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You can Download From the Beginning of Time Questions and Answers, Notes, Plus One History Chapter Wise Questions and Answers Kerala Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Plus One History Chapter Wise Questions and Answers Chapter 1 From the Beginning of Time

Question 1.
As people began to produce food, there were great changes in their lives. What were these changes?
Answer:
During the long history of mankind, until people learned to produce food, they found their food by collecting the flesh of dead animals, hunting animals and collecting roots or fruits from plants. They had learned to make stone weapons and communicate with one another. Later man began to get food by means of agriculture, and domesticating and growing animals. But they continued hunting and gathering food. Even today we find hunter-gatherer communities in some parts of the world.

Question 2.
“The fossils of man, stone weapons and cave drawings or pictures help us to understand the history of man. On the basis of this statement, find out the sources that throw light on the history of early men.
Answer:
The fossils of man, stone weapons and cave drawings or pictures help us to understand the history of man. But in the beginning, many scholars were not willing . to understand or acknowledge the importance of these discoveries. They even refused to acknowledge that they were the fossils of early people. They were doubtful about the capacity of the early people to make stone weapons and to draw pictures, it was after a long time that the scholars recognized the importance of the discoveries.

It is from the fossils that we get the evidence for the evolution of man. The time of the fossil can be determined through chemical testing. Otherwise by examining the remnants of the stone in which the. fossils are found their time can be established. Once the time of the fossils is determined, the order of the human evolution can be found out.

Question 3.
“There are different categories in the Homo Genus Explain.
Answer:
Homo Erectus:
The origin of this species was 1.8 million years ago. Homo Erectus, which means upright man, is the direct predecessor of modem man. It was Homo Erectus that discovered the use of fire and started using clothes. This human race, which appeared in Africa, was interested in migration. From Africa they spread to Europe and Asia.

Homo Sapiens:
This humankind appeared on earth after the homo erectus. Their origin was in Africa some 8 lakh years ago. Just like the homo erectus, the homo sapiens also moved to Europe and Asia. They were known as “Wise or Thinking Man”.

Question 4.
Hominids originated in Africa. Do you agree with this opinion? Explain with examples.
Answer:
Hominids originated in Africa. There are two evidences for this.

1. The apes in Africa are very close to Hominids.
2. Early hominid fossils were discovered from East Africa. The fossils discovered outside Africa are not as old as the ones found in East Africa.

Question 5.
In a classroom discussion, Sheeba opined that there are differences between hominoids and hominids. Do you agree with her opinion? If yes, describe them.
Answer:
Hominoids are apes. All apes which include Gorillas, chimps, humans, orangs gibbons, etc. are hominoids. But Hominids are great apes and exclude gibbons (lesser apes). All hominids are hominoids, but not all hominoids are hominids. Hominids belong to the family called Hominidae. All people come in this group. There are 4 features that make hominids different from others:

1. Bigger brain
2. Upright posture
3. Walkirig.on two legs
4. Special ability in the use of hands.

Question 6.
Hominids are divided into different genus (branches). Mention two important genus.
Answer:
Hominids are divided into different genus (branches) and the ‘two most important of them are Australopithecus and Homo.

Question 7.
Arrange chronologically (in the order of time):

1. homo habilis
2. homo erectus
3. homo sapiens
4. homo sapiens sapiens

Answer:

1. homo sapiens
2. homo erectus
3. homo sapiens sapiens
4. homo habilis

Question 8.
Match the items in column A & B.

A	B
Handy man	Homo erectus
Upright man	Homo sapiens
Intelligent man	Homo sapiens sapiens
Modem man	Homo habilis

Answer:

A	B
Handy man	Homo habilis
Upright man	Homo erectus
Intelligent man	Homo sapiens
Modem man	Homo sapiens sapiens

Question 9.
Babu: “Modern man originated in different places.”
Reena: “No. Modern man originated in one place.”
This is a part of a classroom discussion. With which opinion do you agree? Why?
Answer:
The place of origin of the modem man has been a much-discussed topic. Scholars have put forward two contradictory views on this issue. They are Regional Continuity Model and Replacement Model. Regional Continuity Modeh This model says that modem people originated in different places.

The early homo sapiens in many places slowly evolved as modem people and that is why the modem people in various parts of the world look different from one another at first sight. The regional differences in the features of people are the basis for such a view.

Replacement Model:
This model says that modem man originated in Africa. The spokesmen of this model say that modem people appeared in place of (the old species of people everywhere. As evidence to their claim, they put forward the hereditary and anatomical similarity of modem people.

This model points out that modem people are quite similar everywhere because they originated in the same place – Africa. The first fossils of modem people were discovered from Omo in Ethiopia. This evidence substantiates the Replacement Model.

Question 10.
For procuring food, the early people had used different ways. Explain.
Answer:
The early people got their food by gathering.huriting, taking the flesh from dead animals and fishing. They gathered vegetarian products like seeds, kernel of nuts, fruits arid roots. Some people believe that they stored food but for this, there is ho dear proof.

Although there are many fossils of bones, the fossils of vegetarian stuff have been rare. Remnants of plants and trees that have been burned down by sudden fire last for quite a long time, but archaeologists have not yet found such fossils.

It is natural that the early people collected the flesh of dead animals or the remnants of animals killed by carnivorous beasts. Early hominids ate mammals like rats and squirrels, birds and their eggs, crawling creatures and even insects like termites.

Question 11.
Fire was useful to the early man in many ways. Discuss how they used fire.
Answer:
It gave heat and light in the caves. It was used for cooking. It was used to harden timber. It was helpful in the making of tools. It was also used to drive away dangerous beasts.

Question 12.
“Early man used to make working tools.” Examine the relevance in this statement.
Answer:
The first proofs of man’s making and using working stone tools were got from Ethiopia and Kenya. We don’t know if these tools were made by man, woman or both. Most likely both men and women made such tools. Women must have made these tools to earn their food and also food their children after they had stopped breastfeeding them.

Question 13.
“Among all creatures only man has language.” On the basis of this statement write about the various views regarding the development of language.
Answer:
There are many views regarding the development of language.

1. Scholars say that hominids use first used gestures.
2. Another group argues that sounds (words) of emotional interjections preceded language.
3. There are others who believe that language developed from the calls (sounds) that primates made to one another. In the beginning, there were different sounds. Slowly they developed into a language.

Question 14.
What is anthropology?
Answer:
It is a branch of knowledge that studies human culture and the evolutionary levels of humans/ Anthropos in Greek means man and logos means study. Anthropology is the study of various aspects of humans within past and present societies.

Question 15.
What is the Glacial Age?
Answer:
Once the earth surface was covered with thick ice sheets. This phenomenon is known as the Glacial Age. In the history of the earth there were 4 Such glacial ages. The last glacial age ended some 13,000 years ago. The period between two Glacial Ages is called Inter Glacial Period.

You can Download Nomadic Empires Questions and Answers, Notes, Plus One History Chapter Wise Questions and Answers Kerala Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Plus One History Chapter Wise Questions and Answers Chapter 5 Nomadic Empires

Question 1.
“The term ‘Nomadic Empires’ might look paradoxical.” Examine the validity of this statement.
Answer:
Nomads are wanderers. They are organized as family groups. There is hardly any difference in their; economic life. Their political system is very ancient and uncivilized. But Empires have physical territories . and boundaries. Politically they are stable. The stability of an empire comes from its complex social and economic structures. It rules a vast region. Naturally, an empire should have an administrative system.

Question 2.
The History of Mongols is written by foreign scholars. Explain.
Answer:
It was Russian scholars who made the most valuable researches about Mongols in the 18th arid 19th centuries. The history of Mongols comes in the form of extensive notes prepared by travelers, merchants, warriors, and collectors of antiquity.

Question 3.
The goal of Genghis Khan was not merely a fellowship of the Mongol tribes. What were his other aims?
Answer:
Genghis. Khan unified the Mongol people. He reorganized them into a disciplined military power. It helped him in future conquests. His primary aim was to attack China.

Question 4.
The Mongols did not do any agriculture. On the basis of this statement explain what the means of livelihood of the Mongols were.
Answer:
Many Mongols were shepherds whereas others were hunters and gatherers of food. Shepherds had. domesticated horses and sheep. They also tamed cattle, goats, and camels. In the grassy plains of Central Asia (the modem Mongolia), they lived a nomadic life. It was a beautiful region with snow-capped mountains, plains, rivers and a desert (Gobi). The hunters and the gatherers of food lived in the Northern Siberian forests. Compared to the shepherds they were very humble. During the summer they lived by selling animal hides. The climate in their region was harsh. They had (Orig winter and brief, dry summer.

The Mongols did not do any farming. Their economic system was not capable of maintaining places with high or dense population. Therefore they did not have r any cities.

The Mongols lived in tents. They went on roaming the grassy plains with their flocks and herds both in the winter and also summer.

Question 5.
There were some links that kept the various sections and tribes of the Mongols together. Explain.
Answer:
The Mongols included various sections and tribes. There were Tatars, Khitans, Manchus, and Turkish tribes among them. The main link among them was their common language.

Question 6.
The Great Wall of China was built to protect the agricultural communities of North China from the attacks of the nomads, Based on this statement, describe the relations between the Mongolian nomads and the Chinese agricultural communities.
Answer:
China experienced a lot of problems because of the constant attacks by the nomads and therefore China built fortresses to protect her citizens from these attacks. Connecting these fortresses, China made defensive wall-chain. This is the famous Great Wall of China. This is one of the wonders of the world.

Question 7.
The life of Genghis Khan was full of misery and backlashes. Comment.
Answer:
The real name of Genghis khan is Temujin. He was born in 1162, in a region of the bank of River On on in the Northern side of the present Mongolia. When he was 12, his father was killed. Then it was his mother Hoelun who brought him and brothers up taking a lot of trouble. Temujin faced a lot of problems during the next 12 years. He was caught and was made a slave. Soon after his marriage his wife Borte was kidnapped by some people. Temujin had to fight hard to get his wife back.

By 1209 Genghis Khan defeated the Xi Xia people, in 1213 he crossed the Great Wall and defeated the Chin dynasty. He looted Peking. His fights with the Chin dynasty continued until 1234, Genghis Khan also attacked places like Amu Darya, Transoxiana, and Khwaresmia.

Question 8.
The major part of Genghis Khan’s life was spent on the battlefront. Do you agree with this view? Explain.
Answer:
It is true that the major part of Genghis Khan’s life was spent on the battlefront. His military successes are quite wonderful.
He used new strategies. He also changed traditional strategies that were used in the warfare in the grassy plains. The expertise of the Mongols and Turks in horse-riding gave his army speed and dynamism. These warriors could shoot even as they were riding their horses.

The cavalry in the plains was ready to move with great speed and face any type of weather. The rivers that were frozen during the winter were like highways for the warriors of Genghis Khan and they could easily enter the cities and camps of the enemies.

For the nomads, the fortressed camps of the enemies were hard to conquer. There they suffered huge losses. But Genghis Khan did not mind these obstacles. His engineers made machines to capture the fortresses. They also made firebombs which could be easily carried and used when needed. By using these technically advanced things Genghis Khan was able to defeat his enemies.

Question 9.
Under the successors of Genghis Khan, the Mongol lost their Western world. Give reasons for this.
Answer:
In the decades after 1203, the Mongolian army faced many defeats. In the 1260s, the Mongols lost their desire to maintain their Western regions. Vienna, Western Europe, and Egypt were once with the Mongols. But their withdrawal from the Hungariari Steppes and their defeat from the Egyptian army caused some new political ideas to emerge in their minds. The internal strife among the Mongols themselves and their over-enthusiasm in conquering China made them lose their Western world.

Question 10.
Prepare a seminar paper on the social-political and military arrangements of the Mongols.
Areas to be considered: Military structure, Courier system, Mongols and the Permanent Settlers, Formation of special hereditary system, Yassa.
Answer:
All the healthy males among the Mongols carried arms. In times of need, they served as an army. In short, the Mongolian army was small and uni-tribal. But with the unification of the different Mongolian tribes and with the wars with different peoples, there were changes in the nature and structure of Genghis Khan’s army. The army became big and multi-tribal. In the army, there were soldiers who accepted the authority of Genghis Khan willingly, like the Turkic Uyghurs and defeated the people like the Keraits. Genghis Khan unified the different tribes of Mongols and made them into a confederacy. He tried to destroy the earlier tribal identities of these tribes.

Genghis Khan organized his army on a decimal basis. The units were in multiples of 10. (10,100, 1000,10000, etc.)

The greatest contribution of Genghis Khan was the courier system called Yam he introduced. This Yam system was a relay system that linked different administrative units of his vast empire. At fixed distances, there were horsemen and messengers for carrying messages. To maintain this communication system, the nomadic Mongols had to give one-tenth of their animals (horses or other animals) to the authorities. This was known as Qubkar (kar means tax-likeour’karam’inMalayalam).

After the death of Genghis Khan, the courier system became more efficient. Its speed and reliance had surprised visitors. The great Khans used this relay system to effectively control their far-flung regions. The defeated people were not happy with the new nomadic rulers. After the attacks that took place in the first half of the 13th century, cities were destroyed, farms were left uncultivated and trade and handloom industries were reduced.

Thousands of people were. killed and many more were made prisoners. Right from the top to the bottom, people suffered different kinds of misery and pain. Since the canals in the internal region of the Iranian plateau were not repaired, the desert expanded. This caused environmental min. A good portion of the Khurasan Region never. recovered from this damage.

Once the attacks were over Europe and China were regionally connected. The trade relations between the two were also better. The trade and travel through the silk route reached their height under the Mongols, But the trade routes did not end with China. They extended to Mongolia, the heart of the Empire and Karakoram. Travel and communication were essential for the stability of the Mongol rule. The travelers were given a pass for their easy travel. Merchants had to give a tax (called Baj tax) for this.
In the 13^th century, the contradictions that existed between the nomads and the permanent settlers began to lessen.

The memories of Genghis Khan were cherished by his successors. It was his Yassa (Law) that helped him to be remembered by posterity. In the Assembly of Chiefs (Quriltai) in 1206 Genghis Khan declared his Yassa. It contained administrative controls and laws regarding the organization of hunting, army and postal system. By the middle of the 13^th century, the Mongols began to use the word Yassa to mean the “Laws of Genghis Khan”. Thus the Yassa of Genghis Khan helped him in keeping memory alive.

Question 11.
Sagi: Genghis Khan was an uncivilized attacker.
Usha: He was a great leader.
With whose opinion do you agree? Why?
Answer:
When we think of Genghis khan now, in our imagination, we imagine him to be a violent destroyer of cities and a brute who caused the death of thousands of people. In the 13th century, the city dwellers of China, Iran, and Eastern Europe looked at the Mongols with anger and hatred. But for the Mongols Genghis Khan was their greatest leader. He unified them.

He liberated them from the constant Tribal wars and the exploitation of the Chinese. To the Mongols, he was a great man who brought them prosperity, who formed an intercontinental empire, who recovered the trade routes and markets and who attracted travelers like Marco Polo.

Question 12.
The Mongols have given valuable contributions to world culture. Critically evaluate this statement.
Answer:
The Mongol Khans belonged to different. religious faiths. There were Believers of Shamanism, Buddhists, Christians, and Muslims among them. They did not impose their personal beliefs on others. The Mongol rulers employed members of all races and faiths in their administration and army. Theirs was a multi-racial, multi-lingual and multi-religious rule. Such a rule was quite unusual in those days. The Mongols followed an administrative system that could be imitated by the rulers like the Mughals in India

Students can Download Chapter 4 Biomolecules Notes, Plus One Zoology Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Notes Chapter 4 Biomolecules

How To Analyse Chemical Composition?
Take any living tissue (a vegetable or a piece of liver, etc.) and grind it in trichloroacetic acid (Cl₃CCOOH) using a mortar and a pestle. The thick slurry is formed. Then it is passed through a cheese cloth or cotton getting two fractions.

1. Acid soluble fraction (Filtrate)
2. Acid-insoluble fraction.
   • All the carbon compounds from living tissues are called ‘biomolecules’.
   • The tissue is fully burnt, all the carbon compounds are oxidised to gaseous form (C02, water vapour) and are removed.
   • The remaining is called ‘ash’. It contains inorganic elements (like calcium, magnesium etc).
   • Inorganic compounds like sulphate, phosphate, etc., are also seen in the acid-soluble fraction.

Organic compounds under biological view are classified into

• Amino acids:
• Nucleotide bases
• Fatty acids etc.

Amino acids:

1. They are organic compounds containing four substituent groups occupying the four valency positions.
2. These are hydrogen, carboxyl group, amino group and a variable group designated as R group.

Based on the nature of R group there are many amino acids. However, those which occur in proteins are only of twenty one types.
The R group may be

• Hydrogen (the amino acid is called glycine)
• A methyl group (alanine)
• Hydroxyl methyl (serine), etc.

Based on number of amino and carboxyl groups, there are

1. Acidic (eg: glutamic acid) 2. Basic (lysine) and neutral (valine) amino acids 3. Aromatic amino acids (tyrosine, phenylalanine, tryptophan)

A particular property of amino acids is due to ionizable nature of —NH₂ and —COOH groups

Fatty acid:
It has a carboxyl group attached to an R group.
The R group could be

1. A methyl (—CH₂)
2. Ethyl (—C₂H₅)

Carbon number varies in different fatty acids:

• Palmitic acid – 16 carbon atoms
• Arachidonic acid – 20 carbon atoms

Fatty acids are

1. saturated (without double bond) 2. unsaturated (with one or more C = C double bonds)

• Lipids possess both glycerol and fatty acids.
• They are monoglycerides, or diglycerides or triglycerides.
• These are also called fats and oils based on melting point. Oils have lower melting point eg: gingely oil.
• Some lipids have phosphorous, they are called phospholipids. They are found in cell membrane. eg: lecithin

Nitrogen bases:

• They are (heterocyclic rings) adenine, guanine, cytosine, uracil, and thymine
• If they are found attached to a sugar, they are called nucleosides.
• If a phosphate group is found esterified to the sugar, they are called nucleotides.
• Nucleic acids like DNA and RNA consist of nucleotides only.

Adenosine, guanosine, thymidine, uridine and cytidine are nucleosides. Adenylic acid, thymidylic acid, guanylic acid, uridylic acid and cytidylic acid are nucleotides.

Diagrammatic representation of small molecular weight organic compounds in living tissues.

Primary And Secondary Metabolites:
Primary metabolites:
Organic compounds such as amino acids, sugars, etc.are belongs to primary metabolites. Primary metabolites play important role in normal physiologial processes.

Secondary metabolites:
When analyse plant, fungal and microbial cells the alkaloides, flavonoides, rubber, essential oils, antibiotics, coloured pigments, scents, gums, spices etc are found. These are called secondary metabolites. Many of them are useful to ‘human welfare’ (eg: rubber, drugs, spices, scents and pigments).

Pigments	Carotenoids, Ant.hocyanins, etc.
Alkaloids	Morphine, Codeine, etc.
Terpenoides	Monoterpenes, Diterpenes etc.
Essential oils	Lemon grass oil, etc.
Toxins	Abrin, Ricin
Lectins	Concanavalin A
Drugs	Vinblastin, curcumin, etc.
Polymeric substances	Rubber, gums, cellulose

Biomacromolecules:
The acid insoluble fraction, has only four types of organic compounds i.e., proteins, nucleic acids, polysaccharides and lipids. These compounds, except lipids, have molecularweights in the range often thousand daltons and above.

Lipids, whose molecularweights do not exceed 800 Da, come under acid insoluble fraction. Hence Lipids are not macromolecules.

Biomicromolecules and biomacromolecules:
Molecular weights less than one thousand dalton are referred to as micromolecules or simply biomolecules while those which are found in the acid insoluble fraction are called macromolecules or biomacromolecules.

Component	% of the total cellular mass
Water	70 – 90
Proteins	10 – 15
Carbohydrates	3
Lipids	2
Nucleie acid	5 – 7
Ions	1

Proteins:
Proteins (heteropolymer)are linear chains of amino acids linked by peptide bonds i.e polymer of amino acids There are 21 types of amino acids (eg: alanine, cysteine, proline, tryptophan, lysine, etc.)
Some Proteins and their Function:

• Dietary proteins are the source of essential amino acids.
• Therefore, amino acids are essential or non-essential.
• Essential amino acids obtained through food.

Proteins carry out many functions in living organisms:

1. some transport nutrients across cell membrane
2. some fight infectious organisms
3. Collagen is the most abundant protein in animal world and
4. Ribulose bisphosphate Carboxylase – Oxygenase (RUBISCO) is the most abundant protein in the biosphere.

Protein	Functions
Collagen	Intercellular ground substance
Trypsin	Enzyme
Insulin	Hormone
Antibody	Fights infectious agents
Receptor	Sensory reception (smell, taste, hormone, etc.)
GLUT-4	Enables glucose transport into cells

POLYSACCHARIDES
1. Polysaccharides are long chains of sugars.

2. Forexamplecellulose(homopolymer)is a polysaccharide consist of only one type of monosaccharide i.e. glucose.

3. Starch is store house of energy in plant tissues but animals have glycogen as energy source.

4. Inulin is a polymer of fructose.

5. In a polysaccharide eg glycogen, the right end is called the reducing end and the left end is called the non reducing end.

Starch forms helical secondary structures:

1. Starch can hold l₂ molecules in the helical portion. This reaction product blue in colour.
2. Cellulose does not contain complex helices and hence cannot hold l₂.
3. Cotton fibre is cellulose
4. The complex polysaccharides have as building blocks such as amino-sugars (eg: glucosamine, N— acetyl galactosamine, etc.).
5. Exoskeletons of arthropods have a complex polysaccharide called chitin (heteropolymers)

Nucleic Acids:
Nucleic acids are the another macromolecule that found in the acid insoluble fraction of living tissues. For nucleic acids, the building block is a nucleotide.

Components of nucleic acid:

1. Heterocyclic compound(adenine, guanine, uracil, cytosine and thymine).
2. Monosaccharide and
3. A phosphoric acid or phosphate.

The sugar found in polynucleotides is either ribose (a monosaccharide pentose) or 2’ deoxyribose.

Nature of pentose sugar in DNA and RNA:
A nucleic acid containing deoxyribose is called deoxyribonucleic acid (DNA) while that which contains ribose is called ribonucleic acid (RNA).

Structure Of Proteins (Proteins are heteropolymers containing many amino acids):
Primary structure:
The sequence of amino acid in which the left end represented by the first amino acid (N— terminal amino acid )the right end represented by the last amino acid (C— terminal amino acid). This sequence forms linear structure. It is called the primary structure.

Primary structure of a portion of a hypothetical protein. N and C refer to the two termini of every protein. Single letter codes and three letter abbreviations for amino acids are also indicated.

Secondary structure:
The primary structure have rigid rod like appearance which is folded in the form of a helix (similar to a revolving staircase). It appears as right handed helices. It is called the secondary structure. Secondary structures exhibited by DNA is the Watson-Crick model. In this DNA exists as a double helix.

Tertiary structure:
The long protein chain is also folded upon itself like a hollow wollen ball, it called the tertiary structure. This gives us a 3-dimensional view of a protein. Tertiary structure is necessary for the many biological activities of proteins.

Quaternary structure:
Some proteins are assembled by more than one polypeptide chains .This is called the quaternary structure Adult human haemoglobin consists of 4 subunits. Two of these are identical to each other. Hence, two subunits are of a type and two subunits are of p type together constitute the human haemoglobin (Hb).

Nature Of bond linking Monomers In A Polymer:
1. Peptide bond:
In a protein, amino acids are linked by a peptide bond which is formed when the carboxyl (—COOH) group of one amino acid reacts with the amino (-NH₂) group of the next amino acid with the elimination of a water.

2. Glvcosidic bond:
In a polysaccharide the individual monosaccharides are linked by a glycosidic bond. This bond is also formed by dehydration.

3. Phosphodiester Bond:
In a nucleic acid a phosphate moiety links the 3′-carbon of one sugar of one nucleotide to the 5′-carbon of the sugar of the succeeding nucleotide. The bond between the phosphate and hydroxyl group of sugar is called phosphodiester bond

DNA Structure:

1. The two strands of polynucleotides are antiparallel i.e., run in the opposite direction.
2. The backbone is formed by the sugar-phosphate-sugar chain.
3. The nitrogen bases are A and G of one strand base pairs with T and C, respectively
4. There are two hydrogen bonds between A and T but three hydrogen bonds are present between G and C.
5. Each strand appears like a helical staircase.
6. At each step of ascent, the strand turns 36°.
7. One full turn of the helical strand have ten steps or ten base pairs.
8. The pitch is 34A°. The distance between each base pairs is 3.4A°.
9. This form of DNA is called B-DNA.

Dynamic State Of Body Constituents – Concept Of Metabolism:
Biomolecules are constantly being changed into some other biomolecules and also made from some other biomolecules. This is called turn over. This breaking and making is through chemical reactions constantly occurring in living organisms called as metabolism.

Metabolic reactions and transformation of biomolecules:

1. removal of CO₂ from amino acids making an amino acid into an amine,
2. removal of amino group in a nucleotide base and
3. hydrolysis of a glycosidic bond in a disaccharide
   • Majority of these metabolic reactions are always linked to some other reactions. This series of linked reactions called metabolic pathways.
   • These metabolic pathways are similar to the automobile traffic in a city.
   • Another feature of these metabolic reactions is that every chemical reaction is a catalysed reaction.
   • The catalysts which hasten the rate of a given metabolic conversation are also proteins. These proteins with catalytic power are named enzymes.

Metabolic Basis For Living:

1. Metabolic pathways involves two processes The synthesis step is called anabolic pathways. The degradation step is called catabolic pathways.
2. Catabolic pathways lead to the release of energy.
3. For example, when glucose is degraded to lactic acid in our skeletal muscle, energy is liberated which stored in the form of chemical bonds, when needed, this bond energy is utilized.

Which is the energy currency of a cell?

• The energy currency in living systems is the bond energy in a chemical called adenosine triphosphate (ATP).

The Living State:

• All living organisms exist in a steady-state characterised by concentrations of each of these biomolecules.
• These biomolecules are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium.
• The steady state is a non-equilibirium state. The systems at equilibrium cannot perform work. As living organisms work continuously, they cannot afford to reach equilibrium.

Hence the living state is a non-equilibrium steady-state to be able to perform work.

Metabolism provides a mechanism for the production of energy. Hence the living state and metabolism are synonymous. Without metabolism there cannot be a living state.

Enzymes:

Almost all enzymes are proteins. Some nucleic acids that behave like enzymes are called ribozymes

Enzvme activity:

• The tertiary structure is biologically active, an active site of an enzyme is a crevice or pocket into which the substrate fits.
• Thus enzymes, through their active site, catalyse reactions at a high rate.
• Enzymes are damaged at high temperatures (say above 40°C).
• Some enzymes isolated from organisms who normally live under extremely high temperatures (eg: hot vents and sulphur springs), are stable and retain their catalytic power even at high temperatures (upto 80° – 90°C).
• Thermal stability is thus an important quality of such enzymes isolated from thermophilic organisms.

Chemical Reactions:
Chemical compounds undergo two types of changes.
1. Physical change:
It involves the change in shape without breaking of bonds. eg: when ice melts into water, or when water becomes a vapour.

2. Chemical reaction/change:
When bonds are broken and new bonds are formed during transformation, this will be called a chemical reaction.Eg. Hydrolysis of starch into glucose is an organic chemical reaction. Rate of a physical or chemical process refers to the amount of product formed per unit time.

Role of enzvme in the rate of chemical reaction:
In the absence of any enzyme this reaction is very slow, with about 200 molecules of H₂CO₃ being formed in an hour. But using an carbonic anhydrase, the reaction speeds dramatically with about 600,000 molecules being formed every second.

The enzyme has accelerated the reaction rate by about 10 million times. A multistep chemical reaction, when each of the steps is catalysed by the same enzyme complex or different enzymes, is called a metabolic pathway.

1. In glycolysis glucose becomes pyruvic acid through ten different enzyme catalysed metabolic reactions.
2. Under normal aerobic conditions, pyruvic acid is formed.
3. In yeast, during fermentation, the same pathway leads to the production of ethanol (alcohol).
4. In our skeletal muscle, under anaerobic conditions, lactic acid is formed.

How do Enzymes bring about High Rates of Chemical Conversions?
The chemical which is converted into a product is called a ‘substrate’. Hence enzymes, i.e. proteins with three dimensional structures including an ‘active site’ convert a substrate (S) into a product (P).

What is the transition state?
During the state where substrate is bound to the enzyme active site, a new structure of the substrate called unstable transition state is formed. Then the bond breaking/making is completed, the product is released from the active site. The y-axis represents the potential energy content.

The x-axis represents the progression of the structural transformation or states through the ‘transition state’. If ‘P’ is at a lower level than ‘S’, the reaction is an exothermic reaction one need not to supply energy (by heating) in order to form the product.

However, whether it is an exothermic or spontaneous reaction or an endothermic or energy requiring reaction, the ‘S’ has to go through a much higher energy state or transition state. The difference in average energy content of ‘S’ from that of this transition state is called ‘activation energy’.

Enzymes bring down energy barrier making the transition of ‘S’ to ‘P’ more easy. Catalysed reactions proceed at rates faster than that of uncatalysed ones.

Nature of Enzyme Action:

Each enzyme (E) has a substrate (S) binding site in its molecule so that a highly reactive enzyme-substrate complex (ES) is produced. This complex is short-lived and dissociates into its products.

The catalytic cycle of an enzyme action can be described in the following steps:

1. First, the substrate binds to the active site of the enzyme. 2. The binding of the substrate induces the enzyme to alter its shape. 3. The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme- product complex is formed. 4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate.

Factors Affecting Enzyme Activity:

The activity of an enzyme can be affected by temperature, pH, change in substrate concentration.
1. Temperature and pH:
Each enzyme shows its highest activity at a particular temperature and pH called the optimum temperature and optimum pH. Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.

2. Concentration of Substrate:
With the increase in substrate concentration, the velocity of the enzymatic reaction rises at first. The reaction ultimately reaches a maximum velocity (V_max) which is not increased by further rise in concentration of the substrate because the enzyme molecules are saturated there are no free enzyme molecules to bind with the additional substrate molecules

Enzyme inhibition:
When the binding of the chemical shuts off enzyme activity, the process is called inhibition and the chemical is called an inhibitor. When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. eg: Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure. Such competitive inhibitors are often used in the control of bacterial pathogens.

Classification and Nomenclature of Enzymes:
Enzymes are divided into 6 classes.
1. Oxidoreductases/dehvdroaenases:
Enzymes which catalyse oxidoreduction between two substrates S and S’ eg:

2. Transferases:
Enzymes catalysing a transfer of a group, G (other than hydrogen) between a pair of substrate S and S’ eg:
\(\mathbf{S}-\mathbf{G}+\mathbf{S}^{‘} \longrightarrow \mathbf{S}+\mathbf{S}^{‘}-\mathbf{G}\)

3. Hydrolases:
Enzymes catalysing hydrolysis of ester, ether, peptide, glycosidic, C – C, C – halide or P – N bonds.

4. Lyases:
Enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.

5. Isomerases:
Includes all enzymes catalysing inter-conversion of optical, geometric or positional isomers.

6. Lyases:
Enzymes catalysing the linking together of 2 compounds, eg: enzymes which catalyse joining of C – O, C – S, C – N, P – O etc. bonds.

Co-factors:
Enzymes are composed of one or several polypeptide chains and non-protein constituents called cofactors. They are bound to the enzyme to make the enzyme catalytically active. The protein part of the enzymes is called the apoenzyme.
Three kinds of cofactors are

1. prosthetic groups
2. co-enzymes
3. Metal ions.

1. Prosthetic groups:
They are organic compounds that are tightly bound to the apoenzyme. For example, in peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen. Haem is the prosthetic group and it is a part of the active site of the enzyme.

2. Co-enzymes:
They are also organic compounds loosely bound to apoenzyme for catalysis. Co-enzymes serve as co-factors in a number of different enzyme-catalyzed reactions. Many coenzymes are vitamins eg: coenzyme nicotinamide adenine dinucleotide (NAD) and NADP contain the vitamin niacin.

2. Metations:
Zinc is a cofactor for the proteolytic enzyme carboxypeptidase. Catalytic activity is lost when the co-factor is removed from the enzyme.

Students can Download Chapter 7 Permutation and Combinations Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 7 Permutation and Combinations

Plus One Maths Permutation and Combinations Three Mark Questions and Answers

Question 1.
Find the number of different signals that can be made by arranging at least three flags in order on a vertical pole, if 6 different coloured flags are available.
Answer:

3 flags	6 × 5 × 4 = 120 ways
4 flags	6 × 5 × 4 × 3 = 360 ways
5 flags	6 × 5 × 4 × 3 × 2 = 720 ways
6 flags	6 × 5 × 4 × 3 × 2 × 1 = 720 ways

Hence the number of different atleast 3 flag signals = 120 + 360 + 720 + 720 = 1920.

Question 2.
2. Find the value of n such that

1. nP₅ – 42 × nP₃, n>4
2. (n -1 )P₃: nP₄ = 1 : 9 (3 score each)

Answer:
1. Given; nP₅ = 42 × nP₃
⇒ n(n – 1 )(n – 2)(n – 3)(n – 4) = 42 × n(n – 1)(n – 2)
⇒ (n – 3)(n – 4) = 42
⇒ n₂ – 7n + 12 = 42
⇒ n₂ – 7n – 30 = 0
⇒ (n – 10)(n + 3) = 0
⇒ n = 10; n = -3
The acceptable value is n = 10.

2. Given; 9 × ^{(n – 1)}P₃ = ⁿP₄

Question 3.
Find the value of r if

1. 5 × 4P_r = 6 × 5P_{r – 1}
2. 5P_r = 6P_{r – 1} (3 score each)

Answer:
1. Given; 5 × 4P_r = 6 × 5P_{r – 1}

⇒ 30 – 11r + r² = 6
⇒ r² – 11r + 24 = 0
⇒ (r – 8)(r – 3) = 0
⇒ r = 8, 3
The acceptable answer is r = 3.

2. Given; 5P_r = 6P_{r – 1}

⇒ 42 – 13r + r² = 6
⇒ r² – 13r + 36 = 0
⇒ (r – 9)(r – 4) = 0
⇒ r = 9, 4
The acceptable answer is r = 4.

Question 4.
The letters of the word TUESDAY are arranged in a line, each arrangement ends in S.

1. How many different arrangements are possible? (2)
2. How many of them start with letter D? (1)

Answer:

1. In the word TUESDAY there are 7 letters. When word end in S, there are only 6 possible arrangements. This can be done in 6! = 720
2. The word start with D and end in S, this can be done in 5! = 120.

Question 5.
Consider the word ANNAMALAI

1. How many new words can be formed using the given words? (2)
2. Among the new words how many of them will begin with A and end with I. (1)

Answer:

1. In the word ANNAMALAI there are 9 letters, of which A appears 4 times, N appears 2 times and the rest ail are different. Therefore the total number of ways is \(\frac{9 !}{4! \times 2 !}\) = 7560.
2. The word start with A and end in I, this can be done in \(\frac{7 !}{3! \times 2 !}\) = 420.

Question 6.
Find the rank of the word NAAGI, if these words are written as in a dictionary.
Answer:
The order of the letters will be A, A, G, I, N

Therefore the position of ‘NAAGI’ is 24 + 12 + 12 + 1 = 49.

Question 7.
A committee of 3 persons is to be constituted from a group of 2 men and 3 women.

1. In how many ways can be done? (1)
2. How many of these committees would consist of 1 man and 2 women? (2)

Answer:
1. The required number of ways
= 5C₃ = 5C₂ = \(\frac{5 \times 4}{1 \times 2}\) = 10.

2. One man can be selected in 2C_1. 2 women can be selected in 3C₂
Therefore required number of ways
= 2C₁ × 3C₂ = 2 × 3 = 6.

Question 8.
It was found at a certain dinner meeting that after every member had shaken hand with every other members, 45 handshakes were interchanged. How many members were present at the metting?
Answer:
Let n be the member of person present in the meeting. The total number of handshakes is same as the number of ways of selecting 2 persons from among n persons.
The total number of handshakes = nC₁ = 45
⇒ \(\frac{n(n-1)}{1 \times 2}\) = 45 ⇒ n² – n – 90 = 0
⇒ (n – 10)(n + 9) = 0
⇒ n = 10, -9
n = 10 is acceptable.

Question 9.
In an exam, Arjun has to select 4 questions from each part. There are 6, 7 and 8 question in Part I, Part II and Part III, respectively. What is the number of possible combinations in which he can choose the question?
Answer:
Selecting 4 questions from Part I = 6C₄ = 6C₂
Selecting 4 questions from Part II = 7C₄ = 7 C₃
Selecting 4 questions from Part III = 8C₄
The required number of ways
= 6C₂ × 7C₃ × 8C₄ = 15 × 35 × 70 = 36750.

Plus One Maths Permutation and Combinations Four Mark Questions and Answers

Question 1.
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

1. all vowels occur together. (2)
2. all vowels do not occur together. (2)

Answer:
1. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

2. Number of words all vowels do not occur together = Total number of different words – number of words in which vowels come together
= 8! – 6! × 3! = 6!(8 × 7 – 6) = 2 × 6!(28 – 3)
= 50 × 6! = 36000.

Question 2.
How many permutations are there of the 11 letters in MISSISSIPPI

1. taken all together? (2)
2. all the I’s not come together? (2)

Answer:
1. In the word MISSISSIPPI there are 11 letters, of which S appears 4 times, I appears 4 times, P appears 2 times and the rest all are different.
Therefore the total number of ways is \(\frac{11 !}{4 ! \times 4 ! \times 2 !}\) = 34650.

2. 4 I’s are kept together and should be counted as one unit, then there are 8 units. The number of ways is \(\frac{8 !}{4 \times 2 !}\) = 840. Therefore the I’s not come together = Total arrangements – 4 I’s together.
= 34650 – 840 = 33810.

Question 3.
Find the number of arrangements of 6 boys and 5 girls in a row so that

1. no two girls sit together. (2)
2. boys and girls occupy alternate positions. (2)

Answer:
1. Since no two girls sit together, we have first arrange the 6 boys among themselves. This can be done in 6! ways.

Now no two girls sit together if we place the girls in between boys. There are 7 places and it should be occupied by 5 girls, can be done in 7P₅ ways. Therefore the total number of ways is 6! × 7P₅ = 720 × 7 × 6 × 5 × 4 × 3 = 1814400.

2. Boys and girls occupy alternate position can be done as follows. First place the boys whose number is large.

Boys can be arranged in 6! ways. The place between boys can be filled by 5 girls, can be done in 5! ways. Therefore the total number of ways is 6! × 5! = 720 × 120 = 86400.

Question 4.
If the letters of the word DHRONA be permuted and arranged as in a dictionary, find the rank of the word.
Answer:
The order of the letters will be A, D, H, N, O, R


Therefore the position of ‘DHRONA’ is 120 + 24 + 6 + 6 + 6 + 2 + 2 + 1 + 1 = 168.

Question 5.
If the letters of the word MOTHER be permuted and arranged as in a dictionary, find the rank of the word.
Answer:
The order of the letters will be E, H, M, O, R, T

Therefore the position of ‘MOTHER’ is 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309.

Question 6.
How many

1. Straight line (2)
2. Triangles can be formed by joining 12 points, 4 of which are collinear. (2)

Answer:
1. Number of straight lines that can be formed using 12 points = 12C₂ = \(\frac{12 \times 11}{1 \times 2}\) = 66. Number of straight lines that can be formed using 4 collinear points = 4C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6.  Since the 4 points are collinear, the required number of lines = 66 – 6 + 1 = 61.

2. Number of triangles that can be formed using 12 points = 12C₃ = \(\frac{12 \times 11 \times 10}{1 \times 2 \times 3}\) = 220.
Number of triangles that can be formed using 4 collinear points = 4C₃ = 4. Since the 4 points are collinear, the required number of triangles = 220 – 4 = 216.

Question 7.
From 7 men and 4 ladies a committee of 6 is to be formed. In how many ways can this be done when the committee contains

1. exactly two ladies. (2)
2. at least two ladies. (2)

Answer:
1. Exactly two ladies can be selected from 4 in 4C₂ = \(\frac{4 \times 3}{1 \times 2}\) = 6. The remaining 4 should be selected from 7 men in 7C₄ = \(\frac{7 \times 6 \times 5 \times 4}{1 \times 2 \times 3 \times 4}\) = 35. The required number of ways = 6 × 35 = 210.

2. Atleast 2 ladies can be selected as follows;

The required number of ways = 210 + 140 + 21 = 371.

Question 8.
A box contains 6 apples, 5 oranges, and 8 mangoes.

1. In how many ways a fruit is selected from the box. (1)
2. In how many different ways can an apple and an orange be selected. (1)
3. In how many different ways a person take one apple, one orange, and one mango. (2)

Answer:

1. The box contains 6 + 5 + 8 = 19 fruits, from this one fruit can be selected in 19C₁ = 19 ways.
2. An apple is to be selected from 6 apples and an orange be selected from 5 oranges. The required number of ways = 6C₁ × 5C₁ = 30.
3. An apple is to be selected from 6 apples, an orange is to be selected from 5 oranges and one mango is to be selected from 8 mangoes. The required number of ways = 6C₁ × 5C ₁ × 8C₁ = 240.

Question 9.
A student has to answer 6 out of 10 questions which are divided into two parts containing 5 questions each and he is permitted to attempt not more than 4 from any group. In how many ways can he make up his choice?
Answer:
The different possibilities are mentioned below;

The required number of ways
= 5C₁ × 5C₂ + 5C₃ × 5C₃ + 5C₂ × 5C₁
= 5 × 10 + 10 × 10 + 10 × 5 = 200.

Question 10.

1. How many chords can be drawn through 15 points on a circle? (2)
2. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected? (2)

Answer:

1. Number of chords = ¹⁵C₂ = \(\frac{15 \times 14}{1 \times 2}\) = 105
2. Selection of 2 black and 3 red balls = ⁵C₂ × ⁶C₃ = 10 × 20 = 200.

Question 11.

1. If ⁿC₂ = ⁿC₈ then find n
2. Find n if ⁿP₅ = 42 × ⁿP₃; n> 4

Answer:
1. n = 2 + 8 = 10.

2. DAUGHTER this word has 8 different letters. A, U, E are the vowels. Treat these 3 as one unit, then there are 6 units and can be permuted in 6! ways. The above vowels can be permuted in 3! ways. Hence the total number of words is 6! × 3! = 4320.

Plus One Maths Permutation and Combinations Six Mark Questions and Answers

Question 1.
Find the arrangements of letters of the word INDEPENDENCE. In how many of these arrangements,

1. do the words start with P. (2)
2. do all the vowels always occur together. (2)
3. do the vowels never occur together. (1)
4. do the words begin with I and end in P? (1)

Answer:
1. In the word INDEPENDENCE there are 12 letters, of which N appears 3 times, E appears 4 times, D appears 2 times and the rest all are different.

When the words start with P, then there are 11 letters to be filled in 11 spaces. Therefore the total number of ways is \(\frac{11 !}{3! \times 2 ! \times 4 !}\) = 138600.

2. The vowels EEEEI are to be kept together and should be treated as one unit. Then these vowels can be arranged in \(\frac{5 !}{4 !}\) ways. This single unit together with 7 letter will count to units, can be arranged in \(\frac{8 !}{3! \times 2 !}\). Therefore the total number of ways \(\frac{5 !}{4 !} \times \frac{8 !}{3 ! \times 2 !}\) = 16800.

3. Number of ways of arrangement with vowels do not come together = Total arrangement – vowels coming together.
= \(\frac{12 !}{3 ! \times 2 ! \times 4 !}\) – 16800 = 1663200 – 16800 = 1646400.

4. When the words start with I and ends with P, then there are 10 letters to be filled in 10 spaces. Therefore the total number of ways is \(\frac{10 !}{3 ! \times 2 ! \times 4 !}\) = 12600.

Question 2.
Consider the word ASSASSINATION.

1. How many permutations are there of the letters of the given word? (2)
2. How many different ways can be arranged so that the 4S’s come together? (2)
3. How many different ways can be arranged so that the 4S’s do not come together? (1)
4. How many begin with A? (1)

Answer:
1. In the word ASSASSINATION there are 13 letters, of which A appears 3 times, S appears 4 times, N appears 2 times, I appears 2 times and the rest all are different. Therefore the total number of ways is \(\frac{13 !}{3! \times 4 ! \times 2 ! \times 2 !}\) = 10810800.

2. 4 S’s are kept together and should be counted as one unit, then there are 10 units. The number of ways is \(\frac{10 !}{3! \times 2 ! \times 2 !}\) = 151200.

3. Number of words in which 4S’s do not come together = Total number of words – 4S’s together = 10810800 -151200 = 10659600.

4. The word will start with any one of the 4 A’s. Then total letter arrange will be 12. Number of words in which begin with A \(\frac{12 !}{2! \times 4 ! \times 2 ! \times 2 !}\) = 2494800.

Question 3.
A team of 11 cricket players is to be chosen from 15 players. In how many ways can this be done so as to:

1. Include a particular player A. (2)
2. Exclude a particular player B. (2)
3. Include A and exclude B. (2)

Answer:
1. A particular player A is to be included, then selection of 10 is to be made from 14 players.
The required number of ways 14C₁₀ = 14C₄ = \(\frac{14 \times 13 \times 12 \times 11}{1 \times 2 \times 3 \times 4}\) = 7 × 13 × 11 = 1001.

2. A particular player B is to be excluded, then selection of 11 is to be made from 14 players.
The required number of ways = 14C₁₁ = 14C₃ = \(\frac{14 \times 13 \times 12}{1 \times 2 \times 3}\)
= 14 × 13 × 2 = 364.

3. A particular player A is to be included and player B is to be excluded, then selection of 10 is to be made from 13 players. The required number of ways
= 13C₁₀ = 13C₃ = \(\frac{13 \times 12 \times 11}{1 \times 2 \times 3}\)
= 13 × 2 × 11 = 286.

Question 4.
What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these

1. four cards are of the same suit, (2)
2. are face cards, (1)
3. two are red cards and two are black cards, (2)
4. cards are of the same colour? (1)

Answer:
Selection of 4 cards from 52 = 52C₄
1. There are 4 suits in a pack of 52 playing cards. They are Club, Spade, Diamond, and Heart. Selecting 4 from each can be done in,
= 13C₄ + 13C₄ + 13C₄ + 13C₄
= 4 × 13C₄ = 4 × \(\frac{13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4}\) = 2860.

2. There are 12 face cards in a pack of 52 playing cards. Selection 4 cards can be done in 12C₄ = 495.

3. There are 26 red cards and 26 black in a pack of 52 playing cards. Selection of 2 cards should be done from each colour, this can be done in 26C₂ × 26C₂ = (325)² = 105625.

4. Selection of 4 cards from same colour = 26C₄ + 26C₄ = 29900

Plus One Maths Permutation and Combinations Practice Problems Questions and Answers

Question 1.
There are 3 routes from place A to place B and 2 routes from place B to place C. Find how many different routes are there from A to C.
Answer:
By fundamental principle of counting, there are 2 × 3 = 6 different ways.

Question 2.
How many 3 digit numbers can be formed from the digits 1, 2 and 3, assuming that the repetition of digits is not allowed.
Answer:

By fundamental principle of counting, there are 1 × 2 × 3 = 6 different 3 digit numbers.

Question 3.
How many two-digit even numbers with distinct digits can be formed from the digits 1, 2, 3, 4, 5.
Answer:

filled in 4 ways

filled by 2, 4 – 2ways

Hence by fundamental principle of counting, there are 4 × 2 = 8 different 2 digit even numbers.

Question 4.
Evaluate the following

1. \(\frac{7 !}{5 !}\)
2. \(\frac{12 !}{10 ! \times 2 !}\)
3. 6P₄
4. 9P₄
5. 10P₅

Answer:
1. \(\frac{7 !}{5 !}\) = \(\frac{7 \times 6 \times 5 !}{5 !}\) = 45

2.

3. 6P₄ = 6 × 5 × 4 × 3 = 360

4. 9P₄ = 9 × 8 × 7 × 6 = 3024

5. 10P₅ = 10 × 9 × 8 × 7 × 6 = 30240

Question 5.
Evaluate the following.

1. 10C₄
2. 21C₃
3. 19C₁₅
4. 31C₂₉ (1 score each)

Answer:

1. 10C₄ = \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}\) = 10 × 3 × 7 = 210
2. 21C₃ = \(\frac{21 \times 20 \times 19}{1 \times 2 \times 3}\) = 7 × 10 × 19 = 1330
3. 19C₁₅ = \(\frac{19 \times 18 \times 17 \times 16}{1 \times 2 \times 3 \times 4}\) = = 19 × 3 × 17 × 4 = 3876
4. 31C₂₉ = \(\frac{31 \times 30}{1 \times 2}\) = 465.

Question 6.
How many five digits telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer:
Digits 6, 7 already used, so only 8 digits available to fill remaining 3 places.

Hence by fundamental principle of counting there are 8 × 7 × 6 = 336 different 5 digit telephone numbers.

Question 7.
In how many ways can 5 persons sit in a car, two including the driver in the front seat and 3 in the back seat. If two particular person out of the 5 are to avoid the driver’s seat?
Answer:

Hence by fundamental principle of counting there are 3 × 4 × 3 × 2 × 1 = 72 different ways.

Question 8.
How many numbers can be formed from the digits 1, 2, 3 and 9 if repetition of digits is not allowed?
Answer:

Single digit	4 ways
Two digit	4 × 3 = 12 ways
Three digit	4 × 3 × 2 = 24 ways
Four digit	4 × 3 × 2 × 1 = 24 ways

Hence the number of total numbers = 4 + 12 + 24 + 24 = 64.

Question 9.
How many 3-digit even numbers can be formed from the digit 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer:

Hence by fundamental principle of counting there are 6 × 6 × 3 = 108 different 3 digit even numbers.

Question 10.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

1. repetition of the digits is allowed? (1)
2. repetition of the digits is not allowed (1)

Answer:
1.

Hence by fundamental principle of counting there are 5 × 5 × 5 = 125 different 3 digit numbers.

2.

Hence by fundamental principle of counting there are 5 × 4 × 3 = 60 different 3 digit numbers.

Question 11.
How many 8 letter words, with or without meaning, can be formed using the word EQUATION, using each letter exactly once?
Answer:
EQUATION this word has 8 different letters. Different words that can be made from these letters is 81 = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 =40320.

Question 12.
Find the number of ways in which the letters of the word ASSISTANT can be arranged among themselves.
Answer:
In the word ASSISTANT there are 9 letters, of which S appears 3 times, A appears 2 times, T appears 2 times and the rest all are different. Therefore the total number of ways is
\(\frac{9 !}{3! \times 2 ! \times 2 !}\) = 15120.

You can Download Nationalism Questions and Answers, Notes, Plus One Political Science Chapter Wise Questions and Answers Kerala Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Plus One Political Science Chapter Wise Questions and Answers Chapter 7 Nationalism

Nationalism Questions and Answers

Question 1.
“Nationalism has got two faces.” Explain.
Answer:
Nationalism has two faces – creative nationalism and destructive nationalism. Creative Nationalism is an emotional and spiritual power that unifies all factors that make up a nation. The creative nationalism we see in India is unity in diversity. Militant and aggressive nationalism is the opposite of creative nationalism. The WW II was the result of the militant and aggressive nationalism of the Nazis and Fascists.

Question 2.
What is nationalism?
Answer:
Nationalism is an emotional and spiritual power that unifies all factors that make up a nation. Nationalism includes national styles, national expectations, national goals, national conflicts, national anthem, national flower, national flag and national symbol. Toynbee defines nationalism as the religion of modern nations.

Question 3.
“A nation is an imaginary society.” Do you agree with this opinion? Explain.
Answer:
I agree. Nationalism is the feeling that one is the son/daughter of a particular nation even if he/she is living in another country because of his/her work. For example, an Indian may work and live in Dubai or New York. But nationalism reminds him that he is a son of India and he should be loyal to her. He should be proud of the fact that he is an Indian and do all he can to keep her dignity up. The Palestine nationalism was even above territory. Even though there was no territory for Palestine, Palestine nationalism was very strong. The strong nationalism gave birth to the State of Palestine which now consists of West Bank and Gaza Strip.

Question 4.
Give examples of Indian nationalism.
Answer:

• National Anthem
• National Flag
• National symbol
• Ashoka Pillar

Question 5.
Common political identity is required for nation-building. Explain the factors that contribute towards nationalism.
Answer:
Nationalism is an emotional and spiritual power that unifies all factors of the political theory prevalent in the nation. The nationalism we see in India is unity in diversity. Indian nationalism is the feeling that we all are Indians. Without nationalism, no new nation can originate or survive. It is nationalism that supplies the basis for the building up of a nation. Loyalty to the nation, its strength, its power, and its working and the legality, etc. are emotions that are generated in the minds of people. In the background of nationalism, the common behavior of a nation can be analyzed. Nationalism includes national styles, national expectations, national goals, national conflicts, national anthem, national flower, national flag, and national symbol. Toynbee has said that. Nationalism is the religion of the modern nation. Factors that nourish Nationalism:
a) Shared Beliefs: It is some beliefs that help the formation of a nation. Nation is not a concrete thing like a building, river or forest that we can see and touch. Nation is a concept built around certain beliefs of the people. When we say that a people are a nation, we do not mean their physical behavior. Nation is a vision and a collective fellowship of people who wish to have their own separate identity. It is like a group of a team. They work for a common goal in a common fellowship. A nation can exist only as long as the people are deeply rooted in unity.
b) History: A people who consider themselves as a nation have a sense of a continuous historical existence. It is something that they can look back with pride and look forward to with hope. They have a common history based collective memories, legends and historical documents. Thus they acquire individuality as a nation. We can take our Indian nationalist leaders as examples. We point out our ancient civilization, cultural heritage and other past achievements, thus making a unique identity of our own.
c) Territory: Territory is another essential aspect of a nation. A nation has a separate area of land. There people will have lived together for long, shared their past, bringing a common identity to them. It helps them to think that they are one nation. That is why they can talk about their own country of birth.
d) Shared political ideals: Apart from the sense of a common history and some territory of their own, the people of a nation are persuaded to think of unity because of their common vision and goal for the future. They have some ideals quite peculiar to them and these also make a nation different from others.
e) Common Political Identity: Many people think that just because individuals have a common political vision regarding the society and the government, they can’t form a nation. They feel that something like a common language or common heredity is necessary to have the drive to form a nation. A common language or a common religion might bring about a common cultural Identity. When people celebrate the same feasts on the same day and actively take part in the festivities, they feel some kind of unity. But it is possible that in a democratic setup and democratic values, too much of religion may prove a threat.

Question 6.
Suggest some programs to develop nationalist feelings among students.
Answer:
Celebrating Independence Day, Republic Day and Gandhi Jayanthi in an appropriate manner. The singing of the national anthem every day in school. Organizing Patriotic songs, discussions on important national issues, Debates on issues of national interest, Competitions, etc. Making albums containing the pictures of national leaders, the history of India’s Independent struggle, etc.

Question 7.
What do you understand by the Right to Self-Determination? How has this idea resulted in both the formation of and challenges to nation-states?
Answer:
A nation has the freedom to take decisions on its own and carry them out without being pressurized by any external forces. In the present scenario of globalization and liberalization, there is a threat to the Right to Self-Determination. This will negatively affect nationalism and the continued existence of modem nations.

Question 8.
“Neither descent, nor language, nor religion, nor ethnicity can claim to be a common factor in nationalism all over the word.” Do you agree? Comment.
Answer:
I agree. Much more than the above factors, it is the unity in diversity that influences the nationalism of a nation. Nationalism is strengthened without any difference in religion, tribe, language and region.

Question 9.
Identify the factors leading to the destruction of nationalism.
Answer:
Religious fundamentalism
Casteism
Regionalism
Terrorism
Secessionist tendencies
Economic and Social inequalities
Unemployment
Nepotism
Dictatorship

Question 10.
“We have seen that nationalism can unite people as well as divide them, liberate them as well as generate bitterness and conflict.” Illustrate your answer with suitable examples.
Answer:
Indian nationalism led us to freedom and the integration of the Princely States. It also led to the division of the country. German nationalism led to the unification of Germany and later to World War II.

Question 11.
Prepare a table showing the differences between constructive nationalism and aggressive nationalism.
Answer:

Creative Nationalism	Aggressive (Destructive) Nationalism
Creates loyalty	Creates hatred.
Unites people	Divides people.
Develops sense of freedom & liberates people.	Causes conflicts and wars.
Encourages democracy & self rule	Develops isolation.
Encourages economic growth	Looks at other people with suspicion.
Encourages cultural growth.	Develops instability.
Encourages diversity in social, economic and political spheres.	Creates many small nations.

Question 12.
It was ………… who called nation as an ‘imaginary society’.
Answer:
Benedict Anderson

Question 13.
A ………….. is a society which shares some special characteristics.
Answer:
Nation

Question 14.
Which among the following are the factors that constitute a nation?
a) Territory
b) Political identity
c) Common political principles
d) All of the above
Answer:
All of the above