Class 11

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 6 Biomolecules Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 6 Biomolecules

1. Choose the correct option

Question (A)
Sugar, amino acids, and nucleotides unite to their respective subunits to form ________
(a) bioelements
(b) micromolecules
(c) macromolecules
(d) all of these
Answer:
(c) macromolecules

Question (B)
Glycosidic bond is found in __________ .
(a) Disaccharide
(b) Nucleosides
(c) Polysaccharide
(d) all of these
Answer:
(d) all of these

Question (C)
Amino acids in a polypeptide are joined by _______ bond.
(a) Disulphide
(b) glycosidic
(c) hydrogen bond
(d) none of these
Answer:
(d) none of these

Question (D)
Lipids associated with cell membrane are _________ .
(a) Sphingomyelin
(b) Isoprenoids
(c) Phospholipids
(d) Cholesterol
Answer:
(c) Phospholipids

Question (E)
Linoleic, Linolenic and ________ acids are referred as essential fatty acids since they cannot be synthesized by the body and hence must be included in daily diet.
(a) Arachidonic
(b) Oleic
(c) Steric
(d) Palmitic
Answer:
(a) Arachidonic

Question (F)
Hemoglobin is a type of ________ protein, which plays indispensable part in respiration.
(a) simple
(b) derived
(c) conjugated
(d) complex
Answer:
(c) conjugated

Question (G)
When inorganic ions or metallo-organic molecules bind to apoenzyme, they together form
(a) isoenzyme
(b) holoenzyme
(c) denatured enzyme
(d) none of these
Answer:
(b) holoenzyme

Question (H)
In enzyme kinetics, Km = Vmax/2. If Km value is lower, it indicates _______
(a) Enzyme has less affinity for substrate
(b) Enzyme has higher affinity towards substrate
(c) There will be no product formation
(d) All active sites of enzyme are saturated
Answer:
(b) Enzyme has higher affinity towards substrate

2. Solve the following questions

Question (A)
Observe the following figures and write the differences between them.

Answer:

Saturated fats	Unsaturated fats
1. They contain single chain of carbon atoms with single bonds.	They contain chain of carbon atoms with one or more double bonds.
2. They are solid at room temperature.	They are liquid at room temperature.
3. They increase blood cholesterol level by depositing it in the inner wall of arteries.	They lower the blood cholesterol level and have many health benefits.
4. They do not get spoiled.	They get spoiled easily.
5. Saturated fats are obtained from animal fats, palm oil, etc.	Unsaturated fatty acids are obtained from plant and vegetable oil, etc.

3. Answer the following questions

Question (A)
What are building blocks of life?
Answer:
Life is composed of four main building blocks: Carbohydrates, proteins, lipids and nucleic acids.

Question (B)
Explain the peptide bond.
Answer:
1. The covalent bond that links the two amino acids is called a peptide bond.
2. Peptide bond is formed by condensation reaction.

Question (C)
How many types of polysaccharides you know?
Answer:
There are two types of polysaccharides:
1. Homopolysaccharides: It contains same type of monosaccharides. E.g. Starch, glycogen, cellulose.
2. Heteropolysaccharides: It contains two or more different monosaccharides. E.g. Hyaluronic acid, heparin, hemicellulose.

Question (D)
Enlist the significance of carbohydrates.
Answer:
Significances of carbohydrates are as follows:

1. Carbohydrates provide energy for metabolism.
2. Glucose is the main substrate for ATP synthesis.
3. Lactose, a disaccharide present in the milk provides energy to babies.
4. Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.

Question (E)
What is reducing sugar?
Answer:
1. A sugar that serves as a reducing agent due to presence of free aldehyde or ketone group is called a reducing sugar.
2. These sugars reduce the Benedict’s reagent (Cu²+ to Cu⁺) since they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction.
3. All monosaccharides are reducing sugars.

Question (F)
Enlist the examples of simple proteins and their significance.
Answer:
Examples of simple proteins are: E.g.: Albumins and histones.
Significance:
1. Albumin:
a. % It is the main protein in the blood.
b. It maintains the pressure in the blood vessels.
c. It helps in transportation of substances like hormone and drugs in the body.
2. Histones:
a. It is the chief protein of chromatin.
b. They are involved in packaging of DNA into structural units called nucleosomes.

Question (G)
Describe the secondary structure of protein with examples.
Answer:

1. There are two types of secondary structure of protein: a-helix and P-pleated sheets.
2. The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: a-helix (right handed) and P-helix (left handed).
3. This spiral configuration is held together by hydrogen bonds.
4. The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an a-helix structure.
5. Example of a-helix structure is keratin.
6. In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called P-pleated sheets.
7. Example of P-pleated sheet is silk fibres.
8. Due to formation of hydrogen bonds peptide chains assume a secondary structure.

Question (H)
Explain the induced fit model for mode of enzyme action.
Answer:
1. The induced fit model shows that enzymes are flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. It is also the point at which the final form and shape of the enzyme is determined.
2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

Question (I)
What is RNA? Enlist types of RNA.
Answer:
1. RNA stands for Ribonucleic Acid. It is a long single stranded polynucleotide chain which helps in protein synthesis, functions as a messenger and translates messages coded in DNA into protein.
2. There are three types of RNA:
mRNA (messenger RNA), rRNA (ribosomal RNA) and tRNA (transfer RNA)

Question (J)
Describe the concept of metabolic pool.
Answer:
1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.
2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.

Question (K)
How do secondary metabolites useful for mankind?
Answer:
1. Drugs developed from secondary metabolites have been used to treat infectious diseases, cancer, hypertension and inflammation.
2. Morphine, the first alkaloid isolated from Papaver somniferum is used as pain reliver and cough suppressant.
3. Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.
4. Flavours of secondary metabolites improve our food preferences.
5. Tannins are added to wines and chocolate for improving astringency.
6. Since most secondary metabolites have antibiotic property, they are also used as food preservatives.
7. Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur-containing chemicals. It also offers protection to these plants from many pests.

4. Solve the following questions

Question (A)
Complete the following chart.

Protein	Physiological role
Collagen	(i)
(ii)	Responsible for muscle contraction
Immunoglobulin	(iii)
(iv)	Significant in Respiration
Fibrinogen	(v)

Answer:

Protein	Physiological role
1. Collagen	Provides strength and plays structural role
2. Myosin & Actin	Responsible for muscle contraction
3. Immunoglobulin	Protects the body from infection
4. Haemoglobin	Significant in Respiration
5. Fibrinogen	Responsible for normal clotting of blood.

Question (B)
Answer the following with reference

i. Name the type of bond formed between two polypeptides.
ii. Which amino acid is involved in the formation of such bond?
iii. Amongst I, II, III and IV structural level of protein, which level of structure includes such bond? Answer:
i. Disulfide bond.
ii. Cysteine
iii. Tertiary structure.
[Note: Quaternary structure of protein also have disulfide bond, for stabilization of protein structure.!

Question (C)
Match the following items given in column I and II.

Column I	Column 11
1. RNA	(a) Induced fit model
2. Yam plant	(b) Flax seeds
3. Koshland	(c) Hydrolase
4. Omega – 3 – fatty acid	(d) Uracil
5. Sucrase	(e) Anti-fertility pills

Answer:

Column I	Column 11
1. RNA	(d) Uracil
2. Yam plant	(e) Anti-fertility pills
3. Koshland	(a) Induced fit model
4. Omega – 3 – fatty acid	(b) Flax seeds
5. Sucrase	(c) Hydrolase

5. Long answer questions

Question (A)
What are biomolecules? Explain the building blocks of life.
Answer:
Biomolecules are essential substances produced by our body which are necessary for life.
The building blocks of life are carbohydrates, lipids, proteins and nucleic acids.
1. Carbohydrates:
a. Carbohydrates are biomolecules made from carbon, hydrogen and oxygen.
b. The general formula of carbohydrates is (CH20) n.
c. They contain hydrogen and oxygen in the same ratio as in water (2:1).
d. Carbohydrates can be broken down to release energy.
e. Based on sugar units, carbohydrates are classified into three types: Monosaccharides, disaccharides and polysaccharides.

2. Lipids:
a. These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.
b. In lipids hydrogen to oxygen ration is greater than 2:1.
c. Lipid is a broader term used for fatty acids and their derivatives.
d. They are soluble in organic solvents (non-polar solvents).
e. Fatty acids are organic acids which are composed of hydrocarbon chain ending in carboxyl group (COOH) ….
f. These are divided into: Saturated fatty acids and unsaturated fatty acids.
g. Fatty acids are basic molecules which form different kinds of lipids.
h. Lipids are classified into three types:
Simple lipids, Compound lipids, Derived lipids.

3. Proteins:
a. Proteins are large molecules containing amino acid units ranging from 100 to 3000.
b. They have higher molecular weight.
c. In proteins, amino acids are linked together by peptide bonds which join the carboxyl group of one amino acid residue to the amino group of another residue.
d. A protein molecule consists of one or more polypeptide chains.
e. Proteins contain any or all twenty naturally occurring amino acid types.
f. Proteins have different structures like primary structure, secondary structure, tertiary structure and quaternary structure.
g. Proteins are classified into three types:
Simple proteins: Simple proteins on hydrolysis yield only amino acids. E.g. Histones and albumins. Conjugated proteins: It consists of a simple protein united with some non-protein substance. E.g. Haemoglobin.
Derived proteins: These proteins are not found in nature as such but are derived from native protein molecules on hydrolysis. E.g. Metaproteins, peptones.

4. Nucleic Acids:
a. Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.
b. Each nucleotide is formed of three components i.e. pentose sugar, a nitrogen base and a phosphate (phosphoric acid).
c. When sugar combine with nitrogenous base it forms nucleoside. Nucleotides can be called as nucleoside phosphate.
d. There are two types of nucleic acids, i.e. DNA and RNA.
DNA (Deoxyribonucleic acid) is a genetic material of a cell. It is double stranded helix. Each strand of helix is made up of deoxyribose nucleotides.
RNA (Ribonucleic Acid) is a single stranded structure having fewer nucleotides as compared to DNA. The strands may be straight or variously folded upon itself. It is made up of nucleotides.

Question (B)
Explain the classes of carbohydrates with examples.
Answer:
Based on number of sugar units, carbohydrates are classified into three types namely, monosaccharides, disaccharides and polysaccharides.
1. Monosaccharides:
a. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.
b. They cannot be further hydrolyzed into smaller molecules.
c. They are the building blocks or monomers of complex carbohydrates.
d. They have the general molecular formula (CH20)n, where n can be 3, 4, 5, 6 and 7.
e. They can be classified as triose, tetrose, pentose, etc.
f. Monosaccharides containing the aldehyde (-CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(-C=0) group are classified as ketoses. E.g. ribulose, fructose.

2. Disaccharides:
a. Disaccharide is formed when two monosaccharide react by condensation reaction releasing a water molecule. This process requires energy.
b. A glycosidic bond forms and holds the two monosaccharide units together.
c. Sucrose, lactose and maltose are examples of disaccharides.
d. Sucrose is a nonreducing sugar since it lacks free aldehyde or ketone group.
e. Lactose and maltose are reducing sugars.
f. Lactose also exists in beta form, which is made from P-galactose and p-glucose.
g. Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion.

3. Polysaccharides:
a. Monosaccharides can undergo a series of condensation reactions, adding one unit after the other to the chain till a very large molecule (polysaccharide) is formed. This is called polymerization.
b. Polysaccharides are broken down by hydrolysis into monosaccharides.
c. The properties of a polysaccharide molecule depends on its length, branching, folding and coiling.
d. Examples: Starch, glycogen, cellulose.

Question (C)
Describe the types of lipids and mention their biological significance.
Answer:
Lipids are classified into three main types:
1. Simple lipids:
a. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.
b. Fats are esters of fatty acids with glycerol (CH₂OH-CHOH-CH₂OH).
c. Triglycerides are three molecules of fatty acids and one molecule of glycerol.
d. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.

Biological significance:
a. Fats are a nutritional source with high calorific value and they act as reserved food materials.
b. In plants, fat is stored in seeds to nourish embryo during germination.
c. In animals, fat is stored in the adipocytes of the adipose tissue.
d. Fats deposited in subcutaneous tissue act as an insulator and minimize loss of body heat.
e. Fats deposited around the internal organs act as cushions to absorb mechanical shocks.
f. Wax is another example of simple lipid. They are esters of long chain fatty acids with long chain alcohols.
g. They are found in the blood, gonads and sebaceous glands of the skin.
h. Waxes are not as readily hydrolyzed as fats.
i. They are solid at ordinary temperature.
j. Waxes form water insoluble coating on hair and skin in animals, waxes form an outer coating on stems, leaves and fruits.

2. Compound lipids:
a. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
b. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.
c. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.
d. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).
e. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.
Biological significance:
a. Phospholipids contribute in the formation of cell membrane.
b. Large amounts of glycolipids are found in the brain white matter and myelin sheath.

3. Derived Lipids:
a. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
b. One of the most common sterols is cholesterol.
Biological significance:
a. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.
b. Cholesterol exists either free or as cholesterol ester.
c. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.
d. Cholesterol is not found in plants.
e. Sterols exist as phytosterols in plants.
f. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills, i.e. birth control pills.

Question (D)
Explain the chemical nature, structure and role of phospholipids in biological membrane.
Answer:
Chemical nature: Phospholipids are amphiphilic in nature. As they have hydrophilic head and hydrophobic tail.
Structure: It contains an alcohol, two fatty acid chains and a phosphate group.
Role: Phospholipids forms the membranes around the cells and cellular organelles. They form a lipid bilayer membrane. The phospholipids are arranged tail to tail. It serves as a barrier against movement of any ions or polar compounds into and out of the cell.

Question (E)
Describe classes of proteins with their importance.
Answer:
On the basis of structure, proteins are classified into three categories:
1. Simple proteins:
a. Simple proteins on hydrolysis yield only amino acids.
b. These are soluble in one or more solvents.
c. Simple proteins may be soluble in water.
d. Histones of nucleoproteins are soluble in water.
e. Globular molecules of histones are not coagulated by heat.
f. Albumins are also soluble in water but they get coagulated on heating.
g. Albumins are widely distributed e.g. egg albumin, serum albumin and legumelin of pulses are albumins.
Importance: They are involved in structural components; they also act as a storage kind of protein.
Some are associated with nucleic acids in nucleoproteins of cell.

2. Conjugated proteins:
a. Conjugated proteins consist of a simple protein united with some non-protein substance.
b. The non-protein group is called prosthetic group e.g. haemoglobin.
c. Globin is the protein and the iron containing pigment haem is the prosthetic group.
d. Similarly, nucleoproteins have nucleic acids.
e. Proteins are classified as glycoproteins and mucoproteins.
f. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood.
g. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in brain, plasma membrane, milk etc. Importance: They are involved in structural components of cell membranes and organelles.
They also act as a transporter.
Some conjugated proteins are important in electron transport chain in respiration.

3. Derived proteins:
a. These proteins are not found in nature as such.
b. These proteins are derived from native protein molecules on hydrolysis.
c. Metaproteins, peptones are derived proteins.
Importance: They act as a precursor for many molecules which are essential for life.

Question (F)
What are enzymes? How are they classified? Mention example of each class.
Answer:
1. Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body.
2. Enzymes are classified into six classes:
a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase

b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase

c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase

d. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase.

e. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase.

f. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase.

Question (G)
Explain the properties of enzyme? Describe the models for enzyme actions.
Answer:
1. Proteinaceous Nature:
All enzymes are basically made up of protein.

2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

3. Catalytic property:
a. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged.
b. After completion of the reaction and release of the product they remain active to catalyze again.
c. A small quantity of enzymes can catalyze the transformation of a very large quantity of the substrate
into an end product.
d. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight.

4. Specificity of action:
a. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates.
b. Enzymes are very sensitive to temperature and pH.
c. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH.
d. Any increase or decrease in pH causes decline in enzyme activity e.g. enzyme pepsin (secreted in stomach)shows highest activity at an optimum pH of 2 (acidic)

5. Temperature:
a. Enzymes are destroyed at higher temperature of 60-70°C or below, they are not destroyed but become inactive.
b. This inactive state is temporary and the enzyme can become active at suitable temperature.
c. Most of the enzymes work at an optimum temperature between 20°C and 35°C.

There are two types of models:
1. Lock and Key model:
a. Lock and Key model was first postulated in 1894 by Emil Fischer.
b. This model explains the specific action of an enzyme with a single substrate.
c. In this model, lock is the enzyme and key is the substrate.
d. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

2. Induced Fit model (Flexible Model):
a. Induced Fit model was first proposed in 1959 by Koshland.
b. This model states that approach of a substrate induces a conformational change in the enzyme.
c. It is the more accepted model to understand mode of action of enzyme.
d. The induced fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it.
e. It is also the point at which the final form and shape of the enzyme is determined.
[Note: Temperature is a factor affecting enzyme activity and not a property of enzyme.]

Question (H)
Describe the factors affecting enzyme action.
Answer:
The factors affecting the enzyme activity are as follows:
1. Concentration of substrate:
a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.
b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.
c. Three distinct phases (A, B and C) of the reaction are observed in the graph.
Where V = Measured velocity, V_max = Maximum velocity, S = Substrate concentration,
K_m = Michaelis-Menten constant.
d. K_m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction.
e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the Km value.
f. K_m value is a constant and a characteristic feature of a given enzyme.
g. It is a representative for measuring the strength of ES complex.
h. A low K_m value indicates a strong affinity between enzyme and substrate, whereas a high Km value reflects a weak affinity between them.
1. For majority of enzymes, the K_m values are in the range of 10_-5 to 10_-2 moles.

2. Enzyme Concentration:

a. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate.
b. The rate of reaction is also directly proportional to the square root of the concentration of enzymes.
c. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of enzyme.

3. Temperature:

a. The temperature at which the enzymes show maximum activity is called Optimum temperature.
b. The rate of chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature.
c. Enzymes rapidly denature at temperature above 40°C.
d. The activity of enzymes is reduced at low temperature.
e. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms.

4. Effect of pH:

a. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH.
b. The enzyme cannot perform its function beyond the range of its pH value.

5. Other substances:
a. The enzyme action is also increased or decreased in the presence of some other substances such as co-enzymes, activators and inhibitors.
b. Most of the enzymes are combination of a co-enzyme and an apo-enzyme.
c. Activators are the inorganic substances which increase the enzyme activity.
d. Inhibitor is the substance which reduces the enzyme activity.

Question (I)
What are the types of RNA? Mention the role of each class of RNA.
Answer:
There are three types of cellular RNAs:
1. messenger RNA (mRNA),
2. ribosomal RNA (rRNA),
3. transfer RNA (tRNA). ‘

1. Messenger RNA (mRNA):
a. It is a linear polynucleotide.
b. It accounts 3% of cellular RNA.
c. Its molecular weight is several million. , d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.
e. Size of mRNA is related to the size of message it contains.
f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:
It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.

2. Ribosomal RNA (rRNA):
a. rRNA was discovered by Kurland in 1960.
b. It forms 50-60% part of ribosomes.
c. It accounts 80-90% of the cellular RNA.
d. It is synthesized in nucleus.
e. It gets coiled at various places due to intrachain complementary base pairing.
Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.

3. Transfer RNA (tRNA):
a. These molecules are much smaller consisting of 70-80 nucleotides.
b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf.
c. Each tRNA can pick up particular amino acid.
d. Following four parts can be recognized on tRNA
1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site
2. Amino acid binding site
3. Anticodon loop / codon recognition site
4. Ribosome recognition site.
e. In the anticodon loop of tRNA, three unpaired nucleotides are present called as anticodon which pair with codon present on mRNA.
f. The specific amino acids are attached at the 3′ end in acceptor stem of clover leaf of tRNA.
Role of transfer RNA: It helps in elongation of polypeptide chain during the process called translation.

Question (J)
What is metabolism? How metabolic pool is formed in the cell.
Answer:

1. Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new’ organic material.
2. Metabolic pool in the cell is formed due to glycolysis and Krebs cycle.
3. The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell.
4. These biomolecules can be utilized for synthesis of many important cellular components.
5. The metabolites can be added or withdrawn from the pool according to the need of the cell.

Question 6.
If double stranded DNA has 14% C (cytosine) what percent A (adenine), T (thymine) and G (guanine) would you expect?
Answer:
A purine always pairs with pyrimidine.
Adenine pairs with thymine and cytosine pairs with guanine.
Therefore, as per the given data If cytosine = 14% then guanine = 14%.
According to Chargaff s rule,
(C+G) = 14+ 14 = 28%
Therefore, (A+T) = 72%
So, A= 36%, T= 36%, G = 14%.

Question 7.
Name
1. The reagent used for testing for reducing sugar.
2. The form in which carbohydrate is transported in a plant.
3. The term that describes all the chemical reactions taking place in an organism.
Answer:
1. Benedict’s reagent
2. Sucrose
3. Metabolism

Practical / Project:

Question 1.
Perform an experiment to study starch granules isolated from potato.
Answer:
Isolation of starch granules from potato:

1. Peal the potato with a clean knife.
2. Grind the potato till the homogenous mixture is formed.
3. Then strain the mixture through a cheese cloth into a beaker.
4. Keep it standing for some time.
5. Throw the supernatant and fill the beaker containing starch with water.
6. Stir it well and again allow the starch to settle.
7. After sometime, again through the supernatant.
8. Repeat this for 2-3 times.
9. Collect the white starch in the watch glass and keep it in the oven for drying.

To study the isolated starch granules:
1. Examination under microscope:
Examine starch granules under microscope by using a mixture of equal volumes of glycerol and distilled water.
Result: The potato starch granules appears transparent granules. They are irregularly shaped.
2. Using Iodine solution:
Boil a little amount of starch with water. Cool it. Add iodine solution to it.
Result: The solution changes colour to blue. This indicates the presence of starch.

Question 2.
Study the action of enzyme urease on urea.
Answer:
Urease is an enzyme which exists in a dimer form. It has two active sites which are highly specific and only bind to urea or hydroxy urea. The active sites of urease contain nickel atoms. Urease catalyzes the hydrolysis of substrate urea into carbon dioxide and ammonia. It attacks the nitrogen and carbon bond in amide compounds and forms alkaline product like ammonia.

11th Biology Digest Chapter 6 Biomolecules Intext Questions and Answers

Can you recall? (Textbook Page No. 59)

(i) Which are different cell components?
Answer:
a. The three main components of any cell are: Cell membrane, Cytoplasm, Nucleus.
b. The components present in both plant and animal cells are: Endoplasmic reticulum, ribosomes, golgi apparatus, lysosomes, mitochondria, vacuoles.
c. The components present in plant cell and not in animal cell: Cell wall and plastids.
d. The components present in animal cell and not in plant cell: Cilia and flagella.

(ii) What is the role of each component of cell?
Answer:
The role of each component of a cell is as follows:
a. Cell membrane: Cell membrane separates the cytoplasmic contents from external environment.
b. Cytoplasm: Site for metabolic activities and organelles.
c. Nucleus: It is the control center of the cell. Genetic material is present in the nucleus.
d. Endoplasmic reticulum: It produces, processes and transports proteins and lipids.
e. Ribosomes: Ribosome is the site for protein synthesis.
f. Golgi apparatus: It is involved in modifying, sorting and packing of proteins for secretion. It also transports lipids around the cell.
g. Lysosomes: It is involved in digestion of worn out organelles and waste removal.
h. Mitochondria: It is responsible for production of energy.
i. Vacuoles: It has various functions like storage, waste disposal, protection and growth.
j. Cell wall: It provides strength and support to the cell.
k. Plastids: They are responsible for production and storage of food. It also contains photosynthetic pigments (Chloroplasts).
l. Cilia and flagella: Help in motility.

Can you tell? (Textbook Page No. 62)

What are carbohydrates?
Answer:

1. The word carbohydrates mean ‘hydrates of carbon’.
2. They are also called saccharides.
3. They are biomolecules made from just three elements: carbon, hydrogen and oxygen with the general formula C_x(H₂0)y.
4. They contain hydrogen and oxygen in the same ratio as in water (2:1).
5. Carbohydrates can be broken down (oxidized) to release energy.

Can you tell? (Textbook Page No. 62)

(i) Enlist the natural sources, structural units and functions of the following polysaccharides.
a. Starch
b. Cellulose
c. Glycogen
Answer:
a. Starch:
1. Natural Sources: Cereals (wheat, maize, rice), root vegetables (potato, cassava etc.)
2. Structural units: Starch consist of two types of molecules – Amylose and amylopectin.
3. Functions: It acts as a reserve food and supply energy.

b. Cellulose:
1. Natural sources: Plant fibers (cotton, flax, hemp, jute, etc.), wood.
2. Structural units: It is made from p glucose molecules.
3. Functions: It in a major component of cell wall. It provides structural support.

c. Glycogen:
1. Natural sources: Fruits, starchy vegetables, whole grain foods.
2. Structural units: It consists of linear chains of glucose residues. The glucose is linked linearly by a (1 → 4) glycosidic bonds and branches are linked to the linear chain by a (1 → 6) glycosidic bonds.
3. Functions: It is stored in liver and muscles and it readily provides energy when the blood glucose level decreases.

(ii) The exoskeleton of insects is made up of chitin. This is a ________.
(A) mucoprotein
(B) lipid
(C) lipoprotein
(D) polysaccharide
Answer:
polysaccharide

(iii) List names of structural polysaccharides.
Answer:
Arabinoxylans, cellulose, chitin, pectin.

(iv) Write a note on oligosaccharide and glycosidic bond.
Answer:
Oligosaccharides:
a. A carbohydrate polymer comprising of two to six monosaccharide molecules is called oligosaccharide.
b. They are linked together by glycosidic bond.
c. They are classified on the basis of monosaccharide units:
Disaccharides: These are the sugars containing two monosaccharide units and can be further hydrolysed into smaller components. E.g.: Sucrose, maltose, lactose, etc.
Trisaccharides: These contain three monomers. E.g. Raffmose.
Tetrasaccharides: These contain four monomers. E.g.: Stachyose.

Glycosidic bond:
a. Glycosidic bond is a covalent bond that forms a linkage between two monosaccharides by a dehydration reaction.
b. It is formed when a hydroxyl group of one sugar reacts with the anomeric carbon of the other.
c. Glycosidic bonds are readily hydrolyzed by acid but resist cleavage by base.
d. There are two types of glycosidic bonds: a-glycosidic bond and P-glycosidic bond.

Can you tell? (Textbook Page No. 63)

What are lipids? Classify them and give at least one example of each.
Answer:
Lipids:
Lipids are a group of heterogeneous compounds like fats, oils, steroids, waxes, etc.
They are macro-biomolecules.
These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.

Lipids are classified into:
1. Saturated fatty acids: They contain single chain of carbon atoms with single bonds.
E.g. Palmitic acid, stearic acid
2. Unsaturated fatty acids: They contain one or more double bonds between the carbon atoms of the hydrocarbon chain.
a. Simple lipids: These are esters of fatty acids with various alcohols.
E.g. Fats, wax.
b. Compound lipids: These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
E.g. Lecithin
c. Sterols: They are derived lipids. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
E.g. Cholesterol, phytosterols.

Find out (Textbook Page No. 63)

(i) Why do high cholesterol level in the blood cause heart diseases?
Answer:
a. When there is high level of cholesterol in the blood, the cholesterol builds up on the walls of arteries causing a condition called atherosclerosis (a form of heart disease).
b. Because of this the arteries are narrowed and the blood flow to the heart is slowed down.
c. The blood carries oxygen to the heart, but because of this condition enough blood and oxygen does not reach to the heart and causes heart diseases.
d. If the condition increases, the supply of oxygen and blood is completely cut off to the heart and this can lead to heart attack.

(ii) Polyunsaturated fatty acids are believed to decrease blood cholesterol level. How?
Answer:
a. The liver converts polyunsaturated fatty acids into ketones instead of cholesterol.
b. Therefore, polyunsaturated fatty acids are transported directly to tissues for oxidation without leaving behind any lipoprotein in the form of cholesterol as it is seen in the case of saturated fatty acids.
c. Thus, polyunsaturated fatty acids are believed to decrease blood cholesterol level.

Can you tell? (Textbook Page No. 64)

Which of the following is a simple protein?
(A) nucleoprotein
(B) mucoprotein
(C) chromoprotein
(D) globulin
Answer:
Globulin

Can you tell? Textbook Page No. 64)

What are conjugated proteins? How do they differ from simple ones? Give one example of each.
Answer:
1. Conjugated proteins consist of a simple protein attached with some non-protein substance. The non-protein group is called prosthetic group.
2. The conjugated protein functions in interaction with other chemical group whereas simple proteins contain only amino acids and no other chemical group attached to it.
3. Example of conjugated protein is haemoglobin. Globin is the protein and iron containing pigment and haem is the prosthetic group.

Can you tell? (Textbook Page No. 64)

All Proteins are made up of the same amino acids; then how proteins found in human beings and animals may be different from those of other?
Answer:
The proteins found in human beings and animals may be different from those of others because the ratio of amino acids present in the protein differs.

Can you tell? (Textbook Page No. 67)

What is a nucleotide? How is it formed? Mention the names of all nucleotides.
Answer:
1. Nucleotide is a unit which consists of a sugar, phosphate and a base. Nucleotides are basic units of nucleic acids.
2. The nitrogen base and a sugar form a nucleoside. In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
3. The names of all nucleotides are:

Base	Nucleotides of RNA	Nucleotides of DNA
Adenine	Adenylate	Deoxydenylate
Guanine	Guanylate	Deoxyguanylate
Cytosine	Cytidylate	Deoxy cytidylate
Thymine	–	Deoxythymidylate
Uracil	Uridylate	–

Can you tell? (Textbook Page No. 67)

Describe the structure of DNA molecule as proposed by Watson and Crick.
Answer:

1. According to Watson and Crick, DNA molecule consists of two strands twisted around each other in the form of a double helix.
2. The two strands i.e. polynucleotide chains are supposed to be in opposite direction so end of one chain having 3′ lies beside the 5′ end of the other.
3. One turn of the double helix of the DNA measures about 34A.
4. It consists paired nucleotides and the distance between two neighboring pair nucleotides is 3.4A.
5. The diameter of the DNA molecule has been found be 20A.

Can you tell (Textbook Page No. 70)

Name the chemical found in the living cell which has necessary message for the production of all enzymes required by it.
Answer:
DNA found in the nucleus of a living cell has necessary message for the production of all enzymes required by it. DNA forms mRNA through the process of transcription. This mRNA through the process of translation forms proteins.

Can you tell? (Textbook Page No. 67)

Difference between DNA and RNA is because of
(A) sugar and base
(B) sugar and phosphate
(C) phosphate and base
(D) sugar only
Answer:
Sugar and base

Can you tell? (Textbook Page No. 67)

Differentiate between DNA and RNA.
Answer:

DNA	RNA
1. It is a genetic material of majority of the organisms.	It is a genetic material only of some viruses.
2. It is double stranded.	It is single stranded.
3. Deoxyribose sugar is present.	Ribose sugar is present.
4. Nitrogen bases like Adenine, Guanine, Cytosine, Thymine are present.	Nitrogen bases like Adenine, Guanine, Cytosine, Uracil are present.
5. Specific base pairing is observed.	Nitrogen bases do not form pair.
6. Total number of purines is equal to total number of pyrimidine. Thus, purine to pyrimidine ratio is 1:1.	Amount of purine and pyrimidine may or may not be equal.
7. It is present in nucleus.	It is present in nucleus and cytoplasm.
8 It is responsible for determining hereditary characters and for formation of RNA.	It takes part in protein synthesis.

Can you tell? (Textbook Page No. 70)

Co-enzyme is ________
(A) often a metal
(B) often a vitamin
(C) always as organic molecule
(D) always an inorganic molecule
Answer:
Always as organic molecule

Can you tell? (Textbook Page No. 70)

(i) Which enzyme is needed to digest food reserve in castor seed?
(A) Amylase
(B) Diastase
(C) Lipase
(D) Protease
Answer:
Lipase

(ii) List the important properties of enzymes.
Answer:
a. Proteinaceous Nature
b. Three-Dimensional conformation
c. Catalytic property
d. Specificity of action
e. Temperature

Try this: (Textbook Page No. 70)

To demonstrate the effect of heat on the activities of inorganic catalysts and enzymes.
Answer:
1. Using MnO₂ and Enzymes without any heat treatment:
Mn0₂ and cellular enzymes (catalase/peroxidase) causes breakdown of H₂0₂ and evolution of oxygen.
2. Using Mn0₂ and Enzymes after heat treatment:
Oxygen evolves in the H₂0₂ solution containing boiled and cooled Mn0₂. But oxygen does not evolve in the tube containing the enzyme.
3. This confirms that heat affects the enzyme and inactivates it whereas heat does not have any effect on inorganic catalyst.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 5 Cell Structure and Organization Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 5 Cell Structure and Organization

1. Choose the correct option

Question (A)
Growth of cell wall during cell elongation takes place by ………….
(a) Apposition
(b) Intussusception
(c) Both a & b
(d) Superposition

Question (B)
Cell Membrane is composed of
(a) Proteins and cellulose
(b) Proteins and Phospholipid
(c) Proteins and carbohydrates
(d) Proteins, Phospholipid and some carbohydrates
Answer:
(d) Proteins, Phospholipid and some carbohydrates

Question (C)
Plasma membrane is Fluid structure due to presence of
(A) Carbohydrates
(B) Lipid
(C) Glycoprotein
(D) Polysaccharide
Answer:
(B) Lipid

Question (D)
Cell Wall is present in
(a) Plant cell
(b) Prokaryotic cell
(c) Algal cell
(d) All of the above
Answer:
(d) All of the above

Question (E)
Plasma membrane is
(a) Selectively permeable
(b) Permeable
(c) Impermeable
(d) Semipermeable
Answer:
(a) Selectively permeable

Question (F)
Mitochondria DNA is
(a) Naked
(b) Circular
(c) Double stranded
(d) All of the above
Answer:
(d) All of the above

Question (G)
Which of the following set of organelles contain DNA?
(a) Mitochondria, Peroxysome
(b) Plasma membrane, ribosome
(c) Mitochondria, chloroplast
(d) Chloroplast, dictyosome
Answer:
(c) Mitochondria, chloroplast

2. Answer the following questions

Question (A)
Plants have no circulatory system? Then how cells manage intercellular transport?
Answer:
1. Plant cells show presence of plasmodesmata which are cytoplasmic bridges between neighbouring cells.
2. This open channel through the cell wall connects the cytoplasm of adjacent plant cells and allows water, small solutes, and some larger molecules to pass between the cells.
In this way, though plants have no circulatory system, plant cells manage intercellular transport.

Question (B)
Is nucleolus covered by membrane?
Answer:
A nucleolus is specialized structure present in the nucleus which is not covered by the membrane.

Question (C)
Fluid mosaic model proposed by Singer and Nicolson replaced Sandwich model proposed by Danielli and Davson? Why?
Answer:

1. The Davson-Danielli model of the plasma membrane of a cell, was proposed in 1935 by Hugh Davson and James Danielli.
2. The model describes a phospholipid bilayer that lies between two layers of globular proteins.
3. This model was also known as a Tipo-protein sandwich’, as the lipid layer was sandwiched between two protein layers.
4. But through experimental studies membrane proteins were discovered to be insoluble in water (representing hydrophobic surfaces) and varied in size. Such type of proteins would not be able to form an even and continuous layer around the outer surface of a cell membrane.
5. In case of Fluid-mosaic model, the experimental evidence from research supports every major hypothesis proposed by Singer and Nicolson.

This hypothesis stated that membrane lipids are arranged in a bilayer; the lipid bilayer is fluid; proteins are suspended individually in the bilayer; and the arrangement of both membrane lipids and proteins is asymmetric. Therefore, Fluid mosaic model proposed by Singer and Nicolson replaced Sandwich model proposed by Danielli and Davson.

Question (D)
The RBC surface normally shows glycoprotein molecules. When determining blood group do they
play any role?
Answer:

1. Glycoproteins are protein molecules modified within the Golgi complex by having a short sugar chain (polysaccharide) attached to them.
2. The polysaccharide part of glycoproteins located on the surfaces of red blood cells acts as the antigen responsible for determining the blood group of an individual.
3. Different polysaccharide part of glycoproteins act as different type of antigens that determine the blood groups.
4. Four types of blood groups A, B, AB, and O are recognized on the basis of presence or absence of these antigens.

Question (E)
How cytoplasm differs from nucleoplasm in chemical composition?
Answer:

1. A thick liquid enclosed by cell membrane which surrounds the central nucleus in eukaryotes or nucleoid region in prokaryotes is known as cytoplasm.
2. The cytoplasm shows presence of minerals, sugars, amino acids, t-RNA, nucleotides, vitamins, proteins and enzymes.
3. The liquid or semiliquid substance within the nucleus is called the nucleoplasm.
4. Nucleoplasm shows presence of various substances like nucleic acid, protein molecules, minerals and salts.

3. Answer the following questions

Question (A)
Distinguish between smooth and rough endoplasmic reticulum.
Answer:
Smooth endoplasmic reticulum (SER):
1. Depending on cell type, it helps in synthesis of lipids for e.g. Steroid secreting cells of cortical region of adrenal gland, testes and ovaries.
2. Smooth endoplasmic reticulum plays a role in detoxification in the liver and storage of calcium ions (muscle cells).

Rough Endoplasmic Reticulum (RER):

1. Rough ER is primarily involved in protein synthesis. For e.g. Pancreatic cells synthesize the protein insulin in the ER.
2. These proteins are secreted by ribosomes attached to rough ER and are called secretory proteins. These proteins get wrapped in membrane that buds off from transitional region of ER. Such membrane bound proteins depart from ER as transport vesicles.
3. Rough ER is also involved in formation of membrane for the cell.

The ER membrane grows in place by addition of membrane proteins and phospholipids to its own membrane. Portions of this expanded membrane are transferred to other components of endomembrane system.

Question (B)
Why do we call mitochondria as power house of cell? Explain in detail.
(Hint: Refer chapter Cellular Respiration.)
OR
Mitochondria are power house of the cell. Give reasons.
Answer:
a. Mitochondria possess oxysomes on its inner membrane. These oxysomes take active part in synthesis of ATP molecules.
b. During cellular respiration, ATP molecules are produced and get accumulated in the mitochondria. They play an important role in cellular activities.
c. only mitochondria can convert pyruvic acid to carbon dioxide and water during cell respiration. Therefore, mitochondria are called ‘power house of the cell’.

Question (C)
What are the types of plastids?
Answer:
1. Plastids are classified according to the pigments present in it. Three main types of plastids are – leucoplasts, chromoplasts and chloroplasts.
2. Leucoplasts do not contain any photosynthetic pigments they are of various shapes and sizes. These are meant for storage of nutrients:
a. Amyloplasts store starch.
b. Elaioplasts store oils.
c. Aleuroplasts store proteins.

3. Chromoplasts contain pigments like carotene and xanthophyll etc.
a. They impart yellow, orange or red colour to flowers and fruits.
b. These plastids are found in the coloured parts of flowers and fruits.

4. Chloroplasts are plastids containing green pigment chlorophyll along with other enzymes that help in production of sugar by photosynthesis. They are present in plants, algae and few protists like Euglena.

Question 4.
Label the diagrams and write down the details of concept in your words.


Answer:
A.

B.

C.

D.

Structure of chloroplast:

1. In plants, chloroplast is found mainly in mesophyll of leaf.
2. Chloroplast is lens shaped but it can also be oval, spherical, discoid or ribbon like.
3. A cell may contain single large chloroplast as in Chlamydomonas or there can be 20 to 40 chloroplasts per cell as seen in mesophyll cells.
4. Chloroplasts contain green pigment called chlorophyll along with other enzymes that help in production of sugar by photosynthesis.
5. Inner membrane of double membraned chloroplast is comparatively less permeable.
6. Inside the cavity of inner membrane, there is another set of membranous sacs called thylakoids.
7. Thylakoids are arranged in the form of stacks called grana (singular: granum).
8. The grana are connected to each other by means of membranous tubules called stroma lamellae.
9. Space outside thylakoids is filled with stroma.
10. The stroma and the space inside thylakoids contain various enzymes essential for photosynthesis.
11. Stroma of chloroplast contains DNA and ribosomes (70S).

Question 5.
Complete the flow chart.

Answer:

Question 6.
Identify labels A, B, C in the given diagram. Explain how lysosomes perform intracellular and extracellular digestion.

Answer:
1. A: Food vacuole
B: Golgi complex
C: Lysosome

2. Intracellular digestion:
The intracellular digestion is brought about by autophagic vesicle or secondary lysosomes which contain foreign materials brought in by processes like phagocytosis. E.g. Food vacuole in amoeba or macrophages in human blood that engulf and destroy harmful microbes that enter the body.

3. Extracellular digestion:
Extracellular digestion is brought about by release of lysosomal enzymes outside the cell. E.g. acrosome, a cap like structure in human sperm is a modified lysosome which contain various enzymes like Hyaluronidase. These enzymes bring about fertilization by dissolving protective layers of ovum.

Question 7.
Identify each cell structures or organelle from its description below.

1. Manufactures ribosomes
2. Carries out photosynthesis
3. Manufactures ATP in animal and plant cells.
4. Selectively permeable.

Answer:

1. Nucleolus
2. Chloroplast
3. Mitochondria
4. Plasma membrane

Question 8.
Onion cells have no chloroplast. How can we tell they are plants?
Answer:

1. The bulb of an onion is a modified form of leaves.
2. While photosynthesis takes place in the leaves (present above the ground) of an onion containing chloroplast, the little glucose that is produced from this process is converted in to starch (starch granules) and stored in the bulb.
3. Starch act as reserved food material in plants.
4. Using an iodine solution, we can test for the presence of starch in onion cells. If starch is present, the iodine changes from brown to blue-black or purple. Hence, we can say that though onion cells have no chloroplast they are considered as plants.

Project/ Practical:

Question 1.
Observe the cells of onion root tip under microscope.
Answer:
The cells of onion root tip will show various stages of cell division when observed under micrscope.

Question 2.
Observe the cells from buccal epithelium stained with Giemsa under microscope.
Answer:
The following observations are made when cells from buccal epithelium stained with Giemsa:
1. Cheek cells are flat and irregular in shape.
2. These cells lack cell wall. A distinct blue nucleus can be observed on viewing the cells under the microscope after Geimsa staining.

11th Biology Digest Chapter 5 Cell Structure and Organization Intext Questions and Answers

Can You Recall? (Textbook Page No. 44)

(i) Who observed cells under the microscope for the first time?
Answer:
Robert Hooke observed cells under the microscope for the first time.
[Note. Cell walls were first observed by Robert Hooke (1665) as he looked through a microscope at dead cells from the bark of an oak tree. But Anton van Leeuwenhoek was first to visualize living cells using a single-lens microscope of his own construction.]

(ii) Who made the first microscope?
Answer:
The first microscope was made by two Dutch spectacle makers Hans and Zacharias Janssen.
[Note: The Dutch scientist Anton van Leeuwenhoek made microscopes capable of magnifying single-celled organisms in a drop of pond water.]

Find Out (Textbook Page No. 44)

(i) How do a combination of lenses help in higher magnification?
Answer:
a. In a light microscope, visible light is passed through the specimen and then through two glass lenses.
b. The first lens focuses the magnified image of the object on the second lens, which magnifies it again and focuses it on the back of the eye.
c. The glass lenses bend (refract) the light in such a way that the image of the specimen is magnified.
In this way, a combination of lenses helps in higher magnification.

(ii) When do we use plane and concave mirror and diaphragm?
Answer:
a. Concave mirror is used when low-power objective lenses (useful for examining large specimens or many smaller specimens) or high-power objective lenses (useful for observing fine detail) are used, whereas plane mirror is used when oil immersion objective lens is used.
b. The amount of light passing on to the specimen from the condenser (which concentrates and controls the light that passes through the specimen) is regulated by using iris diaphragm.
c. Light is reduced by closing the diaphragm partially for use with dry objectives.
d. Oil immersion objectives require maximum light and this can be achieved by keeping the iris diaphragm fully open.

(iii) What is the difference between magnification and resolution?
Answer:
a. Magnification is the ratio of an object’s image size to its actual size.
b. Resolution is a measure of the clarity of the image; it is the minimum distance two points can be
separated and still be distinguished as separate points.

Can You Recall? (Textbook Page No. 44)

(i) Why bacterial nucleus is said to be primitive?

(ii) Draw neat and labelled diagram of Prokaryotic cell.
Answer:
1. The DNA-containing central region of bacterial nucleus (prokaryotic cells) i.e. nucleoid, has no nuclear membrane separating it from the cytoplasm. Therefore, bacterial nucleus is said to be primitive.
2. Prokaryotic cell:

Find Out (Textbook Page No. 46)

Why do basal body of bacterial flagella considered as smallest motor in the world?
Answer:
1. The bacterial flagellum is an organelle for motility made up of three parts:
a. The basal body that spans the cell envelope and works as a rotary motor;
b. The helical fdament that acts as a propeller;
c. The hook that acts as a universal joint connecting these two to transmit motor torque to the propeller.
2. The motor i.e. basal body drives the rotation of the long, helical filamentous propeller at hundreds of hertz to produce thrust that allows bacteria to swim in liquid environments.
Therefore, basal body of bacterial flagella considered as smallest motor in the world.

Use your Brainpower (Textbook Page No. 46)

Describe major differences between prokaryotic and eukaryotic cells.
Answer:

Prokaryotic cell	Eukaryotic cell
1. It is a primitive type of cell.	It is an evolved type of cell.
2. Nuclear membrane is absent.	Nuclear membrane is present.
3. Genetic material is in the form of circular coil of DNA without histone proteins.	Genetic material is in the form of a double helix DNA with histone proteins.
4. Membrane-bound cell organelles are absent.	Membrane-bound cell organelles are present.
5. Plasmids are many in number.	Plasmids are absent.
6. Cytoplasm does not show streaming movement.	Cytoplasm shows streaming movement.
7. Ribosomes are smaller and of 70S type.	Ribosomes are larger and of 80S type.
8. Respiratory enzymes are present on the infoldings of the plasma membrane called mesosomes.	Respiratory enzymes are present within mitochondria.
e.g Cyanobacteria (Blue-green algae) and bacteria.	Algae, fungi, plants and animals.

Use Your Brain Power! (Textbook Page No. 52)

Are mitochondria present in all eukaryotic cells?
Answer:
a. Mitochondria are found in nearly all eukaryotic cells, including plants, animals, fungi, and most unicellular eukaryotes.
b. Some of the cells have a single large mitochondrion, but frequently a cell has hundreds of mitochondria.
c. The number of mitochondria correlates with the cell’s level of metabolic activity. For e.g. cells that move or contract have proportionally more mitochondria than metabolically less active cells.
d. However, mature red blood cells in humans lack mitochondria.

Can You Recall? (Textbook Page No. 54)

(i) Consider the following cells and comment about the position, shape and number of nuclei in a eukaryotic cell. Add more examples from your previous knowledge about cell and nucleus. Cuboidal epithelial cell, different types of blood corpuscles, skeletal muscle fibre, adipocyte.
Answer:

Type of cells	Position of nucleus	Shape of Nucleus	Number of nuclei
Cuboidal epithelial cell	Central	Round or spherical	1
Neutrophils	Central	Multilobed/Segmented	1
Basophils	Central	S Shaped / Twisted	1
Eosinophils	Central	Bilobed	1
Monocytes	Central	Kidney Shaped	1
Lymphocytes	Central	Spherical	1
Skeletal Muscle Fibre	Peripheral	Oval	Multinucleate
Adipocytes	Shifted towards periphery	Eccentric	1
Simple squamous epithelium	Central	Flat	1
Ciliated simple columnar epithelium	Near base	Oval	1

(ii) Why nucleus is considered as control unit of a cell?
Answer:
a. Nucleus contains the genetic material of an organism.
b. This genetic material is present in the form of Deoxyribonucleic Acid (DNA) which is responsible for synthesis of various proteins and enzymes.
c. These proteins and enzymes in turn regulate metabolic activities of the cells.
Therefore, nucleus is considered as control unit of a cell.

(iii) Can cells like Xylem or mature human RBCs called living?
Answer:
a. Xylem is a complex tissue consists of tracheids, vessels, xylem parenchyma and xylem fibres. From these components of xylem, tracheids are dead cells and xylem parenchyma is the only living tissue,
b. RBCs do not possess nuclei once they reach maturity as they have to accommodate haemoglobin in them. They do not require a nucleus to function as they do not reproduce but only serve as a vehicle for the transport of oxygen and carbon dioxide in the blood.

(iv) What is a syncytium and coenocyte?
Answer:
Syncytium: It refers to mass of cells formed by fusion of multiple uninuclear cells and followed by dissolution of the cell membrane.
Coenocyte: It is a multinucleate cell resulted from multiple nuclear divisions without undergoing cytokinesis.

Can You Recall? (Textbook Page No. 44)

How do onion peel cells and our body cells differ?
Answer:
1 – (a, b, d, e,f g)

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 4 Kingdom Animalia Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 4 Kingdom Animalia

1. Choose the correct option

Question (A)
Which of the following belongs to a minor phylum?
(a) Comb jelly
(b) Jellyfish
(c) Herdmania
(d) Salpa
Answer:
(a) Comb jelly

Question (B)
Select the animal having venous heart.
(a) Crocodile
(b) Salamander
(c) Rohu
(d) Toad
Answer:
(c) Rohu

Question (C)
In Ascaris, _______ .
(a) mesoglea is present
(b) endoderm is a discontinuous layer
(c) mesoderm is present in patches
(d) body cavity is absent
Answer:
(c) mesoderm is present in patches

Question (D)
Which of the following is INCORRECT in case of birds?
(a) Presence of teeth
(b) Presence of scales
(c) Nucleated RBCs
(d) Hollow bones
Answer:
(a) Presence of teeth

Question (E)
Chitinous exoskeleton is a characteristic of ________ .
(a) Dentalium
(b) Antedon
(c) Millipede
(d) Sea urchin
Answer:
(c) Millipede

2. Answer the following questions.

Question (A)
Reptiles are known for having three chambered heart. Which animal shows a near four chambered condition in reptiles?
Answer:
Crocodiles have a four chambered heart.

Question (B)
The circulatory system has evolved from open to closed type in Animal Kingdom. Which Phylum can be called first to represent closed circulation?
Answer:
Phylum Annelida is the first phylum to represent closed circulation.

Question (C)
Pinna is part of external ear and it is found in mammals. Do Aves and Reptiles show external ear in any form?
Answer:
No, Aves and Reptiles do not show external ear in any form. They possess tympanum which represents the ear.

Question (D)
Fish and frog can respire in water. Can they respire through their skin? If yes, why do they have gills?
Answer:
1. Yes, fishes and frogs can respire through their skin.
2. The larval stage of frog i. e. tadpole respires through gills. During metamorphosis, tadpoles lose their gills and develop lungs.
3. Frogs do not have scales and breathe through their skin underwater.
4. Fishes respire primarily via gills. The body of fishes is covered with scales which limits cutaneous respiration in them.

Question (E)
Birds need to keep their body light to help in flying. Hence, they show presence of some organs only on one side. How their skeleton helps in reducing their weight?
Answer:

1. In birds, the forelimbs are modified into wings for flying.
2. They possess stream-lined body to reduce resistance during flight.
3. Bones are hollow or pneumatic to reduce body weight.
4. In order to reduce body weight, urinary bladder is absent. Also, females possess only left ovary and oviduct.
5. Body is covered by feathers to facilitate flying.

Question (F)
Cnidarians and Ctenophorans are both diploblastic. Which other character do they have in common, which is not found in other phyla?
Answer:
Cnidarians and ctenophorans show tissue level of body organization. They have blind sac body plan and radially symmetrical body.

Question (G)
Crab and Snail both have a protective covering. Is it made up of the same material?
Answer:
No, the protective covering is not made up of same material in crab and snail. The protective covering of crabs is made up of chitin and that of snails is made up of calcium carbonate.

Question (H)
Sponge and sea star show calcareous protective material. Do they belong to the same Phylum?
Answer:
No, they do not belong to same phylum. Sponges belong to phylum Porifera and sea star belongs to phylum Echinodermata.
1. Adult echinoderms are radially symmetrical but larval forms are bilaterally symmetrical.
2. Larvae of echinoderms are free-swimming.

Question (I)
Fish and snake both have scales. How do these scales differ from each other?
Answer:
Fishes have dermal scales covering the body surface whereas snakes have epidermal scales or scutes.

Question (J)
Lower Phyla like Arthropods and Cnidarians show metamorphosis. Is it also found in any class of Phylum Chordata?
Answer:
Yes, it is also found in class Amphibia of phylum Chordata.

Question 3.
Draw neat labelled diagram.
A. Sycon
B. Aurelia
C. Amphioxus
D. Catla
E. Balanoglossus
F. Scolidon
Answer:
A. Sycon

B. Aurelia

C. Amphioxus

D. Catla

E. Balanoglossus

F. Scolidon

Question 4.
Match the following.

Phylum	Characters
1. Annelida	(a) Tube feet
2. Mollusca	(b) Ostia
3. Ctenophora	(c) Radula
4. Porifera	(d) Parapodia
5. Echinodermata	(e) Comb plates

Answer:

Phylum	Characters
1. Annelida	(d) Parapodia
2. Mollusca	(c) Radula
3. Ctenophora	(e) Comb plates
4. Porifera	(b) Ostia
5. Echinodermata	(a) Tube feet

5. Identify the animals given in pictures and write features of its phylum/class.

Question 1.

Answer:
The organism in the given picture is Comb jelly (Red midwater Comb jelly) and it belongs to phylum Ctenophora.

Question 2.

Answer:
The organism in the given picture is Eel and it belongs to phylum Chordata.

Question 3.

Answer:
The given organism in the given picture is Dolphin and it belongs to class Mammalia.

Question 4.

Answer:
The given organism is Snake and it belongs to class Reptilia

Question 5.

Answer:
The given organism is Sea urchin and belongs to phylum Echinodermata.

Question 6.

Answer:
The given organism is flying lizard and belongs to class Reptilia.

Question 7.

Answer:
The organism is Herdmania and belongs to Phylum Chordata (Subphylum Urochordata).

Question 8.

Answer:
The organism in the given picture is Nautilus and it belongs to phylum Mollusca.

Question 9.

Answer:
The organism in the given picture is Amphioxus and it belongs to Phylum Chordata (Subphylum Cephalochordata).

6. Observe and identify body symmetry of given animals.

Question 1.
Observe and identify body symmetry of given animals.

Answer:
Fig i. represents asymmetry
Fig ii. represents radial symmetry
Fig iii. represents bilateral symmetry

Practical/Project:

Question 1.
Study different animals in kingdom Animalia and prepare the chart with detail scientific information.
Answer:
Phylum Porifera (Pori = Pores: feron = bearing): Members of the phylum Porifera are also called sponges. Characteristic features of the phylum:

1. Habitat: They are aquatic, mostly marine but few species are found in fresh water.
2. Forms: They are sedentary animals (attached to substratum or rock).
3. Body shape: They have asymmetrical body. Body of these animals consists of many cells with minimal
   division of labour among cells. Hence, their body is considered as a colony of different types of cells.
4. Body surface: Their body bears minute pores called ‘ostia’ through which water enters the spongocoel (body cavity). Water leaves the body through a large opening called ‘osculum’. Beating of flagella creates water current.
5. Circulation: Water is circulated in the body through the ‘canal system’. When the water enters the body of poriferans, cells absorb the food, exchange respiratory gases and release excretory products.
6. Digestive system: The body cavity of sponges (spongocoel) is lined by unique type of flagellated cells called choanocytes or collar cells for digestion.
7. Endoskeleton: The body of sponges consists of calcareous / siliceous spicules and proteinaceous ‘spongin fibres’.
8. Reproduction: Sponges reproduce asexually as well as sexually. Asexual reproduction takes place by fragmentation and gemmule formation. Sexual reproduction is by formation of gametes. Fertilization is internal and development is indirect through larval stage.
9. Sponges have great power of regeneration.
   e.g. Scypha, Euspongia (Bath sponge), Euplectella (Venus’ flower basket).

Characteristics of members belonging to phylum Cnidaria:

1. Habitat: They are aquatic, mostly marine and few of them are fresh – water forms.
2. Forms: They are sessile or free swimming.
3. Cnidoblasts: Presence of cnidoblasts or stinging cells are present on the tentacles for anchorage, offence and defence.
4. Body Symmetry: They have radially symmetrical body.
5. Germ layer: They are diploblastic.
6. Body cavity: Cnidarians have a central cavity called coelenteron or gastrovascular cavity, which helps in digestion and circulation. They have blind-sac body plan i.e., single pore opening to the exterior in the digestive system.
7. Body form: Members of this phylum exhibit two body forms. The cylindrical form, known as polyp e.g. Hydra and the umbrella – like form (.Aurelia – jelly fish) is known as medusa.
8. Digestion: They have extracellular and intracellular digestion.
9. Reproduction: Cnidarians reproduce asexually and sexually.

Asexual reproduction takes place by budding and regeneration. Sexual reproduction takes place gamete formation. They exhibit metagenesis i.e. alternation of polypoid generation with medusoid generation. Polyps produce medusae asexually and medusae produce polyps sexually, e.g. Obelia
e.g. Hydra, Aurelia (Jellyfish), Physalia (Portuguese man-of-war), Adamsia (Sea anemone), Diploria (Brain coral), Gorgonia (sea fan).

The members of this phylum are commonly known as comb jellies and sea walnuts. They are also known as acnidarians as they lack cnidoblasts. The phylum is considered as one of the minor phyla as it is represented by very few members.

Salient features of phylum Ctenophora:

1. Habitat: They are exclusively marine.
2. Forms: They are free swimming animals.
3. Germ layers: Members of this phylum are diploblastic.
4. Body Symmetry: They are radially symmetrical.
5. Body plan: The animals of this phylum show blind-sac body plan.
6. Body organization: They show tissue level organization.
7. Locomotion: It is earned out by eight rows of ciliated comb plates.
8. Bioluminescence: It is the characteristic feature of the members of this phylum.
9. Digestion: It is extracellular and intracellular.
10. Reproduction: Reproduction is sexual with indirect development.
11. Colloblasts: These sticky cells are used to capture prey, e.g. Pleurobrachia, Ctenoplana

11th Biology Digest Chapter 4 Kingdom Animalia Intext Questions and Answers

Can you recall? (Textbook Page No. 29)

(i) What is the basis for classification?
Answer:
Grades of organization, body symmetry, body cavity, germ layers and segmentation form the basis for classification.

(ii) Who proposed Five kingdom classification system?
Answer:
Robert Whittaker proposed the five kingdom system of classification.

(iii) What is the need and importance of classification?
Answer:
Need and importance for classification:
a. Classification facilitates the identification of animals with great accuracy.
b. The study of animals becomes convenient.
c. It helps in understanding the relationship of animals with other living organisms.
d. It helps to understand the habitat of each animal along with its role in nature.
e. By studying few animals from a group, we can gain a better understanding about the entire group.
f. It helps in understanding different adaptations shown by animals.
g. It gives an idea about evolution of animals.

Observe and discuss. (Textbook Page No. 29)

Discuss the criteria of classification.
Answer:
1. The given diagrams represents the number of germ layers and body symmetry used as criteria for animal classification.
2. Number of germ layers:
(a) When an organism shows only two germ layers, they are called diploblastic animals. In this case, the outer ectoderm is separated from the inner endoderm by a non-living substance called mesoglea.
(b) When an organism shows three germinal layers, they are called triploblastic animals. The three layers are namely – outer ectoderm, middle mesoderm and inner endoderm.
3. Body symmetry:
Body symmetry implies to the similarity in shape, size and number of parts on the opposite sides of a median line when body is divided into two halves by an imaginary line along different plane. Animals may be asymmetrical, radially symmetrical or bilaterally symmetrical.
(a) Asymmetrical animals:
An animal is said to be asymmetrical when its body cannot be divided into two identical halves in any plane.
(b) Radially symmetrical animals:
In certain animals, body can be cut or divided into two similar halves in a number of planes wherein, all the cuts (planes) pass through the centre. This type of symmetry is called radial symmetry.
(c) Bilaterally symmetrical animals:
In this type, the body of the animal can be bisected or divided in two equal or identical halves by a single median or vertical plane.

Internet my friend. (Textbook Page No. 30)

Which are the larval stages of Porifera.
Answer:
Larval stages of Porifera:
Parenchymula – Flagellate larvae of calcinean sponges
Amphiblastula – Free swimming larval stage of Sycon and many other calcareous sponges. Rhagon— Larval stage which give rise to the leuconoid condition in demospongiae.
[Students are expected to find more information about the larval stages of Porifera on internet.]

Find out. (Textbook Page No. 31)

Information about coral reefs and sea fan.
Answer:
Coral reefs:

1. A coral reef is an underwater ecosystem characterized by reef building corals.
2. Coral reefs constitute 25% of all marine species on the planet.
3. They belong to phylum Cnidaria.
4. There are three main types of coral reefs – fringing, barrier and atoll. Coral reefs provide ecosystem services for tourism, fisheries and shoreline protection.
5. They cannot survive in high temperatures, thus due to climate change there is a sharp decline in their population.

Sea fan or Gorgonia:

1. It is a soft coral composed of numerous polyps – cylindrical, sessile (attached) forms that grow together in a flat, fan-like pattern.
2. It belongs to phylum Cnidaria.
3. It does not produce calcium carbonate skeletons.
[Students can find out more information about coral reefs and sea fan using internet ]

Can you tell? (Textbook Page No. 32)

(i) State the parasitic adaptations in Liver fluke and Ascaris.
Answer:
Parasitic adaptations in Liver fluke:
a. Presence of hooks and suckers
b. Body covered with cuticle
c. Lacks digestive system
d. They are hermaphrodites

Parasitic adaptations in Ascaris:
a. Presence of muscular pharynx for sucking the food.
b. Body covered by tough, thick and resistant cuticle.
c. Secretes enzymes against the enzymes secreted by the host.
d. Respiration is anaerobic.
e. Reproductive system is highly developed.

(ii) Give example of free living platyhelminth.
Answer:
Planaria

Find out. (Textbook Page No. 33)

What are the merits and demerits of hermaphroditism?
Answer:
Hermaphroditism is the condition in which an organism possesses reproductive organs of both the sexes.

Merits of hermaphroditism:
a. Assured fertilization which reduces the risk of a species to become extinct due to unavailability of mating partner.
b. Energy required for searching out mating partner is conserved.
c. Frequency of mating is maximized.

Demerits of hermaphroditism:
a. More energy is required to maintain both the reproductive systems.
b. Limited gene diversity.
[Source: http://floydbiology. blogspot. com/2012/06/httpmattc-thinks. html]
[Students are expected to find more information using the internet.]

Why are leeches used in Ayurveda?
Answer:
a. Leeches are used in blood purification therapy to treat many diseases as they suck impure blood from the affected site of the patient’s body.
b. The anticoagulant – hirudin present in saliva of leech, inhibits the coagulation of blood and makes blood thinner. This dissolves the clots found in vessels and facilitates the blood supply.

What is the role of earthworms in agriculture? What is vermicompost?
Answer:
Role of earthworms in agriculture:
a. Earthworms loosen the soil by burrowing deep into it, thus they help to aerate the soil.
b. This continuous digging of soil also helps the water to reach the roots quickly.
c. Earthworms can decompose the organic matter from the soil and convert it into rich manure.
d. This helps in increasing the fertility of soil which ultimately increases the crop production.
e. Earthworm castings are rich in nutrients which act as natural fertilizer.
Vermicompost:
Vermicompost is the product of vermicomposting. It is organic manure produced as vermicast by earthworm feeding on biological waste material and plant residues.

Can you tell? (Textbook Page No. 34)

(i) Explain the term metameric segmentation.
Answer:
In some animals, body consists of many segments arranged along the length of the body. When the external segmentation coincides with the internal segmentation, it is called as metameric segmentation and the phenomenon is called metamerism.

(ii) Give characteristics of Arthropoda.
Answer:
Arthropoda (Arthros: Joint, Podos: leg): Arthropoda forms the largest phylum of kingdom Animalia. Characteristics of Arthropoda:
a. Habitat: Arthropods are omnipresent.
b. Forms: Solitary or colonial, most of them are free-living. Barnacles are sedentary. Few are parasitic and sanguivorous, (e.g. Female mosquito, bed bug.)
c. Body symmetry: Body is bilaterally symmetrical.
d. Germ layers: They are triploblastic.
e. Body cavity: Arthropods are eucoelomates.
f. Body plan: They show tube within tube body plan.
g. Level of body organization: They show organ system level of organization.
h. Special features: The members of this phylum have jointed appendages. Hence, they are known as arthropods. Some insects like honey bee, ants, termites, etc. exhibit polymorphism.
i. Exoskeleton: Body is covered by a tough, non – living chitinous exoskeleton. As the exoskeleton does not allow body growth, arthropods shed off their exoskeleton periodically during growth. This process is called moulting or ecdysis.
j. Body division: Body is divided into head, thorax and abdomen.
k. Segmentation: Body shows metameric segmentation.
l. Digestion: Digestive system is complete and divided into foregut, midgut and hindgut.
m. Circulation: Circulatory system is of open type wherein, blood flows through body cavity called haemocoel.
n. Respiration: Respiration occurs through respiratory organs like gills, trachea, book lungs or book
gills.
o. Excretion: Excretion takes place by green glands, Malpighian tubules or coxal glands.
p. Nervous system: Nervous system consists of nerve ring and double, ventral ganglionated nerve cord.
q. Sense organs:Arthropods have well developed sense organs in the form of antennae, simple or compound eye and various receptors.
r. Sexual reproduction: Sexes are generally separate in arthropods with distinct sexual dimorphism.
s. Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silk worms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

(iii) Enlist the harmful Arthropods.
Answer:
Significance:
Beneficial arthropods: Some arthropods are of economic importance. For example, Honey bees (Apis) are important for their honey and wax, silkworms for the production of silk. Lobsters, prawns, crabs are edible. Harmful arthropods: Some arthropods are harmful and act as vectors to spread various diseases, e.g., Mosquitoes. Locusta (locust) is a gregarious pest. Limulus (King crab) is a living fossil.
Other examples: Cockroach (Periplaneta), butterfly, scorpion (Hottentotta) and millipede (Archispirostreptus) prawn.

Find out. (Textbook Page No. 34)

(i) Why is phylum Arthropoda considered as most successful phylum?
Answer:
Phylum Arthropoda is considered as most successful phylum because of the following reasons:
a. Phylum Arthropoda is the largest phylum of kingdom Animalia. It includes various forms like lobsters, prawns, crabs, insects, millipedes, locust, honeybees, etc.
b. They are omnipresent (present everywhere). Arthropods show great variety of adaptations as their habitat varies from terrestrial to aquatic habitat.
c. Several others factors also contribute to the success of the phylum which includes:
1. The exoskeleton of arthropods is made up of tough chitinous exoskeleton. This enables them to survive on lands in almost all environment and is a great defense against predators.
2. They possess jointed appendages which allow complex movements.
3. They exhibit moulting or eedysis.
4. They have metamerically segmented body helping in movement around diverse environments.

(ii) What do we mean by parthenogenesis?
Answer:
Development of an egg into a complete individual without fertilization is known parthenogenesis. It is found in many non-vertebrates such as bees, rotifers and even some lizards and birds (turkey).

(iii) What do we mean by living fossil?
Answer:
A member of a living animal or plant species that is almost identical to species known from the fossil record (not the recent fossil record), i.e. they have changed very little over a long period.
[Source:https://www. encyclopedia, com/earth-and-environment/ecology-and- environmentalism/environmental-studies/living-fossil]

(iv) How the bees produce honey?
Answer:
a. Bees produce honey using the nectar of flowering plants. A bee sucks the nectar and stores it in a honey sac until it returns to the hive.
b. The nectar is then transferred to worker bees in the hive who suck the nectar from the honey sac through their proboscis. This nectar contains 70% water and 20% honey. Honeybees get rid of excess water by swallowing and regurgitating the nectar again and again. They also fan their wings over filled cells of honeycomb.

When most of the water has evaporated from the honeycomb, the bee seals the comb with a secretion of liquid from its abdomen which eventually hardens into beeswax. This is how the honey bees use nectar to produce a thick, sticky and sweet honey.

(v) What will happen if arthropods do not moult?
Answer:
a. Moulting or eedysis is a periodic shedding of the outer cuticle layer of body in arthropods.
b. The outer layer of body of arthropods is formed of tough, non-living chitinous substance.
c. If arthropods do not moult, they cannot grow and mature into adult forms

Can you tell? (Textbook Page No. 34)

Why do Molluscs have shell?
Answer:
Molluscs are soft-bodied animals. Thus, the calcareous shell provides supports and protects the organisms from predators.

Can you tell? (Textbook Page No. 36)

Give salient features of phylum Echinodermata.
Answer:
Salient features of phylum Echinodermata (Echinus – spines, derma – skin)

1. Habitat: These are exclusively marine.
2. Forms: Members of this phylum are solitary, sedentary or free-living and gregarious, benthic.
3. Body symmetry: These animals are radially symmetrical with pentamerous symmetry.
4. Shape: Members of Echinodermata are spherical, elongated or star-shaped.
5. Body: The endoskeleton is made up of calcareous ossicles. Spines are formed on the body. Hence, they are known as echinoderms. The body has two sides oral and aboral and lacks definite divisions. Mouth is ventrally present on oral surface and anus on aboral surface.
6. Water vascular system: Presence of water vascular system is the peculiar character of echinoderms. The madrepOrite is the opening of water vascular system through which water enters. Water vascular system is useful in locomotion, food capturing, respiration.
7. Digestion: Digestive system is complete.
8. Respiration: Peristomial gills, papillae, respiratory tree, etc. are used for respiration.
9. Circulatory and excretory systems: Absent in echinoderms.
10. Nervous system: Nervous system is simple with a nerve ring around the mouth and radial nerves in arms.
11. Reproduction and development: Sexes are separate (sometimes bisexual). Fertilization is external.
12. Development is indirect, i.e. through larval stages. They show high power of regeneration.

e.g. Sea lily (Antedon), Sea star (Asterias), Sea cucumber (Cucumaria), Brittle star (Ophiothrix), Sea urchin (Echinus).

Can you tell? (Textbook Page No. 36)

Can you tell? (Textbook Page No. 36)
Answer:
1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered as notochord. It is now placed as a separate phylum under Non-Chordata.
2. It possesses certain characteristics of both Chordates and Non-chordates.
3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Find out. (Textbook Page No. 36)

Why Balanoglossus is considered as connecting link between Non-chordates and chordates?
Answer:
Balanoglossus belongs to phylum Hemiehordata. For Explanation:

1. Hemiehordata was earlier considered as sub phylum of Chordata because the buccal diverticulum was considered as notochord. It is now placed as a separate phylum under Non-Chordata.
2. It possesses certain characteristics of both Chordates and Non-chordates.
3. Absence of notochord worm-like body, heart located on the dorsal side are the Non-chordate like characteristics seen in Hemiehordata.
4. Presence of nerve chord, pharyngeal gill slits are some of the Chordate-like characters seen in Hemiehordata. Hence, Hemiehordata is considered as a connecting link between Non-chordata and Chordata.

Observe and discuss. (Textbook Page No. 36)

Compare and contrast between Non-Chordates and Chordates.

Answer:

Non-chordates	enoraates
1. Notochord is absent.	Notochord present at least in the early embryonic life.
2. Nerve cord is ventral, paired and ganglionated.	Nerve cord is single, dorsal and non-ganglionated.
3. The heart, if present is dorsal.	The heart is ventral in position.
4. Pharyngeal gill slits are absent.	Pharyngeal gill slits are present at least in embryonic stage.
5. Post-anal tail is absent.	Post-anal tail is present at least in embryonic stage.

Can you tell? (Textbook Page No. 37)

Herdmania is called a Chordate. Explain.
Answer:
1. Herdmania belongs to phylum Urochordata.
2. It is called a chordate as it shows the following features:
a. Presence of notochord at least in early embryonic life. (In Herdmania, notochord is present in the tail of the larval forms).
b. Presence of hollow, dorsal nerve chord, running throughout the length of the body.
c. Presence of pharyngeal gill slits.
d. Presence of post-anal tail.

Can you tell? (Textbook Page No. 37)

Give characteristics of Petromyzon. Comment on its mode of nutrition.
Answer:
Characteristic features of class Cyclostomata (Cyclos: Circular, stoma-mouth) Lat/Grk

1. Members of class Cyclostomata are jaw-less and eel like organisms.
2. Their skin is devoid of scales, soft and smooth, containing unicellular mucus glands.
3. Median fms are present but paired fins are absent.
4. They are ectoparasites on fishes.
5. They have sucking circular mouth, without jaws.
6. Cranium and vertebral column are made up of cartilage.
7. Their digestive system lacks stomach.
8. Respiration occurs by 6 – 15 pairs of gill slits. Gills slits are without operculum.
9. Heart is two chambered with one auricle and one ventricle.
10. Gonad is single, large and without gonoduct.
11. Fertilization is external. They are anadromous as they migrate for spawning to fresh – water from marine habitat.
12. After spawning, they die within few days. Larvae metamorphosize and migrate to ocean.
    e.g Petromyzon (Lamprey), Myxine (Hagfish).

Can you tell? (Textbook Page No. 38)

(i) What is the lateral line system?
Answer:
a. Lateral line system is the system with mechanoreceptors called neuromasts, for the detection of watei current.
b. These neuromasts are arranged in an interconnected network along the head and body.
c. Lateral line system also known as lateralis system.

(ii) Why Piscian heart is called a venous heart?
Answer:
a. Pisces have two-chambered heart. They have single and closed circulation.
b. Heart of Pisces receives blood only from veins and thus always shows presence of deoxygenated blood which it pumps directly to the gills for oxygenation.
Thus, the heart of Pisces is called a venous heart.

Can you tell? (Textbook Page No. 40)

Amphibians do not have exoskeleton. Give reason.
Answer:
1. Amphibians live in both water and on land.
2. They perform cutaneous respiration (i. e. gaseous exchange across the skin or outer integument.) under water and when on land, they respire through lungs.
Thus, to facilitate cutaneous respiration, amphibians do not have exoskeleton.

Can you tell? (Textbook Page No. 40)

Why are amphibians and reptilians called poikilotherms?
Answer:
Amphibians and reptilians are called poikilotherms as they cannot maintain a constant body temperature. Their body temperature changes according to the change in surrounding temperature.

Can you tell? (Textbook Page No. 41)

Give adaptations in Aves for flying.
Answer:

1. In birds, the forelimbs are modified into wings for flying.
2. They possess stream-lined body to reduce resistance during flight.
3. Bones are hollow or pneumatic to reduce body weight.
4. In order to reduce body weight, urinary bladder is absent. Also, females possess only left ovary and oviduct.
5. Body is covered by feathers to facilitate flying.

Can you tell? (Textbook Page No. 41)

(i) Aves and mammals are homeotherms. Give reason.
Answer:
a. Aves and mammals can generate heat to maintain their body temperature.
b. They keep their body temperature constant, irrespective of fluctuations in environmental temperature. Thus, Aves and mammals are homeotherms.

(ii) How mammals differ from other groups of animals?
Answer:
Features of class Mammalia (mammae: breasts, nipple):

1. Special feature: Presence of mammary glands (milk-producing glands) for the nourishment of young ones. Mammary glands are modified sweat glands.
2. Habitat: Mammals are omnipresent (present everywhere). These are mostly terrestrial, some are aquatic and few are aerial and arboreal (living on trees).
3. Locomotion: Limbs are the organs of locomotion and are modified for walking, climbing, burrowing, swimming, etc.
4. Body division: Body is differentiated into head, neck, trunk and tail. They have external ear (pinna).
5. Body temperature: Mammals are homeotherms or warm-blooded animals.
6. Exoskeleton: It is in the form of hair, fur, nails, hooves, horns, etc.
7. Skin: Skin is glandular and has sweat glands and sebaceous (oil) glands.
8. Mouth cavity: Mammals show heterodont dentition (various types of teeth like incisors, canines, premolars and molars).
9. Circulation: Heart is ventral in position, four chambered with two auricles and two ventricles. RBCs are biconcave and enucleated (except camel). Blood is red in colour.
10. Respiration: Respiration takes place by lungs.
11. Nervous system: Brain is highly developed. Cerebrum shows a transverse band called corpus callosum.
12. Reproduction and development: Only few mammals are oviparous, e.g. Duck billed platypus. Some have pouches for development of immature young ones. These are called marsupials, e.g. Kangaroo. Most of the mammals are placental and viviparous.

Do yourself. (Textbook Page No. 41)

Observe different animals in your surrounding, write a detailed classification and write down the characteristics of animals in the following format.
Answer:

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 3 Kingdom Plantae Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 3 Kingdom Plantae

1. Choose the correct option.

Question (A)
Which is the dominant phase in Pteridophytes?
(a) Capsule
(b) Gametophyte
(c) Sporophyte
(d) Embryo
Answer:
(c) Sporophyte

Question (B)
The tallest living gymnosperm among the following is
(a) Sequoia sempervirens
(b) Taxodium mucronatum
(c) Zamia pygmaea
(d) Ginkgo biloba
Answer:
(a) Sequoia sempervirens

Question (C)
In Bryophytes
(a) sporophyte and gametophyte generation are independent
(b) sporophyte is partially dependent upon gametophyte
(c) gametophyte is dependent upon sporophyte
(d) inconspicuous gametophyte
Answer:
(b) sporophyte is partially dependent upon gametophyte

Question (D)
A characteristic of Angiosperm is
(a) Collateral vascular bundles
(b) Radial vascular bundles
(c) Seed formation
(d) Double fertilization
Answer:
(d) Double fertilization

Question (E)
Angiosperms differ from gymnosperms in having
(a) Vessels in wood
(b) Mode of nutrition
(c) Siphonogamy
(d) Enclosed seed
Answer:
Both (a) Monocotyledons and (d) Enclosed seed

Question 2.
How you place the pea, jowar and fern at its proper systematic position? Draw a flow chart.
Answer:

Question 3.
Complete the following table.

Groups of algae	Chlorophyceae	Phaeophyccac	Rhodophyceae
1. Stored food	Starch
2. Cell wall	—	Cellulose and algin
3. Major pigments	—		Chl-a, d and phycoerythrin

Answer:

Groups of algae	Chlorophyceae	Phaeophyccac	Rhodophyceae
1. Stored food	Starch	Mannitol, laminarin	Floridean starch
2. Cell wall	Cellulose	Cellulose and algin	Cellulose, pectin
3. Major pigments	Chl-a, b	Chl-a, c, fucoxanthin	Chl-a, d and phycoerythrin

Question 4.
Differentiate between Dicotyledonae and Monocotyledonae based on the following characters:
a. Type of roots
b. Venation in the leaves
c. Symmetry of flower
Answer:

Characters	Dicotyledonae	Monocotyledonae
1. Type of roots	Taproots	Fibrous roots
2. Venation in the leaves	Reticulate venation	Parallel venation
3. Symmetry of flower	Tetramerous or Pentamerous symmetry	Trimerous symmetry

Characters Dicotyledonae Monocotyledonae
1. Type of roots Tap roots Fibrous roots
2. V enation in the leaves Reticulate venation Parallel venation
3. Symmetry of flower Tetramerous or Pentamerous symmetry Trimerous symmetry

5. Answer the following questions.

Question (A)
We observe that land becomes barren soon after monsoon. But in the next monsoon it flourishes again with varieties we observed in season earlier. How you think it takes place?
Answer:

1. After monsoon, plants like mosses (bryophytes), ferns (pteridophytes), small herbaceous plants, etc become dry, due to which land becomes barren.
2. However, spores of bryophytes, pteridophytes and seeds of herbaceous plants, grass remain in barren land.
3. During next monsoon, these spores and seeds germinate due to availability of water and other favourable conditions.
4. Bryophytes and pteridophytes require water for reproduction. Hence they flourish during monsoon season.
5. Along with bryophytes and pteridophytes varieties of higher plants like grasses, some seasonal herbs or shrubs grow on barren land during monsoon due to favourable conditions.

Question (B)
Fern is a vascular plant. Yet it is not considered a Phanerogams. Why?
Answer:

1. Fern belongs to sub-kingdom Cryptogamae.
2. Cryptogams produce spores but do not produce seeds.
3. Also, in cryptogams the sex organs are concealed.
4. Phanerogams are seed producing plants and their sex organs are visible.
5. Hence, fern is a vascular plant. Yet it is not considered a Phanerogams.

Question (C)
Chlamydomonas is microscopic whereas Sargassum is macroscopic; both are algae. Which characters of these plants includes them in one group?
Answer:

1.  Both Chlamydomonas and Sargassum belong to division Thallophyta.
2. Members of Thallophyta range from unicellular (e.g. Chlamydomonas) to multicellular (e.g. Sargassum).
3. Both are aquatic plants containing photosynthetic pigments.
4. In both Chlamydomonas and Sargassum plant body is not differentiated into root, stem and leaves.
5. The stored food is mainly in the form of starch and its other forms.
6. Cell wall is made up of cellulose and other components. Due to these characters, both Chlamydomonas and Sargassum are included in one group i.e. Thallophyta.

Question 6.
Girth of a maize plant does not increase over a period of time. Justify.
Answer:

1. Maize plant belongs to class monocotyledonae.
2. In monocotyledonous plants, vascular bundles are closed type.
3. Thus, cambium is absent between xylem and phloem, due to which secondary growth does not occur in these plants.
4. Increase in girth of a stem occurs by secondary growth. Thus, girth of a maize plant does not increase over a period of time.

Question 7.
Radha observed a plant in rainy season on the compound wall of her school. The plant did not have true roots but root like structures were present. Vascular tissue was absent. To which group the plant may belong?
Answer:
The plant observed by Radha belongs may belong to division Bryophyta, as it shows root like structures i.e. rhizoids and absence of vascular tissue.

8. Draw neat labelled diagrams

Question 1.
Draw neat and labelled diagram of:
(A) Spirogyra
(B) Chlamydomonas
Answer:

Question (C)
Draw neat and labelled diagram of Funaria.
Answer:

Question (D)
Draw neat and labelled diagram of Nephrolepis.
Answer:

[Note: Frond: Fern leaf, originating from rhizome. It consists of blade and petiole, Blade: Main part of the frond which is rich in chlorophyll]

Question (E)
Draw neat and labelled diagram of Haplontic and Haplo-diplontic life cycle.
Answer:

Question 9.
Identify the plant groups on the basis of following features:
A. Seed producing plants
B. Spore producing plants
C. Plant body undifferentiated into root, stem and leaves
D. Plant needs water for fertilization
E. First vascular plants
Answer:
1. Phanerogams (Angiospermae and Gymnospermae)
2. Cryptogams (Thallophyta, Bryophyta and Pteridophyta)
3. Thallophyta, Bryophyta
4. Thallophyta, Bryophyta, Pteridophyta
5. Pteridophytes

Practical/Project:

Question 1.
Study the Nephrolepis plant in detail.
Answer:

1. Nephrolepis belongs to division pteridophyta.
2. They grow abundantly in cool, shady, moist places.
3. Roots are adventitious (fibrous) growing from the underground stem.
4. Leaves are well developed on the stem (Rhizome).
5. They show presence of well-developed conducting system for transportation of water and food.
6. They reproduce asexually by spores produced within sporangia, which are present in sori. These sori are located along the posterior surface of leaflets.
7. These plants have neither fruits nor flowers.
8. Some ferms are used as food, medicine or as ornamental plants.

Question 2.
Study the coralloid roots, scale leaf and megasporophyll of Cycas in detail.
Answer:
1. Coralloid roots of Cycas:
Coralloid roots of Cycas show association with blue green algae for nitrogen fixation.
Coralloid roots are coral-like, dichotomously branched and fleshy. They grow upward toward the surface of the soil. These roots arise from the lateral branches of normal roots.
2. Scale leaf of Cycas:
In Cycas leaves are dimorphic i.e. foliage leaves and scale leaves. Scale leaves are minute, membranous and brown. These are non- photosynthetic and provide protection to the stem apex.
3. Megasporophyll of Cycas:
Megasporophylls are usually arranged in compact structures called female cones or female strobili. Megasporophyll contains megasporangia (ovule) which produce megaspores.
[Students are expected to collect more information about coralloid roots, scale leaf and megasporophyll of Cycas.]

Question 10.
Observe the following diagram. Correct it and write the information in your words.

Answer:

1. The given figure indicates alternation of generation.
2. The life cycle of a plant includes two generations, sporophytic (diploid = 2n) and gametophytic (haploid = n)
3. Some special diploid cells of sporophyte divide by meiosis to produce haploid cells.
4. These haploid cells divide mitotically to produce gametophyte.
5. On maturation, gametophyte produces male and female gametes which fuse during fertilization and produce diploid zygote.
6. Diploid zygote divides by mitosis and forms diploid sporophyte.

11th Biology Digest Chapter 3 Kingdom Plantae Intext Questions and Answers

Can you recall? (Textbook Page No. 19)

Why do we call plants as producers on land?
Answer:
Plants can prepare their own food by the process of photosynthesis. Hence, they are called as producers on land.

Can you recall? (Textbook Page No. 19)

What are differences between sub-kingdoms cryptogamae and Phanerogamae?
Answer:

Cryptogamae	Phanerogamae
1. Plants belonging to this sub-kingdom are non­flowering.	Plants belonging to this sub-kingdom are flowering.
2. Sex organs are concealed.	Sex organs are visible.
3. These plants do not produce seeds.	These plants produce fruits and seeds.
4. An ovule is not formed.	An ovule is formed.
5. It is further divided into three divisions, viz.	It is further divided into two divisions, viz.
6. Thallophyta, Bryophyta and Pteridophyta.	Gymnospermae and Angiospermae.

Observe and Discuss (Textbook Page No. 19)

Collect different water samples of fresh water. Mount them on a glass slide and observe under a compound microscope. Try to identify the organisms which are visible under it.
Answer:
Micro-organisms like Paramoecium, Amoeba, blue-green algae, unicellular algae, filamentous algae can be observed under compound microscope.
[Students are expected to observe different water samples of fresh water under compound microscope and identify the organisms.]

Can you tell? (Textbook Page No. 21)

Give salient features of algae.
Answer:
Algae belongs to division Thallophyta.
Salient features of algae:
1. Habitat: Algae are mostly aquatic, few grow on other plants as epiphytes and some grow symbiotically. Some algae are epizoic i.e. growing or living non-parasitically on the exterior of living organisms.
Aquatic algae grow in marine or fresh water. Most of them are free-living while some are symbiotic.

2. Structure: Plant body is thalloid i.e. undifferentiated into root, stem and leaves. They may be small, unicellular, microscopic like Cblorella (non-motile), Chlamydomonas (motile). They can be multicellular, unbranched, filamentous like Spirogyra or branched and filamentous like Chara. Sargassum is a huge macroscopic sea weed which measures more than 60 meters in length.

3. Cell wall: The algal cell wall contains either polysaccharides like cellulose / glucose or a variety of proteins or both.
Reserve food material: Reserve food is in the form of starch and its other forms.

4. Photosynthetic pigments: Photosynthetic pigments like chlorophyll – a, chlorophyll – b, chlorophyll – c, chlorophyll – d, carotenes, xanthophylls, phycobilins are found in algae.

5. Reproduction: Reproduction takes place by vegetative, asexual and sexual method.

6. Life cycle: The life cycle shows phenomenon of alternation of generation, dominant haploid and reduced diploid phases.

Internet my friend (Textbook Page No. 20)

Write different pigments found in algae.
Answer:
Various types of photosynthetic pigments are found in algae.
1. Chlorophyll-a (Essential photosynthetic pigment) is present in all groups of algae.
2. The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Can you tell? (Textbook Page No. 21)

Name the accessory pigments of algae.
Answer:
The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Can you tell? (Textbook Page No. 21)

Differentiate between Chlorophyceae and Phaeophyceae.
Answer:

Chlorophyceae (Green algae)	Phaeophyceae (Brown algae)
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-b.	Photosynthetic pigments are chlorophyll-a, chlorophyll-c and fucoxanthin.
2. Reserve food is in the form of starch.	Reserve food is mannitol and laminarin.
3. e.g. Chlorella, Chlamydomonas, Spirogyra, Chara, I Volvox, Ulothrix	Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Can you tell? (Textbook Page No.21)

Enlist examples of Chlorophyceae and Rhodophvceae.
Answer:
1. Examples of Chlorophyceae:
Chlorella, Chlamydomonas, Spirogyra, Char a, Volvox, Ulothrix, etc.
2. Examples of Rhodophyceae:
Chondrus, Batrachospermum, Porphyra, Gelidium, Gracillaria, Polysiphonia, etc.

Internet my friend (Textbook Page No. 21)

Different forms of green, red, brown and blue green algae.
Answer:
1. Forms of green algae:
Unicellular motile: e.g. Chlamydomonas Unicellular non-motile: E.g. Chlorella Colonial forms: e.g. Volvox Filamentous branched: e.g. Cladophora, Chara Filamentous unbranched: e.g. Ulothrix, Spirogyra

2. Forms of red algae:
The red thalli of most of the red algae are multicellular, macroscopic, e.g. Gracilaria, Gelidium, Porphyra, Polysiphonia, etc. .

3. Forms of brown algae:
Simple, branched and filamentous: Sargassum, Fucus, Ectocarpus Profusely branched: Laminaria, Dictyota, Kelps (Seaweed)

4. Forms of blue-green algae:
Unicellular, colonial or filamentous, freshwater or marine water or terrestrial algae.
[Note: Blue-green algae are cyanobacteria which are photosynthetic autotrophs.]
[Students are expected to collect more information from internet.]

Internet my friend (Textbook Page No. 20)
Enlist the forms of filamentous algae.
Answer: The forms of filamentous algae:
1. Filamentous branched: e.g. Cladophora, Chara, Ectocarpus, Dictyota, etc.
2. Filamentous unbranched: e.g. Ulothrix, Spirogyra, etc.

Internet my friend (Textbook Page No. 21)

Economic importance of algae.
Answer:
(a) Many species of algae are used as food. For e.g. Chlorella (rich in cell proteins hence used as food supplement, even by space travelers), Sargassum, Laminaria, Porphyra, etc.
(b) Alginic acid is produced commercially from Kelps.
(c) Hydrocolloids like algin and carrageen are obtained from brown algae and red algae respectively.
(d) ‘Agar’ which is used as solidifying agent in tissue culture is obtained from red algae like Gelidium and Gracilaria.
(e) Brown algae like sea weeds are used a fodder for sheep, goat, etc.
[Students are expected to collect more information about the economic importance of algae.]
(f) Role of algae in environment.
Answer:
(a) Being photosynthetic, algae help in increasing the level of dissolved oxygen in their immediate environment.
(b) Algae are primary producers of energy rich compounds which forms the basis of food cycles in aquatic animals.
[Students are expected to find out more information about the role of algae in environment on internet.]

Can you recall? (Textbook Page No. 19)

Differentiate between Thallophytes and Bryophytes.
Answer:

Thallophytes	Bryophytes
1. Mostly aquatic in habitat.	Mostly terrestrial, occurs on moist and shady places.
2. Thallus may be unicellular or multicellular.	Thallus is multicellular.
3. Motile and non-motile forms are present.	Non-motile forms present, except male gametes.
4. Rhizoids are absent.	Rhizoids are present.

Can you tell? (Textbook Page No. 23)

Why Bryophyta are called amphibians of Plant Kingdom?
Answer:
Members of Bryophyta are mostly terrestrial plants which depend on water for fertilization and completion of their life cycle. Hence, they are called ‘amphibians of Plant Kingdom’.

Observe and Discuss (Textbook Page No. 21)

You may have seen Funaria plant in rainy season. Why is it called amphibious plant?
Answer:
Funaria belongs to division Bryophyta.
It is a terrestrial plant but requires water for fertilization and completion of its life cycle. Hence, it is called as an amphibious plant.

Observe and Discuss (Textbook Page No. 23)

You may have seen the various plants which do not bear flowers, fruits and seeds but they have well developed root, stem and leaves. Discuss.
Answer:
1. The plants which do not bear flowers, fruits and seeds, but have true roots, stem and leaves belong to division Pteridophyta.
2. These plants are cryptogams as they do not produce seeds and flowers.
3. They have primitive conducting system.

Can you tell? (Textbook Page No. 23)

Pteridophytes are also known as vascular Cryptogams – Justify.
Answer:
1. The reproductive organs of pteridophytes are hidden.
2. Pteridophytes do not produce flowers, fruits and seeds. They reproduce asexually by forming spores and sexually by forming gametes, hence they belong to Cryptogamae.
3. These plants possess a primitive conducting system. Thus, conduction of water and food occurs through vascular tissue.
Hence, Pteridophytes are also known as vascular Cryptogams.

Can you tell? (Textbook Page No. 23)

Give one example of aquatic and xerophytic Pteridophytes.
Answer:
Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).

Can you recall? (Textbook Page No. 19)

Give any two examples of Pteridophyta.
Answer:
Nephrolepis, Selaginella, Azolla, Marsilea, Equisetum, Lycopodium, Psilotum, Dryopteris, Pteris, Adiantum.

Can you tell? (Textbook Page No. 25)

Give general characters of Gymnosperms and Angiosperms.
Answer:
1. General characters of Gymnosperms:
(a) Types: Most of the gymnosperms are evergreen, shrubs or woody trees.
(b) Vascular tissues: They are vascular plants having xylem with tracheids and phloem with sieve cells.
(c) Flower: These are primitive group of flowering plants producing naked seeds.
(d) Body: The plant body is sporophyte. It is differentiated into root, stem and leaves.
(e) Roots: The root system is tap root type. In some gymnosperms, the roots form symbiotic association with other life forms. Coralloid roots of Cycas show association with blue green algae and roots of Pinus show association with endophytic fungi called mycorrhizae.
(f) Stem: In gymnosperms, stem is mostly erect, aerial, solid and cylindrical. Secondary growth is seen in Gymnosperms due to the presence of cambium. In Cycas it is usually unbranched, while in conifers it is branched, (e.g. Pinus, Cedrus).
(g) Leaves: The leaves are dimorphic. The foliage leaves are green, simple needle like or pinnately compound, whereas scale leaves are small, membranous and brown.
(h) Spores: Spores are produced by microsporophyll (Male) and megasporophyll (Female).

(ii) General characters of angiosperms:
(a) Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.
(b) Alternation of generations: Angiosperms show heteromorphic alternation of generation in which the sporophyte is diploid, dominant, autotrophic and independent. The gametophytes (male or female) are haploid, reduced and concealed.
(c) Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.
(d) Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Observe and Discuss (Textbook Page No. 23)

Observe all garden plants like Cycas, Thuja, Pinus, Sunflower, Canna and compare them. Note similarities and dissimilarities among them.
Answer:
1. When we observe garden plants like Cycas, Thuja, Pinus, Sunflower, Canna, following similarities can be observed:
Plant body is divided into root, stem and leaves.
2. When we observe garden plants like Cycas, Thuja, Pinus, Sunflower, Canna, following dissimilarities can be observed:
(a) In Cycas, Thuja and Pinus seeds are not enclosed within a fruit, whereas in Sunflower and Canna seeds are enclosed within a fruit.
(b) Plants like Cycas, Thuja, Pinus show cones bearing microsporophylls and megasporophylls, whereas sunflower and Canna plant bear flowers.
(c) In Cycas, Thuja and Pinus green, simple needle like or pinnately compound foliage leaves and brown, membranous scaly leaves can be observed, whereas in Sunflower, Canna green foliage leaves can be observed.

Can you recall? (Textbook Page No. 24)

What are the salient features of angiosperms?
Answer:
(ii) General characters of angiosperms:
(a) Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.
(b) Alternation of generations: Angiosperms show heteromorphic alternation of generation in which the sporophyte is diploid, dominant, autotrophic and independent. The gametophytes (male or female) are haploid, reduced and concealed.
(c) Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.
(d) Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Can you recall? (Textbook Page No. 24)

What is double fertilization?
Answer:
(a) Double fertilization is a characteristic feature of angiosperms.
(b) In this process one male gamete fuses with egg cell and another male gamete fuses with secondary nucleus, to form an embryo and endosperm respectively.

Can you recall? (Textbook Page No. 24)

Explain in brief the two classes of Angiosperms? Draw and label one example of each class.
Answer:
Two classes of Angiosperms are Dicotyledonae and Monocotyledonae.
а. Dicotyledonae:

1. These plants have two cotyledons in their embryo.
2. They have a tap root system and the stem is branched.
3. Leaves show reticulate venation.
4. Flowers show tetramerous or pentamerous symmetry.
5. Vascular bundles are conjoint, collateral and open type.
6. Cambium is present between xylem and phloem for secondary growth.
7. In dicots, secondary growth is commonly found. e. g. Helianthus annuus (Sunflower)

b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Try This (Textbook Page No. 24)

Study the leaves of Hibiscus, Peepal, Canna, Grass and Tulsi. Classify them as Monocot and Dicot.
Answer:

Monocot leaves	Dicot leaves
Canna. Grass (Parallel venation)	Hibiscus, Peepal, Tulsi (Reticulate venation)

Can you tell? (Textbook Page No. 25)

(i) Distinguish between Dicotyledonae and Monocotyledonae.
Answer:
Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.

(ii) Why do Dicots show secondary growth while Monocots don’t?
Answer:
(a) In dicots, vascular bundles are conjoint, collateral and open type. Cambium is present between xylem and phloem for secondary growth.
(b) Whereas in monocots, vascular bundles are conjoint, collateral and closed type. Thus, due to absence of cambium, secondary growth does not occur in majority of monocots.

Observe and Discuss (Textbook Page No. 23)

Which differences did you notice between Gymnosperms and Angiosperms?
Answer:

Gymnosperms	Angiosperms
1. In gymnosperms, the seeds arc naked.	In angiosperms, the seeds are enclosed within the fruit.
2. Plants are evergreen, shrubs or woody trees.	Plants are annual, biennial or perennial herbs, shrubs or trees, either woody or herbaceous.
3. Xylem is made up of tracheids only.	Xylem is made up of vessels and tracheids.
4. Phloem is with sieve cells only.	Phloem is with sieve tubes and companion cells.
5. Usually two types of leaves are present, i.e. green foliage leaves and scale leaves.	Leaves are of usually one type only, such as green foliage leaves.
6. Double fertilization absent.	Double fertilization occurs.

Can you tell? (Textbook Page No. 26)

What is alternation of generations?
Answer:
The sporophytic and gametophytic generations generally occur alternately in the life cycle of a plant. This phenomenon is called alternation of generations.

Can you tell? (Textbook Page No. 26)

Which phase is dominant in the life cycle of Bryophyta and Pteridophyta?
Answer:
In the life cycle of Bryophyta, gametophyte is the dominant phase whereas in the life cycle of Pteridophyta, sporophyte is the dominant phase.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 2 Systematics of Living Organisms Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 2 Systematics of Living Organisms

1. Choose the correct option.

Question (A)
Which of the following shows single-stranded RNA and lacks protein coat?
(a) Bacteriophage
(b) Plant virus
(c) Viroid
(d) Animal virus
Answer:
(c) Viroid

Question (B)
Causative agent of red tide is ________ .
(a) Dinoflagellate
(b) Euglenoid
(c) Chrysophyte
(d) Lichen
Answer:
(A) Dinoflagellate

Question (C)
Select the odd one out for Heterotrophic bacteria.
(a) Nitrogen-fixing bacteria
(b) Lactobacilli
(c) Methanogens
(d) Cyanobacteria
Answer:
(c) Methanogens or (d) Cyanobacteria

Question (D)
Paramoecium: Ciliated Protist :: Plasmodium: _______ .
(a) Amoeboid protozoan
(b) Ciliophora
(c) Flagellate protozoan
(d) Sporozoan
Answer:
(d) Sporozoan

2. Answer the following

Question (A)
What are the salient features of Monera?
Answer:
Salient features of Kingdom Monera:

1. Size: The organisms included in this kingdom are microscopic, unicellular and prokaryotic.
2. Occurrence: These are omnipresent. They are found in all types of environment which are not generally inhabited by other living beings.
3. Nucleus: These organisms do not have well-defined nucleus. DNA exists as a simple double-stranded circular single chromosome called as nucleoid. Apart from the nucleoid they often show presence of extrachromosomal DNA which is small circular called plasmids.
4. Cell wall: Cell wall is made up of peptidoglycan (also called murein) which is a polymer of sugars and amino acids.
5. Membrane-bound cell organelles: Membrane-bound cell organelles like mitochondria, chloroplast, endoplasmic reticulum are absent. Ribosomes are present, which are smaller in size (70S) than in eukaryotic cells.
6. Nutrition: Majority are heterotrophic, parasitic or saprophytic in nutrition. Few are autotrophic that can be either photoautotrophs or chemoautotrophs.
7. Reproduction: The mode of reproduction is asexual or with the help of binary fission or budding. Very rarely, sexual reproduction occurs by conjugation method.
8. Examples:
   Archaebacteria: e.g. Methanobacillus, Thiobacillus, etc.
   Eubacteria: e.g. Chlorobium, Chromatium, and Cyanobacteria e.g. Nostoc, Azotobacter, etc.

Question (B)
What will be the shape of a bacillus and coccus type of bacteria?
Answer:
The shape of bacillus type of bacteria is rod-shaped and coccus is spherical.

Question (C)
Why is binomial nomenclature important?
Answer:
Binomial nomenclature is important because:

1. The binomials are simple, meaningful and precise.
2. They are standard since they do not change from place to place.
3. These names avoid confusion and uncertainty created by local or vernacular names. The organisms are known by the same name throughout the world.
4. The binomials are easy to understand and remember.
5. It indicates phylogeny (evolutionary history) of organisms.
6. It helps to understand inter-relationship between organisms.

3. Write short notes

Question (A)
Write a note on useful and harmful bacteria.
Answer:
(i) Useful bacteria:
Most of the bacteria act as a decomposer. They breakdown large molecules in simple molecules or minerals. Examples of some useful bacteria:
Lactobacillus’. It helps in curdling of milk.
Azotobacter. It helps to fix nitrogen for plants.
Streptomyces: It is used in antibiotic production such as streptomycin.
Methanogens: These are used for production of methane (biogas) gas from dung.
Pseudomonas spp. and Alcanovorax borkumensis: These bacteria have the ability to destroy the pyridines and other chemicals. Hence, used to clear the oil spills.

(ii)Harmful bacteria:
This includes disease causing bacteria. They cause various diseases like typhoid, cholera, tuberculosis, tetanus, etc. Examples of some harmful bacteria:
Salmonella typhi: It is a causative organism of typhoid.
Vibrio cholerae: It causes cholera.
Mycobacterium tuberculosis’. It causes tuberculosis.
Clostridium tetani: It causes tetanus.
Clostridium spp.: It causes food poisoning.
Many forms of mycoplasma are pathogenic.
Agrobacterium , Erwinia, etc are the pathogenic bacteria causing plant diseases.
Animals and pets also suffer from bacterial infections caused by Brucella, Pastrurella, etc.

Question (B)
Write short note on five kingdom system.
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

1. Kingdom Monera
2. Kingdom Protista
3. Kingdom Plantae
4. Kingdom Fungi
5. Kingdom Animalia

Question (C)
Write short note on useful fungi.
Answer:
Economic importances of fungi are as follows:
1. Role of fungi in medicine:
(a) Antibiotic penicillin is obtained from Penicillium.
(b) Drugs like cyclosporine, immunosuppressant drugs, precursors of steroid hormones, etc are isolated from fungi.

2. Role of fungi in industries:
(a) Yeast is used in bread making. It causes dough to rise and make the bread light and spongy. It is also used in breweries or wine making industries. Sugars present in grapes are fermented by using yeast. This results in production of alcohol which is used for making wine.
(b) Lichen is a symbiotic association of algae and fungi are used in preparation of litmus paper which is used as acid-base indicator.

3. Role of fungi in food:
(a) Fungi like mushrooms are consumed as a food. These are rich source of protein.
(b) Fungi genus Penicillium helps in ripening of cheese.

4. Role of fungi as biocontrol agents:
(a) Fungi help to control growth of weeds.
(b) Pathogenic fungi like Fusarium sp., Phytophthorapalmivora, Alternaria crassa, etc act as mycoherbicides.

Question 4.
Complete tree diagram in detail.

Answer:

5. Draw neat labelled diagrams

Question (A)
Draw neat and labelled diagram of Paramoecium.
Answer:
Characteristics:

1. It belongs to kingdom Protista. It is further classified as animal like protist.
2. It lacks cell wall.
3. It shows heterotrophic and holozoic nutrition.
4. It is a ciliated protozoan where locomotion is due to cilia.
5. It has gullet (a cavity) which opens on the cell surface.

Quesiton (B)
Draw neat and labelled diagram of Euglena.
Answer:
Characteristics:
It belongs to kingdom Protista. It is further classified into euglenoids.

1. Dinoflagellates:

1. They are aquatic (mostly marine) and autotrophic (photosynthetic).
2. They have wide range of photosynthetic pigments which can be yellow, green, brown, blue and red.
3. The cell wall is made up of cellulosic stiff plates.
4. A pair of flagella is present, hence they are motile.
5. They are responsible for famous ‘red tide’. E.g. Gonyaulax. It makes sea appear red.

2. Euglenoids:

1. They lack cell wall but have a tough covering of proteinaceous pellicle.
2. Pellicle covering provides flexibility and contractibility to Euglena.
3. They possess two flagella, one short and other long.
4. They behave as heterotrophs in absence of light but possess pigments, similar to that of higher plants, for photosynthesis.

Question (C)
Draw a neat labelled diagram of TMV.
Answer:

Question 6.
Complete chart and explain in your word.

Answer:

Depending upon the host, viruses are classified into three types as:
1. Plant virus
2. Animal virus
3. Bacterial virus (Bacteriophage)

1. Plant virus:
(a) Generally, they are rod shaped or cylindrical with helical symmetry.
(b) Majority of plant viruses have RNA as their genetic material. (Exception: Cauliflower Mosaic Virus has double stranded DNA as genetic material)
(c) Plant viruses cause disease in plants, e.g. Tobacco Mosaic Virus (TMV).

2. Animal virus:
(a) Generally, they are polyhedral in shape with radial symmetry.
(b) They have either DNA or RNA as genetic material.
(c) It causes disease to majority of animals including human beings, e.g. Influenza virus.

3. Bacteriophage:
(a) They have tadpole-like shape.
(b) They infect bacteria and hence are called as bacteriophage.
(c) Bacteriophages were discovered by Twort.
(d) Bacteriophages have double stranded DNA as the genetic material.
(e) Its body consists of head, collar and tail.

Question 7.
Identify the following diagram, label it and write detail information in your words.



Answer:
The given figure represents Bacteriophage.

A.

B.

c.

D.

E.

F.

Question 8.
The scientific name of sunflower is given below. Identify the correctly written name.
(A) Helianthus annus
(B) Helianthus Annus
(C) Helianthus annuus L.
(D) Helianthus annuus l.
Answer:
The correctly written scientific name of sunflower is Helianthus annuus L.

Question 9.
Match the following.

Kingdom	Examples
1. Monera	a. Riccia
2. Protista	b. Cyanobacteria
3. Plantae	c. Rhizopus
4. Fungi	d. Diatoms

Answer:

Kingdom	Examples
1. Monera	b. Cyanobacteria
2. Protista	d. Diatoms
3. Plantae	a. Riccia
4. Fungi	c. Rhizopus

Question 10.
Complete the following.
1. Plant-like Protista – [ ]
2. [ ] – Entamoeba

Practical /Project:

Question 1.
Make a group of students. Observe living organisms in your school/college campus and try to write their characters with respect to habit, habitat, mode of nutrition, growth- determinate or indeterminate, type of reproduction – vegetative reproduction, asexual reproduction, sexual reproduction. With the help of similarity and dissimilarity, try to classify organisms into different categories. Similar work should implement for animal group.
Answer:
The common living organisms observed near school/college are:
1. Plants
Habit: Herb, shrub, tree, etc.
Habitat: Terrestrial or aquatic
Mode of nutrition: Autotrophic
Growth: Indeterminate
Types of reproduction: Vegetative, asexual and sexual reproduction.

2. Animal e.g. dog, cats, cow, etc.
Habitat: Terrestrial
Mode of nutrition: Heterotrophs
Growth: Determinate
Types of reproduction: Only sexual reproduction

3. Birds e.g. Crow, sparrow, etc.
Habitat: Aviary (shows diverse habitat)
Mode of nutrition: Heterotrophs Growth: Determinate
Types of reproduction: Only sexual reproduction
[Note: Students are expected to collect more information about characteristics of living organisms and classify them into different categories]

Question 2.
Find out types of lichens and its economic importance.
Answer:
Types of lichens are:
1. Based on fungal components:
(a) Ascolichens:
In this category, the fungal partner belongs to Ascomycetes group of fungi.
(b) Basidiolichens:
Here, the fungal partner belongs to Basidiomycetes group of fungi.
(c) Deuterolichens:
In this category, the fungal partner belongs to Deuteromycetes group of fungi.

2. Based on their forms:
(a) Crustose lichen:
These lichens show crust-like growth.
These lichens grow on rocks and bark of the trees,
e. g. Graphis, Lecanora, Haematomma, etc.
(b) Foliose lichen:
These lichens grow on trees in the hilly regions.
The thallus is like a dry forked leaf,
e. g. Parmelia, Collema, Peltigera
(c) Fruticose lichen:
These lichens are seen on the branches of trees hanging down.
They are cylindrical, well branched and pendulous, with hair-like outgrowths,
e. g. Usnea, Cladonia, Alectoria, etc.

3. Economic importance of lichens:

(a) Lichen as food and fodder:
Many species of lichens are used as food by animals including man. Lichens contain a substance lichenin which is similar to carbohydrate making them edible. Parmelia is used in curry powder in India. Lichens like Cladonia, Citraria, Evernia, Parmelia are used as fodder as they form a favourite food for reindeers and cattles.

(b) Lichens in medicine:
Lichens contain usnic acid due to which they are used in medicines. Usnea and Cladonia species are used as an antibiotic against Gram positive bacteria.
Species like Lobaria, Citraria are useful in respiratory disease like T.B., Peltigera is useful in hydrophobia, Parmelia is used in treatment of epilepsy, whereas Usnea is used in urinary disease. Some lichens are also used in medicine due to their anticarcinogenic property.

(c) Industrial use of lichens:
1. Lichens are used in various dyes for colouring fabrics.
2. Species like Rocella and Lasallia are used in preparation of litmus paper which is acid-base indicator.
3. In Sweden and Russia, lichens are used for production of alcohol.
4. Orcein is a biological stain obtained from Orchrolechia androgyna and O. tortaria.
5. Some lichens are also used in tanning process in leather industry.
6. Evernia and Ramalina are the sources of essential oils which are used in preparation of soaps and other cosmetics.

(d) Other uses of lichens:
1. Lichens are used in cosmetics.
2. Some lichens like Everniaprunastri also known as oakmoss is used in making perfumes.
3. Lichen is also used as a preservative for beer.

11th Biology Digest Chapter 2 Systematics of Living Organisms Intext Questions and Answers

Can you tell? (Textbook Page No. 7)

Enlist uses of taxonomy?
Answer:
Uses of taxonomy are as follows:

1. It is used to assign each organism an appropriate place in a systematic framework of classification.
2. It is used to group animals and plants by their characteristics and relationships.
3. It is used to classify organisms based upon their similarities and differences.
4. It is used for nomenclature of an organism. Assigning a name to an organism is essential for its identification without confusion throughout the scientific world.
5. It is used to serve as an instrument for identification of an organisms. A newly isolated organism can be placed to its nearest relative or can be identified as a new organism with unknown characteristics.
6. It becomes easier to understand the evolutionary trends in different groups of organisms.

Can you tell? (Textbook Page No. 7)

Which characters of organisms are visible characters?
Answer:
The visible characters of organisms include habit, colour, mode of respiration, growth, reproduction, etc.

Can you tell? (Textbook Page No. 7)

What is evolution?
Answer:

1. It is believed that the life originated on earth in its very simple form.
2. Constant struggle of the early living beings gave rise to more and more perfect forms of life.
3. This struggle and progress are evolution which led to formation of diverse life forms.

Can you tell? (Textbook Page No. 7)

What is DNA barcoding?
Answer:
DNA barcoding is a new method for identification of any species based on its DNA sequence, which is obtained from a tiny tissue sample of the organism under study.

Can you tell? (Textbook Page No. 7)

Name the recent approaches in taxonomy.
Answer:
The recent approaches in taxonomy includes:

1. Morphological Approach
2. Embryological Approach
3. Ecological Approach
4. Behavioral Approach / Ethological Approach
5. Genetical Approach / Cytological Approach
6. Biochemical Approch
7. Numerical Taxonomy

Can you tell? (Textbook Page No. 9)

Make a flow chart showing taxonomic hierarchy.
Answer:
Kingdom → Sub-kingdom → Division / Phylum → Class → Cohort / Order → Family Genus → Species

Do Yourself (Textbook Page No. 16)

Complete the table (given on textbook Page No.16) through collecting information about sunflower, tiger with characteristic features.
(i) Sunflower:

Category	Taxon	Characteristics
Kingdom	Plantae	Autotrophic, photosynthetic, cell wall present.
Sub-kingdom	Phanerogamae	Seed producing plants, reproductive structures are visible.
Division	Angiospermae	Seeds are enclosed within the fruit.
Class	Dicotyledonae	Two cotyledons, tap root system, reticulate venation, pentamerous symmetry of flower, vascular bundle open.
Order	Asterales	Capitulum inflorescence, showing ray florets and disc florets.
Family	Asteraceae	Aster family
Genus	Helianthus	–
Species	annuus	–

(ii) Tiger:

Category	Taxon	Characteristics
Kingdom	Animalia	Multicellular eukaryotes, cell wall absent, heterotrophic nutrition.
Phylum	Chordata	Notochord present
Class	Mammalia	Presence of mammary gland
Order	Carnivora	Carnivorous in nature
Family	Felidae	Cat-like mammals
Genus	Panthera	Large cats
Species	tigris	–

Can you tell? (Textbook Page No. 9)

Why horse and ass are considered to be two different species or animals?
Answer:
1. Species is a group of organisms that can interbreed under natural conditions to produce fertile offsprings.
2. Horse and ass (donkey) are considered to be two different species or animals, because they cannot interbreed under natural condition to produce fertile offspring.

Internet my friend: (Textbook Page No. 9)

(i) Collect the information about most recent system of classification of living organisms and Kingdom System of Classification, e.g. Search for APG system of classification for Plants.
Answer:
[Note: Students are expected to collect more information about most recent system of classification of living organisms and Kingdom System of Classification from internet on their own.]

(ii) Collect the information about classification systems for all types of organisms.
Answer:
[Note: Students are expected to collect more information about classification systems for all types of organisms from internet on their own.]

Can you recall? (Textbook Page No. 6)

What is Five Kingdom system of classification?
Answer:
Five kingdom system of classification was proposed by R.H. Whittaker in 1969. This system shows the phylogenetic relationship between the organisms.
The five kingdoms are:

1. Kingdom Monera
2. Kingdom Protista
3. Kingdom Plantae
4. Kingdom Fungi
5. Kingdom Animalia

Can you tell (Textbook Page No. 14)

Classify fungi into their types.
Answer:
Fungi are classified into four types on the basis of their structure, mode of spore formation and fruiting bodies as follows:
1. Phycomycetes:
Members of this class are commonly called as algal fungi.
These are consisting of aseptate coenocytic hyphae.
They grow well in moist and damp places on decaying organic matter as well as in aquatic habitats or as parasites on plants.
e.g. Mucor, Rhizopus (bread mold), Albugo (parasitic fungus on mustard).

2. Ascomycetes:
These are commonly called as sac fungi.
These are multicellular. Rarely they are unicellular (e.g. Yeast).
Hyphae are branched and septate.
They can be decomposers, parasites or coprophilous (grow on dung).
Some varieties of this class are consumed as delicacies such as morels and truffles.
Neurospora is useful in genetic and biochemical assays.
e.g. Aspergillus, Penicillium, Neurospora, Claviceps, Saccharomyces (unicellular ascomycetes).

3. Basidiomycetes:
These are commonly called as club fungi.
They have branched septate hyphae.
e.g. Agaricus (mushrooms), Ganoderma (bracket fungi), Ustilago (smuts), Puccinia (rusts), etc.

4. Deuteromycetes:
It is a group of fungi which are known to reproduce only asexually.
They are commonly called imperfect fungi.
They are mainly decomposers, while few are parasitic, e.g. Alternaria.

Can you tell? (Textbook Page No. 14)

Write a note on economic importance of fungi.
Answer:
Economic importances of fungi are as follows:
1. Role of fungi in medicine:
(a) Antibiotic penicillin is obtained from Penicillium.
(b) Drugs like cyclosporine, immunosuppressant drugs, precursors of steroid hormones, etc are isolated from fungi.

2. Role of fungi in industries:
(a) Yeast is used in bread making. It causes dough to rise and make the bread light and spongy. It is also used in breweries or wine making industries. Sugars present in grapes are fermented by using yeast. This results in production of alcohol which is used for making wine.
(b) Lichen is a symbiotic association of algae and fungi are used in preparation of litmus paper which is used as acid-base indicator.

3. Role of fungi in food:
(a) Fungi like mushrooms are consumed as a food. These are rich source of protein.
(b) Fungi genus Penicillium helps in ripening of cheese.

4. Role of fungi as biocontrol agents:
(a) Fungi help to control growth of weeds.
(b) Pathogenic fungi like Fusarium sp., Phytophthorapalmivora, Alternaria crassa, etc act as mycoherbicides.

Can you tell? (Textbook Page No. 14)

Why are fungi considered as heterotrophic organisms?
Answer:
In fungi, chloroplast is absent, thus they cannot synthesize their own food by photosynthesis. Fungi decompose the organic matter by breaking down with the help of enzymes from which they absorb nutrients. Thus, exhibiting heterotrophic mode of nutrition.

Can you tell? (Textbook Page No. 14)

What are coenocytic hyphae?
Answer:
1. In filamentous fungi, body consists of mycelium which is formed by a network of hyphae.
2. When these hyphae are non-septate, multinucleated, they are known as coenocytic hyphae.

Can you tell? (Textbook Page No. 14)

(i) How are fungi different from plants?
Answer:
Fungi are different from plants because:
(a) They lack chloroplast hence, do not perform photosynthesis and are heterotrophic in nutrition. Whereas plants are autotrophic and prepare their own food by photosynthesis.
(b) They are separated from Plantae based on their saprophytic mode of nutrition.
(c) Fungi are decomposers of ecosystem whereas plants are producers of ecosystem.
(d) In fungi, cell wall is made up of fungal cellulose or chitin. Whereas in plants, cell wall is made up of cellulose and pectic compounds.

(ii) Have you seen any diseased plant in your farm?
Answer:
Yes, I have seen some diseased plants in our farm.
There are different pathogens like fungi, bacteria, viruses that cause diseases in plants.
The common plant diseases are:
(a) Leaf rust disease: It is caused by fungus Puccinia triticina. It is the most common rust disease of wheat.
(b) Blight disease in rice: It is caused by harmful bacteria Xanthomonas oryzae. It causes wilting of seedlings and yellowing and drying of leaves.
(c) Early blight of potato: It is caused by fungi Alternaria solani. It causes ‘bulls eye’ patterned leaf spots and tuber blight on potato.
(d) Crown gall disease: It is caused by Agrobacterium tumefaciens. This pathogen infects the plant and forms rough surfaced galls on stem and roots.
[Students are expected to write their observations about diseased plants found informs]

Can you tell? (Textbook Page No. 14)

Complete the following table:
Answer:

Plantae	Animalia
1. Autotrophic mode of nutrition.	Heterotrophic mode of nutrition.
2. They do not show locomotion.	They show locomotion.
3. Cell wall is present.	Cell wall is absent.
4. Chloroplast present.	Chloroplast absent.
5. They do not possess nervous system.	They possess well developed nervous system, i
6. Reproduction can be both sexual and asexual.	Mainly shows sexual reproduction.

Can you tell? (Textbook Page No. 15)

Why are viruses called infectious nucleoproteins?
Answer:
1. Viruses are acellular, highly infectious and ultramicroscopic.
2. Viruses possess their own genetic material in the form of either DNA or RNA, but never both. The genetic material in viruses is covered by a protein coat (capsid), hence called nucleoprotein.
3. They do not show any activity outside the body of host but once they enter their specific host cell, they start multiplying within the living host cells.
4. Viruses lack their own metabolic machinery, they make use of the cellular machinery of the host i.e. ribosome for the synthesis of protein during their reproduction and therefore, they cause severe infection. Thus, they are called infectious nucleoproteins.

Can you tell? (Textbook Page No. 15)

Describe genetic material in plant and animal viruses as well as in bacteriophages.
Answer:
The genetic material in different viruses is as given below:
1. Plant virus: (b) Majority of plant viruses have RNA as their genetic material. (Exception: Cauliflower Mosaic Virus has double-stranded DNA as genetic material)
2. Animal virus: (b) They have either DNA or RNA as genetic material.
3. Bacteriophage: (d) Bacteriophages have double-stranded DNA as the genetic material.

Can you tell? (Textbook Page No. 15)

Differentiate between viruses and viroids.
Answer:

Viruses	Viroids
1. They have high molecular weight.	They have low molecular weight.
2. They are larger in size.	They are smaller in size.
3. They can infect plant, animals and bacteria.	They mainly infect plants.
4. The genetic material can be ss-RNA, ds-RNA or DNA.	The genetic material is single stranded circular RNA.
5. Protein coat is present.	Protein coat is absent.
6. mosaic disease is a plant disease caused by viruses.	Tomato chloric dwarf is a plant disease caused by viroids.

Internet my friend. (Textbook Page No. 15)

In modern medicine, certain infectious neurological diseases were found to be transmitted by abnormally folded proteins. These proteins are called prions. The word prion comes from ‘proteinaceous infectious particle’, e.g. mad cow disease in cattle, Jacob’s disease in human.
Find more information about prions.
Answer:
Prions:
1. A prion is a misfolded form of a protein generally present in brain cells.
2. When the prion gets into a cell containing the normal form of the protein, the prion somehow converts normal protein molecules to the misfolded prion versions.
3. Several prions then aggregate into a complex that can convert other normal proteins to prions.
4. Prions can be transmitted through blood, surgical instruments and contaminated food.
5. Diseases caused by prions are Bovine Spongiform Encephalopathy in cattles, Kuru and Creutzfeldt – Jakob disease in humans.
[Note: Students are expected to search for more information about Prions on internet]

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Miscellaneous Exercise 2

I. Select the correct option from the given alternatives.

Question 1.
The value of the expression
cos1°. cos2°. cos3° … cos 179° =
(A) -1
(B) 0
(C) \(\frac{1}{\sqrt{2}}\)
(D) 1
Answer:
(B) 0

Explanation:
cos 1° cos 2° cos 3° … cos 179°
= cos 1° cos 2° cos 3° … cos 90°… cos 179°
= 0 …[∵ cos 90° = 0]

Question 2.
\(\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}\) is equal to
(A) 2cosec A
(B) 2 sec A
(C) 2 sin A
(D) 2 cos A
Answer:
(A) 2cosec A

Explanation:

Question 3.
If α is a root of 25cos² θ + 5cos θ – 12 = 0, \(\frac{\pi}{2}\) < α < π, then sin 2α is equal to
(A) \(-\frac{24}{25}\)
(B) \(-\frac{13}{18}\)
(C) \(\frac{13}{18}\)
(D) \(\frac{24}{25}\)
Answer:
(A) \(-\frac{24}{25}\)

Explanation:

25 cos² θ + 5 cos θ – 12 = 0
∴ (5cos θ + 4) (5 cos θ – 3) = 0
∴ cos θ = \(-\frac{4}{5}\) or cos θ = \(\frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π,
cos α < 0
∴ cos α = \(-\frac{4}{5}\)
sin² α = 1 – cos² α = 1 – \(\frac{16}{25}=\frac{9}{25}\)
∴ sin α = \(\pm \frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π sin α > 0
∴ sin α = 3/5
sin 2 α = 2 sin α cos α
= \(2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}\)

Question 4.
If θ = 60°, then \(\frac{1+\tan ^{2} \theta}{2 \tan \theta}\) is equal to
(A) \(\frac{\sqrt{3}}{2}\)
(B) \(\frac{2}{\sqrt{3}}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt{3}\)
Answer:
(B) \(\frac{2}{\sqrt{3}}\)

Explanation:

Question 5.
If sec θ = m and tan θ = n, then \(\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}\) is equal to
(A) 2
(B) mn
(C) 2m
(D) 2n
Answer:
(A) 2
Explanation:

Question 6.
If cosec θ + cot θ = \(\frac{5}{2}\), then the value of tan θ is
(A) \(\frac{14}{25}\)
(B) \(\frac{20}{21}\)
(C) \(\frac{21}{20}\)
(D) \(\frac{15}{16}\)
Answer:
(B) \(\frac{20}{21}\)

Explanation:
cosec θ + cot θ = \(\frac{5}{2}\) …………….(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ \(\frac{5}{2}\) (cosec θ – cot θ) = 1
∴ cosec θ – cot θ = \(\frac{2}{5}\) …(ii)
Subtracting (ii) from (i), we get
2 cot θ = \(\frac{5}{2}-\frac{2}{5}=\frac{21}{10}\)
∴ cot θ = \(\frac{21}{20}\)
∴ tan θ = \(\frac{20}{21}\)

Question 7.
\(1-\frac{\sin ^{2} \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}\) equals
(A) 0
(B) 1
(C) sin θ
(D) cos θ
Answer:
(D) cos θ

Explanation:

Question 8.
If cosec θ – cot θ = q, then the value of cot θ is
(A) \(\frac{2 q}{1+q^{2}}\)
(B) \(\frac{2 q}{1-q^{2}}\)
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)
(D) \(\frac{1+q^{2}}{2 q}\)
Answer:
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

Explanation:

cosec θ – cot θ = q ……(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ (cosec θ + cot θ)q = 1
∴ cosec θ + cot θ = 1/q …….(ii)
Subtracting (i) from (ii), we get
2cot θ = \(\frac{1}{\mathrm{q}}-\mathrm{q}\)
∴ cot θ = \(\frac{1-q^{2}}{2 q}\)

Question 9.
The cotangent of the angles \(\frac{\pi}{3}, \frac{\pi}{4}\) and \(\frac{\pi}{6}\) are in
(A) A.P.
(B) G.P.
(C) H.P.
(D) Not in progression
Answer:
(B) G.P.

Explanation:

Question 10.
The value of tan 1°.tan 2° tan 3° equal to
(A) -1
(B) 1
(C) \(\frac{\pi}{2}\)
(D) 2
Answer:
(B) 1

Explanation:

tan1° tan2° tan3° … tan89°
= (tan 1° tan 89°) (tan 2° tan 88°)
…(tan 44° tan 46°) tan 45°
= (tan 1 ° cot 1 °) (tan 2° cot 2°)
…(tan 44° cot 44°) . tan 45°
…tan(∵ 90° – θ) = cot θ]
= 1 x 1 x 1 x … x 1 x tan 45° =1

II. Answer the following:

Question 1.
Find the trigonometric functions of:
90°, 120°, 225°, 240°, 270°, 315°, -120°, -150°, -180°, -210°, -300°, -330°
Solution:
Angle of measure 90° :
Let m∠XOA = 90°
Its terminal arm (ray OA)
intersects the standard, unit circle at P(0, 1).

∴ x = 0 and y = 1
sin 90° = y = 1
cos 90° = x = 0
tan 90° = \(\frac{y}{x}=\frac{1}{0}\), which is not defined
cosec 90° = \(\frac{1}{y}=\frac{1}{1}\) = 1
sec 90° = \(\frac{1}{x}=\frac{1}{0}\), which is not defined
cot 90° = \(\frac{x}{y}=\frac{0}{1}\) = 0

Angle of measure 120° :
Let m∠XOA =120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 2nd quadrant, x < 0, y > 0

[Note: Answer given in the textbook of tan 120° is \(\frac{-1}{\sqrt{3}}\) and cot 120° is \(-\sqrt{3}\). However, as per our \(-\sqrt{3}\) calculation the answer of tan 120° is \(-\sqrt{3}\) and cot 120° is \(-\frac{1}{\sqrt{3}}\)

Angle of measure 225° :
Let m∠XOA = 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 240° :
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 270° :
Let m∠XOA = 270°
Its terminal arm (ray OA) intersects the standard unit circle at P(0, – 1).
x = 0 andy = – 1
sin 270° = y = -1
cos 270° = x = 0
tan 270° = \(\frac{y}{x}\)

Angle of measure 315° :
Let m∠XOA = 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1


[Note: Answer given in the textbook of cot 315° is 1. However, as per our calculation it is -1.]

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (-150°) :
Let m∠XOA = – 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (-180°):
Let m∠XOA = – 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(- 1, 0).
∴ x = – 1 andy = 0
sin (-180°) = y = 0
cos (-180°) = x
= -1

Angle of measure (- 210°):
Let m∠XOA = -210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (- 300°):
Let m∠XOA = – 300° Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 1st quadrant, x>0,y>0
x = OM = \(\frac{1}{2}\) and
y = PM = \(\frac{\sqrt{3}}{2}\)

Angle of measure (- 330°):
Let m∠XOA = – 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Question 2.
State the signs of:
i. cosec 520°
ii. cot 1899°
iii. sin 986°
Solution:
i. 520° =360° + 160°
∴ 520° and 160° are co-terminal angles.
Since 90° < 160° < 180°,
160° lies in the 2nd quadrant.
∴ 520° lies in the 2nd quadrant,
∴ cosec 520° is positive.

ii. 1899° = 5 x 360° + 99°
∴ 1899° and 99° are co-terminal angles.
Since 90° < 99° < 180°,
99° lies in the 2nd quadrant.
∴ 1899° lies in the 2nd quadrant.
∴ cot 1899° is negative.

iii. 986° = 2x 360° + 266°
∴ 986° and 266° are co-terminal angles.
Since 180° < 266° < 270°,
266° lies in the 3rd quadrant.
∴ 986° lies in the 3rd quadrant.
∴ sin 986° is negative.

Question 3.
State the quadrant in which 6 lies if
i. tan θ < 0 and sec θ > 0
ii. sin θ < 0 and cos θ < 0
iii. sin θ > 0 and tan θ < 0
Solution:
i. tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0
sec θ is positive in 1st and 4th quadrants.
∴ θ lies in the 4th quadrant.

ii. sin θ < 0
sin θ is negative in 3rd and 4th quadrants, cos θ < 0
cos θ is negative in 2nd and 3rd quadrants.
.’. θ lies in the 3rd quadrant.

iii. sin θ > 0
sin θ is positive in 1st and 2nd quadrants, tan θ < 0
tan θ is negative in 2nd and 4th quadrants.
∴ θ lies in the 2nd quadrant.

Question 4.
Which is greater?
sin (1856°) or sin (2006°)
Solution:
1856° = 5 x 360° + 56°
∴ 1856° and 56° are co-terminal angles.
Since 0° < 56° < 90°, 56° lies in the 1st quadrant.
∴ 1856° lies in the 1st quadrant,
∴ sin 1856° >0 …(i)
2006° = 5 x 360° + 206°
∴ 2006° and 206° are co-terminal angles.
Since 180° < 206° < 270°,
206° lies in the 3rd quadrant.
∴ 2006° lies in the 3rd quadrant,
∴ sin 2006° <0 …(ii)
From (i) and (ii),
sin 1856° is greater.

Question 5.
Which of the following is positive?
sin(-310°) or sin(310°)
Solution:
Since 270° <310° <360°,
310° lies in the 4th quadrant.
∴ sin (310°) < 0
-310° = -360°+ 50°
∴ 50° and – 310° are co-terminal angles.
Since 0° < 50° < 90°, 50° lies in the 1st quadrant.
∴ – 310° lies in the 1st quadrant.
∴ sin (- 310°) > 0
∴ sin (- 310°) is positive.

Question 6.
Show that 1 – 2sin θ cos θ ≥ 0 for all θ ∈ R.
Solution:
1 – 2 sin θ cos θ
= sin² θ + cos² θ – 2sin θ cos θ
= (sin θ – cos θ)² ≥ 0 for all θ ∈ R

Question 7.
Show that tan² θ + cot² θ ≥ 2 for all θ ∈ R.
Solution:

Question 8.
If sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\), then find the values of cos θ, tan θ in terms of x and y.
Solution:
Given, sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)
we know that
cos²θ = 1 – sin² θ

[Note: Answer given in the textbook of cos θ = \(\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = \(. However, as per our calculation the answer of cos θ = ± [latex]\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = ± \(\frac{x^{2}-y^{2}}{2 x y}\). ]

Question 9.
If sec θ = \(\sqrt{2}\) and \(\frac{3 \pi}{2}\) < θ < 2π, then evaluate \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\)
Solution:
Given sec θ = \(\sqrt{2}\)
We know that,
tan² θ = sec² θ – 1
= (\(\sqrt{2}\)) – 1
= 2 – 1 = 1
∴ tan θ = ±1
Since \(\frac{3 \pi}{2}\) < θ < 2π
θ lies in the 4th quadrant.
∴ tan θ < 0
∴ tan θ = -1

Question 10.
Prove the following:

i. sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1
Solution:
L.H.S. = sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A (cos² B + sin² B) + cos² A (sin² B + cos² B)
= sin²A(1) + cos²A(1)
= 1 = R.H.S.

ii. \(\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-{cosec}^{3} \theta}=\sin ^{2} \theta \cos ^{2} \theta\)
Solution:

iii. L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)\)
Solution:
L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}\)
= (tanθ + secθ)² + (tanθ – secθ)²
= tan² θ + 2 tan θ sec θ + sec² θ
+ tan² θ – 2 tan θ sec θ +.sec² θ
= 2(tan² θ + sec² θ)

iv. 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ
Solution:
LHS.
= 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ =  = 2 sec² θ – (sec² θ)² – 2cosec² θ + (cosec² θ)²
= 2(1+ tan² θ) – (1+ tan² θ)² – 2(1+ cot² θ)
+ (1+ cot² θ)²
= 2 + 2tan² θ – (1 + 2tan² θ + tan⁴ θ)
– 2 – 2cot² θ + 1 + 2cot² θ + cot⁴ θ
= 2 + 2.tan² θ – 1 – 2 tan² θ – tan⁴ θ – 2
– 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ
= cot⁴ θ – tan⁴ θ = R.H.S.

v. sin⁴ θ + cos⁴ θ = sin⁴ θ + cos⁴ θ
Solution:
L.H.S. = sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2sin² θ cos² θ
… [ v a² + b² = (a + b)² – 2ab]
= 1 – 2sin² θ cos² θ
= R.H.S.

vi. 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 = 0
L.H.S =
2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1=0
= sin⁶ θ + cos⁶ θ
= (sin² θ)³ + (cos² θ)³ = (sin² θ + cos² θ)³
– 3 sin² θ cos² θ (sin2 0 + cos2 0)
…[••• a³ + b³ = (a + b)³ – 3ab(a + b)]
= (1)³ – 3 sin² θ cos² θ(1)
= 1-3 sin² θ cos² θ sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2 sin² θ cos² θ
…[Y a² + b² = (a + b)² – 2ab]
= 1-2 sin² θ cos² θ
L.H.S.= 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2(1-3 sin² θ cos² θ) -3(1 – 2 sin² θ cos² θ) + 1
= 2-6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1 = c
= R.H.S.

vii. cos⁴ θ – sin⁴ θ + 1 = 2cos²θ
L.H.S. = cos⁴ θ – sin⁴ θ + 1
= (cos² θ)² – (sin² θ)² + 1 = (cos²θ + sin²θ) c(os² θ – sin²θ) +1
= (1) (cos²θ – sin²θ) + 1 = cos² θ + (1 – sin²θ)
= cos² θ + cos²θ = 2cos²θ = R.H.S.

viii. sin⁴θ + 2sin²θ cos²θ = 1 – cos⁴θ
L.H.S. = sin⁴θ + 2sin²θ cos²θ = sin²θ(sin²θ + 2cos²θ)
= (sin²θ) (sin²θ + cos²θ + cos²θ) = (1 – cos²θ) (1 + cos²θ)
= 1 – cos⁴θ = R.H.S.

ix. \(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta-\cos \theta}=2\)
Solution:

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ)
= 2 (sin² θ + cos² θ)
= 2(1)
= 2 = R.H.S.

x. tan² θ – sin² θ = sin⁴ θ sec² θ
Solution:
L.H.S. = tan² θ – sin² θ
= \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) – sin²θ
= sin² θ (\(\frac{1}{\cos ^{2} \theta}-1 \))
= \(\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\)
= (sin² θ) (sin² θ)sec² θ
= sin⁴ θ sec² θ
= R.H.S

xi. (sinθ + cosecθ)² + (cos θ + see θ)² = tan² θ + cot² θ + 7
Solution:
L.H.S. = (sinθ + cosecθ)² + (cos θ + see θ)²
= sin ² θ + cosec² θ + 2sinθ cosec θ
+ cos² θ + sec² θ + 2sec0 cos0
= (sin² θ + cos² θ) + cosec² θ + 2 + sec² θ + 2
= 1 + (1 + cot² θ) + 2 + (1 + tan² θ) + 2 = tan² θ + cot² θ + 7
= R.H.S.

xii. sin⁸θ – cos⁸θ = (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
Solution:
L.H.S. = sin⁸θ – cos⁸θ
= (sin⁴θ)² – (cos⁴θ)²
= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)
= [(sin² θ)² – (cos² θ)² ]
. [(sin² θ)² + (cos² θ)² ]
= (sin² θ + cos² θ) (sin² θ – cos² θ). [(sin² θ + cos² θ)² – 2sin² θ.cos² θ] …[Y a² + b² = (a + b)² – 2ab]
= (1) (sin² θ – cos² θ) (1² – 2sin² θ cos² θ)
= (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
= R.H.S.

xiii. sin⁶A + cos⁶A = 1 – 3 sin²A + 3sin⁴A
Soluiton:
L.H.S. = sin⁶A + cos⁶A
= (sin² A)³ + (cos² A)³
= (sin² A + cos² A)³
– 3sin²A cos²A(sin² A + cos² A)
…[ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3sin²A cos²A (1)
= 1 – 3sin²A cos²A
= 1 – 3 sin²A (1 – sin²A)
= 1 – 3 sin²A + 3sin⁴A
= R.H.S.

xiv. (1 + tanA tanB)² + (tanA – tanB)² = sec ²A sec²B
Solution:
L.H.S. = (1 + tanA tanB)² + (tanA – tanB)²
= 1 + 2tanA tanB + tan²A tan² + tan² A- 2tanA tanB + tan²B
= 1 + tan²A + tan² B + tan²A tan²B
= 1(1+ tan²A) + tan² B(1 + tan²A)
= (1 + tan²A) (1 + tan²B)
= sec²A sec²B = R.H.S.

xv. \(\frac{1+\cot \theta+{cosec} \theta}{1-\cot \theta+{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-1}{\cot \theta-{cosec} \theta+1}\)
Solution:
We know that cosec²θ – cot² θ = 1
∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

xvi. \(\frac{\tan \theta+\sec \theta-1}{\tan \theta+\sec \theta+1}=\frac{\tan \theta}{\sec \theta+1}\)
Solution:
We know that
tan²θ = sec² θ – 1
∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

xvii. \(\frac{{cosec} \theta+\cot \theta-1}{{cosec} \theta+\cot \theta+1}=\frac{1-\sin \theta}{\cos \theta}\)
Solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

xviii. \(\frac{{cosec} \theta+\cot \theta+1}{\cot \theta+{cosec} \theta-1}=\frac{\cot \theta}{{cosec} \theta-1}\)
solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.2

Question 1.
If 2sin A = 1 = \(\sqrt{2}\) cos B and \(\frac{\pi}{2}\) < A < π, \(\frac{3 \pi}{2}\)
Solution:
Given, 2sin A = 1
∴ sin A = 1/2
we know that,
cos² A = 1 – sin² A = 1 – \(\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}\)
∴ cos A = \(\pm \frac{\sqrt{3}}{2}\)
Since \(\frac{\pi}{2}\) < A < π
A lies in the 2nd quadrant.

We know that,
Sin² B = 1 – cos² B = 1 – \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)\(\frac{1}{2}=\frac{1}{2}\)
∴ sin B = \(\pm \frac{1}{\sqrt{2}}\)
Since \(\frac{3 \pi}{2}\) < B < 2π
B lies in the 4th quadrant,

Question 2.
If \(\) and A, B are angles in the second quadran, then prove that 4cosA + 3 cos B = -5
Solution:
Given, \(\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}\)
∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)
We know that,
cos² A = 1 – sin² = 1 – \(\left(\frac{3}{5}\right)^{2}\) = 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ Cos A = ± \([{4}{5}\)
Since A lies in the second quadrant,
cos A < 0
∴ Cos A = –\(\frac{4}{5}\)
Sin B = 4/5
We know that,
cos²B = 1 – sin²B = 1 – \(\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}\)
∴ Cos B = ±\(\frac{4}{5}\)
Since B lies in the second quadrant, cos B < 0

Question 3.
If tan θ = \(\frac{1}{2}\), evaluate \(\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}\)
Solution:

Question 4.
Eliminate 0 from the following:
i. x = 3sec θ, y = 4tan θ
ii. x = 6cosec θ,y = 8cot θ
iii. x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
iv. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4tan θ,3y = 5 + 3sec θ
Solution:
i. x = 3sec θ, y = 4tan θ
∴ sec θ = \(\frac{x}{3}\) and tan θ= \(\frac{y}{4}\)
We know that,
sec²θ – tan²θ = 1

∴ 16x² – 9y² = 144

ii. x = 6cosec θ and y = 8cot θ
.’. cosec θ = \(\) and cot θ = \(\)
We know that,
cosec² θ – cot² θ =

16x² – 9y² = 576

iii. x = 4cos θ – 5 sin θ … (i)
y = 4sin θ + 5cos θ .. .(ii)
Squaring (i) and (ii) and adding, we get
x² + y² = (4cos θ – 5sin θ)² + (4sin θ + 5cos θ)²
= 16cos²θ – 40 sinθ cosθ + 25 sin²θ + 16 sin² θ + 40sin θ cos θ + 25 cos² θ
= 16(sin² θ + cos² θ) + 25(sin² θ + cos² θ)
= 16(1) + 25(1)
= 41

iv. x = 5 + 6cosec θ andy = 3 + 8cot θ
∴ x – 5 = 6cosec θ and y – 3 = 8cot θ
∴ cosec θ = \(\frac{x-5}{6}\) and cot θ = \(\frac{y-3}{8}\)
We know that,
cosec² θ – cot² θ = 1
∴ \(\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}\) = 1

v. 2x = 3 – 4tan θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ and 3y – 5 = 3sec θ
∴ tan θ = \(\frac{3-2 x}{4}\) and sec θ = \(\frac{3 y-5}{3}\)θ
We know that, sec² θ – tan² θ = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}\) = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}\) = 1

Question 5.
If 2sin² θ + 3sin θ = 0, find the permissible values of cosθ.
Solution:
2sin² θ + 3sin θ = 0
∴ sin θ (2sin θ + 3) = 0
∴ sin θ = 0 or sin θ = \(\frac{-3}{2}\)
Since – 1 ≤ sin θ ≤ 1,
sin θ = 0
\(\sqrt{1-\cos ^{2} \theta}\) = 0 …[ ∵ sin² θ = 1- cos² θ]
∴ 1 – cos² θ = 0
∴ cos² θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Question 6.
If 2cos² θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.
Solution:
2cos²θ – 11 cos θ + 5 = 0
∴ 2cos² θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2cos θ – 1 = 0
∴ cos θ = 5 or cos θ = 1/2
Since, -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

Question 7.
Find the acute angle θ such 2cos² θ = 3sin θ.
Solution:
2cos²0 = 3sin θ
∴ 2(1 – sin² θ) = 3sin θ
∴ 2 – 2sin² θ = 3sin θ
∴ 2sin² θ + 3sin 9-2 = θ
∴ 2sin² θ + 4sin θ – sin θ – 2 = θ
∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ
∴ (sin θ + 2) (2sin θ – 1) = 0
∴ sin θ + 2 = 0 or 2sin θ – 1 = 0
∴ sin θ = -2 or sin θ = 1/2
Since, -1 ≤ sin θ ≤ 1
∴ Sin θ = 1/2
∴ θ = 30° …[ ∵ sin 30 = 1/2]

Question 8.
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5tan² θ + 3 = 9sec θ
∴ 5(sec² θ – 1) + 3 = 9sec θ
∴ 5sec² θ – 5 + 3 = 9sec θ
∴ 5sec² θ – 9sec θ – 2 = 0
∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0
∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0
∴ (sec θ – 2) (5sec θ + 1) = 0
∴ sec θ – 2 = 0 or 5sec θ + 1 = 0
∴ sec θ = 2 or sec θ = -1/5
Since sec θ ≥ 1 or sec θ ≤ -1,
sec θ = 2
∴ θ = 60° … [ ∵ sec 60° = 2]

Question 9.
Find sin θ such that 3cos θ + 4sin θ = 4.
Solution:
3cos θ + 4sin θ = 4
∴ 3cos θ = 4(1 – sin θ)
Squaring both the sides, we get .
9cos²θ = 16(1 – sin θ)²
∴ 9(1 – sin² θ) = 16(1 + sin² θ – 2sin θ)
∴ 9 – 9sin² θ = 16 + 16sin² θ – 32sin θ
∴ 25sin² θ – 32sin θ + 7 = 0
∴ 25sin² θ – 25sin θ – 7sin θ + 7 = 0
25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0
∴ (sin θ – 1) (25sin θ – 7) = 0
∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0
∴ sin θ = 1 or sin θ = \(\frac{7}{25}\)
Since, -1 ≤ sin θ ≤ 1
∴ sin θ = 1 or \(\frac{7}{25}\)
[Note: Answer given in the textbook is 1. However, as per our calculation it is 1 or \(\frac{7}{25}\).]

Question 10.
If cosec θ + cot θ = 5, then evaluate sec θ.
Solution:
cosec θ + cot θ = 5
∴ \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5\)
∴ \(\frac{1+\cos \theta}{\sin \theta}=5\)
∴ 1 + cos θ = 5.sin θ
Squaring both the sides, we get
1 + 2 cos θ + cos² θ = 25 sin² θ
∴ cos² θ + 2 cos θ + 1 = 25 (1 – cos² θ)
∴ cos² θ + 2 cos θ + 1 = 25 – 25 cos² θ
∴ 26 cos² θ + 2 cos θ – 24 = 0
∴ 26 cos² θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ or 26 cos θ – 24 = 0
∴ cos θ = -1 or cos θ = \(\frac{24}{26}=\frac{12}{13}\)
When cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cos θ ≠ -1
∴ cos θ = \(\frac{12}{13}\)
∴ sec θ = \(\frac{1}{\cos \theta}=\frac{13}{12}\)
[Note: Answer given in the textbook is -1 or \(\frac{13}{12}\).
However, as per our calculation it is only \(\frac{13}{12}\).]

Question 11.
If cot θ = \(\frac{3}{4}\) and π < θ < \(\frac{3 \pi}{2}\), then find the value of 4 cosec θ + 5 cos θ.
Solution:
We know that,
cosec²θ = 1 + cot² θ = \(\left(\frac{3}{4}\right)^{2}\) = 1 + \(\frac{9}{16}\)
∴ cosec² θ = \(\frac{25}{16}\)
∴ cosec θ = \(\pm \frac{5}{4}\)
Since π < θ < \(\frac{3 \pi}{2}\)
θ lies in the third quadrant.
∴ cosec θ < 0
∴ cosec θ = –\(\frac{5}{4}\)
cot θ = \(\frac{3}{4}\)
tan θ = \(\frac{1}{\cot \theta}=\frac{4}{3}\)
We know that,
sec² θ = 1 + tan² θ = 1 + \(\left(\frac{4}{3}\right)^{2}\)
= 1 + \(\frac{16}{9}=\frac{25}{9}\)
∴ sec θ = ±\(\frac{5}{3}\)
Since θ lies in the third quadrant,
sec θ < 0
∴ sec θ = –\(\frac{5}{3}\)
cos θ = \(\frac{1}{\sec \theta}=\frac{-3}{5}\)
∴ 4cosec θ + 5cos θ
= \(4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)\)
= -5 – 3 = -8
[Note: The question has been modified.]

Question 12.
Find the Cartesian co-ordinates of points whose polar co-ordinates are:
i. (3, 90°) ii. (1, 180°)
Solution:
i. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 3cos 90° and y = 3sin 90°
∴ x = 3(0) = 0 and y = 3(1) = 3
∴ the required cartesian co-ordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 1(cos 180°) and y = 1(sin 180°)
∴ x = -1 and y = 0
∴ the required cartesian co-ordinates are (-1, 0).

Question 13.
Find the polar co-ordinates of points whose Cartesian co-ordinates are:
1. (5, 5) ii. (1, \(\sqrt{3}\))
ii. (-1, -1) iv. (-\(\sqrt{3}\), 1)
Solution:
i. (x, y) = (5, 5)
∴ r = \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{25+25}\)
\(=\sqrt{50}=5 \sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{5}{5}\) = 1
Since the given point lies in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar co-ordinates are (\(5 \sqrt{2}\), 45°).

ii. (x, y) = ( 1, \(\sqrt{3}\))
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}\)
Since the given point lies in the 1st quadrant,
θ = 60° …[∵ tan 60° = \(\sqrt{3}\)]
∴ the required polar co-ordinates are (2, 60°).

iii. (x, y) = (-1, -1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{-1}{-1}=1\)
∴ tan θ = tan \(\frac{\pi}{4}\)
Since the given point lies in the 3rd quadrant,
tan θ = tan \(\left(\pi+\frac{\pi}{4}\right)\) …[∵ tan (n + x) = tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{4}\right)\)
∴ θ = \(\frac{5 \pi}{4}\) = 225°
∴ the required polar co-ordinates are (\(\sqrt{2}\), 225°).

iv. (x, y) = (-\(\sqrt{3}\) , 1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{1}{-\sqrt{3}}\) = -tan \(\frac{\pi}{6}\)
Since the given point lies in the 2nd quadrant,
tan θ = tan \(\left(\pi-\frac{\pi}{6}\right)\) …[∵ tan (π – x) = – tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{6}\right)\)
∴ θ = \(\frac{5 \pi}{6}\) = 150°
∴ the required polar co-ordinates are (2, 150°)

Question 14.
Find the values of:
i. sin\(\frac{19 \pi^{e}}{3}\)
ii. cos 1140°
iii. cot \(\frac{25 \pi^{e}}{3}\)
Solution:
i. We know that sine function is periodic with period 2π.
sin \(\frac{19 \pi}{3}\) = sin \(\left(6 \pi+\frac{\pi}{3}\right)\) = sin \(\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

ii. We know that cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = \(\frac {1}{2}\)

iii. We know that cotangent function is periodic with period π.
cot \(\frac{25 \pi}{3}\) = cot \(\left(8 \pi+\frac{\pi}{3}\right)\) = cot \(\frac{\pi}{3}\) = \(\frac{1}{\sqrt{3}}\)
dhana work.txt
Displaying dhana work.txt.

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 10 Single Entry System Textbook Exercise Questions and Answers.

Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 10 Single Entry System

1. Answer in One sentence only.

Question 1.
What do you mean by a Single Entry System?
Answer:
A system of bookkeeping in which an accountant or businessman records only one aspect of business transaction (either debit or credit and ignores the other aspect is called ‘Single Entry System’.

Question 2.
What is a Statement of Affairs?
Answer:
A list of all assets and liabilities prepared under a single entry system to find out capital balance is called a statement of affairs.

Question 3.
Which type of accounts are normally not kept under the Single Entry System?
Answer:
Under a single entry system, records of impersonal accounts i.e. real accounts and nominal accounts are not maintained.

Question 4.
Which statement is prepared under the Single Entry system to ascertain the capital balances?
Answer:
A statement of Affairs is prepared under a single entry system to ascertain capital balances.

Question 5.
How Opening Capital is calculated under the Single Entry System?
Answer:
Under a single entry system, opening capital is ascertained by preparing the opening statement of affairs.

Question 6.
Which types of accounts are maintained under the Single Entry System?
Answer:
Under a single entry system, all personal accounts and cash accounts are maintained.

Question 7.
Can a Trial Balance be prepared under a Single Entry System?
Answer:
A trial balance cannot be prepared under a single entry system.

Question 8.
Which type of organizations generally follow the Single Entry System?
Answer:
Organizations having small sizes of business such as sole trading concerns and partnership firms follow a single entry system.

2. Write a word, term, or phrase which can substitute each of the following statements.

Question 1.
A statement that is similar to the Balance Sheet.
Answer:
Statement of Affairs

Question 2.
The system of Accounting is normally suitable for small business organizations.
Answer:
Single Entry System

Question 3.
A statement similar to the Balance Sheet is prepared to find out the amount of opening capital.
Answer:
Opening Statement of Affairs

Question 4.
An excess of assets over liabilities.
Answer:
Capital

Question 5.
Excess of closing capital over opening capital of proprietor under Single Entry System.
Answer:
Profit

Question 6.
Name of the method of accounting suitable to firms having limited transactions.
Answer:
Single Entry System

Question 7.
A System of accounting that is unscientific.
Answer:
Single Entry System

Question 8.
Further capital introduced by the proprietor in the business concern over and above his existing capital.
Answer:
Additional Capital

3. Select the most appropriate answer from the alternatives given below and rewrite the sentence.

Question 1.
The capital balances are ascertained by preparing _______________
(a) Statement of Affairs
(b) Cash Account
(c) Drawings Accounts
(d) Debtors Accounts
Answer:
(a) Statement of Affairs

Question 2.
Under Single Entry System, Opening Capital = Opening Assets less _______________
(a) Opening Liabilities
(b) Closing Liabilities
(c) Debtors Account
(d) Creditors Account
Answer:
(a) Opening Liabilities

Question 3.
Additional Capital introduced during the year is _______________ from closing capital in order to find out the correct profit.
(a) Added
(b) Deducted
(c) Divided
(d) Ignored
Answer:
(b) Deducted

Question 4.
Single Entry System may be useful for _______________
(a) Sole traders
(b) Company
(c) Government
(d) None of these
Answer:
(a) Sole traders

Question 5.
In order to find out the correct profit, drawings is _______________ from closing capital.
(a) Multiplies
(b) Divided
(c) Deducted
(d) Added
Answer:
(d) Added

Question 6.
The difference between assets and liabilities is called _______________
(a) Capital
(b) Drawings
(c) Income
(d) Expenses
Answer:
(a) Capital

Question 7.
When Closing Capital is greater than the Opening Capital, the difference is _______________
(a) Profit
(b) Loss
(c) Assets
(d) Liabilities
Answer:
(a) Profit

Question 8.
Opening Capital is ₹ 30,000; Closing Capital is ₹ 60,000; Withdrawals are ₹ 5,000; and further capital brought in is ₹ 3,000; Profit is _______________
(a) ₹ 45,000
(b) ₹ 35,000
(c) ₹ 32,000
(d) ₹ 22,000
Answer:
(c) ₹ 32,000

4. State True or False with reasons:

Question 1.
The double Entry System of Book-keeping is a scientific method of books of accounts.
Answer:
This statement is True.
In the double-entry system of book-keeping, there are two-fold effects. Both the effects are recorded simultaneously with an equal amount. This system also follows principles and rules of debit and credit. Due to this, there are very fewer chances of mistakes. So double entry system of Book-Keeping is a scientific method of the book of accounts.

Question 2.
Preparation of Trial Balance is not possible under the Single Entry System.
Answer:
This statement is True.
Under the single entry system, only cash and personal accounts of debtors and creditors are open. So it is not possible to prepare. Trail balance under single entry system as it has incomplete information of Accounting.

Question 3.
Statement of Affairs and Balance Sheet are one and the same.
Answer:
This statement is False.
There is a difference between a statement of Affairs and the Balance sheet. Statement of Affair shows estimated values of assets and liabilities.

Question 4.
The single Entry System is not useful for large organizations.
Answer:
This statement is True.
Under the Single Entry System, only the cash book and personal account of Debtor and Ciygditor are maintained. Real and Nominal accounts are not maintained. It has no proper set of rules to be followed. It is useful for small organisations and not for a large organisations.

Question 5.
Only Cash and Personal accounts are maintained under the Single Entry System.
Answer:
This statement is True.
The single Entry System is an ancient and unscientific method of recording business transactions. This system maintains minimum accounts so it is easy for traders to write books of accounts. This system does not follow any accounting rules. To know the cash collections and amount payable or receivable only cash and personal accounts are maintained under a single entry system.

5. Do you agree with the following statements?

Question 1.
Further capital introduced during the year increases profit.
Answer:
Disagree

Question 2.
Interest in Drawings decreases the amount of profit under the Single Entry System.
Answer:
Disagree

Question 3.
Real and Nominal accounts are not maintained under the Single Entry System.
Answer:
Agree

Question 4.
The single Entry System is based on certain rules and principles.
Answer:
Disagree

Question 5.
Statement of Profit is just like Profit and Loss Account.
Answer:
Disagree

6. Fill in the Blanks.

Question 1.
Statement of Affairs is just like _______________
Answer:
Balance Sheet

Question 2.
Under Single Entry System, Profit = Closing Capital Less _______________
Answer:
Opening Capital

Question 3.
In order to find out the correct profit, drawings are _______________ to the closing capital.
Answer:
Added

Question 4.
In _______________ Book Keeping System, in every business transactions we find two effects.
Answer:
Double Entry System

Question 5.
The difference between Assets and Liabilities is called _______________
Answer:
Capital

Question 6.
Single Entry System is more popular for _______________
Answer:
Sole Trader

Question 7.
Additional Capital introduced during the year is _______________ from Closing Capital in order to find out the correct profit.
Answer:
Deducted

Question 8.
Single Entry System is Suitable for _______________ business.
Answer:
Small

7. Find the odd one:

Question 1.
Interest on Drawings, Outstanding Expenses, Undervaluation of Assets, Prepaid Expenses.
Answer:
Outstanding Expenses

Question 2.
Interest on Capital, Interest on Loan, Overvaluation of Liabilities, Depreciation on Assets.
Answer:
Overvaluation of Liabilities

Question 3.
Creditors, Bills Payable, Bank Overdraft, Stock in Trade.
Answer:
Stock in Trade

8. Complete the following table:

Question 1.

Answer:
₹ 5,000

Question 2.

Answer:
₹ 30,000

Question 3.

Answer:
₹ 5,000

Question 4.

Answer:
₹ 25,000, ₹ 20,000

Question 5.

Answer:
₹ 19,000

9. Complete the following table. Put Proper mark in Box.

Question 1.

Answer:

1. Add
2. Add
3. Add
4. Less
5. Add
6. Less
7. Add
8. Less
9. Less
10. Less

Practical Problems

Question 1.
Mr. Poonawala keeps his books under the Single Entry System and gives the following information.
Capital as of 31.3.2017 – ₹ 60,000
Capital as on 31.3.2018 – ₹ 1,00,000
Drawings made during the year ₹ 2,000
Additional capital introduced during the year ₹ 12,000
Calculate Profit or Loss during the year.
Solution:
In the books of Mr. Poonawala
Statement of Profit or Loss for the year ended 31st March 2018

Question 2.
Sujit a small trader provides you with the following details of his business.

Additional information:
1. Sujit withdraws ₹ 5,000 for his personal use, on 1st Oct. 2017.
2. He had also withdrawn ₹ 30,000 for rent of his residential flat.
3. Depreciation Furniture by 10% p.a. and writes off ₹ 1,000 from Motor Van.
4. Charge interest on Drawings ₹ 3,000.
5. 10% Govt. Bonds were purchased on 1st Oct. 2017.
6. Allow interest on capital at 10% p.a.
7. ₹ 1,000 is written off as bad debts and provides 5% p.a. R.D.D. on Debtors.
Prepare Opening Statement of Affairs, Closing Statement of Affairs, and Statement of Profit or Loss for the year ending 31st March 2018.
Solution:
In the books of Sujit
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 3.
Anjali keeps her books by the Single Entry System. Her position on 1.4.2016 was as follows.
Cash at Bank ₹ 4,000, Cash in Hand ₹ 1,000, Stock ₹ 6,000; Sundry Debtors ₹ 8,400, Plant and Machinery ₹ 7,500, Bill Receivable ₹ 2,600, Creditors ₹ 3500; Bills Payable ₹ 4,000
On 31.3.2017 her position was as follows; cash at Bank ₹ 3,900, Cash in Hand ₹ 2,000. Stock ₹ 9000, Sundry Debtors, ₹ 7,500; Plant and Machinery ₹ 7,500; Bills Payable ₹ 2,200, Bills Receivable ₹ 3,400; Creditors ₹ 1,500.
During the year Anjali introduced further Capital of ₹ 1,500 and she spent ₹ 700 per month for her personal use.
Depreciation Plant and Machinery by 5% p.a. and create Reserve for Doubtful debts @ 5% p.a. on the debtor. Prepare Opening and Closing Statement of Affairs and Statement of Profit or Loss for the year ended 31.3.2017.
Solution:
In the books of Anjali
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2017

Question 4.
Mr. Vijay is dealing in the business of fruits. He maintains an accounting record with a single entry. The following figures are taken from his record.

Additional information:
1. Mr. Vijay introduced ₹ 7,000 as fresh capital.
2. He spent ₹ 40,000 from his business for his daughter’s marriage.
3. Depreciate Building by ₹ 6,000.
4. Create a 5% reserve for doubtful debts on Sundry Debtor.
Prepare:
1. Opening Statement of Affairs.
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Mr. Vijay
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 5.
Miss. Fiza keeps her books on the Single Entry System and disclosed the following information about her business.

Additional information:
1. Miss. Fiza transferred ₹ 2,000 per month during the first half-year and ₹ 1000 per month for the second half-year from a business account to her personal account.
2. She sold her private asset for ₹ 40,000 and brought the proceeds into her business.
3. She also took goods worth ₹ 12,000 for private use.
4. Plant and Machinery is to be depreciated by 10% p.a.
5. Provide R.D.D. on debtors at 5% p.a.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018
Solution:
In the books miss Fiza
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 6.
Miss. Sanika keeps her books on the Single Entry System. The statement of affairs is given on 31st March 2018.

On 31st March 2018 their Assets and Liabilities were as follows:
Plant and Machinery ₹ 42,000, Stock ₹ 38,000, Cash in Hand ₹ 10,000, Creditors ₹ 7,000, Debtors ₹ 25,000, Bills Payable ₹ 6,000
Drawings during the year were ₹ 5,500, Plant and Machinery were found Overvalued by 5% p.a. and Stock was found Undervalued by 20% p.a., R.D.D. was to be created at 10% p.a. on Debtors, Interest on Capital was allowed at 10% p.a.
Prepare:
1. Closing Statement of Affairs.
2. Statement of Profit or Loss for the year ended 31st March 2018.
Solution:
In the books of miss Sanika
Closing statement of Affairs as on 31.03.2018

Statement of Profit or Loss for the year ended 31st March 2018.

Question 7.
Mr. Suhas commenced his business with the Capital of ₹ 1,50,000 on 1st April 2017. His financial position was as follows as on 31st March 2018, Cash ₹ 20,000, Stock ₹ 15,000, Debtors ₹ 30,000, Premises ₹ 90,000, Vehicles ₹ 45,000, Creditors ₹ 18,500, Bills Payable ₹ 10,000.
Additional information:
1. He brought additional capital ₹ 10,000 on 30th Sept. 2017, Interest on capital is to be provided at 5% p.a.
2. He withdrew ₹ 15,000 for personal use on which interest is to be charged at 5% p.a.
3. Write off Bad debts ₹ 500.
Prepare:
1. Closing Statement of Affairs
2. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Mr. Suhas
Closing statement of Affairs as on 31.3.2018

Statement of Profit or Loss for the year ended 31st March 2018.

Question 8.
Ganesh keeps his books by the Single Entry Method. Following are the details of his business:

During the year he has withdrawn ₹ 25,000 for his private purpose and goods of ₹ 3,000 for household use. On 1st Oct. 2016. He sold his household furniture for ₹ 4,000 and deposited the same amount in a business Bank Account.
Provide Depreciation on Plant and Machinery at 10% p.a. (assuming additions were made on 1st Oct. 2016) and Furniture at 5%.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2017.
Solution:
In the books of Ganesh
Opening and Closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2017

Question 9.
Peter keeps his books on the Single Entry System. From the following particulars, Prepare Opening and Closing Statement of Affairs and Statement of Profit or Loss for the year ending 31st March 2018.

Additional Information:
1. Peter has withdrawn ₹ 15,000 from the business for his personal use.
2. He has introduced additional capital of ₹ 10,000 in the business on 1st January 2018.
3. Depreciate furniture @ 10% p.a.
4. Maintain reserve for doubtful debts @ 5% on Sundry Debtors.
5. Closing Stock is overvalued by 25% in the books.
Solution:
In the books of Peter
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 10.
Suresh keeps his books by the Single Entry System. His position on 1.4.2017 was as follows.
Cash at Bank ₹ 4,000, Cash in Hand ₹ 3,000; Stock ₹ 8,000; Sundry Debtors ₹ 9,000; Plant & Machinery ₹ 10,000; Bills Receivable ₹ 3000; Creditors ₹ 1500; Bills Payable ₹ 2000.
On 31st March 2018, his position was as follows:
Cash at bank ₹ 6,400; Cash in Hand ₹ 1,800; Stock ₹ 10000; Sundry and Debtors ₹ 8,000; Plant & Machinery ₹ 10,000; Bills Payable ₹ 4,000; Bills Receivable ₹ 5,200; Creditors ₹ 2,000 During the year Suresh introduced further capital of ₹ 3,000 and his drawings were ₹ 700 per months. Depreciate Plant & Machinery by 5% and create a reserve for bad doubtful debts @ 5%.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Suresh
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 9 Final Accounts of a Proprietary Concern Textbook Exercise Questions and Answers.

Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 9 Final Accounts of a Proprietary Concern

1. Answer in One Sentence.

Question 1.
What is Trading Account?
Answer:
An account in which direct expenses are compared with direct incomes to find out gross profit or gross loss for a given period is known as Trading Account.

Question 2.
What do you mean by Profit and Loss Account?
Answer:
An account in which indirect expenses are compared with indirect incomes to find out net profit or net loss for a given period is known as the Profit and Loss Account.

Question 3.
Why Balance Sheet is prepared?
Answer:
The balance sheet is prepared to ascertain the financial position of the business on a specific date usually at the end of the accounting year.

Question 4.
State the meaning of Final Accounts.
Answer:
Final Accounts are the group of Trading Account, Profit and loss account and Balance sheet prepared to know the results of business for a given period.

Question 5.
What is Net Profit?
Answer:
When the total credit side of Profit and Loss A/c is greater than the total of debit side, it indicates credit balance which is known as net profit.

Question 6.
What do you mean by Gross Profit?
Answer:
When the total credit side of Trading A/c is greater than the total of debit side, it indicates credit balance, which is called gross profit.

Question 7.
State the meaning of Accrued Income?
Answer:
Income that is due and accumulated but not yet actually received during the current accounting year is called accrued income.

Question 8.
State the meaning of Outstanding Expenses?
Answer:
The expenses which are incurred in the current year, but not paid partly or fully during the current accounting year are termed as outstanding expenses.

Question 9.
What is Depreciation?
Answer:
Depreciation means a continuous reduction in the value of property or asset due to wear and tear, accident, fall in the market price, the passage of time, etc.

Question 10.
What do you mean by Prepaid Expenses?
Answer:
The expense which is paid in advance before they are due for payment is called prepaid expenses.

2. Give a word, term, or phrase which can substitute each of the following statements:

Question 1.
Expenses are paid before it is due.
Answer:
Prepaid Expenses

Question 2.
Income due but not yet received.
Answer:
Accrued Income

Question 3.
Carriage paid on the sale of goods.
Answer:
Carriage Outwards

Question 4.
Statement of Assets and liabilities.
Answer:
Balance Sheet

Question 5.
Account prepared to know Net Profit or Net loss.
Answer:
Profit and Loss A/c

Question 6.
Value of goods remaining unsold at the end of the year.
Answer:
Closing Stock

Question 7.
The provision was made to compensate the loss on account of likely debts.
Answer:
Provision for Bad and Doubtful Debts

Question 8.
The accounts are prepared at the end of the accounting year to know the profit or loss and financial position of the business.
Answer:
Final Accounts

Question 9.
An amount spent on promoting the sale of goods.
Answer:
Selling Expenses

Question 10.
Additional information is provided below the Trial Balance.
Answer:
Adjustments

3. Select the most appropriate alternatives given below and rewrite the sentence:

Question 1.
_____________ is excess of assets over liabilities.
(a) Goodwill
(b) Capital
(c) Investments
(d) Drawings
Answer:
(b) Capital

Question 2.
Discount earned is transferred to credit side of _____________ account.
(a) Current
(b) Profit and Loss
(c) Trading
(d) Capital
Answer:
(b) Profit and Loss

Question 3.
_____________ is a statement which shows the financial position of business on a specific date.
(a) Trading Account
(b) Trial Balance
(c) Profit and Loss A/c
(d) Balance Sheet
Answer:
(d) Balance Sheet

Question 4.
Outstanding expenses are shown on the _____________ side of Balance sheet.
(a) Assets
(b) Liability
(c) Both
(d) None of these
Answer:
(b) Liability

Question 5.
Interest on Drawing is credited to _____________ Account.
(a) Trading
(b) Profit and Loss
(c) Capital
(d) All
Answer:
(b) Profit and Loss

Question 6.
Debit balance of Trading Account means _____________
(a) Gross Loss
(b) Net Loss
(c) Net Profit
(d) Gross Profit
Answer:
(a) Gross Loss

Question 7.
Carriage Inward is debited to _____________ Account.
(a) Trading
(b) Profit and Loss
(c) Capital
(d) Bank
Answer:
(a) Trading

Question 8.
Excess of credit over to debit in Profit and Loss A/c indicates _____________
(a) Net Profit
(b) Gross Profit
(c) Gross Loss
(d) Net Loss
Answer:
(a) Net Profit

Question 9.
Closing stock is always valued at cost or market price which is _____________
(a) more
(b) less
(c) zero
(d) equal
Answer:
(b) less

Question 10.
When specific date is not given, in that case interest on drawings is charged for _____________ month.
(a) Four
(b) Six
(c) Eight
(d) Nine
Answer:
(b) Six

4. State True or False with reasons:

Question 1.
In every adjustment at least there are three effects.
Answer:
This statement is False.
There are at least two effects in every adjustment of final accounts.

Question 2.
Every item of Trial Balance has only one effect.
Answer:
This statement is True.
Every transaction is recorded through journal or subsidiary books with the principle of the double-entry book-keeping system. Journal and subsidiary books are posted to the ledger account and trial balance is prepared from the balances of the ledger so there are already two effects passed. So every item of Trial Balance has only one effect.

Question 3.
Income due but not received is a liability.
Answer:
This statement is False.
Income due but not received is an Asset and not a liability.

Question 4.
Goodwill is not a fictitious asset.
Answer:
This statement is True.
Goodwill is the reputation or name and fame of a business organization in the market. It is the money value of a business reputation earned by a business. It is an intangible asset.
Fictitious assets are created by accounting entry in the books of accounts it doesn’t have any realizable value.
E.g.: Share issue expenses.

Question 5.
The credit balance of the Profit & Loss account shows a net profit.
Answer:
This statement is True.
The credit side of profit and loss A/c represents incomes when the credit side is greater than the debit side (expenses) it shows the Net Profit of the year.

5. Fill in the blanks:

Question 1.
Gross Profit is transferred to _____________ account.
Answer:
Profit and Loss A/c

Question 2.
Debit Balance of Trading Account indicates _____________
Answer:
Gross Loss

Question 3.
Income Receivable appears on _____________ side of Balance Sheet.
Answer:
Asset

Question 4.
Interest on Bank Loan is debited to _____________ A/c.
Answer:
Profit and Loss A/c

Question 5.
Profit and Loss account is prepared to find out _____________ results of the business.
Answer:
Net Working

Question 6.
All indirect/operating expenses are transferred to _____________ account.
Answer:
Profit and Loss A/c

Question 7.
Interest of proprietor’s drawing is credited to _____________ account.
Answer:
Profit and Loss A/c

Question 8.
An excess of debit over credit in the Profit & Loss A/c represents the _____________
Answer:
Net Loss

Question 9.
All direct expenses are transferred to _____________ account.
Answer:
Trading A/c

Question 10.
Balance Sheet is _____________ of assets & liabilities.
Answer:
Statement

6. Find the odd one:

Question 1.
Rent, Salary, Insurance, Plant, and Machinery.
Answer:
Plant and Machinery

Question 2.
Purchases, Closing stock, Debtors, Factory Rent.
Answer:

Question 3.
Capital, Bills Payable, Debtors, Outstanding wages.
Answer:
Debtors

Question 4.
Advertisement, Travelling Expenses, Factory Rent, Insurance.
Answer:
Factory Rent

Question 5.
Cash in Hand, Debtors, Outstanding Income, Reserve for Doubtful Debts.
Answer:
Reserve for Doubtful Debts

7. Do you agree or disagree with the following statements:

Question 1.
Reserve for bad debts is created by debiting Profit and Loss Account.
Answer:
Agree

Question 2.
A balance Sheet is a statement as well as an account.
Answer:
Disagree

Question 3.
Indirect Expenses are debited to Trading Account.
Answer:
Disagree

Question 4.
Bank Overdraft is treated as an Internal Liability.
Answer:
Disagree

Question 5.
Capital is excess of Liabilities over Assets.
Answer:
Disagree

8. Correct and Rewrite the following statements:

Question 1.
The balancing figure of the Trading Account is Net Profit or Net Loss.
Answer:
The balancing figure of the Trading Account is Gross Profit or Gross Loss.

Question 2.
All direct expenses are debited to the Profit and Loss Account.
Answer:
All direct expenses are debited to Trading Account.

Question 3.
When the credit side of the Profit and Loss Account is greater than the debit side, it is called Net Loss.
Answer:
When the credit side of the Profit and Loss account is greater than the debit side, it is called Net Profit.

Question 4.
Capital A/c……………..Dr.
To Profit and Loss Account
(Being Net Profit transferred to Capital A/c)
Answer:
Profit and Loss Account…………….Dr.
To Capital A/c
(Being Net Profit transferred to Capital A/c)

Question 5.
Trading A/c……………Dr.
To Sales A/c
(Being Sales transferred to Trading A/c)
Answer:
Sales A/c…………Dr
To Trading A/c
(Being sales transferred to Trading A/c)

9. Calculate the following.

Question 1.
Calculate the Capital

Solution:
Capital = Assets – Liabilities
= ₹ 1,00,000 – ₹ 39,950
= ₹ 60,050

Question 2.
The machinery of ₹ 35,500 is purchased on 1st July 2018 and on the same day ₹ 4,500 are spent on the installation of the Machinery. The proprietor has decided to Depreciate Machinery at the rate of 7% p.a. Calculate the amount of depreciation, assuming that accounting year is ending on 31st March every year.
Solution:
Cost of Machinery = Purchase Price + Installation Charges
= 35,500 + 4,500
= ₹ 40,000
Depreciation for 9 months = 40,000 × \(\frac{7}{100} \times \frac{9}{12}\) = ₹ 2,100

Question 3.
Mr. Pramod borrowed a Loan from the State Bank of India ₹ 3,50,000 on 1st Oct. 2018 at the rate of interest of 12% p.a. Calculate the Interest on a bank loan for the year 2018-19, assuming that the financial year-end on 31st March every year.
Solution:
Interest on Bank loan for 6 months = 3,50,000 × \(\frac{12}{100} \times \frac{6}{12}\) = ₹ 21,000

Question 4.
Annual Insurance Premium ₹ 8,000 is paid on 1st Dec 2018. Calculate the amount of Insurance Premium for the accounting year ending on 31st March 2019.
Solution:
Annual Insurance Premium for 12 months = ₹ 8,000
Less: Prepaid for 8 months = ₹ 5,333
Insurance for 4 months (01.12.18 to 31.03.19) = ₹ 2,667

Question 5.
Calculate the Gross Profit/Gross Loss
Purchases A/c ₹ 15,500, Sales A/c ₹ 30,000, Carriage Inward ₹ 1,200, Opening Stock ₹ 5,000, Purchases Returns ₹ 500, Closing Stock ₹ 18,000
Solution:
Cost of Goods Sold = Opening Stock + Purchases – Purchases Returns + Carriage Inward – Closing Stock
= 5,000 + 15,500 – 500 + 1,200 – 18,000
= ₹ 3,200
Gross Profit = Sales – Cost of goods sold
= 30,000 – 3,200
= ₹ 26,800

Practical Problems

Question 1.
From the following Balances of Jayashri Traders, you are required to prepare Trading Account for the year ended 31/03/2019.

Solution:
In the books of Jayashri Traders

Question 2.
Prepare Profit and Loss Account of Sanjay Brothers for the year ended 31st March 2018 from the following balances.
1. Bank charges ₹ 22,000
2. Interest (Cr.) ₹ 16,000
3. Sundry expenses ₹ 42,000
4. Insurance ₹ 35,000
5. Salaries ₹ 40,000
6. Rates and Taxes ₹ 13,000
7. Postage ₹ 8,000
8. Advertisement ₹ 40,000
9. Rent paid ₹ 32,000
10. Bad debts ₹ 10,000
11. Commission (Cr) ₹ 17,500
12. Printing & Stationery ₹ 21,000
13. Loss by fire ₹ 18,000
14. Discount (Dr) ₹ 23,000
15. Discount (Cr) ₹ 37,000
16. Misc. Income ₹ 14,000
17. Depreciation ₹ 34,000
18. Carriage Outwards ₹ 60,000
19. Godown Expenses ₹ 40,000
Note: Gross Profit ₹ 4,07,500
Solution:
In the books of the Sanjay Brothers.

Question 3.
From the following Trial Balance of Sanjiv & Sons. Prepare Trading Account and Profit & Loss Account for the year ending on 31st March 2019 and a Balance Sheet as on that date.

Additional information:
1. Closing stock on 31st March 2019, was at cost ₹ 60,000 and Market Price ₹ 70,000.
2. Outstanding expenses: Wages ₹ 4,000, Salary ₹ 2,400
3. Provide depreciation at 10% on Motor Van and 5% on Furniture.
4. Write off ₹ 2,000 for bad debts and create R.D.D. at 5% on debtors.
5. Provide 10% p.a. interest on capital.
Solution:
In the books of Sanjiv & Sons

Balance Sheet as of 31st March, 2019

Question 4.
From the following Trial Balance of Nandini & Co. as of 31st March 2019. Prepare Final Accounts after considering the adjustments given below.

Adjustments:
1. Closing stock valued at ₹ 1,00,000.
2. Write off ₹ 2,000 as bad debts and create a provision for doubtful debts @ 5% on Sundry Debtor.
3. Depreciate Machinery by 10% p.a. and Loose Tools is valued at ₹ 1,00,000.
4. Charge Interest on Capital @ 2% p.a.
Solution:
In the books of Nandini & Co.

Balance Sheet as of 31st March 2019

Question 5.
Prepare Final accounts of Abdul Traders for the year ending 31st March 2019 with the help of the following Trial Balance and Adjustments.
Trial Balance as of 31st March 2019.

Adjustments:
1. Closing stock valued at ₹ 89,600
2. Outstanding expenses Salaries ₹ 2,000, Wages ₹ 4,000
3. Charge depreciation on Machinery @ 10%
4. Bad debts are written off ₹ 2,000 and create a provision for bad and doubtful debts 5% on Sundry Debtors.
Solution:
In the books of Abdul Traders.


Balance Sheet as of 31st March 2019

Question 6.
Following is the Trial Balance of Geeta Enterprises. You are required to prepare a Trading and Profit & Loss Account for the year ended 31st March 2019 and the Balance Sheet as of that date after taking into account the additional information provided to you.
Trial Balance as of 31st March, 2019

Additional information:
1. Closing stock of goods on 31st March 2019 valued at ₹ 7,100 at cost price and ₹ 7,500/- as market price.
2. Travelling expenses include ₹ 125 spent on personal traveling.
3. ₹ 175 is to be written off as bad debts which were due from Mr. Ashok, a debtor, and 5% R.D.D. is to be maintained on debtors.
4. Reserve for discount on debtors as well as on creditors is to be maintained at 2% and 3% respectively.
5. Provide 10% depreciation on Plant & Machinery and Furniture.
Solution:
In the books of Geeta Enterprises.

Balance Sheet as of 31st March 2019

Question 7.
Following are the closing ledger balances of Deepak & Co. Prepare Trading Account and Profit & Loss Account for the year ended 31st March 2019 and Balance sheet as of that date.
Ledger Balances of Mr. Deepak and Co. as of 31st March, 2019

Adjustments:
1. Closing stock was valued at ₹ 60,000
2. An amount of ₹ 3,000 is still to be received on account of commission.
3. Provision for discount on debtors and Provision for discount on Creditors are to be created 2% and 3% respectively.
4. Amount of Furniture is to reduce by ₹ 4,500 and Building by 10%.
5. Outstanding expenses Salaries ₹ 4,500 and Wages ₹ 1,500.
Solution:
In the books of Deepak & Co.

Balance Sheet as of 31st March 2019

Question 8.
Following is the Trial Balance extracted from the books of Raju Traders. You are required to prepare Trading A/c, Profit & Loss A/c for the year ending on 31st March 2019 and Balance Sheet as of that date after Considering the additional information given below.
Trial Balance as of 31st March 2019

Adjustments:
1. Closing stock is valued at ₹ 40,000 at Cost Price and ₹ 44,000 as Market Price.
2. Provide Depreciation on Plant & Machinery, Furniture, Computers @ 5%, 10%, 15% respectively.
3. Salaries are paid for 10 months only.
4. Further Bad debts amounted to ₹ 400 and provide 10% R.D.D. on Sundry Debtors.
5. Advertisement is paid for 2 years.
Solution:
In the books of Raju Traders.


Balance Sheet as of 31st March 2019

Question 9.
From the following Trial Balance of Shradha Enterprises, you are required to prepare Final Accounts for the year ending on 31st March 2019.
Trial Balance as of 31st March 2019

Adjustments:
1. Insurance is prepaid to the extent of ₹ 2,250
2. Closing stock is valued at ₹ 3,80,000 Cost price and ₹ 4,00,000 as Market price.
3. Outstanding Expenses are Wages ₹ 6,000 and Rent ₹ 5,000
4. Write off further had debts ₹ 1,500 and provide 5% Reserve for doubtful debts.
5. Depreciation on Furniture and Plant & Machinery at 10% p.a. and on Freehold Premises at 15% p.a.
Solution:
In the books of Shradha Enterprises

Balance Sheet as of 31st March 2019

Question 10.
From the following Trial Balance of Ayub & Co. as of 31st March 2019, you are required to prepare Trading Account, Profit and Loss Account for the year ending 31st March 2019, and Balance Sheet as of that date after making necessary adjustments.
Trial Balance as of 31st March 2019

Adjustments:
1. Stock on hand on 31st March 2019 valued at ₹ 60,000
2. Rent amounting to ₹ 600 Prepaid.
3. Bad Debts ₹ 600 and create a Provision for Doubtful Debts 5%
4. Depreciation on Plant & Machinery by 10% and Furniture is valued at ₹ 4,500
5. Outstanding Salaries ₹ 900
Solution:
In the books of Ayub and Co.

Balance Sheet as of 31st March 2019

Question 11.
From the following Trial Balance of Rajnish & Sons and the additional information given below prepare Trading & Profit and Loss Account for the year ending on 31st March 2018 and Balance Sheet as on that date.
Trial Balance as of 31st March 2018

Adjustments:
1. Closing Stock valued at ₹ 3,00,000 cost price and ₹ 3,20,000 at Market price.
2. Salaries were paid for 10 months only.
3. Insurance is paid for one year ending on 30.06.2018
4. One of the debtors Mr. Amit became insolvent, from whom ₹ 10,000 was not received.
5. 5% R.D.D. is to be maintained on Debtors.
6. Depreciate Machinery & Furniture @ 10% and 5% respectively.
Solution:
In the books of Rajnish & Sons


Balance Sheet as of 31st March 2018

Question 12.
From the following Trial Balance of John & Sons, you are required to prepare Trading Account, Profit and Loss Account for the year ending 31st March 2019 and Balance Sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Closing Stock ₹ 27,000
2. Charge Depreciation on Machinery and Motor car @ 10% and 5% respectively.
3. Create R.D.D. 5% on Sundry Debtors
4. Interest on Drawings @ 5% p.a.
5. Create Discount on Sundry Creditors 3%
6. Advertisement ₹ 1,000 is prepaid.
7. Outstanding Rent ₹ 1,500
Solution:
In the books of John and Sons

Balance Sheet as of 31st March 2019

Question 13.
From the following Trial Balance of Pushkraj, you are required to prepare Trading Account and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Stock on 31st March 2019 was valued at ₹ 28,000
2. Create a Provision for doubtful debts on Sundry Debtors @ 5%
3. Depreciate Motor car by 5% p.a. and Machinery by 7% p.a.
4. Outstanding expenses Rent ₹ 800 & Wages ₹ 1,000
5. Charge interest on Capital @ 3% p.a.
6. Goods of ₹ 4,000 withdrawn by the proprietor for personal use.
Solution:
In the books of Pushkraj


Balance Sheet as of 31st March 2019

Question 14.
From the following Trial Balance of Jyoti, Trading Co. Prepare the Trading Account and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Closing stock valued at ₹ 58,000 Cost Price while the Market price is ₹ 60,000
2. Write off ₹ 1,200 as Bad debts and create provision for doubtful debts 2% on Sundry Debtors and also create provision for discount on Creditors 5%.
3. Loose Tools is valued at ₹ 52,000 and depreciate Furniture by 10% p.a.
4. Outstanding expenses Salary ₹ 1,000 and Wages ₹ 225
5. Charge interest on Capital 2% and on Drawings 10%.
Solution:
In the books of Jyoti Trading Co.

Balance Sheet as of 31st March 2019

Question 15.
From the following Trial Balance of Manish Enterprise, Prepare the Trading Account and Profit and Loss Account for the year ended 31st March 2019 and Balance sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Closing Stock was ₹ 32,000.
2. Write off 50% of patents, depreciate Plant & Machinery by 10% p.a and Office Equipment by 20%.
3. Reserve for bad debts is to be created 5% and discount on Debtors 2%.
4. Outstanding expenses Mobile charges ₹ 300 and Freight ₹ 500
5. Charge Interest on Capital @ 5%.
6. Goods of ₹ 2,000 distributed on free samples.
Solution:
In the books of Manish Enterprise

Balance Sheet as of 31st March 2019

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 8 Rectification of Errors Textbook Exercise Questions and Answers.

Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 8 Rectification of Errors

1. Answer in One Sentence:

Question 1.
What is meant by rectification of errors?
Answer:
The correction of accounting errors in a systematic manner is called the rectification of errors.

Question 2.
What is meant by the error of principle?
Answer:
An error committed by the accountant by not following accounting principles properly is called an error of principle.

Question 3.
What is meant by the error of partial omission?
Answer:
An error in which a transaction is correctly recorded in the books of account but one of the postings is wrong is known as partial omission. If will affect the trial balance.

Question 4.
What is meant by the error of complete omission?
Answer:
Failure on the part of an accountant to record the business transactions in the books of account is called an error of complete omission. It does not affect the agreement of the trial balance.

Question 5.
What are compensating errors?
Answer:
The error which is committed on one side of the ledger account compensates for an error committed on the other side of some other leader account is called compensating error.

2. Give one word/term or phrase for each of the following statements.

Question 1.
Errors that affect the agreement of Trial Balance.
Answer:
One-sided errors

Question 2.
Taking the total more while closing books of accounts.
Answer:
Overcasting

Question 3.
The error arises when a transaction is partially or completely omitted to be recorded in the books of accounts.
Answer:
Error of omission

Question 4.
Transactions recorded due to violating of the accounting principles.
Answer:
Error of principle

Question 5.
Accounts to which difference in Trial Balance is transferred.
Answer:
Suspense account

Question 6.
Error in which the effect of one mistake is nullified by another mistake.
Answer:
Compensating error

Question 7.
Errors that are not disclosed by the Trial Balance.
Answer:
Two-sided errors

Question 8.
Errors of incorrect entries or wrong posting.
Answer:
Errors of commission

3. Select the most appropriate alternative from those given below and rewrite the sentence.

Question 1.
Rectification entries are passed in ______________
(a) Journal Proper
(b) Ledger
(c) Balance Sheet
(d) Cash Book
Answer:
(a) Journal Proper

Question 2.
The type of error for which journal entry is always required for rectification ______________
(a) over casting
(b) one sided error
(c) under casting
(d) two sided error
Answer:
(d) two-sided error

Question 3.
Error occurred due to wrong posting are called error of ______________
(a) principal
(b) commission
(c) compensating
(d) omission
Answer:
(b) commission

Question 4.
If transaction is totally omitted from the books, it is called ______________
(a) Error of recording
(b) Error of omission
(c) Error of principle
(d) Error of commission
Answer:
(b) Error of omission

Question 5.
Suspense Account is opened when ______________ does not tally.
(a) Balance sheet
(b) Trading Account
(c) Profit and loss
(d) Trial Balance
Answer:
(d) Trial Balance

4. State whether the following statements are True or False with reasons.

Question 1.
Trial Balance is prepared from the balance of ledger accounts.
Answer:
This statement is True.
A Trial balance is a statement of debit and credit balances extracted from the various accounts in the ledger. All business transactions are recorded first in Journal or in subsidiary books and subsequently, they are posted to respective ledger accounts. At the end of the year, they are balanced and transferred to the Trial balance.

Question 2.
A Trial Balance can agree in spite of certain errors.
Answer:
This statement is True.
The error of principle or error of complete omission or compensatory error is not disclosed by the Trial Balance. It will be agreed with debit and credit balances but there may be a certain error.

Question 3.
Rectification entries are passed in Cash Book.
Answer:
This statement is False.
Rectification entries are passed in the Journal Proper book. Cashbook is mainly used for cash transactions and not for rectification of errors.

Question 4.
There is no need to open a Suspense Account if the Trial Balance agrees.
Answer:
This statement is True.
When the Trial Balance does not tally a temporary account called ‘Suspense Account’ is opened to balance the trial balance. So when the trial balance is agreed there is no need to open ‘Suspense Account’.

Question 5.
All the errors can be rectified only through Suspense Account.
Answer:
This statement is False.
The errors of principle and errors of complete omission can be rectified by passing entries. So all the errors can not be rectified by the Trial Balance.

5. Do you agree or disagree with the following statements.

Question 1.
The unintentional omission or commission of amounts and accounts while recording the transactions is known as an error.
Answer:
Agree

Question 2.
The errors committed due to wrong recording, wrong posting, wrong totaling, wrong balancing, wrong calculations are known as Arithmetical errors.
Answer:
Disagree

Question 3.
When one or more debit errors happen to equal one or more credit errors it is said to be a Compensating error.
Answer:
Agree

Question 4.
The agreement of Trial balance is not affected when a transaction is not recorded at all in the original Books.
Answer:
Agree

Question 5.
When a transaction is not recorded according to the principles of accounting it is known as Compensating error.
Answer:
Disagree

6. Complete the following sentence.

Question 1.
______________ is assured only when there are no errors in the books of accounts.
Answer:
Accuracy

Question 2.
Transactions recorded in contravention of the accounting principles are known as ______________
Answer:
errors of principle

Question 3.
______________ entry depends generally on when the error is detected.
Answer:
Rectifying

Question 4.
Temporary account opened to rectify the entry is known as ______________
Answer:
suspense account

Question 5.
Errors are caused due to ______________ recording of transactions.
Answer:
wrong

Practical Problems

Question 1.
Rectify the following errors:
1. Salary paid to Pravin was wrongly debited to his personal account ₹ 6,500/-
2. Cash Purchases ₹ 12,000/- from Siddhant Traders was debited to Siddhant Trader Account.
3. Paid Rent ₹ 5,000 to landlord Shantilal was debited to his personal account.
4. Received interest ₹ 700 from Bank was wrongly credited to Bank Account.
5. Advertisement expenses ₹ 5,000/- paid to Times of India was debited to Times of India.
Solution:
Journal Proper

Working Note:

Question 2.
Rectify the following errors:
1. Machinery purchased for ₹ 9,000/- has been debited to Purchase Account.
2. ₹ 15,000/- paid to Indus Company for Machinery purchased stand debited to Indus Company Account.
3. Printer Purchased for ₹ 10,000/- was wrongly passed through Purchase Book.
4. ₹ 800/- paid to Mohan as Legal Charges were debited to his personal account.
5. Cash paid to Ramesh ₹ 500/- was debited to Suresh.
Solution:
Journal Proper

Working Note:

Question 3.
Rectify the following errors:
1. A credit sale of goods to Sanjay ₹ 3,000/- has been wrongly passed through the ‘Purchase Book’.
2. A credit purchase of goods from Sheetal amounting to ₹ 2,000/- has been wrongly passed through the ‘Sales Book’.
3. A return of goods worth ₹ 500/- to Umesh was passed through the ‘Sales Return Book’.
4. A return of goods worth ₹ 900/- by Ganesh was entered in ‘Purchase Return Book’.
5. Credit Purchases from Neha ₹ 10,000/- were recorded as ₹ 11,000/-
Solution:
Journal Proper

Working Note:

Question 4.
Rectify the following errors:
1. Paid Rent ₹ 2,000/- to Nikhil has been debited to his personal account.
2. Total of the Sales Return Book is wrongly taken more by ₹ 200/-
3. Goods sold to Dhanraj ₹ 6,500/- on credit were not posted to his personal account.
4. Old Computer purchased was debited to Repairs account ₹ 8,000/-
5. Repairs to Furniture of ₹ 500/- have been debited to Furniture account.
Solution:
Journal Proper

Working Note:

Question 5.
Rectify the following errors:
1. Wages paid for the construction of Building ₹ 10,000/- was wrongly debited to Wages Account.
2. Cash received from Patel ₹ 5,000/- though recorded in Cash Book was not posted to his personal account in the Ledger.
3. Sold goods worth ₹ 9,000/- to Rohini has been wrongly debited to Mohini’s Account.
4. Material purchased for construction of Building was debited to Purchase Account ₹ 5,000/-
Solution:
Journal Proper

Working Note:

Question 6.
There was a difference of ₹ 1230/- in a Trial Balance. It was placed on the Debit side of Suspense A/c. Later on, the following errors were discovered. Pass rectifying entries and prepare Suspense A/c.
1. Sales Book was overcast by ₹ 1,000/-
2. Goods sold to Aarti for ₹ 4,400/- has been posted to her account as ₹ 4,000/-
3. Purchases Book was overcast by ₹ 100/-
4. An amount of ₹ 500/- received from Ranjeet, has not been posted to his account.
5. Goods sold to Sameer for ₹ 750/- were recorded in Purchase Book.
6. An amount of ₹ 500/- has been posted to the credit side of the Commission Account instead of ₹ 570/-
Solution:
Journal Proper

Question 7.
A bookkeeper finds that the debit side of the Trial Balance is short of ₹ 308/- and so for the time being, the balances of the side by putting the difference to Suspense Account. The following errors were disclosed.
1. The debit side of the purchases account was undercast by ₹ 100/-
2. ₹ 100/- is the monthly total of discount allowed to customers were credited to the discount account in the ledger.
3. An entry for goods sold of ₹ 102/- to Mihir was posted to his account as ₹ 120/-
4. ₹ 26/- appearing in the Cash Book as paid for the purchase of Stationery for office use have not been posted to Ledger.
5. ₹ 275/- paid by Mihir were credited to Mithali’s Account.
You are required to make the necessary Journal Entries and the Suspense Account.
Solution:
Journal Proper

Question 8.
The trial Balance of Anurag did not agree. It showed an excess credit of ₹ 6,000/-. He put the difference to Suspense Account. He discovered the following errors.
1. Cash received from Ramakant ₹ 8,000/- posted to his account as ₹ 6,000/-
2. Credit purchases from Naman ₹ 7,000/- were recorded in Sales Book. However, Naman’s Account was correctly credited.
3. Return Inwards Book overcast by ₹ 1,000/-
4. Total of Sales Book ₹ 10,000/- was not posted to Sales Account.
5. Machinery purchased for ₹ 10,000/- was posted to Purchases Account as ₹ 5,000/-.
Rectify the errors and prepare Suspense Account.
Solution:

Question 9.
There was an error in the Trial Balance of Mr. Yashwant on 31st March 2019, and the difference in Books was carried to a Suspense Account. Ongoing through the Books you found that.
1. ₹ 1,000/- being purchases return were posted to the debit of Purchase Account.
2. ₹ 4,000/- paid to Badrinath was debited to Kedarnath’s Account.
3. ₹ 5,400/- received from Kishor was posted to the debit of his account.
4. Discount received ₹ 2,000/- was posted to the debit of Discount Allowed Account.
5. ₹ 2,740/- paid to Repairs to Motor Cycle was debited to Motor Cycle Account ₹ 1,740/-
Give Journal Entries to rectify the above errors and ascertain the amount transferred to Suspense Account on 31st March 2019 by showing the Suspense Account, assuming that the Suspense Account is balanced after the above corrections.
Solution:
Journal Proper

Question 10.
Rectify the following errors.
1. Goods purchased from Kishor ₹ 700/- were passed through Sales Book.
2. An item of ₹ 120/- in respect of purchase returns, has been wrongly entered in the Purchase Book.
3. Amount payable to Subhash for repairs done to Printer ₹ 180/- and new Printer supplied for ₹ 1,920/-, were entered in the Purchase Book as ₹ 2,000/-
4. Returned goods to Nitin ₹ 1,500/- was passed through Returns Inward Book.
5. An item of ₹ 450/- relating to the Prepaid Rent account was omitted to be brought forward.
Solution:
Journal Proper

Note: In entry No. 5 Suspense A/c is not used as the problem is silent about the opening of Suspense A/c.