"Some basic concepts of Chemistry" Important questions-answers

 Some basic concepts of Chemistry

Assignment -1

Definition:-

1. Chemistry :-Chemistry is the science of atoms, molecules and their transformations.

It can also be defined as that branch of science that deals with the study of the composition, properties and interaction of matter.

2. Pure substance :-Pure substances have fixed composition.

For example: - Copper, silver, gold, water, glucose are some examples of pure substances.

Also, the constituents of pure substances cannot be separated by simple physical methods.

3. Elements:-An element consists of only one type of particles. These particles may be atoms or molecules.

For example: - Sodium (Na), copper (Cu), silver (Ag), hydrogen (H), oxygen (O) atoms etc. They contain only one type of atoms.

4. Compounds:-When two or more atoms of different elements combine, the molecule of a compound is obtained.

For example: - Water (H20), Ammonia (NH3), Sugar, carbon dioxide (CO2).

Constituents of a compound cannot be separated into simpler substances by physical methods. They can be separated by chemical methods.

5. Mixtures:-A mixture contains two or more substances present in it (in any ratio) which are called its components.

6. Mass:-Mass is defined as the amount of matter present in a substance.It is constant

7. Volume:- Space occupied by matter is called its volume. It has the units of (length) 3. So in SI system, volume has units of m3.

8. Meter:- The meter is the length of the path travelled by light in vaccum during a time interval of 1/299792458 of a second.

9. Homogeneous mixture:-In homogeneous mixture, the components completely mix with each other and its composition is uniform throughout.

For example: - Air, Sugar solution.

10. Heterogeneous mixture:-In heterogeneous mixtures, the composition is not uniform throughout and sometimes the different components can be observed.

For example: - Mixture of dal mot and bhujia, badam etc.

The components of a mixture can be separated by using physical methods such as simple hand picking, filtration, crystallisation, distillation etc.

11. Physical Properties:- They are those properties which can be measured or observed without changing the identity or the composition of the substance.

Some examples of physical properties are colour, odour, melting point, boiling point, density etc.

12. Chemical properties:- They are those properties which can be measured or observed without changing the identity or the composition of the substance.

Some examples of physical properties are colour, odour, melting point, boiling point, density etc.

13. Weight:-Weight is defined as the force exerted by the gravity on an object. It may vary from one place to another due to change in gravity.

14. Density:-Density of a substance is its amount of mass per unit volume.

SI unit of density = kg/m3.

15. Temperature:- Measure of hottest or coldness of an object is called temperature. There are three common scales to measure temperature — °C (degree Celsius), °F (degree Fahrenheit) and K (kelvin). Here K is the unit SI unit.

The temperatures on two scales are related to each other by the following relationship:-

°F = (9/5) (°C) + 32

The Kelvin scale is related to Celsius scale as follows:-

K =°C + 273.15

16. Atomic mass:-Atomic mass is the mass of the atom.

17. Average atomic mass:-  mass of an atom based upon the existence of its isotopes is called average atomic mass.

18. Molecular mass:-Molecular mass is the sum of atomic masses of the elements present in a molecule.

For example:-Molecular mass of methane CH4 = (12.011 u) + 4 (1.008 u) = 16.043 u.

19. Formula mass:- Formula mass is the sum of all the atomic masses of all the constituents of the formula.

20. Molar mass :-The mass of one mole of a substance in grams is called its molar mass.

The molar mass in grams is numerically equal to atomic /molecular/formula mass in u.

21. Limiting reagent:-In a chemical reaction, reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent.

22. Mass % or weight % :-Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture.

Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.

23. Mole fraction :-It is the ratio of number of moles of a particular component to the total number of moles of the solution.

Mole fraction of A

= (No. of moles of A)/(No. of moles of solution)

(nA )/( nA + nB)

Mole fraction of B

=(No. of moles of B)/(No. of moles of solution)

(nB )/( nA + nB)

24. Molarity :-Molarity is defined as the number of moles of the solute in 1 litre of the solution.

It is widely used unit and is denoted by ‘M’.

Molarity (M) = (No. of moles in solute)/(Volume of solution in litres)

25. Molality:-It is defined as the number of moles of solute present in 1 kg of solvent.

It is denoted by m.

Thus Molality(m) =

(No. of moles of solute)/(Mass of solvent in kg)

26. % composition:-The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100

27. Empirical formula:-An empirical formula represents the simplest whole number ratio of various atoms present in a compound.

Write advantages of Chemistry and mention two problems that is still challenge to scientists.

Chemistry plays a central role in our life some advantages are listed below:-

Advantages of studying chemistry

With a better understanding of chemical principles it has now become possible to design and synthesise new materials having specific magnetic, electric and optical properties.

Safer alternatives to environmentally hazardous refrigerants like CFCs (chlorofluorocarbons), responsible for ozone depletion in the stratosphere, have been successfully synthesised.

Problems which can be tackled with the study of chemistry:-

Management of Green House gases like methane, carbon dioxide etc.

Understanding of bio-chemical processes, use of enzymes for large-scale production of chemicals and synthesis of new exotic materials are some of the intellectual challenges for the future generation of chemists.

Understanding of bio-chemical processes, use of enzymes for large-scale production of chemicals.

Write drugs for treatment of

1. Cancer 

2. AIDS

Ans:- 1. Cis-platin and taxol

2. AZT (Azidotgymidine)

Define matter and it's physical states.

Anything that has mass and occupies space is defined as matter.

For example:-soap, notebook, pencil, apple, battery etc.

Physical States of matter

Matter can exist in 3 physical states namely solid, liquid and gas.

Solids

In case of solids, the constituent particles are held very close to each other in an orderly fashion and there is not much freedom of movement.

Solids have definite shape and definite volume.

For example: - Apple it has definite shape as well as definite volume.

Liquids

In liquids, the particles are close to each other but they can move around.

Liquids have definite volume but not definite shape. They take the shape of the container in which they are placed.

For example: Water it takes the shape of the tumbler in which it is poured but has volume.

Gases

In gases, the particles are far apart as compared to those present in solid or liquid states and their movement is easy and fast.

It takes the shape of  container in which they are placed.

For example: - Smoke does not have definite shape or volume.

Define 5 laws of chemical combination.

The combination of elements to form compounds is governed by the following five basic laws:-

Law of Conservation of Mass.

Law of Definite Proportions.

Law of Multiple Proportions.

Gay Lussac’s Law of Gaseous Volumes.

Avogadro Law.

Law of Conservation of Mass

Law of conservation of mass states that the matter cannot be created nor be destroyed.

This law was put forth by Antoine Lavoisier in 1789.

He performed careful experimental studies for combustion reactions for reaching to the above conclusion.

Law of Multiple Proportions

According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

For example: - Hydrogen combines with oxygen to form 2 compounds, water and hydrogen peroxide.

H2(2g) + (1/2)O2 (16g)  -- >H2O(18g)

H2(2g) + O2 (32g) -- > H2O2 (34g)

The masses of oxygen O (16g and 32g) combine with the fixed mass of (2g) hydrogen H. Therefore the simple ratio is 16:32 or 1:2.

This law was given by Dalton in 1803.

Law of Definite Proportions

According to this law, a given compound always contains exactly the same proportion of elements by weight.

This law was given by French chemist, Joseph Proust in 1806.

Gay Lussac’s Law of Gaseous Volumes

Gay Lussac’s law was given by Gay Lussac in 1808.

He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

For example: -

 H (Hydrogen) (100mL) + O (Oxygen) (50mL) -- >Water(100mL).

The volumes of hydrogen (H) and oxygen (O) which combine together (i.e. 100mL and 50mL) bear a simple ratio of 2:1.

Avogadro Law

In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules.

He made distinction between atoms and molecules

Write Dalton's atomic theory.

Matter consists of indivisible atoms.

All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.

Compounds are formed when atoms of different elements combine in a fixed ratio.

Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.

Electricity

 Introduction:-

In an atom, there are 3  sub-atomic particles.

Electrons-- > carry negative charge -- > -1.6*10^-19C

Proton -- > carry positive charge -- > 1.6*10^19 C

Neutron-- > carry no charge.

All matter is made up of atoms

In this lesson we will consider only two types of charges; positive and negative.

Like charges (similar charge) repel each other 

Unlike charges(opposite charges) attract each other.

SI unit of charge (electric charge) is coulomb C

If a net charge Q , flow across any cross section of a conductor in time t, then the current I through cross section is 

I = Q/t

Where I = Electric current,

Q = Electric charge,

t = time

S I unit of charge is coulomb,C

One coulomb of charge is that quantity of charge which flows through a circuit when one ampere of current flows through it in one second.

1C = 1A * 1s

What is electric current (I)?

General :- Flow of charge.

Definition:- Amount of charge flowing through a particular area in unit time.

Direction of current is similar to the direction of flow of proton.

Or

Direction of current is opposite to the direction of flow of electron 

S I unit of current is ampere,A

One ampere is the amount of current flowing through a circuit when one coulomb of charge flows through it in one second.

1A = 1C/1s

Instrument which measures current is called ammeter (Ammeter has low resistance)

Electric circuit:- A continuous and closed path of an electric current is called electric circuit.

In outer circuit direction of flow of electric circuit is from positive terminal of the cell to negative terminal and cell.

Conductor:- Substances through which charges can easily pass. Example:- Silver, copper etc 

Numerical:- Calculate the number of electrons constituting one coulomb of charge.

Solution:- 

1e- = -1.6* 10^-19 C

So 1C = 1/1.6*10^-19e-

           = 0.625 * 10^19  e-

          = 6.25 * 10^18 e-

Draw a schematic diagram of an electric circuit comprising - cell, electric bulb, ammeter and plug key.

Electric Potential:-

Definition 1:- Electric Potential at a point in an electric field is defined as work done in moving a unit positive charge from infinity to that point.

Electric Potential = Electric potential energy (work)/ charge

Definition 2:- Electric Potential energy per unit charge is called electric Potential.

Electric potential difference:- 

For flow of charge in a conducting metallic wire, the gravity has no role to play, electrons move only if there is a difference of electric potential along the conductor.

This difference is called electric Potential difference.

Definition:- The potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit positive charge from one point to another point.

Potential difference = work done (W)/ Quantity of charge moved(Q)

V = W/Q

SI unit of potential difference is volt.

One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to other.

1volt = 1joule/1coulomb

1V = 1J/C

Voltmeter measure the potential difference.

Voltmeter is always connected in parallel across the points between which potential difference is to be measured.

Voltmeter has a high resistance so that it takes a negligible current from the circuit.

Symbols of some commonly used components in circuit diagrams:-

Ohm's law:-

At constant temperature, the current flowing through a conductor is directly proportional to potential difference across its ends.

Mathematical expression:- 

I directly proportional to V

V directly proportional to I

V = IR 

Where V = potential difference,

I = Current,

R = resistance.

Resistance :- It is defined as the ratio of potential difference applied between the ends of a conductor and the current flowing through it 

It is a constant quantity.

SI unit is ohm

1ohm = 1volt/1Ampere.

Define 1 ohm .

The resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is applied across its ends.

Definition:-

Good conductors:- Those substances which have very low electrical resistance are called good conductors. Example:- Silver is the best conductor.

Resistor:- Those substance which have comparatively high electrical are called resistors. Example :- Nichrome (Alloy of nickel, chromium, manganese and iron metals)

Insulators:- Those substances which have infinitely high electrical resistance are called insulators. Example:- Rubber.

Draw a graph which show relationship between potential difference and current.

Variable resistance:- A component used to regulate current without changing the voltage source is called variable resistance.

Rheostat:- A device used to change resistance in the circuit.

Factors on which the resistance of a conductor depends:-

Resistance depends upon:-

Length of conductor

Area of cross section of conductor and 

On nature of material.

How it depends?

Resistance is directly proportional to length of conductor 

Resistance is inversely proportional to area of cross section.

Where rho is resistivity or specific resistance

SI unit of resistivity is ohm-m

Resistance also depends upon nature of material.

Resistivity range:-

Good conductors; eg metals and alloys; resistivity= 10^-8 ohm-m to 10^-6 ohm-m

Insulators; eg rubber and glass, resistivity= 10^12 ohm-m to 10^17 ohm-m

Note:- Resistance and resistivity of material vary with temperature and nature of substance.

Resistivity of alloy is generally higher than that of its constituent metals.

Applications of alloys:-

Heating elements of electrical heating appliances such as electric iron and toaster etc are made up of alloys

Reasons:- 

High resistivity of alloy.

Doesn't undergo oxidation (burn) even at high temperature.

Application of metals based upon resistivity:-

Tungsten is used for filaments of electric bulbs.

Copper and aluminium re generally used for electrical transmission lines.

Combination of resistors in series:-

When two or more resistance are combined end to end consecutively they are said to be connected in series.

Example:-

Resistors connected in series.

In series combination of resistors, current flow in each part of circuit is same. { I1 = I2 = I3 = I }

Potential difference V is equal to sum of potential difference; 

Vtotal = V1 + V2 + V3....

According to Ohm's law:-

Vital = IRtotal  ;.  V1 = IR1   ;.  

  V2 = IR2 ;     V3 = IR3

So put values in above equation , we get:-

IRtotal = IR1 + IR2 + IR3.....

IRtotal= I (R1 + R2 + R3.....)

Rtotal = R1 + R2 + R3.....

Thus resistance of combination is equal to sum of their individual resistances; R1, R2 and R3

Combination of resistance in parallel:-

When two or more resistances are connected between same two points, they are sad to be connected in parallel.

Example:-

Resistors connected in parallel.

In parallel combination, total current I is equal to sum of separate currents through each branch of combination.

I = I1 + I2 + I3......

According to Ohm's law:-

I1= V/R1.       , I2 =V/R2.      ,

I3 = V/R3.      ,I = V/R

Put in 1 , we get

I = I1 + I2 + I3..

V/R = V/R1 + V/R2 + V/R3

V/R = V (1/R1 +1/R2 +1/R3)

1/R =1/R1 +1/R2 +1/R3

So Reciprocal of equivalent resistance joined in parallel is equal to sum of reciprocal of individual.

Note :-In parallel combination potential difference is same.

Disadvantages of series circuit for domestic wiring:-

In series circuit, if one electrical appliance stop working due to some defect, then all other appliances also stop working.

All appliances have only one switch due to which they cannot be turned on or off separately.

Appliances do not get the same voltage as that of power supply.

Overall resistance of circuit increases too much due to which current from power supply is low.

Advantages of parallel circuits in Domestic wiring:-

If one electrical appliance stops working due to some defects then all other appliances keep working normally.

Each electrical appliance has its own switch due to which it can be turned on or off independently without affecting other appliances.

Each electrical appliance get same voltage (220V) as that of power supply line.

In the parallel connection of electrical appliances the overall resistance of household circuit is reduced due to which current from power supply is high.

Heating effect of current:-

When an electric current is passes through a high resistance wire, like Nichrome wire, the resistance wire becomes very hot and Produce heat. This is called the heating effect of current.

When an electric charge Q moves against a potential difference V, the amount of work done is given by:- W = Q * V

We know:-

 I = Q/t 

Q = It

From Ohm's law:-

V/I = R

V= IR

Put the values of V and Q in 

W= Q*V we get:-

W = It*IR

W = I2Rt

Assuming all electric work done is converted into heat energy:-

So H = W = I2Rt

H = I2Rt ( joules law of heating)

SI unit of heat Produced , H is joules (J)

According to joules law of heating given by the formula:-

H = I2Rt

Heat Produced in a wire is directly proportional to

Square of current (I2)

Resistance of wire (R)

Time for which current is passed.

Note:- total work done by a current in an electric circuit is called electric energy.

Applications of heating Effect of current:-

Used in working of electrical heating appliances such as electric iron, electric kettle, electric oven etc.

Used in Electric bulbs for producing light 

Used in electric fuse for protecting household wiring and electrical appliance.

Electric power, P and Interrelation between P, V, I, R

The rate at which work is done by an electric current is called electric power. Or

The rate at which electric energy is consumed or dissipated is called electric power.

W = VQ

Now Q = It and V = IR

W= VIt = I2Rt

So Power = work/time

P = W/t

P = VQ/t =VIt/t = VI

P =VI

P = I2R

P= V2/R where I =V/R in eq P=VI

Note :- 

P = I2R is used when I is constant;

P =V2R is used when V is constant

P = VI is used in all cases 

Commercial unit of electric energy:-  kilowatt hour

The electric energy consumed when an electric appliance of power one kilowatt works for one hour is called one kilowatt hour.

1 kWh = 1kW * h

1 kWh = 1000W * 3600s

1 kWh = 3.6 *10^6 joule (W-s=J)

1kWh = 3.6 MJ

Property -- > unit --- > symbol

Current -- >ampere (A) ---- > I

Charge -- > coulomb (C) --- > Q

Time -- > sec -- > t

Temperature -- > °k -- > T

Potential difference -- > volt -- > V

Work done --- > joule -- > W

Resistance --- > ohm -- > R

Resistivity  -- >   ohm-m --> 

Power -- > watt --- > P

Heat --- > joule -- > H

Formula used = >

I = Q/t

Current = charge / time

V= W/Q

Potential 

difference = work done /charge

V = IR (Ohm's law)

Potential

difference = current * Resistance

R = rho *L/A

Resistance = resistivity * length of conductor /area of cross section

For series connection=

R = R1 + R2 + R3.....

V = V1 + V2 + V3 ....

Here current flow is same

For parallel connection =

1/R = 1/R1 + 1/R2 + 1/R3

I = I1 +I 2+ I3......

Here potential difference is same

P = VI

Power = Potential difference * current

H = I^2Rt (joule's law of heating)

Heat = current ^2 *resistance*time

Class 7 back exercise NCERT

Back exercise 

Chap 1

Q.6. Fill in the blanks:

(a) Green plants are called ________________ since they synthesise their own food.

(b) The food synthesised by the plants is stored as ________________ .

(c) In photosynthesis solar energy is captured by the pigment called ________________

(d) During photosynthesis plants take in ________________ and release ________________ .

Ans. (a) autotrophs (b) starch (c) chlorophyll (d) carbon dioxide, oxygen

Q.7. Name the following:

(i) A parasitic plant with yellow, slender and tubular stem.

(ii) A plant that has both autotrophic and heterotrophic mode of nutrition.

(iii) The pores through which leaves exchange gases.

Ans. (i) cuscuta (ii) Insectivorous plant (iii) Stomata

Q.8. Tick the correct answer:

(a) Amarbel is an example of:

(i) Autotroph (ii) Parasite (iii) Saprotroph (iv) Host

(b) The plant which traps and feeds on insects is:

(a) Cuscuta (ii) China rose {iii) Pitcher plant (iu) Rose

Ans. (a) (ii) Parasite (b) (iii) Pitcher plant

Q.9. Match the items given in Column I with those in Column II:

Chlorophyll --- > Leaf

Nitrogen ----- > Rhizobium

Cuscuta ------ > Parasite

Animals ------ > Heterotrophs

Insects -------- > Pitcher plant

Q.10. Mark T’ if the statement is true and ‘F’ if it is false:

(i) Carbon dioxide is released during photosynthesis. (T/F)

(ii) Plants which synthesise their food themselves are called saprotrophs. (T/F)

(iii) The product of photosynthesis is not a protein. (T/F)

(iv) Solar energy is converted into chemical energy during photosynthesis. (T/F)

Ans. (a) F (ii) F (iii) T (iu) T

Q.11.Choose the correct option from the following.

Which part of the plant takes in carbon dioxide from the air for photosynthesis?

(i) Root hair (ii) Stomata (iii) Leaf veins (iv) Sepals

Ans. (ii) Stomata

Q.12. Choose the correct option from the following:

Plants take carbon dioxide from the atmosphere mainly through their:

(i) Roots (ii) Stem (iii) Flowers (iv) Leaves

Ans. (iv) Leaves

Chapter :- 2

Q.1.Fill in the blanks:

(a) The main steps of nutrition in humans are ___________, ___________, ___________ , ___________ and ___________

(b) The largest gland in the human body is ___________.

(c) The stomach releases hydrochloric acid and ___________ juices which act on food.

(d) The inner wall of the small intestine has many finger-like outgrowths called ___________

(e) Amoeba digests its food in the ___________ .

Ans.(a) ingestion, digestion, absorption, assimilation, egestion

(b) liver (c) digestive (d) villi (e) food vacuole.

Q.2. Mark ‘T’ if the statement is true and ‘F’ if it is false:

(a) Digestion of starch starts in the stomach. (T/F)

(b) The tongue helps in mixing food with saliva. (T/F)

(c) The gall bladder temporarily stores bile. (T/F)

(d) The ruminants bring back swallowed grass into their mouth and chew it for sometime. (T/F)

Ans. a) F (b) T (c) T (d) T

Q.3.Tick (S) mark the correct answer in each of the following:

(a) Fat is completely digested in the

(i) stomach (ii) mouth (in) small intestine (iv) large intestine

(b) Water from the undigested food is absorbed mainly in the:

(i) Stomach (ii) Food pipe (iii) Small intestine (iv) Large intestine

Ans.(a) (iii) Small intestine (b) (iv) Large intestine

Q.4.Match the items of column I with those given in column II:

Food components ---- > Products of digestion

Carbohydrates ------ > Sugar

Proteins --------> Amino acids

Fats ---------- > Fatty acids and glycerol

Q.9. Which part of the digestive canal is involved in:

(i) Absorption of food ________ .

(ii) Chewing of food ________ .

(iii) Killing of bacteria ________ .

(iv) Complete digestion of food ________ .

(v) Formation of faeces ________ .

Ans. (i) Small intestine (ii) Mouth (iii) Stomach (iv) Small intestine (v) Large intestine

Q.11. Match the items of Column I with suitable items in Column II.

Salivary gland----- > Saliva secretion

Stomach------ > Acid release

Liver ------- > Bile juice secretion

Rectum ----- > Storage of undigested food

Small intestine ---- > Digestion is completed

Large intestine ------ > Absorption of faeces.

Chap 3:-

Q.2. The silkworm is (a) a caterpillar (b) a larva. Choose the correct option.

(i) (a) (ii) (b) (iii) both (a) and (b) (iv) neither (a) nor (b)

Ans. (iii) both (a) and (b).

Q.3. Which of the following does not yield wool?

(i) Yak (ii) Camel (iii) Goat (iv) Woolly dog

Ans. (iv) Woolly dog

Q.5. Given below is a sequence of steps in the processing of wool. Which are the missing steps? Add them.

Shearing, ________ , sorting,________, __________

Ans. Shearing, scouring, sorting, picking out of burrs, colouring, rolling.

Q.7. Out of the following, which are the two terms related to silk production? Sericulture, floriculture, moriculture, apiculture and silviculture.

Hints: (i) Silk production involves cultivation of mulberry leaves and rearing silkworms. (ii) Scientific name of mulberry is Morus alba.

Ans. (i) Sericulture (ii) Moriculture

Q.8 Match the words of Column 1 with those given in Column 2:-

Scouring -------->  cleaning sheared skin

Mulberry leaves ------> Food of silk worm

Yak -------- > Wool yielding animal.

Cocoon--------- >Yields silk fibres

Chap :- 4

Q.3.Fill in the blanks

The hotness of an object is detetmined by its ____________ .

(b) Temperature of boiling water cannot be measured by a ____________ thermometer.

(c) Temperature is measured in degree ____________ .

(d) No medium is required for transfer of heat by the process of ____________.

(e) A cold steel spoon is dipped in a cup of hot milk. It transfers heat to its other end by the process of ____________

(f) Clothes of ___________ colours absorb heat better than clothes of light colours.

Ans. (a) temperature (b) clinical (c) Celsius (d) radiation (e) conduction (f) dark

4. Match the following :-

(i) Land breeze blows during - > night

(ii) Sea breeze blows during - > day

(iii) Dark coloured clothes are preferred during ------- > winter

(iv) Light coloured clothes are preferred during  ----- > summer

Q.8. One litre of water at 30°C is mixed with one litre of water at 50°C. The temperature of the mixture will be:

(a) 80°C (b) More than 50°C but less than 80°C

(c) 20°C (d) Between 30°C and 50°C

Ans.(d) Between 30°C and 50°C.

Q.9. An iron ball at 40°C is dropped in a mug containing water at 40°C. The heat will:

(a) flow from iron ball to water.

(b) not flow from iron ball to water or from water to iron ball.

(c) flow from water to iron ball.

(d) increase the temperature of both.

Ans. (b) not flow from iron ball to water or from water to iron ball

Q.10. A wooden spoon is dipped in a cup of ice-cream. Its other end:

(a) becomes cold by the process of conduction

(b) becomes cold by the process of convection

(c) becomes cold by the process of radiation

(d) does not become cold

Ans.(d) does not become cold.

Q.11.Stainless steel pans are usually provided with copper bottoms. The reason for this could be that:

(a) copper bottom makes the pan more durable

(b) such pans appear colourful

(c) copper is a better conductor of heat than the stainless steel

(d) copper is easier to clean than the stainless steel

Ans.(c) copper is better conductor of heat than the stainless steel

Chap :-5

Q.2.Ammonia is found in many household products, such as window cleaners. It turns red litmus blue. What is its nature?

Ans.Ammonia has basic nature

Q.6.Mark ‘T’ if the statement is true and ‘F’ if it is false:

(i) Nitric acid turns red litmus blue. (T/F)

(ii) Sodium hydroxide turns blue litmus red. {T/F)

(iii) Sodium hydroxide and hydrochloric acid neutralise each other and form salt and water. (T/F)

(id) Indicator is a substance which shows different colours in acidic and basic solutions. . (T/F)

(v) Tooth decay is caused by the presence of a base. (T/F)

Ans.(1) F (ii) F (iii) T (iv) T (V) F

Q. 11. Consider the following statements:

(a) Both acids and bases change colour of all indicators.

(b) If an indicator gives a colour change with an acid, it does not give a change with a base.

(c) If an indicator changes colour with a base, it does not change colour with an acid.

(d) Change of colour in an acid and a base depends on the type of the indicator. Which of these statements are correct?

(i) All four (ii) (a) and (d) (iii) (b) and (c) (iv) only (d)

Ans. (ii) (a) and (d)

Chap 6

Q.1.Classify the changes involved in the following processes as physical or chemical changes:

(a) Photosynthesis

(b) Dissolving sugar in water

(c) Burning of coal

(d) Melting of wax

(e) Beating aluminium to make aluminium foil

(f) Digestion of food

Ans.(a) Chemical change (b) Physical change

(c) Chemical change (d) Physical change

(e) Physical change (/) Chemical change

Q.2.(a) Cutting a log of wood into pieces is a chemical change. (True/ False)

(b) Formation of manure from leaves is a physical change. (True/ False)

(c)Iron pipes coated with zinc do not get rusted easily. (True/ False)

(d)Iron and rust are the same substances. (True/ False)

(e)Condensation of steam is not a chemical change. (True/ False)

Ans. (a)False

Correct statement: Cutting a log of wood into pieces is an irreversible physical change.

(b)False

Correct statement: Formation of manure from leaves is a chemical change.

(c) True

(d)False

Correct statement: Iron and rust are two different chemical substances.

(e)True

Q.3.Fill in the blanks in the following statements:

(a) When carbon dioxide is passed through lime water, it turns milky due to the formation of _________.

(b) The chemical name of baking soda is .

(c) Two methods by which rusting of iron can be prevented are __________ and __________

(d) Changes in which only ____________ properties of a substance change are called physical changes.

(e) Changes in which new substances are formed are called _____________ changes..

Ans. (a)calcium carbonate

(b) sodium hydrogen carbonate

(c) painting or greasing, galvanisation

(d) physical

(e) chemical

Q.11. The gas we use in the kitchen is called liquified petroleum gas (LPG). In the cylinder it exists as a liquid. When it comes out from the cylinder it becomes a gas (Change- A) then it bums (Change-B). The following statements pertain to these changes. Choose the correct one.

(i) Process-A is a chemical change.

(ii) Process-B is a chemical change.

(iii) Both processes A and B are chemical changes.

(iv) None of these processes is a chemical change.

Ans. (ii) Process-B is a chemical change.

Q.12.Anaerobic bacteria digest animal waste and produce biogas (Change-A). The biogas is then burnt as fuel (Change-B). The following statements pertain to these changes. Choose the correct one.

(i) Process-A is a chemical change.

(ii) Process-B is a chemical change.

(iii) Both processes A and B are chemical changes.

(iv) None of these processes is a chemical change.

Ans.(iii) Both processes A and B are chemical change

Chap :- 7

Q.3.Fill in the blanks

(i) The average weather taken over a long time is called __________

(ii) A place receives very little rainfall and the temperature is high throughout the year, the climate of that place will be ________ and _________

(iii) The two regions of the earth with extreme climatic conditions are __________ and ____________

Ans. climate of the place (ii) hot, dry (iii) polar, tropical regions

Q.4.Indicate the type of climate of the following areas:

 (a) Jammu and Kashmir:

(b) Kerala:,

(c) Rajasthan:

(d) North-east India:

Ans.(a) Jammu and Kashmir—moderately hot and moderately wet climate.

(b) Kerala—very hot and wet climate.

(c) Rajasthan—hot and dry climate.

(d) North-east India—The north eastern India receives rain for a major part of the year, hence wet climate.

Q.5.Which of the two changes frequently, weather or climate?

Ans.Weather

Q.6.Followings are some of the characteristics of animals:

(i) Diets heavy on fruits 

(ii) White fur 

(iii) Need to migrate 

(iv) Loud voice

(v) Sticky pads on feet

 (vi) Layer of fat under skin

(vii) Wide and large paws

 (viii) Bright colours

(ix) Strong tails

 (x) Long and large beak

For each characteristic indicate whether it is adaptation for tropical rainforests or polar regions. Do you think that some of these characteristics can be adapted for both regions?

Ans:-

(i) Diets heavy on fruits --------------- > Tropical rainforest

(ii) White fur --- > Polar region

(iii) Need to migrate --->Polar region

(iv) Loud voice----------- >Tropical rainforest

(v) Sticky pads on feet------------- >Tropical rainforest

 (vi) Layer of fat under skin------------------ >Polar region

(vii) Wide and large paws----------------- >Polar region

 (viii) Bright colours ------ >Tropical rainforest

(ix) Strong tails --------- >Tropical rainforest

 (x) Long and large beak ----------------------- >Tropical rainforest

Q.10.Choose the correct option which answers the following question:

A carnivore with stripes on its body moves very fast while catching its prey. It is likely to be found in:

(i) polar regions (ii) deserts

(iii) oceans (iv) tropical rainforests

Ans. (iv) tropical rainforests

Q.11.Which features adapt polar bears to live in extremely cold climate?

(i) A white fur, fat below skin, keen sense of smell.

(ii) Thin skin, large eyes, a white fur.

(iii) A long tail, strong claws, white large paws.

(iv) White body, paws for swimming, gills for respiration.

Ans.(iv) A white fur, fat below skin, keen sense of smell.

Q.12.Which option best describes a tropical region?

(i) hot and humid

(ii) moderate temperature, heavy rainfall (iii) cold and humid (iv) hot and dry

Ans.(i) Hot and humid

Chap :-8

1.Fill the missing word in the blank spaces in the following statements:

(a) Wind is ___________air.

(b) Winds are generated due to _________ heating on the earth.

(c) Near the earth’s surface _________ air rises up whereas _________ air comes down.

(d) Air moves from a region _________of pressure to a region _________of pressure.

Ans.(a) moving (b) uneven

(c) warm, cooler (d) high, low

Q.8. Which one of the following places is unlikely to be affected by cyclone?

(i) Chennai (ii) Mangalore

(iii) Amritsar (iv) Puri

Ans.(iv) Amritsar

Q.9.Which of the statements given below is correct?

(i) In winter the winds flow from the land to the ocean.

(ii) In summer the winds flow from the land towards the ocean.

(iii) A cyclone is formed by a very high-pressure system with very high-speed winds revolving around it.

(iv) The coastline of India is not vulnerable to cyclones.

Ans.(i) In winter the winds flow from the land to the ocean.

Chapter 9

Tick the most suitable answer in question 1 and 2.

In addition to the rock particles, the soil contains

(i) Air and water

(ii) Water and plants

(iii) Minerals, organic matter, air and water

(iv) Water, air and plants

Answer:

In addition to the rock particles, the soil contains

(iii) Minerals, organic matter, air and water.

Q2. The water holding capacity is the highest in

(i) Sandy soil

(ii) Clayey soil

(iii) Loamy soil

(iv) Mixture of sand and loam

Answer:

The water holding capacity is the highest in

(ii) Clayey soil

Q3. Match the items in Column I with those in Column II:

1. A home for living organisms----------> All kinds of soil

2. Upper layer of soi----- > Dark in colour

3. Sandy soil----------> Large particles

4. Middle layer of soil--------> lesser amount of humus

5. Clayey soil -----> small particles and packed light.

Chapter 10

 Tick the correct answer:

(a) In cockroaches, air enters the body through

(i) lungs

(ii) gills

(iii) spiracles

(iv) skin

Answer: (iii) spiracles

(b) During heavy exercise, we get cramps in the legs due to the accumulation of

(i) carbon dioxide

(ii) lactic acid

(iii) alcohol

(iv) water

Answer: (ii) lactic acid

(c) Normal range of breathing rate per minute in an average adult person at rest is:

(i) 9 – 12

(ii) 15 – 18

(iii) 21 – 24

(iv) 30 – 33

Answer: (ii) 15 – 18

(d) During exhalation, the ribs

(i) move outwards

(ii) move downwards

(iii) move upwards

(iv) do not move at all

Answer: (ii) move downwards

Mark T if the statement is true and F if it is false:

(i) During heavy exercise the breathing rate of a person slows down. (T/ F)

(ii) Plants carry out photosynthesis only during the day and respiration only at night. (T/ F)

(iii) Frogs breathe through their skins as well as their lungs. (T/ F)

(iv) The fishes have lungs for respiration. (T/ F)

(v) The size of the chest cavity increases during inhalation. (T/ F)

Answer:

(i) F

(ii) F

(iii) T

(iv) F

(v) T

Chapter 11

Q.1.Match structures given in Column I with functions given in Column II

1. Stomata. -------> Transpiration 

2. Xylem-----> Transport of water

3. Root hairs----- > Absorption of water

4. Phloem-----> Transport of food

Q.2.-Fill in the blanks :

(i) The blood from the heart is transported to all parts of the body by the _____________

(ii) Haemoglobin is present in ___________ cells.

(iii) Arteries and veins are joined by a network of ___________ .

(iv) The rhythmic expansion and contraction of the heart is called ___________

(v) The main excretory product in human beings is ___________

(vi) Sweat contains water and .

(vii) Kidneys eliminate the waste materials in the liquid form called .

(viii) Water reaches great heights in the trees because of suction pull caused by .

Ans.(i) arteries (ii) red blood cells • (in) capillaries (iv) heartbeat (v) urea (vi) salts (vii) urine (viii) transpiration.

Q.3.Choose the correct options:

(a) In plants, water is transported through

(i) Xylem (ii) Phloem (iii) Stomata (iv) Root hair

(b) Water absorption through roots can be increased by keeping the plants

(i) In the shade (ii) in dim light (iii) under the fan (iv) covered with a polythene bag

Ans.(a) (i) Xylem {b) (iii) under the fan

Chap 12

Q.1.Fill in the blanks:

(a) Production of new individuals from the vegetative part of parent is called ___________. 

(b) A flower may have either male or female reproductive parts. Such a flower is called _______.

(c) The transfer of pollen grains from the anther to the stigma of the same or of

another flower of the same kind is known as __________.

(d) The fusion of male and female gametes is termed as ____________ .

(e) Seed dispersal takes place by means of ________ and __________.

Ans. (a) vegetative reproduction (b) unisexual flower (c) pollination (d) fertilization (e) wind, water

Q.9.Match items in Column I with those in Column II

1. Bud ---> yeast

2. Eyes----> potato

3. Fragmentation-----> Spirogyra

4. Wings------ > Maple

5. Spores ------ > Bread mould

Tick  the correct answer:

(a) The reproductive part of a plant is the

(i) leaf (ii) stem (iii) root (iv) flower

(b) The process of fusion of the male and the female gametes is called

(i) fertilisation (ii) pollination (iii) reproduction (iv) seed formation

c) Mature ovary forms the

(i) seed (ii) stamen (iii) pistil (iv) fruit

(d) A spore producing plant is

(i) rose (ii) bread mould (iii) potato (iv) ginger

(e) Bryophyllum can be reproduced by its

(i) stem (ii) leaves (iii) roots (iv) flower

Ans.(a) (iv) flower (b) (i) fertilisation (c) (iv) fruit (d) (ii) bread mould (e) (ii) leaves

Chap 13

Q.1.Classify the following as motion along a straight line, circular or oscillatory motion :

(i) Motion of your hands while running.

(ii) Motion of a horse pulling a cart on a straight road.

(iii) Motion of a child in a merry-go-round. •

(iv) Motion of a child on a see-saw.

(v) Motion of the hammer of an electric bell.

(vi) Motion of a train on a straight bridge.

Ans.(i) Oscillatory motion

(ii) Linear motion

(iii) Circular motion

(iv) Oscillatory motion

(v) Oscillatory motion

(vi) Linear motion

Q.2.Which of the following are not correct?

(i) The basic unit of time is second.

(ii) Every object moves with a constant speed.

(iii) Distances between two cities are measured in kilometers.

(iv) The time period of a given pendulum is not constant.

(v) The speed of a train is expressed in m/h

Ans. (ii), (iv), (v)

.8. Which of the following relations is correct?

(i) Speed = Distance Time

(ii) Speed = Distance/Time

(iii)Speed – Time/Distance

(iv)Speed = 1/Distance Time

Ans. (ii) Speed = Distance/Time is correct

Q.9.The basic unit of speed is:

(i) km/min (ii) m/min (iii) km/h (iv) m/s

Ans. (iv) m/s

10.A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:

(i) 100 km (ii) 25 km (iii) 15 km (iv) 10 km

Ans.(ii) 25 km

Q.11.Suppose the two photographs, shown in fig. 13.1 and fig. 13.2 of NCERT had been taken at an interval of 10 seconds. If a distance of 100 metres is shown by 1 cm in these photographs, calculate the speed of the blue car.

Ans.0.1 cm/s or 10 m/s

Q.12.Fig 13.10 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Ans. ‘A’ is moving faster.

Q.13.Which of the following distance-time graphs shows a truck moving with speed which is not constant?

Ans. (iii)

Chap 14

Q.8.Fill in the blanks:

(a) Longer line in the symbol for a cell represents its ___________terminal

(b) The combination of two or more cells is called a ____________.

(c) When current is switched ‘on’ in a room heater, it ___________.

(d) The safety device based on the heating effect of electric current is called a _____________.

Ans.(a) positive (b) battery (c) becomes red hot and emits heat (d) fuse.

Q.9.Mark ‘T’ if the statement is true and ‘F’ it is a false:

(a) To make a battery of two cells, the negative terminal of one cell is connected to the negative terminal of the other cell. [T/F]

(b) When the electric current through the fuse exceeds a certain limit , the fuse wire melts and breaks. [T/F]

(c) An electromagnet does not aatract a piece of iron. [T/F]

(d) An electric bell has an electromagnet. [T/F]

Ans. (a) F (b) T (c) F (d) T

Chap 15

Q.1.Fill in the blanks:

(a) An image that cannot be obtained on a screen is called ___________ .

(b) Image formed by a convex is __________ always virtual and smaller in size.

(c) An image formed by a __________ mirror is always of the same size as that of the object.

(d) An image which can be obtained on a screen is called a __________ image.

(e) An image formed by a concave __________ cannot be obtained on a screen.

Ans.(a) virtual image (b) mirror (c) plane (d) real (e) lens

Q.2.Mark ‘T’ if the statement is true and ‘F’ if it is false:

(a) We can obtain an enlarged and erect image by a convex mirror. (T/F)

(b) A concave lens always form a virtual image. (T/F)

(c) We can obtain a real, enlarged and inverted image by a concave mirror. (T/F)

(d) A real image cannot be obtained on a screen. (T/F)

(e) A concave mirror always form a real image. (T/F)

Ans. a) F (b) T (c) T (d) F (e) F

Q.3. Match the items given in Column I with one or more items of Column II

1. A plane mirror ---- >The image is erect and of the same size as the object

2. A convex mirror ---- > Can form image of objects spread over a large area

3. A convex lens --- > Used as magnifying glass

4. A concave mirror---- >used by dentist to see enlarged image of teeth

5. A concave lens----- > The image is erect and smaller in size than the object.

Q.9. Which type of mirror can form a real image?

Ans. Concave mirror can form a real image.

Q.10. Which type of lens forms always a virtual image?

Ans. Concave lens always forms a virtual image.

Choose the correct option in Questions 11-13:

Q.11. A virtual image larger than the object can be produced by a

(i) concave lens (ii) concave mirror (iii) convex mirror (iv) plane mirror

Ans. (ii) concave mirror

Q.12. David is observing his image in a plane mirror. Die distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, then the distance between David and his image will be

(i) 3 m (ii) 5 m (iii) 6 m (iv) 8 m

Ans. (iii) 6 m

Q.13. The rear view mirror of a car is a plane mirror. A driver is reversing his car at a speed of 2 m/s. The driver sees in his rear mew mirror the image of a truck parked behind his car. The speed at which the image of the truck appears to approach the driver will be

(i) 1 m/s (ii) 2 m/s (iii) 4 m/s {iv) 8 m/s

Ans. (ii) 4 m/s

Chap 16

Mark ‘T’ if the statement is true and ‘F’ if it is false:

(a) The freshwater stored in the ground is much more than that present in the rivers and lakes of the world. (T/F)

(b) Water shortage is a problem faced only by people living in rural areas. (T/F)

(c) Water from rivers is the only source for irrigation in the fields. (T/F)

(d) Rain is the ultimate source of water. (T/F)

Ans. (a) T (b) F (c) F (d) T

Q.6.Fill in the blanks with the appropriate answers: .

(a) People obtain groundwater through and .

(b) Three forms of water are solid, and .

(c) The water bearing layer of the earth is .

(d) The process of water seepage into the ground is called .

Ans.(a) wells, hand pumps

(b) liquid, gas

(c) aquifer

(d) infiltration

Q.7.Which one of the following is not responsible for water shortage?

(i) Rapid growth of industries

(ii) Increasing population

(iii) Heavy rainfall

(iv) Mismanagement of water resources

Ans.(iii) heavy rainfall

Q.8.Choose the correct option. The total water

(i) in the lakes and rivers of the world remains constant.

(ii) under the ground remains constant.

(iii) in the seas’and oceans of the world remains constant.

(iv) of the world remains constant.

Ans.(iv) of the world remains constant.

Chap 17

Q.7.Fill in the blank:

(a) The insects, butterflies, honeybees and birds help flowering plants in .

(b) A forest is a purifier of and .

(c) Herbs form the layer in the forest.

(d) The decaying leaves and animal droppings in a forest enrich the .

Ans. (a) pollination (b) water, air (c) lowest (d) soil as humus.

Q.11.Which of the following is not a forest product?

(i) Gum (ii) Plywood (iii) Sealing wax (iv) Kerosene

Ans.(iv) Kerosene

Q.12.Which of the following statements is not correct?

(i) Forests protect the soil from erosion.

(ii) Plants and animals in a forest are not dependent on one another.

(iii) Forests influence the climate and water cycle.

(iu) Soil helps forests to grow and regenerate.

Ans.(ii) Plants and animals in a forest are not dependent on one another.

Q.13.Micro-organisms act upon the dead plants to produce

(i) sand (ii) mushrooms (iii) humus (iu) wood

Ans. (i) Humus

Chap 18

.1. Fill in the blanks:

(a) Cleaning of water is process of removing __________

(b) Wastewater released by houses is called __________.

(c) Dried __________ is used as manure.

(d) Drains get blocked by __________ and __________.

Ans.(a) pollutants (b) sewage (c) sludge (d) (d) chemicals, kitchen waste

.7. Name two chemicals used to disinfect water.

Ans. Chlorine and ozone

Study the following statements about ozone:

(a) It is essential for breathing of living organisms.

(b) It is used to disinfect water.

(c) It absorbs ultraviolet rays.

(d) Its proportion in air is about 3%.

Which of these statements are correct?

(i) (a), (b) and (c) (ii) (b) and (c) (iii) (a) and (d) (iv) All four

Ans. (ii) (b) and (c)

Class 7 chap 15 light

 Chap 15 Light notes

Introduction:-

Light is a form of energy which enable us to see world.

Speed of light = 30000Km/sec

Travels along a straight path.

Luminous objects:- Produce their own light .

Eg -- > Sun

Illuminous objects:- Donot Produce their own light . Eg:- Chair

Capable of reflecting light to our eyes coming from luminous objects.

Uses of Light:- Images formed by Mirror and lens

Rainbow formation

Ray of light:- Straight line path of light is called ray of light.

Beam of light:- Bunch of rays of light is called beam of light.

Reflection of light:-

Bouncing back of ray of light into the same medium after striking a polished surface such as mirror is called reflection of light.

Incident ray :- Ray of light that strike the surface.

Reflected ray :- Ray of light sent back by the surface after reflection.

Normal:- Perpendicular to the surface at the point of incident . It is an imaginary line.

Angle of incidence= angle between the Incident ray and normal.

Angle of reflection= Angle between the reflected ray and normal.

Law of reflection:- 

1. Angle of incidence= angle of reflection

2. Incident ray, reflected ray and the normal all lie on the same plane.

Types of mirror:- two types

1. Plane mirror

2. Spherical mirror

Plane mirror:- mirror with a flat reflective surface.

Characteristics of image formed by plane mirror:-

1. Virtual and erect image.

2. Same size as that of object.

3. Image formed behind mirror.

4. Distance of the image behind mirror is same as distance of object from mirror.

Virtual image:- Image formed when reflected rays appear to meet. It can't obtained on screen.

Real image:-Image formed when reflected rays actually meet. It can obtained on screen.

Lateral inversion:- In mirror left appears right and right appears left . This is called lateral inversion.

Spherical mirror:- Mirror having spherical reflecting surface.

Two types :-

1. Convex mirror 

2. Concave mirror

CONCAVE MIRROR:- mirror whose reflecting surface is curved inward. It is a converging mirror.

CONVEX MIRROR:- Mirror whose reflecting surface is curved outward. It is a diverging mirror.

Use of Convex mirror:- 

1. Can use as rear view mirror to get wider field view to help driver to see traffic behind them.

Used of concave mirror:-

1. Doctor use concave mirror for examining eyes, nose, ears and throat.

2. Dentists use concave mirror to see enlarge image of teeth.

3. Reflectors of torches, headlights of cars and scooters are concave in shape.

Lens:- Transparent optical medium bounded by two surfaces at least one being spherical is called lens.

Types of lens:- 

1. Convex lens( converging lens)

Thicker at middle and thinner at edges

2. Concave lens( diverging lens)

Thicker at edges and thinner at middle.

Uses:-

1. Magnifying glass.(Convex lens)

2. Used in Spectacles, telescopes and microscopes.

Sunlight:- It is a white light that is made up of seven colours.

Prism:- Transparent element with three rectangular and two triangular surface.

Dispersion of light:- 

Splitting of white light into its constituent colours is called dispersion of light. For example :- rainbow is formed when sunlight split into its 7 coloured band (VIBGYOR)

This pattern of colour obtained is called spectrum.

Obelia

 Subject: Coelenterata : Obelia (sea fur)

Systematic position:- 

• Phylum :- Coelenterata or cnidaria ( Diploblastic, tissue organisation, coelenteron and cnidoblasts)
•  Class:- Hydroids (fixed polyps; medusa with ectodermal sense organs)
•   Order:- Hydroida( Fixed polyps; medusa with ectodermal sense organs)
•    Family:- Campanularidae( cup shaped hydrotheca)
•     Genus:- Obelia ( Free swimming and discoidal medusae)
•      Species:- geniculata (Colourless)

Habitat:- 

• Obelia is a marine, cosmopolitan, polymorphic, colonial and sedentary hydrozoan Coelenterate found attached on the stones, rocks, molluscan shells, ships, weeds, etc. In shallow coastal waters upto the depth of 80 metres.
• It is branched like a tree, so called zoophyte (zoo= animal ; phyte = plant) or sea fur.
• There are many species of Obelia but the most common species is O. geniculata.

Habits :- 

(1) Feeding :- 

• Obelia is carnivores and feeds upon the living insect larvae, small crustaceans and worms.
• Food is captured and paralysed with the help of stinging cells.
• Digestion is partly intercellular and partly intracellular.

(2) Reproduction:-

•  Obelia shows asexual as well as sexual reproduction.
• Asexual reproduction occurs by external budding.
• In Obelia reproductive zooids are meduse which are inside or dioecious but don't show sexual dimorphism.
• Fertilization is external.
• Development is indirect and includes a free swimming planula larva which brings about dispersal.
• Life history of Obelia shows alternation of generation.

(3) Respiration and Excretion :- 

• These occur through general body surface.

Morphology :-

(A) Shape, size , colour and symmetry:-  

• Obelia appears as a network of branched threads giving fur-like appearance, so called sea fur.
• It is about 25-30 mm in height, creamish- white or greyish in colour and radially symmetrical.

(B) Two types of filaments:-

• Obelia colony is formed of 2 types of filaments:-
• (a) Hydrorhizae:- These are horizontal branched which creep over the substratum and fix the colony to the substratum.
• (b) Hydrocauli:- These are upright branches.
• Each hydrocaulus grows in a scorpioid cymose manner.
• Each branch stops growing ends into a zooid and gives rise a new branch.
• Adjoining branches are on alternate sides.
• So it is formed of lower parts of all the true branches and grows in a zig zag manner.
• Each side branch and main hydrocaulus end into a terminal bud which later forms a polyp.

(C) Coelenteron:- 

• Coelenteron also called gastro-vascular cavity, Is the central cavity running through both types of filaments.
• It is bounded by 2 coats:-
• (a) Coenosarc:- It is inner and living coat.
• It is formed of two layers:-
• Inner gastrodermis lining the Coelenteron, and outer epidermal.
• Between the two cellular layers is a non- cellular and gelatinous mesoglea.
• (b) Perisarc :- It is outer and non-living coat.
• It is chitinous in nature and is secreted by myoepithelial cells is epidermis.
• In early stages, it is in contact with Coenosarc but later two are separated by a fluid-filled space  except at certain points.
• It is supportive and protective in function,so acts as exoskeleton.
• At the base of zooids and near the bends  of the hydrocaulus, Perisarc forms ring like grooves, called perisarcal annuli, which give flexibility to it , so it sways in water to withstand the wave action.

Polymorphism in Obelia :-

• Polymorphism us the phenomenon in which  a colonial organism is formed of many types of organisms (called zooids in coelenterates) which Differ in structure and function for the benefit of whole Colony.
• Obelia is a polymorphic (trimorphic) Coelenterate as the colony has three types of zooids which through are homologous zooids but are specialized in their morphology and physiology. 

So polymorphism is associated with division of labour.

(1) Hydrants or polyps ( nutritive and protective zooids)

(2) Blastostyles or Gonozooids(budding zooids)

(3) Medusae or Gonophores( sexual zooids)

(1) Hydranth or Polyp:-

(i) Position :-

•  These are more numerous in number than other zooids.
• These are distributed throughout the colony and are terminal in position.

(ii) Structure :-

(a) External structure:-

•  Polyp or Hydranth is a hollow and cylindrical zooid which is proximally continuous with the hydrocaulus.
• It's free distal end is produced into a conical process called oral cone or manubrium or hypostome having a termination and circular mouth.
• Oak cone is about one third of total length of zooid.
• Arising from the base of Hydranth, there is a ring of 24-30 solid, tapering and fillifo tentacles, each having batteries of  stinging cells.
• The tentacles are radially arranged to capture the food from all directions.
• Hydranth is protected by a cup-shaped and transparent cover called hydrotheca.
• So Obelia belongs to sub order Thecata while the hydranths of Bougainvillea and Tubularia have no theca, so placed in sub-order Athecata.
• It is formed by the expansion of Perisarc.
• Hydrotheca is open at the tip so Hydranth can be protruded.
• At its lower end, hydrotheca is produced inward to form a centrally- perforated circular shelplf which keeps a limit on the retraction of polyp.

(b) Coelenteron:- 

• Mouth of Polyp leads into a small cavity of manubrium, sometimes called gullet, which opens in the Coelenteron of body of Hydranth which in turn is continuous with that of hydrocaulus.

(c) Histology :- 

• Histologically , living Coenosarc of Hydranth is formed of 2 layers with a non-living and gelatinous mesoglea in between.
• So Obelia is a Diploblastic animal.

(A) Epidermis:- 

• It is outer and thinner layer.
• It is formed of 5 types of cells:-

(1) Myoepithelial or epitheliomuscular cells:-

• Position and structure:- 
• These forms the bulk of epidermis.
• These are colonial-shaped with their broader side outward.
• Their broader sides meet and form a continuous layer except when interrupted by sensory cells and cnidoblasts.

Each Myoepithelial cell is differentiated into 2 parts:-

Epithelial part and muscular part:-

• Epithelial part is outer and nucleated part having vacuolated cytoplasm, a layer of secretary granules in its outer part and two supporting fibrils called tonofibrils.
• Muscular part is inner and is produced into two or more tail or muscle processes having a contractile myoneme formed of myofilaments.
• Myonemes of these cells are arranged along the longitudinal axis so forming a kind of longitudinal muscles which when contract, make the  Hydranth small and thick.
• Functions:- These are protective, secretory and contractile in function.

(2) Interstitial cells:- 

Position and structure:- 

• These are groups of cells present in the spaces between the inner narrow ends of Myoepithelial cells.
• These are mainly present in the basal part tthe polyp.
• Each is spherical-shaped, about 5micrometer, with clear cytoplasm and nucleus having 1 to 2 nucleoli.
• Position:- 
• These act as reserve or undifferentiated cells which can differentiate into any type of cell depending upon their requirements, these are called totipotent cells.
• So these help in regeneration, budding and reproduction.
• Breton showed that these can renew cells within 45 days 

(3) Cnidoblasts:-

Position and structure:- 

• Cnidoblasts ,also called nematoblasts or stinging cells, are present among the Myoepithelial cells on the hypoblast and tentacles.
• These are peculiar cell-type of Coelenterates which are also called cnidarians while their  study is called Cnidology.
• These form ring-like groups called annular batteries on the tentacles.
• Each cnidoblast is pear shaped and about 16 micrometer in size.
• It has contractile cytoplasm, eccentric nucleus, a sensory trigger hair called cnidocil at it's free distal end and a sac like nematocyst inside.

• A namatocyst is differentiated into two parts :- 
• Capsule and thread tube.
• Capsule is oval-shaped, double walled , chitinous sac externally covered by a lid or operculum.
• It is filled with a poisonous fluid called hypnotoxic (a mixture of proteins and phenol).
• Thread tube is differentiated into basal butt and distal thread tube proper.
• Butt is again formed of proximal wider and funnal shaped part with no spines, called shaft, and distal and tapering part with three large-sized stylets or barbs and three spiral rows of spines or barbules, called spinneret.
• Thread tube proper is long tube of uniform diameter which ends into a hypodermic needle.
• Such nematocyst of Obelia are called penetrants or stenoteles.
• On the outer surface of capsule, there are a number of refractile rods, each giving rise to a contractile fibril extending upto the base of cnidoblast.
• Another coiled thread, called lasso, also extends the outer surface of the capsule to its base.
• These both prevent the protrusion of capsule.

Function:-

•  when the stinging. Cells are undischarged,spines and stylets of the butt are inward directed and thread tube proper is coiled around the butt as a watch spring.
• When stimulated, stimulus is first received by cnidocil and is conducted to contractile cytoplasm, the contraction of which causes the eversion of thread tube, so the spines and styles are now outward directed while the hypodermic needle is injected in the prey to release the hypnotoxin, which paralyses or kills the prey.So the stinging cells of Obelia are called penetrants or stenosis.
• It is now proposed that discharge of nematoblasts is probably due to rapid increase in Calcium level and intracapsular pressure inside the capsule of these cells.
• Their coordinated discharge is controlled by neural connections .
• So the nematoblasts are cells for offence and defence and are involved in food capturing and protection.
• Nematoblasts are independent effectors.
• These are stimulated by partly mechanical and partly Chemical stimuli.
• Once discharged, these are never withdrawn so these move in the Coelenteron and are digested.
• New nematoblasts are regenerated from interstitial cells in about 48 hours in the basal parts of the polyps .

4) Sensory cells:-

Position and structure:- 

• These are present among the Myoepithelial cells of epidermis, especially in hypostome and tentacles.
• Each is a columnar and neuro-sensonry cell.
• It has a sensory hair on other side and a few modulated nerve fibres on inner side which join the nerve fibres of nerve cells.

Functions:- 

• These are sensitive to externally stimuli like touch, temperature, light , Chemicals tc so also called receptors.

(5) Nerve cells:-.

Position and structure:- 

• Nerve cells, also called protoneurons, from an epidermal nerve net on outer border of mesoglea . 
• These are more developed in the tentacles.
• Each is an irregular cell and is differentiated into a central nucleated part called cell body and a number of nerve processes called neuritis .
• Depending upon the number of neurites, nerve cells may be bipolar or multipolar.
• But neurites are not differentiated into axon and dendron so nerve impulse can travel in any direction so form diffuse type nervous system.
• Neurites of adjacent nerve cells coil around one another to form a net , so called plexiform nervous system.
• But there is no continuity of their cytoplasm and nerve impulses are conducted only through synapses, so called synaptic nervous system.
• Function:- These form a part of sensory-neuro-motor system and conduct the nerve impulses.

(B) Gastrodermis:- 

• It is inner and thicker .
• It forms the bulk of body wall and is digestive, circulatory and muscular in function.

It is formed of following five types of cells:-

(1) Myoendothelial or Endothelio-muscular cells:-

Position and structure:- 

• These form the bulk of Gastrodermis .
• Each is long and columnar cell which is differentiated into 2 parts:-
• Endothelial part and muscular part.
• Endothelial part is inner and  nucleated part having vacuolated cytoplasm and either pseudopodia  or 2 flagella .
• Their inner free ends also have microvilli to absorb the digested food from the coelenteron.
• It's muscular part is outer and is produced into a muscle process having contractile fibrils called myonemes.
• Their myonemes are arranged perpendicular to longitudinal axis of hydranthand act as circular muscle fibres.
• When contract, these make the polyp long and thin .
• Around the mouth these muscle processes act as sphincter.
• Function:- These are nutritive, absorptive, muscular and circulatory in function.

(2) Gland cells:- 

Position and structure:- 

• These are modified Myoendothelial cells present only on manubrium and body of hydranth while these are absent on the tentacles.
• Each is columnar-shaped like the Myoendothelial cell but is smaller in size and lacks the muscle process.

Function:- 

• These secrete either mucus (those in manubrium and called mucous gland cells) or digestive enzymes ( those in the body and. Called enzymatic gland cells) so help in ingestion and intercellular digestion of food in coelenteron.

(3) Interstitial cells:-

• These are present in groups in the spaces between the outer ends of Myoendothelial cells.
• These resemble those of epidermis in their structure and function.

(4) Sensory cells:-

• These are present between the Myoendothelial cells of Gastrodermis.
• These resemble those e epidermis in their structure and function except that are less in number.

(5) Nerve cells:- 

• These form a gastrodermal nerve net on the inner border of mesoglea.
• These also resemble those of epidermis in their structure and function but are less in number
• Epidermal and gastrodermal nerve nets are interconnected by nerve fibres.

(C) Mesoglea:-

• Mesoglea , also called mesenchyme, is the middle homogeneous, non-cellular and gelatinous layer of uniform thickness.
• It is joint secretary product of both epidermis and Gastrodermis.

It helps in following functions:-

• It acts as middle lamella and provides binding surface for the attachment of muscle processes of both epidermis and Gastrodermis.
• It is elastic in nature and helps in extension of hydranth .
• It acts as the skeleton so provides mechanical support to hydranth

Plasmodium Vivax ( the malarial parasite)

Subject: Protozoa:-  Plasmodium Vivax ( the malarial parasite)

Systematic position:- 

• Kingdom:- Protists (Acellular eukaryote)
•  Phylum:- Protozoa (Heterotrophic and undergoes fission)
•   Super-Class:- Sporozoa (No organelles of locomotion)
•   Class- Coccidia (Mature trophozoite is intracellular)
•     Genus- Plasmodium (True malarial parasite)
•        Species- vivax ( causes benign tertian malaria as erythrocytic schizogony is completed in 48 hours)

Introduction:-

• Malaria is a widely known human disease. It is caused by infection with a pathogenic Protozoan parasite of blood, the plasmodium, whose 60 known species cause malaria in man and other animals.
• Four species of Plasmodium are known to cause different types of malaria fever in man.
• They are P. vivax , P. malariae and P. falciparum. Female Anopheles mosquitoes transmit Plasmodium from person to person , thus serving as carrier or vector host.
• It derives certain proteins from human blood and are important in its egg formation.
• Male feeds upon plant juices.
• P. vivax is mostly widely studied of Plasmodium. 
• Hence, the following description is mainly based upon it.

Distribution:-  

• Plasmodium vivax is cosmopolitan parasite.
• It is wide spread in tropical and temperate countries.
• Some species infect migratory birds.

Habitat:-

• Plasmodium is endoparasite, found in the blood of man:-
• Intracellular in R.B.Cs and parenchymal cells of liver.
• Also found in the alimentary canal and salivary glands of female Anopheles Mosquitoes.

Habits:-

• Being intracellular parasite, it depends for all vital activities on host.

Life cycle:-

• Life cycle of Plasmodium is complicated.
• It comprises several stages and requires two hosts for completion- a primary, definitive or principal host and a secondary, intermediate or vector host
• Such a two host life cycle is called digenetic.
• In human species of Plasmodium, intermediate host is always female Anopheles. 
• In human body, the parasite multiplies asexually, while is female Anopheles, it undergoes a sexual cycle followed by an Asexual multiplication called sporogony.

Asexual cycle and schizogony in man:- 

• The adult phase of Plasmodium is called trophozoite. 
• It occurs in R.B.Cs of man but before invading the R.B.Cs the paradises invade liver cells and undergoes extensive asexual multiplication by Schizogony. The life cycle of Plasmodium in man consists of two phases:-

(1) Exoerythrocytic or hepatic Schizogony

(2) Erythrocytic schizogony

Exoerythrocytic Schizogony:- 

(a) Inoculation or Infection:- 

• A healthy person acquires infection when a female Anopheles Mosquito, containing infective stage of parasite in their salivary glands.
• They feed upon human blood which they such, usually during night, by means of their piercing and sucking mouth parts.
• The mosquito punctures the host's skin by its proboscis and first introduces some saliva into the wound.
• Saliva contains an anticoagulant which prevents clotting of blood during sucking.
• This saliva carries thousands of sporozoites into the human blood.

(b) Structure of Sporozoites:- 

• A sporozoite has a spindle shaped, slightly curved or sickle shaped body tapering at either end .
• It is uninucleate organism, measuring 6 micro to 15 micro in length and 0.5-1 micro in width 
• Each has a covering of firm, but elastic pellicle, containing longitudinal contractile microtubules for its wriggling movements.
• It's Anterior end is marked by a minute aperture, the micropyle, followed an apical cap of connective rings.
• A pair of long and slender secretory organelles open into the cap 
• These is a single vesicular nucleus with a centrally located nucleolus and a mitochondrion with tubular cristae, scattered endoplasmic reticulum is also present.

(C) liver Schizogony:- 

• Within half an hour or so after inoculation into human blood, all Sporozoites disappear from the blood.
• As they reach into the liver with blood circulation, they make their way out from Blood capillaries and enter into the liver cells.
• Here they multiply asexually by Schizogony. 
• Liver Schizogony has two phases:- pre- erythrocytic and exo- erythrocytic.

(1) Pre- erythrocytic phase:- 

• In the liver cells, the parasite become spherical and are known as the cryptoschizoites.
• In the liver cell, the cryptozoite feeds on cytoplasm and grows into a rounded structure, the pre-erythrocytic schizont also known as cryptoschizont.
• The later divides into about 1000 minutes cryptomerozoites by a special type of fission called schizogony ( multiple fission).
• This Schizogony is called exo- erythrocytic, because it occurs away from blood R.B.Cs. 
• The schizont and the liver cell break up and liberate cryptomerozoites into the liver sinusoids.
• From the sinusoids, some of these cryptomerozoites invade fresh liver cells to continue Exoerythrocytic Schizogony, while other remain in blood stream and invade R.B.Cs to initiate the erythrocytic cycle.

(iii) Exo-erythrocytic phase:-

• Cryptomerozoites enter fresh liver cells to become metacryptozoites or phanerozoites. 
• The later are of two types- micrometscyptozoites and macrometacryptozoites.
• Each micro-metacryptozoite undergoes Schizogony, producing 100 to 10000 minute micro-metacryptozoite which pass into blood stream and invade R.B.Cs for fresh erythrocytic phase.
• Schizogony in each macrometacryptozoites, on the other hand, produces only about 64, relatively large macrometacryptomerozoites which invade fresh liver cells to continue Exo-erythrocytic cycle.
• Pre and exo-erythrocytic phases of parasite remain immune to the resistance of host and parasitea are not susceptible to the action of any anti-malarial drug.
• Also little damage to the host is done during these stages 

(iii) Pre-patent and incubation periods:- 

• The pre- erythrocytic phase is completed in about 8 to 10 days .
• This inverval between inoculation and initiation of erythrocytic phase is called Pre-patent period.
• The period between infection and appearance of first malarial symptom is the incubation period which is about 10-17 days.

2.Erythrocytic Schizogony:- 

• This cycle occurs in R.B.Cs and begins when a cryptomerozoites or micro-metacryptomerozoite enters into an R.B.Cs . 

It includes the following stages:-

(i) Trophozoite stage:-  

• Inside the corpuscle, the parasite become rounded and grow.
• During the growth period it is called trophozoite.

(ii) Signet ring stage:- 

• As the trophozoite grows in size, a large non-contractile vacuole appeara in the centre, pushing the cytoplasm and nucleus to a thin peripheral layer.
• This stage is clinically referred to as signet ring stage as it resembles the signet ring.

(iii) Amoeboid stage:-  

• The signet ring trophozoite developes into an active amoeboid trophozoite.
• Now, the trophocytes starts feeding more actively upon cytoplasm of host R.B.Cs with the help of pseudopodia.
• It secretes out digestive enzymes which liquify, the cytoplasm of R.B.Cs by its partial digestion.
• The haemoglobin is not digested 
• It gets decomposed into a brownish black pigmentz the haematin, which lies in the cytoplasm of trophozoite.
• In about 36 hours, the young trophozoite grows into an adult, occupying almost the entire R.B.Cs.
• Besides plasmolemma, endoplasmic reticulum, mitochondria, food vacuole, Golgi complex, nucleus etc, a concentric structure of unknown significance has been recently observed in this trophozoite by electron microscopy by Garnham, 1966 and Zinska , 1969.
• The host R.B.Cs gets somewhat enlarged and irregular in shape.
• At this time, small red eosinophilic granules appear in the cytoplasm of the host corpuscle which are known as Schuffner's granules.

(iv) Schizont :- 

• The amoeboid trophozoite, after active feeding becomes rounded, grows in size and becomes Schizont.
• It now undergoes Schizogony.
• It's nucleus undergoes repeated mitotic divisions, forming. 12-24 daughter nuclei, which get arranged at the periphery and cytoplasm masses surrounding them.
• Each cytoplasmic mass with one nucleus becomes an oval shaped merozoite.
• They are shorter and thicker than the Sporozoite.
• A part of the cytoplasm of the Schizont is not used in the formation of merozoite.
• This residual cytoplasm contains the haematin and metabolic wastes or toxins produced by the parasites.
• The merozoite arranged themselves around the residual cytoplasm like the petals of rose flower.
• This stage is called rosette stage.
• One complete erythrocytic cycle takes 48 hours in P. vivax.
• The residual cytoplasm and haematin are eaten up by  special cells, the phagocytes in the spleen, kidney and liver.
• The interval between inoculation of Sporozoites into human blood and first attack of the fever is called incubation period.
• Repeated erythrocytic cycles results, not only in intermittent attack of fever, but also in a large scale destruction of R.B.Cs, weakening the patient to a considerable extent.
• In P.vivax, the erythrocytic Schizogony takes 48 hours and therefore the fever recurs rhythmically every third day.
• Destruction of R.B.Cs in erythrocytic Schizogony releases toxins and makes the patient feel chilly.

(v) Formation of gametocytes:- 

• After a number of erythrocytic cycles, some merozoite, invading fresh R.B.Cs, grow not into normal schizont, but into a different kind of rounded form called gametocytes or gamonts.
• These show sexual dimorphism, being of two types.
• The male or microgametocyte is smaller and contains a large diffused nucleus.
• The gametocytes donot divide, but remain as intracellular parasitrs within their host's blood corpuscles, until they either die or are ingested by the vector, in which they continue their Development.

(2) Sexual cycle of Plasmodium in mosquito :-

(i) Infection of mosquito:-

• When a female Anopheles Mosquito sucks blood of the infected person, it receives  R.B.Cs containing different stages of erythrocytic cycle, including gametocytes.
• In its gut, all stages except the gametocytes are digested
• Gametocytes however survive.
• They break the corpuscle and become free in the stomach of mosquito.
• Their becoming active depends upon temperature.
• While the trophozoite and merozoites are active at high temperature (98.6°F) of the human body, the gametocytes need low temperature for further change and therefore, become active only in cold blooded mosquito host.
• The Development of the gametocytes in the mosquito's stomach involves two imo events.

(1) Sexual cycle and

(2)  Sporogony

Sexual cycle:-

(a) Gametogony or Gametogenesis:-

• Gametogony or Gametogenesis means the Development of gametes.
• Male and female gametes are dissimilar. 
• Such gametes are called heterogametes.

Formation of male gametes:-

•  The male or microgametocyte undergoes Spermatogenesis.
• The nucleus of each rapidly divides by meiosis, into 6 to 8 haploid nuclei.
• The later migrate to the periphery.
• The cytoplasm gathers around these nuclei.
• Suddenly each cytoplasmic mass, together with nucleus shoots out from the surface of the microgametocyte in the form of a 20 micro to 25 micro long, whip-like microgametes or sperms.
• This process is called as exflagellation. 
• Soon they become free from the residual cytoplasm and start creeping by lashing movements.

Formation of female gametes:-

• A macrogametocyte forms a single female gamete or macrogamete or ovum.
• It's nucleus casts off one or  two small pieces, which leave the Gametocyte with a bit or cytoplasm.
• This converts the macrogametocyte into a microgamete.

(b) Fertilization:- 

• The megagamete developes a small cytoplasmic projection, the come of reception, on one side.
• The nucleus moves into this cone.
• One male gamete penetrates the female gametes through this cone.
• The nuclei and cytoplasm of the gametes fuse to produce a diploid zygotic nucleus or synkaryon.
• Syngamy is heterogamy  or anisogamy as the uniting male and female gametes are dissimilar.
• Zygotes are formed in the stomach of mosquito about 9 or 10 days after the blood meal.

(iii) Ookinete:-

• The zygote, when formed is rounded and motionless.
• After remaining inactive for sometime, each zygote starts elongating and undergoes certain changes so that, in about 9to 10 hours, it becomes 15 micro to 20 micro long and 3 micro thick, worm-like motile organism called vermicular or Ookinete.
• The later has dense cytoplasm, a single pseudopodium -like blunt extension, one irregular nucleus, mitochondria, ribosomes, microtubules and 2 or more, non-contractile vacuoles.
• The Ookinete pierces through the wall of stomach and comes to rest just beneath it's outermost layer.
• Here , it reassumed sphere shape and soon becomes enclosed inside a cyst, about 8 micro to 10 micro in diameter.
• The cyst-wall is derived partly from zygote and partly by tissues of stomach wall.
• Encysted zygote is called oocyst or sporont.
• Just one or two days after fertilization, 50 or more oocysts appear upon the surface of the stomach of the infected mosquito.

(3) Sporogony:- 

• Each oocyte undergoes asexual multiplication known as Sporogony.
• It's nucleus divides repeatedly by mitosis, forming about 10000 minute daughter nuclei within 2-3 days.
• At the same time ,the cytoplasm developes large vacuoles and takes up a sponge- like structure, in which numerous irregular cytoplasmic developes large vacuoles and takes up a sponge -like structure, in which numerous irregular cytoplasmic mass are formed connected by protoplamic strands.
• Each cytoplasmic mass becomes elongated and spindle shaped.
• Together with its nucleus, it now projects into the adjacent vacuole.
• These spindle shaped bodies are called sporozoites.
• Each sporozoite has tapering ends and a broad middle part containing a single nucleus.
• By their pressure, the oocyst ruptures.
• Floating freely in haemocoel or body cavity, these reach the salivary glands of the host and become lodged in its common salivary duct in large numbers.
• In mosquito, whole sexual cycle is completed with 10-20 days depending upon temperature.
• 
  

Some characters of P.vivax:- 

• 
  
• Geographical distribution:- Tropical and sub-tropical regions
• Prepatent period:- 8 days
• Duration of erythrocytic cycle :- 48 days
• No. Of metacryptozoites formed per Schizont:- 10000
• Incubation period :-14 days
• Signet ring form in R.B.Cs:- large ring with vacuole and usually one chromatin dot.
• Haematin:- Dark brown granules and rodless
• Schizont:- Larger than normal erythrocyte.
• Gametocyte in R.B.Cs:- Rounded to oval
• Microgametes formed:- 4 to 8
• Duration of mosquito cycle:- 10 days
• Type of malaria:- Benign tertian, death rate low.

Some characters of P.malariae:- 

• Geographical distribution:- Tropical and sub-tropical regions
• Prepatent period:- 7-12 days
• Duration of erythrocytic cycle:-  72 hours
• No. Of metacryptozoites formed per Schizont :- 2000
• Incubation period:-  18-24 days
• Signet ring form in R.B.Cs:- Usually one ring and one chromatin dot.
• Haematin:- Dark brown to black as coarse granules
• Schizont:- Slightly smaller than normal erythrocyte.
• Gametocyte in R.B.Cs:- Oval or circular
• Microgametes formed:- 2 to 5
• Duration of mosquito cycle:- 25-28 days
• Type of malaria:- Quartan, severe.

Some characters of P.ovale:- 

• Geographical distribution:- West Africa and South America
• Prepatent period:- 9 days
• Duration of erythrocytic cycle:- 48 hours
• No. Of metacryptozoites formed per Schizont:- 15000
• Incubation period:- 14 days
• Signet ring form in R.B.Cs:- Usually one ring and one dot
• Haematin:- Dark brown less aboundant coarse granules
• Schizont:- Much smaller than a normal erythrocyte.
• Gametocyte in R.B.Cs:- Rounded or oval
• Duration of mosquito cycle:- 16 days
• Type of malaria:- Ovale or mild tertian severe

Some characters of P. falciparum:-

• Geographical distribution:- Tropical, sub-tropical and warmer temperate regions.
• Prepatent period:- 5-6 days
• Duration of erythrocytic cycle:- 36-48 hours
• No. Of metacryptozoites formed per Schizont:- 40000
• Incubation period:- 12 days
• Signet ring form in R.B.Cs:- Small ring situated at edge of R.B.Cs. sometimes 2 chromatin dots.
• Haematin:- Dark brown coarse granules in a compact mass.
• Schizont:- Smaller than a normal erythrocyte. 
• Gametocyte in R.B.Cs:-Sausage or crescent shaped
• Microgametes formed:- 4 to 8
• Duration of mosquito cycle:- 10-20 days
• Type of malaria:-Malignant tertian, death rate high