1styear CHEMISTRY Notes Chapter-1

Chapter-1
INTRODUCTION TO FUNDAMENTAL CONCEPTS OF CHEMISTRY

Introduction to Fundamental Concepts of Chemistry 
Atom
It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.


Molecule
When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.

Formula Weight
It is the sum of the weights of the atoms present in the formula of a substance.

Molecular Weight
It is the sum of the atomic masses of all the atoms present in a molecule.

Chemistry
It is a branch of science which deals with the properties, composition and the structure of matter.

Empirical Formula
Definition
It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.
Examples
The empirical formula of benzene is "CH". It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.
The empirical formula of glucose is "CH2O". This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.

Determination of Empirical Formula
To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.

Illustrated Example of Empirical Formula
Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.

Step I - Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.

Step II - Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm

Step III - Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%

Step IV - Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5

Step V - Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO

Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.

Molecular Formula
Definition
The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.
OR
It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.
Examples
The molecular formula of benzene is "C6H6". It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.

The molecular formula of glucose is "C6H12O6". The formula represents that


1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.

Determination of Molecular Formula
The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula)n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
n = Molecular Mass / Empirical Formula Mass

Example
The empirical formula of a compound is CH2O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH2O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH2O)6
= C6H12O6

Molecular Mass
Definition
The sum of masses of the atoms present in a molecule is called as molecular mass.
OR
It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.
Example
The molecular mass of CO2 may be calculated as
Molecular mass of CO2 = Mass of Carbon + 2 (Mass of Oxygen)
= 12 + 2 x 16
= 44 a.m.u
Molecular mass of H2O = (Mass of Hydrogen) x 2 + Mass of Oxygen
= 1 x 2 + 16
= 18 a.m.u
Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
= 1 + 35.5
= 36.5 a.m.u

Gram Molecular Mass
Definition
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.
Examples
1. The molecular mass of H2O is 18. If we take 18 gm H2O then it is called 1 gm molecular mass of H2O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.

Mole
Definition
It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.
OR
The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.
Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.

From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO2 is 44. If we take 44 gm of CO2 it is called one mole of CO2 or the molecular mass of H2O is 18. If we take 18 gm of H2O it is called one mole of H2O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contain 6.02 x 10(23) atoms
While
One gram molecule contain 6.02 x 10(23) molecules.

Avagadro's Number
An Italian scientist, Avagadro's calculated that the number of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10(23). This number is known as Avogadro's number and represented as N(A).

Example
1 gm mole of Na contain 6.02 x 10(23) atoms of Na.
1 gm mole of Sulphur = 6.02 x 10(23) atoms of Sulphur.
1 gm mole of H2SO4 = 6.02 x 10(23) molecules H2SO4
1 gm mole of H2O = 6.02 x 10(23) molecules of H2O
On the basis of Avogadro's Number "mole" is also defined as
Mass of 6.02 x 10(23) molecules, atoms or ions in gram is called mole.

Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance
Example 1
Calculate the number of atoms in 9.2 gm of Na.
Solution
Atomic mass of Na = 23 a.m.u
If we take 23 gm of Na, it is equal to 1 mole.
23 gm of Na contain 6.02 x 10(23) atoms
1 gm of Na contain 6.02 x 10(23) / 23 atoms
9.2 gm of Na contain 9.2 x 6.02 x 10(23) /23
= 2.408 x 10(23) atoms of Na

Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance
Example 2
Calculate the mass in grams of 3.01 x 10(23) molecules of glucose.
Solution
Molecular mass of glucose = 180 a.m.u
So when we take 180 gm of glucose it is equal to one mole So,
6.02 x 10(23) molecules of glucose = 180 gm
1 molecule of glucose = 180 / 6.02 x 10(23) gm
3.01 x 10(23) molecules of glucose = 3.01 x 10(23) x 180 / 6.02 x 10(23)
= 90 gm

Stoichiometry
(Calculation Based On Chemical Equations)
Definition
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called stoichiometry.
In this study we always use a balanced chemical equation because a balanced chemical equation tells us the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass - Mass Relationship
2. Mass - Volume Relationship
3. Volume - Volume Relationship

Mass - Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.
Example
Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.
Solution
Step I - Write a Balanced Equation
CaCO3 ----> CaO + CO2
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO3 ----> CaO + CO2

Method I - MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO3 gives 1 mole of CO2
0.5 mole of CaCO3 will give 0.5 mole of CO2
Mass of CO2 = Moles x Molecular Mass
= 0.5 x 44
= 22 gm

Method II - FACTOR METHOD
From equation we may write as
100 gm of CaCO3 gives 44 gm of CO2
1 gm of CaCO3 will give 44/100 gm of CO2
50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2
= 22 gm of CO2

Mass - Volume Relationship
The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm3 volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.
Example
Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.
Solution
Step I - Write a Balanced Equation
CH4 + 2 O2 ----> CO2 + 2 H2O
Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CH4 + 2 O2 ----> CO2 + 2 H2O

Method I - MOLE METHOD
Convert the given mass of CH4 in moles
Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4
From Equation
1 mole of CH4 gives 1 moles of CO2
1.25 mole of CH4 will give 1.25 mole of CO2
No. of moles of CO2 obtained = 1.25
But 1 mole of CO2 at S.T.P occupies 22.4 dm3
1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25
= 28 dm3

Method II - FACTOR METHOD
Molecular mass of CH4 = 16
Molecular mass of CO2 = 44
According to the equation
16 gm of CH4 gives 44 gm of CO2
1 gm of CH4 will give 44/16 gm of CO2
20 gm of CH4 will give 20 x 44/16 gm of CO2
= 55 gm of CO2
44 gm of CO2 at S.T.P occupy a volume 22.4 dm3
1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3
55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44
= 28 dm3

Volume - Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac's law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH4 + 2 O2 ----> CO2 + 2 H2O
In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to give one volume of CO2 and two volumes of H2O
Examples
What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4 (ethylene)?
Solution
Step I - Write a Balanced Equation
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O

Step II - Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O

According to Equation
1 dm3 of C2H4 requires 3 dm3 of O2
500 dm3 of C2H4 requires 3 x 500 dm3 of O2
= 1500 dm3 of O2

Limiting Reactant
In stoichiometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stoichiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.

Problem
We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N2 + 3 H2 ----> 2 NH3
Solution
In this problem moles of N2 and H2 are as follows
Moles of N2 = Mass of N2 / Mol. Mass of N2
= 50 / 28
= 1.79
Moles of H2 = Mass of H2 / Mol. Mass of H2
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N2 + 3 H2 ----> 2 NH3
1 mole of N2 gives 2 moles of NH3
1.79 mole of N2 gives 2 x 1.79 moles of NH3
= 3.58 moles of NH3
Mass of NH3 = Moles of NH3 x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH3

1styear BIOLOGY Notes Chapter-14

Chapter-14
TRANSPORT


"BOTONY "

DIFFUSION
The movement of ions or molecules from the region of higher concentration to the region of lower concentration is known as diffusion.
EXAMPLES
1. If a bottle of perfume is opened in a corner of a room, it can be smelt in the entire room.
2. Leakage of gas pipes can be smelt from a farther point.
3. If we drop a KMNO4 crystal in clean water, then after sometime the crystals will dissolve and colour of water changes from colorless to purple.

FACTORS ON WHICH RATE OF DIFFUSION DEPENDS


1-SIZE
Small molecules move faster than larger ones.

2-TEMPERATURE
Rate of diffusion will be high at high temperatures.

3-CONCENTRATION GRADIENT
Greater the difference in concentration and shorter the distance between two regions, greater will be the rate of diffusion.

FACILITATED DIFFUSION
Diffusion of the substances across the cell membrane through the specific carrier proteins is known as facilitated diffusion. These membrane transport proteins are channel proteins, receptors, cell pumps or carriers, made up of usually proteins and don’t require energy for transport.

PASSIVE TRANSPORT
Movement of substances in and out of the cell, caused by simple kinetic motion of molecules, doesn’t require energy of ATP is known as passive transport, e.g. Simple diffusion and facilitated diffusion.

OSMOSIS
The movement of water molecules from the region of higher concentration to the region of lower concentration through a semi-permeable membrane, is known as osmosis.

TYPES OF OSMOSIS
A- ENDOSMOSIS
The movement of water molecules into the cell, when it is placed in hypotonic solution is called as Endosmosis.

B- EXOMOSIS
The movement of water molecules out of the cell when the cell is placed in a hypertonic solution.

ACTIVE TRANSPORT
The movement of ions or molecules across the cell membrane against the concentration gradient i.e. from lower concentration to higher concentration with the help of specific transport proteins in the cell membrane, at the expense of cell’s metabolic energy – ATP is called active transport.
EXAMPLES
1. Sodium-Potassium pump in nerve cells which pump Na+ out of the nerve cell, and K+ into the cell against the concentration gradient.
2. Cells lining the intestine can transport glucose actively from a lower concentration in the intestinal contents to higher concentration in blood.
3. In plants phloem loading is an ex. Of active transport.

IMBIBITIONS
Adsorption of water and swelling up of hydrophilic (water loving) substances is known as imbibitions.

HYDROPHILIC SUBSTANCES
Those which have great affinity for water are hydrophilic e.g. starch, gum, protoplasm, cellulose, proteins, e.g. seeds swell up when placed in water.
• Wrapping up of wooden framework during rainy seasons.
• Dead plant cells are hydrophilic colloids.
• The chemical potential of water is a quantitative expression of the free energy associated with the water.
• UNIT: Joules/mole
• This term has been replaced by water potential

WATER POTENTIAL (PSI)
It is the difference between the fee energy of water molecules in pure water and energy of water in any other system, or solution. Water potential is a relative quantity, depends upon gravity and pressure.
Q = Q* + f (concentration) + f (pressure) + f (gravity)
?* is standard water potential or pure water potential of valve O Mpa.
Unit : Megapascal’s – MPa
(1 Mpa = 9.87 atmospheres)

USES
The direction of water flow across cell membrane can be determined. It is a measure of water status of the plant.

OSMOTIC PRESSURE
The pressure exerted upon a solution to keep it in equilibrium with pure water when the two are separated by a semi permeable membrane is known as Osmotic pressure.
It prevents the process of osmosis.

OSMOTIC POTENTIAL
The tendency of a soln to diffuse into another, when two solutions of different concentrations are separated by a differentially permeable membrane.
• It is represented by ßs for pure water ßs = 0
• The ßs decrenses as the osmotic concentration increases.
• Osmotic concentration is the number of osmotic-ally active particle per unit volume.
• Osmotic potential has been replaced by solute potential.
• The concentration of solute particles in a solution is know as solute potential ßs. It value is always negative.

PRESSURE POTENTIAL ?P
When a cell is placed in pure water or in aqueous solution with higher water potential than the cell sap water follows into the vacuole by endosmosis thru cell membrane and tonoplast. Due to this inflow of water, the tension developed by the cell wall causes an internal hydrostatic pressure to develop, which is called as pressure potential.
? = ßs + ßp or Qp = Q – Qs
In turgid cells ßp is equal and opposite to ßs

TURGID CELL
When the cell is fully stretched with maximum pressure potential, the water cannot flow into it. This condition is called turgidity and the cell is turgid.

PLASMOLYSIS
If a cell is placed in a hypertonic solution, which has more negative solute and water potentials then water will come out of the cell, by exosmosis and protoplasm starts separating from cell wall leaving a gap between cell wall and cell membrane. This withdrawal of protoplasm from cell wall is known as plasmolysis.
The point where protoplasm just starts separating from cell wall is known as “Incipient plasmolysis” when it is completely separated, full plasmolysis occurs.
In plasmolysis cell ßp = 0 therefore ßw = ßs

DEPLASMOLYSIS
When a cell is placed is a hypotonic solution or pure water, there will be an inflow of water by endosmosis. Protoplasm starts expanding and presses cell wall due to which pressure potential develops and water potential becomes less negative. This swelling of cell is known as deplasmolysis.

WATER AND MINERALS UPTAKE BY ROOTS
1. Absorption of water and mineral salts takes place through root system.
2. Roots are provided with enormous number of tiny root hairs.
3. These root hairs are more in number in tap root system.
4. Roots hairs are out growths of epidermal cells.
5. Roots hairs increase the surface area for absorption.
6. Most of the absorption takes place at root tips.
7. From hairs and epidermal cells water flows thru cortex, endodermis, pericycle and them enters xylem.
There are 3 pathways for water to enter xylem.

A- CELLULAR PATHWAY
In this route water flows through cell to cell. Water enters the root hairs or epidermal cells down a concentration gradient: it flows through cell wall and cell membrane and enters the adjacent cell from where water may again flow towards the deeper cells by osmosis.

B- SYMPLAST PATHWAY
Cytoplasm of the cortical cells are interconnected by small pores in the cell wall known as plasmodesmata.
These pores provide another way of transporting water and solutes across the plasma membrane at root hairs.

C- APOPLAST PATHWAY
The cell walls of cortical and epidermal cells are hydrophilic and form a continuous matrix. Soil solution flows freely through these hydrophilic walls. The movement of soil soln.through extra cellular pathway provided by continuous matrix of cell walls is known as “Apoplast pathway”.
Simplast and apoplast usually both occur concurrently.
Endodermis forms a waxy barriers against the flow of water and salts known as “casparion strip”. So, water cannot enter endodermis via apoplast pathway. Symplast is the only way to cross the barrier. Endodermal cells actively transport salts to pericycle resulting in high osmotic potential which causes inflow of water by osmosis salts. Form pericycle water flows in to xylem via both symplast and apoplast pathways.

TRANSPIRATION
The loss of water in the form of vapours from aerial parts of the plant is called transpiration.

TYPES OF TRANSPIRATION
Following are the three types of transpiration.

A- STOMATAL TRANSPIRATION
It is a type of transpiration in which the water vapours escape through the stomata. 90% of the total transpiration occur thru this method. In isobilateral leaves the stomata are present in both upper and lower epidermis e.g. lily and maize leaves. In dorsiventral leaves, the stomata are only confined to lower epidermis e.g. Brassica and sunflower.

B- CUTICULAR TRANSPIRATION
The loss of water in the form of vapours through the cuticle of leaves is called Cuticular Transpiration. About 5-7% of total transpiration takes place thru this route cuticle is a waxy layer which covers the leaves and tis is not completely impermeable to water.

C- LENTICULAR TRANSPIRATION
It is the loss of water vapours through lenticles present in the stems of dicot plants. Lecticles are aerating pores present in the bark formed as a result of secondary growth. It accounts for only 1-2% of total transpiration.

MECHANISM OF STOMATAL RESPIRATION
STRUCTURE OF STOMATA
Stomata are microscopic pores present in the epidermis of leaves and herbaceous stems. Number of stomata are variable in different leaves and depend upon the availability of water and climate of the region. Each stomata is surrounded by 2 specialized epidermal cells, as guard cells, they are bean shaped or kidney shaped and unlike other epidermal cells, they contain chlorophyll, hence perform photo-synthesis. The inner wall of guard cell is thick while the outer wall is thin and elastic. This structural difference is important for opening and closing of stomata.

STAGES OF TRANSPIRATION
There are two processes involved in stomata transpiration.

+ EVAPORATION
In the first step, water evaporates from the wet surfaces of turgid mesophyll cells and collected in the intercellular air spaces.

+ DIFFUSION
In this stage water vapours diffuse out from intercellular spaces where they are in higher concentration to the outer atmosphere where they are in lower concentration through the stomata.

MECHANISM OF OPENING AND CLOSING OF STOMATA
The opening and closing of stomata depends upon the turgidity of guard cells, which is due to increase or decrease in the osmotic potential of the guard cells. When water enters the guard cells by osmosis, they swell up. Since their outer walls are thin and elastic, they stretch and bulge out. The inner thick walls cannot stretch and so arch in and become crescent shaped thus the gap between the two guard cells widens, opening the stomata when the guard cell lose water, they become flaccid and the inner wall of two guard cells meet each other, closing the stomata.
Generally the stomata remain open during day time and close at night. Thus light appears as the primary factor which control the opening and closing of stomata.

FACTORS REGULATING OPENING AND CLOSING OF STOMATA
There are two main factors which greatly influence the opening and closing of stomata these are

1- LIGHT
In the presence of light, chlorophyll containing guard cells synthesize sugars which is turn increase the osmotic potential of guard cells. This increase Qs results in endosmosis and ultimately to turgidity. While in darkness these guard cells consume carbohydrates (sugars) by respiration for energy production or transported to other neighbouring cells for respiration and different purposes. This decreases the osmotic potential of guard cells leading to flaccidity because of exomosis of water.

2- CONCENTRATION OF K+ IONS
Turgidity of guard cells of many plants is regulated by K+ ion concentration. During daytime, guard cells actively transport K+ions into them from neighbouring cells. Accumulation of K+ ions lowers the water potential of guard cells. This causes on inflow of water by endosmosis from epidermal cells. During night when they lose K+ ion, water potential increases. Water flows out of the guard cells by exosmosis causing them to become flaccid which result in closure of pore.

FACTORS AFFECTING TRANSPIRATION
Rate of transpiration is very important for a plant because transpiration stream is necessary to distribute dissolved mineral salts through out the plants. Water is transported to photosynthesizing cells of leaves. Transpiration is also very important as it cools the plant. This is especially important in higher temperatures. If the rate of transpiration is very high, there would be much loss of water from the plant. So at high temperatures the stomata almost close and reduction in the rate of transpiration is effected. This stops witting of the leaves and of herbaceous stems of plants.

Following are some important factors which affect the rate of transpiration.

1. LIGHT
Light affects the transpiration in two ways:
a. Light regulates the opening and closing of stomata. During sunshine the stomata are open, losing water vapours thus rate of transpiration is high and during night, the stomata are closed, so the rate of transpiration is low.
b. Greater intensity of light, increases the temperature and warms the leaf, so leaves lose heat by evaporating water molecules to cool themselves.

2. TEMPERATURE
Plants transpire more rapidly at higher temperature than at low. Rise in temperature has two effects:
i. It increases kinetic energy of water molecules, which results in rapid evaporation of water and decreases the rate of transpiration.
ii. High temperature reduces the humidity of surrounding air. Due to this, evaporation from surfaces of mesophyll cells increase and hence rate of transpiration.

3. WIND
The air in motion is called wind. The area around the stomata is saturated with water vapours due to transpiration. During high velocity wind the area around leaves is quickly replaced by fresh drier air which increases diffusion of water molecules from air spaces to outside atmosphere and increases the rate of transpiration.
When air is still, the rate of diffusion of water molecules is reduced and the rate of transpiration is also reduced.

4. HUMIDITY
When air is dry, the rate of diffusion of water molecules, from the surfaces of mesophyll cells, air spaces and through stomata, to outside the leaf increases. So more water is lost, increasing the rate of transpiration.
In humid air, the diffusion of water molecules is reduced. This decreases the rate of transpiration.

5. SOIL WATER
A plant can’t continue to transpire rapidly if its moisture loss is not made up by absorption of fresh supplies of water from the soil. When absorption of water by roots fails to keep up with rate of transpiration, loss of turgor occurs and wilting of leaf takes place.

DISADVANTAGES OF TRANSPIRATION
1. Transpiration is said to be necessary evil because it is an inevitable, but potentially harmful, consequence of the existence of wet cell surfaces from which evaporation occurs.
2. High rate of transpiration causes water deficiency and thus the excessive transpiration leads to wilting and death of plants.
3. There is good evidence that even mild water deficiency results in reduced growth rate of plants.
4. Excessive transpiration effects the protein synthesis, sugar synthesis and other metabolic activities of plants.

ADVANTAGES OF TRANSPIRATION
1. Water is conducted in most parts of plants due to transpiration pull or ascent of sap.
2. It causes absorption of water and minerals from the soil.
3. Minerals dissolved in water are conducted throughout the plant body by transpiration stream.
4. Evaporation of water from the exposed surface of cells of leaves has cooling effect on plant.
5. Excess water is removed.
6. Wet surface of leaves allow gaseous exchange.

GUTTATION
It is the loss of water in the form of droplets from the ends of large leaf-veins. It take place through special openings called hydathodes.

DIFFERENCES BETWEEN TRANSPIRATION AND GUTTATION
TRANSPIRATION
• Water escapes in the form of wapours.
• Escape water is pure and does not contain solutes.
• It takes place through stomata, and cuticle.
• It is regulated by stomata.
• Normally takes place in light

GUTTATION
• Water escapes as liquid.
• Escaped water contain solutes.
• It takes place through hydathodes and end of veins.
• It is not a regulated process.
• Takes place at night.

TRANSLOCATION OF ORGANIC SOLUTES
Transport of organic products of photosynthesis, like sugars from mature leaves to the growing and storage organs in plants is called translocation. This movement of photo assimilates and other organic materials takes place via the phloem and is therefore called “Phloem Translocation.”
The phloem is generally found on the outer side of xylem and constitutes the bark. The cells of phloem that take part in phloem translocation are called sieve elements. Phloem tissue also contains companion cells, parenchyma cells, fibres like sclereids latex containing cells. But only sieve tube cells are directly involved in tansport of organic solutes.

SOURCE TO SINK MOVEMENT
The translocation of photosynthesis always takes place from source to sink tissues, therefore, the phloem transport is also referred as “source to sink movement.”

SOURCE
The part o plant which forms the sugars or photoynthates is known as source. For example Mature Leaves.

SINK
Sinks are the areas of active metabolism or storage of food e.g: Roots, Tubers developing fruits, immature leaves, growing tips of roots and shoots. Some source and sinks are interconvertible during the process of development of plants. For example: developing and mature leaves, developing and germinating seeds, root of sugar beets etc.

MUNCH HYPOTHESIS (MECHANISM OF PHLOEM TRANSLOCATION)
Phloem translocation is mainly explained by a theory called the “Pressure flow hypothesis” proposed by Ernest munch in 1930 which explains the steps involved in the movement of photosynthates from mesophyll chloroplasts to the sieve elements of phloem of mature leaves.

STEPS
The following steps explain flow theory:
1. The glucose formed during photosynthesis in mesophyll cells, is used in respiration or converted into non-reducing sugar i.e. sucrose.
2. The sucrose is actively transported to bundle sheath cells and then to companion cell of the nearest smallest vein in the leaf. This is called “short distance transport” because solutes cover only a distance of two or three cells.
3. Sucrose diffuse into sieve tube cell or sieve elements by symplast pathway or apoplast pathway. This is called phloem loading, this raises the conc. of sugars in sieve elements, which causes osmosis of water from nearby xylem in the leaf. It causes an increase in the hydrostatic pressure or tugor pressure.
4. The increase hydrostatic pressure moves the sucrose and other substances in the sieve tube cells, and moves to sinks. The photo-assimilates (sugars etc) can be moved a long distance i.e. of several meters, therefore this is known as “Long distance transport.”
5. In the sink tissues, present at the other end of pathway, sugars are delivered by phloem by an active process called “Phloem Unloading.” It produces a low osmotic pressure in sieve elements of sink, as a result of this water potential begins to rise in the phloem and causes an exosmosis of water molecules from the sieve tubes. This causes a decrease in turgor pressure of the sieve tubes (phloem).
6. The presence of sieve plates in the sieve elements greatly increases the resistance along the pathway and results in the generation and maintenance of a substantial pressure gradient in the sieve elements between source and sink. The sieve elements contents are physically pushed along the traslocation pathway by bulk flow, much like water flowing through a garden house.

SIGNIFICANCE OF TRANSLOCATION
1. Food can be formed or stored as in sugar beet’s root or stem of sugar cane.
2. Sucrose is transported to sink where it is converted to glucose and used as energy.
3. Productivity of crop can be increased by accumulation of photo-synthates in edible sink tissues like cereal grains, pulses, ground nuts etc.
4. Fruit is forme by this process e.g. Apples, Mango etc.

ASCENT OF SAP
The upward movement of water and dissolved mineral salts from the roots to the leaves agains the downward pull of gravity is known as “Ascent of Sap.”

PATH OF MOVEMENT
The distance traveled by water is small and easy in plans like herbs and shrubs and longest in tall trees like pinus, red wood, eucalyptus etc. For transport different tissues of xylem is used for conduction of water in different plants. These are open ended cells called “Vessels” and porous cells called “tracheids” (Fig. From book).

A. VESSELS
1. These are thick walled tube like structures which extend through several feet of xylem tissue.
2. They range in diameter from 20µm to 70µm.
3. Their walls are lignified and perforated by pits. At the pit, cell wall is only made up of cellulose. Pits of adjacent cells match up with each other, so that their cavities are interconnected.
4. Xylem vessels arise from cylindrical cells, which placed end to end. They die at maturity forming a continuous duct, providing a channel for long-distance transport of water.
5. Rate of flow of water is 10 times faster than tracheids.
OCCURANCE
VESSELS are mostly found in Angiospermic plants.

B. TRACHEIDS
1. These are individual cells about 30µm in diameter. They are several mm long and tapered.
2. Like vessels, they are also dead, made up of thick lignified walls.
3. Their walls are perforated by small pits, which are of two types, simple and bordered.
4. The Tracheids are connected by pits and forming a long channel for conduction of water.
OCCURANCE
In Ferns and Conifers.

MECHANISM OF ASCENT OF SAP
Water and dissolved mineral salts present in xylem, flow in upward direction at the rate of 15m/hour. Xylem sap ascends because of two reasons:
1. Push from below – Root Pressure Theory
2. Pull from above – Dixon’s Theory

1. ROOT PRESSURE THEORY
According to Stephen Hales:
“The force which is responsible for the upward movement of water molecules in xylem is by the pushing effect from below (i.e. roots) and is known as “Root Pressure.” Root Pressure is created by active secretion of sals and other solutes from the other cells into xylem sap.
This lowers the water potential of xylem sap. Water enters by osmosis, thus increasing the level of sap. Water also take apoplast or symplast pathway to enter the xylem cells, this increased level causes a pressure effect in xylem and pushes the water upwards.

OBJECTIONS/FAILURE OF THEORY
1. This force is unable to push water in tall plants.
2. It is seasonal.
3. Completely absent from Cycads and Conifers, so how they transfer water.
4. When a cut shoot is placed in water, the water rises in shoots although roots are absent.
5. It is also present in plant which donot have well developed root system.

2. TRANSPIRATION PULL (DIXON’S THEORY) OR ADHESION-COHESION-TENSION THEORY
Dixon and Jolly proposed this theory for ascent of sap. It provides a reasonable explanation of flow of water and minerals from the roots to leaves of plants. It depends on:

ADHESION
Adhesion is the sticking together of molecules of different kinds. Water molecules adhere to the cell walls of xylem cells, so that the column of water in xylem tissue doesn’t break. The cellulose of cell wall has great affinity for water, which helps in the process.

COHESION
Cohesion is the attraction among molecules of same kind, which holds water molecules together, forming a solid chain-like column within the xylem tubes. Extensive hydrogen bonding in water gives rise to property of cohesion. The molecules of water in xylem tube form a continuous column.

TRANSPIRATION PULL
The loss of water from the aerial parts of the plant especially through stomata of leaves is called transpiration.
During daytime the leaf after absorbing sunlight, raising its temperature starts transpiration. When a leaf transpires, the water potential of its mesophyll cells drop. This drop causes water to move by osmosis from the xylem cells of leaf into dehydrating mesophyll cells.
The water molecules leaving the xylem are attached to other water molecules of tube by H-bonding.
Therefore, when one water molecules moves up the xylem, the process continues all the way to the root, where water is pulled from the xylem cells, i.e. tracheids or vessels.
Due to this pulling force or transpiration pull, water in xylem is placed under tension which is transmitted to root through vessels. Tension is due to H-bonding and strong cohesive forces between water molecules, and is strong enough to pull water upto 200 metres or even more.

ASCENT OF SAP IS SOLAR POWERED
To transport water over a long distance, plants do not use their metabolic energy or ATPs. It is done only by forces like adhesion, cohesion, evaporation and presence of sunlight. Thus ascent of sap is “Solar Powered.”

SIGNIFICANCE OF ASCENT OF SAP
• Water can be transported to the different parts of the plant.
• Transpiration is regulated.
• Food is formed in presence of water.
• Photosynthesis requires water.
• Salts and minerals are also absorbed along water by roots.

1styear BIOLOGY Notes Chapter-13

Chapter-13
GASEOUS EXCHANGE

RESPIRATORY ORGANS OF COCKROACH TRACHEAL SYSTEM
Cockroach has evolved a special type of invaginated respiratory system called Tracheal system, especially adopted for terrestrial mode of life and high metabolic rate of insects.



STRUCTURAL CONSTITUENTS OF TRACHEAL SYSTEM
1. TRACHEA
2. SPIRACLES
3. TRACHEOLES

1.TRACHEA
Tracheal system consists of number of internal tube called Trachea which are the connection between the spiracles and tracheal fluid.

2. SPIRACLES
Laterally, trachea open outside the body through minute, slit like pores called as spiracles.
• There are 2 pairs of spiracles on lateral side of cockroach.
• 2 lie in thoracic segments and 8 in first abdominal segments.

3. TRACHEOLES
On the other side, trachea ramify throughout the body into fine branches or tracheols.
• Tracheoles, finally end as blind, fluid filled fine branches which are attached with cells of tissue.
• Both the trachea and tracheoles are lined internally by thin layer of cuticle.

MECHANISM OF RESPIRATION “INFLOW OF OXYGEN”
The cockroach takes in air directly from the atmosphere into the trachea through spiracles. This air diffuses directly into fluid filled tracheoles through which diffuses into the cells of tissues. Hence the blood vascular system of cockroach is devoid of haemoglobin.

OUTFLOW OF CARBONDIOXIDE
Removal of CO2 from cells of body is largely depended upon plasma of blood, which takes up CO2 for its ultimate removal through body surface via the cuticle.

RESPIRATORY SYSTEM OF FISH
MAIN RESPIRATORY ORGAN
In fish, main respiratory organs are “Gills”. They are out growth of pharynx and lie internally with in the body so that they are protected from mechanical injuries.

INTERNAL STRUCTURE OF GILLS


Each gill is highly vascularized structure. It is composed of
1. Filaments
2. Gill bar or Gill arch
3. Lamella

1. FILAMENTS
Each gill is composed of two rows of hundreds of filaments, which are arranged in V-shape.

2. GILL BAR OR GILL ARCH
Filaments are supported by a cartilage or a long curved bone the gill bar or gill arch.

3. LAMELLA
Lamella is a plate like structure which is formed by infolding of filaments. Lamella greatly increase the surface area of the gill. Each lamella is provided by a dense network of capillaries.

OPERCULUM (IN BONY FISHES)
Gills are covered on each side by gill cover called “operculum”

MECHANISM OF VENTILATION
In bony fishes, ventilation is brought about by combined effect of mouth and operculum.
• Water is drawn into the mouth. It passes over the gills through pharynx and ultimately exists at the back of operculum through open operculur valve.
• Water is moved over the gills in a continuous unidirectional flow by maintaining a lower pressure in operculur cavity than in buccopharynx cavity.

COUNTER CURRENT FLOW OF WATER AND BLOOD
• Gaseous exchange is facilitated in gills due to counter current flow of H2O and blood.
• In the capillaries of each lamella, blood flows in direction opposite to the movement of water across the gill. Thus the most highly oxygenated blood is brought to water that is just entering the gills and has even high O2 content than the blood. As the H2O flows over the gills, gradually loosing its oxygen to the blood, it encounter the blood that is also increasingly low in oxygen. In this way a gradient is establishment which encourages the oxygen to move from water to blood

IMPORTANCE
Counter current flow is very effective as it enables the fish to extract upto 80–90% of the oxygen from water that flows over the gills.

RESPIRATORY SYSTEM OF MAN
MAIN FUNCTION OF RESPIRATION
The main function of respiratory system is inflow of O2 from the atmosphere to the body and removal of CO2 from body to the atmosphere.

COMPONENTS OF RESPIRATORY SYSTEM
(1) PAIRED LUNGS
The respiratory (gas exchange) organs.

(2) AIR PASSAGE WAYS
Which conduct the air

(3) THORACIC CAVITY
Which lodges the lungs

(4) INTERCOSTAL MUSCLES AND DIAPHRAGM
Which decreases and increase the diameters of thoracic cavity


(5) RESPIRATORY CONTROL CENTRES
Areas in brain which control the respiration.

DETAILS OF COMPONENTS
+ THORACIC CAVITY
Paired lungs with in the pleural sacs are situated in the thoracic cavity. Separating the thoracic cavity from the abdominal cavity is a dome-shaped musculo-tendinuous partition called as Diaphragm.

BOUNDARIES OF CAVITY
Thoracic cavity is supported by bony cage (thoracic cage) which is made up of
• Sternum -> in front
• Vertebral column -> at the back
• 12 pairs of ribs -> on each side
• Ribs are supported by Intercostal muscles

FUNCTION
Increase in thoracic cavity diameter is responsible for inspiration. While decrease in diameter is responsible for expiration.

AIR PASSAGE WAYS
Air is drawn into the lungs by inter-connected system of branching ducts called as “Respiratory tract” or “Respiratory passage ways”
Air passage ways consists of

AIR CONDUCTING ZONE(which only conducts the air)
1. Nostrils
2. Nasal Cavity
3. Pharynx (nasopharynx and oropharynx)
4. Larynx
5. Trachea
6. Bronchi
7. Bronchioles (also called terminal Branchioles)
RESPIRATORY ZONE(Where gaseous exchange takes place)
8. Respiratory Bronchioles
9. Alveolar duct
10. Alveolar sacs or alveoli

GENERAL FUNCTIONS OF CONDUCTING AIR PASSAGES
1. Conduction of air from atmosphere to the lungs
2. Humidification of inhaled dry air.
3. Warming / cooling of air to body temp.
4. The injurious particles are entrapped by mucous and removed by ciliary movements.
5. Lymphoid tissues of pharynx provide immunological functions
6. Cartilages prevent the passages from collapse but are not present in Bronchioles which remains expanded by same pressure that expand the alveoli.

CONDUCTING ZONE
1. NASAL CAVITY
Atmospheric air enters the respiratory tract through a pair of openings called external nares (Nostrils), which lead separately into nasal cavity. Nasal cavity opens into naso pharynx through posterior nares (choanae).
• Nasal cavity is lined internally by Pseudostratified columnar ciliated epithelium containing mucous secreting cells.
• Hairs, sweat and sebaceous glands are also present.

SPECIALIZED FUNCTIONS
• Warming of air
• Humidification or moistening of air
• Filteration of air with the help of hairs
• All these together called as Air conditioning function of upper respiratory passages
• Olfaction ( sense of smell)

2. PHARYNX
Air enters from Nasal cavity into pharynx through internal nostrils. The openings of nostrils are guarded by soft palate. It is internally lined by Pseudostratified ciliated epithelium, mucous glands are also present.
FUNCTION
Pharynx is responsible for conduction of air as well as food

3. LARYNX (VOICE BOX)
Pharynx leads air into larynx through an opening called glottis. Glottis is guarded by flap of tissue called epiglottis. During swallowing, soft palate and epiglottis close the nostrils opening and glottis respectively so that food is prevented to go either into nasal cavity or glottis. Larynx, a small chamber consists of pair of vocal cords
FUNCTION
During speech, vocal cords move medially and their vibration produce sound

4. TRACHEA (WIND PIPE)
Larynx leads the air into a flexible air duct or trachea. It bears C-shaped tracheal cartilages which keep its lumen patent during inspiration. Its internal lining is pseudostratified columnar ciliated epithelium containing mucous secreting goblet cells.
FUNCTION
• Conduction of air
• Due to mucous and upward beating of cilia, any residues of dust and germs are pushed outside the trachea towards the pharynx.

5. BRONCHI
“At its lower end, trachea bifurcates into two smaller branches called Principle Bronchi? which leads the air into lung of its side. They are also supported by C-shaped cartilage rings upto the point where they enter the lungs”.
• In all areas of trachea and bronchi, not occupied by cartilage plates, the walls are composed mainly of smooth muscles.

6. BRONCHIOLES
On entering the lungs, each bronchus divide repeatidly. As the bronchi become smaller, U-shaped bars of cartilage are replaced by irregular plates of cartilages. The smallest bronchi divide and give rise to Bronchioles (less than 1.5 mm in diameter).

7. TERMINAL BRONCHIOLES
Bronchioles divide and give rise to terminal bronchioles (less than 1 mm in diameter). Walls possess no cartilages and are almost entirely the smooth muscles. These are the smalled airways without alveoli.

RESPIRATORY ZONE
In this zone of respiratory tract, gaseous exchange between capillary blood and air takes place.

1. RESPIRATORY BRONCHIOLES
Terminal bronchioles show delicate outpouchings from their walls, which explains the name Respiratory Bronchioles (less than 0.5 mm in diameter). They bear the pulmonary alveoli.

2. ALVEOLAR DUCTS AND SACS
Each respiratory bronchioles terminates at a tiny hollow sac like alveolar duct that lead into tabular passages with numerous thin walled out pouchings called Alveolar sacs.

3. PULMONARY ALVEOLI
The alveolar sacs consists of several alveoli openings into a single chamber. Alveoli are the site of exchange of respiratory gases so they are considered as Respiratory surfaces of lungs. Each alveolus is surrounded by a network of blood capillaries.

INTERNAL STRUCTURE OF ALVEOLI
The alveolar lining cells consists of
1. Type I cells
2. Type II cells
They are also called pneumocytes.
“Bifurcation of trachea is called Carina”.

TYPE I PNEUMOCYTES
Squamous shaped cells which form the epithelial lining of alveoli

TYPE II PNEUMOCYTES
Irregular and cuboidal shaped cells which secretes a substance called Surfactant

SURFACTANT
The internal area of an alveoli is provided with a thin layer of fluid called as Surfactant secreted by type II cells.
FUNCTION OF SURFACTANT
1. It reduces the internal surface tension of alveoli which prevent it collapsing during expiration.
2. It increases the compliance.
3. It stabilize the alveoli.
4. It also helps to keep the alveoli dry.

LUNGS
Lungs are paired, soft, spongy, elastic and highly vascularized structures, which occupy most of thoracic cavity. In child they are pink, but with age they become dark and mottled due to inhalation of dust.

RIGHT LUNG
Partitioned into 3 lobes by two fissures.

LEFT LUNG
Divided into 2 lobes by one fissures.

PLEURAL MEMBRANES
Each lung is enclosed by two thin membranes called as Visceral and parietal pleural membranes.

PLEURAL CAVITY
In between the membranes there is a narrow cavity, the pleural cavity filled with pleural fluid which acts as lubricant.

FUNCTION OF CAVITY
1. Cardinal function is to exchange gases.
2. Phagocytosis of air borne particles
3. Temperature regulation
4. Removal of water
5. Maintainence of acid-base balance (by elemination of CO2)
6. Acts as Reservoir of blood.

BREATHING
DEFINITION
“Breathing is the process of taking in (inspiration or inhalation) and giving out of air (expiration or exhalation) from the atmosphere up to the respiratory surface and vice versa”

TYPES OF BREATHING
There are two types of Breathing
• Negative pressure Breathing
• Positive pressure Breathing

NEGATIVE PRESSURE BREATHING
Normal breathing in man is termed as negative pressure breathing in which air is drawn into the lungs due to negative pressure (decrease in pressure in thoracic cavity in relation to atmospheric pressure).

POSITIVE PRESSURE BREATHING
“In this kind of breathing, lungs are actively inflated during inspiration under positive pressure from cycling valve”.
EXAMPLES
Frog uses positive pressure breathing.

PHASES OF BREATHING
1. INSPIRATION OR INHALATION
2. EXPIRATION OR EXHALATION

(1) INSPIRATION
DEFINITION
“Inspiration is an energy consuming process in which air is drawn into the lungs due to negative pressure in thoracic cavity”

MECHANISM
During inspiration volume of thoracic cavity increases which creates a pressure (intra thoracic) that ****s the air into the lungs.

INCREASE IN VOLUME OF THORACIC CAVITY
Volume of thoracic cavity increases due to
1. Inc. in Anterio-posterior diameter
2. Inc. in Vertical diamter.

INCREASE IN ANTERIO-POSTERIOR DIAMETER During contraction of external intercostals muscle, the ribs as well as the sternum move upward and outward, which causes the increase in anterior-posterior diameter of thoracic cavity.

INCREASE IN VERTICAL DIAMETER
Vertical diameter of thoracic cavity inc. due to Contraction (descent) of Diaphragm which makes it flat.
• As a consequence thoracic cavity enlarges and the pressure is developed inside the thoracic cavity and ultimately in the lungs. So the air through the respiratory tract rushes into the lungs upto the alveoli where gaseous exchange occurs.

(2)EXPIRATION
DEFINITION
“It is reserve of inspiration. The passive process in which air is given out of lung due to increased pressure in thoracic cavity is called “Expiration”

MECHANISM
During expiration, elastic recoil of pulmonary alveoli and of the thoracic wall expels the air from the lungs.


DECREASE IN VOLUME OF THORACIC CAVITY
Volume of thoracic cavity ? due to
1. DECREASE IN ANTERIO-POSTERIOR DIAMETER
2. DECREASE IN VERTICAL DIAMETER

(1) DECREASE IN ANTERIO-POSTERIOR DIAMETER
It is caused by relaxation of external intercostals muscles and contraction of internal intercostals muscles which moves the ribs and sternum inward and downward.

(2) DECREASE IN VERTICAL DIAMETER
It is caused by relaxation of diapharagm which makes it dome shaped thus reducing the volume of thoracic cavity.
• As a consequence, the lungs are compressed so the air along with water vapours is exhaled outside through respiratory passage.

CONTROL OF RATE OF BREATHING
Rate of breathing can be controlled by two modes.
• VOLUNTARY CONTROL
• INVOLUTARY CONTROL

VOLUNTARY CONTROL
Breathing is also under voluntary control by CEREBRAL CORTEX
EXAMPLES
We can hold our breath for short time or can breath faster and deeper at our will.

INVOLUNTARY CONTROL
Mostly, rate of breathing is controlled automatically. This is termed as Involuntary control which is maintained by coordination of respiratory and cardio-vascular system.

TWO MODES OF INVOLUNTARY CONTROL
A. NERVOUS CONTROL (through respiratory centers in brain)
B. CHEMICAL CONTROL (through chemoreceptors)

(A) NERVOUS CONTROL
• Control of rate of breathing by nervous control is through the Respiratory centers in Medulla oblongata which are sensory to the changes in Conc. of CO2 and H+ present in the cerebro-spiral fluid (CSF).

RESPIRATORY CENTRES IN MEDULLA
Two center are present

(1) DORSAL GROUP OF NEURONS
Medulla contains a dorsal group (Inspiratory group) of neurons responsible for inspiration
FUNCTION
In response to increase conc. of CO2 and H+ (decreased pH), it sends impulses to the intercostals muscles to increase the breathing rate

(2) VENTRAL GROUP OF NEURONS
Another area in the medulla is ventral (expiratory) group of neurons.
FUNCTION
It inhibits the dorsal group and mainly responsible for expiration

(B) CHEMICAL CONTROL
Chemical control of rate of breathing is through chemoreceptors.

LOCATION OF CHEMORECEPTORS
• AORTIC BODIES
• CAROTID BODIES

AORTIC BODIES
The peripheral chemoreceptors which are located above and below the arch of aorta are called Aortic bodies. It sends impulses to medulla through Vagus nerve.

CAROTID BODIES
Chemoreceptors which are located at the bifurcation of carotid arteries are called Carotid bodies. It sends impulses to medulla through Glossopharyngeal nerve.
FUNCTION
Inc. in concentration of CO2 and H+ in blood are basic stimuli to increase the rate of breathing which are monitered by these chemoreceptors and then send the impulses to medulla oblongata which produce action potential in inspiratory muscles.

DISORDERS OF RESPIRATORY TRACT
(1) LUNG CANCER (BRONCHIAL CARCINOMA)
CAUSES
• Smoking is a major risk factor either acitively or passively.
• Asbestos, nickel, radioactive gases are associated with increased risk of bronchial cascinoma

PHYSIOLOGICAL EFFECTS
+ LOSS OF CILIA
The toxic contents of smoke such as nicotine and SO2 cause the gradual loss of cilia of epithelical cells so that dust and germ are settled inside the lungs.

+ ABNORMAL GROWTH OF MUCOUS GLANDS
Tumor arises by uncontrolled and abnormal growth of bronchial epithelium mucous glands. The growth enlarges and some times obstruct a large bronchus.
• The tumours cells can spread to other structures causing cancer.

SYMPTOMS
• Cough- due to irritation
• Breath lessness – due to obstruction.

(2) TUBERCLOSIS (KOCH’S DISEASE)(INFECTIOUS DISEASE OF LUNG)
CAUSE
Caused by a Bacterium called as “MYCOBECTERIUM TUBERCLOSIS”

PHYSIOLOGICAL EFFECTS
• Tuber Bacili causes
• Invasion of infected region by macrophages
• Fibrosis of lungs thus reducing the total amount of functional lung tissues
These effects cause
• Increased work during breathing
• Reduced vital and breathing capacity
• Difficulty in diffusion of air from alveolar air into blood.

SYMPTOMS
• Coughing (some time blood in sputum)
• Chest pair
• Shortness of breath
• Fever
• Sweating at night
• Weight loss
• Poor apetite

PREVENTION
A live vaccine (BCG) provides protection against tuberclosis.

3. COPD-(CHRONIC OBSTRUCTIVE PULMONARY DISEASE)
They include
A. Emphysema
B. Asthma

(3-A)EMPHYSEMA
CAUSES
It is a chronic infection caused by inhaling Smoke and other toxic substances such as Nitrogen dioxide and Sulphur dioxide

PHYSIOLOGICAL EFFECTS
• Long infection – Irritants deranges the normal protective mechanisms such as loss of cilia, excess mucus secretion causing obstruction of air ways
• Elasticity of lung is lost
• Residual volume increases while vital capacity decreases.
• Difficulty in expiration due to obstruction
• Entrapment of air in alveoli
• All these together cause the marked destruction of as much as 50-80% of alveolar walls.
• Loss of alveolar walls reduces the ability of lung to oxygenate the blood and remove the CO2
• Oxygen supply to body tissues especially brain decreases.

SYMPTOMS
• Victim’s breathing becomes labored day by day.
• Patient becomes depressed, irritable and sluggish.
• Concentration of CO2 increases which may cause death.

(3-B) ASTHAMA
“Respiratory tract disorder in which there are recurrent attacks of breathlessness, characteristically accompanied by wheezing when breathing out.”
CAUSES
It is usually caused by Allergic hypersensitivity to the plant pollens, dust, animal fur or smoke or in older person may be due to common cough.
Heridity is major factor in development of Asthma.

PHYSIOLOGICAL EFFECTS
• Localized edema in walls of small bronchioles.
• Secretion of thick mucus.
• Spastic Contraction of bronchial smooth muscles (so the resistance in air flow increases).
• Residual volume of lung increases due to difficulty in expiration.
• Thoracic cavity becomes permanently enlarged.

SYMPTOMS
• The asthmatic patient usually can inspire quite adequately but has great difficulty in expiring.

LUNG CAPACITIES
1. TOTAL AVERAGE LUNG CAPACITY
DEFINITION
“It is the maximum volume in which the lung can be expanded with greatest possible inspiratory efforts.”
Or
“Total lung capacity is the combination of residual volume and vital capacity.

VALUE
Total lung capacity = 5000 cm3 or 5 lit of air.

2. TIDAL VOLUME
“The amount of air which a person takes in and gives out during normal breathing is called Tidal Volume.”

VALUE
450cm3 to 500 cm3 (1/2 litre)

3. INSPIRATORY RESERVE VOLUME
DEFINITION
'“Amount of air inspired with a maximum inspiratory effort in excess of tidal volume.”

VALUE
200 cm3 or 2 lit. (Average value)

4. EXPIRATORY RESERVE VOLUME
DEFINITION
“Amount of air expelled by an active expiratory effort after passive expirations.”

VALUE
1000 cm3 or 1 litre.

5. VITAL CAPACITY
DEFINITION
“After an extra deep breath, the maximum volume of air inspired and expired is called Vital capacity.”
Or
“It is the combination of inspiratory reserve volume, expiratory reserve volume and tidal volume.”

VALUE
Averages about 4 litre.

6. RESIDUAL VOLUME
DEFINITION
“Amount of air which remains in lung after maximum expiratory effort is called Residual volume.”

VALUE
Approximately 1 litre or 1000 cm3.

IMPORTANCE OF LUNG CAPACITY
• Residual volume prevent the lung from collapsing completely.
• Responsible for gaseous exchange in between breathing.
• It is not stagnant since inspired air mixes with it each time.
• Aging or Emphysema, etc can increase the residual volume at the expense of vital capacity.

HAEMOGLOBIN
INTRODUCTION
“Haemoglobin is an iron containing respiratory pigment present in the red blood cells of vertebrates and responsible for their red colour.”

STRUCTURE
Haemoglobin consists of
1. Heme
2. Protein (globin like chains)

1. HEME
One Haemoglobin molecule consists of 4 molecules of Heme. Each Heme molecule contains an iron (Fe++) binding pocket. Thus one molecule of Haemoglobin can combine with 4 iron atoms.

2. GLOBIN
Each Hb molecule contains four globin like chains (Two a chains and Two ß chains).

ROLE OF HB DURING RESPIRATION
Two major functions are performed by Hb.
1. Transport of O2 from lung to tissues.
2. Transport of CO2 from tissues to lungs.

1. “TRANSPORT OF O2 FROM LUNGS TO TISSUES”
“Nearly 97% of O2 is transported from the lungs to the tissues in combination with Hb.”

ATTACHMENT OF O2 WITH HB
It is the iron of Hb molecule which reversibly binds with oxygen. One Hb molecule can bind 4 molecules of O2. Thus due to Hb, blood could carry 70 times more oxygen than plasma.

MECHANISM OF TRANSPORT
• Due to high O2 concentration in alveolar air, the O2 moves from air to the venous blood where O2 concentration is low.
• It combines loosely with Hb to form Oxyhemo Globin.
• In this form, O2 is carried to the tissues where due to low oxygen concentration in tissues, oxy Hb dissociates releasing oxygen, which enters in tissues.
Whole process can be represented by following equation.

2. “TRANSPORT OF CO2 FROM TISSUES TO LUNGS”
“Haemoglobin is also involved in 35% of transport of CO2 from tissues to alveolar blood capillaries in alveoli.”

ATTACHMENT OF CO2 WITH HB
CO2 binds reversibly with NH2 group of Hb to form loose compound called “Carboamino Haemoglobin.”

MECHANISM OF TRANSPORT
• Carbon dioxide due to its higher concentration in tissue diffuses out into the blood where it combines with Hb to form Carboamino Hb.
• In the alveoli it breaks and CO2 diffuses out into the Alveoli from where it is expired.

MYOGLOBIN
INTRODUCTION
“Myoglobin is a heme protein, smaller than Hb, found in muscles and giving red colour to them.

STRUCTURE
Myoglobin consists of one heme molecule and one globin chain. It can combine with one iron (Fe++) atom and can carry one molecule of O2.

FUNCTION OF MYOGLOBIN
• Myoglobin has high affinity for O2 as compared to Haemoglobin so it binds more tightly.
• It stores the O2 within the muscles.
• It supplies the O2 to the muscles when there is severe oxygen deficiency (During exercise)
It can be represented as follows:
Mb + O2 ? MbO2

TRANSPORT OF GASES
Oxygen and carbondioxide are exchanged in, Alveoli by Diffusion.

O2 TRANSPORT
Blood returning into the lungs from all parts of body is depleted from oxygen. This deoxygenated blood is dark maroon in colour to appear bluish through skin. It becomes oxygenated in the lungs.

TWO FORMS OF O2 IN BLOOD
O2 is transported in the blood in two forms:
• Dissolved form (3%)
• Combination with Hb (97%) ® Oxyhaemoglobin

MECHANISM OF O2 TRANSPORT
+ DIFFUSION OF O2 FROM ALVEOLUS INTO PULMONARY BLOOD
The air inhaled into the lungs has high concentration of oxygen while venous blood in pulmonary capillaries has low in concentration. Due to this difference in concentration across the respiratory surface, oxygen diffuses into the blood flowing into capillaries around the Alveoli. Now blood becomes oxygenated which is bright red in colour.

+ DIFFUSION OF O2 FROM CAPILLARIES INTO CELLS
Concentration of O2 in the arterial end of capillaries is much more greater than concentration of O2 in the cells. So O2 diffuses from the blood to the body cells. Since the blood takes in oxygen much more rapidly than water. Thus it can transport enough oxygen to the tissues to meet their demand.

CO2 TRANSPORT
Blood returning from tissues contain excess of CO2 as a respiratory by-product, which is eliminated from the body during expiration in the lungs.”

THREE FORMS OF CO2 IN BLOOD
• Dissolved form (in plasma) – 5%
• In form of HCO3- (in RBC’s) – 60%
• In combination with Hb (Carboamino Hb) – 35%

+ DISSOLVED FORM
Only 5% of CO2 is transported in dissolved form in plasma. Here it combines with H2O of plasma to form H2CO3. But this reaction is very slow as plasma does not contain Carbonic Anhydrase to accelerate this reaction.
Reactions can be represented by following equations.
CO2 + H2O ? H2CO3
H2CO3 ?HCO3- + H+
HCO3- + k+ ? KHCO3

+ IN FORM OF HCO3-
60% of CO2 is transported in the blood in form of HCO3- in RBC’s. Here it combines with water to form H2CO3. But this reaction occurs rapidly in RBC’s due to presence of Carbonic Anhydrase.
Reactions can be represented by following equations
CO2 + H2O ? H2CO3
H2CO3 ? HCO3- + H+
HCO3- + Na+ ? NaHCO3

+ IN COMBINATION WITH HB
As discussed previously in role of Hb.

MECHANISM OF CO2 TRANSPORT
+ DIFFUSION OF CO2 FROM CELLS INTO CAPILLARIES
CO2 is continuously synthesizing in the tissues as a result of metabolism. Thus due to its higher concentration. CO2 diffuses from the tissues into blood, which becomes deoxygenated.

+ DIFFUSION OF CO2 FROM PULMONARY BLOOD INTO ALVEOLUS
Blood returning from tissues contain high concentration of CO2. This blood is brought to lungs, where CO2 diffuses from the blood into alveolus where its concentration is lower.

FACTORS EFFECTING THE TRANSPORT OF GASES
Following are some factors, which influence the transport of respiratory gases across the alveolar wall.
1. Concentration Gradient
2. Presence of competitor such as CO
3. Moisture
4. Surfactant
5. pH