CHAPTER 2: CELL STRUCTURE AND FUNCTION

No Mark Scheme Sub Mark

Total Mark

1 (a) Able to name Q,R and S.

Answer :

Q: Golgi apparatus / jasad golgi

R: Rough endoplasmic reticulum / jalinan endoplasma kasar

S: Mitochondria / mitokondrion

1

1

1

3

(b) 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

1

1

1

3

(c) 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

1

1

2

(e) Able to state and explain the number of P if the cell

undergoes mitosis

Sample answers :

F : 6

E1: mitosis maintain the diploid number of the cell

Mitosis mengekalkan bilangan diploid bagi sel

1

1

2

TOTAL

12

CHAPTER 3: MOVEMENT OF SUBSTANCES ACROSS THE

PLASMA MEMBRAN

No Mark Scheme Sub Mark

Total Mark

1 (a)(i)

Able to name structure labeled P ,Q and R. Answer P: Cell wall Dinding sel Q: Cytoplasma Sitoplasma R: Vacoule /Cell sap Vakuol

1

1

1

3

(a)(ii)

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

1

1

(b) 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

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

4

(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. ( any 2)

1

1

1

2

2

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

1 1

2

CHAPTER 4: CHEMICAL COMPOSITION OF THE CELL

No Mark Scheme Sub

Mark Total Mark

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

1 1

2

(b) (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 1

2

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

1 1 1

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 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 [Any three] [Mana-mana tiga]

1

1

1

1

3

(ii) Able to explain another treatment to avoid sliced apples from turning brown Answer 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 [Any 1F+2P] [Mana-mana 1F+2P ]

1

1

1

1

1

1

3

JUMLAH 12

CHAPTER 5: CELL DIVISION

NO MARK SCHEME MARKS

3 (a)(i)

Able to state the type of cell division

Answer:

Mitosis

1

1

(a)(ii)

Able to give two reasons for the type of the cell division

Sample answer:

P1: each daughter cells has identical number of chromosome to

the parent cell / setiap sel anak mempunyai bilangan kromosom

yang seiras/serupa dengan induknya

P2: chromosomes in daughter cells have same genetic material to

the parent cell./ kromosom di dalam sel anak mempunyai

maklumat genetik sama dengan induk

P3: two daughter cells are produced by each parent cell / dua sel

anak dihasilkan bagi setiap induk

1

1

1

Any 2

2

(b) Able to explain the importance of mitosis

Sample answer:

F1: increases the number of cells /

menambahkan bilangan sel

P1. replacing dead cell/repair the damaged tissue/organ /

menggantikan tisu mati/ membaik pulih tisu rosak/organ

P2. for growth/development in living organisme /

untuk pertumbuhan/ perkembangan organisma hidup

F2: to produce genetically identical for daugther cells / untuk

menghasilkan sel anak yang seiras

P3. asexual reproduction (for unicellular organisms) /

pembiakan asexual

P4. maintain the chromosomal number(of daughter cells) /

1

1

1

1

1

1

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

Sample answer

1

1

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:

P1: benzo-alfapyrene

P2: nicotine

P3: x-rays

P4: ultra-violet rays

P5: tar ( in tobacco )

P6: formaldehyde

P7: gamma rays

1

1

1

1

1

1

1

Any

two

2

(e) Able to name the disease cause by uncontrolled meiosis

Sample answer:

Down / Turner/Klinefelter syndrom

1

1

TOTAL

12

CHAPTER 6: NUTRITION

No Mark Scheme Sub

Mark

Total

Mark

3(a) Able to name X, Y and Z

X = glucose

Y = amino acid

Z = fatty acid and glycerol

1

1

1

3

3(b) Able to draw one vilus that show the parts and following

label.

R- Able to show following parts:

- blood capillaries, epithelium, lacteal.

L- Able to label any of two from the following:

Blood capillaries, epithelium, lacteal,

1

1

2

3(c ) (i) Able to state the process that occurs in the projection

- absorption / simple diffusion / facilitated diffusion / active

transport (reject diffusion)

1

1

(ii) Able to state one characteristic of the projections

- one cell thick wall / very thin wall

- a lot of blood capillaries

1

1

1

3(d) Able to name the correct process.

Deamination

1

1

3(e) (i) Able to name the circulatory system involved in

transporting nutrient X and nutrient Y

Blood circulatory system

1

1

(ii) 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 Mark Scheme Sub

Mark Total Mark

1 (a)(i)

Process R : Anaerobic respiration Process S : Aerobic respiration

1 1

2

(b) Reactant : Glucose + oxygen Product : Carbon dioxide + water + 2898 kJ

1 1

2

( c ) R S

D1 Absent of oxygen Present of oxygen D2 Glucose is partially

oxidised Glucose is completely oxidised

D3 Produce lactic acid Do not produce lactic acid

D4 Produce less energy / 150kJ/2ATP

Produce more energy/2898kJ/36ATP

ANY 2D

1 1 1 1

(d) i)gills 1 1 ii)

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

1 1

2

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

3

TOTAL 12

2

CHAPTER 8: DYNAMIC ECOSYSTEM

ANSWER MARKS

a. J : ammonium compound

K : nitrites

L : nitrates

1

1

1

b. X : Nitrosomonas sp.

Y: Nitrobacter sp.

1

1

c. Nostoc sp and Rhizobium sp. 2

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

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

decomposers (such as fungi)

E2: Converted into ammonium compound

1

1

1

e. Z : Denitrifying bacteria

E1: They breakdown nitrates into nitrogen and oxygen

E2 : Returned to the atmosphere

1

1

1

TOTAL 12

CHAPTER 9: ENDANGERED ECOSYSTEM

No Mark Scheme Sub Mark

Total Mark

1 (a) Able to state two substances X Answer P1- Carbon dioxide// Karbon dioksida P2-Nitrogen monoxide / nitrogen dioxide// Oksida nitrogen / nitrogen dioksida P3-Sulphur dioxide / oxides of sulphur// Sulfur dioksida / oksida sulfur P4- Carbon monoxide / soot // Karbon monoksida / jelaga [any two] [mana-mana dua]

1 1 1 1

2

(b) Able to give the name of phenomenon as a result by substance X Answer Greenhouse effect Kesan rumah hijau

1

1

(c) 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

P2-most of the radiation is absorb by earth sebahagian/kebanyakan pancaran matahari diserap oleh bumi P3- this warm the surface of Earth Menyebabkan permukaan bumi panas. 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 1

1 1

1

P5-cause increasing of Earth’s temperature/ global warming menyebabkan suhu bumi meningkat / pemanasan global [1F + 2P]

1

3

(d) Able to state two implication of greenhouse effect Answer P1- acid rain hujan asid P2- increasing of Earth’s temperature peningkatan suhu bumi P3- climate change perubahan iklim P4-polar ice caps melt pencairan ais / paras air laut meningkat P5- loss of biodiversity / loss of habitats of many of flora and fauna kepupusan flora dan fauna / kehilangan biodiversity P6- Any correct answers Mana-mana jawapan yang betul

[any two] [mana-mana dua]

1 1 1 1 1 1

2

(e) Able to state the steps to reduce greenhouse effect Answer P1- Limit deforestation Hadkan penerbangan hutan P2-replant the plants after logging tanam semula pokok/ tumbuhan hijau selepas pembalakan 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 1 1

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 [any two] [mana-mana dua]

1 1

2

JUMLAH

10

2 (a) (i) 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.

1

1

(ii) 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.

1

1

(b) (i) Able to state which river water in which zone most polluted Answer Zone T / Zon T

1

1 (ii) 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

1

1

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

1

3

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

1 1

(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

P2- Treat sewage before enter the rivers

Rawatan sisa kumbahan

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 [any two] [mana-mana dua]

1 1 1 1 1 1

2

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

P2- can cause water-borne disease / cholera

Boleh menyebabkan penyakit taun

1 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 73 15 71 30 68 45 66

3

Dapat merekod 3 data betul 2 Dapat merekod 2 data betul 1 Dapat merekod 1 data betul atau respon salah 0

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

Kriteria Markah 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%

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. Jisim akhir telur asin bertambah / berkurang. 2. Kepekatan larutan garam semakin tinggi.

1

Tiada respon atau repon salah 0

Penskoran

correct innacurate idea Wrong Score

2 - - - 3

1 1 - -

2 - 2 - -

1 - 1 -

1

- -- 2 -

1 - - 1

- 1 1 -

- 1 - 1

0 - - 1 1

B (ii) Membuat Inferens

Dapat membuat inferens dgn betul berdasarkan kriteria di bawah (mana-

mana 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. Jisim telur asin berubah.

1

Tiada respon atau respon salah 0

( c ) pembolehubah

Dapat menyatakan semua pembolehubah dan cara mengendalikannya

dengan betul

Contoh jawapan:

Pembolehubah Cara mengendalikan pembolehubah

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

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

Dimalarkan

1. Jisim awal

2. Isipadu larutan

garam yg

digunakan

3. Suhu

4. Jenis larutan

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

3

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

P3: hubungan : tinggi / rendah

Contoh :

1. Semakin tinggi kepekatan larutan garam semakin berkurang jisim

akhir telur asin.

3

Dapat menyatakan hipotesis dengan menghubungkan 2 pembolehubah

dengan kurang tepat.

Contoh:

1. Kepekatan larutan garam yg berbeza menghasilkan jisim akhir telur

asin yang berbeza.

2. Kepekatan larutan garam mempengaruhi perubahan peratusan jisim

akhir telur asin

2

Dapat menyatakan satu idea bg hipotesis

Contoh :

1. Jisim akhir telur asin semakin berkurang.

1

Tiada respon atau respon salah 0

( e ) (i) 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

larutan

garam %

Jisim awal

telur asin, g

Jisim akhir

telur asin, g

Perbezaan

jisim

Peratusan

perubahan

jisim, %

5 70 73 3 4.3

3

15 70 71 1 1.4

30 70 68 -2 -2.9

45 70 66 -4 -5.7

Mana-mana 2 kriteria betul 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 - 1

markah

E1: molekul air banyak / sedikit / zat terlarut sedikit/ banyak

berbanding telur asin

- 1 markah

3

E2: molekul air meresap masuk / keluar secara osmosis - 1

markah

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

2

Boleh menerangkan hubungan diperingkat idea atau tanpa

penerangan

R 1 sahaja

1

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

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.

3

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

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.

3

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 Lacteal Glukosa Vitamin B Asid amino

Vitamin D Gliserol Asid lemak

Semua 6 betul

3

4-5 betul 2 2-3 betul 1 0-1 betul 0

CHAPTER 4: CHEMICAL COMPOSITION OF THE CELL

Q1 Mark Scheme Score 1(a) Able to record the data correctly

pH of buffer solution Time taken for iodine solution to remain

yellow (min)

5 28

6 6

7 2

8 6

9 26

3

(b) (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.

3

(c) Able to state the variables and the method to handle the variables: Variables

Pemboleh ubah

Method to handle the variable

Cara mengendali pemboleh ubah

Manipulated variable

pH of buffer solution

Experiment is repeated by using

different pH values (5,6,7,8,9)

Responding variable

Time taken for the iodine solution to

remain yellow//

Record the time taken using the

stopwatch //

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

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

(d) 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

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.

3

(e) (i) Able to construct a table and record all the data collected in this experiment correctly :

Title + UNIT (T) = 1 mark

pH Time taken for iodine solution

to remain yellow (minute) Rate of amylase activity on

starch (minute-1)

5 28 0.04

6 6 0.17

7 2 0.50

8 6 0.17

9 26 0.04

Data transfer correctly (D) = 1 mark Calculation (C) = 1 mark

3

(ii) Able to draw a graph correctly

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: 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.

3

(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

3

Q2 Mark Scheme Score

PS 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…?) Sample answer

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

3

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.

**accept wrong hypothesis

3

V Able to state the variables correctly Manipulated variable : the temperature (of water) Responding variable : the time taken for the iodine solution remain yellow//the time taken for the hydrolysis of starch// the rate of enzyme reaction Constant variable : concentration/volume of enzyme//pH value//concentration of substrate

1

1

1

AM Able to list the materials and apparatus correctly

Material Apparatus Starch suspension Saliva Water bath Iodine solution

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) Average time

taken (min)

Rate of enzyme

reaction (min-1)

1st reading 2nd reading

10

25

37

50

1 mark 1 mark

2

CHAPTER 6: NUTRITION

QUESTION 1

1 (a) [KB0603 – Measuring Using Number]

Score Criteria

3

Able to record the increase in water temperature correctly.

Sample answer:

Type of food sample Increase in water

temperature

PI Bread 32- 29 03

Q / Anchovy 38- 29 09

R / Cashew nut 44- 29 15

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

Score Criteria

3

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 Materials

Needle

Retort stand

Boiling tube

Thermometer

Burning bread

Distilled water

1(d) [KB0610 – Controlling Variables]

Variables Method to handle the variable

Manipulated:

Type of food sample

Used different types of food sample ( bread , anchovy

and cashew nut)

Responding variable:

Increase in water temperature //

Energy Value

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 //

Constant variable:

Initial water temperature

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 : Type of food sample (manipulated variable)

P2 : Increase in water temperature // Energy Value

(responding variable)

P3 : 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 Criteria

3

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

food

sample

Mass of

food

sample (g)

Increase in water

temperature

( °C)

Energy

Value (Jg-1)

PI Bread 0.6 03 420

Q/ Anchovy 0.8 09 I 945

R/ Cashewnut 1.2 15 1,050

1(g) (i) [KB0607 – Using spatial and time relationship]

Score Criteria

3 Able to plot the bar chart with all criteria:

(Paksi) P : all axis with uniform scale and correct units

(Titik) T : all points is transferred correctly

(Bentuk) B : correctly bars are plotted

1(g)(ii) [KB0608 – Interpreting Data]

Score Criteria

3 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 Criteria

3 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

P1

P3

P5

2 Fill a test tube with 1ml 0.1% DCPIP solution (using a syringe)

P1 P2 P2

P1

P2

3 Do not shake the test tube. P5

4 Fill up a syringe with 0.1% ascorbic acid solution.

P1 P2

P1

P2

5 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

P1

P5

6 Record the volume of ascorbic acid that decolourised the DCPIP solution

using a syringe.

P4

7 Repeat steps 2 until 5 by replacing the ascorbic acid with watermelon and

pineapple (juices).

P3

8 Use clean / different syringe P5

9 Calculate the concentration / percentage of vitamin C in the fruit (juices).

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 Tabulate the result. P1

11 Repeat the experiment to get average readings. 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( %)

0.1% ascorbic acid

Watermelon

Pineapple

Wrong or no response

0

(i)

CHAPTER 7: RESPIRATION

QUESTION 1

a) Score Explanation

Answer;

Apparatus Material 1. J-tube 1. Potassium hydroxide 2. Ruler 2. Water 3. Rubber tube 3. The boy 4. Beaker 5. Test tube

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

2 Able to list all material and 2 or 3 apparatus correctly. 1 Able to list any one material and one apparatus correctly. 0 No response or wrong response

1 (b) Score Explanation

Answer; Data 1: 9.7 cm Data 2: 9.3 cm Data 3: 8.9 cm

3 Able record all three data correctly. 2 Able record any two data correctly. 1 Able record only one data correctly. 0 No response or wrong response

1 (c) (i)

Score Explanation

3

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

2 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.

1 Able to state one correct observation or two inaccurate response or idea. 1. Different distances result in different length of air column.

0 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

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 2 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

1 Able to state one correct inference or two inaccurate inference or idea. 1. inference like hypothesis

0 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 Able to get all 6 (√) 2 Able to get 4 – 5 (√) 1 Able to get 2 – 3 (√) 0 No response or wrong response

1(e) Score Explanation

3 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 2 Able to state the hypothesis but less accurate.

Running at a farther distance increases the cellular respiration.

1 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

0 No response or wrong response

1(f) (i) Score Explanation

3

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

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

The difference in air column

Percentage of carbon dioxide

(%)

100 0.3 3.0 400 0.7 7.0 800 1.2 12.0

2 Able to construct a table and record any two criteria 1 Able to construct a table and record any one criteria 0 No response or wrong response

1 (f) (ii) Score 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) (√)

3 Able to get all criteria correct 2 Able to get any two criteria correct 1 Able to get any one criteria correct 0 No response or wrong response

1(g) Score Explanation

3

Able to interpret data correctly and explain with the following aspect. 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 Able to interpret data correctly with two aspect correctly. 1 Able to interpret data correctly with one aspect correctly. The

water absorb is higher/increase. 0 No response or wrong response

1(h) Score Explanation

3 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 Able to predict based on any two criteria. 1 Able to predict based on any one criteria. 0 No response or wrong response

1(h) Score Explanation

3 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.

2 Able to state the definition of expired air operationally based on any two criteria.

1 Able to state the definition of expired air operationally based on any one criterion or an ideal or hypothesis form.

0 No response or wrong response

SOALAN 2

SOALAN KRITERIA SKOR PERNYATAAN MASALAH

Dapat menyatakan pernyataan masalah dengan betul merujuk kriteria berikut: C 1: Pembolehubah dimanipulasi C 2: Pembolehubah bergerakbalas C 3: Hubungan (dalam bentuk soalan) (?) Contoh jawapan: Adakah kehadiran yis menyebabkan doh/adunan mengembang//Ketinggian

3

Hipotesis

Dapat menyatakan hipotesis dengan betul merujuk kriteria berikut: C 1: Pembolehubah dimanipulasi C 2: Pembolehubah bergerakbalas C 3: Hubungan Contoh jawpan: Kehadiran yis akan menyebabkan doh/adunan mengembang//Ketinggian doh/adunan //Diameter doh/adunan// ukurlilit doh/adunan meningkat .

3

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. 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

3

Jdual

Kandungan doh/adunan Ketinggian doh/adunan //Diameter

doh/adunan// ukurlilit doh/adunan (cm)

Doh + yis Doh tanpa yis

*unit adalah wajib betul

2

CHAPTER 8: DYNAMIC ECOSYSTEM QUESTION 1 No ANSWERS MARKS a. Tray

Dulang Dry mass of 10 rice seedlings (g) Jisim kering 10 anak benih padi

(g) A 1.3 B 1.7 C 2.3 D 2.8 E 3.1

3

b P1: amount of fertilizer P2 : dry mass of 10 rice seedlings P3 : reading 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

3

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

3

d. Variables Method to handle the variable Amount of fertilizer used

Use different amount of fertilizer in each tray : 2g, 4g, 6g, 8g and 10g

Dried mass / growth of rice seedlings

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

of soil, amount of water, duration of growth, number of rice seedling, size of tray

fix the amount of water // fix the duration of growth// fix the number of rice seedlings // use the same size of tray.

e P1: The amount of fertilizer used P2 : Growth rate of rice seedling // dry mass P3: Relationship 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.

3

f Tray Amount of fertilizer (g)

Dry mass of 10 rice seedlings (g)

Average dry mass of rice seedlings (g)

Growth rate (g/day)

A 2 1.3 0.13 0.019 B 4 1.7 0.17 0.024 C 6 2.3 0.23 0.033 D 8 2.8 0.28 0.040 E 10 3.1 0.31 0.044

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

3

g

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

3

h P1: the amount of fertilizer used increase, the growth rate increase

P2 : fertilizer contain more macronutrients//micronutrients P3: form new cells for growth 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.

3

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

3

j. Manipulated variables Responding variables Temperature Type of nutrients Duration of watering

Length of leaves Volume Height

3

QUESTION 2

ANSWER MARKS

Problem statement:

P1 : distance between the green pea plants

P2 : dry mass of the plant // height of the plants

R : relationship

How does the distance between the green pea plants affect the height of the plants?

3

Hypothesis

P1 : distance between the green pea plants

P2 : dry mass of the plant // height of the plants

R : relationship

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

plants

3

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

APPARATUS

Nursery box // tray

Meter ruler

Electronic balance // weighing balance

Oven (dry mass)

MATERIALS

Garden soil

Green pea seedlings

Distilled water

3A + 3M

3

PROCEDURE:

K1 – steps

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)

5K – 3M

3-4K –

2M

1-2K –

1M

0K - OM

PRESENTATION OF DATA

Nursery box Initial dry mass

(g)

Final dry mass (g) Dry mass (g)

A

B

2M

TOTAL 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: Sample answers

Places where slide is located

Number of solid particles as seen under microscope (10x10)

Set A 5 Set B 8

Set C 12

Set D 20

3

Able to measure and record two readings correctly 2 Able to measure and record one readings correctly 1 No response or wrong response. 0

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

Criteria Score 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) 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.

3

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. 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).

1

Not able to response or wrong response. 0

1 (b) (ii) [KB0604 - Making inferences]

Criteria Score 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

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.

3

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. 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.

2

Able to make an idea of inference with one criterion. Sample answers

1. Air pollution is lowest / higher 2. Clean/ dirty place.

Or any other suitable answer

1

Not able to response or wrong response.

0

1(c) [KB061001 – Controlling Variables]

Criteria Score Able to state 3 variables and methods to handle each variable Sample answers

Variables How the variables are operated Manipulated: Location where glass slide is placed

Glass slides are put at different location.

Responding: Number of solid particles as seen under microscope (10x10)

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

Fixed: Time exposure // size of cellophane tape on the glass slide

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 2 Able to state 2-3 ticks 1 No response or incorrect response of 1 tick only 0 1(d) KB0611- Making Hypothesis]

Criteria Score 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

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.

3

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. Sample answer: 1. Number of solid particles as seen under a microscope (10x10) varied / different 2. Level of air pollution is varied

1

Not able to response or wrong response.

0

1(e) (i) [KB0606 – Communicating]

Criteria Score 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 Number of solid particles as seen under microscope (10x10)

Air-conditioned 5 Class room 8

School canteen 12 School car park 20

3

Able to draw a table with incomplete data 2 Able to draw a table without data 1 Not able to response or wrong response. 0

1(e) (ii) KB0607 – Space and time relationship

Criteria Score Able to draw a bar chart with 3 criteria:

P Correct title of x-axis and y-axis with unit and uniform scale on the axis x-axis: places where the glass slide is located y-axis: Number of solid particles as seen under microscope (10x10)

1 mark

T P (point) : correct data transferred correctly / all points plotted 1 mark B S (Shape): Correct shape (bar graph) 1 mark

Sample answers:

Able to plot a graph with any 2 criteria 2 Able to plot a graph with any 1 criterion 1 Not able to response or wrong response. 0

1 (f) [KB0608 – Interpreting Data]

Criteria Score 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.

3

R+any 2 E’s

Air-

conditioned

room

5

8

12

Places where

slide is located

Number of solid

particles as seen under

microscope (10x10)

20

Graph of the number of solid particles as seen under microscope

(10x10) against the places where slide is located

School

car park

Classroom

School

canteen

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.

2

R+any 1 E

Able to state the idea of the relationship using one criteria. Sample answer: 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).

1

R only

Not able to response or wrong response.

0

(g) [KB0609] [Define operationally]

Criteria Score Able to state the definition of air pollution operationally, complete and correctly based on the following criteria:

D1: Number of solid particles seen under microscope (10x10). D2: Places where the glass slide is located D3: 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.

3

Able to state air pollution operationally base on 2 criteria. 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.

2

Able to state air pollution operationally base on 1 criterion or able to state the idea of the air pollution 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)

1

Not able to response or wrong response.

0

(h) [KB0605 – Predicting] Criteria Score

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

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.

3

P+2C

Able to predict the result less accurate base on 1 criterion Sample answer: 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. Sample answer: The number of solid particles seen under microscope (10x10) is 50 (more compared to set D)

1

P only

Not able to response or wrong response.

0

(h) [KB0602 – Classifying]

Criteria Score Able to list down all material and apparatus correctly

Variable Pembolehubah

Apparatus Radas

Material Bahan

Manipulated Dimanipulasi

reagent bottle water sample from various rivers

Responding Bergerak balas

stop watch

Fixed Dimalarkan

syringe

0.1% methylene blue solution

3

Able to list down materials and apparatus 4 or 3 correctly. 2

2

P+1C

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

Not able to response or wrong response. 0 QUESTION 2

Question Mark scheme Marks

(i)Identify problem statement

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

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?

3

Able to state a problem statement less accurately

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.

2

No “?”

No “question

form”

Able to state a problem statement at idea level

Sample answers:

1. Water affect the level of water pollution

2. Water pollution is influenced by location of water.

1

No response or incorrect response

0

(ii)

Hypothesis

Able to state the hypothesis based on the following aspects.

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.

3

Able to state the hypothesis less accurately

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

2

Able to state the hypothesis at idea level / based on P1/P2/ wrong concept

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

1

No response or incorrect response 0

(iii)

variables

Able to state all the three variables correctly.

Sample answers:

1. MV :

(Different) sources of water

3

*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

Able to state any two variables correctly. 2

Able to state any one variable correctly. 1

No response or incorrect response 0

(iv)

Materials and apparatus

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:

Diagram with 4 labels

K1

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:

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

K1

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

K5

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

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.

The level of water pollution is calculated using formula:

Rate of the pollution = 1 Time taken methylene blue to decolourise

K3

(vi)

Able to present the table with unit correctly.

P1- Titles for operating MV and operating RV with units

P2- Responding variable with unit

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

From station Y

From station Z

SCORING:

All titles with unit = 1m

Name of sources of water from the river = 1m

Any one criteria 1

No response or incorrect response 0

END OF ANSWER SCHEME