BIOLOGYLOVE

(1)

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 the

collection of biology essays©

 MORE ATTRACTIVE

 MORE NOTES

 MORE ESSAYS

 SMART EXAM TIPS

form 4

(4551)

(2)

2 

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

-

Both

have 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 BOTH

(3)

3 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

(4)

4 

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

(5)

5 

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.

(6)

6 

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 osmosis

through

a 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 to

the 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 OSMOSIS

(7)

7 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

The concentration of sodium

ions 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.

(8)

8 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

(9)

9 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

This how you answer the question

Percent change in mass

ofpotato

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 potato

Water diffuse into and out of the cell via

osmosis at equal rate

P

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.

(10)

10 ‘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

(11)

11

BIOLOGYLOVE second edition 2.0/2012 2 types of nucleic acid

 Deoxyribonucleic acid (DNA)

 Ribonucleic acid (RNA)

DNA found in

 Nucleus of a cell

 Chloroplast

 Mitochondria

DNA contains 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.

LIPID

Energy 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.

(12)

12 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 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

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

(13)

13 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 at

maximum rate.



Beyond the optimum temperature, rate of reaction

will not increase.



Bonds that hold enzyme molecules begin to break.



Actives sites destroyed.

(14)

14 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,

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.

Reducing sugar is

All monosaccharides, maltose and lactose

Non reducing sugar is

All polysaccharides and sucrose

Give negative result on Benedict’s 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 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.

Exam tips :

Addition of water is

necessary. Must be

written!!

(15)

15 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

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

(16)

16 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 are still 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 extended

state(chromatin).

Exam Tips :

You can use this note to answer question about chromosome behaviour

(17)

17

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.

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 half the number of chromosome (n=haploid)

(18)

18 Application of Mitosis

Animal Cloning

1 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 :

(19)

19 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 embryo

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

Meiosis

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

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.

(20)

20 PROPHASE I

METAPHASE I

ANAPHASE I

TELOPHASE I

PROPHASE II

METAPHASE II

ANAPHASE II

TELOPHASE II

Stages in Meiosis I

(21)

21 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

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

(22)

22 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

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

Homologous chromosomes synapse

Chiasma forms and crossing over takes place

No replication No synapsis No chiasma and no crossing over

Metaphase Paired homologous

chromosomes align at the equator

Sister chromatids

align at the equator

Anaphase Separation of homologous

chromosomes to 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

46

(23)

23 Energy value of food (kJ g

-1

)

Test on food samples

Test for

Reagent

Observation

Conclusion

Starch

Iodine solution

Colour change

from

yellow

to

blue-black

Food sample

contains starch

Reducing sugar

(refer chapter 4)

Benedict’s

solution

Change from

blue

to

brick red

precipitate

Food contain

reducing sugar

Protein

Biuret’s test

(20% of sodium

hydroxide

solution and 1%

copper(II)

sulphate

solution

Change from

blue

to

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.

(24)

24 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 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. Salivary amylase

(25)

25 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

Starch + water

maltose

Trypsin

7.1-8.2

Polypeptides + water

peptides

Lipase

7.1-8.2

Lipid droplets + water

fatty acid + glycerol

Mouth

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

Starch + water maltose Salivary amylase

Stomach

Have gastric gland to 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 gland which 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 trypsin always need alkaline medium (pH > 7)

Pepsin and renin need acidic medium (pH < 7) Acidic medium is in Stomach

Alkaline medium is in Duodenum

Digestion is a popular

question in SPM!!!

(26)

26 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. F6

F7

The content of the reticulum, called the cud, is then regurgitated bit by bit into the mouth to 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 harder

and drier.

F6 Allows rodent (give example) to recover the nutrients initially lost with the faeces.

(27)

27 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 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.

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 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

(28)

28 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

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

(29)

29 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

(30)

30

BIOLOGYLOVE second edition 2.0/2012 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.

(31)

31

BIOLOGYLOVE second edition 2.0/2012 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.

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

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 glucose Produce 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 energy

(32)

32 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

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 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

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!!!

(33)

33 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 capillaries

E3 For efficient transport of gases

F4 The lungs consist of a pair of thin walled sacs connected to the mouth through an opening called glottis

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.

(34)

34

Breathing mechanisms in man

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.

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.

(35)

35 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

(36)

36

Food chain

Sequence of organism which energy is transferred/flow from trophic level to another trophic level by eating process.

Food web is 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

(37)

37 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 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

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 it play 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 a defined 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 :

Avicennia sp. (sea) Sonneratia sp. (river)

(38)

38 Essays

Explain the process of colonisation and succession in mangrove swamp

F1 The pioneer species of a mangrove swamp are the Sonneratia sp. and Avicennia 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 Rhizophora sp. 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 Bruguiera sp., which replaces the Rhizophora sp.

F9 The buttress root system of the Bruguiera sp. 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 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) 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

(39)

39 Colonisation and succession in pond

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

Essays

Explain 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 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

Population ecology Frequency :

x 100%

Density :

Percentage coverage :

x 100%

The capture, mark, release and recapture technique

Population size :

(40)

40 Essay

Nitrogen Cycle

P1 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 to animals

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, thus

nitrogen 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

(41)

41 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 the atmosphere.

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)

(42)

42 Eutrophication

P1 Excessive fertilizer/animal organic waste from agricultural land /farming area flows into river nearby when it rains

P2 The presence of more minerals/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

(43)