BIOLOVE SPM

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)

BIOLOGYLOVE©


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

for similarities


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


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

Chapter 3 : Movement of Substances Across the Plasma Membrane


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.


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


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.

P6 This cause the carrier protein to release the sodium ions outside of the cell


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


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. - That is why the mass decrease


‘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


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.


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


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. Enzyme denatured.


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!!


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

Chapter 5 : Cell Division


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

Pro Meta Ana Telo phase


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)


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 : Explain the technic used in animal cloning


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.


PROPHASE I METAPHASE I ANAPHASE I TELOPHASE I

PROPHASE II METAPHASE II ANAPHASE II TELOPHASE II

Stages in Meiosis I

Stages in Meiosis II


Synapsis, Crossing Over and Chiasmata(singular : chiasma)

Synapsis is the process where

the chromosomes pairing up

Crossing over is the process

where non sister chromatids

exchange segment of genetic

material (DNA)

Chiasmata is the point where

the crossing over process occur

Exam tips : If the question ask for the type of division, you must answer whether

1. Mitosis or 2. Meiosis

If the question ask for stage, then you can answer

1. Prophase//Prophase I//Prophase II 2. Metaphase and so on.....

Before you answer the question, look at the diagram//question whether it is mitosis or meiosis!!!


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

Similarities between Meiosis I and Meiosis II

Both consist of four stages Both involve nuclear division Both involve cytokinesis Both have chromatids

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 46 96


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.


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

pepsin


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!!!


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.


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


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 LIPID.


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


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.


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

S4 Both catalysed by enzyme


Respiratory structures and breathing mechanism in human and animals Respiratory structue is the organ for respiration Respiratiry surface is the site where the exchange of gases occur

Organisms Respiratory structure Respiratory surface

Human Lungs Alveoli

Grasshopper/insects Tracheal system Trachiole

Amoeba sp. No specific structure Plasma membrane

Fish Gills Lamella/filament

Frog Skin and lungs Skin/walled sac in the lungs

Four common characteristics(adaptation) of the respiratory surface

1) Large surface area to maximize the exchange of gases by diffusion 2) Moist respiratory surface for gases to dissolve 3) Thin as one-cell thick for effective diffusion of gases 4) Network of blood capillaries for effective transportation of gases

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!!!


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.


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.


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


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


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)


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


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 :

Hierarchy in the classification of organisms


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


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)

Chapter 9 : Endangered Ecosystem


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

Essays (SPM Trial Johor 2011)

BIOLOVE SPM

Education

Transcript of BIOLOVE SPM