BIOLOVE SPM

BIOLOGYLOVE© form 4(4551)
THIS IS A COMPILATION OF BIOLOGY ESSAYS AND NOTES
COLLECTED FROM VARIOUS OF SOURCES
I HOPE THIS COMPILATION OF ESSAYS AND NOTES CAN HELP YOU
TO ACHIEVE BETTER RESULT IN BIOLOGY FOR
SPM (SIJIL PELAJARAN MALAYSIA)

THIS IS A REVISED VERSION
of the collection of biology essays©
 MORE ATTRACTIVE
 MORE NOTES
 MORE ESSAYS
 SMART EXAM TIPS

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BIOLOGYLOVE

second edition 2.0/2012

Chapter 2 : Cell Structure and Cell Organisation
 Phagocytosis
The pseudopodia are also used for feeding.
Amoeba sp. engulfs food by phagocytosis.
Amoeba sp. is a holozoic organisms which feed on microscopic organisms such as bacteria.
The presence of food causes Amoeba sp.to advance by extending its pseudopodia.
The pseudopodia encloses the food which is then packaged in food vacoule.
The food vacoule fuses with lysosome and the food is digested by hydrolitic enzyme called lysozyme.
The resulting nutrients are absorbed into the cytoplasm.

 Comparison between the structure of animal and plant cell
Similarities

Smart Exam Tips !
- Comparison include
similarities and differences

- Both have a nucleus, cytoplasm, a plasma membrane, Golgi apparatus, mitochondria, endoplasmic
reticulum and ribosomes.

Smart Exam Tips !
- Use word BOTH
for similarities

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Differences
Animal cells
Do not have fixed shape
Do not have cell wall
Do not have chloroplast
Do not have vacoule (if have,
vacoule is only small and
numerous)
Carbohydrate is stored in the form
of glycogen
Have centrioles

But
Smart Exam Tips !
- Use word BUT for
differences

Plant cells
Have fixed shape
Have cell wall
Have chloroplast
Mature plant cell have a large
central vacoule
Carbohydrate stored as starch
Do not have centrioles

 The density of organelles in specific cells
Type of cells
Organelles found abundantly (high density)
Sperm cells
Mitochondria
Smart Exam Tips !
Muscle cells
- This question
Meristematic cells
always been
asked in Paper 1
Palisade mesophyll cells
Chloroplast
and 2
Pancreatic glands
Ribosome/RER/Golgi apparatus
- SPM Question
Cell in salivary gland
Extra notes : please memorise all the cells ( shape and the function). Please know how to differentiate all the
cells based on the structure and function. – SPM Questions

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Chapter 3 : Movement of Substances Across the Plasma Membrane

 Simple Diffusion
Net movement of molecules or ions from a region of higher concentration to a region of lower
concentration.
Going down concentration gradient until an equilibrium is achieved.
The particles are distibuted equally throughout the system.
 Osmosis : Diffusion of water
Net movement of freely moving water from a region of lower solute concentration to a region of higher
solute concentration through a semi-permeable membrane.//
Net movement of water from region higher water concentration to a region of lower water concentration
through a semi-permeable membrane.//
Net movement of water from hypotonic region to hypertonic region through a semi-permeable
membrane.

**Choose any one

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Facilitated Diffusion
For water soluble molecules//molecules which are not soluble in lipids (ions, nucleic acid, amino acids and glucose)
Carrier Protein
The carrier protein function by binding to the molecules to pass through the plasma membrane.
The molecules move to the carrier protein which is specific for the molecules.
Molecules bind with the carrier protein at the active site.
Carrier protein changes its shape and pass the molecules through the plasma membrane.

 Active Transport
Movement of molecules or ions against the concentration gradient across the plasma membranes.
Requires both carrier proteins and expenditure of energy.
Energy from ATP (adenosine triphosphate) that is generated during respiration in the mitochondria.
Has active sites which bind to the ATP molecules.
The carrier protein changes shape when the phosphate group from the ATP molecule binds to it
Then the solute is moved across the plasma membrane.
 Animal and plant cells in an isotonic solution
Solution in which the solute concentration is equal to that of the cytoplasmic fluid.
Water diffuse in and out of the cells at equal rate.
No net movement of water.
Cells retain its normal shape.

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 Hypotonic solution

Hypertonic
 Hypertonic solution solution

 Concentration of solute outside a cell is lower than
 The concentration of solute in the
concentration of solute inside cell.
solution is higher than the concentration
Animal cells
of solutes within the cell sap.
 Cell placed in hypotonic solution.
 Hypertonic to the cell sap of the cell.
 Solution is hypotonic to the cell sap of the cell.
Animal cells
 Net movement of water into the cells via osmosis.
 Net movement of water from inside to
 Cell swells up.
the outside of the cell.
 When extremely hypotonic, cells will eventually burst
 Cells shrink//shrivel, internal pressure
 Cannot withstand the osmotic pressure because of thin plasma decrease.
membrane.
 Red blood cells immersed in hypertonic
 E.g : red blood cells (haemolysis)
solution , the cell shrink and the plasma
Plant cells
membrane crinkles up.
 Do not burst
 Cell undergone crenation.
 Rigid cell wall.
Plant cells
 Water diffuse into vacoule of cell via osmosis through
 Water diffuse out via osmosis.
a semi-permeable membrane.
 Vacoule and cytoplasm shrink and
 Cell swells up and becomes turgid
plasma membrane pulls away from the
 Tugor pressure in plant.
cell wall.
 Supporting the plant.
 This process called plasmolysis.
Exam tips : To answer Question on Osmosis
1. Mention about the solution (whether HYPERTONIC or HYPOTINIC) to
the cell sap of the cell.
2. Water diffuse into/out of the cell via OSMOSIS
3. What happen to cell (plasmolysis, crenation, turgid, haemolysis)

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Similarities between facilitated diffusion and active transport
S1- Both (ways of transportation)need carrier protein.
E1- To bind with molecules/ion/substrate/examples
S2- Both transport specific molecules only.
E2- Because the carrier protein have specific site to certain molecules.
S3- Both processes occur in living cell.
E3- Because carrier protein need/can change shape to allow substances to move across.

Sodium Potassium Pump

P1
P2
P3
P4
P5
P6

The concentration of sodium ions is higher on the inside of the cell
The sodium ions approach the carrier protein. The carrier protein has a site for the ions to bind with
The carrier protein binds to the sodium ions. The ATP molecule is split into Adenosine diphosphate (ADP) and
phosphate(P).
The phosphate group the attach itself to the carrier protein. The splitting of ATP releases energy to the carrier
protein.
Energy from the ATP changes the shape of carrier protein.
This cause the carrier protein to release the sodium ions outside of the cell

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Examples of transport of substances
Transport
process
Simple diffusion

Examples
-

Facilitated
diffusion

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Ions intake by root hairs of a plant

Gaseous exchange in the alveoli and
blood capillaries

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Osmosis

Active transport

Absorption of digested food in the
villus

Absorption of water by root hair
cell

Example of question

Explain how red blood cell burst
F1 the solution outside the cell is hypotonic to the cell sap
E1 water molecules diffuse into the cell by osmosis
E2 the plasma membrane is too thin to withstand the osmotic
pressure inside the cell
E3 So the cell burst/haemolysis occured


This how you answer the question
Flaccid cell
F1 : the cell sap is hypotonic to the solution
F2 : water diffuse out from the cell via osmosis
F3 : cytoplasm shrink//plasmolysis occured
F4 : cell becomes flaccid
Turgid cell
F1 : the cell sap is hypertonic to the solution
F2 : water diffuse into the cell via osmosis
F3 : the cell swells up//vacoule becomes bigger
F4 : the cell becomes turgid

Percent change in mass
ofpotato
25
20

At this stage, the sucrose solution is isotonic to
the cell sap of the potato
Water diffuse into and out of the cell via
osmosis at equal rate

P

15
10

At Q
-

5
0
-5

At P
-

Sucrose molarity (mol)
0.2

0.4

0.6

-10

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0.8

1.0
Q

-

The solution is hypotonic to the cell
sap.
Water diffuse into the cell via osmosis.
Cell becomes turgid.
That is why the mass increased.

The solution is hypertonic to the cell
sap.
Water diffuse out from the cell via
osmosis.
Cell plasmolysed.
That is why the mass decrease


Chapter 4 : Chemical Composition of the Cell
Phosphate group

carbohydrate

Nitrogenous base
Adenine (A), Guanine (G), Cytosine (C),
Thymine (T)
monosaccharides

Nitrogenous base

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polysaccharides

disaccharides

cellulose

glycogen

SPM 2011

Pentose sugar

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Complementary base pairing
A--------T
C--------G
G--------C
T--------A

starch

‘Lock and key’ hypothesis
The substrate molecule fits into the active site of the enzyme
molecule.
The substrate is the ‘key’ that fits into the enzyme ‘lock’.
Various types of bonds such as hydrogen and ionic bonds hold
the substrate
in the active site forming the enzyme-substrate complex.
Once the complex is formed, the enzyme changes the substrate
to its product.
The product leaves the active site.
The enzyme is not altered by the reaction and it can be reused.

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2 types of nucleic acid
 Deoxyribonucleic acid (DNA)
 Ribonucleic acid (RNA)
DNA found in
 Nucleus of a cell
 Chloroplast
 Mitochondria

LIPID
Energy rich organic compound.
Contains phosphorus and nitrogen.
Insoluble in water.
Includes fats, oil, waxes, phospholipids, steroids.
TRIGLYCERIDES

DNA contains genetic information about an organism
RNA found in
 Cytoplasm
 Ribosomes
 Nucleus

condensation

hydrolysis
The importance of Nucleic acids
Store genetic information
The stored genetic information can be duplicated/copied
for transmission
Stable in storing genetic information within the lifetime
of organism
Enable the transmission of genetic information from on
egeneration to next generation

glycerol

3 molecules
of fatty acids

triglycerides

PROTEINS
Amino acid as the basic unit (monomer)
Build new cell for growth.

CARBOHYDRATES
Provide energy during respiration.
Build cell wall (cellulose) in plants.
External skeleton of insects.

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Protein Structure

Aspect

Unsaturated fats

Type s of chemical
bonds

All covalent bonds
between carbon
atoms single C - C

Existence of double
covalent bonds
between carbon
atoms C = C

Reactivity

Less reactive

More reactive
because of double
bond

State of matter at
room temperature

Solid (fats)

Liquid (oil)

Source

Mainly from animal
products : red meat,
chicken fat, buuter
and coconut oil

Mainly from plant :
Vegetable,
palm/corm/olive

Effects on blood
cholestrol level
Tertiary structure
 Enzymes
 Hormones
 Antibodies
 Plasma proteins

Saturated fats

Increase level of bad
cholestrol
Contains more
cholestrol

Contains less
cholestrol

Quartenary structure
 Haemoglobin
 Pore protein

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General characteristics of enzymes
Effects of temperature on enzyme activity
Alter or speed up the rates of chemical
reactions
 At low temperature, reaction takes place slowly.
Remain unchanged at the end of reaction.
 As temperature increases, movement of substrate
increase.
Do not destroyed by reactions they catalysed.
 Increase their chances of colliding with each other
Have specific sites called active site to bind
and with the active site of the enzymes.
with specific substrates.
 At optimum temperature, the reaction is at
Needed in small quantities.
maximum rate.
Reaction are reversible
 Beyond the optimum temperature, rate of reaction
Can be slowed down or stopped by inhibitors.
will not increase.
E.g: lead and mercury
 Bonds that hold enzyme molecules begin to break.
Require helper molecules, called cofactors.
 Actives sites destroyed.
Inorganic cofactor : ferum, copper
 Enzyme denatured.
Organic cofactor: water soluble vitamins, B
vitamins .
Extracellular enzyme
Extracellular enzyme is produced in a cell, then packed and secreted from the
cell.
It catalyses its reaction outside the cell. An example is amylase.
The nucleus contains DNA which carries the information for synthesis of
enzymes.
Protein that are synthesised at the ribosomes are transported through the
spaces within the rough ER.
Proteins that depart from the rough ER wrapped in vesicles then bud off from
the membrane of the rough ER.
These transport vesicle then fuse with the membranes of the golgi apparatus
and empty their contents into the membranous space.
The proteins are further modified during their transport in the Golgi apparatus.
For example, carboohydrates are added to protein to form glycoproteins.
Secretory vesicles containing these modified protein bud off from the Golgi
apparatus and travel to the plasma membrane.
Enzymes are released.

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What is monosaccharides?
Glucose
Fructose
Galactose

Exam tips :

Glucose + monosaccharides

Exam tips :
Addition of water is
necessary. Must be
written!!

What is disaccharides?
Maltose
Sucrose
Lactose
Glucose + glucose
Glucose + fructose
Glucose + galactose

condensation

condensation
condensation
condensation

maltose + water
sucrose + water

disaccharides + water

Enzymes for substrates
Starch
Sucrose
Lactose
Maltose

: amylase
: sucrase
: lactase
: maltase

lactose + water

Reducing Sugar and Non-Reducing sugar
To test for reducing sugar, Benedict’s test must be carried out.
If the colour of Benedict’s solution changes from BLUE to BRICK-RED PRECIPITATE, that’s mean the solution
contains Reducing sugar.
Sometime sucrose can show positive result. Why? Because sucrose can be
Reducing sugar is
hydrolysed into glucose and fructose.
All monosaccharides, maltose and lactose
Non reducing sugar is
All polysaccharides and sucrose
Give negative result on Benedict’s test
(colour does not change)

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BIOLOGYLOVE

Steps:
1. Add dilute hydrochloric acid and boil the solution.
2. When the solution is cooled, put a spoon of calcium carbonate to
neutralise the solution.
3. Then test with benedict’s solution.
4. Then the colour will change.

Chapter 5 : Cell Division
Where do mitosis occur?
In plants, mitosis occur in meristematic tissues.
What is meristematic tissues?
Roots tips
Shoot tips
Bud tips
Terminal buds
Cambium
In animal?
All parts of the body except TESTES and OVARY
CHROMOSOMES AND CHROMOSOMAL NUMBER

This is a chromosome
With sister
chromatids
1 chromosome

This also chromosome
But single chromatids
1 chromosome also

n = haploid
2n = diploid
Human has diploid number (2n) of chromosomes which is 46
Sperm (n=23) + ovum (n=23)  zygote (2n=46)

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Stages in mitosis

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Pro  Meta  Ana  Telo

Prophase
Chromosomes in the
nucleus condense.
Chromosomes appear
shorter and thicker.
Consist of sister
chromatid joined at the
centromere.
Spindle fibres begin to
form.
Centrioles migrate at
opposite poles.
At the end, nucleolus
disappears and the
nuclear membrane
disintegrates.

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Metaphase
Chromosomes align at the
metaphase
plate//equatorial plate.
Mitotic spindle are fully
formed.
Two sister chromatids are
still attached to one
another at the
centromere.
Ends when the centromere
divides.

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phase

Anaphase
Two sister
chromatids separate
at the centromere.
Sister chromatids
pulled apart at
opposite poles.
Chromatids are
referred to as
daughter
chromosomes.

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Telophase
Chromosomes reach
the opposite poles of
the cell.
Chromosomes uncoil
and revert to their
extended
state(chromatin).

Exam Tips :
You can use this note to
answer question about
chromosome behaviour


Cleavage furrow

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Cytokinesis in animal cell
Process of cytoplasmic division.
Begins before nuclear division is completed.
Actin filament formed contractile ring.
Contracts and constrict pull aring of plasma membrane inwards.
Groove of cleavage furrow pinches at the equator between two nuclei.
Cytokinesis in plant cell
Vesicles join to form a cell plate.
Cell plate grows until it edges fuse with the plasma membrane of the
cell. Cell divides.
Cellulose are produced by the cell to strengthen the new cell walls.

Exam Tips :
Chromosome :
Gamete (ovum and sperm) contain half the number of
chromosome (n=haploid)

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Uncontrolled mitosis
Cell divides through mitosis repeatedly without control.
Produce cancerous cells.
Cancer is a genetic disease caused by uncontrolled mitosis.
Disruption of cell cycle.
Cancerous cells divides freely and uncontrollably not according to
the cell cycle.
These cells compete with surrounding normal cells for energy and
nutrients.
Cancer cells formed tumour.
Tumour invade and destroy neighbouring cells.

Application of Mitosis
Animal Cloning

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1

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4

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3

2
5

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Advantages of cloning
Biotechnologists to multiply copies of
useful genes or clones.
Clones can be produced in a shorter time
and in large numbers.
Cloned plants, however, can produced
flowers and fruits within a shorter period.
Clones are better quality.
Delayed ripening.
Does not need polinating agents.
Propagation can take place at any time.

Disadvantages of cloning
Long-term side effects are not yet known.
May undergo natural mutations. Disrupt
the natural equilibrium of an ecosystem.
Clones do not show any genetic
variations.
Has the same level of resistance towards
certain disease.
Certain transgenic crops contain genes
that are resistant to herbicides.
These genes may be transferred to weeds
through viruses. These weeds would then
become resistant to herbicides.
Cloned animals has shorter lifespan.

Example of Question :
Explain the technic used in animal cloning

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Tissue culture

Meiosis
hormone

Small pieces of
tissue is cut (e.g :
root/shoot)

Plant cell divide by
mitosis to form callus
an undifferentiate
mass of tissue

Meiosis I
1. During prophase I, homologous
chromosomes pair up (synapsis) and
crossing over between non sister
chromatids occurs.
2. During Metaphase I, homologous
chromosomes align at the metaphase
plate (equator, middle) of the cell.
3. During Anaphase I, homologous
chromosomes separates and move to
opposite poles. Sister chromatids are still
attached together and move as a unit.
4. At the end of Telophase I, two haploid
daughter cells are formed. Each daughter
cell has only one of each type of
chromosomes, either the paternal or
maternal chromosomes.

Meiosis II

Plantlet are then
transferred to soil
where they grow into
adult plant

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Cell in the callus develop
into embryo

1. During Prophase II, synapsis of homologous
chromosomes and crossing over between non-sister
chromatids do not take place.
2. During Metaphase II, chromosomes consisting of
two sister chromatids align at the metaphase plate
(equator/middle) of cell.
3. During Anaphase II, sister chromatids separate,
becoming daughter chromosomes that move to
opposite poles.
4. At the end of Telophase II, four haploid daughter
cells are formed. Each daughter cell has the same
number of chromosomes as the haploid cell
produced in Meiosis I, but each has only one of the
sister chromatids.

Stages in Meiosis I

PROPHASE I

METAPHASE I

ANAPHASE I

TELOPHASE I

Stages in Meiosis II

PROPHASE II

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METAPHASE II

ANAPHASE II

TELOPHASE II


Synapsis, Crossing Over and Chiasmata(singular : chiasma)
Synapsis is the process where
the chromosomes pairing up
Crossing over is the process
where non sister chromatids
exchange segment of genetic
material (DNA)
Chiasmata is the point where
the crossing over process occur

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Exam tips :
If the question ask for the type of division, you must
answer whether
1. Mitosis or
2. Meiosis
If the question ask for stage, then you can answer
1. Prophase//Prophase I//Prophase II
2. Metaphase and so on.....
Before you answer the question, look at the
diagram//question whether it is mitosis or meiosis!!!


Exam tips :
- Process in Meiosis II is likely same as
Mitosis
- The term use for Meiosis I is
Homologous chromosome while in
Meiosis II, the term used is sister
Chromatids

Similarities between Meiosis I and Meiosis II
Both consist of four stages
Both involve nuclear division
Both involve cytokinesis
Both have chromatids

Why Meiosis is needed?
Meiosis is needed to produce haploid
gamete
Meiosis only occur in the GONAD (TESTES
and OVARY)

Why gamete must be haploid?
Aspect

Meiosis I
Reduce 2n chromosome to n

Prophase

Metaphase

Anaphase

Telophase

Chromosome already
replicated
Homologous chromosomes
synapse
Chiasma forms and crossing
over takes place
Paired homologous
chromosomes align at the
equator
Separation of homologous
chromosomes to opposite
poles
Single cytokinesis
2 identical cell produced

Meiosis II
Divides the remaining
set of chromosome in
a mitosis like process
No replication
No synapsis
No chiasma and no
crossing over

If gamete is not haploid, the number of
chromosome in the organism will be double
from the real number!!

Sister chromatids
align at the equator

46

Separation of sister
chromatids to
opposite poles
Two cytokinesis
4 identical cell
produced

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46

23

Exam Tips :
You have to master all the Comparison between Meiosis I and Meiosis II
also between Mitosis and Meiosis

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96

46

Chapter 6 : Nutrition
Energy value of food (kJ g-1)
( )

( )

Percentage of vitamin C in fruit juice =

( )

x 0.1%

Test on food samples

Concentration of vitamin C in fruit juice =

Test for

Reagent

Observation

Conclusion

Starch

Iodine solution

Colour change
from yellow to
blue-black
Change from
blue to brick red
precipitate
Change from
blue to purple

Food sample
contains starch

Translucent
mark
Oily mixture on
the surface of
water

Food contain
lipid
Food contain
lipid

Reducing sugar Benedict’s
(refer chapter 4) solution
Protein

Lipid

Biuret’s test
(20% of sodium
hydroxide
solution and 1%
copper(II)
sulphate
solution
Filter paper

Lipid

Emulsion test

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Food contain
reducing sugar
Food contain
protein

x 1.0 mg cm-3

Exam tips:
-

The above formula always been used in the exam.
So, you have to remember the formula.
No formula provided in the exam paper unless it is given in the
question.


Examples of essays
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Digestion in mouth
Secretion of saliva by three pairs of
salivary glands
Saliva contains the enzyme salivary
amylase
Begins the hydrolysis of starch to maltose.
Salivary amylase

Starch + water
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maltose

An additional digestive process occurs
further along the alimentary canal to
convert maltose to glucose.
pH is maintained at 6.5-7.5

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Digestion in stomach
Epithelial lining of the stomach contains
gastric glands.
These glands secrete gastric juice.
Consists of mucus, HCL and enzyme
pepsin and renin.
HCL make the pH around 2.0.
High acidity destroy bacteria.
Acidity stop the activity of salivary
amylase enzyme.

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pepsin
Protein + water

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polypeptides

Renin coagulate milk by converting the
soluble milk protein, caseinogen into
soluble caesin.
Stomach contents become a semi-fluid
called chyme.
Chyme gradually enter the duodenum.

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Digestion in small intestine
Duodenum received chyme from stomach
and secretion from the gall bladder and
pancreas.
Starch, protein and lipids are digested.
Bile which produced by the liver and
stored in the gall bladder enter the
duodenum via the bile duct.
Bile helps neutralise the acidic chyme and
optimise the pH for enzyme action in
duodenum.
Bile salts imulsify lipids, breaking them
down into tiny droplets.
Providing high TSA for digestion.
Pancreas secrete pancreatic juice into
duodenum via pancreatic duct.
Pancreatic juice contains pancreatic
amylase, trypsin and lipase.
Pancreatic amylase complete the
digestion of starch to maltose.
Trypsin digests polypeptides into
peptides.
Lipase complete the digestion of lipid into
fatty acid and glycerol.
Glands in the ileum (small intestine)
secrete intestinal juice which contain
digestive enzyme needed to complete the
digestion of peptides and disaccharides.
Peptides digested by erepsin into amino
acids.
Maltose digested by maltase into glucose.
Disaccharides digested by its own enzyme
into monosaccharides and glucose.

Summary of the digestion
Ileum (small intestine)
Have intestinal gland which secrete intestinal juice
that contain enzymes :

Mouth
Have salivary gland to secrete saliva
Saliva contain salivary amylase
Salivary amylase will digest

maltase
Maltose + water

Stomach
Have gastric gland to secrete gastric juice
Gastric juice contain enzyme pepsin and renin
pH is acidic

Salivary amylase

glucose
Starch + water

pepsin
maltose

protein + water

polypeptides

lactase
Lactose + water

renin

glucose + galactose
Exam tips :

sucrase
Sucrose + water

glucose + fructose
erepsin

Peptides + water

caseinogen + water

amino acids

Please remember that enzyme trypsin always
need alkaline medium (pH > 7)
Pepsin and renin need acidic medium (pH < 7)

Digestion is a popular
question in SPM!!!

Acidic medium is in Stomach
Alkaline medium is in Duodenum

Site of digestion : duodenum
Digestive organ Digestive juice
Liver
Bile, bile salts
Pancreas
Pancreatic juice

enzyme
None
Pancreatic amylase

pH
7.6-8.6
7.1-8.2

Substrates and products
Emulsification of fats
Pancreatic amylase

Starch + water
Trypsin

7.1-8.2

maltose
trypsin

Polypeptides + water
Lipase

7.1-8.2

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BIOLOGYLOVE

peptides
lipase

Lipid droplets + water

casein

fatty acid + glycerol


Digestion in Rodent and Ruminant (essay)
F1
F2
F3
F4
F5
F6
F7
F8
F9
F10
F11
F12
F13
F14

F1
F2
F3
F4
F5
F6

Digestion of cellulose by ruminant
Partially chewed food is passed to the rumen (largest compartment of the stomach).
Cellulose is broken down by cellulase produced by bacteria.
Part of the breakdown products are absobed by bacteria, the rest by the host.
Food enters the reticulum.
Cellulose undergoes further hydrolysis.
The content of the reticulum, called the cud, is then regurgitated bit by bit into the mouth
to be thoroughly chewed.
Helps soften and break down cellulose, making it more accessible to further microbial
action.
The cud is reswallowed and moved to the omasum.
Here, the large particles of food are broken down into smaller pieces by peristalsis.
Water is removed from the cud.
Food particles moved into obamasum, the true stomach of the ruminant. (e.g : cow).
Gastric juice complete the digestion of protein and other food substances.
The food then passes through the small intestine to be digested and absorbed in the
normal way.
Digestion of cellulose by rodent
Caecum and appendix are enlarged to
store the cellulose-digesting bacteria.
The breakdown products pass through the
alimentary canal twice.
The faeces in the first batch are usually
produced at night.
Faeces are then eaten again. To absorb
the products of bacterial breakdown.
The second batch of the faeces are harder
and drier.
Allows rodent (give example) to recover
the nutrients initially lost with the faeces.

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Essays
Explain three structural adaptation of intestinal for effective absorption of food
Exam tips
- When the question asked for
adaptation, your answer must be in the
form of STRUCTURE + FUNCTION
- When the question asked for function,
start your answer with the word TO

P1
P2
P3
P4
P5
P6

The villi have microscopic projection in the epithelial cell called
microvilli
Both the villi and microvilli increases the total surface area of the ileum
for the absorption of soluble end product of digestion
A dense blood capillary network at each villus
Enable the food substances absorbed to be carried away quickly
The epithelial lining of the villus is one-cell thick
Allows soluble food molecules to diffuse quickly into the villus

Explain what happen to the product of digestion after they are absorbed in the
Small intestine
F1
Absorbed by blood capillaries at the villus
P1
Blood capillaries at the villus absorb galactose, amino acid, minerals, vitamin
P2
by simple diffusion through the epithelium of the villus
P3
These substances are carried by the hepatic portal vein to the liver and then
distributed to body cell by the circulatory system (CS)
F2
Absorbed by lacteal at the villus
P4
The products of fats digestion such as glycerol and fatty acid as well as
vitamins are absorbed into the lacteal of the villus
P5
From the ileum, the thoracic duct carries the content of the lacteal into the
bloodstream via the left subclavian vein and is then distributed into body
cells by the CS.

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Similarities between the digestive system and digestion process
of rodent and ruminant
Similarities
S1
Both alimentary canal contain bacteria/protozoa
P1
To secrete extracellular enzyme//to digest
P2
To digest cellulose into glucose
S2
Both have large surface area
P3
To increase rate of diffusion
Differences
D1
Ruminant has 4 stomach chamber but rodent has 1
stomach chamber
P1
Because ruminant have to digest glucose//rodent don’t
have to digest cellulose
D2
Ruminant has a small/short caecum but rodent has a
big/long size caecum
P2
Because ruminant do not digest cellulose
D3
Most bacteria in reticulum but rodent most bacteria in
caecum
P3
To secrete cellulase enzyme
D4
Ruminant, the food passes through the stomach
chamber twice but for rodent, the food passes the
stomach chamber once
P4
To complete thedigestion//to absorb digested food
D5
The food is regurgitated twice in mouth cavity(ruminant)
but the food is regurgitated once in mouth
cavity(rodent)
P5
Food that enter in mouth cavity, oesophagus, rumen and
reticulum are then regurgitated back in mouth cavity for
ruminant


Assimilation of digested food
In the Liver
F1 Amino acids is needed to synthesis new plasma protein
F2 When a short supply of glucose and glycogen occurs, the liver
converts amino acids into glucose
F3 Excess amino acid cannot be stored, so amino acids is broken
down in the liver through process of deamination
F4 During deamination, urea is produced and transported to the
kidney to be excreted
F5 Glucose in the liver is used for respiration
F6 Excess glucose in body is converted into glycogen
F7 by hormone insulin and stored in the liver
F8 Once the glycogen store in the liver is full, excess glucose is
converted into lipid by the liver
F9 Lipids which enter the subclavian vein are transported in the
bloodstream to body cells
In the cell
F1 Amino acids which enter the cells are used for synthesis of
new protoplasm and the repair of damaged tissues
F2 Also important to synthesis of enzymes and hormones
F3 Also used in the synthesis of proteins of plasma membrane
F4 Glucose is oxidised to produce energy during cellular
respiration
F5 Energy is used for various chemical process
F6 Excess glucose is stored as glycogen in the muscle
F7 Lipids such as phospholipids and cholestrol are major
components of plasma membrane
F8 Fats are stored around organ and act as a cushion that protect
the organ
F9 Excess fats are stored in the adipose tissue underneath the
skin as reserve energy
F10 When the body lacks of glucose, fats is oxidised to release
energy

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F1
F2
F3
F4
F5

Formation faeces
Faeces which contain dead cells that are shed from intestinal linings.
toxic substances and bile pigments enter the colon by action of
peristalsis.
In colon, more water is absorbed.
The undigested food residues harden to become faeces.
Faeces contain undigestible residues that remain after the process of
digestion and absorption of nutrients that take place in the small
intestine.

Exam tips
There are only three types of food classes assimilated in the liver and
the body cell, AMINO ACID(MONOMER OF PROTEIN), GLUCOSE AND
LIPID.


Photosynthesis Mechanism
P1
P2

Photosynthesis mechanism
The formation of starch in plants is by the process ofphotosynthesis which occurs in chloroplasts.
The two stages in photosynthesis are the light and dark reactions.

P3

Light reaction:
takes place in grana.

P4

Chlorophyll captures light energy which excites the electrons of chlorophyll molecules to higher
energy levels.
In the excited state, the electrons can leave the chlorophyll molecules.
Light energy is also used to split water molecules into hydrogen ion (H +) and hydroxyl ions (OH-)
(Photolysis of water).
The hydrogen ions then combine with the electrons released by chlorophyll to form hydrogen
atoms.
The energy from the excited electrons is used to form energy-rich molecules of adenosine
triphosphate /ATP.
Hydroxyl ion loses an electron to form a hydroxyl group. This electron is then received by
chlorophyll.
The hydroxyl groups then combine to form water and gaseous oxygen.

P5
P6
P7
P8

Exam tips:
- You have to memorise and
understand the mechanism.
- You also have to know
about the structure of
chloroplast
- Each of the structure of the
chloroplast plays important
role

P9
P10
P11
P12
P13
P14
P15

Dark Reaction:
take place in stroma.
Do not require light energy.
The hydrogen atoms are used to fix carbon dioxide in a series of reactions catalysed by
photosynthetic enzymes
and caused the reduction of carbon dioxide into glucose.
The glucose monomers then undergo condensation to form starch which is temporarily stored as
starch grains in the chloroplasts.

Extra :
- In another words, carbon dioxide is reduced into glucose by the
hydrogen atom

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BIOLOGYLOVE


More essays
Explain the diet for the following people
A lady athlete:
F1

An athlete is a very active person and has high rate of metabolism to produce energy.

E1

The diet should include more carbohydrates to supply enough energy to carry out the vigorous activity in

Exam tips:

sports.// She needs to contract
and relax her muscles frequently for her vigorous activities. //Energy is needed to contract the muscles.
E2

You must be able to
relate the diet with the
needs of the people.

The diet should include more protein to build new tissues to replace tissues that are dead or damaged.

E3

-

She also needs calcium, sodium and potassium to strengthen the bones and to prevent muscular cramp.

A pregnant lady:
F2

A pregnant lady has a high rate of metabolism to provide energy for herself and the baby.

E4

The pregnant lady also needs more iron and calcium to build red blood cells to avoid anemia.

E5

She needs a high quantity of calcium and phosphate to form strong teeth and bones for the baby.

An old lady:
F3

An old lady has low rate of metabolism as she does not need energy to grow. (age)

E6

An old lady needs less carbohydrates and fats because she is less active and thus do not need much energy.

E7

she needs more proteins, vitamins and minerals to replace dead tissues and maintain her daily activities

E8

She needs calcium and phosphorus to prevent osteoporosis

E9

She should avoid food that contains a lot of fats, sugar and salt because excess fat can lead to heart diseases,
excess sugar can cause diabetes mellitus and excess salt can cause high blood pressure.

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Chapter 7 : Respiration
Respiration is the process of oxidation of complex
organic substances with the release of energy utilizes
oxygen an dremoving carbon dioxide in living cells
There are two types of respiration
Aerobic respiration (presence of oxygen)
Anaerobic respiration (absence of oxygen)
Aerobic Respiration
(complete breakdown of glucose)
Glucose + oxygen

carbon dioxide + water + energy

Anaerobic respiration in human muscle
When doing vigorous activities
E.g : running
Need more energy
Glucose

Lactic acid + energy (150 kJ//2 ATP)

Oxygen debt is said to have been paid when all the lactic acid has been
Eliminated through increased breathing.

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Anaerobic respiration in yeast
Glucose

ethanol + carbon dioxide + energy
also called as fermentation and is catalysed by the enzyme zymase

Essays
Compare the aerobic respiration and anaerobic respiration
Differences
Aerobic respiration
Need oxygen
Complete oxidation of glucose
Produce water, carbon dioxide and
energy
Produce 36 ATP (2898 kJ energy)
Occur in mitochondria

But

Anaerobic respiration
No tion of need oxygen
Not complete oxidation glucose
Animal : lactic acid and energy
Plant/yeast : ethanol, carbon dioxide
and energy
Produce 2 ATP. Some of the energy
stored in lactic acid or ethanol
Occur in cytoplasm

Similarities
S1 Both involve cell respiration
S2 Both involve oxidation of glucose
S3 Both produce energy
S4 Both catalysed by enzyme


Respiratory structures and breathing mechanism in human and animals
Respiratory structue is the organ for respiration
Respiratiry surface is the site where the exchange of gases occur
Organisms
Respiratory structure
Respiratory surface
Human
Lungs
Alveoli
Grasshopper/insects Tracheal system
Trachiole
Amoeba sp.
No specific structure
Plasma membrane
Fish
Gills
Lamella/filament
Frog
Skin and lungs
Skin/walled sac in the lungs
Four common characteristics(adaptation) of the respiratory surface
1) Large surface area to maximize the exchange of gases by diffusion
2) Moist respiratory surface for gases to dissolve
3) Thin as one-cell thick for effective diffusion of gases
4) Network of blood capillaries for effective transportation of gases

Adaptation of the filament
F1 Have network of blood capillaries
E1 To transport respiratory gases effeiciently
F2 One-cell thick wall
E2 To nesure diffusion of gases occured easily
F3 Has numerous lamella
E3 To increase total surface area (TSA) for diffusion of gases
F4 Has counter current exchange mechanism
E4 To allow the gaseous exchange efficiently
Countercurrent exchange

Exam tips :
- Memorising the four common characteristics is important because
you can use it to answer question on adaptions of all organisms

Essays
Adaptation of tracheal system
F1
E1
F2
E2
F3
E3
F4

Have spiracle
To allow oxygen and carbon dioxide to get in and out of the cell
The spiracle have valve
To allow the opening and closing of the trachea so that air can go in and out
The trachea are reinforced with chitin(made up of protein)
To prevent the trachea from collapsing
The trachea branched into finer tubes called tracheole which are in direct
contact to the cell and organ
E4 To transport the respiratory gases quickly
F5 The tips of the tracheole is one-cell thick wall and contain fluid(moist)
E5 To allow the respiratory gases to dissolve
F6 The tracheal system has air sacs
E6 To speed up the movement of gases to and from the insect’s tissues

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P1
P2
P3
P4

Blood and water flow in opposite direction
Maintains diffusion gradient
Maximizing oxygen transfer from water to blood
It is significant because ensure oxygen concentration is always higher
in the water
P5 So that oxygen will always diffuse to the blood capillaries
Exam tips:
- Respiratory gases is Oxygen and Carbon dioxide
- For fish, the adaptation of moist respiratory surface is
not suitable because fish is already in the water!!!


The adaptation of respiratory structure of amphibians(frog)

Essays
Anaerobic respiration in human muscle
P1
During a vigorous exercise (running), the breathing rate is increased.
P2
This is to supply more oxygen to the muscles for rapid muscular contraction.
P3
P4

However, the supply of oxygen to muscles is still insufficient.
and the muscles have to carry out anaerobic respiration to release energy.

P5

The glucose is converted into lactic acid, with only a limited amount of energy
being produced.
An oxygen debt builds up in the body, when no oxygen use in energy
production.
High level of lactic acid in the muscles cause them to ache.
After running, the athlete breathes more rapidly and deeply than normal for
twenty minutes.
There is recovery period after 10 minutes until it reaches 20 minutes when
oxygen is paid back during aerobic respiration.
About 1/6 lactic acid is oxidized to carbon dioxide, water and energy.

P6

F1
E1
F2
E2
F3
E3
F4
E4
F5
E5

The skin of the frog is thin
highly permeable to respiratory gases
The skin/membrane of the lung is moist
To dissolve respiratory gases
The skin has alrge number of blood capillaries under the
skin/ lungs have network of blood capillaries
For efficient transport of gases
The lungs consist of a pair of thin walled sacs connected to
the mouth through an opening called glottis
To allow gases from mouth move to the lungs
The membrane of the lungs are thin
To allow diffusion of gases to occur easily

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P7
P8
P9
P10

Anaerobic respiration in yeast
P1 Yeast normally respires aerobically.
P2 Under anaerobic condition, yeast carry out anaerobic respiration.
P3 Produces ethanol.
P4 Process known as fermentation.
P5 Catalysed by the enzyme zymase.
P6 Ethanol produced can be used in making wine and beer.
P7 In bread making, the carbon dioxide released during fermentation of yeast
causes the dough to rise.


Breathing mechanisms in man
P1
P2
P3
P4
P5
P6
P7
P8
P9
P10
P11
P12
P13

Diaphragm is a muscular sheet in the body cavity separating the thorax from the abdomen.
At the start of inhalation, the muscles of the diaphragm contract , making it less arched.
This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs.
When the muscles of the diaphragm relax , it returns to its arched condition , reducing the volume of the thoracic cavity and increasing the
pressure of the thoracic cavity. Air is forced out of the lungs.
The muscles between the ribs are known as intercostals muscles.
During inhalation the external intercostals muscle contracts and raise the lower ribs.
This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs.
During exhalation the external intercostals muscles contract , the ribs return to their original position , reduce the pressure of the thoracic
cavity.
Air is forced out of the lungs.
The alveoli are thin-walled air sacs with the lungs.
These sacs are surrounded by a network of capillaries.
During inhalation the alveoli are filled with air and gaseous exchange occurs between the alveoli and the capillaries.
Oxygen from the alveoli diffuses into the capillaries while carbon dioxide diffuses from the capillaries into the alveoli.

Transport of oxygen and carbon dioxide in the body
P1
Gaseous exchange across the alveolus occurs by diffusion.
P2
Diffusion of gas depends on differences in partial pressure between two regions.
P3
The partial pressure/ concentration of oxygen in the air of the alveoli is higher compared to the partial pressure/ concentration of
oxygen in the blood capillaries.
P4
Therefore, oxygen diffuse across the surface of the alveolus and blood capillaries into blood.
P5
The transport of oxygen is carried out by the blood circulatory system.
P6
Oxygen combines with respiratory pigment called haemoglobin in the red blood cells.
P7
To form oxyhaemoglobin.
P8
When the blood passed the tissue with low partial pressure of oxygen,
P9
Oxyhaemoglobin dissociates to release oxygen.
P10 Carbon dioxide released by repairing cells can be transported by dissolve carbon dioxide in the blood plasma.
P11 Bind to the haemoglobin.
P12 As carbaminohaemoglobin.
P13 In form of bicarbonate ions.
P14 Carbon dioxide is expelled with water vapour from the lung.

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Essays
Describe how the change of oxygen and carbon dioxide content are
regulated by the body
F1 The higher level of carbon dioxide in the blood cause the drop of
the pH value
F2 The drop in pH is detected by central chemoreceptor in medulla
oblongata
F3 Then the central chemoreceptor send nerve impulses tto the
diaphgram and intercoastal muscle
F4 Causing (respiratory muscle) to contract and relax
F5 Finally, increases the breathing and ventilation rate
concentration of carbon dioxide
F6 And pH value of the blood return to normal level

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Chapter 8 : Dynamic Ecosystem
Food chain
Sequence of organism which energy is transferred/flow
from trophic level to another trophic level by eating
process.

In food chain, the energy received by the organism in each trophic level is only 10%
from the previous organism. 90% energy lost as heat.
Example : the producer get 25000J energy from the sun, then how much energy is
received by the secondary consumer?
1st consumer get : 10% from 25000J = 2500J
2nd consumer get : 10% from 2500J = 250J

Food web is the interconnection of many food chains
Why most food chain havenot more than 4/5 links?
- Because animals at the end of the food chain would not get
enough food/energy.
Interaction between biotic components
Parasitism (+ -)
Mutualism (+ +)
Commensalism (+ 0)
The organism which always get negative effect or did not get any effect
is always the host

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Colonisation and succession in an ecosystem
Term
Species

Definition
A group of organisms that look alike and capable
of interbreeding and producing fertile offspring
The natural environment in which organism can
get food, shelter, living space, nesting and
breeding sites
1. The function of an organism or the role it
play in an ecosystem
2. And the space it occupies

Habitat

Niche

Example : the grasshopper eats grass in the field

Population
Community
Ecosystem

So the idea of an ecological niche is very simple.
You just need to know where the animal or plant
and what it does
A group of organism of the same species living in
the same habitat at the same time
All the plant and animals species living within a
defined area or habitat in an ecosystem
A community of living organism interacting whith
each other and with the non-living environment

The role of pioneer species :
Modify the environment, creating conditions which are less
favourable to themselves.
Make the condition more conducive to other species that called
successor species.
(in other words, the pioneer will sacrifice themselves for the successor
species).

Colonisation and succesion in mangrove swamp
Pioneer species :
Avicennia sp. (sea)
Sonneratia sp. (river)

Colonisation takes place in newly formed area where no life previously
existed.
The first colonizer is called pioneer species
Adaptation of pioneer species
 Have dense root system to bind the sand particle, hold water
and humus.
 Have root nodules containing nitrogen fixing bacteria to fix
Nitrogen from atmosphere to form nitrate as fertilizer.
 Have short life cycle/colonize open space faster.
 When they die, their remains add to the humus content of the soil

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Essays
Explain the process of colonisation and succession in mangrove swamp
F1
F2
F3
F4
F5
F6
F7
F8
F9
F10
F11
F12

The pioneer species of a mangrove swamp are the Sonneratia sp. and Avicennia sp.
The presence of this species gradually changes the physical environment of the habitat.
The extensive root systems of these plants trap and collect sediments, including organic matter from decaying plant parts.
As time passes, the soil becomes more compact and firm. This condition favours the growth of Rhizophora sp
Gradually the Rhizophora sp. replaces the pioneer species.
The prop root system of the Rhizophora sp. traps silt and mud, creating a firmer soil structure over time.
The ground becomes higher. As a result, the soil is drier because it is less submerged by sea water.
The condition now becomes more suitable for the Bruguiera sp., which replaces the Rhizophora sp.
The buttress root system of the Bruguiera sp. forms loops which extend from the soil to trap more silt and mud.
As more sediments are deposited, the shore extends further to the sea.
The old shore is now further away from the sea and is like terresterial ground.
Over time, terrestrial plants like nipah palm and Pandanus sp. begin to replace the Bruguiera sp.

Adaptation of the pioneer mangrove species to survive and colonised their habitat
(to overcome problem during colonisation)
Problem faced by mangrove plant (Fact)
F1
F2

Soft muddy soil/strong coastal wind
Waterlogged condition of the soil//very
little oxygen for root transpiration

F3

The high content of salt makes the water
soil hypertonic compared to the cell sap of
the root cell(so water diffuse out from the
plant and make the plant dehydrated)
Excessive exposure to the sunlight//high
rate of transpiration

F4

F5

High mortality rate//low survival rate of
seedlings

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P1
P2
P3
P4
P5
P6
P7
P8

Adaptive characteristics of pioneer mangrove plants
(explaination)
Highly branched root system to support themselves
(avicennia) have breathing roots/pneumatophore to absorb
oxygen from the atmosphere
Gaseous exchange occcurs through pores/lenticel
Cell sap of the root cells are hypertonic compared to the soil
water
Excess salt in the plant is eliminated by the salt gland
(hydathode)
The leaves have a thick cuticle/sunken stomata to reduce
transpiration
The leaves are thick/succulent to store water
Have vivaporous seedling//the seeds are able to germinate
while still attached to the mother plant


Colonisation and succession in pond

Frequency :

Pioneer : Elodea sp. and Hydrilla sp. (submerged)
Successor 1 : Lemna sp. and Pistia sp. (floating)
Successor 2 : Sedges and cattails (emergent)

x 100%
Density :

Essays
Explain how colonization and succession bring about the formation of primary
forest
P1
Activities of pioneer species (submerged plant) causes change in the
environment/ habitat, make it more suitable for other species
P2
The remains of plant/decayed bodies sinked/deposited in the pond
bed
P3
Water level in the pond decreases//pond becomes shallower
P4
Also add nutrients to pond water/soil//changes water/soil pH
P5
Favours the growth of floating plants(any example) to replace the
pioneer species
P6
Floating plants cover the water surface, preventing light from
penetrating the water/cause less rate of photosynthesis in the pond
P7
Results in greater rate of plant death which sink to the bottom/bed of
pond
P8
Raising the pond bed/making the pond shallower
P9
Floating plants are gradually replaced by emergent plants (example of
plant)
P10
The successor causes further changes to the habitat/pond make it
more favourable for emergent plants to grow
P11
Finally, emergent plant are replaced by land/terrestrial community
which dominates the area

Percentage coverage :

x 100%
The capture, mark, release and recapture technique
Population size :

Hierarchy in the classification of organisms

Population ecology
Quadrat sampling technique
This technic can be used to determine
 Frequency
 Density
 Percentage coverage

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BIOLOGYLOVE


Essay

Nitrogen Cycle
P1
P2
P3
P4
P5
P6
P7
P8
P9
P10
P11

Nitrogen fixing bacteria/Rhizobium sp. in the root nodules of
legumes plant//Azotobacter sp/Nostoc
Use nitrogen in the air to make nitrates/carries out nitrogen
fixation
Nitrates produced by the bacteria are absorbed by plants to
make proteins
When animal eats plants, the protein is transferred to
animals
Excretory nitrogenous substances/urea/waste
material/faeces
When plants/animal die
The plants/animals are decomposed by decaying
bacteria/saprophytic bacteria/fungi
Breaks them down to ammonia/ammonium compounds
Nitrifying bacteria/Nitrosomonas converts ammonium
compound/nitrates into nitrites
Nitrifying bacteria/Nitrobacter converts nitrites to nitrates
Denitrifying bacteria converts nitrates into nitrogen, thus
nitrogen content in the atmosphere is maintained

What happen if there is no microorganism?
P1
P2
P3
P4

No breakdown/decomposition of the dead organism
Mineral ions, for example nitrates cannot be released/
Nitrogen cycle is stopped
Soil become infertile/less nutrient in the soil
Plants will die/photosynthesis cannot takes place

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Chapter 9 : Endangered Ecosystem
Green house effects
F1
F2
F3
F4
F5
F6

Ultra violet(uv) from solar radiation is absorbed by the earth
and some of them is reflected back to the atmosphere in the
form of heat/infra red.
Heat or infrared radiation cannot be reflected back to the
atmosphere.
Because it is trapped by green house gases such as CO2,
nitrogen dioxide and methane.
Heat/infrared warmed the surface of earth.
Earth temperature increases.

Essays (negeri Perak 2010)

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Eutrophication
P1
P2
P3
P4
P5
P6
P7
P8
P9

Excessive fertilizer/animal organic waste from agricultural land
/farming area flows into river nearby when it rains
The presence of more minerals/organic substances
Promotes algal growth /growth of aquatic plants in the river/ alga
bloom
The surface of the river is covered up by the algae(which grow
extensively)
The plants in the lower depths of the water cannot obtain sunlight
They are unable to carry out photosynthesis
Hence, the plant die
The number of aerobic bacteria / decompose the dead plants also
increase
They use more of the oxygen (in the water) during the composition

P10 This reduces the concentration of oxygen in the water
P11 Causes the death of more aquatic organisms
P12 The biochemical oxygen demand (BOD) increases

Essays (SPM Trial Johor 2011)

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