisim, % 4.3

15 70 30 70 45 70 Mana-mana 2 kriteria betul

71 68 66

1 -2 -4

1.4 -2.9 -5.7 2

Mana-mana 1 kriteria betul

1

( e ) (ii) memplot graf Dapat memplot graf dengan lengkap dan betul Paksi (P) : paksi X dan Y berlabel dan berunit,skala seragam - 1 markah Titik (T) : plot 4 titik betul

- 1 markah

Bentuk (B) : menyambung semua titik - 1 markah 3

( f) Hubungan Dapat menyatakan hubungan antara kepekatan larutan garam dengan perubahan jisim telur asin berdasarkan kriteria: R1: menyatakn hubungan markah

-1

E1: molekul air banyak / sedikit / zat terlarut sedikit/ banyak berbanding telur asin - 1 markah

3

E2: molekul air meresap masuk / keluar secara osmosis markah

-1

Contoh : Semakin tinggi kepekatan larutan garam, semakin tinggu perubahan jisim telur asin kerana kandungan zat terlarut dalam larutan garam lebih tinggi berbanding telur asin maka air meresap keluar secara osmosis. Boleh menerangkan hubungan tetapi kurang lengkap R 1 + E1 / E2 Boleh menerangkan hubungan diperingkat idea atau tanpa penerangan

2

1

R 1 sahaja Tiada respon atau respon salah

0

(g) mendefinisi secara operasi Dapat mendefinisikan secara operasi proses osmosis berdasarkan kriteria: P1 : pergerakan /resapan molekul air masuk / keluar merentasi membrane telur asin. P2 : perubahan jisim telur asin / jisim akhir berkurang / bertambah

3

P3 : dipengaruhi oleh kepekatan larutan garam (hipotesis) Contoh : Osmosis ialah pergerakan / resapan molekul air merentasi membrane telur asin yang menyebabkan jisim akhir telur asin berubah dan dipengaruhi oleh kepekatan larutan garam. Mana-mana 2 P yang betul

2

1 P betul

1

Tiada respons atau respons salah

0

(h) Meramal Dapat meramal kepekatan larutan garam yang bersifat isotonic terhadap kepekatan zat terlarut dalam telur asin dgn betul dan menerangkan jawapan berdasarkan kriteria: P1: larutan garam 18% P2: Tiada perubahan jisim P3: jumlah / kadar air yang meresap keluar / masuk merentasi membrane adalah sama

3

Contoh : Larutan garam 18% bersifat isotonic terhadap kepekatan zat terlarut dalam telur asin kerana tiada perubahan jisim berlaku. Hal ini kerana jumlah kadar molekul air yang meresap masuk dan keluar dari sel adalah sama. 2 P betul termasuk P1

2

Hanya P1

1

Tiada respons atau respons salah

0

(i)

Mengelas

Dapat mengelaskan kesemua nutrient berdasarkan penyerapan nutrient yang berlaku dalam vilus dengan betul: Kapilari darah Glukosa Vitamin B Asid amino Semua 6 betul 4-5 betul 2-3 betul 0-1 betul

Lacteal Vitamin D Gliserol Asid lemak

3

2 1 0

CHAPTER 4: CHEMICAL COMPOSITION OF THE CELL Q1 1(a)

Mark Scheme

pH of buffer solution 5 6 7 8 9 (b)

Score

Able to record the data correctly Time taken for iodine solution to remain yellow (min) 28 6 2 6 26

3

(i) Able to state two observations correctly according to 2 criteria: P1 : pH value P2 : time taken for iodine solution to remain yellow Sample answer: 1. At pH 5/7, the time taken for iodine solution to remain yellow is 28/2 minutes. 2. The time taken for iodine solution to remain yellow is 28/2 minutes when use buffer solution at the pH of 5/7. 3. At pH 5/7, the time taken for iodine solution to remain yellow is the longest/shortest.

3

(ii) Able to state two inferences correctly according to 2 criteria: P1 : pH medium P2 : the rate of enzyme/amylase reaction Sample answer: 1. At (slightly) acidic/alkaline medium, the rate of enzyme/amylase reaction is low. 2. At neutral medium, the rate of enzyme/amylase reaction is high/maximum. (c)

Able to state the variables and the method to handle the variables: Variables Method to handle the variable Pemboleh ubah Cara mengendali pemboleh ubah Manipulated variable Experiment is repeated by using pH of buffer solution different pH values (5,6,7,8,9) Responding variable Record the time taken using the Time taken for the iodine solution to stopwatch // remain yellow//

3

3

The rate of enzyme reaction

Calculate the rate of the enzyme reaction using formula: The rate of enzyme = 1 reaction Time taken

Constant variable Concentration /volume of amylase solution// Concentration / volume of starch solution// Temperature of water bath

(d)

Fix the same concentration amylase at 1% / the same volume amylase at 3 ml// Fix the same concentration starch at 1% / the same volume starch at 4 ml // Fix the temperature of water bath at 37°C

Able to state the hypothesis correctly based on criteria: P1 : pH value P2 : time taken for iodine solution to remain yellow//the rate of enzyme reaction/amylase activity P3 : relationship Sample answer

3

1. At pH of buffer solution 7, the time taken for iodine to remain yellow is the shortest. 2. At pH 7, the rate of enzyme reaction/ amylase activity is maximum. (e)

(i) Able to construct a table and record all the data collected in this experiment correctly : Title + UNIT (T) = 1 mark pH 5 6 7 8 9

Time taken for iodine solution to remain yellow (minute) 28 6 2 6 26

Data transfer correctly (D) = 1 mark (ii) Able to draw a graph correctly

Rate of amylase activity on starch (minute-1) 0.04 0.17 0.50 0.17 0.04

3

Calculation (C) = 1 mark 3

(f)

Able to explain the relationship between the rate of amylase activity on starch and the pH values correctly: Sample answer The rate of amylase activity on starch increase with the pH value of the mixture solution until pH of 7. After pH 7, the rate of amylase activity starts to decrease.

(g)

Able to state the operational definition for the hydrolysis of starch by amylase enzyme: 3 Sample answer: Hydrolysis of starch by amylase enzyme refers to the process of breaking down the starch into simple substances when the time for iodine solution to remain yellow is affected by the pH value.

(h)

Able to predict the outcome of this experiment correctly based on criteria: P1 : time taken is longer than 2 minutes P2 : amylase (activity) less active P3 : temperature is low Sample answer: The time taken is longer than 2 minutes because amylase activity becomes less active at low temperature.

3

(i)

Able to classify the materials and apparatus correctly: Variables Pemboleh ubah

Material Bahan

Apparatus Radas

Manipulated

Buffer solution

Boiling tube

Responding

Iodine solution

stopwatch

Constant

Amylase solution

thermometer

Q2 PS

Mark Scheme

3

Score

Able to state the problem statement relating MV to RV correctly based on criteria: P1 (MV) : the temperature (of water) P2 (RV) : the rate of enzyme reaction P3 : question form (What….?)(How…?)(Do…?)

3

Sample answer 1. How to the temperature affects the rate of enzyme reaction? 2. Does the temperature affect the rate of enzyme reaction? H

Able to state the hypothesis by relating MV and RV correctly based on criteria: P1 (MV) : the temperature P2 (RV) : the rate of enzyme reaction P3 (R) : relationship between P1 and P2 Sample answer 1. As the temperature increase, the rate of enzyme reaction also increases. 2. At 37°C/optimum temperature, the rate of enzyme reaction is the highest. 3. The temperature is different, the rate of enzyme reaction also different. 4. As the temperature increase, the rate of enzyme reaction decreases.

3

**accept wrong hypothesis V

Able to state the variables correctly Manipulated variable : the temperature (of water)

1

Responding variable : the time taken for the iodine solution remain yellow//the time taken for the hydrolysis of starch// the rate of enzyme reaction 1 Constant variable

: concentration/volume of enzyme//pH value//concentration of substrate

1

AM

Able to list the materials and apparatus correctly

   

Material Starch suspension Saliva Water bath Iodine solution

      

Apparatus Beaker Tile with grooves Test tube Thermometer Stopwatch Bunsen burner + tripod stand + wire gauze Syringe ALL apparatus and material √= 3 At least 3A + 2M √ = 2 1A + 1M √= 1 1A + 0M // 0A + 1M = 0

Pr

Able to describe the steps of the experiment procedure correctly: K1 : How to set up the apparatus ( at least 3 steps) K2 : How to operate the constant variable (any one) K3 : How to operate the responding variable (any one) K4 : How to operate the manipulated variable (any one) K5 : Precaution//Steps to increase accuracy (any one) Sample answer: 1. Saliva is collected (K1) in a beaker and diluted with an equal volume of distilled water. 2. 5 ml (K2) 1 %(K2) of starch suspension is added (K1) to each test tube, labelled A1, B1, C1, and D1 respectively with a new syringe. 3. 2 ml (K2) of saliva is added to another set of test tubes, labelled A2, B2, C2 and D2. 4. All the test tubes are immersed (K1) respectively into 4 different water baths which are kept at the temperatures of 10°C, 25°C,37°C and 50°C(K4) 5. The test tubes are left (K1) for 10 minutes (K2). 6. The starch suspension in test tube A1 is poured(K1) into the saliva in the test tube A2. 7. A drop of mixture from A2 is dropped (K2) in first groove of the tile containing

the iodine solution. 8. The iodine test is repeated every minute for 10 minutes and the time taken for the hydrolysis of starch to be completed is recorded by using stopwatch (K3). 9. Steps 5 until 8 is repeated (K1) for test tube B, C and D 10. Repeat the experiment twice to get the average reading of time taken (K5) for the hydrolysis of starch to be completed. 11. Record (K1) all the data in a table. List ALL 5K √ = 3 List 3K – 4K √ = 2 List 1K – 2K √= 1 0K = 0 D

Able to present all the data with UNIT correctly based on criteria:

Temperature (°C)

Time taken for hydrolysis of starch to completed (min) 1st reading

Average time taken (min)

Rate of enzyme reaction (min-1)

2nd reading 2

10 25 37 50 1 mark

1 mark

CHAPTER 6: NUTRITION QUESTION 1 1 (a) [KB0603 – Measuring Using Number] Score 3

Criteria Able to record the increase in water temperature correctly. Sample answer: Type of food sample

Increase in water

PI Bread Q / Anchovy R / Cashew nut

temperature 03 09 15

32- 29 38- 29 44- 29

1 (b) (i) [KB0601 - Observation] Score 3

Criteria Able to state correct observations based on the manipulated and responding variables: P1 :Type of foo sample ( Manipulated Variable) P2 : Increase in water temperature (Responding Variable) P3 : Value/data Sample answers: Observation 1 : 1. For P, the increase in water temperature is 3 oC 2. For P or Bread, the final water temperature is 32°C 3. For Q or Anchovy, the increase in water temperature is 9° C. 4. For Q / Anchovy , the final water temperature is 38 C Observation 2: 1. For R / Cashew nut, the increase in water temperature is 15° C. 2. For R / Cashew nut, the final water temperature is 44° C.

1 (b) (ii) [KB0604 - Making inferences] Score Criteria 3 Able to state correct inferences which corresponds to the observation . Sample answer: Inference 1: 1.P / Bread is carbohydrate classes of food which has lowest energy value 2.P / Bread released the least /lowest heat energy which is absorbed by water / has lowest energy value. 3. Q/ Anchovy is protein classes of food which has low energy value. 4. Q / Anchovy release lower / less heat energy which is absorbed by water / has low energy value. Inference 2; 1. R/ Cashew nut is lipid classes of food which has highest energy value. 2. R/ Cashew nut released more /most heat energy which is absorbed by water./ has highest energy value

1(c ) [KB0602 – Classifying] Score Criteria 3 Able to classify the apparatus and materials correctly based on Diagram 2 . Sample answer: Apparatus Needle Retort stand Boiling tube Thermometer

Materials Burning bread Distilled water

1(d) [KB0610 – Controlling Variables] Variables Manipulated: Type of food sample Responding variable: Increase in water temperature // Energy Value

Constant variable: Initial water temperature

Method to handle the variable Used different types of food sample ( bread , anchovy and cashew nut) Measure and record the increase in water temperature using thermometer // Calculate the energy value using formula: 4.2 x water mass x temperature increase Mass of food // Used the same water temperature (29 o C).

** IF VARIABLE IS WRONG, METHOD TO HANDLE IS REJECTED.

1(e) KB0611- Hypothesis Score Criteria 3 Able to state a hypothesis to show a relationship between the manipulated variable and responding variable: P1 : P2 : P3 :

Type of food sample (manipulated variable) Increase in water temperature // Energy Value (responding variable) Relationship

Sample answer : Bread / P has the lowest energy value / increase in water temperature compare to Anchovy(Q) and Cashew nut (R)// vice versa

1(f) [KB0606 – Communicating] Score 3

Criteria Able to draw and fill a table and show all the criteria: T : Title with correct unit D : All data correct L : The level of water pollution Sample answers : Type of

Mass of

Increase in water

food

temperature

sample

sample (g)

( °C)

PI Bread

0.6

03

420

Q/ Anchovy

0.8

09 I

945

R/ Cashewnut

1.2

15

food

Energy Value (Jg-1)

1,050

1(g) (i) [KB0607 – Using spatial and time relationship] Score Criteria 3 Able to plot the bar chart with all criteria: (Paksi) (Titik) (Bentuk)

P : all axis with uniform scale and correct units T : all points is transferred correctly B : correctly bars are plotted

1(g)(ii) [KB0608 – Interpreting Data] Score 3

Criteria Able to explain the relationship between energy value and the type of food sample based on: P1: Hypothesis statement P2: Classes of food P3: Heat energy absorbed by water to increased the temperature Sample answer:

1.Bread /P has the lowest energy value compare to anchovy and cashew nut 2. because it is carbohydrate classes of food 3.least heat energy is absorbed by water to increased the temperature / 3° C of Water OR 1. Cashew nut / R has the highest energy value compare to anchovy and bread 2.because it is protein and lipid classes of food 3. most/ highest heat energy is absorbed by water to increase the temperature /19° C of water.

1 (h) [KB0605] [predicting ] Score 3

Criteria Able to explain prediction of the outcome correctly based on: P1. Name classes of food P2: Energy value P3: Highest heat energy released / absorbed by water Sample answer: 1.S is cobra which has more lipid 2. Its energy value is more than cashew nut / more than 1050 Jg'1 3. Heat energy released is the highest / absorbed by water

(h) [KB0609 – define operationally] Score Criteria 3 Able to define operationally based on: P1: What is energy value P2: How it is determine P3: What factor cause them. Sample answer. 1.Energy value is the quantity of heat (energy) produce by bread /anchovy/ cashew nut/food sample. 2.which is absorbed by water to increase them to 03° C /09 C /15 C. OR determine / shown by the increase in water temperature. 3.The energy value is affected by the type of food sample.

QUESTION 2 Problem Statement Score Criteria 3 Able to state the problem statement of the experiment correctly that include criteria:  Manipulate variables  Responding variables  Relation in question form and question symbol [?] Sample answers: 1. What is the concentration of vitamin C in watermelon (juice) and pineapple (juice)? 2. Does pineapple (juice) contain more concentration of vitamin C than watermelon (juice)?

Hypothesis Score Criteria 3 Able to state the hypothesis correctly according to the criteria:  Manipulate variables  Responding variables  Relationship of the variables Sample answers: 1. Pineapple (juice) contains more concentration of vitamin C than watermelon ( juice). 2. The concentration of vitamin C in pineapple (juice) is higher than in watermelon( juice ).

Variables Score Criteria 3 Able to state the 3 variables correctly. Sample answers: Manipulated variable: Type of fruits // Watermelon and pineapple (juices) Responding variable: concentration of vitamin C Controlled variable: Concentration of DCPIP solution / ascorbic acid solution // Temperature // Volume of DCPIP solution.

Materials and Apparatus Score Criteria 3 Able to state all functional materials= 3 materials and 2 apparatus Materials: Watermelon and pineapple fruits, DCPIP solution, and 0.1% ascorbic acid solution Apparatus: Syringe (with needle), test tube/ beaker, measuring cylinder, muslin cloth / filter paper

Procedure Score Criteria 3 Able to state five procedures P1, P2, P3, P4 and P5 correctly. P1 : How to Set Up The Apparatus (3P1) P2 : How to Make Constant The Control Variable (1P2) P3 : How to Manipulate The Manipulated Variable (1P3) P4: How to Record The Responding Variable (1P4) P5 : Precaution (1P5) 2 Able to state 3-4 of any procedures P1, P2, P3, P4 and P5 correctly 1 Able to state 2 of any procedures P1, P2, P3, P4 and P5 correctly 0 Not able to response or wrong response. Example of Procedure: 1

Prepare fresh juices of watermelon and pineapple. P1 P5 P3

2

Fill a test tube with 1ml 0.1% DCPIP solution (using a syringe) P1 P2 P2 Do not shake the test tube. Fill up a syringe with 0.1% ascorbic acid solution. P1 P2 Add the ascorbic acid into the DCPIP solution, drop by drop / immersed the needle of the syringe in the DCPIP solution / stir with the needle slowly Record the volume of ascorbic acid that decolourised the DCPIP solution using a syringe. Repeat steps 2 until 5 by replacing the ascorbic acid with watermelon and P3 pineapple (juices). Use clean / different syringe P5 Calculate the concentration / percentage of vitamin C in the fruit (juices). P4

3 4 5 6 7 8 9

P1 P3 P5 P1 P2 P5 P1 P2 P1 P5 P4

Concentration of vitamin C = Volume of 0.1% ascorbic acid used mgcm-3 Volume of fruit juice OR Percentage of vitamin C = Volume of 0.1% ascorbic acid used x 0.1% Volume of fruit juice 10 11

Tabulate the result. Repeat the experiment to get average readings.

P1 P5

Presentation of data Score Criteria 2 Able to construct a table of data with 2 criteria:  (i) Correct title and units  (ii) Manipulated variable Sample answers: Fruit (juice) / Solution

Volume needed to decolourise 1ml DCPIP solution ( ml )

Concentration of vitamin C (mgcm-3) OR Percentage of vitamin C( %)

(i)

0.1% ascorbic acid Watermelon Pineapple

Wrong or no response

0

CHAPTER 7: RESPIRATION QUESTION 1 a) Score

Explanation Answer; Apparatus 1. J-tube 2. Ruler 3. Rubber tube 4. Beaker 5. Test tube

Material 1. Potassium hydroxide 2. Water 3. The boy

3

Able to list all material and 4 or 5 apparatus used in the experiment correctly.

2 1 0

Able to list all material and 2 or 3 apparatus correctly. Able to list any one material and one apparatus correctly. No response or wrong response

1 (b) Score

3 2 1 0

Explanation Answer; Data 1: 9.7 cm Data 2: 9.3 cm Data 3: 8.9 cm Able record all three data correctly. Able record any two data correctly. Able record only one data correctly. No response or wrong response

1 (c) (i) Score

Explanation Able to state any two correct observation based on following criteria.

3

2

1 0

P1 – length of air column P2 – sportsman activities 1. After running for 100 metres ,the length of the air column is 9.7 cm. 2. After running for 400 metres, the length of the air column is 9.3 cm. 3. After running for 800 metres, the length of the air column is 8.9 cm. Able to state any one correct observation or two inaccurate response. 1. Running for 100 metres produces higher length of air column. 2. Running faster produces the lower length of air column. Able to state one correct observation or two inaccurate response or idea. 1. Different distances result in different length of air column. No response or wrong response (response like hypothesis)

1(c) (ii) 3

Able to state two reasonable inferences for the correspond to the observation. P1 – amount of air / carbon dioxide P2 – absorbed by potassium hydroxide

2

1 0

1. The longer air column is a result of little amount of air / carbon dioxide being absorbed by potassium hydroxide 2. The shorter air column is a result of more air / carbon dioxide being absorbed by potassium hydroxide Able to state one correct inference and one inaccurate inference. 1. Little air has lost from the air column. 2. Less water has lost from the air column Able to state one correct inference or two inaccurate inference or idea. 1. inference like hypothesis No response or wrong response.

1(d) Score

Explanation Able to state the variable and the method to handle variable correctly (√) for each variable and method Manipulated Variable: The distance taken by the boy to run (√) Method to handle: The boy ran at different distances which were 100 m, 400 m and 800 m (√) Responding Variable: Length of air column (√) Method to handle: Measure and Record the length of air column in J-tube (√)

by using a ruler

Controlled variable : Initial length of air column (√) Method to handle: Measure the initial distance of air column which was 10 cm. (√) 3 2 1 0

Able to get all 6 (√) Able to get 4 – 5 (√) Able to get 2 – 3 (√) No response or wrong response

1(e) Score 3

2

1

0

Explanation Able to state the hypothesis correctly based on the following criteria: P1 (manipulated) – the distance P2 (responding) – length or air column. R - State the relationship between P1 and P2. 1. The farther the distance taken by the boy, the shorter the length of the air column . 2. The content of carbon dioxide increases when the boy ran at a farther distance Able to state the hypothesis but less accurate. Running at a farther distance increases the cellular respiration. Able to state the idea of the hypothesis. The carbon dioxide produced is different when running at different distances. Running at different distance produces different amount of carbon dioxide No response or wrong response

1(f) (i) Score

Explanation Able to construct a table and record the result of the experiment which the following criteria:

3 C – State the distance taken by the boy to run (√) D – Transfer all data correctly / the difference in air column (√) T – calculate percentage of carbon dioxide(unit %) (√) The distance

2 1 0

The difference in air column

Percentage of carbon dioxide (%)

100 0.3 3.0 400 0.7 7.0 800 1.2 12.0 Able to construct a table and record any two criteria Able to construct a table and record any one criteria No response or wrong response

1 (f) (ii) Score

3 2 1 0

Explanation Able to draw the graph for relationship between the distance taken by the boy to run against the percentage of carbon dioxide. P1 – right y-axis and x-axis (√) P2 – Percentage of carbon dioxide (√) P3 – Smooth curve (didn’t tough X-axis or/and Y-axis) (√) Able to get all criteria correct Able to get any two criteria correct Able to get any one criteria correct No response or wrong response

1(g) Score

Explanation Able to interpret data correctly and explain with the following aspect.

3 Relationship: P1 - Able to state the relationship between manipulated and responding variable Explanation: P2 - Able to state the percentage of carbon dioxide released. P3 Able to state the distance taken by the boy to run. Sample Answer: When the distance taken by the boy to run increases, the percentage of carbon dioxide in the exhaled air increases 2 1 0

Able to interpret data correctly with two aspect correctly. Able to interpret data correctly with one aspect correctly. The water absorb is higher/increase. No response or wrong response

1(h) Score 3

Explanation Able to predict and explain the outcome of the experiment correctly with the following aspect. Prediction: P1 – Able to predict the length of air column// percentage of carbon dioxide (12 % or more) Explanation: P2 – Able to state the increase of cellular respirations / most active P3 – Able to state more carbon dioxide produced / anaerobic respiration Sample answer: The length of air column is 8.9 cm (less ) //The percentage of carbon dioxide released by the boy is 12 % / or more / because cellular respiration increases and more carbon dioxide is produce// an anaerobic respiration takes place.

2 1 0

Able to predict based on any two criteria. Able to predict based on any one criteria. No response or wrong response

1(h) Score 3

2 1 0

Explanation Able to state the definition of expired air completely and correctly, based on the following criteria. P1 – contain carbon dioxide P2 – carbon dioxide is absorbed by potassium hydroxide P3 – amount of carbon dioxide produced is influeced by the distance takenSample answer The expired air contains carbon dioxide which can be absorbed by potassium hydroxide and the amount of carbon dioxide produced is influenced by the distance taken by the boy. Able to state the definition of expired air operationally based on any two criteria. Able to state the definition of expired air operationally based on any one criterion or an ideal or hypothesis form. No response or wrong response

SOALAN 2 SOALAN KRITERIA PERNYATAAN Dapat menyatakan pernyataan masalah dengan betul merujuk kriteria MASALAH berikut: C 1: Pembolehubah dimanipulasi C 2: Pembolehubah bergerakbalas C 3: Hubungan (dalam bentuk soalan) (?)

SKOR 3

Contoh jawapan: Adakah kehadiran yis menyebabkan doh/adunan mengembang//Ketinggian

Hipotesis

Dapat menyatakan hipotesis dengan betul merujuk kriteria berikut: C 1: Pembolehubah dimanipulasi C 2: Pembolehubah bergerakbalas C 3: Hubungan

3

Contoh jawpan: Kehadiran yis akan menyebabkan doh/adunan mengembang//Ketinggian doh/adunan //Diameter doh/adunan// ukurlilit doh/adunan meningkat .

Pemboleh ubah

Dapat menyatakan 3 pemboleh ubah dengan betul Contoh jawpan: 1. Dimaipulasi : Kehadiran yis 2. Bergerak balas : Ketinggian doh/adunan//Diameter doh/adunan// ukurlilit doh/adunan 3. Dimalarkan : jisim tepung gandum//masa //jisim gula

3

Radas dan bahan Dapat menyatakan 11 - 13 item ( termasuk * ) BAHAN : tepung gandum, gula, yis, air ( suling) , RADAS : bikar, tabung uji, besen kecil, jam randik, penimbang elektronik, silinder penyukat, tuala lembab, spatula dan pembaris meter

Prosedur

1. Timbang (SA) 50 g tepung gandum(FV) dan masukkan ke dalam sebuah besen kecil . 2. Timbang (SA)10g gula (FV), sukat (SA) 20 ml air suling(FV), 2 spatula yis(FV) dan masukkan kedalam tabung uji. 3. Goncang tabung uji tersebut(SA) supaya kesemua bahan itu terlarut. 4. Masukkan kandungan tabung uji (SA) ke dalam tepung gandum di dalam besen kecil tadi. 5. Maukkan air suling (SA) sedikit demi sedikit dan uli tepung sehingga menjadi doh. 6. Pastikan air yang di tambah tidak berlebihan (PreC) untuk mengelakkan penghasilan doh yang terlalu lembik/cair/melekit. 7. Masukkan doh (SA)ke dalam bikar 250 cm3 dan labelkan sebagai A. 8. Tutup dengan tuala lembab yang bersih(SA) 9. Ukur dan rekod ketinggian awal doh (RV) dalam bikar A 10.Biarkan selama 30 minit (FV) dengan menyukat masa menggunakan jam randik. 11.Ukur ketinggian doh (RV) dalam bikar tersebut selepas 30 minit dan rekodkan dalam jadual. 12 .Ulang langkah 1 hingga 11 untuk adunan yang kedua(MV), tetapi kali ini tanpa menambahkan yis dan labelkan sebagai B. 13. Jika menggunakan besen yang sama, pastikan pelajar mencuci dengan bersih terlebih dahulu besen selepas doh yang pertama dihasilkan (PreC), bagi menghilangkan kesan yis pada besen tersebut sebelum menjalankan eksperimen untuk doh yang kedua.

3

Dapat menyatakan semua berikut: 5 k1 : Penyediaan alat radas 1 k2 : Pembolehubah dimalarkan 1 k3 : Pembolehubah bergerak balas 1 k4 : Pembolehubah dimanipulasi 1 k5 : Langkah berjaga-jaga

Jdual

Kandungan doh/adunan

Doh + yis Doh tanpa yis *unit adalah wajib betul

Ketinggian doh/adunan //Diameter doh/adunan// ukurlilit doh/adunan (cm)

2

CHAPTER 8: DYNAMIC ECOSYSTEM QUESTION 1 No a.

b

ANSWERS Tray Dulang

Dry mass of 10 rice seedlings (g) Jisim kering 10 anak benih padi (g) 1.3 1.7 2.3 2.8 3.1

A B C D E P1: amount of fertilizer P2 : dry mass of 10 rice seedlings P3 : reading

MARKS 3

3

Observation 1 When 2 g of fertilizer is used in tray A, the dry mass af 10 rice seedlings is 1.3g Observation 2 When 10g of fertilizer is used in tray E, the dry mass of 10 rice seedlings is 3.1 g C.

P1 ; Dry mass of 10 rice seedling is the lowest//highest P2 : Amount of fertilizer used is the least//most

3

Inference 1 The dry mass of 10 rice seedlings is the lowest when the amount the amount of fertilizer used is the least Inference 2 The dry mass of 10 rice seedlings is the highest when the amount of fertilizer used is the most d.

Variables Amount of fertilizer used Dried mass / growth of rice seedlings

Method to handle the variable Use different amount of fertilizer in each tray : 2g, 4g, 6g, 8g and 10g Measure and record the dry mass of rice seedlings in each tray using a weighing scale Calculate and record the growth rate of rice seedling using the formula:

Type of soil, the mass

Use same type of soil // fix the mass of soil //

3

e

of soil, amount of fix the amount of water // fix the duration of water, duration of growth// fix the number of rice seedlings // growth, number of rice use the same size of tray. seedling, size of tray P1: The amount of fertilizer used P2 : Growth rate of rice seedling // dry mass P3: Relationship

3

The greater the amount of fertilizer used, the higher the growth rate of rice seedlings // the greater the amount of fertilizer used, the higher the dried mass of rice seedlings. f

Tray

Amount of fertilizer (g)

A B C D E

2 4 6 8 10

Dry mass of 10 rice seedlings (g) 1.3 1.7 2.3 2.8 3.1

Average dry mass of rice seedlings (g) 0.13 0.17 0.23 0.28 0.31

Growth rate (g/day)

3

0.019 0.024 0.033 0.040 0.044

P1: Title with unit P2 : Data transferred P3: Correct calculation g

3

P1 : Axes with unit ; P2: Point; P3: Shape of the graph

h

P1: the amount of fertilizer used increase, the growth rate increase P2 : fertilizer contain more macronutrients//micronutrients P3: form new cells for growth

3

When the amount of fertilizer used is increase, the growth rate is increase. This is because the fertilizer contains more macronutrients and micronutrients. The nutrients are used for plant to form new cells and tissue for growth. i.

P1: Dry mass of rice seedling P2 : is shown by the mass after 10 days P3 : Affected by amount of fertilise

3

Growth rate is the dried mass of rice seedlings per day. The growth is shown by the mass of rice seedling after 10 days. The dried mass is affected by the amount of fertilizer used j.

Manipulated variables Temperature Type of nutrients Duration of watering

Responding variables Length of leaves Volume Height

3

QUESTION 2 ANSWER Problem statement: P1 : distance between the green pea plants P2 : dry mass of the plant // height of the plants R : relationship

MARKS 3

How does the distance between the green pea plants affect the height of the plants? Hypothesis P1 : distance between the green pea plants P2 : dry mass of the plant // height of the plants R : relationship

3

The greater the distance between the green pea plants, the greater the height of the plants

VARIABLES : Manipulated variable: the distance between green pea plants Responding variable : the height // the mass of green pea plants Constant variable : type of soil // size of tray // mass of soil // duration //amount of water

3

3A + 3M APPARATUS Nursery box // tray Meter ruler Electronic balance // weighing balance Oven (dry mass) MATERIALS Garden soil Green pea seedlings Distilled water

PROCEDURE: K1 – steps

3

K2 – constant variable K3 – manipulated variable K4 – responding variable K5 – precaution step 1. soak 100g green peas seedling overnight (K1) 2. Fill two boxes A and B of the same size (2 x2 m) with an equal amount of garden soil (K1 and K2) 3. use certain number of seedlings in the 2 boxes, weight and get the average initial reading (K1) 4. Plant the green pea seedlings as follows Box A : At interval of 2cm apart Box B : At interval of 8cm apart (K3) 5. Water all the boxes with distilled water everyday.(K1) 6. Leave the seeds to germinate and grow for 5 weeks under bright condition.(K1 and K5) 7. After 5 weeks, clean the roots of the seedlings and dry at temperature 100 C in an oven.(K1) 8. Weigh and record the weight of the dried seedlings and get the average reading for Box A and Box B. (K4) 9. Record the result in a table (K1)

2M

PRESENTATION OF DATA Nursery box A B TOTAL

Initial dry mass (g)

5K – 3M 3-4K – 2M 1-2K – 1M 0K - OM

Final dry mass (g) Dry mass (g)

17 M

CHAPTER 9: ENDANGERED ECOSYSTEM QUESTION 1 1 (a) [KB0603 – Measuring Using Number] Criteria Score Able to record the number of solid particles as seen under microscope (10x10) in Table 1 correctly: 3 Sample answers Places where slide is Number of solid particles as seen under microscope located (10x10) Set A Set B Set C Set D

5 8 12 20

Able to measure and record two readings correctly Able to measure and record one readings correctly No response or wrong response.

2 1 0

1 (b) (i) [KB0601 - Observation] Criteria Able to state any two observations correctly according to 2 criteria: P1 - Place where slide is located (Manipulated Variable) P2 - Number of solid particles as seen under microscope (10x10) (Responding Variable)

Score 3

Note: observation must match with inference Sample answers: 1. In set A / air-conditioned room, the number of solid particle as seen under microscop (10x10) is 5. 2. In set B / classroom, the number of solid particle as seen under microscop (10x10) is 8. 3. In set C / school canteen, the number of solid particle as seen under microscop (10x10) is 12. 4. In set D / school car park, the number of solid particle as seen under microscop (10x10) is 20. Able to state any one observation correctly and one inaccurate observation or Able to state any two inaccurate observations ( any 2 criteria) Sample answers: 1. In set A / set D, the number of solid particle as seen under microscope (10x10) is less/more. 2. In set D, the number of solid particle as seen under microscope (10x10) is more than the number of solid particle as seen under microscope (10x10) in set A / set B / set C. .

2

Able to state any one idea of observation.(any 1criteria) Sample answers:

1

1. The number of solid particle as seen under microscope (10x10) in each sets are different 2. Different set give different number of solid particle as seen under microscope (10x10). Not able to response or wrong response.

0

1 (b) (ii) [KB0604 - Making inferences] Criteria Able to make one logical inference for each observation based on the criteria C1: Number of solid particles as seen under microscope (10x10) is less/ more C2: level of air pollution is lower / higher C3 : cleanest / dirtiest environment

Score 3

Note: inference must match with observation Sample answers: 1. In set A, the number of solid particle as seen under microscope (10x10) is less because the level of air pollution is lower indicate that it was the cleanest environment. 2. In set D, the number of solid particle as seen under microscope (10x10) is more because the level of air pollution is high indicate that it was the dirtiest environment. Able to make one logical inference for any one observation. or Able to make one logical and incomplete inference base on one criterion for each observation.

2

Sample answer: 1. In set A/ set D, the level of air pollution is lowest/ higher to show the environment is clean / dirty. 2. In set A/ set D, the number of solid particle as seen under microscope (10x10) is less / more because the level of air pollution is lowest / higher.

Able to make an idea of inference with one criterion.

1

Sample answers 1. Air pollution is lowest / higher 2. Clean/ dirty place. Or any other suitable answer Not able to response or wrong response.

0

1(c) [KB061001 – Controlling Variables] Criteria Able to state 3 variables and methods to handle each variable

Score

Sample answers Variables Manipulated: Location where slide is placed

How the variables are operated Glass slides are put at different location. glass

Responding: Number of solid particles as seen under microscope (10x10) Fixed: Time exposure // size of cellophane tape on the glass slide

Count and record the number of solid particles as seen under microscope (10x10) by using light microscope.

Fix two days for exposure for each set // Use the same size of cellophane tape on each of glass slide

Able to state 4-5 ticks Able to state 2-3 ticks No response or incorrect response of 1 tick only

2 1 0

1(d) KB0611- Making Hypothesis] Criteria Able to state a hypothesis to show a relationship between the manipulated variable and responding variable and the hypothesis can be validated, based on 3 criteria: P1 : Manipulated variable (Places where slide is located) P2 : Responding variable (Number of solid particles as seen under microscope (10x10)) P3 : Relationship

Score 3

Sample answers: 1. Air sample is school park is more // less polluted than air sample in an air-conditioned room / school canteen. 2. The number of solid particles in school park air sample is higher // lower than air sample in an air-conditioned room / class room / school canteen. Able to state less accurate hypothesis to show a relationship between manipulated variable and responding variable base on 2 criteria. Sample answer: 1. Different location of slides have different number of solid particles as seen under a microscope (10x10) 2. Different location of slides influence / affect the number of solid particles as seen under a microscope (10x10

2

Able to state idea of hypothesis to show a relationship between manipulated variable and responding variable base on 1 criterion.

1

Sample answer: 1. Number of solid particles as seen under a microscope (10x10) varied / different 2. Level of air pollution is varied Not able to response or wrong response.

0

1(e) (i) [KB0606 – Communicating] Criteria

Score 3

Able to draw and fill a table with all columns and rows labeled with complete unit T: Tittles 1 mark S: Places where glass slide is located 1 mark D: Number of solid particles as seen under a microscope (10x10) 1 mark Sample answers: Places where slide is located Air-conditioned Class room School canteen School car park

Number of solid particles as seen under microscope (10x10) 5 8 12 20

Able to draw a table with incomplete data Able to draw a table without data Not able to response or wrong response.

2 1 0

1(e) (ii) KB0607 – Space and time relationship Criteria Able to draw a bar chart with 3 criteria: P

Correct title of x-axis and y-axis with unit and uniform scale 1 mark on the axis x-axis: places where the glass slide is located y-axis: Number of solid particles as seen under microscope (10x10) T P (point) : correct data transferred correctly / all points plotted 1 mark B S (Shape): Correct shape (bar graph) 1 mark

Score

Sample answers: Graph of the number of solid particles as seen under microscope (10x10) against the places where slide is located Number of solid particles as seen under microscope (10x10)

20 12 8 5 AirClassroom conditioned room

School canteen

School car park

Places where slide is located

Able to plot a graph with any 2 criteria Able to plot a graph with any 1 criterion Not able to response or wrong response.

2 1 0

1 (f) [KB0608 – Interpreting Data] Criteria Able to state clearly and accurately the relationship between the places where slide is located and the number of solid particles as seen under microscope (10x10):   

P1- places where slide is located P2- number of solid particles as seen under microscope (10x10) P3- reasoning / air pollutants (dust, smoke, soot)

Sample answer: 1. In set A/air-conditioned room, the number of solid particles seen under microscope (10x10) is lowest thus the existence of air pollutants (dust, soot, smoke) also less and not polluted. 2. In set D/ school car park, the number of solid particles seen under microscope (10x10) is highest. This is because the exhaust fumes emit large amount of soot and particles as a result of combustion of fossil fuels. 3. In set B/classroom contain few number of solid particles seen under microscope (10x10) is because the doors and the windows are closed thus less dust and particles in the environment. 4. In set C/ school canteen, the number of solid particles seen under microscope (10x10) is higher compared to in set B/ set A because it was an open air which contains more particulate matter.

Score 3 R+any 2 E’s

Able to state clearly but less accurate the relationship between any two criteria Sample answer: 1. In set A/air-conditioned room / set D/ school car park, the number of solid particles seen under microscope (10x10) is lowest. Able to state the idea of the relationship using one criteria. Sample answer:

2 R+any 1 E

1 R only

1. Number of solid particles seen under microscope (10x10) is depends on places 2. Different places affect the number of solid particles seen under microscope (10x10). Not able to response or wrong response.

0

(g) [KB0609] [Define operationally] Criteria Able to state the definition of air pollution operationally, complete and correctly based on the following criteria: D1: D2: D3:

Score 3

Number of solid particles seen under microscope (10x10). Places where the glass slide is located Level of air pollution

Sample answer: 1. Air pollution is an air sample with the presence of solid particles and the level of air pollution is affected by the location where the glass slides are placed. The higher the number of the solid particles, the higher the level of air pollution. Able to state air pollution operationally base on 2 criteria.

2

Sample answer: 1. Air pollution is the number of solid particles seen under microscope (10x10) which influence / affected by different place. 2. Air pollution is shown by the place where the glass slide is located and the number of solid particle. Able to state air pollution operationally base on 1 criterion or able to state the idea of the air pollution

1

Sample answer: 1. Air pollution is shown by the place where the glass slide is located 2. Air pollution is the number of solid particles seen under microscope (10x10) Not able to response or wrong response.

0

(h) [KB0605 – Predicting] Criteria Able to predict the result accurately base on 2 criteria. P: Expected the number of solid particles seen under microscope (10x10). C1: Condition at the construction site. C2: Level of pollution

Score 3 P+2C

Sample answer: P: The number of solid particles seen under microscope (10x10) is 50 (more compared to set D) C1: Because in the construction area there will be more particulate matter (soot, dirt, dust) compared to set D. C2: This indicates that area has highest level of air pollution compare to other place. Able to predict the result less accurate base on 1 criterion Sample answer:

2 P+1C

The number of solid particles seen under microscope (10x10) is 50 (more compared to set D) because have more dust. Able to give idea of the result.

1 P only

Sample answer: The number of solid particles seen under microscope (10x10) is 50 (more compared to set D) Not able to response or wrong response.

0

(h) [KB0602 – Classifying] Criteria Able to list down all material and apparatus correctly Variable Pembolehubah

Apparatus Radas

Material Bahan

Manipulated Dimanipulasi Responding Bergerak balas

reagent bottle

water sample from various rivers

Fixed Dimalarkan

Score 3

stop watch syringe

0.1% methylene blue solution

Able to list down materials and apparatus 4 or 3 correctly.

2

Able to list down materials and apparatus 2 or 1 correctly.

1

Not able to response or wrong response.

0

QUESTION 2 Question (i)Identify problem statement

Mark scheme Able to write a problem statement relating the manipulated variable to the responding variable correctly.

Marks 3

P1 (MV) : different sources of water P2 (RV) : the level of water pollution P3 : relationship in question form Sample answers: 1. What are the effect different sources of water on the level of water pollution? 2. Does a different source of water affect the level of water pollution? 3. How does a different source of water affect the level of water pollution? Able to state a problem statement less accurately

2

Sample answers: 1. What are the effect sources of water on the water pollution. 2. The level of water pollution affected by different sources of water.

Able to state a problem statement at idea level

No “?” No “question form” 1

Sample answers: 1. Water affect the level of water pollution 2. Water pollution is influenced by location of water. No response or incorrect response

0

(ii)

Able to state the hypothesis based on the following aspects.

3

Hypothesis P1 (MV) : (Different) sources of water P2 (RV) : the time taken for the time taken for the methylene blue solution to decolourise H : Relationship: Level of water pollution, Highest / lowest compare to Sample answers: 1. The water source from Station B has highest level of water pollution compare to water sample from station A and C. 2. The water source from Station C has lowest level of water pollution compare to water sample from station A and B. Able to state the hypothesis less accurately

2

Sample answers: 1. When the water sources collected in B is increases the level of water pollution / time taken of methylene blue to decolurise increase 2. The higher the water sources, the higher the level of water pollution / time taken of methylene blue to decolourise 3. The water sources affects the the level of water pollution / time taken of methylene blue to decolourise Able to state the hypothesis at idea level / based on P1/P2/ wrong concept

1

Sample answers: 1. Water sources causes the water pollution / time taken of methylene blue to decolourise 2. Water pollution affects the water sources 3. The time taken of methylene blue to decolourise increase the level of water pollution No response or incorrect response

0

(iii)

Able to state all the three variables correctly.

3

variables

Sample answers: 1. MV :  (Different) sources of water

*each variable 1M

2. RV :  The time taken for the methylene blue solution to decolourise  The level of water pollution 3. CV :  Concentration / Volume of methylene blue solution  Volume of water sample  Place to keep the water sample after adding methylene blue

(iv) Materials and apparatus

Able to state any two variables correctly.

2

Able to state any one variable correctly.

1

No response or incorrect response

0

Apparatus (A): 1. Reagent bottles (250 ml) with stoppers 2. Syringes 3. Beakers 4. Measuring cylinder 5. Stopwatch 6. Aluminium foil / black paper Materials (M): 1. Methylene blue solution 2. Water samples 3. Distilled water SCORING: 6A + 3M = 3 Any 4-5A + 3M = 2 Any 2-3A + 2M = 1 Less than 2A / 2M = 0 No water sample / methylene blue solution = 0

(v) Procedure

Able to describe the steps of the experiment correctly based on following criterias: K1 : preparation of materials and apparatus (stated 5 times) K2 : operating the constant variable K3 : operating the responding variable K4 : operating the manipulated variable K5 : precaution steps / steps taken to get accurate results / readings

SCORING: 5K = 3 3-4K = 2 1-2K = 1 0K = 0

Sample answers:

K1

Diagram with 4 labels 1. Water samples are collected from the locations of station A, B and C of the river

K1

2. The three reagent bottles are labeled as A, B and C.

K1

3. The reagent bottles are filled with the following water samples with the same volume of:

K1

A :250 ml of water source from location X B: 250 ml of water source from location Y C: 250 ml of water source from location Z by using measuring cylinder

K2 K4

4. The three reagent bottles are wrapped with aluminium foil / black paper to shield the samples from light to prevent the photosynthesis carried out by the algae in the water samples

K1

5. Same volume of 1ml of methylene blue solution is added to the base of each water sample.

K2, K1, K5

6. The reagent bottles are quickly closed by stoppers

K5

7. The test is run for all the water samples on the same day.

K2

8. The three bottles are placed inside a cupboard and the stopwatch is started.

K5 , K3

9. The reagent bottles areexamined from time to time

K3

K5

10. The time taken for methylene blue solution to decolourise for all the three samples is recorded.

K3

11. All the data are tabulated // are recorded into the table.

K3

12. The faster the time taken for methylene blue solution to decolourise for the sample the higher the level of water pollution.

K3

The level of water pollution is calculated using formula: Rate of the pollution =

(vi)

1 Time taken methylene blue to decolourise

Able to present the table with unit correctly. P1- Titles for operating MV and operating RV with units

SCORING:

P2- Responding variable with unit

All titles with unit = 1m

Sample answer: Sources of water from the river

Time taken for methylene blue solution to decolourise, t (hour) or (min)

Rate of water pollution = 1/t (hour -1) or (min -1)

From station X

Name of sources of water from the river = 1m

From station Y From station Z

Any one criteria

1

No response or incorrect response

0

END OF ANSWER SCHEME