Module & More Bio Tg 4

  Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma

BAB 3 Hipotesis 1. Sel tumbuhan menjadi segah dalam larutan hipotonik.
Hypothesis Plant cells become turgid in hypotonic solution.
2. Sel tumbuhan menjadi flasid dalam larutan isotonik.
Pemboleh Plant cells become flaccid in isotonic solution.
ubah 3. Sel tumbuhan mengalami plasmolysis dalam larutan hipertonik.
Variables Plant cells undergone plasmolysis in hypertonic solution.

Bahan (a) Dimanipulasikan: kepekatan larutan
Materials Manipulated: the concentration of solutions
Radas (b) Bergerak balas: keadaan sel tumbuhan
Apparatus Responding: the condition of plant cells
Prosedur (c) Dimalarkan: isi padu larutan, suhu persekitaran
Procedure Constant: Volume of solutions , surrounding temperature

Pemerhatian Larutan sukrosa 0.5 M, larutan sukrosa 1.0 M, air suling , bawang, kertas turas
Observation 0.5 M sucrose solution, 1.0 M sucrose solution, distilled water, onion bulb, filter papers

Mikroskop cahaya, slaid kaca, penutup kaca, forseps, jarum tenggek, penitis, pisau kecil
Light microscope, glass slides, cover slips, forceps, mounting needle, droppers, scalpel

1. Sedia dan labelkan tiga slaid kaca sebagai A, B dan C.
Prepare and label three slides as A, B and C.
2. Kupil satu lapisan epidermis dari permukaan dalam daun sisik bawang.
Peel off a single layer of the epidermal cells from the inner surface of an onion scale leaf.
3. Letakkan lapisan epidermis pada slaid A dan tambah setitis air suling. Tutup dengan penutup kaca.
Place the epidermis on slide A and cover it with a drop of distilled water. Add a cover slip.
4. Perhatikan slaid di bawah mikroskop. Lukis dan labelkan sel-sel epidermis.
Examine the slide under the microscope. Draw and label the epidermal cells.
5. Titis setitis larutan sukrosa 0.5 M di sisi penutup kaca dan biarkan larutan meresap di bawah penutup kaca

dengan meletakkan sekeping kertas turas di sisi kaca bertentangan (Slaid B).
Put a drop of 0.5 M sucrose solution at one side of the coverslip and let the solution to spread out beneath the coverslip

by placing a piece of filter paper at the opposite edge (Slide B).

6. Perhatikan bentuk sel epidermis di bawah mikroskop cahaya. Catat pemerhatian anda.

Observe the shape of the epidermal cells under a light microscope. Record your observations

7. Ulangi langkah 2 hingga 6 dengan menggunakan larutan sukrosa 1.0 M pada slaid C.

Repeat steps 2 to 6 using 1.0 M sucrose solution on slide C.

Slaid Larutan Keadaan Sel Lukisan sel
Slide Solution Condition of Cells Drawing of the cell

A Larutan hipotonik Sel menjadi segah, vakuol
(Sel dalam air suling / Hypotonic solution membesar.
Cells in distilled water)
The cells appear turgid, the
vacuoles swell up

B Larutan isotonik Sel menjadi flasid dan
(Sel dalam 0.5 M larutan Isotonic solution struktur yang normal
The cells flaccid and structure
sukrosa / Cells in 0.5 M
sucrose solution) remain normal

C Larutan hipertonik Sel menjadi plasmolisis,
(Sel dalam 1.0 M larutan Hypertonic solution vakuol menjadi kecil.
The cells plasmolysed,
sukrosa / Cells in 1.0 M vacuoles become smaller.
sucrose solution)

42

Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma 

Perbincangan 1. Apakah jenis larutan yang berikut?
Discussion What is the types of the following solution?

Larutan / Solution Jenis larutan / Types of solution

Air suling / Distilled water Hipotonik / Hypotonic

Larutan sukrosa 0.5 M / 0.5 M sucrose solution Isotonik / Isotonic

Larutan sukrosa 1.0 M / 1.0 M sucrose solution Hipertonik / Hypertonic


2. Terangkan apa yang berlaku di BAB 3

Explain the process in

(a) Slaid A / Slide A

Kesegahah berlaku / Turgidity occurs

Air meresap masuk secara osmosis ke dalam sel tumbuhan . Vakuol mengembang dan sel menjadi segah.

Water diffuses into the cells by osmosis. The vacuoles swell and the cells become turgid.

(b) Slaid B / Slide B
Sel tumbuhan menjadi flasid. / Plant cells become flaccid.
Peresapan air masuk dan keluar sel secara osmosis adalah sama.
Diffusion of water into and out of the cells by osmosis is equal.

(c) Slaid C / Slide C
Plasmolisis berlaku / Plasmolysis occurs
Air meresap keluar daripada sel tumbuhan secara osmosis. Vakuol mengecil dan sitoplasma mengecut.
Water diffuses out from the cell by osmosis. The vacuole shrink and the cytoplasm constrict.

Kesimpulan 1. Sel tumbuhan menjadi segah dalam larutan hipotonik.
Conclusion Plant cells become turgid in hypotonic solution.

2. Sel tumbuhan menjadi flasid dalam larutan isotonik.

Plant cells become flaccid in isotonic solution.

3. Sel tumbuhan mengalami plasmolysis dalam larutan hipertonik.
Plant cells undergone plasmolysis in hypertonic solution.

Eksperimen 3.4 Menentukan kepekatan larutan yang isotonik dengan sap sel tumbuhan

Determine concentration of cell sap of plant which is isotonic to the cell sap of plant

TP 2

Tujuan Mengkaji dan menentukan kepekatan larutan luar yang isotonik kepada sap sel tumbuhan.
Aim To study and determine the concentration of an external solution which is isotonic to the cell sap of plant.

Pernyataan Apakah kepekatan larutan luar sel yang isotonik kepada sap sel tumbuhan?
masalah What is the concentration of an external solution which is isotonic to the cell sap of plant?
Problem
statement Apabila kepekatan larutan luar sel isotonik kepada sap sel ubi kentang, jisim jalur kentang tidak berubah.
The concentration of the solution which is isotonic to the cell sap of potato has no effect on the mass of potato strips.
Hipotesis
Hypothesis a) Dimanipulasikan: kepekatan larutan sukrosa
Manipulated: concentration of sucrose solutions
Pemboleh b) Bergerak balas: perubahan jisim jalur kentang
ubah Responding: the changes of the mass of potato strips
Variables c) Dimalarkan: isi padu larutan, jenis ubi kentang, masa
Constant: volume of the solution, type of potato, time

43

  Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma

Bahan Ubi kentang, air suling, larutan sukrosa 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M dan 0.6 M, kertas turas
Materials Potato, distilled water, sucrose solutions with concentrations of 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M and 0.6 M, filter papers
Radas
Apparatus Piring Petri, penebuk gabus, tabung didih, skalpel, forseps, jam randik, penimbang elektronik, silinder penyukat,
Prosedur pembaris, bikar 50 ml
Procedure Petri dishes, cork-borer, boiling tubes, scalpel, forceps, stopwatch, electronic balance, measuring cylinder, ruler, 50 ml beaker

BAB 3 Pemerhatian 1. Sediakan tujuh tabung didih dan dilabel A, B, C, D, E, F dan G.
Observation Prepare seven boiling tubes and labelled A, B, C, D, E, F and G.

2. Isikan setiap tabung didih dengan larutan masing-masing seperti di bawah.
Fill each of the boiling tube with the following solutions respectively.

A 10 ml air suling / 10 ml of distilled water

B 10 ml larutan sukrosa 0.1 M / 10 ml of 0.1 M sucrose solution

C 10 ml larutan sukrosa 0.2 M / 10 ml of 0.2 M sucrose solution

D 10 ml larutan sukrosa 0.3 M / 10 ml of 0.3 M sucrose solution

E 10 ml larutan sukrosa 0.4 M / 10 ml of 0.4 M sucrose solution

F 10 ml larutan sukrosa 0.5 M / 10 ml of 0.5 M sucrose solution

G 10 ml larutan sukrosa 0.6 M / 10 ml of 0.6 M sucrose solution


3. Peroleh jalur ubi kentang dengan menggunakan penebuk gabus.
Obtain strips of potato using a cork borer.

4. Ukur dan potong tiap jalur kentang sepanjang 50 mm.
Measure and cut each strip into a length of 50 mm.

5. Lap jalur kentang dengan kertas turas.
Wipe dry the strips with a piece of filter paper.

6. Timbang dan rekod jisim setiap jalur kentang.

Weigh and record the mass of each potato strip.

7. Rendam satu jalur kentang ke dalam setiap tabung didih A hingga G, dan biarkan selama 1 jam.

Place one strip of potato into each of the boiling tubes A to G, and left to stand for an hour.

8. Selepas 1 jam, keluarkan jalur kentang dari tabung didih dengan forsep dan lap dengan menggunakan
kertas turas.

After an hour, remove each potato strip from the boiling tube using the forceps and wiped dry with filter paper.

9. Timbang jalur kentang semula dan catat jisim akhir.
Weigh the mass of each strip again and recorded.

10. Perhatikan tekstur dan rupa jalur kentang.
Observe the texture and appearance of the potato strip.

1 1. Rekod keputusan dalam jadual.
Record the results in a table.

12. Lukis graf peratus perubahan jisim melawan kepekatan larutan sukrosa.

Draw a graph of the percentage changes in mass against the concentration of sucrose solution.

Tabung Kepekatan Jisim jalur kentang (g) Perubahan Peratus perbezaan Tekstur dan
didih larutan Mass of the potato strip (g) jisim rupa
Boiling jisim (%)
tube Concentration Awal Akhir Difference in Percentage difference in Texture and
of solution Initial End mass (g) appearance
A mass (%)
air suling Keras
distilled water Firm

larutan Keras
sukrosa 0.1 M Firm
B 0.1 M sucrose

solution

44

Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma 

larutan
sukrosa 0.2 M Keras
C 0.2 M sucrose Firm
Keras
solution Firm
Keras
larutan Firm
sukrosa 0.3 M Lembut
D 0.3 M sucrose soft
Lembut
solution soft

larutan BAB 3
sukrosa 0.4 M
E 0.4 M sucrose

solution

larutan
sukrosa 0.5 M
F 0.5 M sucrose

solution

larutan
sukrosa 0.6 M
G 0.6 M sucrose

solution

Perbincangan 1. Bagaimanakah kepekatan sap sel kentang dapat ditentukan?
Discussion
How can the concentration of cell sap of potato be determined?

Kepekatan di mana persilangan graf pada paksi X.
The concentration where the curve crosses the X-axis.

2. Berdasarkan graf, tentukan kepekatan larutan dalam sap sel kentang.
Based on the graph, determine the solute concentration of the potato cell sap.
Berdasarkan graf murid / Depends on student’s graph

3. Apakah hubungan antara perubahan jisim jalur kentang dengan kepekatan sap sel kentang?
What is the relationship between the changes of mass of the potato strips and the cell sap?
Apabila perubahan purata jisim jalur kentang adalah sifar, tidak ada pergerakan bersih air masuk dan
keluar dari sel-sel kentang. Kepekatan larutan adalah isotonik kepada sap sel kentang.
When the average change in mass of the potato strips is zero, there is no net movement of water in and out of the potato
cells. The concentration of the solution is isotonic to the plant cell sap.

4. Apakah hubungan antara perubahan jsism jalur kentang dan teksturnya?
What is the relationship between the changes in mass of the potato strips and their texture?
Jisim meningkat, tekstur menjadi lebih keras (sel segah) / Jisim mengurang, tekstur menjadi lembut (sel flasid).
Increase in mass, the texture becomes harder (turgid) / decrease in mass, the texture becomes softer (flaccid).

Kesimpulan Berdasarkan graf, kepekatan sap sel tisu kentang adalah......., iaitu kepekatan larutan sukrosa yang tidak
Conclusion mempunyai kesan ke atas jisim jalur kentang.

Based on the graph, the concentration of the cell sap of potato tissue is……., that is the concentration of sucrose solution
which has no effect on the mass of potato strips.

45

  Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma

3.4 Pergerakan Bahan Merentasi Membran Plasma dalam Kehidupan Harian

Movement of Substances Across a Plasma Membrane and Its Application in Daily Life

1. Hubung kait kepekatan sap sel suatu tisu tumbuhan dengan fenomena kelayuan tumbuhan: TP 2

Correlate the concentration of cell sap in a plant tissue with the phenomenon of plant wilting:.

Fenomena kelayuan tumbuhan / The phenomenon of plant wilting

BAB 3 (a) Penggunaan baja (d) Tumbuhan menjadi layu .
The plant wilts .
berlebihan menyebabkan
air tanah menjadi (c) Sel tumbuhan mengalami
plasmolisis .
hipertonik terhadap sap
Plant cells become plasmolysed .
sel akar.
The excessive use of fertilisers

causes the soil solution
to become hypertonic to the

cell sap of root cells.

(b) Air meresap keluar daripada sel ke dalam tanah secara osmosis .
Water diffuses out from the cell into the soil by osmosis .

2. Contoh aplikasi konsep pergerakan bahan merentasi membran plasma dalam kehidupan harian:

Examples of the application of the concept of movement of substances across a plasma membrane in daily life:

(a) Ikan masin (b) Jeruk buah-buahan dan sayur-sayuran
Salted fish Pickled fruits and vegetables

(i) Kepekatan garam yang tinggi menyebabkan air

meresap keluar dari sel makanan secara osmosis . (i) Kepekatan gula yang tinggi menyebabkan larutan
The high concentration of salt causes the water difuses pengawet hipertonik terhadap makanan yang
diawet.
out of the food cells by osmosis .
The high concentration of sugar makes the preservative
(ii) Makanan menjadi kering dan pada masa yang solution hypertonic to the preserved food.

sama bakteria serta kulat akan kehilangan air dan (ii) Air meresap keluar dari sel makanan secara
osmosis.
akhirnya mati.
The food becomes dehydrated, while bacteria and fungi The water diffuses out of the food cells by osmosis.

will lose water at the same time and eventually die.

46

Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma 

Contoh Aplikasi Konsep Pergerakan Bahan Merentasi Membran Plasma dalam Kehidupan Harian

Examples of the Application of the Concept of Movement of Substances Across a Plasma Membrane in Daily Life

1 Minuman isotonik bagi atlet 3 Minuman penghidratan semula bagi pesakit cirit birit TP 3

Isotonic drinks for athletes Rehydration drinks for diarrhea patients

(a) Dapat mengganti air , elektrolit dan (a) Dapat mengembalikan kehilangan air dan elektrolit dalam pesakit BAB 3
tenaga badan yang hilang melalui loss.
perpeluhan . cirit-birit.
To help diarrhea patients replace water and electrolytes
To help athletes replace water , electrolytes  ,
and energy loss by sweating. 4 Liposom dalam bidang perubatan, kosmetik dan lain-lain.

2 Larutan saline dalam bidang perubatan Liposome in the field of medicine, cosmetic industry, etc.

Saline solution in the medical field

Ubat Bahan
Drugs hidrofilik aktif
Hydrophilic
actives substance

(a) Dapat membekalkan air dan garam (a) Untuk menghantar ubatan . (b) Untuk mengangkut bahan aktif
To deliver drugs . penting ke dalam kulit . active
(NaCl) kepada badan.
To supply air and salt (NaCl) to the body. To transport important
substances into the skin .

Proses Osmosis Berbalik dalam Penulenan Air

Reverse Osmosis in Water Purification

(a) Aplikasi tekanan digunakan untuk mengatasi tekanan osmosis .
An applied pressure is used to overcome osmotic pressure.
RO

Aplikasi tekanan UF TUS Air tulen
Applied pressure UV Purified water
(b) Membran telap memilih digunakan Proses osmosis berbalik Air turasan
untuk menyingkir ion, molekul Air turasan The reverse osmosis process
Unfiltered Membran telap memilih Filtered
tercemar dan zarah besar daripada water Selectively permeable membrane water

air tanpa ditapis. Bahan tercemar
A selectively permeable membrane Contaminants

is used to remove ions, unwanted
molecules and larger particles from
unfiltered water.

Water Flow / Aliran air

47

  Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma 3

PRAKTIS SPM

Soalan Objektif

1. Rajah 1 menunjukkan satu eksperimen ringkas. IV Mengikut kecerunan kepekatan
Diagram 1 shows a simple experiment. KBAT Mengaplikasi Follow concentration gradient
2018
Awal Keseimbangan A I dan II / I and II C III dan IV / III and IV
At beginning At equilibrium B II dan III / II and III D I dan IV / I and IV
BAB 3
3. Rajah 3 menunjukkan satu eksperimen.
Diagram 3 shows an experiment.
X 2018
Bahan terlarut
Pelarut Solute
Solvent

Larutan pekat
Concentrated

solution

Rajah 1/ Diagram 1 Pelarut
Solvent
Berdasarkan keputusan eksperimen tersebut, apakah
ciri yang ditunjukkan oleh X? Rajah 3/ Diagram 3

Based on the resultsof the experiment,what is thecharacteristic Apakah proses yang berlaku dalam eksperimen ini?
shown by X?
A Ketelapan penuh C Kurang telap What is the process that occurs in the experiment?

Fully permeable Less permeable A Pengangkutan aktif C Resapan berbantu
B Telap memilih D Tidak telap Active transport Facilitated diffusion
B Resapan ringkas D Osmosis
Selectively permeable Non-permeable
Simple diffusion Osmosis

2. Rajah 2 menunjukkan pergerakan suatu bahan 4. Rajah 4 menunjukkan pergerakan ion mineral (K+, H+,
merentasi membran plasma. NH4+, HPO42-) dalam rambut akar tumbuhan.

2 018 Diagram 2 shows the movement of a substance across a Diagram 4 shows the movement of mineral ions (K+, H+, NH4+,
plasma membrane. HPO42-) in the root hair of a plant.

Di luar sel
Outside of cell

ATP
Di dalam sel
Inside of cell

Rajah 2/ Diagram 2 Rajah 4/ Diagram 4

Antara yang berikut, pernyataan yang manakah Apakah proses pergerakan ion mineral tersebut?
benar tentang pergerakan bahan itu? What is the movement process of the mineral ions?

Which of the following statements are correct about the A Pengangkutan aktif C Resapan berbantu
movement of that substance? Active transport Facilitated diffusion
I Melalui protein liang /Across pore protein B Resapan ringkas D Osmosis
II Melalui protein pembawa /Across carrier protein
III Memerlukan tenaga /Energy is required Simple diffusion Osmosis

48

Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma 

5. Rajah 5 menunjukkan keadaan sel tumbuhan B Hipertonik Plasmolisis
sebelum dan selepas direndam dalam larutan P. Hypertonic Plasmolysis
Deplasmolisis
Diagram 5 shows the condition of plant cells before and after C Hipotonik Deplasmolysis
being immersed in solution P. Hypotonic Deplasmolisis
Deplasmolysis
D Hipertonik
Hypertonic

Sebelum Selepas 6. Apakah proses yang berlaku apabila sel darah merah BAB 3
direndam direndam direndam dalam larutan hipotonik?
Before immersion After immersion
What is the process that occurs when a red blood cell is
Rajah 5/ Diagram 5 immersed in a hypotonic solution?

Apakah larutan P dan proses yang terlibat? A Hemolisis / Haemolysis
What is solution P and the process involved? B Krenasi / Crenation
C Deplasmolisis / Deplasmolysis
Larutan P / Solution P Proses / Process D Plasmolisis / Plasmolysis

A Hypotonic Plasmolisis
Hypotonic Plasmolysis

Bahagian A
1. Rajah 1.1 menunjukkan struktur membran plasma.
Diagram 1.1 shows the structure of plasma membrane.

2019

Vitamin K

Vitamin K Vitamin K Na+
Na+
Vitamin K
Na+

Vitamin K ATP ADP Y X

Vitamin K Na+ +Pi [2 markah / marks]

Na+ Na+
Na+

Rajah 1.1 / Diagram 1.1

(a) Apakah X dan Y?
What are X and Y?
X : Kolesterol / Cholesterol
Y : Protein pembawa / Carrier protein

49

BAB 3   Biologi  Tingkatan 4  Bab 3 Pergerakan Bahan Merentasi Membran Plasma

(b) Rajah 1.2 menunjukkan molekul fosfolipid dalam membran plasma.
Diagram 1.2 shows the structure of phospholipid molecule in the plasma membrane.

A

B

Rajah 1.2 / Diagram 1.2

Nyatakan ciri A dan B.
State the characteristics of A and B.

A : Hidrofilik / Hydrophilic

B : Hidrofobik / Hydrophobic

[2 markah / marks]

(c) Berdasarkan Rajah 1.1, terangkan pergerakan vitamin K merentasi membran plasma.
Based on Diagram 1.1, explain the movement of vitamin K across the plasma membrane.

Vitamin K adalah kecil dan larut lemak / Vitamin K is small and lipid soluble

Meresap melalui dwilapisan fosfolipid secara resapan ringkas

Diffuse across the phospholipid bilayers through simple diffusion

Dari kawasan berkepekatan tinggi ke kawasan berkepekatan rendah

From the area of high concentration to the area of lower concentration

[2 markah / marks]

(d) Kehadiran merkuri merencat pengangkutan ion natrium merentasi membran plasma. Terangkan.
The presence of mercury inhibits the transport of sodium ions across plasma membrane. Explain.

Merkuri merencat respirasi sel / Mercury inhibits cellular respiration

Tiada tenaga / ATP dijana oleh mitokondrion / No energy / ATP is generated by mitochondrion

Protein pembawa tidak berubah bentuk / Carrier protein does not change its shape

Pengangkutan aktif tidak berlaku / Active transport does not occur

[3 markah / marks]

Kuiz 3

50

BAB Komposisi Kimia dalam Sel

4 Chemical Composition in a Cell

PETA Konsep

Komposisi Kimia dalam Sel BAB 4
Chemical Composition in a Cell

Air (H2O) Asid Nukleik
Water Nucleic acid

Sifat-sifat air Protein DNA RNA
Properties of water
Karbohidrat Lipid
Kekutuban air Carbohidrate Lipid
Polarity of water Monosakarida
Monosaccharide Lemak
Muatan haba Disakarida Fats
tentu air Disaccharide
Specific heat Polisakarida Lilin
capacity of water Polysaccharide Wax

Daya lekitan Fosfolipid
dan daya Phospholipid
lekatan air
Cohesive and Steroid
adhesive forces Steroids
of water

5511

  Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel

4.1 Air

Water

Sifat-sifat Air TP 2

The Properties of Water

Formula kimia air adalah H2O .

The chemical formula of water is H2O .

O Atom oksigen Cas separa negatif δ− Ikatan hidrogen
Oxygen atom Partial negative charge δ+ Hydrogen bond
δ– O δ–
H

BAB 4 HH δ+ δ− O
H
δ+ H H δ+ δ− δ+
104.5°

Atom hidrogen Cas separa positif δ+ δ−
Hydrogen atom Partial positive charge

a Satu atom oksigen b Molekul air adalah berkutub ; satu c Atom hidrogen satu molekul air tertarik

bergabung dengan dua hujung molekul mempunyai cas kepada atom oksigen daripada molekul
atom hidrogen oleh ikatan separa positif dan satu lagi cas air jiran dan membentuk ikatan hidrogen .
kovalen . separa negatif .
One oxygen atom combines The hydrogen atom of one water molecule is
with two hydrogen atoms by Water molecules are polar ; one attracted to the oxygen atom of a neighbouring
water molecule forming a hydrogen bond.
covalent bond. end of the molecule bears a partial
positive charge and the other a

partial negative charge.

Sifat-sifat Air dan Kepentingannya dalam Sel TP 2

The Properties of Water and Its Importance in the Cell

1 Air mempunyai kekutuban dan Sifat-sifat 2 Air mempunyai muatan haba tentu
air dan
adalah pelarut bagi kebanyakan yang amat tinggi . Ini membolehkan air
molekul biologi. kepentingannya menyerap banyak tenaga haba dengan
Due to its polarity , water is a solvent dalam sel kenaikan suhu yang kecil. Tasik dan
for many biological molecules.
• medium untuk tindak balas The properties of water lautan mempunyai suhu yang stabil
and its importance in
biokimia. dan menjadikannya persekitaran yang
a medium for biochemical reactions. the cell sesuai untuk organisma akuatik .

3 Air ialah medium pengangkutan Water has a high specific heat capacity. A lot
of energy is required to raise the temperature
untuk membawa bahan terlarut of water. The lakes and ocean have
relatively constant temperatures which
seperti karbon dioksida, urea, creating a thermally stable environment for
nutrien dan hormon . aquatic organisms.
Water acts as a transport medium
for many dissolved substances, 4 Molekul air mempunyai daya lekitan dan lekatan yang tinggi. Ini membolehkannya
such as carbon dioxide, urea, nutrients
and hormones . bergerak dalam salur xilem tumbuhan bagi memastikan pengaliran air yang

berterusan dari akar ke daun.
Water molecules have large cohesive and adhesive forces. This enables movement through the

xylem vessel in plants to maintain a continuous flow of water from the roots to the leaves.

52

Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel 

4.2 Karbohidrat

Carbohydrates

TP 1

1. Karbohidrat adalah sebatian organik. Bahan ini terdiri daripada CH2OH
unsur karbon (C), hidrogen (H) dan oksigen (O).
Karbohidrat C O
Carbohydrates are organic compounds. They contain the elements HH OH
Carbohydrates carbon (C), hydrogen (H) and oxygen (O).
CC
2. Nisbah H:O ialah 2:1 .
Ratio of H:O is 2:1 . HO OH HH

CC

H OH

Jenis karbohidrat 3. Struktur molekul glukosa BAB 4

Types of carbohydrate Structure of a glucose molecule
C6H12O6

a Monosakarida b Disakarida c Polisakarida

Monosaccharides Disaccharides Polysaccharides

(i) Bentuk paling ringkas atau (i) Terbentuk daripada kondensasi (i) Rantai polimer monosakarida yang digabung
monomer karbohidrat. dua molekul monosakarida . secara kondensasi .

The simplest form or Formed by condensation of two Polymers of monosaccharides joined by
monomers of carbohydrates. condensation .
monosaccharides molecules.
(ii) Sebagai sumber tenaga . (ii) Boleh dihidrolisis kepada monosakarida
As a source of energy . (ii) Boleh diuraikan kepada secara pemanasan dengan asid atau
melalui tindak balas enzim .
(iii) Manis dan larut dalam monosakarida melalui
air . hidrolysis . Can be hydrolysed into monosaccharides by heating

Taste sweet and are soluble in Can be broken down into with acids or by enzymatic reactions.

water . monosaccharides by hydrolysis . (iii) Tidak manis dan tidak larut dalam air .

(iv) Sejenis gula penurun . (iii) Manis dan larut dalam Do not taste sweet and are insoluble in water .
air .
A reducing sugar.
Sweet and are soluble in
water .

(iv) Adalah gula penurun kecuali .
sukrosa .

Are reducing sugars except sucrose

Formula am / General formula: Formula am / General formula: Formula am / General formula:
C6H12O6 C12H22O11 (C6H10O5)n

53

  Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel

Monosakarida Disakarida Polisakarida
Monosaccharides Disaccharides Polysaccharides

Contoh Sumber Contoh Sumber makanan Contoh Sumber makanan
makanan
Example Example Food source Example Food source
Food source
(f) Kanji Nasi dan Ubi kentang
(a) Glukosa dan Buah dan (c) Maltosa Barli yang bercambah
Fruktosa madu Maltose Germinating barley Starch Rice and potatoes

Glucose and Fruits and honey (g) Glikogen -
fructose Hasil tumbuhan
(d) Sukrosa Tebu Glycogen
Plant products
Sucrose Sugar cane (h) Selulosa

BAB 4 (b) Galaktosa Susu dan hasil (e) Laktosa Susu dan hasil tenusu Cellulose
tenusu Lactose Milk and dairy products
Galactose
Milk and dairy
products

Kondensasi / Condensation Kondensasi / Condensation

(i) Glukosa + Glukosa Maltosa + air (j) Glukosa + Fruktosa Sukrosa + air

Glucose + Glucose Maltose + water Glucose + Fructosa Sucrose + water

Hidrolisis / Hydrolysis Hidrolisis / Hydrolysis

Kondensasi / Condensation

(k) Glukosa + Galaktosa Laktosa + air

Glucose + Galactose Lactose + water

Hidrolisis / Hydrolysis

Kepentingan Karbohidrat dalam Sel TP 2

The Importance of Carbohydrates in Cells

2 Selulosa adalah sumber serat diet. Kepentingan karbohidrat 1 Sebagai sumber utama tenaga .
dalam sel
Cellulose is a source of dietary fibre. As the main energy source.
The importance of carbohydrates
4 Glikogen disimpan dalam hati in cells 3 Glukosa diuraikan semasa respirasi

atau sel otot. untuk membebaskan tenaga.
Glycogen is stored in the liver or Glucose is directly broken down in

muscle cells as energy reserved. respiration to release energy.

5 Kanji adalah karbohidrat simpanan dalam tumbuhan.

Starch is the storage form of carbohydrates in plants.

54

Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel 

4.3 Protein

Proteins

a Struktur molekul satu asid amino (iii) Protein adalah polimer organik yang terdiri daripada rantai panjang TP 1
asid amino . Protein
Molecular structure of an amino acid Proteins
Proteins are large organic polymers comprised of long chains of amino acids.
R
(iv) Monomer amino asid mempunyai unsur karbon (C), hidrogen (H),
HO oksigen (O) dan nitrogen (N). Sesetengah asid amino mengandungi

N CC sulfur (S) dan fosforus (P).
Amino acid, the monomer contains elements carbon (C), hydrogen (H),
H OH
H oxygen (O) and nitrogen (N). Some amino acids contain sulphur (S)
and phosphorus (P).
(i) Kumpulan (ii) Kumpulan BAB 4
amino karboksil b Jenis asid amino Diperoleh daripada diet .

Amino group Carboxyl group Types of amino acid Have to be obtained from the diet .

(iv) Asid amino perlu
Essential amino acids

(iii) Asid amino tak perlu Dapat disintesis oleh badan .
Non-essential amino acids Can be synthesised by our body .

(i) Dipeptida (ii) Polipeptida
Dipeptides Polypeptides

• Dipeptida terbentuk daripada kondensasi dua • Kombinasi dipeptida dengan asid amino lain membentuk polipeptida.
asid amino. Combination of a dipeptida dengan asid amino lain membentuk polipeptida.

A dipeptide is formed from condensation of two • Protein adalah terdiri daripada satu atau lebih polipeptida.
amino acids. A protein may consist of one or more polypeptides.

• Ia boleh diuraikan kepada dua asid amino secara
hidrolisis .

It can be broken down into amino acids by hydrolysis .

Kondensasi / Condensation

Asid amino + Asid amino Dipeptida + air

Amino acid + Amino acid Dipeptides + water
Hydrolysis
Hidrolisis /

Kepentingan Protein dalam Sel TP 2

The Importance of Proteins in Cells

1 Untuk pertumbuhan dan membaiki 3 Asid amino diurai untuk membebaskan

tisu badan yang rosak. tenaga sekiranya karbohidrat dan

For growth and repair of damaged lipid simpanan badan telah digunakan.
body tissues. Amino acids are broken down to release

Kepentingan protein dalam sel energy if the carbohydrates and lipids
stored in the body are used up.
The importance of
proteins in cells

2 Membentuk enzim , antibodi dan beberapa 4 Kekurangan protein akan menyebabkan penyakit kwashiorkor .

jenis hormon. A deficiency of protein may lead to kwashiorkor.
Form enzymes , antibodies and some hormones.

55

  Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel

4.4 Lipid

Lipids

1. Lipid terdiri daripada unsur karbon (C), hidrogen (H) dan oksigen (O). TP 1

Lipid Lipids contain the elements carbon (C), hydrogen (H) and oxygen (O).
Lipids
2. Nisbah H:O adalah lebih besar daripada 2:1

The H:O ratio is much greater than 2:1

3. Lipid tidak larut dalam air tetapi boleh larut dalam pelarut organik .

Lipids are insoluble in water but soluble in organic solvents.

BAB 4 Jenis lipid b Lilin c Fosfolipid d Steroid
Type of lipids
Wax Phospholipids Steroids
a Lemak

Fats

(i) Merupakan trigliserida yang (i) Molekul berantai panjang (i) Komponen utama (i) Sebatian organik kompleks,
terbentuk melalui kondensasi yang kalis air. membran plasma . seperti kolesterol ,
satu molekul gliserol dan testosteron , estrogen dan
tiga molekul asid lemak . Long chain molecules which are Main components of progesteron .
water proof . plasma membrane .
A triglyceride formed from the Complex organic compound, such
condensation of one glycerol (ii) Terdapat pada kutikel as cholesterol , testosterone ,
molecule and three fatty acids oestrogen and progesterone .
epidermis daun, buah dan biji.
molecules. Found on the epidermis cuticlc

of leaves, fruits & seeds.

Pembentukan dan Penguraian Trigliserida

The Formation and Breakdown of a Triglyceride

COOH 3 H2O (a) Kondensasi
HO
Condensation
COOH + HO + 3 H2O
(b) Hydrolysis
COOH HO
Hidrolisis

(c) 3 molekul asid lemak (d) 1 molekul gliserol (e) 1 molekul trigliserida (f) 3 molekul air
+ molecule + molecules
3 fatty acids molecules 1 glycerol 1 triglyceride molecule 3 water

56

Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel 

Lemak Tepu dan Lemak Tak Tepu

Saturated Fats and Unsaturated Fats

TP 4

1. Komponen asas ialah gliserol dan asid lemak .
Basic components are glycerol and fatty acids .

2. Dibentuk melalui kondensasi antara satu molekul gliserol dengan tiga molekul asid lemak.
Formed through the condensation of one glycerol molecule and three fatty acid molecules.

3. Diurai semula kepada gliserol dan asid lemak melalui tindak balas hidrolisis .
Can be broken down to glycerol and fatty acids by hydrolysis reactions.

Persamaan / Similarities BAB 4

Lemak tepu / Saturated fats Lemak tak tepu / Unsaturated fats

HH HH
CC CC
HH

Perbezaan / Differences

(a) Bersifat pepejal pada suhu bilik. Contoh: mentega (b) Bersifat cecair pada suhu bilik. Contoh: minyak kelapa sawit .
Solids at room temperature. Example: butter Liquids at room temperature. Example: palm oil

(c) Mengandungi asid lemak tepu yang tidak mempunyai (d) Mengandungi asid lemak tak tepu yang mempunyai
ikatan ganda dua antara atom karbon (C=C). sekurang-kurangnya satu ikatan ganda dua antara atom
karbon .
Contains saturated fatty acids which do not have any double
bonds between the carbon atoms (C=C). Contains unsaturated fatty acids which have at least one

double bond between the carbon atoms.

(e) Semua ikatan antara atom karbon mempunyai bilangan (f) Atom karbon dalam rantai hidrokarbon tidak terikat
atom hidrogen yang maksimum .
kepada jumlah atom hidrogen yang maksimum.
All the bonds between the carbon atoms have the maximum The carbon atoms in the hydrocarbon chain are not bonded to
number of hydrogen atoms.
the maximum number of hydrogen atoms.

57

  Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel TP 2

Kepentingan Lipid dalam Sel dan Organisma Multisel

The Importance of Lipids in Cells and Multicellular Organisms

1 Sumber tenaga: 2 Terlibat dalam pembentukan membran plasma dan hormon .

Trigliserida diurai untuk membebaskan tenaga Involve in making plasma membranes and some hormones .
dan menjana ATP semasa respirasi sel.

Energy source: 3 Bekalan tenaga:

Triglycerides can be broken down in cellular respiration Disimpan di tisu adipos lemak

to release energy and generate ATP . sebagai bawah kulit atau

sekitar organ dalaman sebagai:

(a) penebat haba

(b) penyerap hentakan untuk melindungi organ-organ dalaman.

BAB 4 Kepentingan lipid dalam sel Energy reserves: fat or around
dan organisma multisel Stored in adipose tissues under the skin as subcutaneous

The importance of lipids in cells and internal organs, which act as:
multicellular organisms (a) heat insulator
(b) shock absorbers to protect the internal organs.

5 Terlibat dalam pengangkutan dan penyimpanan 4 Keapungan:

vitamin larut lemak (A, D, E, K). Disebabkan lemak adalah kurang tumpat daripada air, ini
Involve in transporting and storing lipid -soluble
dapat membantu mamalia akuatik untuk terapung .
vitamins (A, D, E, K)
Buoyancy: dense than water, it is used by aquatic mammals
Because fat is less afloat .
to help them stay

4.5 Asid Nukleik

Nucleic Acids

1. Asid nukleik adalah rantai polimer yang terdiri daripada monomer nukleotida . TP 2

Asid nukleik Nucleic acids are polymers made of monomers called nucleotides .

Nucleic acids 2. Terdiri daripada unsur karbon , hidrogen , oksigen , nitrogen , dan fosforus .

Contain the elements carbon , hydrogen , oxygen , nitrogen , and phosphorus .

Polinukleotida / Polynucleotide
Polinukleotida / Polynucleotide

Struktur nukleotida Petunjuk / Key :
StrukTtuhreTnhsuetrkuslcntertuuuoccrteltieduooraetfidoef Petunjuk K/ Kuemyp:ulan fosfat / Phosphate group

nucleotide Kumpulan fosfat / Phosphate group
Jenis asid nukleik
JeTnyipseassoidf nnuuclkeliecikacid Gula 5-karbon / 5-carbon sugar
Types of nucleic acid Gula 5-karbon / 5-carbon sugar

Bes bernitrogen / Nitrogenous
Bebsabseernitrogen / Nitrogenous
base

Asid deoksiribonukleik (DNA) Asid ribonukleik (RNA)
Deoxysibonucleic acid (DNA) Ribonucleic acid (RNA)

58

Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel 

a Asid deoksiribonukleik (DNA) b Asid ribonukleik (RNA)

Deoxyribonucleic acid (DNA) Ribonucleic acid (RNA)

(i) Molekul DNA terdiri daripada dua rantai polinukleotida (i) Molekul RNA terdiri daripada satu rantai polinukleotida .

membentuk heliks ganda dua. RNA molecules consist of one polynucleotide strands
DNA molecules consist of two polynucleotide (ii) Gula pentose: ribosa
strands formed
double helix .
(ii) Gula pentose: deoksiribosa The pentose sugar: ribose .
The pentose sugar: deoxyribose .
(iii) Bes bernitrogen: Sitosina C), urasil (U), adenina (A) atau
(iii) Bes bernitrogen: Sitosina (C), timina (T), adenina (A) atau guanina (G).
guanina (G).
The bases: Cytosine (C), uracil (U), adenine (A) or guanine (G).
The bases: Cytosine (C), thymine (T), adenine (A) or guanine (G).

Kepentingan Asid Nukleik dalam Sel Struktur Polinukleotida BAB 4
Structure of Polynucleotide
The Importance of Nucleic Acids in Cells
VIDEO 5 TP 2

• Asid nukleik sebagai pembawa maklumat pewarisan dan penghasilan protein .
Nucleic acids store genetic information and enable protein production.

a Kromosom → DNA b c Protein → trait organisma
Protein → trait of an organism
Chromosome → DNA DNA → gen / DNA → gene

Nukleus Gen 1 Gen 2 Gen 3
Sel Nucleus Gene 1 Gene 2 Gene 3
Cell
Kromosom
Chromosome

DNA

Tisu hidup / Living tissues

Protein 1 Protein 2 Protein 3 Sel hidup / Living cell

d Gen → protein / Gene → protein TP 2
Pembentukan Kromosom daripada DNA dan Protein

The Formation of Chromosomes from DNA and Proteins

Nukleus
Nucleus

a Setiap molekul DNA berpintal dengan protein khusus Sel
histon membentuk kromosom . Cell

Each large molecule of DNA is tightly wound around Kromosom
special histone proteins into chromosomes . Chromosome

Gen ACG T C A
Gene T

59 G
C AGT

DNA

  Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel SPM 4

PRAKTIS

Soalan Objektif Apakah karbohidrat P?
1. Sebatian manakah adalah lipid? What is carbohydrate P?
Which compounds are lipids?
2018 A Glikogen dan fosfolipid A Maltosa C Sukrosa

Glycogen and phospholipid Maltose Sucrose
B Selulosa dan steroid
Cellulose and steroid B Laktosa D Galaktosa
C Fosfolipid dan steroid
Phospholipid and steroid Lactose Galactose
D Fosfolipid dan selulosa
BAB 4 Phospholipid and cellulose 5. Pernyataan berikut menerangkan tentang sebatian
2. Rajah 1 menunjukkan hidrolisis protein. Antara yang kimia M.
berikut, yang manakah mewakili urutan hidrolisis
2018 protein yang lengkap? The following information describes about chemical
Diagram 1 shows the hydrolysis of protein. Which of the compound M.
following represents the sequence for the complete hydrolysis
of protein? M mempunyai kelikatan yang rendah yang
menjadikannya sebagai bahan pelincir yang baik.
Rajah 1/ Diagram 1 M has low viscosity which makes it a useful lubricant.

A PPrrootteeinin → ddiippeepptitdied a → aamsidinoamaciindo → polipeptida Apakah sebatian kimia M?
polypeptide What is chemical compound M?

A Air
B PPrrootteeinin → ppoollyippeepptitdide a → ddiippeepptitdied a → asid amino Water
amino acid B Lipid
Lipids
C Asid amino
Amino acids
D Asid nukleik
Nucleic acid

C PPrrootteeinin → ppoollyippeepptitdide a → aamsidinoamaciindo → peptid 6. Persamaan berikut menunjukkan suatu tindak balas
peptide kimia.

The following equation shows a chemical reaction.
D PPrrootteeinin → ddiippeepptitdied a → ppoollyippeepptitdide a → asid amino
amino acid


3. Molekul X dihidrolisis menjadi asid amino. Apakah Asid amino + Y kondensasi dipeptida + Z
Amino acid condensation dipeptide

molekul X?
Molecule X is hydrolysed into amino acids. What is molecule X? Apakah bahan Y dan Z?
What is substance Y and Z?
A Protein C Kanji

Protein Starch Y Y
A Asid amino Air
B Lemak D Maltose Water
Amino acid Asid amino
Fat Maltosa B Protein Amino acid
Protein
4. Persamaan berikut menunjukkan hidrolisis sejenis Protein Protein
karbohidrat, P. C Polipeptida Protein
Protein
The following equation shows the hydrolysis of a type of Polypeptides
carbohydrate, P. D Air

P + water ➞ glucose + fructose Water
P + air ➞ glukosa + fruktosa

60

Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel 

Bahagian A
1. Rajah 1.1 menunjukkan makanan yang mengandungi karbohidrat.
Diagram 1.1 shows the foods which contain carbohydrates.

2018

X: Madu / Honey Y: Nasi / Rice Z: Susu / Milk BAB 4
[2 markah / marks]
Rajah 1.1/ Diagram 1.1

(a) Namakan jenis karbohidrat dalam X dan Y.
Nama the type of carbohydrates in X and Y.
X : Monosakarida / Monosaccharides

Y : Polisakarida / Polysaccharide

H CH2OH H H CH2OH H
O O OH O
CH2OH OH
H H H

O OH H OH CH2OH OH H OH

OH H OH H O

H OH H H H OH + H2O H OH H H OH
Air galaktosa
Water galactose

H OH Laktosa H OH
glukosa
Lactose glucose

Rajah 1.2/ Diagram 1.2

(b) Rajah 1.2 menunjukkan penguraian molekul laktosa dalam susu. Terangkan.
Diagram 1.2 shows the breakdown of lactose molecule in milk. Explain.
(i) Penguraian laktosa.
The breakdown of lactose.

Ikatan molekul laktosa diurai secara hidrolisis dengan menggunakan satu molekul air untuk menghasilkan

satu molekul glukosa dan satu molekul galaktosa.

The bond in lactose molecule is broken down by hydrolysis by using a water molecule to produce a glucose molecule and a

galactose molecule.

[2 markah / marks]

61

  Biologi  Tingkatan 4  Bab 4 Komposisi Kimia dalam Sel

(ii) Pembentukan laktosa.
The formation of lactose.

Satu molekul glukosa bergabung dengan satu molekul galaktosa secara kondensasi untuk membentuk
molekul laktosa. Satu molekul air disingkirkan.
A glucose molecule binds with a galactose to form lactose molecule by condensation. One molecule of water is removed.

[2 markah / marks]
(c) Rajah 1.3 menunjukkan keputusan ujian Benedict.
Diagram 1.3 shows Benedict’s test results.

BAB 4 Larutan biru Mendakan hijau Mendakan jingga Mendakan merah bata
Blue solution Green precipitate Orange precipitate Brick red precipitate

Keputusan negatif / Negative result Keputusan positif / Positive result

Rajah 1.3/ Diagram 1.3

Apabila larutan A dipanaskan dengan larutan Benedict, warna biru larutan itu tidak berubah. Terangkan mengapa.
When solution A is heated with Benedict’s solution, the blue solution remains unchanged. Explain why. KBAT Menganalisis
Ujian Benedict memberi keputusan negatif pada larutan A. Larutan A adalah sukrosa, iaitu sejenis gula bukan

penurun. Sukrosa tidak bertindak balas dengan larutan Benedict dan mendakan merah bata tidak terbentuk.

Benedict’s test gives negative result on A. Solution A is sucrose which is a non-reducing sugar. Sucrose does not react with

Benedict's solution and no brick red precipitate is formed.

[3 markah / marks]

Kuiz 4

62

BAB Metabolisme dan Enzim

5 Metabolism and Enzymes

PETA Konsep

Metabolisme dan Enzim
Metabolism and Enzymes

Metabolisme Enzim BAB 5
Metabolism Enzymes

Anabolisme Definisi Sifat umum Faktor yang Aplikasi enzim
Anabolism Definition General mempengaruhi Application of
Katabolisme characteristics aktiviti enzim enzymes
Catabolism Factors that
Penamaan Jenis affect enzyme
Tindak balas enzim enzim activities
penguraian Naming of Types of
Breakdown enzymes enzymes Suhu
reactions Temperature
Molekul kompleks Tindak balas Enzim
Complex molecules pembinaan intrasel pH
Build up reactions Intracellular pH
enzymes

Molekul ringkas Enzim Kepekatan
Simple molecules ekstrasel substrat
Extracellular Substrate
enzymes concentration

Molekul ringkas Molekul kompleks Kepekatan
Simple molecules Complex molecules enzim
Enzyme
concentration

63

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim

5.1 Metabolisme

Metabolism

1. Definisi metabolisme: TP 1

Definition of metabolism:

BAB 5 Metabolisme ialah kesemua tindak balas biokimia yang berlaku dalam organisma. Ini termasuk
katabolisme dan anabolisme .

Metabolism is the sum of biochemical reactions that take place in an organism. It includes catabolism and
anabolism .



2. Jenis metabolisme dalam sel: TP 2

The types of metabolism in a cell:

(a) Katabolisme / Catabolism (b) Anabolisme / Anabolism

(i) (i) Tenaga

Tenaga Energy

Energy

Molekul kompleks Molekul ringkas Molekul ringkas Molekul kompleks
Complex molecules Simple molecules Simple molecules Complex molecules

(ii) Tindak balas penguraian dalam organisma. (ii) Tindak balas pembinaan dalam organisma.

The breakdown reactions in an organism. The build-up reactions in an organism.

Tenaga dibebaskan . Tenaga diperlukan .
Energy is released . Energy is required .

(iii) Contoh / Example: (iii) Contoh / Example:
Pencernaan makanan kompleks Sintesis protein daripada asid amino
Digestion of complex foods The synthesis of a protein from amino acids

Penguraian gula semasa respirasi sel Fotosintesis : sintesis glukosa daripada karbon
Breakdown of sugar in cellular respiration dioksida dan air

Photosynthesis : the synthesis of glucose from carbon
dioxide and water.

64

Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim  

5.2 Enzim

Enzymes

1. Definisi enzim: TP 1

Definition of enzymes:

(a) Enzim ialah mangkin biologi yang boleh mempercepatkan tindak balas metabolisme dalam BAB 5
organisma.

Enzymes are biological catalysts which can speed up metabolic reactions in living organism.

(b) Bukan semua enzim disintesis daripada protein. Ribozim (terdiri daripada RNA) juga bertindak
sebagai mangkin biologi.

Not all enzymes are proteins. Ribozymes (made of RNA) can also serve as biological catalysts.


2. Keperluan enzim dalam proses metabolisme: TP 2

The necessity of enzymes in metabolism:

(a) Enzim membolehkan kehidupan berlaku dengan mempercepat tindak balas kimia sel tanpa mengubah
keadaan sitoplasma .

Enzymes make life possible by speeding up the chemical reactions in cells without changing the conditions in
the cytoplasm .

(b) Tanpa enzim, tindak balas penghasilan tenaga dan bahan sel baharu akan berlaku dengan sangat perlahan .

Without enzyme, the reactions that provide energy and produce new biological material would take place very slow .

Penamaan Enzim

Naming of Enzymes

1. Kebanyakan enzim dinamakan berdasarkan nama substrat yang dimangkinkan, dengan menambahkan
'-ase' di hujung nama berdasarkan International of Biochemistry and Molecular Biology (IUBMB). TP 1

Most enzymes named based on the name of their substrate catalysed, with the ending '-ase' added according to

International of Biochemistry and Molecular Biology (IUBMB).

Enzim (a) Sukrase (b) Laktase (c) Maltase (d) Selulase (e) Lipase

Enzyme Sucrase Lactase Maltase Cellulase Lipase

Substrat Sukrosa Laktosa Maltosa Selulosa Lipid

Substrate Sucrose Lactose Maltose Celliblose Lipids

2. Sesetengah enzim telah dinamakan sebelum cara penamaan enzim yang sistematik dibentukkan sebelum

cara penamaan enzim yang sistematik. a systematic way of naming enzymes was formulated.

There are some enzymes that were named before

Enzim (a) Pepsin (b) Tripsin (c) Renin

Enzyme Pepsin Trypsin Rennin

Substrat Polipeptida Protein Kasinogen

Substrate polypeptide Protein Caseinogen


65

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim TP 2

Sifat-sifat Umum Enzim

General Properties of Enzymes

(a) Enzim ialah protein yang disintesis (b) Enzim ialah mangkin biologi. (c) Tindakan enzim adalah sangat
oleh sel hidup. Enzymes are biological catalysts . spesifik .

Enzymes are proteins which are Mempercepatkan tindak balas The actions of enzymes are highly
synthesised by living cells. metabolik dengan merendahkan specific .
tenaga pengaktifan tindak balas.
Struktur dan aktiviti enzim adalah peka Enzim hanya bertindak pada substrat
kepada perubahan suhu dan pH. Speed up metabolic reactions by
Kebanyakan enzim ternyahasli pada lowering the activation energy yang dapat bergabung dengan
suhu tinggi. of the reactions.
The structure and activity of enzymes are tapak aktif nya.
sensitive to the changes in temperature and Sifat umum enzim
The properties of An enzyme only acts on substrates
pH . Most enzymes are denatured enzymes that fit into its active site .
at high temperatures.
BAB 5 (e) Enzim diperlukan dalam kuantiti (f) Aktiviti enzim dipengaruhi oleh
(d) Enzim boleh diguna semula . yang sedikit . perencat .
Enzymes are reusable .
Enzymes are needed in relatively Enzyme activity is affected by
Enzim kekal tidak berubah di akhir small amounts. inhibitors .
tindak balas. Ia boleh bergabung kepada
molekul substrat lain selepas tindak balas. Enzim dapat diguna semula . Perencat (contoh: merkuri)
Enzymes remain unchanged after reactions. Enzymes can be reused . memperlahan atau menghentikan
They can bind to other substrate molecules aktiviti enzim.
after the reaction is complete. Inhibitors (example: mercury) slow
or stop the enzyme activity.

Enzim Intrasel dan Enzim Ekstrasel TP 2

Intracellular and Extracellular Enzymes

Enzim / Enzyme

(a) Enzim intrasel (b) Enzim ekstrasel
Intracellular enzymes Extracellular enzymes

(i) Enzim yang disintesis dan diguna (i) Enzim yang disintesis dalam sel tetapi
dirembes ke luar sel untuk berfungsi di
dalam sel sendiri. luar sel.
Enzymes which are synthesised and work
Enzymes which are synthesised inside the cell but
inside the cells.
secreted from the cell to function externally .
(ii) Terdapat dalam sitoplasma , nukleus ,
,
mitokondrion dan kloroplas . (ii) Contoh: enzim pencernaan ( amilase ,
tripsin , lipase ) dan lisozim .
Found in the cytoplasm , nucleus ,
Example: digestive enzymes ( amylase
mitochondrion and chloroplast .
trypsin , lipase ) and lysozyme .
(iii) Contoh : Oksidoreduktase
Example: Oxidoreductase

66

Penghasilan Enzim Intrasel dan Ekstrasel Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim  
TP 2
Production of Intracellular and Extracellular Enzymes

2. Protein disintesis di 1. Nukleus mengandungi DNA yang membawa maklumat 7. Vesikel ini kemudian
ribosom . Protein ini untuk sintesis enzim . bercantum dengan
membran plasma
The nucleus contains DNA which carries the information for
enzyme synthesis. sebelum membebaskan
diangkut melalui ruang
dalam jalinan protein ke luar sel sebagai
enzim luar sel .
endoplasma
Enzim dibebaskan The vesicles then fuse
kasar . Jalinan endoplasma kasar Enzyme expelled with the plasma
Rough endoplasmic reticulum membrane before releasing
Membran plasma
Proteins are synthesised 7 Plasma membrane the proteins outside the cell
at the ribosomes . The 6 as extracellular enzymes.
Vesikel rembesan
proteins are transported Secretory vesicle

through the spaces 1
rough DNA
within the Ribosom
endoplasmic 2 Ribosome 6. Vesikel rembesan BAB 5
bergerak menuju ke
reticulum . membran plasma .

3. Protein yang dibungkus 4 5 The secretory vesicles
dalam vesikel 3 Jasad Golgi travel to the plasma
dibebaskan daripada Vesikel angkutan Golgi apparatus membrane .
membran jalinan Transport vesicle
endoplasma kasar sebagai 5. Vesikel rembesan yang
vesikel angkutan . 4. Vesikel angkutan kemudian bercantum dengan membran mengandungi protein
jasad Golgi . Dalam jasad Golgi, protein diubah suai . yang diubah suai
Proteins wrapped in meninggalkan jasad Golgi.
vesicles bud off from The transport vesicles then fuse with the membranes of the
Golgi apparatus. The proteins are further modified in Secretory vesicles
the membranes of rough containing modified
the Golgi apparatus. proteins bud off from
endoplasmic reticulum as the Golgi apparatus.
transport vesicles.

Mekanisme Tindakan Enzim: Hipotesis ‘Mangga dan Kunci’ TP 2

Mechanism of Enzyme Action:‘Lock and Key’ Hypothesis

1. Mekanisme tindakan enzim yang khusus boleh diterangkan melalui hipotesis ‘ mangga dan kunci ’.

The specificity of enzyme actions can be explained by the lock and key ’ hypothesis.

2. Molekul substrat mewakili ‘ kunci ’ manakala molekul enzim mewakili ‘ mangga ’.

Substrate molecule acts as the ‘ key ’ while the enzyme molecule acts as the ‘ lock ’.

a Substrat

Substrate (i) Setiap molekul enzim mempunyai tapak aktif .

Each enzyme molecule has an active site .

Tapak aktif (ii) Molekul enzim hanya bergabung dengan molekul substrat yang boleh
Active site melengkapi bentuk tapak aktifnya.

Enzim An enzyme molecule only binds to substrate molecules that fit the shape
Enzyme
of its active site.

67

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim

BAB 5 b (i) Tindak balas metabolik bermula apabila molekul substrat bergabung dengan
tapak aktif enzim untuk membentuk kompleks enzim-substrat .
Kompleks enzim-substrat
Enzyme substrate complex A metabolic reaction begins when the substrate molecule binds to the active site
of an enzyme to form an enzyme-substrate complex .
c
(ii) Pembentukan ini merendahkan tenaga pengaktifan tindak balas.
Hasil
Product Its formation lowers the activation energy of the reaction.

Enzim (i) Substrat ditukar membentuk hasil tindak balas.
Enzyme
The substrates are converted into products .

(ii) Hasil tindak balas meninggalkan tapak aktif enzim.

The product then leaves the active site of the enzyme.

(iii) Molekul enzim dibebaskan dalam bentuk asal.

The enzyme molecule is released in its original form.

Rajah Tenaga untuk Mekanisme Tindakan Enzim TP 2

Diagram of Energy for the Mechanism of Enzyme Action

1. Bagi setiap tindak balas, sejumlah tenaga perlu dibekalkan kepada bahan tindak balas sebelum tindak
balas boleh berlaku. Ini dikenali sebagai tenaga pengaktifan .

For every reaction, a certain amount of energy must be supplied to the reactants before the reaction can occur. This is
called the activation energy.

2. Enzim berfungsi dengan menurunkan tenaga pengaktifan supaya tindak balas kimia dapat berlaku pada

suhu badan dengan kadar yang lebih cepat .

Enzymes work by lowering the activation energy so that chemical reactions can take place at body temperature at a

faster rate.

Aras tenaga / Energy level Bahan Tenaga pengaktifan tanpa enzim
tindak balas Activation energy without enzyme
Reactants Tenaga pengaktifan dengan enzim
Activation energy with enzyme

Hasil tindak
balas
Products

Tindak balas / Progress of reaction

68

Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim  

Kesan Perubahan Faktor ke atas Mekanisme Tindakan Enzim TP 2

Factors Affecting the Mechanism of Enzymatic Reactions

( A) Suhu / Temperature

1. Pada suhu rendah, enzim adalah tidak Kadar tindak balas / Reaction rate 3 3. Kadar tindak balas enzim adalah .
aktif. 2 maksimum pada suhu optimum

At low temperatures, enzymes are inactive. The rate of enzymatic reaction is
maximum at optimum
(a) Molekul enzim dan molekul
substrat bergerak perlahan dan temperature.
frekuensi pelanggaran berkesan

antara molekul adalah rendah. 1 4 4. Suhu tinggi menyebabkan
The enzyme and substrate molecules perubahan tapak aktif molekul

move slowly and the frequency of enzim.
collision with each other is low.
0 10 20 30 40 50 Suhu High temperatures may cause change in
(b) Kadar tindak balas adalah the active side of the enzyme molecule.
rendah . Suhu optimum Temperature
Optimum temperature (a) Enzim ternyahasli .
Therefore, the reaction rate is low . BAB 5
The enzyme is denatured .

2. Apabila suhu meningkat, molekul-molekul enzim dan substrat bergerak dengan lebih (b) Molekul substrat tidak dapat
pantas. bergabung dengan tapak aktif
enzim.
As the temperature rises, both enzyme and substrate molecules move around more rapidly.
The substrate molecule can no longer
(a) Frekuensi pelanggaran berkesan antara satu sama lain lebih tinggi . fit into the active site of the
The frequency of collision with each other increases.
enzyme.
(b) Ini meningkatkan peluang pembentukan kompleks enzim-substrat .
This increases the chance of forming enzyme-substrate complexes . (c) Akibatnya, kadar tindak balas
enzim menurun .
(c) Akibatnya, kadar tindak balas enzim meningkat .
As a result, the rate of the enzymatic reaction increases . As a result, the rate of the enzymatic
reaction decreases .

(B) pH 3. pH yang tidak sesuai akan menyebabkan penyahaslian enzim.
An unsuitable pH will cause denaturation of the enzyme.
1. Dalam sel, kebanyakan enzim berfungsi dalam
julat pH yang kecil. Maka kadar tindak balas enzim menurun .
Therefore the rate of the enzymatic reaction decreases .
Most enzymes work in a narrow range of pH in cells.
(a) Kadar tindak balas adalah maksimum pada Pepsin Amilase liur Tripsin
Pepsin Salivary amilase Trypsin
pH optimum .
The rate of reaction is maximum at optimum Kadar tindak balas / Reaction rate

pH.

(b) Contoh: Pepsin 1.5 hingga 2.5 , amilase
6.8 , tripsin 8.5

Example: Pepsin 1.5 to 2.5 , amylase 6.8 ,
trypsin 8.5

2. Perubahan pH boleh mengubah cas tapak 2 4 6 8 10 pH
aktif enzim dan permukaan substrat.

A change in pH value can alter the charges on
the active site of an enzyme and the surface of

the substrate.

69

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim

(C) Kepekatan substrat / Substrate concentration

Di A / At A: Kadar tindak balas B
1. Apabila kepekatan substrat semakin meningkat , kadar tindak balas meningkat . Reaction rate A
As the substrate concentration increases , the rate of reaction increases .
2. Kepekatan substrat menjadi faktor pengehad yang mengehadkan tindak balas. Kepekatan substrat
Substrate concentration is the limiting factor which limits the reaction. Substrate concentration

BAB 5 Di B / At B:
3. Peningkatan kepekatan substrat seterusnya, tindak balas akan mencapai kadar maksimum .
As the substrate concentration increases even further, the reaction will reach its maximum rate.
4. Peningkatan kepekatan substrat tidak akan meningkatkan kadar tindak balas.
The increase of substrate concentration will not increase the rate of reaction.
5. Ini kerana semua tapak aktif enzim dipenuhi oleh molekul substrat .
This is because all the enzyme’s active sites are occupied with substrate molecules.
6. Kadar tindak balas menjadi malar dan kepekatan enzim menjadi faktor pengehad .
The rate of reaction becomes constant and the enzyme concentration becomes the limiting factor.

( D) Kepekatan enzim / Enzyme concentration

Di C / At C: Kadar tindak balas C D
1. Apabila kepekatan substrat semakin meningkat , kadar tindak balas meningkat . Reaction rate Kepekatan enzim
As enzyme concentration increases , the rate of reaction increases . Enzyme concentration
2. Kepekatan enzim menjadi faktor pengehad yang mengehadkan tindak balas.
Enzyme concentration is the limiting factor which limits the reaction.

Di D / At D:
3. Pada kepekatan substrat tertentu, tindak balas adalah pada kadar maksimum .
The reaction is at its maximum rate for the fixed substrate concentration.
4. Jika peningkatan kepekatan enzim terus meningkat, ini tidak akan meningkatkan

kadar tindak balas.
If the enzyme concentration increases further, they will be no increase in the rate of reaction.
5. Ini kerana tapak aktif molekul enzim berlebihan tidak akan dipenuhi oleh molekul substrate .
This is because active sites of the extra enzyme molecules will not be occupied by substrate molecules.
6. Kadar tindak balas menjadi malar dan kepekatan substrat menjadi faktor pengehad .
The rate of reaction becomes constant and the substrate concentration becomes the limiting factor.

Eksperimen 5.1 Kesan suhu terhadap aktiviti enzim amilase

Effects of temperature on the activities of amylase

Tujuan Mengkaji kesan suhu terhadap aktiviti enzim amilase liur. TP 2
Aim To investigate the effect of temperature on the activity of salivary amylase.
Pernyataan masalah
Problem statement Apakah kesan suhu ke atas kadar tindak balas amilase liur ke atas kanji?
Hipotesis What are the effects of temperature on the activity of salivary amylase on starch?
Hypothesis
Semakin tinggi suhu, semakin tinggi aktiviti enzim amilase liur ke atas kanji sehingga mencapai suhu optimum.
Pemboleh ubah As the temperature increases, the rate of activity of salivary amylase on starch increases until it reaches the optimum
Variables temperature.

a) Dimanipulasikan: Suhu
Manipulated: Temperature of the medium
b) Bergerak balas: Kadar tindak balas enzim amilase
Responding: The rate of reaction catalysed by salivary amylase
c) Dimalarkan: Kepekatan ampaian kanji, isi padu amilase
Constant: Concentration of starch suspension, volume of salivary amylase

70

Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim  

Bahan: Ampaian kanji 1%, air liur, larutan iodin, ais, dan air suling. BAB 5
Materials 1% of starch suspension, saliva, iodine solution, ice cubes and distilled water
Radas:
Apparatus: Bikar, tabung uji, termometer, picagari, penitis, rod kaca, jam randik, jubin berlekuk, kukus air
Prosedur: Beakers, test tube, thermometer, syringe, droppers, glass rods, stopwatch, white tiles with grooves, water bath
Procedure:
1. Kumur mulut dengan air suam dan kumpulkan air liur. Cairkan air liur dengan isi padu air suling yang
Pemerhatian: sama.
Observation:
Rinse the mouth with warm water and collect the saliva. Dilute the saliva with equal volume of distilled water.
2. Masukkan 5 ml ampaian kanji 1% ke dalam tabung uji berlabel A1, B1, C1, D1 dan E1 masing-masing

dengan menggunakan picagari.
Add 5 ml of 1% starch suspension into each test tube, labelled A1, B1, C1, D1, and E1 respectively with a syringe.
3. Masukkan 2 ml larutan air liur ke dalam tabung uji berlabel A2, B2, C2, D2 dan E2 dengan

menggunakan picagari.
Add 2 ml of saliva each into another set of test tubes, labelled A2, B2, C2, D2 and E2 using a new syringe.
4. Rendam tabung uji A1 dan A2, B1 dan B2, C1 dan C2, D1 dan D2, E1 dan E2 masing-masing di dalam

kukus air pada suhu 0 oC, 20 oC, 30 oC, 40 oC dan 60 oC.
Immerse test tubes A1 and A2, B1 and B2, C1 and C2, D1 and D2, E1 and E2 respectively into 5 different water

baths with temperature which are kept constant at 0 oC, 20 oC, 30 oC, 40 oC and 60 oC.
5. Biarkan tabung uji ini selama 5 minit.
Left the test tubes for 5 minutes.
6. Sementara itu, titiskan 2 titis larutan iodin pada setiap lekuk jubin putih berlekuk.
Meanwhile, place a drop of iodine solution into each groove of a white tile.
7. Selepas 5 minit rendaman, tuang ampaian kanji dalam tabung uji A1 ke dalam tabung uji A2 yang

mengandungi larutan air liur.
After five minutes, pour the starch suspension in test tube A1 into the saliva solution in test tube A2.
8. Kacau campuran ini dengan menggunakan rod kaca. Hidupkan jam randik dengan serta merta.
Stir the mixture using a glass rod. Activate the stopwatch immediately.
9. Dengan menggunakan penitis, masukkan setitis campuran daripada A2 ke dalam larutan iodin di

lekuk pertama jubin putih. Lekuk pertama dianggap 0 minit.
Place immediately a drop of mixture from test tube A2, in the first groove of the tile containing the iodine solution.

The first groove is considered as 0 minute.
10. Ulang ujian iodin setiap 10 minit. Cuci penitis setiap kali mengambil sampel.
Repeat the iodine test every minute for 10 minutes. Rinse the dropper in a beaker of distilled water after each sampling.
1 1. Rekod masa yang diambil untuk hidrolisis kanji menjadi lengkap, iaitu selepas larutan iodin tidak

berubah menjadi biru tua apabila ditambah larutan kanji.
Record the time taken for the complete hydrolysis of starch, that is, when the mixture no longer turned blue-black

when tested with the iodine solution.
1 2. Biarkan tabung uji yang mengandungi campuran dalam kukus air masing-masing sepanjang eksperimen.
Keep the test tubes containing the mixture in their respective water bath throughout the experiment.
13. Ulang langkah 7 – 9 bagi tabung uji B1, C1, D1 dan E1.
Repeat the steps 7 to 10 for test tubes B1,C1,D1 and E1
14. Gunakan termometer untuk memastikan suhu adalah tetap kekal sepanjang eksperimen.
Use the thermometer to ensure that the temperature remain constant throughout the experiment
1 5. Rekod keputusan dan plot graf kadar tindak balas enzim melawan suhu.
Record the results and plot a graph of the rate of enzymatic reaction against temperature.

Tabung uji Suhu (oc) Masa yang diambil untuk hidrolisis Kadar tindak balas (min-1)
Test tube Temperature lengkap kanji (minit) Rate of reaction (min-1)

A1 (oC) Time taken for the hydrolysis of starch to be 0.00
B1 completed (minutes) 0.50
C1 0 1.00
D1 20 Masa maksimum / Maximum time 0.33
E1 30 2 0.00
40 1
60 3

20 (masa maksimum / maximum time)

71

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim

Graf: Kadar tindak balas (min-1)
Graph: Rate of reaction (min-1)

Suhu (°C)
Temperature (°C)

Perbincangan: 1. Mengapakah tabung uji diletakkan dalam kukus air selama 5 minit? Terangkan.
Discussion:
Kesimpulan: Why the test tubes are placed in a water bath for 5 minutes? Explain.
Conclusion:
Untuk membolehkan kedua-dua larutan mencapai suhu sama yang disetkan.
To allow both solutions to reach the temperature set.

BAB 5 2. Terangkan kadar tindak balas enzim pada / Explain the rate of reaction at
(a) suhu 37oC / temperature 37oC

Kadar tindak balas adalah maksimum. Enzim adalah paling aktif.

The rate of reaction is maximum. The enzyme is active.

(b) suhu 60oC / temperature 60oC

Kadar tindak balas menurun. Enzim ternyahasli.

The rate of reaction decreases. The enzyme has denatured.

3. Apakah suhu optimum bagi amilase bertindak ke atas kanji?

What is the optimum temperature for amylase to act on starch?
37oC

Hipotesis diterima. Semakin tinggi suhu, semakin tinggi aktiviti enzim amilase liur ke atas kanji sehingga
mencapai suhu optimum 37oC.
The hypothesis is accepted. As the temperature increases, the rate of activity of salivary amylase on starch increases
until it reaches the optimum temperature of 37oC.

Eksperimen 55..22 Kesan pH terhadap aktiviti enzim amilase dan pepsin

Effects of pH on the activities of amylase and pepsin

Tujuan Mengkaji kesan pH terhadap aktiviti enzim pepsin.
Aim To study the effect of pH on the activity of pepsin.
Pernyataan masalah
Problem statement Apakah pH optimum bagi tindak balas pepsin?
Hipotesis What is the optimum pH of pepsin activity?
Hypothesis
Pemboleh ubah pH optimum untuk aktiviti pepsin adalah dalam medium berasid.
Variables The optimum pH for the activity of pepsin is in an acidic medium.

(a) Dimanipulasikan: pH medium tindak balas
Manipulated: pH of medium
(b) Bergerak balas: Kejernihan larutan tindak balas
Responding: Clarity of the solution
(c) Dimalarkan: Kepekatan ampaian albumen, suhu
Constant: Concentration of albumen suspension, temperature

72

Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim  

Bahan: Ampaian albumen 1%, larutan pepsin 1%, asid hidroklorik 0.1 M, natirum hidroksida 0.1 M, air suling,
Materials kertas pH
Radas: 1% of albumen suspension, 1% of pepsin solution, 0.1 M of hydrochloric acid, 0.1 M of sodium hydroxide, distilled
Apparatus: water, pH papers
Prosedur:
Procedure: Bikar, tabung uji, termometer, picagari, penitis, jam randik, kukus air
Beakers, test tube, thermometer, syringe, droppers, stopwatch, water bath
Pemerhatian:
Observation: 1. Sediakan tiga tabung uji berlabel P, Q dan R seperti dalam jadual.
Prepare three test tubes labelled P, Q and R as shown in the table.
Perbincangan:
Discussion: Tabung uji / Test tube Kandungan / content

Kesimpulan: P 5 ml ampaian albumen + 1 ml asid hidroklorik 0.1 M + 1 ml larutan pepsin 1%
Conclusion: 5 ml of albumen suspension + 1 ml of 0.1 M hydrochloric acid + 1 ml of 1% pepsin solution

Q 5 ml ampaian albumen + 1 ml air suling + 1 ml larutan pepsin 1%
5 ml of albumen suspension + 1 ml of distilled water + 1 ml of 1% pepsin solution
BAB 5
R 5 ml ampaian albumen + 1 ml natrium hidroksida 0.1 M + 1 ml larutan pepsin 1%
5 ml of albumen suspension + 1 ml of 0.1 M sodium hydroxide + 1 ml of 1% pepsin solution

2. Simpan semua tabung uji tersebut dalam kukus air 37 oC selama 20 minit.
Keep the test tubes in water bath at 37 oC for 20 minutes.

3. Perhatikan kekeruhan dalam setiap tabung uji pada permulaan eksperimen dan selepas 20 minit.
Observe the turbidity of each test tube at the beginning of the experiment and after 20 minutes.

Keadaan kandungan / Condition of content

Tabung uji pH 0 minit Selepas 20 minit
Test tube 0 minute After 20 minutes

P3 Keruh / Cloudy Jernih / Clear
Q7
R8 Keruh / Cloudy Keruh / Cloudy

Keruh / Cloudy Keruh / Cloudy

1. Mengapakah tabung uji perlu diletakkan dalam kukus air pada suhu 37 oC?

Explain why the test tubes are placed in a water bath maintained at 37 oC?

Suhu optimum untuk tindakan pepsin.
The optimum temperature for the action of pepsin.

2. Mengapakah kandungan tabung uji Q dan R kekal keruh selepas 20 minit?

Why does the content of test tube Q and R remain cloudy after 20 minutes?

Albumen tidak dihidrolisis oleh pepsin kepada polipeptida dalam medium neutral dan beralkali.
Albumen adalah tidak larut air.
The albumen has not been hydrolysed by pepsin into polypeptides in neutral and alkali medium. Albumen is
insoluble in water.

Hipotesis diterima. pH optimum bagi aktiviti pepsin adalah dalam medium berasid.
The hypothesis is accepted. The optimum pH for the activity of pepsin is in an acidic medium.

73

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim

5.3 Aplikasi Enzim dalam Kehidupan Harian

Applications of Enzymes in Daily Life

Contoh penggunaan enzim dalam kehidupan harian:

Examples of the application of ezymes in daily life:

Enzim / Enzyme Kegunaan / Uses

1. Protease • Melembutkan daging dalam industri makanan.
Protease
Tenderises meat in food industry.

• Menanggalkan kulit ikan dalam industri ikan.

Removes the skin of fish in fish industry.

• Sebagai bio detergen untuk melarutkan kesan protein dan kanji di pakaian.

As a bio detergent to dissolve protein and starch stains in clothes.

BAB 5 2. Amilase • Menukar kanji kepada gula dalam pembuatan sirap.
Amylase
Changes starch into sugar in the making of syrup.

• Menyingkir kanji yang digunakan sebagai pengukuh daripada kain.

Removes starch that is used as stiffeners from fabrics.

3. Lipase • Pematangan keju dalam produk tenusu.
Lipase
Ripening of cheese in dairy products.

4. Selulase • Mengurai selulosa dan membuang kulit biji benih daripada bijirin.
Cellulase
Breaks down cellulose and removes seed coats from cereal grains.

• Mengekstrak agar daripada rumpai laut.

Extracts agar from seaweed.

PRAKTIS SPM 5

Soalan Objektif 2. Aktiviti enzim dalam salur pencernaan dipengaruhi oleh
suhu. Antara yang berikut, apakah yang akan berlaku
1. Puan Mariam biasanya memerap daging dengan apabila seseorang mengalami demam panas?
kepingan betik muda untuk melembutkan daging
tersebut. Antara yang berikut, yang manakah dapat Enzyme activity in the alimentary canal is affected by
membantu untuk meminimumkan masa bagi temperature. Which of the following will occur when someone
melembutkan daging? has high fever?
A Ketidakhadaman
Mrs. Mariam usually marinates meat with slices of unripe Indigestion
papaya to tenderise the meat. Which of the following can help B Cirit-birit
to minimise the time for tendering the meat? Diarrhea
A Simpan pada suhu 40oC C Gastritis
Keep at 40oC Gastric
B Simpan pada suhu 80oC D Sembelit
Keep at 80oC Constipation
C Simpan dalam peti sejuk selama dua jam
Keep in refrigerator for two hours
D Tambahkan sedikit garam
Add some salt

74

3. Rajah 1 menunjukkan tindakan enzim E ke atas Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim  
substrat S.
4. Rajah 2 menunjukkan sehelai baju yang mempunyai
2 018 Diagram 1 shows the action of enzyme E on substrate S. kotoran makanan di atasnya. Baju tersebut dicuci
Substrat S menggunakan serbuk pencuci biologi yang
mengandungi lipase.
Substrate S
Diagram 2 shows a shirt which has a food stain on it. The
shirt is washed using some biological washing powder which
contains lipase.

Enzim E Kotoran makanan
Enzyme E Food stain
Rajah 1/ Diagram 1
Rajah 2/ Diagram 2
Apakah yang akan berlaku pada akhir tindak balas?
What will happen at the end of the reaction? Apakah jenis kotoran yang melekat di baju itu? BAB 5
What type of food stain is on the shirt?
A Hanya sebahagian substrat S akan dihidrolisis.
Only a part of substrate S will be hydrolysed. A Lipid C Protein
B Tapak aktif enzim E adalah spesifik kepada Lipids Protein

substrat S. B Karbohidrat D Selulosa
Active site of enzyme E is specific to substrate S. Carbohydrate Cellulose
C Pembentukan kompleks enzim-substrat
Formation of enzyme-substrate complex
D Substrat S tidak akan dihidrolisis.
Substrate S will not be hydrolysed.

Bahagian A

1. (a) Rajah 1 menunjukkan kesan enzim ke atas tenaga pengaktifan bagi suatu tindak balas biokimia.
Diagram 1 shows the effect of enzyme on the activation energy of a biochemical reaction.

Aras tenaga
Energy level

Substrat Ea1
Substrate E

a2

Kompleks enzim-substrat Hasil
Enzyme-substrate complex Products

Tindak balas biokimia
Biochemical reaction

Rajah 1/ Diagram 1

(i) Beri maksud tenaga pengaktifan.
Define activation energy.

Tenaga pengaktifan ialah tenaga minimum yang diperlukan untuk memulakan suatu tundak balas biokimia.
Activation energy is the minimum energy required to start a biochemical reaction.

[1 markah / mark]

75

  Biologi  Tingkatan 4  Bab 5 Metabolisme dan Enzim

(ii) Nyatakan graf yang menunjukkan [2 markah / marks]
State the graph which shows
Tindak balas tanpa enzim / Reaction without enzyme: Ea1
Tindak balas dengan enzim / Reaction with enzyme: Ea1

(iii) Berdasarkan Rajah 1, terangkan bagaimana enzim mempengaruhi kadar tindak balas biokimia.
Based on Diagram 1, explain how enzyme affects the rate of biochemical reaction.

Enzim bertindak sebagai pemangkin biologi / Enzyme acts as biological catalyst

Mempercepat kadar tindak balas / Speed up the rate of reaction

Untuk mengurangkan tenaga pengaktifan / To reduce the activation energy

Substrat bergabung dengan enzim dengan lebih mudah / Substrate binds to enzyme easier

BAB 5 Lebih banyak enzim-substrat kompleks terbentuk / More enzyme-substrate complex formed

Produk yang terhasil adalah pada kadar lebih cepat / Product produced is at faster rate

[3 markah / marks]

(c) Pulpa mangga mengandungi pektin yang tinggi dan memerlukan rawatan pektinase untuk penjernihan jus.
Pektin adalah bahan dalam buah-buahan yang membantu untuk melekatkan sel-sel tumbuhan bersama. Jus
mangga yang diekstrak dengan menggunakan pektinase adalah jernih manakala jus mangga buatan sendiri
keruh.
Terangkan mengapa enzim pektinase digunakan dalam penghasilan jus mangga. KBAT Mengaplikasi
Mango pulp contains high pectin and needs pectinase treatment for clarification of the juice. Pectin is a substance in fruits which
helps to stick plant cells together. Mango juice which is extracted using pectinase is clear while homemade mango juice is cloudy.
Explain why the enzyme pectinase is used in the mango juice production.

Pektinase bertindak ke atas pektin / Pectinase act on pectin

dengan hidrolisiskan pectin / which hydrolyse pectin

Tanpa pektin, sel mangga tidak melekat / Without pectin, mango cells not stick together

Memudahkan pengekstrakan jus dari buah / Easier to extract juice from the fruit

Menambah hasil jus / Increase juice yield

[3 markah / marks]

Kuiz 5

76

BAB Pembahagian Sel

6 Cell Division

PETA Konsep

Pembahagian Sel
Cell Division

Kitar sel Keperluan mitosis dan meiosis dalam
Cell cycle kehidupan
The necessity of mitosis and meiosis in life
BAB 6
Interfasa Fasa M Sitokinesis
Interphase M Phase Cytokinesis

Mitosis Meiosis
Mitosis Meiosis

Profasa Meiosis I
Prophase Profasa I / Prophase I
Metafasa Metafasa I / Metaphase I
Metaphase Anafasa I / Anaphase I
Anafasa Telofasa I / Telophase I
Anaphase Sitokinesis / Cytokinesis
Telofasa
Telophase Meiosis II
Profasa II / Prophase II
Metafasa II / Metaphase II
Anafasa II / Anaphase II
Telofasa II / Telophase II
Sitokinesis / Cytokinesis

77

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

6.1 Pembahagian Sel

Cell Division

1. Semua organisma hidup terdiri daripada sel . Sel wujud daripada sel yang sedia ada , melalui proses
pembahagian .
. The cells arise from existing cells, by the process of division .
All living organisms are composed of cells

2. Pembahagian sel adalah proses di mana sel induk membahagi untuk membentuk sel anak baharu, yang
berlaku dalam dua peringkat utama:

Cell division is a process in which a parent cell divides to form new daughter cells, which occurs in two main stages:

(a) Kariokinesis: Pembahagian nukleus (Mitosis dan meiosis).

Karyokinesis : Nuclear division (Mitosis and meiosis).

(b) Sitokinesis: Pembahagian sitoplasma .

Cytokinesis: Cytoplasm division.

BAB 6 3. Pembahagian sel melibatkan pewarisan maklumat genetik daripada sel induk kepada sel anak . TP 1

Cell division involves the passing on of genetic information from parent cells to daughter cells.

4. Maklumat ini penting untuk menentukan struktur dan fungsi sel dan dibawa dalam asid deoksiribonukleik
(DNA). TP 2

This information is important in determining the structure and functions of the cells and is carried in deoxyribonucleic acid

(DNA).

5. Di dalam nukleus, DNA dan protein membentuk struktur yang dikenali sebagai kromosom .

Inside the nucleus, DNA and proteins together form structures called chromosomes .

6. Apabila sel tidak membahagi, individu kromosom tidak kelihatan tetapi wujud sebagai kromatin .

When a cell is not dividing, individual chromosomes are not visible but exist as chromatin .

7. Semasa pembahagian sel, setiap kromosom kelihatan terdiri daripada dua kromatid yang seiras.

At the time cell division begins, each chromosome is seen to consist of two identical chromatids .

8. Lengkapkan struktur kromosom di bawah.

Complete the structure of chromosome as below.

Molekul DNA
DNA molecule

Satu kromatid
A chromatid

Bebenang kromatin Protein
Chromatin fibre

Sentromer / Centromere

Dua kromatid kembar membentuk satu kromosom.
Two sister chromatids make up one chromosome.

78

Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  

Kromosom

Chromosomes

1. Dalam sel badan, kromosom wujud secara berpasangan , iaitu kromosom homolog .

Chromosomes exist in pairs called homologous chromosomes in body cells.

2. Kromosom yang berasal daripada induk lelaki dikenali sebagai kromosom paternal manakala kromosom
yang berasal dari induk perempuan pula adalah kromosom maternal .

The chromosome comes from the male parent is called the paternal chromosome while the other one from the female
parent is called the maternal chromosome.

3. Setiap sel badan mempunyai dua set kromosom, iaitu diploid (2n).

Each body cell has two sets of chromosomes and is described as diploid (2n).

4. Gamet mempunyai satu set kromosom, iaitu diploid (n).

A gamete has one set of chromosomes and is described as haploid (n).

Haploid (n) Diploid (2n)

6.2 Kitar Sel dan Mitosis BAB 6

The Cell Cycle and Mitosis

1. Sel badan boleh membahagi secara mitosis .

Many body cells can divide by mitotic cell division.

2. Kitar sel ialah tempoh bermulanya sel baharu dihasilkan sehingga sel melengkapkan pembahagian sel itu.

The cell cycle is the period a new cell is produced until the cell completes a cell division.

3. Fasa utama dalam kitar sel:

The phases of the cell cycle:

Kitar sel / Cell

Interfasa / Interphase Fasa M / M phase

Pertumbuhan sel Pembahagian nukleus dan sitoplasma

Cell growth Nuclear and cytoplasmic division

G1 S G2 Mitosis / Mitosis Sitokinesis / Cytokinesis
Kitar sel / Cell cycle
Label
B
A : Interfasa / Interphase
J B : Fasa G1 / G1 phase
C : Fasa S / S phase
H A C D : Fasa G2 / G2 phase
IG A E : Profasa / Prophase
F : Metafasa / Metaphase
F A G : Anafasa / Anaphase
E H : Telofasa / Telophase
I : Mitosis / Mitosis
D J : Sitokinesis / Cytokinesis

79

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

Fasa G1: Fasa pertumbuhan 1 / G1 phase: Growth phase 1
Sel mula mendapatkan dan mensintesis bahan yang diperlukan untuk pembahagian sel.
The cell begins to acquire and synthesise the materials required for cell division.

• Protein dan organel baharu (mitokondrion dan kloroplas) disintesiskan.
Proteins and new organelles (mitochondria and chloroplast) are synthesised.

• Kadar metabolisme sel adalah tinggi. / Metabolic rate of the cell is high.

• Kromosom tidak jelas kelihatan dan muncul seperti struktur bebenang yang dikenali sebagai kromatin .
Chromosomes are extremely fine and appear as thread-like structure known as chromatin .

Fasa M : Mitosis & sitokinesis G1 FGa1 spahaGse1 Fasa S: Fasa sintesis

M phase : Mitosis & cytokinesis AATnCTneSaeylaopittlopfhooakhafkisaasninsaeseeesaissisPembahagian S phase : Synthesise phase

Mitosis: profasa → metafasa → Sintesis DNA berlaku.
Synthesis of DNA occurs.
anafasa → telofasa
• DNA menjalani replikasi .
Mitosis : prophase → metaphase → DNA undergoes replication .
anaphase → telophase .
• Setiap kromosom yang mengganda
• Sitokinesis: sitoplasma membahagi mengandungi dua kromosom yang
untuk membentuk dua sel anak, seiras disebut kromatid kembar .
setiap satu mengandungi satu
nukleus yang memiliki bilangan A duplicated chromosome consists
kromosom diploid .
of two identical chromosomes called
Cytokinesis: the cytoplasm divides to sister chromatids .
form two daughter cells, each having
one nucleus which contains the • Dalam sel haiwan, sentriol
diploid number of chromosomes. mengganda .
BAB 6 MMeettaapfhaassae sel
MitosisCell division In animal cells, the centrioles duplicate .
sa / Interphase
PPrroofpahsaase Interfa Fasa S
S phase

GFa2 pshaaGse2

Fasa G2: fasa pertumbuhan 2 / G2 Phase: Growth phase 2
Sel terus membesar dan kekal aktif secara metabolik .
The cell continues to grow and remains metabolically active.

• Sel mengumpulkan tenaga dan membuat persiapan terakhir untuk pembahagian sel yang seterusnya.
The cell accumulates energy and completes its final preparations for the next division.

Peringkat-peringkat Mitosis dalam Sel

Stages of Mitosis in a Cell

(a) Profasa / Prophase 1. Kromosom menjadi pendek dan tebal yang boleh dilihat di bawah mikroskop cahaya.
The chromosomes shorten and thicken that are visible under a light microscope.

2. Setiap kromosom terdiri daripada dua kromatid kembar yang bercantum di sentromer .
Each chromosome consists of two sister chromatids joined at the centromere .

3. Setiap pasangan sentriol bergerak ke kutub sel bertentangan .
Each pair of the centrioles move to the opposite poles of the cell.

4. Gentian gelendong mula terbentuk antara sentriol.
Spindle fibres begin to form between the centrioles.

5. Dalam sel tumbuhan, gentian gelendong terbentuk tanpa sentriol .
In plant cells, the spindle fibres form without the presence of centrioles .

6. Pada akhir profasa, nukleolus hilang. / At the end of prophase, the nucleolus disappears.

7. Membran nukleus terurai. / The nuclear membrane breaks down.

80

(b) Metafasa / Metaphase Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  

1. Kromosom tersusun sebaris di satah khatulistiwa sel.
The chromosomes line up along the equator of the cell.
2. Gentian gelendong melekat pada kromosom di sentromer .
Spindle fibres attach to the chromosomes by centromer .
3. Metafasa berakhir apabila sentromer mula membahagi.
Metaphase ends when the centromeres divide.

(c) Anafasa / Anaphase 1. Gentian gelendong mengecut .
The spindle fibres contract .
.
2. Kromatid kembar terpisah dan bergerak ke hujung kutub sel bertentangan.
The sister chromatids separate and move to opposite poles of the cell. BAB 6

3. Sitoplasma mula membahagi
The cytoplasm starts to divide.

4. Pada akhir anafasa, setiap kutub sel mempunyai satu set kromosom yang lengkap dan seiras
At the end of anaphase, each pole of the cell has a set of complete and identical chromosomes.

(d) Telofasa / Telophase 1. Telofasa bermula apabila kedua-dua set kromosom tiba di kutub sel yang bertentangan.

Telophase begins when the two sets of chromosomes reach the opposite poles of the cell.

2. Kromosom membuka lingkaran dan membentuk semula kromatin .

The chromosome uncoil to become chromatin again.

3. Gentian gelendong menghilang .

Spindle fibres disappear .

4. Membran nukleus baharu terbentuk menyelaputi setiap kumpulan kromosom.

New nuclear membranes form around each group of chromosomes.

Sitokinesis: Pembahagian Sitoplama

Cytokinesis: Cytoplasmic Division

• Sitoplasma sel induk membahagi untuk membentuk dua sel anak .

The cytoplasm of the parent cell divides into two equal halves, forming two daughter cells .

(a) Sitokinesis dalam sel haiwan / Cytokinesis in animal cells (b) Sitokinesis dalam sel tumbuhan / Cytokinesis in plant cells

1. Membran plasma di tengah sel mencerut ke dalam untuk 1. Plat sel terbentuk di antara dua nukleus.
membentuk belahan pencerutan . A cell plate is formed between the two daughter nuclei.

The plasma membrane around the centre of the cell constricts
inwards, forming a cleavage furrow .



2. Belahan pencerutan mencerut beransur-ansur sehingga sel 2. Plat sel berkembang ke luar dari tengah sel, membahagikan
terpisah menjadi dua .
sitoplasma kepada dua sel anak.
The cleavage furrow deepens progressively until the cell separates into The cell plate grows outwards from the centre of the cell, dividing the
two daughter cells .
cytoplasm into two halves.

81

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

Perbandingan antara Mitosis Sel Haiwan dengan Sel Tumbuhan

Comparison between Mitosis in Animal and Plant Cells

(a) Mitosis dalam sel haiwan / Mitosis in animal cells (b) Mitosis dalam sel tumbuhan / Mitosis in plant cells

1. Sel haiwan mempunyai sentriol . 1. Sel haiwan tiada sentriol .
Centrioles are found in animal cells. Centrioles are absent in animal cells.
2. Mitosis berlaku di sel stem . 2. Mitosis berlaku di tisu meristem .
Mitosis occurs in stem cells. Mitosis occurs in meristematic tissues.

Keperluan Mitosis TP 3

BAB 6 The Necessity of Mitosis

(a) Perkembangan embrio / Development of embryo
(b) Pertumbuhan organisma / Growth of organisms
(c) Penyembuhan luka pada kulit / Healing of wounds on the skin
(d) Pertumbuhan semula / Regeneration
(e) Pembiakan aseks / Asexual reproduction

6.3 Meiosis

Meiosis

1. Meiosis adalah satu proses pembahagian nukleus yang mengurangkan bilangan kromosom sel anak kepada
separuh . TP 1

Meiosis is a process of nuclear division that reduces the chromosome number in daughter cells by half .

2. Meiosis hanya berlaku di sel pembiakan dan membawa kepada pembentukan gamet.

Meiosis occurs only in reproductive cells and results in the formation of gametes .

Jenis-jenis Sel yang Menjalankan Meiosis TP 2

The Types of Cell that Undergo Meiosis

Organsima / Organisms Haiwan / Animals Tumbuhan / Plants
Jantina / Sex
Jantan / Male Betina / Female Jantan / Male Betina / Female
Organ pembiakan
Reproductive organ Testis Ovari Anter Ovari
Testes Ovaries Anthers Ovary
Gamet
Gametes Sperma Ovum Butir debunga Sel telur
Sperms Ovum Pollens Egg cell

Keperluan Meiosis TP 3

The Necessity of Meiosis

(a) Menghasilkan gamet haploid / Producing haploid gametes.

• Pembahagian sel secara meiosis menghasilkan gamet haploid (n) untuk pembiakan secara seks.
Meiotic cell division produces haploid (n) gametes for sexual reproduction.

(b) Mengekalkan bilangan kromosom diploid dari generasi ke generasi.
Maintains diploid number of chromosomes from generation to generation.

82

Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  

• Semasa persenyawaan , nukleus gamet jantan dan gamet betina bergabung membentuk zigot

diploid . Bilangan kromosom yang normal dapat dikekalkan dari generasi ke generasi.

When the male and female gamete nuclei fuse during fertilisation , a diploid zygote is produced. The

normal chromosome number is maintained through the generations.

(c) Menghasilkan variasi genetik dalam gamet.
Producing genetic variations in gametes.

• Variasi genetik di kalangan suatu spesies meningkatkan peluang kemandirian spesies jika keadaan
persekitaran berubah.

Genetic variations within a population increases its chances of survival when environment changes.

Pembahagian Sel Secara Meiosis / Meiotic Cell Division TP 1

1. Pembahagian sel secara meiosis berlaku selepas interfasa .
Meiotic cell division occurs after interphase .

2. Ia terdiri daripada meiosis I dan meiosis II .
It consists of first meiotic division and second meiotic division.

3. Setiap proses bermula dengan pembahagian nukleus , diikuti oleh pembahagian sitoplasma . BAB 6
Each of them starts with nuclear division, followed by cytoplasmic division.

4. Pembahagian meiosis I melibatkan pasangan dan perpisahan kromosom homolog .
First meiotic division involves the pairing and separation of homologous chromosomes .

5. Pembahagian meiosis II melibatkan perpisahan kromatid kembar setiap kromosom.

Second meiotic division involves the separation of the sister chromatids of each chromosome.

Peringkat Meiosis / Stages of meiotic division TP 2

(a) IInntteerrpfahsaaseGG11 (c) IInntteerrpfahsaaseGG22 (d) Profasa I
Prophase I

(b) Interfasa S (e) Metafasa I
Interphase S Metaphase I

(l) Sitokinesis Meiosis (f) Anafasa I
Cytokinesis Meiosis Anaphase I

(k) Telofasa II (g) Telofasa I
Telophase II Telophase I
(h) Profasa II
(j) Anafasa II Prophase II
Anaphase II

(i) Metafasa II
Metaphase II

83

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

Meiosis I

(a) Profasa I / Prophase I

• Kromosom memendek , menebal dan jelas kelihatan.
Chromosomes shorten , thicken and become visible.
• Kromosom homolog berpasangan dan membentuk bivalen melalui proses sinapsis .
Homologous chromosomes pair up and form bivalents through synapsis .
• Proses pindah silang berlaku di antara kromatid tak seiras di kiasma dan menghasilkan

kombinasi gen yang baharu pada kromosom.
The crossing over between non-sister chromatids occurs in chiasmata and produces new combinations

of genes along the chromosome.
• Pada akhir profasa I, nukleolus hilang dan membran nuklues terurai.
At the end of prophase I, the nucleolus disappears and the nuclear membrane breaks down.

BAB 6 (b) Metafasa I / Metaphase I

• Pasangan kromosom homolog tersusun di satah khatulistiwa .
Pairs of homologous chromosomes line up along the equator of the cell.
• Gentian gelendong melekat pada kromosom.
Spindle fibres attach to the chromosomes.

(c) Anafasa I / Anaphase I

• Gentian gelendong mengecut dan menarik kromosom homolog. Setiap kromosom
pada pasangan homolog itu terpisah dan bergerak ke kutub sel yang bertentangan .

The spindle fibres contract and pull the homologous chromosomes apart. The chromosomes of
each homologous pair separate and move to opposite poles of the cell.

• Sitoplasma mula membahagi.
The cytoplasm starts to divide.

(d) Telofasa I / Telophase I

• Setiap kumpulan kromosom yang mengandungi bilangan kromosom haploid tiba di kutub sel.
Each group of chromosomes containing haploid number of chromosomes arrive at the poles .

• Gentian gelendong mula hilang.

Spindle fibres disappear.

• Membran nukleus baharu terbentuk mengelilingi tiap kumpulan kromosom.

New nuclear membranes form around each group of chromosomes.

(e) Sitokinesis / Cytokinesis

• Dua sel anak haploid terbentuk.
Two haploid daughter cells formed.

84

Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  

Meiosis II
(a) Profasa II / Prophase II

• Membran nuklues menghilang.

The nuclear membranes breaks down again.

• Gentian gelendong mula terbentuk.

Spindle fibres begin to form.

(b) Metafasa II / Metaphase II

• Kromosom tersusun pada satah khatulistiwa .
Chromosomes line up along the equator .
• Gentian gelendong yang baharu terikat kepada kromosom.
New spindle fibres attach to the chromosomes.

(c) Anafasa II / Anaphase II BAB 6

• Gentian gelendong mengecut. Setiap kromatid kembar berpisah dan
tertarik ke kutub sel yang bertentangan.

The spindle fibres contract. The sister chromatids separate and move to opposite
poles of the cells.

• Sitoplasma mula membahagi.
The cytoplasm starts to divide.

(d) Telofasa II / Telophase II

• Kromosom membuka pintalan dan menjadi kromatin semula.
The chromosomes uncoil to become chromatin again.

• Gentian gelendong menghilang.

The spindle fibres disappear.

• Membran nuklues baharu terbentuk mengelilingi tiap

kumpulan kromosom.

New nuclear membranes form around each group of
chromosomes.

(e) Sitokinesis / Cytokinesis

• Empat sel anak haploid .
Four haploid daughter cells formed.

Perbandingan antara Meiosis I dengan Meiosis II

Comparison between Meiosis I and Meiosis II

Persamaan / Similarities nukleus dan sitoplasma.
1. Kedua-duanya merupakan proses pembahagian
Both involve division of nucleus and cytoplasm.

85

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

2. Kedua-duanya terdiri daripada empat peringkat: profasa , metafasa , anafasa dan telofasa .
Both consist of four stages: prophase , metaphase , anaphase and telophase .

3. DNA bereplikasi hanya sekali .
DNA replicates only once .

Perbezaan / Differences TP 2

Meiosis I Meiosis II

1. Semasa profasa I, kromosom homolog berpasangan 1. Semasa profasa II, sinapsis kromosom homolog dan pindah
menerusi sinapsis dan pindah silang di antara kromatid silang di antara kromatid tidak seiras tidak berlaku.
tidak seiras berlaku.
During prophase II, synapsis of homologous chromosomes and
During prophase I, homologous chromosomes pair up through crossing over between non-sister chromatids does not take
synapsis and crossing over between non-sister chromatids place.

occurs.

BAB 6 2. Semasa metafasa I, pasangan kromosom homolog tersusun 2. Semasa metafasa II, kromosom tersusun pada satah
pada satah khatulistiwa sel. khatulistiwa sel. equator of

During metaphase I, paired homologous chromosomes align at the During metaphase II, chromosomes align at the
equator of the cell. the cell.

3. Semasa anafasa I, pasangan kromosom homolog terpisah 3. Semasa anafasa II, kromatid kembar terasing, menjadi
dan bergerak ke kutub sel bertentangan. Kromatid kembar
kromosom anak yang bergerak ke kutub sel bertentangan.
masih melekat bersama dan bergerak sebagai satu unit. During anaphase II, sister chromatids separate, becoming daughter
During anaphase I, paired homologous chromosomes separate and
chromosomes that move to the opposite poles.
move to opposite poles. Sister chromatids are still attached together
and move as a unit.

4. Pada akhir telofasa I, dua sel anak haploid terbentuk 4. Pada akhir telofasa II, empat sel anak haploid terbentuk.
two haploid daughter cells are
At the end of telophase I, At the end of telophase II, four haploid daughter cells are
formed. formed.

5. Setiap sel anak mempunyai satu jenis kromosom; kromosom 5. Setiap sel anak mempunyai bilangan kromosom yang sama
paternal atau kromosom maternal . seperti sel haploid yang dihasilkan dalam meiosis I, tetapi
setiap sel mempunyai hanya satu daripada kromatid kembar.
Each daughter cell has only one of each type of chromosomes;
either the paternal or maternal . Each daughter cell has the same number of chromosomes as the
haploid cell produced in meiosis I, but each cell has only one of the
sister chromatids.

Perbandingan antara Pembahagian Sel Secara Meiosis dengan Mitosis

Comparison between Meiotic and Mitotic Cell Divisions

Persamaan / Similarities pembahagian sitoplasma .
1. Kedua-duanya melibatkan pembahagian nukleus dan

Both involve nuclear division and cytoplasmic division.

2. DNA bereplikasi hanya sekali sebelum kedua-dua proses bermula.
DNA has replicated only once before the beginning of both processes.

3. Kedua-duanya melibatkan pergerakan kromosom dan pengagihan kromosom di sel-sel anak.
Both involve the movement of chromosomes and the distribution of chromosomes among daughter cells.

86

Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  

Perbezaan / Differences Meiosis Mitosis

1. Tempat berlaku (a) Berlaku di organ pembiakan semasa (b) Berlaku di sel badan (sel soma ).
Place of occurrence cells ( somatic
pembentukan gamet . Occurs in normal body
Occurs in reproductive organs during gamete cells).

production.

2. Proses (a) Bilangan (i) Sel induk membahagi dua kali dalam (ii) Sel induk membahagi satu kali dalam
Process pembahagian sel
satu kitar. satu kitar.
Number of cell Involves two nuclear divisions in one cycle. Involves only one nuclear division in one cycle.
division

(b) Pemasangan (i) Kromosom homolog berpasangan menerusi (ii) Pemasangan kromosom homolog melalui
kromosom sinapsis semasa profasa I . sinapsis tidak berlaku.
homolog
Homologous chromosomes pair up through Pairing of homologous chromosomes through
Pairing of synapsis during prophase I . synapsis does not occur.
homologous
chromosomes

(c) Pemisahan (i) Berlaku / Occurs (ii) Tidak berlaku.
kromosom Does not occur.
homolog ke
nukleus sel anak BAB 6

Separation of
homologous
chromosomes into
daughter nuclei

(d) Pindah silang (i) Pindah silang antara kromatid bukan seiras (ii) Pindah silang tidak berlaku.
Crossing over berlaku semasa profasa I . Crossing over does not occur.

Crossing over between non-sister chromatids may
occur during prophase I .

3. Sel anak (a) Bilangan sel anak (i) Empat sel anak dihasilkan. (ii) Dua sel anak dihasilkan.
Four daughter cells produced. Two daughter cells produced.
(gamet) terhasil
Daughter Number of
cell
daughter cell
(gamete) produced

(b) Bilangan (i) Separuh daripada sel induk (  haploid / n) (ii) Sama dengan sel induk (  diploid / 2n).
kromosom Half of parent cell ( haploid /n). Same as parent cell (  diploid / 2n).

Chromosomal
number

(c) Kandungan (i) Berbeza daripada sel induk dan antara (ii) Sama dengan sel induk dan antara satu
genetik
satu sama lain. sama lain.
Genetic Different from parent cell and among daughter Same as parent cell and among daughter

make-up cells. cells.

(d) Jenis sel (i) Gamet (ii) Sel badan
Type of cell Gametes Body cell.

4. Kepentingan (a) Menghasilkan gamet haploid untuk (c) Menghasilkan sel-sel yang seiras secara
Importance genetik untuk pertumbuhan , penggantian
pembiakan seks. sel rosak dan pembiakan aseks .
Forms haploid gametes for sexual reproduction.
(b) Menghasilkan variasi genetik untuk Forms genetically identical cells for growth ,
repair and asexual reproduction.
kemandirian spesies.
Produces genetic variations that enhance the

survival of the species.

87

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

6.4 Isu Pembahagian Sel Terhadap Kesihatan Manusia

Issues of Cell Divison on Human Health

Kesan Ketidaknormalan Mitosis terhadap Kesihatan Manusia

The Effects of Abnormal Mitosis on Human Health

1. Kerosakan pada gen yang mengawal mitosis boleh mengakibatkan pembahagian sel tidak terkawal .

Damage to the genes that regulate mitosis can lead to uncontrolled cell division.

2. Ini mengakibatkan perkembangan tumor , sekumpulan sel abnormal.

As a consequence, a group of abnormal cells called a tumour develops.

3. Tumor boleh berkembang di mana-mana organ badan dan biasanya dijumpai di peparu , kelenjar prostat
(lelaki), payu dara dan ovari (perempuan), usus besar , perut , esofagus dan pankreas .

Tumours can develop in any organ of the body which most commonly found in the lungs , prostate gland (male),
breast and ovaries (female), large intestine , stomach , oesophagus and pancreas .

BAB 6 1. Sel-sel normal terdedah kepada bahan 2. Bahan karsinogen mengakibatkan 3. Sel yang mengalami mutasi membahagi secara
karsinogen yang berpanjangan . mutasi dalam sel. mitosis dengan pesat dan tanpa kawalan .

Normally dividing cells subjected to The carcinogen causes mutation The mutated cell undergoes repeated, rapid
mitosis without control .
prolonged exposure to carcinogens . in a cell.

Sinar ultra ungu Formaldehid Mutasi / Mutation
Ultraviolet light Formaldehyde

Asap rokok Sel abnormal / abnormal cells Mitosis pesat
Tobacco smoke Rapid mitosis

Sinar-X
X-rays

Sel-sel normal Sel normal Kekal di tapak asal
Normal cells Normal cells Remain at the
original site

5. Tidak mengancam nyawa, boleh 4. Tumor benigna menyerap Tumor benigna Kumpulan sel abnormal- tumor
disingkir melalui pembedahan . nutrien , membesar , tetapi Benign tumour Abnormal mass of cells- tumour

Are not life-threatening, can be tidak merebak dari tapak asal . Menyerang,
removed by surgery . A benign tumour absorbs nutrients , merebak dan
memusnahkan tisu
7. Biasanya merosakkan fungsi enlarges , but does not spread from bersebelahan.
organ yang diserang dan akan its original site . Intrude, spread
membawa kepada maut . and destroyed
6. Tumor maligna bersaing dengan sel neighbouring
The functions of invaded organs are
typically impaired and lead to tissues.
death .
normal di sekitarnya untuk mendapatkan
nutrien dan tenaga yang terhad bagi

pertumbuhannya.

A malignant tumour competes with surrounding Tumor maligna atau sel-sel kanser
normal cells to obtain sufficient nutrients and Malignant tumour or cancer cells
energy for growth.

88

Kesan Ketidaknormalan Meiosis terhadap Kesihatan Manusia Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  
TP 4
The Effects of Abnormal Meiosis on Human Health

1. Jika meiosis tidak berlaku dengan betul, tak-disjunsi berlaku; gamet terbentuk akan mempunyai bilangan
kromosom yang abnormal .

If meiosis does not occur properly, non-disjunction occurs; the gametes formed will have an abnormal number of
chromosomes

2. Hasilnya, zigot yang terbentuk akan menjadi abnormal.

As a result, the zygote that is formed would later become abnormal.

3. Contoh kesan ketidaknormalan meiosis: Sindrom Down

Example of abnormal meiosis: Down Syndrome

Ibu ( 46 kromosom) + 46 Bapa ( 46 kromosom)
Mother ( 46 chromosomes) 46 Father ( 46 chromosomes)

Tak disjunsi Meiosis BAB 6
Non-disjunction

Ovum ( 24 kromosom) Sperma ( 23 kromosom)
22 Ovum ( 24 chromosomes) 24 23 Sperm ( 23 chromosomes)

Persenyawaan
Fertilisation

Zigot ( 47 kromosom)
47 Zygote ( 47 chromosomes)

1 2 3 4 5 1 2 3 4 5

Kromosom manusia normal – 6 7 8 9 10 6 7 8 9 10
23 pasang (46 kromosom)
Typical human chromosomes 1 1 12 13 14 15 11 12 13 14 15
23 pairs (46 chromosomes)
1 6 17 18 19 20 16 17 18 19 20
4. Simptom Sindrom Down
21 22 23 21 22 23
Symptoms of Down syndrome:
Lebih satu kromosom pada kromosom nombor 21
An extra chromosome number 21

Mata sepet Kecacatan mental
Slanted eye Mental retardation

Hidung rata Lidah tebal
Flat nose Protruding tongue

89

  Biologi  Tingkatan 4  Bab 6 Pembahagian Sel

PRAKTIS SPM 6

Soalan Objektif Berapakah bilangan kromosom
1. Rajah 1 menunjukkan tempoh masa dalam jam bagi fasa dalam sel anak selepas meiosis
lengkap?
kitar sel. What is the number of chromosomes
2 018 Diagram 1 shows a period in hour for the phase of the cell cycle. found in a daughter cell after completing

BAB 6 FGa1 spahaGse1 Fasa S FGa2 spahaGse2 meiosis? Rajah 3/ Diagram 3
MITOSISS phase
A 24 C 6

B 12 D 3

0 10 20 4. Rajah 4 menujukkan satu fasa dalam pembahagian sel.
(Jam / Hours) Rajah 1/ Diagram 1 Diagram 4 shows a phase in the cell division.

Berapakah jumlah tempoh masa bagi fasa G2? 2018

What is the total period for G2 phase?
A 6 jam C 16
jam
6 hours 16 hours
B 14 jam D 24 jam

14 hours 24 hours Rajah 4/ Diagram 4

2. Rajah 2 menunjukkan kariotip manusia yang abnormal. Apakah fasa yang ditunjukkan?
Tiga kromosom hadir dalam kromosom 21 dalam semua What is the phase shown?
A Anafasa I C Metafasa I
2018 sel akibat mutasi. Anaphase I Metaphase I
Diagram 2 shows the karyotype of an abnormal human. Three B Anafasa II D Metafasa II

copies of chromosome 21 are present in all cells as a result of Anaphase II Metaphase II
mutation.

12 3 45 5. Rajah 5 menunjukkan satu fasa P
dalam pembahagian sel.
6 7 8 9 10 11 12
2 018 Diagram 5 shows a phase in a cell
division.

Apakah kemungkinan yang
akan berlaku jika struktur P
tidak terbentuk?
13 14 15 16 17 18 What are the possibilities that
happen if structure P is not formed?
Rajah 5/ Diagram 5

19 20 21 22 X Y I Pindah silang yang tidak lengkap

Rajah 2/ Diagram 2 Incomplete crossing over

Pada peringkat pembahagian sel yang manakah II Kromosom homolog tidak berpisah
Homologous chromosomes do not separate
mutasi ini berlaku? III Boleh menyebabkan mutasi gen
At which stage of cell division did this mutation occur?
A Profasa I C Anafasa I Can cause gene mutation
IV Salah satu sel mempuyai bilangan kromosom
Prophase I Anaphase I yang lebih
B Metafasa I D Telofasa I
One of the cells has more number of chromosomes
Metaphase I Telophase I
A I dan III / I and III
3. Rajah 3 menunjukkan kromosom homolog dalam satu B II dan III / II and III
peringkat meiosis. C II dan IV / II and IV
D III dan IV / III and IV
Diagram 3 shows homologous chromosomes in a stage of
meiosis.

90

Biologi  Tingkatan 4  Bab 6 Pembahagian Sel  

Bahagian A
1. Rajah 1 menunjukkan peringkat-peringkat yang berbeza dalam suatu pembahagian sel.
Diagram 1 shows the different stages during cell division.

PQR S T
Rajah 1/ Diagram 1
[1 markah / mark]
(a) Namakan jenis pembahagian sel dalam Rajah 1. [1 markah / mark]
Name the type of cell division in Diagram 1. [1 markah / mark]
[2 markah / marks]
Mitosis

(b) Namakan jenis sel pada manusia di mana proses dalam 1 (a) berlaku. BAB 6
Name the type of cells in humans where the process in 1 (a) takes place.

Sel soma / Somatic cells

(c) (i) Susun peringkat pembahagian sel itu dalam urutan yang betul.
Arrange the stages of the cell division in the correct sequence.
P→Q→T→S→R

(ii) Namakan peringkat P hingga T. / Name stages P to T.

P : Interfasa / Interphase Q : Profasa / Prophase

(d) Berdasarkan Rajah 1, terangkan perlakuan kromosom dalam peringkat
Based on Diagram 1, explain the chromosomal behaviour in stage

(i) T : Kromosom tersusun pada satah khatulistiwa.

Chromosomes align at the metaphase plate.

(ii) S : Kromatid beradik berpisah dan bergerak ke kutub yang bertentangan.

(e) Sister chromatids separate and move towards the opposite poles.
[2 markah / marks]

Terangkan kepentingan untuk mengekalkan bilangan kromosom diploid dalam organisma. KBAT Mengaplikasi
Explain the significance of maintaining the diploid chromosome number in organisms.

Untuk membolehkan anak mempunyai bilangan kromosom yang sama seperti induk.

In order for the offspring to have the same chromosomal number as their parent.

[2 markah / marks]

Kuiz 6

91