CHAPTER 1 : MOLECULES OF LIFE - WordPress.com · •Chemical formula : H 2 O •Shape –V shape with angle 104.50 •2 hydrogen atoms joined to 1 oxygen atom by single covalent bond

CHAPTER 1 :MOLECULES OF LIFE

Subtopics

1.1 Water

1.2 Carbohydrates

1.3 Lipids

1.4 Proteins

1.5 Nucleic acids

1.1 : WATER At the end of this topic, students should be able to:

• Explain the structure of water molecule.

• Describe the properties of water & its importance:

Learning Outcomes

Polar1. moleculeLow2. viscosityHigh3. specific heat capacity High4. latent heat of vaporizationHigh5. surface tension Maximum6. density at 4 ̊C

Figure 3.1

How does the habitat of a polar bear depend on the chemistry of water?

• Water is important as an internal constituent.

• Cell shape & internal structure depend on water.

• A large part of the mass of most organism is water.

• Water is the solvent for most biological reactions.

• The cytoplasm of a cell is a water-based solution that contains a variety of ions, salts & molecules which makes life possible.

A large part of human mass is water.

Learning outcome : 1.1 - WATER a) Explain the structure of water molecule

Water is the substances which exist in natural environment in all three physical state matter :

solid, liquid & gas.

• Chemical formula : H2O

• Shape – V shape with angle 104.50

• 2 hydrogen atoms joined to 1 oxygen atom by single covalent bond.

Covalent bond in water molecule


Structure of Water Molecule

Structure of water moleculeHow to draw ????

Drawing must include:

1. Atoms : 1 oxygen atom & 2 hydrogen atoms.

2. Bond : Covalent bond.

3. Polarity : δ+ at hydrogen atom & δ- at oxygen atom.

4. Shape : V shape with 104.50

Structure of water molecule




How covalent bond form in water molecule?

• Each O atom has six electrons in its outer shell.

• Each H atom has one electron.

• So each two H atoms share an electron with the O atom.

• Covalent bond : Bond that formed when two atoms share one or more pair of valence electrons so that each atom has complete outer shell.


• Water is a polar molecule (e.g: two ends of water molecule have opposite partial charges).

• Which is due to unequal distribution of shared electrons.

• O atom is more electronegative compare to H atoms.

• So shared electrons of the covalent bond are closer at O atom than to H atoms.

• Thus H atoms have partial positive charge (δ+) & O atom has partial negative charge (δ-).


Polar covalent bond



Electronegativity:

Tendency of an atom to attract shared electron

to itself in a chemical bond.

The more electronegative

an atom (e.g: oxygen atom), the more strongly it

pulls shared electrons toward itself.

• Polar property allows water molecules to form hydrogen bonds with each other which is constantly break & reformed.

• Hydrogen bond : A chemical bond that formed when the δ+

charge H atom of a polar covalent bond in one molecule is attracted to δ– charge atom of a polar covalent bond in another molecule, causing them to stick together.

Hydrogen bond within water molecule


Each water molecule can form hydrogen bond maximally with four

other water molecules


How to draw ????

Each water molecule can form hydrogen bond maximally with four other water molecules

Drawing must include:

1. Atoms : One oxygen atom & two hydrogen atoms.

2. Bonds : Covalent bond & hydrogen bond.

3. Polarity : δ+ at hydrogen atom & δ-

at oxygen atom.

4. Shape : V shape with 104.50


• Water molecules are attracted to other polar molecules, ionic compounds & other organic compounds due to the hydrogen bonds.

• In fact, hydrogen bonds largely responsible for the unique properties of water.

Hydrogen bond within water molecule


1. Polar molecules

2. Low viscosity

3. High specific heat capacity

4. High latent heat of vaporization

Maximum5. density at 4 ̊C

6. High surface tension : cohesive & adhesive forces

Learning outcome : 1.1 - WATER b) Describe the properties of water & its importance

6 Properties of Water

• Water is a polar molecule due to unequal distribution of shared electrons.

• Thus, they have ability to form hydrogen bonds with other molecules.

• Give them as powerful universal solvent for :

i. Other polar substances e.g: sugar which have slightly charged hydroxyl (-OH) groups.

ii. Ionic compounds (e.g: sodium chloride, NaCl) : Na+

& Cl- ions are exposed to the solvent.

iii. Organic molecules with ionized groups (such as the carboxyl group –COO- & amino group –NH3

+).


1. Polar molecule

Water is a powerful solvent for ionic compounds. E.g.: when salt (NaCl) is placed in water.

✓ The oxygen atoms has a partially negative charged which are attracted to the positively charged sodium ions, Na+

✓ The hydrogen atoms has a partially positive charged which are attracted to the negatively charged chloride ions, Cl-

✓ Thus, water molecules neutralized electrostatic interactions between the oppositely charged ions.

✓ As a result, water molecules surround the individual Na+ & Cl-, separating them from one another.

✓ The sphere of water molecules around each dissolved ions known as a hydration shell.


Partial negative oxygen regions of polar water

molecules are attracted to sodium cations (Na+)

Partial positive hydrogen regions of water molecules

are attracted to chloride anions (Cl-)

Table salt dissolve in water


E.g: when salt is placed in water

✓ Ionic bonds between NaCl molecules are become weaker.

✓ This will caused these ions to separate.

✓ Thus, salt are dissolved.

Hydration shell


Table salt dissolve in water

• Water is a powerful solvent for organic molecules with ionized groups ; such as the carboxyl group – COO- & amino group –NH3

+.

Ionized group in amino acid


Polar molecule - Importance

• Water acts as universal solvent which is able to dissolve many substances such as ionic compounds, other polar molecules & organic molecules.

• Water acts as medium for biochemical reaction (solutes are more reactive when dissolved).

Blood plasma as medium for biochemical reaction in living organism.


• Weak hydrogen bondsbetween water molecules are constantly break & reform.

• Causing the water molecules slide easily over each other & flow with less friction.

• Thus, water has low viscosity.

Viscosity of water is considered lower than the honey


2. Low Viscosity

• Act as medium of transportation in living organisms (e.g.: blood easily flow in the circulatory system through narrow vessels).

• Act as a good lubricant to reduce friction on cell surface (e.g.: mucus facilitates movement of feces through the bowel). Flow of blood in the circulatory system

through narrow vessels


Low Viscosity – Importance

3. High specific heat capacity

Water has a high specific heat capacity (• 1 cal/g/ ̊C). E.g: a lot of energy required to increase the temperature of 1 g water by 1 ̊C.

Water can • resist the changing of its temperature; when water changed its temperature, it will absorbed or lost a relatively large quantity of heat for each degree of change.

“The large amount of heat that must be absorbed/ lost for 1 g of water to change its temperature

by 1 ̊C”.


• Due to the hydrogen bonds

✓ Heat is absorbed to break hydrogen bonds. Much of heat is used to disrupt hydrogen bonds before the water molecules can begin to move faster.

✓ Heat is released when hydrogen bonds form. Slightly drops of temperature will caused additional hydrogen bonds to form & released a considerable amount of heat energy.


High specific heat capacity - Importance

• Able to protect water from rapid temperature changes; water temperature do not change easily.

✓ E.g : Aquatic environments like ponds & lakes ~ very slow to change temperature when the surrounding air temperature changes.

✓ It stabilize aquatic temperature ~ suitable for aquatic organisms to live in.

✓ Organisms made up primarily of water ~ able to resist changes in their own temperature.


4. High latent heat of vaporization (vaporization = transformation from a liquid to a gas)

“The large amount of heat required to convert 1 g of liquid water to vapor”

Water has a high latent heat of •

vaporization (580 cal needed to evaporate 1 g water at 25 ̊C).

A lot of energy is needed to break the •

hydrogen bonds between water molecules to change water from liquid state into vapor.

Due to hydrogen bonds ~ difficult to •

separate & vaporized.

Prevent water from easily evaporate.• Evaporation of water


High latent heat of vaporization - Importance

Release excess body heat efficiently • - body heat will vaporized sweat gives cooling effect because the sweat absorb a lot of heat energy from our bodies.

Transpiration• ~ the vaporization process needs a lot of energy ~ cause cooling effect to leaves & helps prevent overheating.

Sweat give cooling effect to our body Cooling effect in plant


5. Maximum density at 4 ̊C

Water is • less dense as a solid than as a liquid.

Unlike other materials, •

water molecule expand when they solidify.

At above • 4 ̊C, water behaves like other liquids; expand as it warms & contract as its cool. Water in the form of ice


As • temperature falls from 4 ̊C to 0 ̊C ;

Water begin to freeze & solidify when its molecules ✓

are no longer moving vigorously, thus, not enough to break their hydrogen bonds.

Then, water is locked into ✓ crystalline lattice ; each molecule hydrogen bonded to maximum 4 other molecules.

The ✓ ice formed is less dense than the cold wateraround it.

It is because the ✓ hydrogen bonds in ice create space & far apart the water molecules.


Comparison of hydrogen bonds in liquid water & in ice.

• In the liquid water, hydrogen bonds are not stable; constantly break & reform.

• In ice, the hydrogen bonds are more stable & a regular lattice crystal forms in which each water molecule hydrogen bonded with 4 other water molecules.

• Ice is less dense than water because the hydrogen bonds in ice space the water molecules relatively far apart, making them to have 10% fewer molecules than an equal volume of liquid water at 40C. This gives ice to float in water.

Comparison of hydrogen bonds in liquid water and in ice


• Ice floats on the water surface for survival aquatic organisms.

• The floating layer of ice insulates the water below, preventing it from freezing allowing aquatic life to exist under the frozen surface.

Maximum density at 4 ̊C - Importance

Ice insulate the water below


6. High surface tension : Cohesive & Adhesive forces (mainly involve cohesive)

COHESION ADHESION

The sticking together of particles of the same

substance.

The force of attraction between molecules of different substances

Water molecules stick to each other by the hydrogen bonds

Water molecules cling to the surface of another

substance


Hydrogen bond between water molecules create large •

cohesive force.

Related to cohesion is the • surface tension, a measure of how difficult it is to stretch or break the surface of liquid.

At the interface between water & air, there is an •

arrangement of water molecules, which hydrogen bonded to one another & to the water below it.


Direction : W = water, F = paper fiber

Allow • some aquatic organisms to move on the water surface.

E.g.: aquatic •

insect such as water strider.

Water strider on water surface


High Surface Tension - Importance

Capillary action of •

cohesive & adhesive forces helps in transportation of water & dissolved mineral ions in plants.

Water adheres strongly •

to xylem surfaces & can be drawn upwards, without breaking the water column.

Capillary action in plant transportation


Transportation of water in plant


4/30/13

Property Reason of property Importance of property

Water as polar

molecule

• Unequal distribution of shared electron because O more electronegative than H, shared electrons are closer at O atom than to H atoms. Causing H has δ+ and O has δ-

• As universal solvent (has ability to dissolved many compounds by forming hydrogen bonds). Ability to dissolve salt

• As medium of biochemical reaction.

Water has low

viscosity

• Weak hydrogen bonds are continuously break & reformed. Causing water molecules to slide each other easily without friction.

• As lubricant on cell surface• As transportation medium

Water has high specific

heat capacity

• Large amount heat energy need to be absorbed to break the hydrogen bonds to increase temperature of water by 1oC.

• Protect from rapid temperature change.• E.g: water stabilize the temperature of

ocean, our body.

Water has high latent

heat of vaporization

• Large amount heat energy need to be absorbed to break the hydrogen bonds before water molecules vaporized.

• Release excess body heat by sweating (animals) or by transpiration (plants) to give cooling effect.

Water has maximumdensity at

4oC

• Ice (solid form) - locked in crystalline latticestructure, each water hydrogen bonded with four another, space the water molecules far apart each other, & become less density.

• Ice float on water surface for survival of aquatic organisms, by preventing water below become freeze.

Water has high surface

tension

• Hydrogen bond between water moleculescreate large cohesive forces.

• Allow some aquatic insect to move on water surface.

• Combination cohesive & adhesive forces create capillary action for transportation water in plants.

Large organic

molecules of all living thing

Carbohydrates

Proteins

Lipids

Nucleic acids

Macromolecules –chain like molecules (polymer).

Polymer – long molecule consisting of many similar building block (monomer) linkedby covalent bonds.

Synthesis & breakdown of polymers

Monomers • are covalently bonded by condensation / dehydration process & removing a H2Omolecule.

Polymers• are breakdown to monomers by hydrolysis by adding the water molecule.

Synthesis & breakdown of polymer

a) Describe various forms & classes of carbohydrates : 1. Monosaccharides2. Disaccharides3. Polysaccharides

b) Describe the formation & breakdown of maltose.

c) Describe the structures & functions of starch, glycogen & cellulose.

Learning Outcomes

1.2 : CARBOHYDRATE

:LegumesCorn & pumpkin

Dairy products

Cereals

Learning outcome : 1.2 CARBOHYDRATES a) Describe various forms & classes of carbohydrates.

Energy source •

~ e.g: glucose.

As energy •

source, energy storage & to build necessary cell components.

Source of energy

COMPOSED OFCARBON (C) , HYDROGEN (H) & OXYGEN (O) ATOMS

CLASSES

MONOSACCHARIDES Single/simple sugars

DISACCHARIDES Consists of two monosaccharides joined by a glycosidic bond in

condensation reaction

POLYSACCHARIDESPolymers of many

monosaccharides joined by glycosidic bonds in condensation

reaction

Carbohydrates

Monosaccharide(monomer)

Disaccharide(dimer)

Polysaccharide(polymer)

Glucose

Fructose

Galactose

Maltose

Sucrose

Lactose

Starch

Glycogen

Cellulose

Monosaccharides

• Simple sugar units (monomers) which make up all other carbohydrates; cannot be further hydrolyzed into smaller units.

• Molecular formula = CH2O

• Monosaccharides contains :

✓a carbonyl group (C=O) either at C1 atom (aldose) or at the C2 (ketose) atom.✓multiple hydroxyl groups

(-OH).Aldose & ketose sugar


Classification of Monosaccharides:

Number of carbon atom• in the carbon skeleton (3C to 6C long).

Position of functional group•

(carbonyl group), (C = O).


From linear to ring structure :

Linear & ring forms of glucose.


HH

H

H

H

β – glucose α – glucose Fructose

Ring structure of monosaccharides


• In aqueous solutions, glucose as well as most other sugars form a ring structure.

• E.g. : Monosaccharides (hexose).

Small1. molecule

Sweet2. tasting

Readily 3. soluble in water

A result of the hydroxyl groups attached to the carbon •

chains are polar & readily form hydrogen bonds with water molecules.

Reducing sugar4.

Glucose acts as the reducing agent can react with •

Benedict’s solution when heated & reduce the copper compound in the solution to yield a red precipitate. Carbonyl group oxidized into carboxyl group.

Can be 5. crystallized (white crystal).

Characteristic of monosaccharides:Learning outcome : 1.2 CARBOHYDRATES a) Describe various forms & classes of carbohydrates.

Characteristic of Monosaccharides:

Reducing sugar ….. how to identify???It can be classified based on functional group.

Free – OH group at functional carbon can react with Benedict

reagent Reducing sugar


It can be classified based on position of functional group.


Both functional group have reducing properties ~ reducing sugars

ALDOSES KETOSES

Carbonyl group (C=O) at C1

is the aldehyde group Carbonyl group (C=O) at C2

is the ketone group

Sugar is known as aldose (aldehyde sugar)

Sugar is known as ketose (ketone sugar)

E.g: glucose E.g: fructose

It can be classified based on the number of carbon atom.

Trioses Pentoses Hexoses

contains 3 carbon atoms contains 5 carbon atoms contains 6 carbon atoms

C3 H6 O3 C5 H10 O5 C6 H12 O6

Glyceraldehydes Dihydroxyacetone

Ribose Deoxyribose

Glucoseimmediate source of energy

for cellular respiration

Building blocks to form larger molecules

Components of nucleic acids

Galactosesugar found in

milk and yogurt

Fructosesugar found in honey



TRIOSES

Contains 3 carbon atoms

C3H6O3

Glyceraldehydes, Dihydroxyacetone

Building blocks to form larger molecules.

Triose sugar


PENTOSES


C5H10O5

Ribose, Deoxyribose

Components of nucleic acids

Pentose sugar




HEXOSES


C6H12O6

Glucose immediate source of energy for cellular

respiration

Galactose sugar found in milk &

yogurt

Fructose sugar found in honey

Position OH = ABBAB

Hexose sugar


GLUCOSEprimary source of energy

α - glucose β - glucose

Hydroxyl (-OH) group of the C1 projects

belowthe plane of the ring

Hydroxyl (-OH) group of the C1 projects

upwardthe plane of the ring

Position OH = BBABA


Importance of Monosaccharides:

• Glucose

1. Main/major energy source.

2. Main/major respiratory substrate in plants & animals – in the cellular respiration, cells extract energy in a series of reactions starting with glucose molecules.

3. Monomer/ raw materials for synthesis of disaccharides & polysaccharides.


Reaction between 2 molecules of monosaccharides.

Formed by the condensation reaction between the reducing group on one monosaccharide & the hydroxyl group on another.

The bond between 2 molecules of monosaccharides is called a glycosidic bond

The binding between the 2 sugars results in the loss of a hydrogen atom from one molecule & a hydroxyl group from the other ~ release water.

All condensation reactions can be reversed by adding

water to the glycosidic bond – a.k.a. hydrolysis.


Disaccharides

Example Monomers Structure

Maltose α – glucose &

α – glucose

Sucrose Glucose &

fructose

Lactose Glucose &

galactose

Bond: α-1,4 glycosidic bond

Bond: α-1,2 glycosidic bond

Bond: β-1,4 glycosidic bond

H H

Characteristic of Disaccharides:

Small1. molecule (made up from two monosaccharides)

Sweet2. tasting

Water soluble3.

Reducing sugar4. (except sucrose)

Can be 5. crystallized


Describe the formation & breakdown of maltose

Describe the formation & breakdown of maltose.

Formation

• Maltose is form by combination of 2 molecules of α – glucose.

• Through condensation process.

• It combined by α – 1, 4 glycosidic bond.

• One water molecule will be released in order to form α – 1, 4 glycosidic bond.

• Water removed from hydroxyl group (-OH) of one α –glucose & hydrogen (H) atom from another α – glucose.

• The end product are 1 molecule of maltose & 1 water.

Learning outcome : 1.2 CARBOHYDRATE b) Describe the formation and breakdown of maltose.

Breakdown

Breakdown of maltose is called hydrolysis process.•

One water molecule will be added in order to break the•

α – 1, 4 glycosidic bond.

The • hydroxyl group (-OH) of both α – glucose is reformed.

The end product is • 2 molecules of α – glucose.

Learning outcome : 1.2 CARBOHYDRATE b) Describe the formation and breakdown of maltose.

Describe the formation & breakdown of maltose.


Polysaccharides• - macromolecules that formed from hundreds to a few thousand monosaccharides joined together by glycosidic bonds.

Polymers that :•

formed from the polymerization of many units of ✓

monosaccharides through condensation process.

breakdown by hydrolysis into monomers.✓

Polysaccharides

STARCH (plant)GLYCOGEN

(animal)CELLULOSE (plant)

Storage polysaccharide Structural polysaccharide


Polysaccharides

AMYLOSE AMYLOPECTIN


1. Large molecules

2. Not taste sweet

3. Insoluble in water(form colloidal solutions when added to water)

General characteristic of polysaccharides :

Learning Outcomes :(c)describe the structures and functions of starch, glycogen and cellulose

▪ Used by plants to store energy.

▪ Excess glucose is converted into starch granules within chloroplast & other organelles.

▪ Insoluble in water.

Storage polysaccharide - STARCH

Learning outcome : 1.2 CARBOHYDRATE c) Describe the structure & functions of starch.

Maize starch, pea starch, potato starch, wheat starch

Plants store starch, a polymer of glucose monomers, as granules within chloroplasts & other plastids such as amyloplast.

Learning Outcomes :(c)describe the structures and functions of starch, glycogen and cellulose

Learning outcome : 1.2 CARBOHYDRATE c) Describe the structure & functions of starch.

• Consists entirely of α - glucose molecules linked together in both linear & branched forms.

• Starch is a mixture of amylose & amylopectin.

Figure 1.2.12 : A mixture of amylose

(simple form of starch) & amylopectin

(complex form of starch).

AMYLOSE AMYLOPECTIN

α - glucose molecules

-1,4 glycosidic

bond

C1 of one glucose

molecule to C4 of another

- 1,4 glycosidic

bond

C1 of one glucose molecule to C4

of another

α-1,6 glycosidic

bond

C1 of one glucose molecule to C6

of another

Forming linear unbranched helical chain.

200-1500 units in length

Branched helical chains. Branches occur every 30 units

2000 to 200,000 α-glucose

Storage polysaccharide - GLYCOGEN

A storage polysaccharide in animals.

Found as granules mainly in liver & muscle cells.

Structure is similar to amylopectin but more branches ~ macromolecule made from α-glucose.

Insoluble in water ~ no effect on water potential.

Learning outcome : 1.2 CARBOHYDRATE c) Describe the structure & functions of glycogen.

Glycogen

Structural polysaccharides - CELLULOSE

Structural polysaccharide in cell wall of plants ~ stable structure.

Monomer is β-glucose.

Linked by β-1,4 glycosidic bonds.

14

3 2

5

6

Learning outcome : 1.2 CARBOHYDRATE c) Describe the structure & functions of cellulose.

β-glucose Cellulose

Straight chain (no branch).

Long chains of 2000-3000 units of β-glucose units which are arranged in a straight parallel strands & cross linked by hydrogen bonds.

Each β-glucose is related to the next monomer by a rotation of 180°. Making every glucose monomer upside down with respect to its neighbors.


1 4 1 4 1 4

β-1,4 glycosidic bonds

Cellulose microfibrils

Hydrogen bond


Parallel cellulose molecules are held together by hydrogen bonds between hydroxyl groups which attached to C3 & C6.

Parallel cellulose molecules held together into microfibrils.

Cellulose fibers

4/30/13

Starch Glycogen Cellulose

Monomer composition α-glucose α-glucose β-glucose

Bond that linked

monomers together

Amylose:

α-1,4 glycosidic bond

Amylopectin:


(at linear site) &


(at branch site)


(at linear site) &


(at branch site)

β-1,4 glycosidic bond

Shape

Amylose:

Linear unbranched helix

chains

Amylopectin:

Branched helix chains

Highly branched

helix chains

Unbranched but in

straight parallel

chains that form cross

linkage hydrogen

bond with each other

Function

As storage

polysaccharide in plants

(stored as granules in

chloroplast & other

plastids)

As storage polysaccharide in

animals & fungi

(stored as granules in muscle

& liver cells)

*why suitable as storage

carbohydrate: large compact

molecule & insoluble in

water

As component of

plant cell wall that

give structural

support to plant cell

Glucose Glucose

4

1

4

1

41

41

Polymers

with

- glucose

are straight

Polymers with

- glucose

are helical

and glucosering structures

Starch: 1–4 linkage of glucose monomers Cellulose: 1–4 linkage of glucose monomers

a) State the types of lipid:• Triglycerides (fat & oil)• Phospholipids• Steroids

b) Describe the structure of :• Fatty acids• Glycerols

c) Describe the formation & breakdown of triglycerides.

Learning Outcomes

1.3 : LIPID

Lipid

Triglycerides (eg: oil, fat)

Phospholipids(eg: lecithin)

Steroids (eg: cholesterol,

testosterone)

1 glycerol 1 glycerol

3 fatty acid 2 fatty acid

Phosphate group

carbon skeleton - 4 fused carbon ring

(with variable length of side

chain)

Greek ~ • lipos ~ fat

Not include true polymers• – lipid such as triglycerides (fat & oil) made up from 1 glycerol & 3 fatty acids only.

Insoluble in water • (hydrophobic) but soluble/dissolve in organic solvents such as acetone, ethers, benzene & other non-polar solvents.

E.g. many natural oils, waxes & steroids.•

Organic compounds that contain : carbon (C), hydrogen

(H) & oxygen (O).

The ratio of oxygen atoms to hydrogen atoms is much lower than the 1:2 ratio found in carbohydrates.

Learning outcome : 1.3 LIPIDS a) State the types of lipid : Triglycerides, Phospholipids & Steroids

Energy storage 1. – twice energy content compared to carbohydrates due to long hydrocarbon chain which contain many C – H bonds in fatty acid lipid compare to carbohydrates.

Protection of vital organs 2. – adipose tissue cushion the organs.

Heat insulator 3. - layer of fat beneath skin insulates the body.

Transport medium for fat soluble vitamins 4. – A, D, E, K.

Component of cell membrane5. .

LIPIDS ~ role


Triglycerides - Fats

Made up from • 1 glycerol & 3 saturated fatty acid.

Most animal fats are made up from saturated fatty acid •

& solid at room temperature. E.g: butter

Triglycerides (building block)

1 Glycerol 3 fatty acid

Saturated Unsaturated

Learning outcome : 1.3 LIPIDS a) State the types of lipid : Triglycerides (fat and oil)

Saturated fatty acids • = no double bond between C atoms of hydrocarbon chain fatty acid (carbon atom is saturated with hydrogen atoms).


Saturated fatty acids

Unsaturated fatty acids

• Made up from 1 glycerol & 3 fatty acids.

• Unsaturated fatty acid – have one/ more double bond between C atoms of hydrocarbon chain of fatty acid.

• Fatty acid will have kink in its hydrocarbon chain wherever double bond occur – to prevent the molecules from packing closely together to solidify at room temperature.

• Most fat of plants & fish are unsaturated = liquid at room temperature E.g: olive oil, cod liver oil.

Triglycerides - Oil


Fats OilsMost fats made up from

saturated fatty acidsMost oils made up from one or more unsaturated fatty acids

Solid at room temperaturehydrocarbon chain fatty acid can closely packed

together

Oil at room temperaturepresence of double bond

between C atoms ofhydrocarbon chain fatty acid causing it cannot be closely

packed together

Most fats origin from animal

Most oils origin from plant & fish

E.g: butter E.g.: olive oil, cod liver oil

Differences between fats & oils.

Lipids : Phospholipids

Learning outcome : 1.3 LIPIDS a) State the types of lipid : Phospholipids

• 3 components : ✓ 1 glycerol✓ 2 fatty acids ✓ 1 phosphate group.

Phospholipids structure

Phospholipid symbol

The phospholipids molecule has :

i. A polar “head” at one end (the charged phosphate group).

Two long, ii. non-polar “tails” at the other.

Phospholipids • have both

hydrophilic region✓ (due to a charge at polar phosphate head).

hydrophobic regions✓ (no charge at non polar tails of long hydrocarbon fatty acid chain).

Known as • amphipathic molecules.


Micelle structure

When phospholipids are •

added to water, they self-assemble into aggregates with hydrophobic tails pointing toward the center, excluded from water & the hydrophilic heads on the outside, facing the water.

This type of structure is •

called a micelle.

At the surface of a cell, •

phospholipids are arranged in a double layer.


Hydrophilichead

Hydrophobictail

WATER

WATER

• The hydrophilic heads are on the outside of the bilayer, in contact with the aqueous solutions inside & outside of the cell.

• The hydrophobic tails point toward the interior of the membrane bilayer, away from the water.

• The phospholipid bilayer forms a barrier between the cell & its external environment.Phospholipids are arranged in a double

layer


Lipid with • carbon skeleton that consists of four fused carbon ring , with variable length of side chain.

No fatty acids•

Cholesterol• : precursor to synthesis other steroidincluding sex hormones

Learning outcome : 1.3 LIPIDS a) State the types of lipid : Steroids

Steroid hormones

Lipids : Steroids

Triglycerides Phospholipids Steroids

• Insoluble in water -due to long hydrocarbon chain that contain non polar C-H bonds

• Has amphipathic

properties (hydrophilic

head & hydrophobic tail)

• Insoluble in

water

• Made up from :

✓ 1 glycerol

✓ 3 fatty acid

• Made up from

✓ 1 glycerol

✓ 2 fatty acids

✓ 1 phosphate group

• Made up from

✓ fused carbon

ring without

fatty acid


. A.k.a. glycerin

An alcohol

Consists of a three carbon skeleton with a hydroxyl group attached to the end.

LIPIDS : - Glycerol

Learning outcome : 1.3 LIPIDS b) Describe the structure of : Glycerol and Fatty Acids

Structure of glycerol

.

Learning Outcomes :(b) describe the structure of fatty acids and glycerols

Structure of fatty acid


LIPIDS : - Fatty Acids

• Chemical formula of fatty acids : RCOOH

• R = hydrocarbon chain

• Consists of a carboxyl group (-COOH) at the end of chain attached to a long hydrocarbon chain, often 16 to 18 carbons long.

• The many non-polar C-H bonds in the long hydrocarbon chain make lipid become hydrophobic.

Fatty Acid classification

Based on presence of double bond in hydrocarbon chain fatty acid

Saturated fatty acid Unsaturated fatty acid

No double bond between carbon atoms of hydrocarbon

chain (maximum no of hydrogen atoms)

One / more double bonds between carbon atoms of

hydrocarbon chain

EG: STEARIC ACID EG: OLEIC ACID

.


LIPIDS : – saturated fatty acid

Saturated fatty acid

If there are • no double bond between C atom of hydrocarbon chain » the molecule is a saturated fatty acid - a hydrogen at every possible position.

Has the • maximum number of hydrogen atoms.

Straight chains.•

Fats are mostly made up from saturated •

fatty acid.


.

Unsaturated fatty acid

• If there are one or more double bondbetween C atoms of hydrocarbon chain » the molecule is an unsaturated fatty acid – each of the carbon involved in the formation of double bond can still accept ONE HYDROGEN ATOM.

• Oils are mostly made up from one or more unsaturated fatty acid.


LIPIDS : – unsaturated fatty acid

Saturated fatty acid Unsaturated fatty acid

No double bond betweenC atoms of hydrocarbon

chain. Each C are saturated with H atoms.

Presence of double bond betweenC atoms of hydrocarbon chain.

C are not saturated with H atoms when double bond occurs.

Hydrocarbon chain can be closely packed together. It

tend to give a straight chain.

Hydrocarbon chain cannot be closely packed together due to

presence of double bond betweenC atoms. It tend to give a bend shape & form kink structure.

Solid at room temperature Oil at room temperature

Differences between saturated & unsaturated fatty acids



The • synthesis of triglycerides involves one molecule of 3C glycerol containing three hydroxyl (-OH) groups.

Hydroxyl group on the glycerol molecule may form • ester bond with three fatty acid to form a triglycerides or triacylglycerol.

Each fatty acid contains a • hydrocarbon chain & a terminal carboxyl (-COOH) group.

In order to form an ester bond between a fatty acid & a •

glycerol molecule, one OH at the carboxyl groups on fatty acids & one H at the hydroxyl group on glycerol are liberated as a water molecule.

Formation & breakdown of triglycerides

Learning outcome : 1.3 LIPIDS c) Describe the formation and breakdown of triglycerides

Formation of triglyceride


Describe the formation & breakdown of triglycerides

Formation

Triglycerides form from • 1 molecule of glycerol and 3 molecules of fatty acids

Joined by • 3 ester bond / linkage

• Through the condensation/ esterification reaction

Removed • 3 water molecules

Each water remove from • - OH at the carboxyl groups on fatty acids and -H at the hydroxyl group on glycerol

The end products is • 1 molecule of triglycerides and 3 molecules of water


Breakdown of triglyceride

Describe the formation and breakdown of triglycerides

Breakdown

• Breakdown of triglyceride is called hydrolysis reaction

• 3 water molecules will be added to break 3 ester bonds / linkages

• The -OH water react with carboxyl end of fatty acid and -H water react with hydroxyl end of glycerol

• Carboxyl group of each fatty acid is reformed

• Hydroxyl group of glycerol is also reformed

• The end product is 1 molecule of glycerol and 3 molecules of fatty acids


a. Describe the basic structure & classes of amino acids.

b. Explain how amino acids are grouped.

c. Describe the formation & breakdown of dipeptide.

d. Explain primary, secondary, tertiary & quaternary levels of proteins & the types of bonds involved.

e. Classify proteins according to their structure.

f. Explain the effect of pH & temperature on the structure of protein.

Learning Outcomes

1.4 : PROTEIN

Learning outcome : 1.4 PROTEIN a) Describe the basic structure and classes of amino acids

• A large, complex molecule made up from monomers of amino acids.

• Joined together by peptide bonds into long polypeptide chains.

• A protein consists of one or more polypeptides that folded & coiled into a specific 3D shape or conformation.

Source of protein & structure of amino acid

Structure of Amino AcidsAmino acids • consist of four components attached to a central carbon.

It can be distinguished from fats •

& carbohydrates by containing nitrogen (N).

Other elements •

include carbon (C), hydrogen (H), oxygen (O), sulphur (S) & sometimes phosphorus (P).

Amino group

Hydrogen atom

Carboxyl group

Side chain

Structure of amino acid


Amino acids are • amphoteric ~ having the characteristics of an acid & a base.

Amino acid are capable of • reacting chemically either as an acid or a base because it has acidic carboxyl group (-COOH) & basic amino group (-NH2).



In • aqueous solution, the carboxyl & amino groups interact resulting in the transfer of proton from carboxyl group to amino group.

Thus the • amino acid exists in ionized form known as zwitterion.

Zwitterion of amino acid can act as • bufferto neutralize the solution.

Structure of zwitterion

Ionization States of Amino Acid Depend on pH

Ionization states of amino acid in different pH


Learning outcome : 1.4 PROTEIN b) Explain how amino acid are grouped

Groups of amino acid

All • 20 amino acids are classified under specific group based on properties of their side chain ( R group).

Groups of amino acid – non polar amino acids

Amino acids – non polar side chains

Groups of amino acid – polar amino acids

Amino acids –

polar side chains

Groups of amino acid – basic and acidic amino acids

Amino acids – basic & acidic side chains

Learning outcome : 1.4 PROTEIN c) Describe the formation and breakdown of dipeptide

** during formation or breakdown of dipeptide, use the basic structure amino acid (without ions) not the zwitterion form.

Formation of Dipeptide

Formation of dipeptide


Formation of dipeptide

Formation

Dipeptide formed from combination of two amino acids.•

Joined by peptide bond.•

Through the condensation reaction.•

Removed • 1 water molecule.

Water remove from • -OH at the carboxyl group of one amino acid & -H at the amino group of another amino acid.

The end products is • 1 molecule of dipeptide & 1 molecule of water.

Describe the formation & breakdown of dipeptide


Breakdown of dipeptide


Breakdown of Dipeptide

** during formation or breakdown of dipeptide, use the basic structure amino acid (without ions) not the zwitterion form.

Breakdown

Breakdown of dipeptide is called hydrolysis reaction.•

1 • water molecule will be added to break the peptide bond.

The • -OH water react with carboxyl end of one amino acid & -H water react with amino end of another amino acid.

Carboxyl group of amino acid is reformed.•

Amino group of another amino acid is also reformed.•

The end product is • 2 molecules of amino acid.


Describe the formation & breakdown of dipeptide

Learning Outcomes :(c) describe the formation and breakdown of dipeptide

Formation of the polypeptide

Polypeptide • –many amino acidsare joined together by peptide bonds.

The • repeated sequence (-N-C-C) is the polypeptide backbone..

Learning outcome : 1.4 PROTEIN d) Explain primary, secondary, tertiary & quaternary levels of

proteins; the types of bonds involved

Formation of polypeptide

Learning Outcomes :(c) describe the formation and breakdown of dipeptide

Polypeptide backbone

Peptide bond

Primary level

Secondary level

Tertiary level

Quaternary level

Classification of Protein

Learning outcome : 1.4 PROTEIN d) Explain primary, secondary, tertiary & quaternary levels of

proteins; the types of bonds involved

Level of structure

Base on protein level

Level of structure

Learning outcome : 1.4 PROTEIN d) Explain primary levels & the types of bonds involved

Primary structure

The • linear sequence of amino acidsin a polypeptide chain.

Amino acids joined together by •

peptide bonds.

The primary structure of a protein •

is its unique sequence of amino acids.

E.g.: Primary structure of lysozyme •

made up of a single strand containing 129 amino acids.

Primary structure

Learning outcome : 1.4 PROTEIN d) Explain secondary levels & the types of bonds involved

Secondary structure

Most proteins have segments of their polypeptide chains •

that repeatedly coiled or folded that contributed into protein’s overall shape.

Are the result • of hydrogen bonds within polypeptide backbone.

Hydrogen bonds • between H of the -NH groups & O of the -CO groups within polypeptide backbone hold the secondary structure together.

Coils to form theα helix

The coil held together by hydrogen bonding

between every 4th

amino acid

E.g: Keratin in hair, nails, horn & feathers

Folds to form theβ pleated sheet

Two or more region of polypeptide chain lie parallel to each other

E.g: Silk protein

Learning outcome : 1.4 PROTEIN d) Explain secondary levels & the types of bonds involved

Secondary structure

Learning outcome : 1.4 PROTEIN d) Explain tertiary levels & the types of bonds involved

Tertiary structure

Overall shape of a polypeptide •

resulting from interactions between side chains (R groups) of the various amino acids.

A polypeptide may be further •

coiled & folded into a globular 3D shape call globular protein.

E.g: enzymes, hormones, •

antibodies, myoglobin & plasma protein.

Tertiary structure

Types of bonds in tertiary structure of protein

• Disulfide bridge Between side chains with sulfhydryl groups (-SH).

• Ionic bonds Between positively & negatively charged side chains.

• Hydrogen bonds Between polar side chains.

• Hydrophobic

• van der Waals interactions

Between nonpolar side chains.

Types of bonds in primary, secondary & tertiary structure of

proteinFeatures Primary Secondary Tertiary

Type of bond exist

1. Peptide bond

1. Peptide bond.

2. Hydrogen bond (between C=O & NH group of backbone)

1. Peptide bond.2. Hydrogen bond

(between C=O & NH group of backbone).

3. Disulfide bridge.4. Ionic bond.5. Hydrogen bond

(between R groups).6. Hydrophobic (van der

Waals) Interaction.

Hemoglobin

Learning outcome : 1.4 PROTEIN d) Explain quaternary levels & the types of bonds involved

Quaternary structure Structure of a protein which consists of • two or more polypeptide chains.

Overall protein structure that results from interactions •

among polypeptide subunits.

Interaction by •

hydrogen bond, ionic bond, van der Waals interactions & disulfide bridges.

E.g : Haemoglobin. It consists •

of 4 polypeptides chains.

Primary level Secondary level Tertiary level Quaternary level

Refers to

linear

sequence of

amino acid in

polypeptide

chain.

Refers to the coiling or

folding of polypeptide

chain that contribute to

the protein’s overall

shape.

•α-helix (coiling)

•β-pleated sheet (folding)

due to interaction of

hydrogen bonds within

polypeptide backbone

Refers to overall globular

3D shape of polypeptide

that resulting from

interaction between side

chain of amino acids.

Refers to protein that

consist two or more

polypeptide chains.

• Peptide

bond join

amino acids

• Peptide bond join

amino acids

• Hydrogen bonds within

polypeptide backbone


amino acids

• Hydrogen bonds

• Ionic bonds

• Disulfide bridge

• Hydrophobic

interaction / van der

Waals interaction


amino acids

• Hydrogen bonds

• Ionic bonds

• Disulfide bridge

• Hydrophobic interaction

/ van der Waals

interaction

Any protein

has primary

basic level

E.g: keratin (α-helix) : silk

protein (β-pleated sheet)

E.g: enzyme, hormone,

myoglobin

E.g: haemoglobin (consist

four polypeptide chains)

Learning outcome : 1.4 PROTEIN e) Classify proteins according to their structures

Classification of Protein

Based on Structure

Fibrous Globular Conjugated

4/30/13

CONJUGATED PROTEINS

1. A protein portion attached to a non – protein components(prosthetic group).

2. E.g: Lipoprotein, glycoprotein, haemoglobin.3. Haemoglobin made up from protein component (globin) & non

protein component (haem group).

GLOBULAR PROTEIN

1. Unstable protein.2. Made up from compactly folded spherical shape polypeptide. 3. Generally soluble in water/form colloidal solution.4. Function : involve in chemical processes in organism.5. Protein level : tertiary or quaternary.6. E.g: Enzymes, hormones, antibodies & haemoglobin.

FIBROUS PROTEIN

1. Stable protein.2. Made up from long fiber strands/ sheets polypeptide.3. Insoluble in water.4. Function: involve in mechanical & structural support.5. Protein level : secondary.6. E.g: Keratin, collagen, myosin, actin.

Conjugated protein

Type conjugated protein

Prosthetic group components

Location found

Lipoprotein Lipid Plasma membrane

Glycoprotein Carbohydrates Plasma membrane

Haemoglobin Haem group with Fe Red blood cell

Myoglobin Haem group with Fe Muscle

Nucleoprotein Nucleic acid Chromosome

Fibrous protein Globular protein

Made up from long fiber strand polypeptide

Made up from compact folded spherical shape polypeptide

Insoluble in water Generally soluble in water

Function for mechanical & structural support

Function for chemical process in organism

Protein made up from secondary level

Protein made up from tertiary or quaternary level

Stable structure protein Unstable structure protein

E.g: Keratin, collagen, myosin, actin

E.g: Enzyme, hormone, antibody, haemoglobin

Factor Effect of protein structure

Extreme pH • 3D structure of protein is relatively unstable ~ due to weak bond such as ionic bond & hydrogen bond disrupted.

• Extreme pH causes these bonds break ~ 3D shape will change/ alteration ~ protein denatured.

• Protein (enzyme, antibodies, hormone) will lose the biological activities.

• Protein not function

High temperature

(e.g: temperature

> 400C)

• 3D structure of protein is relatively unstable ~ due to weak bond such as hydrogen, ionic bonds & hydrophobic interaction.

• High temperature causes these bonds break ~ 3D shape will change/ alteration ~ protein denatured.

• Protein (enzyme, antibodies, hormone) will lose the biological activities.

• Protein not function.• Protein (e.g: enzyme) only work best at optimum pH or

optimum temperature only. If deviation occurs from optimum pH or optimum temperature, protein may be denatured.

Example of Protein DENATURATION

Denaturation of enzyme

Learning outcome : 1.4 PROTEIN f) Explain the effect of pH & temperature on the structure of protein

Protein denaturation

Learning outcome : 1.4 PROTEIN f) Explain the effect of pH & temperature on the structure of protein

Example of Protein DENATURATION

a) Describe the structure of nucleotide as the basic components of nucleic acid (DNA & RNA).

b) Describe the structure of DNA based on the Watson & Crick Model.

c) State the types & function of RNA.

d) State the differences of DNA and RNA.

Learning Outcomes

1.5 : NUCLEIC ACID

• Nucleic acids are in the form of :

✓DNA (deoxyribonucleic acid)

✓RNA (ribonucleic acid)

• A group of complex compounds found in all living cells & viruses that is essential for life.

• They function in controlling cellular function & heredity (encoding, transmitting & expressing genetic information).

Learning outcome : 1.5 NUCLEIC ACID a) Describe the structure of nucleotide as the basic composition of

nucleic acid (DNA and RNA)

Nucleotide made up from

Phosphate group Pentose sugar Nitrogenous bases

In DNA : Deoxyribose sugar

In RNA :Ribose sugar

Purine (double ring)Adenine & Guanine

Pyrimidine (single ring)Cytosine, Thymine

(in DNA) &Uracil (in RNA)

Learning outcome : 1.5 NUCLEIC ACID a) Describe the structure of nucleotide as the basic composition of

nucleic acid (DNA and RNA)

Types of nucleic acid – RNA and DNA

Nitrogenous BasePURINE PYRIMIDINE

ADENINE GUANINE CYTOSINE THYMINE (DNA) URACIL (RNA)

Chargaff’s Rule : to figure out how the FOUR BASES match up in pairs

Base pairing between nitrogenous bases

• Pentose sugar in DNA is deoxyribose while for RNA is ribose.

Pentose sugar in DNA and RNA

Learning outcome : 1.5 NUCLEIC ACID a) Describe the structure of nucleotide as the basic composition

of nucleic acid (DNA and RNA)

Nucleotide Formation

Formation of nucleotide



Nucleotide = An organic compound consists :

Nitrogen1. containing purine/pyrimidine base

Sugar2. (ribose/ deoxyribose)

Phosphate group3. .

1 5

In a nucleotide : the nitrogenous •

base is joined to the C1 of the pentose.the phosphate •

group is joined to the C5 of the pentose.

Structure of nucleotide

Nucleotides are combined by 1. condensation reaction.

2. 2 nucleotides are joined by a phosphodiester bond.

Many nucleotides3. joined together formed polynucleotides.

Forming a backbone with 4. repeating sugar-phosphate units.

Breakdown by 5. hydrolysis ~ into nucleotide.



Polynucleotide Formation

Formation of polynucleotide

Covalent bond (N-glycosidic bond)

Phosphoester bond

Phosphoester bond

Phosphodiester bond

HOW TO DRAW POLYNUCLEOTIDE?

**diagram above only show the joining of two

nucleotide. However, polynucleotide is large

molecule consist of many nucleotides.

Structure of polynucleotide

Formation of polynucleotide

Learning outcome : 1.5 NUCLEIC ACID b) Describe the structure of DNA based on the Watson &

Crick Model

1. Nucleotide of DNA consists of deoxyribose sugar, nitrogenous bases & phosphate group.

2. Two strands of polynucleotide are twisted forming double helix structure. Each full turn of helix consist of 10 base pairs.

Watson and CrickProposed the double helix DNA structure

3. Two strands of polynucleotide are anti-parallel (in opposite direction). One strand from 5’to 3’ end & another strand from 3’ to 5’ end.

4. Each strand made up from deoxyribose sugar & phosphate group, repeating to form sugar phosphate backbone.


Crick Model


Crick Model

The bond between 5. 3’ hydroxyl group of deoxyribose sugar & the phosphate group is known as phosphodiester bond.

The sugar phosphate backbones are on the outside of 6.helix structure.

Which protected the 7. paired nitrogenous bases inside the helix structure.

Polynucleotide strands linked together by 8. hydrogen bonds between the pair of nitrogenous bases.


9. The base pairing between 2 strands must be complementary.

✓ Where purine base pairing with pyrimidine base.

✓ Adenine (A) base pairing with Thymine (T) joined by 2 hydrogen bonds.

✓ Guanine (G) base pairing with Cytosine (C) joined by 3 hydrogen bonds.



Crick Model

Learning outcome : 1.5 NUCLEIC ACID c) State the types and function of RNA (mRNA, tRNA, rRNA)

Structure of RNA

RNA

Polynucleotide : Single stranded

Subunit :

1. Pentose sugar: ribose

2. Phosphate group

3. Nitrogenous bases :

Adenine, Guanine, ✓

Cytosine & Uracil

RNA ~ 3 types

Messenger RNA

(mRNA)

Ribosomal RNA

(rRNA)

Transfer RNA (tRNA)

Types of RNA


RNA ~ 3 types

Messenger RNA

(mRNA)

Ribosomal RNA

(rRNA)

Transfer RNA

(tRNA)

• Copies genetic information

from DNA to the ribosome

• Serve as template during

translation

• Assemble with protein to form

ribosomal subunits in the

cytoplasm

• Involve in protein

synthesis

• Transfer specific

amino acids to the

ribosome during

translation


Function of mRNA, rRNA and tRNA

Function of mRNA, rRNA and tRNA


Function of mRNA, rRNA

and tRNA

DNA RNA

Double stranded polynucleotides Single stranded polynucleotide

Helix form Linear form

Deoxyribose as pentose sugar Ribose as pentose sugar

Nitrogenous bases: Adenine (A), Thymine (T),Cytosine (C) & Guanine

(G). A pairs with T, C pairs with G.

Nitrogenous bases: Adenine (A), Uracil (U),Cytosine (C) & Guanine (G)

Found only in nucleus Manufactured in nucleus but found in cytoplasm

Only one type (DNA) 3 types (mRNA, rRNA, tRNA)

Larger molecular mass Smaller molecular mass

Stable molecule Less stable molecule

Ratio of A and T is the same, ratio of C and G is the same

Ratio of A and U is the not same, ratio of C and G is not the same

Structure of DNA and RNA

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CHAPTER 1 : MOLECULES OF LIFE - WordPress.com · •Chemical formula : H 2 O •Shape –V shape with angle 104.50 •2 hydrogen atoms joined to 1 oxygen atom by single covalent bond