quick revision notes form 4 chap 2_ cell and organization

8/13/2019 Quick Revision Notes Form 4 Chap 2_ Cell and Organization

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1 BIOLOGYLOVE second edition 2 0/2012

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 thecollection of biology essays

MORE ATTRACTIVE

MORE NOTES

MORE ESSAYS

SMART EXAM TIPS

form 4(4551)BIOLOGYLOVE


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1 BIOLOGYLOVE second edition 2 0/2012

2BIOLOGYLOVE second edition 2.0/2012

Phagocytosis

The pseudopodia are also used for feeding.Amoeba sp.engulfs food by phagocytosis.

Amoebasp. is a holozoic organisms which feed on microscopic organisms such as bacteria.

The presence of food causesAmoeba 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

- Bothhave a nucleus, cytoplasm, a plasma membrane, Golgi apparatus, mitochondria, endoplasmic

reticulum and ribosomes.

Chapter 2 : Cell Structure and Cell Organisation

Smart Exam Tips !

- Comparison include

similarities and differences

Smart Exam Tips !

- Use word BOTHfor similarities


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Differences

Animal cells

But

Plant cells

Do not have fixed shape Have fixed shape

Do not have cell wall Have cell wall

Do not have chloroplast Have chloroplast

Do not have vacoule (if have,

vacoule is only small and

numerous)

Mature plant cell have a large

central vacoule

Carbohydrate is stored in the form

of glycogen

Carbohydrate stored as starch

Have centrioles Do not have centrioles

The density of organelles in specific cells

Type of cells Organelles found abundantly (high density)Sperm cells

Muscle cells

Meristematic cells

Mitochondria

Palisade mesophyll cells Chloroplast

Pancreatic glands

Cell in salivary gland

Ribosome/RER/Golgi apparatus

Smart Exam Tips !

- Use word BUT for

differences

Smart Exam Tips !

- This question

always been

asked in Paper 1

and 2

- SPM Question

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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Simple DiffusionNet 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

Chapter 3 :Movement of Substances Across the Plasma Membrane


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6 BIOLOGYLOVE second edition 2.0/2012

Hypotonic solution Concentration of solute outside a cell is lower than

concentration of solute inside cell.

Animal cells

Cell placed in hypotonic solution.

Solution is hypotonic to the cell sap of the cell.

Net movement of water into the cells via osmosis. Cell swells up.

When extremely hypotonic, cells will eventually burst

Cannot withstand the osmotic pressure because of thin plasma

membrane.

E.g : red blood cells (haemolysis)

Plant cells

Do not burst

Rigid cell wall.

Water diffuse into vacoule of cell via osmosisthrougha semi-permeable membrane.

Cell swells up and becomes turgid

Tugor pressure in plant.

Supporting the plant.

Hypertonic solution

Hypertonic solution The concentration of solute in the

solution is higher than the concentration

of solutes within the cell sap.

Hypertonic to the cell sap of the cell.

Animal cells

Net movement of water from inside tothe outside of the cell.

Cells shrink//shrivel, internal pressure

decrease.

Red blood cells immersed in hypertonic

solution , the cell shrink and the plasma

membrane crinkles up.

Cell undergone crenation.

Plant cells

Water diffuse out via osmosis.

Vacoule and cytoplasm shrink and

plasma membrane pulls away from the

cell wall.

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 OSMOSIS3. What happen to cell (plasmolysis, crenation, turgid, haemolysis)


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Similarities between facilitated diffusion and active transportS1- Both (ways of transportation)need carrier protein.E1-To bind with molecules/ion/substrate/examples

S2- Bothtransport specific molecules only.

E2-Because the carrier protein have specific site to certain molecules.

S3- Bothprocesses occur in living cell.

E3-Because carrier protein need/can change shape to allow substances to move across.

Sodium Potassium Pump

P1 The concentration of sodiumions is higher on the inside of the cell

P2 The sodium ions approach the carrier protein. The carrier protein has a site for the ions to bind with

P3 The carrier protein binds to the sodium ions. The ATP molecule is split into Adenosine diphosphate (ADP) and

phosphate(P).

P4 The phosphate group the attach itself to the carrier protein. The splitting of ATP releases energy to the carrier

protein.

P5 Energy from the ATP changes the shape of carrier protein.

P6 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

Examples

Simple diffusion - Gaseous exchange in the alveoli and

blood capillaries

Facilitated

diffusion

- Absorption of digested food in the

villus

Osmosis - Absorption of water by root hair

cell

Active transport - Ions intake by root hairs of a plant

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


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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 cellF1 : 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

This how you answer the question

Percent change in mass

Sucrose molarity (mol)

25

20

15

10

5

0

-5

-10

0.2 0.4 0.6 0.8 1.0

At this stage, the sucrose solution is isotonic to

the cell sap of the potatoWater diffuse into and out of the cell via

osmosis at equal rateP

Q

At P

- The solution is hypotonic to the cell

sap.

- Water diffuse into the cell via osmosis.

- Cell becomes turgid.

- That is why the mass increased.

At Q

- The solution is hypertonic to the cell

sap.

- Water diffuse out from the cell via

osmosis.

- Cell plasmolysed.

- That is why the mass decrease


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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.

Chapter 4 : Chemical Composition of the Cell

Phosphate group

Nitrogenous base

Pentose sugar

Nitrogenous base

Adenine (A), Guanine (G), Cytosine (C),

Thymine (T)

Complementary base pairing

A--------T

C--------G

G--------C

T--------A

SPM 2011

carbohydrate

monosaccharides polysaccharides

glycogen

starch

cellulose

disaccharides


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2 types of nucleic acid

Deoxyribonucleic acid (DNA)

Ribonucleic acid (RNA)

DNAfound in

Nucleus of a cell

Chloroplast

Mitochondria

DNAcontains genetic information about an organism

RNA found in

Cytoplasm

Ribosomes

Nucleus

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

CARBOHYDRATES

Provide energy during respiration.

Build cell wall (cellulose) in plants.External skeleton of insects.

LIPIDEnergy rich organic compound.

Contains phosphorus and nitrogen.

Insoluble in water.

Includes fats, oil, waxes, phospholipids, steroids.

TRIGLYCERIDES

condensation

hydrolysis

glycerol 3 molecules

of fatty acids

triglycerides

PROTEINS

Amino acid as the basic unit (monomer)

Build new cell for growth.


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Aspect Saturated fats 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 doublebond

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

Increase level of bad

cholestrol

Contains more

cholestrol

Contains less

cholestrol

Tertiary structure

Enzymes

Hormones

Antibodies

Plasma proteins

Quartenary structure Haemoglobin

Pore protein

Protein Structure


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General characteristics of enzymes

Alter or speed up the rates of chemical

reactions

Remain unchanged at the end of reaction.

Do not destroyed by reactions they catalysed.

Have specific sites called active site to bind

with specific substrates.

Needed in small quantities. Reaction are reversible

Can be slowed down or stopped by inhibitors.

E.g: lead and mercury

Require helper molecules, called cofactors.

Inorganic cofactor : ferum, copper

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.

Effects of temperature on enzyme activity

At low temperature, reaction takes place slowly.

As temperature increases, movement of substrate

increase.

Increase their chances of colliding with each other

and with the active site of the enzymes.

At optimum temperature, the reaction is atmaximum rate.

Beyond the optimum temperature, rate of reaction

will not increase.

Bonds that hold enzyme molecules begin to break.

Actives sites destroyed.

Enzyme denatured.


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What is monosaccharides?

Glucose

Fructose

Galactose

What is disaccharides?

Maltose

Sucrose

Lactose

Glucose + glucose maltose + water

Glucose + fructose sucrose + water

Glucose + galactose lactose + water

Reducing Sugar and Non-Reducing sugar

To test for reducing sugar, Benedicts testmust be carried out.

If the colour of Benedicts solution changes from BLUEto BRICK-RED PRECIPITATE, thats mean the solution

contains Reducing sugar.

Reducing sugar is

All monosaccharides, maltose and lactose

Non reducing sugar isAll polysaccharides and sucrose

Give negative result on Benedicts test

(colour does not change)

condensation

condensation

condensation

Exam tips :

Glucose + monosaccharides condensation disaccharides + water

Enzymes for substrates

Starch : amylase

Sucrose : sucrase

Lactose : lactase

Maltose : maltase

Sometime sucrose can show positive result. Why? Because sucrose can be

hydrolysed into glucose and fructose.

Steps:

1. Add dilute hydrochloric acidand boil the solution.2. When the solution is cooled, put a spoon of calcium carbonate to

neutralise the solution.

3. Then test with benedicts solution.

4. Then the colour will change.

Exam tips :

Addition of water is

necessary. Must bewritten!!


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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 exceptTESTES and OVARY

CHROMOSOMES AND CHROMOSOMAL NUMBER

n = haploid

2n = diploid

Human has diploid number (2n) of chromosomes which is 46

Sperm (n=23) + ovum (n=23) zygote (2n=46)

This is a chromosome

With sister

chromatids

1 chromosome

This also chromosome

But single chromatids

1 chromosome also

Chapter 5 : Cell Division


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Stages in mitosis

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.

Metaphase

Chromosomes align at the

metaphase

plate//equatorial plate.

Mitotic spindle are fully

formed.

Two sister chromatids arestill attached to one

another at the

centromere.

Ends when the centromere

divides.

Anaphase

Two sister

chromatids separate

at the centromere.

Sister chromatids

pulled apart at

opposite poles.

Chromatids are

referred to as

daughter

chromosomes.

Telophase

Chromosomes reach

the opposite poles of

the cell.

Chromosomes uncoil

and revert to their

extendedstate(chromatin).

Exam Tips:

You can use this note to

answer question about

chromosome behaviour

Pro Meta Ana Telo phase


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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 tothe cell cycle.

These cells compete with surrounding normal cells for energy and

nutrients.

Cancer cells formed tumour.

Tumour invade and destroy neighbouring cells.

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.

Cleavage furrow

Exam Tips:Chromosome :

Gamete (ovum and sperm)contain halfthe number of

chromosome (n=haploid)


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Application of Mitosis

Animal Cloning1

2

3

4

5

6

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

Small pieces of

tissue is cut (e.g :

root/shoot)

hormone

Plant cell divide by

mitosis to form callus

an undifferentiate

mass of tissue

Cell in the callus develop

into embryoPlantlet are then

transferred to soil

where they grow into

adult lant

Meiosis

Meiosis I

1. During prophase I,homologous

chromosomespair up (synapsis)and

crossing overbetween 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 separatesand 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

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 daughtercells 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.


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PROPHASE I METAPHASE I ANAPHASE I TELOPHASE I

PROPHASE II METAPHASE II ANAPHASE II TELOPHASE II

Stages in Meiosis I

Stages in Meiosis II


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Synapsis, Crossing Over and Chiasmata(singular : chiasma)

Synapsisis the process where

the chromosomes pairing up

Crossing overis the process

where non sister chromatidsexchange segment of genetic

material (DNA)

Chiasmatais the point where

the crossing over process occur

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!!!


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Exam tips:

- Process in Meiosis II is likely same as

Mitosis

- The term use for Meiosis Iis

Homologous chromosomewhile in

Meiosis II, the term used is sister

Chromatids

Similarities between Meiosis I and Meiosis II

Bothconsist of four stages

Bothinvolve nuclear division

Bothinvolve cytokinesis

Bothhave chromatids

Aspect Meiosis I Meiosis II

Reduce 2n chromosome to n Divides the remaining

set of chromosome in

a mitosis like process

Prophase Chromosome already

replicated

Homologouschromosomes

synapse

Chiasma forms and crossingover takes place

No replication

No synapsis

No chiasma and no

crossing over

Metaphase Paired homologous

chromosomesalign at the

equator

Sister chromatids

align at the equator

Anaphase Separation of homologous

chromosomesto opposite

poles

Separation of sister

chromatids to

opposite poles

Telophase Single cytokinesis

2 identical cell produced

Two cytokinesis

4 identical cell

produced

Exam Tips :

You have to master all the Comparison between Meiosis I and Meiosis II

also between Mitosis and Meiosis

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?

If gamete is not haploid, the number of

chromosome in the organism will be double

from the real number!!

23

23 46

4646 96


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Energy value of food (kJ g-1

)

Test on food samples

Test for Reagent Observation Conclusion

Starch Iodine solution Colour change

from yellowto

blue-black

Food sample

contains starch

Reducing sugar

(refer chapter 4)

Benedicts

solution

Change from

blueto brick red

precipitate

Food contain

reducing sugar

Protein Biurets test

(20% of sodium

hydroxide

solution and 1%

copper(II)

sulphate

solution

Change from

blueto purple

Food contain

protein

Lipid Filter paper Translucent

mark

Food contain

lipid

Lipid Emulsion test Oily mixture on

the surface of

water

Food contain

lipid

() ()

()

Chapter 6 : Nutrition

Percentage of vitamin C in fruit juice =

x 0.1%

Concentration of vitamin C in fruit juice =

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.


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Examples of essays

Digestion in mouth Digestion in stomach Digestion in small intestine

Secretion of saliva by three pairs of

salivary glands

Saliva contains the enzyme salivary

amylase

Begins the hydrolysis of starch to maltose.

Starch + water maltose

An additional digestive process occurs

further along the alimentary canal to

convert maltose to glucose.

pH is maintained at 6.5-7.5

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.

Protein + water 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.

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 theduodenum 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.

Salivary amylase

pepsin


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Summary of the digestion

Digestive organ Digestive juice enzyme pH Substrates and products

Liver Bile, bile salts None 7.6-8.6 Emulsification of fats

Pancreas Pancreatic juice Pancreatic amylase 7.1-8.2


Trypsin 7.1-8.2

Polypeptides + water peptides

Lipase 7.1-8.2

Lipid droplets + water fatty acid + glycerol

Mouth

Have salivary glandto secrete saliva

Saliva contain salivary amylase

Salivary amylase will digest


Salivary amylase

Stomach

Have gastric glandto secrete gastric juice

Gastric juice contain enzyme pepsin and renin

pH is acidic

protein + water polypeptides

caseinogen + water casein

pepsin

renin

Ileum (small intestine)

Have intestinal glandwhich secrete intestinal juice

that contain enzymes :

Maltose + water glucose

Lactose + water glucose + galactose

Sucrose + water glucose + fructose

Peptides + water amino acids

maltase

lactase

sucrase

erepsin

Site of digestion : duodenum

lipase

Pancreatic amylase

trypsin

Exam tips :

Please remember that enzyme trypsinalways

need alkaline medium (pH > 7)

Pepsin and reninneed acidic medium (pH < 7)

Acidic medium is in Stomach

Alkaline medium is in Duodenum

Digestion is a popular

question in SPM!!!


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Digestion in Rodent and Ruminant (essay)

Digestion of cellulose by ruminant

F1 Partially chewed food is passed to the rumen (largest compartment of the stomach).

F2 Cellulose is broken down by cellulase produced by bacteria.

F3 Part of the breakdown products are absobed by bacteria, the rest by the host.

F4 Food enters the reticulum.

F5 Cellulose undergoes further hydrolysis.

F6F7

The content of the reticulum, called the cud, is then regurgitated bit by bit into the mouthto be thoroughly chewed.

F8 Helps soften and break down cellulose, making it more accessible to further microbial

action.

F9 The cud is reswallowed and moved to the omasum.

F10 Here, the large particles of food are broken down into smaller pieces by peristalsis.

F11 Water is removed from the cud.

F12 Food particles moved into obamasum, the true stomach of the ruminant. (e.g : cow).

F13 Gastric juice complete the digestion of protein and other food substances.

F14 The food then passes through the small intestine to be digested and absorbed in the

normal way.

Digestion of cellulose by rodent

F1 Caecum and appendix are enlarged to

store the cellulose-digesting bacteria.

F2 The breakdown products pass through the

alimentary canal twice.

F3 The faeces in the first batch are usually

produced at night.

F4 Faeces are then eaten again. To absorb

the products of bacterial breakdown.

F5 The second batch of the faeces are harderand drier.

F6 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

P1 The villi have microscopic projection in the epithelial cell called

microvilli

P2 Both the villi and microvilli increases the total surface area of the ileum

for the absorption of soluble end product of digestion

P3 A dense blood capillary network at each villus

P4 Enable the food substances absorbed to be carried away quickly

P5 The epithelial lining of the villus is one-cell thick

P6 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 asvitamins 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.

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

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 areaP3 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 dont

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 andreticulum are then regurgitated back in mouth cavity for

ruminant


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Assimilation of digested food

In the Liver

F1 Amino acidsis 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 thekidney to be excreted

F5 Glucosein 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 Lipidswhich 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 ofnew 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 Glucoseis 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 Lipidssuch 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

Formation faeces

F1 Faeces which contain dead cells that are shed from intestinal linings.

F2 toxic substances and bile pigments enter the colon by action of

peristalsis.

F3 In colon, more water is absorbed.

F4 The undigested food residues harden to become faeces.

F5 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.


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Photosynthesis Mechanism

Photosynthesis mechanism

P1 The formation of starch in plants is by the process ofphotosynthesis which occurs in chloroplasts.

P2 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.

P5 In the excited state, the electrons can leave the chlorophyll molecules.

P6 Light energy is also used to split water molecules into hydrogen ion (H+) and hydroxyl ions (OH

-)

(Photolysis of water).

P7 The hydrogen ions then combine with the electrons released by chlorophyll to form hydrogen

atoms.

P8 The energy from the excited electrons is used to form energy-rich molecules of adenosine

triphosphate /ATP.

P9 Hydroxyl ion loses an electron to form a hydroxyl group. This electron is then received by

chlorophyll.

P10 The hydroxyl groups then combine to form water and gaseous oxygen.

P11 Dark Reaction:

take place in stroma.

P12 Do not require light energy.

P13 The hydrogen atoms are used to fix carbon dioxide in a series of reactions catalysed by

photosynthetic enzymes

P14 and caused the reduction of carbon dioxide into glucose.

P15 The glucose monomers then undergo condensation to form starch which is temporarily stored as

starch grains in the chloroplasts.

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

Extra :

- In another words, carbon dioxide is reduced into glucose by the

hydrogen atom


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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

sports.// She needs to contract

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

E2 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.

Exam tips:

- You must be able to

relate the diet with the

needs of the people.


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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 muscleWhen 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.

Chapter 7 : Respiration

Respirationis the process of oxidation of complex

organic substances with the release of energy utilizes

oxygen an dremoving carbon dioxide in living cells

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

But

Anaerobic respiration

Need oxygen No tion of need oxygen

Complete oxidation of glucose Not complete oxidation glucoseProduce water, carbon dioxide and

energy

Animal : lactic acid and energy

Plant/yeast : ethanol, carbon dioxide

and energy

Produce 36 ATP (2898 kJ energy) Produce 2 ATP. Some of the energy

stored in lactic acid or ethanol

Occur in mitochondria Occur in cytoplasm

Similarities

S1 Both involve cell respiration

S2 Both involve oxidation of glucose

S3 Both produce energyS4 Both catalysed by enzyme


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Respiratory structures and breathing mechanism in human and animalsRespiratory structueis the organ for respiration

Respiratiry surfaceis 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

Essays

Adaptation of tracheal system

F1 Have spiracle

E1 To allow oxygen and carbon dioxide to get in and out of the cell

F2 The spiracle have valve

E2 To allow the opening and closing of the trachea so that air can go in and out

F3 The trachea are reinforced with chitin(made up of protein)

E3 To prevent the trachea from collapsing

F4 The trachea branched into finer tubes called tracheole which are in direct

contact to the cell and organ

E4 To transport the respiratory gases quicklyF5 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 insects tissues

Exam tips :

- Memorising the four common characteristics is important because

you can use it to answer question on adaptions of all organisms

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

P1 Blood and water flow in opposite direction

P2 Maintains diffusion gradient

P3 Maximizing oxygen transfer from water to blood

P4 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!!!


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The adaptation of respiratory structure of amphibians(frog)

F1 The skin of the frog is thin

E1 highly permeable to respiratory gases

F2 The skin/membrane of the lung is moist

E2 To dissolve respiratory gases

F3 The skin has alrge number of blood capillaries under the

skin/ lungs have network of blood capillariesE3 For efficient transport of gases

F4 The lungs consist of a pair of thin walled sacs connected to

the mouth through an opening called g lottis

E4 To allow gases from mouth move to the lungs

F5 The membrane of the lungs are thin

E5 To allow diffusion of gases to occur easily

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 However, the supply of oxygen to muscles is still insufficient.

P4 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.

P6 An oxygen debt builds up in the body, when no oxygen use in energy

production.

P7 High level of lactic acid in the muscles cause them to ache.

P8 After running, the athlete breathes more rapidly and deeply than normal for

twenty minutes.

P9 There is recovery period after 10 minutes until it reaches 20 minutes when

oxygen is paid back during aerobic respiration.

P10 About 1/6 lactic acid is oxidized to carbon dioxide, water and energy.

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.


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Breathing mechanisms in man

Transport of oxygen and carbon dioxide in the bodyP1 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.

P1 Diaphragm is a muscular sheet in the body cavity separating the thorax from the abdomen.

P2 At the start of inhalation, the muscles of the diaphragm contract , making it less arched.

P3 This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs.

P4 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.

P5 The muscles between the ribs are known as intercostals muscles.

P6 During inhalation the external intercostals muscle contracts and raise the lower ribs.

P7 This helps to increase the volume of the thoracic cavity and reduce the pressure of the thoracic cavity. Air rushes into the lungs.

P8 During exhalation the external intercostals muscles contract , the ribs return to their original position , reduce the pressure of the thoracic

cavity.

P9 Air is forced out of the lungs.

P10 The alveoli are thin-walled air sacs with the lungs.

P11 These sacs are surrounded by a network of capillaries.

P12 During inhalation the alveoli are filled with air and gaseous exchange occurs between the alveoli and the capillaries.

P13 Oxygen from the alveoli diffuses into the capillaries while carbon dioxide diffuses from the capillaries into the alveoli.


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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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Food chain

Sequence of organism which energy is transferred/flow

from trophic level to another trophic level by eating

process.

Food webis the interconnection of many food chains

Chapter 8 : Dynamic Ecosystem

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

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

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 fixNitrogen 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

Term Definition

Species A group of organisms that look alike and capable

of interbreeding and producing fertile offspring

Habitat The natural environment in which organism can

get food, shelter, living space, nesting and

breeding sites

Niche 1. The function of an organism or the role itplay in an ecosystem

2. And the space it occupies

Example : the grasshopper eats grass in the field

So the idea of an ecological niche is very simple.

You just need to know where the animal or plant

and what it does

Population A group of organism of the same species living in

the same habitat at the same time

Community All the plant and animals species living within adefined area or habitat in an ecosystem

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 :

Avicenniasp. (sea)

Sonneratia sp. (river)


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Essays

Explain the process of colonisation and succession in mangrove swamp

F1 The pioneer species of a mangrove swamp are the Sonneratia sp. andAvicennia sp.

F2 The presence of this species gradually changes the physical environment of the habitat.

F3 The extensive root systems of these plants trap and collect sediments, including organic matter from decaying plant parts.

F4 As time passes, the soil becomes more compact and firm. This condition favours the growth of Rhizophora sp

F5 Gradually the Rhizophora sp. replaces the pioneer species.

F6 The prop root system of the Rhizophorasp. traps silt and mud, creating a firmer soil structure over time.F7 The ground becomes higher. As a result, the soil is drier because it is less submerged by sea water.

F8 The condition now becomes more suitable for the Bruguierasp., which replaces the Rhizophorasp.

F9 The buttress root system of the Bruguierasp. forms loops which extend from the soil to trap more silt and mud.

F10 As more sediments are deposited, the shore extends further to the sea.

F11 The old shore is now further away from the sea and is like terresterial ground.

F12 Over time, terrestrial plants likenipah palm and Pandanussp. begin to replace the Bruguierasp.

Adaptation of the pioneer mangrove species to survive and colonised their habitat

(to overcome problem during colonisation)

Problem faced by mangrove plant (Fact) Adaptive characteristics of pioneer mangrove plants

(explaination)

F1 Soft muddy soil/strong coastal wind P1 Highly branched root system to support themselves

F2 Waterlogged condition of the soil//very

little oxygen for root transpiration

P2 (avicennia) have breathing roots/pneumatophore to absorb

oxygen from the atmosphere

P3 Gaseous exchange occcurs through pores/lenticel

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)

P4 Cell sap of the root cells are hypertonic compared to the soil

water

P5 Excess salt in the plant is eliminated by the salt gland

(hydathode)

F4 Excessive exposure to the sunlight//high

rate of transpiration

P6 The leaves have a thick cuticle/sunken stomata to reduce

transpiration

P7 The leaves are thick/succulent to store water

F5 High mortality rate//low survival rate of

seedlings

P8 Have vivaporous seedling//the seeds are able to germinate

while still attached to the mother plant


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Colonisation and succession in pondPioneer : Elodea sp. and Hydrilla sp. (submerged)

Successor 1 : Lemna sp. and Pistiasp. (floating)

Successor 2 : Sedges and cattails (emergent)

EssaysExplain how colonization and succession bring about the formation of primary

forest

Quadrat sampling technique

This technic can be used to determine

Frequency

Density

Percentage coverage

P1 Activities of pioneer species (submerged plant) causes change in theenvironment/ 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

pondP8 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

Population ecology

Frequency :

x 100%

Density :

Percentage coverage :

x 100%

The capture, mark, release and recapture technique

Population size :

Hierarchy in the classification of organisms


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Essay

Nitrogen CycleP1 Nitrogen fixing bacteria/Rhizobium sp. in the root nodules of

legumes plant//Azotobacter sp/Nostoc

P2 Use nitrogen in the air to make nitrates/carries out nitrogen

fixation

P3 Nitrates produced by the bacteria are absorbed by plants to

make proteins

P4 When animal eats plants, the protein is transferred toanimals

P5 Excretory nitrogenous substances/urea/waste

material/faeces

P6 When plants/animal die

P7 The plants/animals are decomposed by decaying

bacteria/saprophytic bacteria/fungi

P8 Breaks them down to ammonia/ammonium compounds

P9 Nitrifying bacteria/Nitrosomonas converts ammonium

compound/nitrates into nitrites

P10 Nitrifying bacteria/Nitrobacter converts nitrites to nitrates

P11 Denitrifying bacteria converts nitrates into nitrogen, thusnitrogen content in the atmosphere is maintained

What happen if there is no microorganism?P1 No breakdown/decomposition of the dead organism

P2 Mineral ions, for example nitrates cannot be released/

Nitrogen cycle is stopped

P3 Soil become infertile/less nutrient in the soil

P4 Plants will die/photosynthesis cannot takes place


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Green house effects

F1 Ultra violet(uv) from solar radiation is absorbed by the earth

F2 and some of them is reflected back to the atmosphere in the

form of heat/infra red.

F3 Heat or infrared radiation cannot be reflected back to theatmosphere.

F4 Because it is trapped by green house gases such as CO2,

nitrogen dioxide and methane.

F5 Heat/infrared warmed the surface of earth.

F6 Earth temperature increases.

Essays (negeri Perak 2010)

Chapter 9 : Endangered Ecosystem


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Eutrophication

P1 Excessive fertilizer/animal organic waste from agricultural land

/farming area flows into river nearby when it rains

P2 The presence of moreminerals/organic substances

P3 Promotes algal growth /growth of aquatic plants in the river/ alga

bloom

P4 The surface of the river is covered up by the algae(which grow

extensively)

P5 The plants in the lower depths of the water cannot obtain sunlight

P6 They are unable to carry out photosynthesis

P7 Hence, the plant die

P8 The number of aerobic bacteria / decompose the dead plants also

increase

P9 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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quick revision notes form 4 chap 2_ cell and organization

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