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Chapter 14 - Biomolecules - 12th Chemistry Guide Samacheer Kalvi Solutions - Pdf Download [2024-2025]

Chapter 14 - Biomolecules - 12th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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2. Vitamin C
Sources

1. Yeast. milk, vegetables.
2. Citrus fruits, amia and green leafy vegetables.
Deficiency Diseases
1. Ben - ben
2. Scurvy (bleeding gums)
3 mark Question and Answers
VII. Answer the following questions.
Question 1.

Explain the methods of preparation of glucose.
Answer:
1. When sucrose is boiled with diI- $\mathrm{H}_2 \mathrm{SO}_4$ in alcoholic solution, hydrolysis take place and glucose and fructose are formed.

2. Glucose is produced commercially by the hydrolysis of starch with dilute $\mathrm{HCL}$ at high temperature and pressure.
Question 2.
What happens when glucose reacts with
1. $\mathrm{Br}_2 / \mathrm{H}_2 \mathrm{O}$
2. Conc. $\mathrm{HNO}_3$
Answer:
1.

2.

Question 3.
How will you prove the presence of aldehyde group in glucose?
Answer:
Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen's reagent) and alkaline copper sulphate (Fehling's solution). Tollen's reagent is reduced to metallic silver and Fehling's solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.

Question 4.
Define
1. Epimers
2. Epimerisation.
Answer:
1. Sugar differing at an asymmetric centre is known as epimers.
2. The process by which one epimer is converted into other is called epimerisation and it requires the enzyme epimersase.
3. Galactose is converted to glucose by this manner in our body.

Question 5.
Explain the methods of preparation of fructose with equations.
Answer:
1. Fructose is obtained from sucrose by heating with dilute $\mathrm{H}_2 \mathrm{SO}_4$ (or) with the enzyme invertase.

2. Fructose is prepared commercially by the hydrolysis of Inulin (a polysaccharide) in acidic medium

Question 6.
What happens when fructose is treated with sodium amalgam and water?
Answer:
When fructose is treated with sodium amalgam and water, partial reduction take place and the products formed are epimers of sorbitol and mannitol. New asymmetric carbon is formed at $\mathrm{C}-2$. This reaction
confirms the presence of keto group in fructose.

Question 7.
Explain about the cyclic structure of fructose?
Answer:
Fructose forms a five membered ring similar to furan. Hence it is called furanose form. When fructose is a component of a saccharide as in sucrose, it usually occurs in furanose form.

Question 8.
Explain about the structure, nature and properties of sucrose.
Answer:
1. Sucrose commonly known as table sugar is the most abundant disaccharide. It is obtained mainly from juice of sugar cane and sugar beets. Insects such as honey bees have the enzyme mvetase that catalyses the hydrolysis of sucrose into glucose and fructose mixture.
2. Honey is primarily a mixture of glucose, fructose and sucrose. On hydrolysis sucrose yields equal amount of glucose and fructose units.

3. Sucrose $\left(+66.6^{\circ}\right)$ and glucose $\left(52.5^{\circ}\right)$ are dextrorotatory compounds while fructose is levo rotatory $(-$ $\left.92.4^{\circ}\right)$
4. During hydrolysis of sucrose the optical rotation of the reaction mixture changes from dextro to levo. Hence sucrose is also as invert sugar.
5. Structure:

In sucrose $C_1$ of $\alpha-D$ glucose is joined to $C_2$ of $D$ - Fructose. The glycosidic bond thus formed is called $\alpha$, 1,2 - glycosidic bond. Since both the carbonyl carbons are involved in the glycosidic bonding, sucrose is a non - reducing sugar.
Question 9.
Prove that sucrose is
1. invert sugar
2. non - reducing sugar.
Answer:
1. Sucrose is an invert sugar: Sucrose $\left(+66.6^{\circ}\right)$ and glucose $\left(+52.5^{\circ}\right)$ are dextro rotatory compounds while fructose is levo rotatory $\left(-92.4^{\circ}\right)$. During hydrolysis of sucrose, the optical rotation of the reaction mixture changes from dextro to levo. Hence sucrose is also called invert sugar.
2. In sucrose, $C_1$ of $\alpha-D$ glucose is joined to $C_2$ of $\alpha-D$ fructose. The glycosidic bond thus formed is called $\alpha-1,2$ - glycosidic bond. Since both the carbonyl carbons (reducing groups) are involved in the glycosidic bonding, sucrose is a non - reducing sugar.
Question 10.
Write a note about lactose
Answer:
1. Lactose is a disaccharide found in milk of mammals and hence it is referred to as milk sugar.
2. On hydrolysis, it yeilds galactose and glucose. The $\beta-D$ galactose and $\beta-D$ glucose are linked by $\beta-1$, 4 - glycosidic bond. the aldehyde carbon is not involved in the glycosidic linkage. Hence it retains its reducing property and is called a reducing sugar.

Question 11.
Lactose act as reducing sugar. Justify this statement.
Answer:
1. Lactose is a disaccharide and contains one galactose unit and one glucose unit.
2. In lactose, the $\beta-\mathrm{D}$ galactose and $\beta-\mathrm{D}$ glucose are linked by $\beta-1,4-$ glycosidic bond.
3. The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.
Question 12.
Write about maltose with its structure.
Answer:
1. Maltose is extracted from malt and it is called malt sugar. Malt from sprouting barely is the major source of maltose. Maltose is produced during digestion of starch by the enzyme $\alpha$-analyse.

3. Maltose consists of two molecules of $\alpha-\mathrm{D}$ glucose units linked by an $\alpha-1,4-$ glycocidic bond between anomeric carbon ofone unit and $\mathrm{C}-4$ of the other unit. Since one of the glucose has the carbonyl group intact it act as a reducing sugar.
Question 13.
Sucrose and maltose are disaccharides but sucrose Is a non reducing sugar while maltose Is a reducing sugar. Give reason.
Answer:
1. Sucrose is a disaccharide that composed of $\alpha-\mathrm{D}$ glucose and $\beta-\mathrm{D}$ fructose. In sucrose $\mathrm{Cl}$ of $\alpha-\mathrm{D}$ glucose is joined to $\mathrm{C}_2$ of $\mathrm{D}$ - fructose. The glycosidic bond thus formed is called $\alpha-1,2$ - glycosidic bond. Since both the carbonyl carbons (reducing carbons) are involved in the glycosidic bonding, sucrose is a non-reducing sugar.
2. But maltose contains two molecules of $x-D$ glucose units that are linked by an $\alpha-1,4-$ glycosidic bond. Anomeric carbon of one unit and $\mathrm{C}-4$ of other unit are connected together. Smce one of the glucose bas the carbonyl group intact it acts as reducing sugar.
Question 14.
Give brief account of nature and structure of cellulose.
Answer:
1. Cellulose is the major constituent of plant cell walls. Cotton is almost pure cellulose. On hydrolysis, cellulose yields D - glucose molecules
2. Cellulose is a straight chain polysaccharide. The glucose molecules are linked by $\beta(1,4-)$ glycoside bond.
3. Structure of cellulose.

Question 15 .
What are the uses of cellulose?
Answer:
1. Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
2. Gun cotton - nitrated ester of cellulose an explosive is prepared from cellulose.
3. Cellulose act as food for animals
Question 16.
Human cannot use cellulose as food - Why?
Answer:
Human cannot use cellulose as food because our digestive systems do not contain the necessary enzymes such as glycosidases (or) cellulases that can hydrolyse the cellulose. But animals contain cellulose enzyme in their digestive system and they can digest cellulose. So cellulose can used as food for animals but not for human.
Question 17.
What are the major classification of proteins? Give example?
Answer:
1. Proteins are classified based on their structure into two major types. They are fibrous protein and glubular protein.
2. Fibrous proteins are linear molecules similar to fibres. They are generally insoluble in water and are held together by disulphide bridges and weak inter molecular hydrogen bonds. These proteins often used as structural proteins. Example, Keratin, Collagen.
3. Globular proteins have an overall spherical shape. The poly peptide chain is folded into a spherical shape. These proteins are usually soluble in water and have many functions including catalysis.
Question 18 .
Explain the mechanism of enzyme action?
Answer:
1. Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.
2. In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme sybstate complex. During this stage the substrate is converted into product and

Question 19.
Explain about the nature, classification and properties of lipids (or) write a note about lipids.
Answer:
1. Lipids are organic molecules that are soluble in organic solvents such as chloroform and methanol and are insoluble in water. Lipids means fat. They are the principal component of cell membrane including cell walls.
2. Lipids act as energy source for living systems. Fat provide $2-3$ fold higher energy compared to carbohydrates or proteins.
3. Based on their structure, lipids can be classified as simple lipids, compound lipids and derived lipids.
4. simple lipids can be further classified into fats, which are esters of long chain fatty acids fats, with glycerol (triglycerides) and waxes which are esters of fatty acid with long chain monohydric alcohols (Bees wax)
5. Compound lipids are the esters of simple fatty acid with glycerol which contain additional group. Based on the groups attached, they are further classified into phospholipids, glycolipids and lipproteins. Phospho lipids contain a phospho ester linkage while the glycolipids contain a sugar molecule attached. The iipo proteins are complexes of lipid with proteins.
Question 20.
Write the chemical name, sources and deficient disease of the following.
1. Vitamin D
2. Vitamin $\mathrm{E}$
3. Vitamin $\mathrm{K}$
Answer:

Question 21.
What are the biological functions of nucleic acids?
Answer:
1. Energy carriers (ATP)
2. Components of enzyme cofactors. Example - Co enzyme A, NAD, FAD
3. Chemical messengers. Example-Cyclic AMP, CAMP
Question 22.
What happens when $\mathrm{D}-$ glucose is treated with the following reagents?
1. $\mathrm{HI}$
2. Bromine water
3. $\mathrm{HNO}_3$
Answer:

1.

2.

3.

Question 23.
Define the following as related to proteins.
1. Peptide linkage
2. Primary structure
3. Denaturation
Answer:
1. Peptide linkage:
Amino acids are bifunctional molecules with $\mathrm{NH}_2$ group at its one end and $\mathrm{COOH}$ at the other. Therefore, the $\mathrm{COOH}$ of one molecule and $\mathrm{NH}_2$ of another molecule can interact with elimination of a $\mathrm{H}, \mathrm{O}$ molecule to form an amide like linkage called peptide bond or peptide linkage.
2. Primary structure:
The sequence in which amino acids are linked together in a polypeptide chain forms the primary structure.
3. Denaturation:
The process by which secondary and tertiary structure of proteins get disturbed by change of $\mathrm{pH}$ or temperature, so that they are not able to perform their functions, is called denaturation of proteins.
Question 24.
Difference between Globular and fibrous proteins.
Answer:
Globular proteins
1. They form $\alpha$-helix structure
2. They are soluble in water.
3. They are cross lined condensation polymers of acidic and basic amino acids.
Fibrous Proteins
1. They have - plcated structure.
2. They are insoluble in water.
3. They are linear condensation polymeric proteins.
Question 25.
Explain what is meant by
1. a peptide linkage
2. a glycosidic linkage.
Answer:
1. Peptide linkage:

Polymers of $\mathrm{x}$ - amino acids are connected to each other by peptide bond or peptide linkage. Chemically peptide linkage is an amide formed between - $\mathrm{COOH}$ group and $\mathrm{NH}_2$ group.
2. Glycosidic linkage:
The two monosaccharide units are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two mono saecharide units through oxygen atom is called glycosidic linkage.
Question 26.
What are essential and non-essential amino acids? Give one example of each type.
Answer:
1. Essential amino acids:
Amino acids which are not synthesised by the human body are called essential amino acids. ExampleValine, Leucinc.
2. Non - essential amino acids:
Amino acids which are synthesised by human body are called non - essential amino acids. Example Glycine, Aspartic acid, etc.
Question 27.
Mention the type of linkage responsible for the formation of the following.
1. Primary structure of proteins
2. Cross linking of polypeptide chains.
3. $\alpha$-helix formation
4. $\beta$-sheet structure.
Answer:
Type of Structure
1. Primary structure of proteins
2. Cross linking of polypeptide chains
3. $\alpha$-helix
4. $\beta$-sheet structure
Type of linkages
1. Peptide bond or peptide linkage
2. Polypeptide linkage
3. Hydrogen bond
4. Intermolecular hydrogen bond
Question 28.
Name the chemical components which constitute nucleotides. Write any two functions of nucleotides in a cell.
Answer:

Question 29.
Define the following and give one example of each.
1. Isoelectric point
2. Mutarotation
3. Enzymes
Answer:
1. Isoelectric point:
It is the $\mathrm{pH}$ at which +ve and - ve charges on zwitter ion are equal. Example amino acid exists as zwitter ion at $\mathrm{pH}=5.5$ to 6.3 .
2. Mutarotation:
It is a spontaneous change in optical rotation when an optically active substance is dissolved in water. Example $\alpha$ - glucose, when dissolved in water, then its optical rotation changes from 1110 to $52.5^{\circ}$.
3. Enzymes:
Enzymes are biocatalysts which speeds up the reactions in biosystems. They are highly specific and selective in their action. Chemically all enzymes are proteins.
Question 30.
What is denaturation and renaturation of proteins? Give reason: Amylose present in the saliva becomes inactive in the stomach.
Answer:
1. The process of disruption of $2^{\circ}$ and $3^{\circ}$ structure of proteins without changing its primary structure is called denaturation.
2. In stomach the $\mathrm{pH}$ is acidic, therefore amylose becomes inactive. That is why digestion of carbohydrates does not take place in stomach.
Question 31.
Define the following terms.
1. Nucleotide
2. Anomers
3. Essential amino acids
Answer:
1. Nucleotide:
It is the monomer unit of DNA which is formed by a nitrogenous base, deoxyribose sugar and phosphoric acid.

2. Anomers:
Anomers are cyclic monosaccharides which are differing from each other in the configuration of $\mathrm{C}-1$ if it is an aldose or in the configuration at $\mathrm{C}-2$ if it is a ketose.
3. Essential amino acids:
The amino acids cannot be synthesised by the body and are essential for the body.
Question 32.
Which one of the following is disaccharide.
1. Starch, Maltose, Fructose, Glucose.
2. Write the name of vitamin whose deficiency causes bone deformities in children.
Answer:
1. Maltose
2. Vitamin D
Quesiton 33.
Write the major classes in which the carbohydrates are divided depending upon whether thee undergo hydrolysis and if so, the number of products formed.
Answer:
On the basis of hydrolysis, carbohydrates are divided into three major classes:
1. Monosaccharides. These cannot be hydrolysed into simpler molecules. These are further classified as aldoses, and ketoses.
2. Oligosaccharides. These carbohydrates on hydrolysis give 2-10 units of mono-saccharides. For example - Sucrose.
3. Polysaccharides. These are high molecular mass carbohydrates which give many molecules of monosaccharides on hydrolysis. Form example: Cellulose, starch.
Question 34.
1. What changes occur in the nature of egg proteins on boiling?
2. Name the type of bonding which stabilises the a-helix structure in proteins.
Answer:
1. On boiling, protein of egg gets denatured. Thus, due to coagulation water get absorbed.
2. Hydrogen bonding between

Question 35.
Answer the following questions briefly.
1. What are reducing sugars?
2. What is meant by denatu ration of a protein?
3. How is oxygen replenished in our atmosphere?
Answer:
1. Reducing sugar:
All those carbohydrates which reduce Fehling's solution and Tollens' reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.
2. Denaturation of a protein:
When $2^{\circ}$ and $3^{\circ}$ structure of a protein is destroyed due to the physical changes like temperature, change in $\mathrm{pH}$, it is called denaturation of a protein.
Example: Coagulation of egg white on boiling.
3. We take oxygen from atmosphere and release $\mathrm{CO}_2$. Plants take up $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ from the atmosphere to prepare their food in the presence of sunlight and release $\mathrm{O}_2$ thus $\mathrm{O}_2$ is replenished in atmosphere.
5 mark Questions Answers
VIII.Answer the following questions.
Question 1.

How would you prove the structure of glucose? (OR) Elucidate the structure of glucose.
Answer:
1. Elemental analysis and molecular weight determination show that the molecular formula of glucose is $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2. On reduction with Conc. $\mathrm{HI}$ and red $\mathrm{P}$ at $373 \mathrm{~K}$, glucose gives a mixture of $\mathrm{n}$ - hexane and 2 - iodohexane indicating that the six carbon atoms are bonded linearly.

3. Glucose reacts with hydroxylamine to form oxime and with $\mathrm{HCN}$ to form cyanohydrin. These reactions indicate the presence of carbonyl group in glucose.

4. Glucose gets oxidised to gluconic acid with bromine water with bromine water suggests that the carbonyl group is an aldchyde group and it occupies one end of the carbon chain. When glucose is oxidised by
conc. $\mathrm{HNO}_3$, glucaric acid is formed and it suggest that the other end Id occupied by a primary alcoholi.

5. Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Toiles reagent) and alkaline copper sulphate (Fehling's solution). These reactions further confirm the presence of an aldehyde group.

6. Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol groups.

7. Glucose is a stable compound and does not undergo dehydration easily. It indicates that not more than one hydroxyl group is bonded to a single carbon atom. Thus the five hydroxyl groups are attached to five different carbon atoms and the sixth carbon is an aldehyde group.
8. The exact special arrangement of $-\mathrm{OH}$ groups was given by Emil Fischer as follows.

9. $\mathrm{D}(+)$ glucose has D configuration and it is dextro rotatoly.
Question 2.
Explain about the cyclic structure of Glucose.
Answer:
1. Fischer identified that the open chain Penta hydroxyl aldehyde structure of glucose that he proposed did not completely explain its chemical behaviour.
2. Unlike simple aldehyde, glucose did not form crystalline hisuiphite compound with sodium bisuiphite. Glucose does not give Schifis test and pent.a acetate derivative of glucose was not oxidised by Tollen's reagent. This behaviour could not be explained by open chain structure.

3.

4. In order to explain these it was proposed that one of the hydroxyl group reacts with aldehyde group to form a cyclic structure (hemiacetal form). This also results in the conversation of the achiral aldehyde carbon into a chiral one leading the possibility of two isomers. These two isomers differ only in the configuration of $\mathrm{C}_1$ carbon. These isomers are called anomers.
5. The two anomeric forms of glucose are called - and $\beta$-forms. This cyclic structure of glucose is similar to pyran, a cyclic compound with 5 carbon and one oxygen atom, and hence is called pyranose form.
6. The specific rotation of pure $\alpha-$ and $3-$ (D) glucose are $112^{\circ} \& 18.7^{\circ}$ respectively. However, when pure form any one of these sugars dissolved in water, slow interconversion of $-\mathrm{D}$ glucose and $\beta-\mathrm{D}$ glucose via open chain form until equilibrium is established giving constant specific rotation $+53^{\circ}$. This phenomenon is called mutarotation.
Question 3.
Explain about the structure of Fructose. (OR) Elucidate the structure of Fructose.
Answer:
1. Elemental analysis and molecular weight determination of fructose show that it has the molecular formula $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2. Fructose on reduction with $\mathrm{Hl}$ and red phosphoms gives a mixture of $\mathrm{n}$-hexane (major product) and 2 iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.

3. Fructose reacts with $\mathrm{NH}_2 \mathrm{OH}$ and $\mathrm{HCN}$. It shows the presence of carbonyl groups in the molecules of fructose.
4. Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of' five hydroxyl groups in a fructose molecule.
5. Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde ($\mathrm{CHO})$ group.
6. Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at $\mathrm{C}-2$. This confirms the presence of keto group.

7. On oxidation with nitric acid, it gives glycolic acid and tartane acids which contain smaller number of carbon atoms than in fructose. This shows that a keto group is present in $\mathrm{C}-2$. It also shows the presence
of 10 alcoholic groups at C- 1 and C- 6 . From the above reaction the structure of fructose is

This shows that a keto group is present in $\mathrm{C}-2$. It also shows the lpresence of 10 alcoholic groups at $\mathrm{C}-1$ and $\mathrm{C}-6$. From the above reaction the structure of fructose is

Question 4.
Describe about the structure, nature and properties of starch.
Answer:
1. Starch is used for energy storage in plants. Potatoes, corn, wheat and rice are the rich sources of starch.
2. It is a polymer of glucose in which glucose molecules are lined by $\alpha(1,4)$ glycosidic bonds.
3. Starch can be separated into two fractions namely, water soluble amylose and water insoluble amylo pectin. Starch contain about $20 \%$ of amylase and about $80 \%$ amylopectin.
4. Amylose is composed of unbranched chains upto $4000 \alpha$-D glucose molecules joined by $\alpha(1,4)$ glycosidic bonds.
5. At branch points, new chains of 24 to 30 glucose molecules are linked by $\alpha(1,6)$ glycosidic bonds with iodine solution amylose gives blue colour while amylo pectin gives a purple colour.

Question 5.
Explain about the structure of proteins.
Answer:
1. Proteins are polymers of amino acids. Their three dimensional structure depends mainly on the sequence of amino acids. The protein structure can be described at four hierarchal levels called primary, secondary, tertiary and quaternary structures.
Primary structure of proteins:
Proteins are polypeptide chains made up of amino acids connected through peptide bonds. The relative arrangement of the amino acids in the polypeptide chain is called the primary structure of the protein.

Secondary structure of proteins:
The amino acids in the polypeptide chain forms highly, regular shapes through the hydrogen bond between the carbonyl oxygen and the
neighbouring amine hydrogen of the main chain, $\alpha$-Helix and $\beta$ - strands or sheets are two most common substructures formed by proteins.
$\alpha-$ Helix:
In the $\alpha$-helix sub-structure, the amino acids are arranged in a righthanded helical (spiral) structure and are stabilised by the hydrogen bond. The side chains of the residues protrude outside of the helix. Each turn of an $\alpha$ - helix contains about 3.6 residues and is about $5.4 \mathrm{~A}$ long.
$\beta$ - Strands:
$\beta$ - Strands are extended peptide chain rather than coiled. The hydrogen bond occur between main chain carbonyl group one such strand and the amino group of the adjacent strand resulting in the formation of a sheet like structure. This arrangement is called $\beta$-sheets.
Tertiary structure:
The secondary structure elements ( $\alpha$-helix \& $\beta$-sheets) further folds to form a three dimensional arrangement. This tertiary structure of proteins are stabilised by the interactions between the side chains of the amino acids. These interactions include the disulphide bridges between cysteine residues, electrostatic, hydrophobic, hydrogen bonds and van der Waals interactions.
Quaternary Structure:
The oxygen transporting protein, haemoglobin contains four polypeptide chains while DNA polymerase enzyme that make copies of DNA, has ten polypeptide chains. In these proteins the individual polypeptide chains interacts with each other to form the multimeric structure which known as quaternary structure. The interactions that stabilises the tertiary structures also stabilises the quaternary structures.
Question 6.
What are the biological importance of proteins?
Answer:
Proteins are the functional units of living things play vital role in all biological processes
1. All biochemical reactions occur in the living systems are catalysed by the catalytic proteins called enzymes.
2. Proteins such as keratin, collagen acts as structural back bones.
3. Proteins are used for transporting molecules (Haemogiobin), organelles (Kinesins) in the cell and control the movement of molecules in and out of the cells (Transporters).
4. Antibodies help the body to fight various diseases.
5. Proteins are used as messengers to coordinate many functions. Insulin \& glucagon controls the glucose level in the blood.
6. Proteins act as receptors that detect presence of certain signal molecules and activate the proper response.
7. Proteins are also used to store metals such as iron (Femtin) etc.
Question 7.
Write the chemical name, source and deficient disease of the following
1. Vitamin $A$
2. Vitamin $B_1$
3. Vitamin $B_2$
4. Vitamin $B_3$
5. Vitamin $\mathrm{B}_5$
Answer:

Question 8.
Write the chemical name, source and deficient disease of the following
1. Vitamin $\mathrm{B}_6$
2. Vitamin $B_7$
3. Vitamin $\mathrm{B}_9$
4. Vitamin $B_{12}$
5. Vitamin C
Answer:

Question 9.
Explain about the composition and structure of nucleic acids.
Answer:
1. Nucleic acids are biopolymers of nucleotides. Controlled hydrolysis of DNA and RNA yields three components namely a nitrogeneous base, a pentose sugar and phosphate group.
2. Nitrogen base.
(a) These are nitrogen containing organic compounds which are derivatives of two parent compounds, pyrimidine and purine.
(b) Both DNA and RNA have two major purine bases, adenine (A) and guanine (G). In both DNA and RNA, one of the pyrimidines is cytosinc (C), but the second pyrimidine is thymine (T) in DNA and uracil (U) in RNA.
3. Pentose Sugar.
Nucleic acids have two types ofpentoses. The recurring deoxyribonucleotide units of DNA contain 2' dcoxy - D - ribose and the ribonucleotide units of RNA contain D - ribose. In nucleotides, both types of pentoses are in their - furanose form.
4. Phosphate group.
Phosphoric acid forms phosphor diester bond between nucleotides. Based on the number of phosphate group present in the nucleotides, they are classified mono nucleotide, dinucleotide and trinucleotide.
5. The molecule without the phosphate group is called a nucleoside. A nucleotide is derived from a nucleoside by the addition of a molecule of phosphoric acid.
6. Sugar + Base $\rightarrow$ Nucleoside
Nucleoside + Phosphate $\rightarrow$ Nucleotidc
Nucleotide $\rightarrow$ Polynucleotide (Nucleic Acid)
Question 10.
Describe about the double strand helix structure of DNA.
Answer:
1. Watson and Crick postulated a 3 - dimensional model of RNA structure which consisted of two antiparallel helical DNA chains wound around the same axis to form a right handed double helix.
2. Th e hydrophilic backbones of alternating deoxyribose and phosphate groups are on the outside of the double helix, facing the surrounding water. Th e purine and pyrimidine bases of both strands are stacked inside the double helix,
with their hydrophobic and ring structures very close together and perpendicular to the long axis, thereby reducing the repulsions between the charged phosphate groups. The offset pairing of the two strands creates a major groove and minor groove on the surface of the duplex.

3. The model revealed that there are 10.5 pairs $\left(36 \mathrm{~A}^{\circ}\right)$ per turn of the helix and $3.4 \mathrm{~A}^{\circ}$ between the stacked bases. They also found that each base is hydrogen bonded to a base in opposite strand to form a planar base pair.
4. Two hydrogen bonds are formed between adenine and thymine and three hydrogen bonds are formed between guanine and cytosine. Other pairing tends to destablize the double helical structure. This specific association of the two chains of the double helix is known as complementary base pairing.
5. The DNA double or duplex is held together by two forces.
- Hydrogen bonding between complementary base pairs.
- Base - stacking interactions.
The complementary between the DNA strands is attributable to the hydrogen bonding between base pairs but the base stacking interactions are largely non-specific, make the major contribution to the stability of the double helix.

Question 11.
Explain about the types of RNA molecules.
Answer:
1. Ribonucleic acids are similar to DNA. Cells contain upto eight times high quantity of RNA than DNA. RNA is found in large amount in the cytoplasm and a lesser amount in the nucleus.
2. RNA molecules are classified according to their structure and function into three major types.
1. Ribosomal RNA (r - RNA)
2. Messenger RNA (m-RNA)
3. Transfer RNA ( $\mathrm{t}-\mathrm{RNA})$
3. $\mathrm{r}-\mathrm{RNA}$ :
$\mathrm{r}$ - RNA is mainly found in cytoplasm and in ribosomes, which contain $60 \%$ RNA and $40 \%$ protein. Ribosomes are the sites at which protein synthesis takes place.
4. $\mathrm{t}-\mathrm{RNA}$ :
$\mathrm{t}$ - RNA molecules have lowest molecular weight of all nucleic acids. They consist of $73-94$ nucleotides in a single chain. The function of $t-$ RNA is to carry amino acids to the sites of protein synthesis on ribosomes.
5. $\mathrm{m}-\mathrm{RNA}$ :
$\mathrm{m}$ - RNA is present in small quantity and very short lived. They are single stranded and their synthesis take place on DNA. The synthesis m-RNA from DNA strand is called transcription. $\mathrm{m}-\mathrm{RNA}$ carries genetic information from DNA to the ribosomes for protein synthesis.
Question 12.
Explain about DNA finger printing process.
Answer:
1. DNA finger printing is one of the most accurate methods for placing an individual at the scene of a crime has been a finger print.
2. The DNA finger print is unique for every person and can be extracted from traces of sample from blood, saliva, hair etc. By using this method, we can detect the individual specific variation in human DNA.
3. In this method, the extracted DNA is cut at specific points of varying lengths in the formation of DNA fragments of varying lengths which were analysed by technique called gel electrophoresis. This method separates the fragments based on their size. The gel containing the DNA fragments are then transferred to a nylon sheet using a technique called blotting. Then the fragments will undergo autoradiography in which they were exposed to DNA probes.

4. A piece of X-ray film was then exposed to the fragments, and a dark mark was produced at any point where a radioactive probe had become attached. The resultant pattern of marks could then be compared with other samples.
5. DNA finger printing is based on slight sequence differences between individuals. These methods are providing decisive in court cases world wide.
Common Errors
1. Carbohydrates \& Enzyme names are different.
2. Peptide bond and amide bond are different.
3. Vitamins: Solubility based classification
4. Main source of vitamin
Rectifications
1. Carbohydrates name should end as - ose.
Eg. Diatose, maltose
Enzyme name should end as - ase.
Eg. Diastase, maltase

2.

3. Vitamin A, D, E, K-fat soluble. Vitamin C, B complex are water soluble.
4. Milk:
Vitamin A $\mathrm{A}_1, \mathrm{~B}_2, \mathrm{~B}_6, \mathrm{~B}_7, \mathrm{D}$ (6 Vitamins)
Green Vegetables:
$\mathrm{B}_1, \mathrm{~B}_3, \mathrm{~B}_7, \mathrm{~K}$ (4 vitamins)
Liver:
$
\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3, \mathrm{~B}_7
$

-->
$
2. Vitamin C
Sources

1. Yeast. milk, vegetables.
2. Citrus fruits, amia and green leafy vegetables.
Deficiency Diseases
1. Ben - ben
2. Scurvy (bleeding gums)
3 mark Question and Answers
VII. Answer the following questions.
Question 1.

Explain the methods of preparation of glucose.
Answer:
1. When sucrose is boiled with diI- $\mathrm{H}_2 \mathrm{SO}_4$ in alcoholic solution, hydrolysis take place and glucose and fructose are formed.

2. Glucose is produced commercially by the hydrolysis of starch with dilute $\mathrm{HCL}$ at high temperature and pressure.
Question 2.
What happens when glucose reacts with
1. $\mathrm{Br}_2 / \mathrm{H}_2 \mathrm{O}$
2. Conc. $\mathrm{HNO}_3$
Answer:
1.

2.

Question 3.
How will you prove the presence of aldehyde group in glucose?
Answer:
Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen's reagent) and alkaline copper sulphate (Fehling's solution). Tollen's reagent is reduced to metallic silver and Fehling's solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.

Question 4.
Define
1. Epimers
2. Epimerisation.
Answer:
1. Sugar differing at an asymmetric centre is known as epimers.
2. The process by which one epimer is converted into other is called epimerisation and it requires the enzyme epimersase.
3. Galactose is converted to glucose by this manner in our body.

Question 5.
Explain the methods of preparation of fructose with equations.
Answer:
1. Fructose is obtained from sucrose by heating with dilute $\mathrm{H}_2 \mathrm{SO}_4$ (or) with the enzyme invertase.

2. Fructose is prepared commercially by the hydrolysis of Inulin (a polysaccharide) in acidic medium

Question 6.
What happens when fructose is treated with sodium amalgam and water?
Answer:
When fructose is treated with sodium amalgam and water, partial reduction take place and the products formed are epimers of sorbitol and mannitol. New asymmetric carbon is formed at $\mathrm{C}-2$. This reaction
confirms the presence of keto group in fructose.

Question 7.
Explain about the cyclic structure of fructose?
Answer:
Fructose forms a five membered ring similar to furan. Hence it is called furanose form. When fructose is a component of a saccharide as in sucrose, it usually occurs in furanose form.

Question 8.
Explain about the structure, nature and properties of sucrose.
Answer:
1. Sucrose commonly known as table sugar is the most abundant disaccharide. It is obtained mainly from juice of sugar cane and sugar beets. Insects such as honey bees have the enzyme mvetase that catalyses the hydrolysis of sucrose into glucose and fructose mixture.
2. Honey is primarily a mixture of glucose, fructose and sucrose. On hydrolysis sucrose yields equal amount of glucose and fructose units.

3. Sucrose $\left(+66.6^{\circ}\right)$ and glucose $\left(52.5^{\circ}\right)$ are dextrorotatory compounds while fructose is levo rotatory $(-$ $\left.92.4^{\circ}\right)$
4. During hydrolysis of sucrose the optical rotation of the reaction mixture changes from dextro to levo. Hence sucrose is also as invert sugar.
5. Structure:

In sucrose $C_1$ of $\alpha-D$ glucose is joined to $C_2$ of $D$ - Fructose. The glycosidic bond thus formed is called $\alpha$, 1,2 - glycosidic bond. Since both the carbonyl carbons are involved in the glycosidic bonding, sucrose is a non - reducing sugar.
Question 9.
Prove that sucrose is
1. invert sugar
2. non - reducing sugar.
Answer:
1. Sucrose is an invert sugar: Sucrose $\left(+66.6^{\circ}\right)$ and glucose $\left(+52.5^{\circ}\right)$ are dextro rotatory compounds while fructose is levo rotatory $\left(-92.4^{\circ}\right)$. During hydrolysis of sucrose, the optical rotation of the reaction mixture changes from dextro to levo. Hence sucrose is also called invert sugar.
2. In sucrose, $C_1$ of $\alpha-D$ glucose is joined to $C_2$ of $\alpha-D$ fructose. The glycosidic bond thus formed is called $\alpha-1,2$ - glycosidic bond. Since both the carbonyl carbons (reducing groups) are involved in the glycosidic bonding, sucrose is a non - reducing sugar.
Question 10.
Write a note about lactose
Answer:
1. Lactose is a disaccharide found in milk of mammals and hence it is referred to as milk sugar.
2. On hydrolysis, it yeilds galactose and glucose. The $\beta-D$ galactose and $\beta-D$ glucose are linked by $\beta-1$, 4 - glycosidic bond. the aldehyde carbon is not involved in the glycosidic linkage. Hence it retains its reducing property and is called a reducing sugar.

Question 11.
Lactose act as reducing sugar. Justify this statement.
Answer:
1. Lactose is a disaccharide and contains one galactose unit and one glucose unit.
2. In lactose, the $\beta-\mathrm{D}$ galactose and $\beta-\mathrm{D}$ glucose are linked by $\beta-1,4-$ glycosidic bond.
3. The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.
Question 12.
Write about maltose with its structure.
Answer:
1. Maltose is extracted from malt and it is called malt sugar. Malt from sprouting barely is the major source of maltose. Maltose is produced during digestion of starch by the enzyme $\alpha$-analyse.

3. Maltose consists of two molecules of $\alpha-\mathrm{D}$ glucose units linked by an $\alpha-1,4-$ glycocidic bond between anomeric carbon ofone unit and $\mathrm{C}-4$ of the other unit. Since one of the glucose has the carbonyl group intact it act as a reducing sugar.
Question 13.
Sucrose and maltose are disaccharides but sucrose Is a non reducing sugar while maltose Is a reducing sugar. Give reason.
Answer:
1. Sucrose is a disaccharide that composed of $\alpha-\mathrm{D}$ glucose and $\beta-\mathrm{D}$ fructose. In sucrose $\mathrm{Cl}$ of $\alpha-\mathrm{D}$ glucose is joined to $\mathrm{C}_2$ of $\mathrm{D}$ - fructose. The glycosidic bond thus formed is called $\alpha-1,2$ - glycosidic bond. Since both the carbonyl carbons (reducing carbons) are involved in the glycosidic bonding, sucrose is a non-reducing sugar.
2. But maltose contains two molecules of $x-D$ glucose units that are linked by an $\alpha-1,4-$ glycosidic bond. Anomeric carbon of one unit and $\mathrm{C}-4$ of other unit are connected together. Smce one of the glucose bas the carbonyl group intact it acts as reducing sugar.
Question 14.
Give brief account of nature and structure of cellulose.
Answer:
1. Cellulose is the major constituent of plant cell walls. Cotton is almost pure cellulose. On hydrolysis, cellulose yields D - glucose molecules
2. Cellulose is a straight chain polysaccharide. The glucose molecules are linked by $\beta(1,4-)$ glycoside bond.
3. Structure of cellulose.

Question 15 .
What are the uses of cellulose?
Answer:
1. Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
2. Gun cotton - nitrated ester of cellulose an explosive is prepared from cellulose.
3. Cellulose act as food for animals
Question 16.
Human cannot use cellulose as food - Why?
Answer:
Human cannot use cellulose as food because our digestive systems do not contain the necessary enzymes such as glycosidases (or) cellulases that can hydrolyse the cellulose. But animals contain cellulose enzyme in their digestive system and they can digest cellulose. So cellulose can used as food for animals but not for human.
Question 17.
What are the major classification of proteins? Give example?
Answer:
1. Proteins are classified based on their structure into two major types. They are fibrous protein and glubular protein.
2. Fibrous proteins are linear molecules similar to fibres. They are generally insoluble in water and are held together by disulphide bridges and weak inter molecular hydrogen bonds. These proteins often used as structural proteins. Example, Keratin, Collagen.
3. Globular proteins have an overall spherical shape. The poly peptide chain is folded into a spherical shape. These proteins are usually soluble in water and have many functions including catalysis.
Question 18 .
Explain the mechanism of enzyme action?
Answer:
1. Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.
2. In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme sybstate complex. During this stage the substrate is converted into product and

Question 19.
Explain about the nature, classification and properties of lipids (or) write a note about lipids.
Answer:
1. Lipids are organic molecules that are soluble in organic solvents such as chloroform and methanol and are insoluble in water. Lipids means fat. They are the principal component of cell membrane including cell walls.
2. Lipids act as energy source for living systems. Fat provide $2-3$ fold higher energy compared to carbohydrates or proteins.
3. Based on their structure, lipids can be classified as simple lipids, compound lipids and derived lipids.
4. simple lipids can be further classified into fats, which are esters of long chain fatty acids fats, with glycerol (triglycerides) and waxes which are esters of fatty acid with long chain monohydric alcohols (Bees wax)
5. Compound lipids are the esters of simple fatty acid with glycerol which contain additional group. Based on the groups attached, they are further classified into phospholipids, glycolipids and lipproteins. Phospho lipids contain a phospho ester linkage while the glycolipids contain a sugar molecule attached. The iipo proteins are complexes of lipid with proteins.
Question 20.
Write the chemical name, sources and deficient disease of the following.
1. Vitamin D
2. Vitamin $\mathrm{E}$
3. Vitamin $\mathrm{K}$
Answer:

Question 21.
What are the biological functions of nucleic acids?
Answer:
1. Energy carriers (ATP)
2. Components of enzyme cofactors. Example - Co enzyme A, NAD, FAD
3. Chemical messengers. Example-Cyclic AMP, CAMP
Question 22.
What happens when $\mathrm{D}-$ glucose is treated with the following reagents?
1. $\mathrm{HI}$
2. Bromine water
3. $\mathrm{HNO}_3$
Answer:

1.

2.

3.

Question 23.
Define the following as related to proteins.
1. Peptide linkage
2. Primary structure
3. Denaturation
Answer:
1. Peptide linkage:
Amino acids are bifunctional molecules with $\mathrm{NH}_2$ group at its one end and $\mathrm{COOH}$ at the other. Therefore, the $\mathrm{COOH}$ of one molecule and $\mathrm{NH}_2$ of another molecule can interact with elimination of a $\mathrm{H}, \mathrm{O}$ molecule to form an amide like linkage called peptide bond or peptide linkage.
2. Primary structure:
The sequence in which amino acids are linked together in a polypeptide chain forms the primary structure.
3. Denaturation:
The process by which secondary and tertiary structure of proteins get disturbed by change of $\mathrm{pH}$ or temperature, so that they are not able to perform their functions, is called denaturation of proteins.
Question 24.
Difference between Globular and fibrous proteins.
Answer:
Globular proteins
1. They form $\alpha$-helix structure
2. They are soluble in water.
3. They are cross lined condensation polymers of acidic and basic amino acids.
Fibrous Proteins
1. They have - plcated structure.
2. They are insoluble in water.
3. They are linear condensation polymeric proteins.
Question 25.
Explain what is meant by
1. a peptide linkage
2. a glycosidic linkage.
Answer:
1. Peptide linkage:

Polymers of $\mathrm{x}$ - amino acids are connected to each other by peptide bond or peptide linkage. Chemically peptide linkage is an amide formed between - $\mathrm{COOH}$ group and $\mathrm{NH}_2$ group.
2. Glycosidic linkage:
The two monosaccharide units are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two mono saecharide units through oxygen atom is called glycosidic linkage.
Question 26.
What are essential and non-essential amino acids? Give one example of each type.
Answer:
1. Essential amino acids:
Amino acids which are not synthesised by the human body are called essential amino acids. ExampleValine, Leucinc.
2. Non - essential amino acids:
Amino acids which are synthesised by human body are called non - essential amino acids. Example Glycine, Aspartic acid, etc.
Question 27.
Mention the type of linkage responsible for the formation of the following.
1. Primary structure of proteins
2. Cross linking of polypeptide chains.
3. $\alpha$-helix formation
4. $\beta$-sheet structure.
Answer:
Type of Structure
1. Primary structure of proteins
2. Cross linking of polypeptide chains
3. $\alpha$-helix
4. $\beta$-sheet structure
Type of linkages
1. Peptide bond or peptide linkage
2. Polypeptide linkage
3. Hydrogen bond
4. Intermolecular hydrogen bond
Question 28.
Name the chemical components which constitute nucleotides. Write any two functions of nucleotides in a cell.
Answer:

Question 29.
Define the following and give one example of each.
1. Isoelectric point
2. Mutarotation
3. Enzymes
Answer:
1. Isoelectric point:
It is the $\mathrm{pH}$ at which +ve and - ve charges on zwitter ion are equal. Example amino acid exists as zwitter ion at $\mathrm{pH}=5.5$ to 6.3 .
2. Mutarotation:
It is a spontaneous change in optical rotation when an optically active substance is dissolved in water. Example $\alpha$ - glucose, when dissolved in water, then its optical rotation changes from 1110 to $52.5^{\circ}$.
3. Enzymes:
Enzymes are biocatalysts which speeds up the reactions in biosystems. They are highly specific and selective in their action. Chemically all enzymes are proteins.
Question 30.
What is denaturation and renaturation of proteins? Give reason: Amylose present in the saliva becomes inactive in the stomach.
Answer:
1. The process of disruption of $2^{\circ}$ and $3^{\circ}$ structure of proteins without changing its primary structure is called denaturation.
2. In stomach the $\mathrm{pH}$ is acidic, therefore amylose becomes inactive. That is why digestion of carbohydrates does not take place in stomach.
Question 31.
Define the following terms.
1. Nucleotide
2. Anomers
3. Essential amino acids
Answer:
1. Nucleotide:
It is the monomer unit of DNA which is formed by a nitrogenous base, deoxyribose sugar and phosphoric acid.

2. Anomers:
Anomers are cyclic monosaccharides which are differing from each other in the configuration of $\mathrm{C}-1$ if it is an aldose or in the configuration at $\mathrm{C}-2$ if it is a ketose.
3. Essential amino acids:
The amino acids cannot be synthesised by the body and are essential for the body.
Question 32.
Which one of the following is disaccharide.
1. Starch, Maltose, Fructose, Glucose.
2. Write the name of vitamin whose deficiency causes bone deformities in children.
Answer:
1. Maltose
2. Vitamin D
Quesiton 33.
Write the major classes in which the carbohydrates are divided depending upon whether thee undergo hydrolysis and if so, the number of products formed.
Answer:
On the basis of hydrolysis, carbohydrates are divided into three major classes:
1. Monosaccharides. These cannot be hydrolysed into simpler molecules. These are further classified as aldoses, and ketoses.
2. Oligosaccharides. These carbohydrates on hydrolysis give 2-10 units of mono-saccharides. For example - Sucrose.
3. Polysaccharides. These are high molecular mass carbohydrates which give many molecules of monosaccharides on hydrolysis. Form example: Cellulose, starch.
Question 34.
1. What changes occur in the nature of egg proteins on boiling?
2. Name the type of bonding which stabilises the a-helix structure in proteins.
Answer:
1. On boiling, protein of egg gets denatured. Thus, due to coagulation water get absorbed.
2. Hydrogen bonding between

Question 35.
Answer the following questions briefly.
1. What are reducing sugars?
2. What is meant by denatu ration of a protein?
3. How is oxygen replenished in our atmosphere?
Answer:
1. Reducing sugar:
All those carbohydrates which reduce Fehling's solution and Tollens' reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.
2. Denaturation of a protein:
When $2^{\circ}$ and $3^{\circ}$ structure of a protein is destroyed due to the physical changes like temperature, change in $\mathrm{pH}$, it is called denaturation of a protein.
Example: Coagulation of egg white on boiling.
3. We take oxygen from atmosphere and release $\mathrm{CO}_2$. Plants take up $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ from the atmosphere to prepare their food in the presence of sunlight and release $\mathrm{O}_2$ thus $\mathrm{O}_2$ is replenished in atmosphere.
5 mark Questions Answers
VIII.Answer the following questions.
Question 1.

How would you prove the structure of glucose? (OR) Elucidate the structure of glucose.
Answer:
1. Elemental analysis and molecular weight determination show that the molecular formula of glucose is $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2. On reduction with Conc. $\mathrm{HI}$ and red $\mathrm{P}$ at $373 \mathrm{~K}$, glucose gives a mixture of $\mathrm{n}$ - hexane and 2 - iodohexane indicating that the six carbon atoms are bonded linearly.

3. Glucose reacts with hydroxylamine to form oxime and with $\mathrm{HCN}$ to form cyanohydrin. These reactions indicate the presence of carbonyl group in glucose.

4. Glucose gets oxidised to gluconic acid with bromine water with bromine water suggests that the carbonyl group is an aldchyde group and it occupies one end of the carbon chain. When glucose is oxidised by
conc. $\mathrm{HNO}_3$, glucaric acid is formed and it suggest that the other end Id occupied by a primary alcoholi.

5. Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Toiles reagent) and alkaline copper sulphate (Fehling's solution). These reactions further confirm the presence of an aldehyde group.

6. Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol groups.

7. Glucose is a stable compound and does not undergo dehydration easily. It indicates that not more than one hydroxyl group is bonded to a single carbon atom. Thus the five hydroxyl groups are attached to five different carbon atoms and the sixth carbon is an aldehyde group.
8. The exact special arrangement of $-\mathrm{OH}$ groups was given by Emil Fischer as follows.

9. $\mathrm{D}(+)$ glucose has D configuration and it is dextro rotatoly.
Question 2.
Explain about the cyclic structure of Glucose.
Answer:
1. Fischer identified that the open chain Penta hydroxyl aldehyde structure of glucose that he proposed did not completely explain its chemical behaviour.
2. Unlike simple aldehyde, glucose did not form crystalline hisuiphite compound with sodium bisuiphite. Glucose does not give Schifis test and pent.a acetate derivative of glucose was not oxidised by Tollen's reagent. This behaviour could not be explained by open chain structure.

3.

4. In order to explain these it was proposed that one of the hydroxyl group reacts with aldehyde group to form a cyclic structure (hemiacetal form). This also results in the conversation of the achiral aldehyde carbon into a chiral one leading the possibility of two isomers. These two isomers differ only in the configuration of $\mathrm{C}_1$ carbon. These isomers are called anomers.
5. The two anomeric forms of glucose are called - and $\beta$-forms. This cyclic structure of glucose is similar to pyran, a cyclic compound with 5 carbon and one oxygen atom, and hence is called pyranose form.
6. The specific rotation of pure $\alpha-$ and $3-$ (D) glucose are $112^{\circ} \& 18.7^{\circ}$ respectively. However, when pure form any one of these sugars dissolved in water, slow interconversion of $-\mathrm{D}$ glucose and $\beta-\mathrm{D}$ glucose via open chain form until equilibrium is established giving constant specific rotation $+53^{\circ}$. This phenomenon is called mutarotation.
Question 3.
Explain about the structure of Fructose. (OR) Elucidate the structure of Fructose.
Answer:
1. Elemental analysis and molecular weight determination of fructose show that it has the molecular formula $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2. Fructose on reduction with $\mathrm{Hl}$ and red phosphoms gives a mixture of $\mathrm{n}$-hexane (major product) and 2 iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.

3. Fructose reacts with $\mathrm{NH}_2 \mathrm{OH}$ and $\mathrm{HCN}$. It shows the presence of carbonyl groups in the molecules of fructose.
4. Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of' five hydroxyl groups in a fructose molecule.
5. Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde ($\mathrm{CHO})$ group.
6. Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at $\mathrm{C}-2$. This confirms the presence of keto group.

7. On oxidation with nitric acid, it gives glycolic acid and tartane acids which contain smaller number of carbon atoms than in fructose. This shows that a keto group is present in $\mathrm{C}-2$. It also shows the presence
of 10 alcoholic groups at C- 1 and C- 6 . From the above reaction the structure of fructose is

This shows that a keto group is present in $\mathrm{C}-2$. It also shows the lpresence of 10 alcoholic groups at $\mathrm{C}-1$ and $\mathrm{C}-6$. From the above reaction the structure of fructose is

Question 4.
Describe about the structure, nature and properties of starch.
Answer:
1. Starch is used for energy storage in plants. Potatoes, corn, wheat and rice are the rich sources of starch.
2. It is a polymer of glucose in which glucose molecules are lined by $\alpha(1,4)$ glycosidic bonds.
3. Starch can be separated into two fractions namely, water soluble amylose and water insoluble amylo pectin. Starch contain about $20 \%$ of amylase and about $80 \%$ amylopectin.
4. Amylose is composed of unbranched chains upto $4000 \alpha$-D glucose molecules joined by $\alpha(1,4)$ glycosidic bonds.
5. At branch points, new chains of 24 to 30 glucose molecules are linked by $\alpha(1,6)$ glycosidic bonds with iodine solution amylose gives blue colour while amylo pectin gives a purple colour.

Question 5.
Explain about the structure of proteins.
Answer:
1. Proteins are polymers of amino acids. Their three dimensional structure depends mainly on the sequence of amino acids. The protein structure can be described at four hierarchal levels called primary, secondary, tertiary and quaternary structures.
Primary structure of proteins:
Proteins are polypeptide chains made up of amino acids connected through peptide bonds. The relative arrangement of the amino acids in the polypeptide chain is called the primary structure of the protein.

Secondary structure of proteins:
The amino acids in the polypeptide chain forms highly, regular shapes through the hydrogen bond between the carbonyl oxygen and the
neighbouring amine hydrogen of the main chain, $\alpha$-Helix and $\beta$ - strands or sheets are two most common substructures formed by proteins.
$\alpha-$ Helix:
In the $\alpha$-helix sub-structure, the amino acids are arranged in a righthanded helical (spiral) structure and are stabilised by the hydrogen bond. The side chains of the residues protrude outside of the helix. Each turn of an $\alpha$ - helix contains about 3.6 residues and is about $5.4 \mathrm{~A}$ long.
$\beta$ - Strands:
$\beta$ - Strands are extended peptide chain rather than coiled. The hydrogen bond occur between main chain carbonyl group one such strand and the amino group of the adjacent strand resulting in the formation of a sheet like structure. This arrangement is called $\beta$-sheets.
Tertiary structure:
The secondary structure elements ( $\alpha$-helix \& $\beta$-sheets) further folds to form a three dimensional arrangement. This tertiary structure of proteins are stabilised by the interactions between the side chains of the amino acids. These interactions include the disulphide bridges between cysteine residues, electrostatic, hydrophobic, hydrogen bonds and van der Waals interactions.
Quaternary Structure:
The oxygen transporting protein, haemoglobin contains four polypeptide chains while DNA polymerase enzyme that make copies of DNA, has ten polypeptide chains. In these proteins the individual polypeptide chains interacts with each other to form the multimeric structure which known as quaternary structure. The interactions that stabilises the tertiary structures also stabilises the quaternary structures.
Question 6.
What are the biological importance of proteins?
Answer:
Proteins are the functional units of living things play vital role in all biological processes
1. All biochemical reactions occur in the living systems are catalysed by the catalytic proteins called enzymes.
2. Proteins such as keratin, collagen acts as structural back bones.
3. Proteins are used for transporting molecules (Haemogiobin), organelles (Kinesins) in the cell and control the movement of molecules in and out of the cells (Transporters).
4. Antibodies help the body to fight various diseases.
5. Proteins are used as messengers to coordinate many functions. Insulin \& glucagon controls the glucose level in the blood.
6. Proteins act as receptors that detect presence of certain signal molecules and activate the proper response.
7. Proteins are also used to store metals such as iron (Femtin) etc.
Question 7.
Write the chemical name, source and deficient disease of the following
1. Vitamin $A$
2. Vitamin $B_1$
3. Vitamin $B_2$
4. Vitamin $B_3$
5. Vitamin $\mathrm{B}_5$
Answer:

Question 8.
Write the chemical name, source and deficient disease of the following
1. Vitamin $\mathrm{B}_6$
2. Vitamin $B_7$
3. Vitamin $\mathrm{B}_9$
4. Vitamin $B_{12}$
5. Vitamin C
Answer:

Question 9.
Explain about the composition and structure of nucleic acids.
Answer:
1. Nucleic acids are biopolymers of nucleotides. Controlled hydrolysis of DNA and RNA yields three components namely a nitrogeneous base, a pentose sugar and phosphate group.
2. Nitrogen base.
(a) These are nitrogen containing organic compounds which are derivatives of two parent compounds, pyrimidine and purine.
(b) Both DNA and RNA have two major purine bases, adenine (A) and guanine (G). In both DNA and RNA, one of the pyrimidines is cytosinc (C), but the second pyrimidine is thymine (T) in DNA and uracil (U) in RNA.
3. Pentose Sugar.
Nucleic acids have two types ofpentoses. The recurring deoxyribonucleotide units of DNA contain 2' dcoxy - D - ribose and the ribonucleotide units of RNA contain D - ribose. In nucleotides, both types of pentoses are in their - furanose form.
4. Phosphate group.
Phosphoric acid forms phosphor diester bond between nucleotides. Based on the number of phosphate group present in the nucleotides, they are classified mono nucleotide, dinucleotide and trinucleotide.
5. The molecule without the phosphate group is called a nucleoside. A nucleotide is derived from a nucleoside by the addition of a molecule of phosphoric acid.
6. Sugar + Base $\rightarrow$ Nucleoside
Nucleoside + Phosphate $\rightarrow$ Nucleotidc
Nucleotide $\rightarrow$ Polynucleotide (Nucleic Acid)
Question 10.
Describe about the double strand helix structure of DNA.
Answer:
1. Watson and Crick postulated a 3 - dimensional model of RNA structure which consisted of two antiparallel helical DNA chains wound around the same axis to form a right handed double helix.
2. Th e hydrophilic backbones of alternating deoxyribose and phosphate groups are on the outside of the double helix, facing the surrounding water. Th e purine and pyrimidine bases of both strands are stacked inside the double helix,
with their hydrophobic and ring structures very close together and perpendicular to the long axis, thereby reducing the repulsions between the charged phosphate groups. The offset pairing of the two strands creates a major groove and minor groove on the surface of the duplex.

3. The model revealed that there are 10.5 pairs $\left(36 \mathrm{~A}^{\circ}\right)$ per turn of the helix and $3.4 \mathrm{~A}^{\circ}$ between the stacked bases. They also found that each base is hydrogen bonded to a base in opposite strand to form a planar base pair.
4. Two hydrogen bonds are formed between adenine and thymine and three hydrogen bonds are formed between guanine and cytosine. Other pairing tends to destablize the double helical structure. This specific association of the two chains of the double helix is known as complementary base pairing.
5. The DNA double or duplex is held together by two forces.
- Hydrogen bonding between complementary base pairs.
- Base - stacking interactions.
The complementary between the DNA strands is attributable to the hydrogen bonding between base pairs but the base stacking interactions are largely non-specific, make the major contribution to the stability of the double helix.

Question 11.
Explain about the types of RNA molecules.
Answer:
1. Ribonucleic acids are similar to DNA. Cells contain upto eight times high quantity of RNA than DNA. RNA is found in large amount in the cytoplasm and a lesser amount in the nucleus.
2. RNA molecules are classified according to their structure and function into three major types.
1. Ribosomal RNA (r - RNA)
2. Messenger RNA (m-RNA)
3. Transfer RNA ( $\mathrm{t}-\mathrm{RNA})$
3. $\mathrm{r}-\mathrm{RNA}$ :
$\mathrm{r}$ - RNA is mainly found in cytoplasm and in ribosomes, which contain $60 \%$ RNA and $40 \%$ protein. Ribosomes are the sites at which protein synthesis takes place.
4. $\mathrm{t}-\mathrm{RNA}$ :
$\mathrm{t}$ - RNA molecules have lowest molecular weight of all nucleic acids. They consist of $73-94$ nucleotides in a single chain. The function of $t-$ RNA is to carry amino acids to the sites of protein synthesis on ribosomes.
5. $\mathrm{m}-\mathrm{RNA}$ :
$\mathrm{m}$ - RNA is present in small quantity and very short lived. They are single stranded and their synthesis take place on DNA. The synthesis m-RNA from DNA strand is called transcription. $\mathrm{m}-\mathrm{RNA}$ carries genetic information from DNA to the ribosomes for protein synthesis.
Question 12.
Explain about DNA finger printing process.
Answer:
1. DNA finger printing is one of the most accurate methods for placing an individual at the scene of a crime has been a finger print.
2. The DNA finger print is unique for every person and can be extracted from traces of sample from blood, saliva, hair etc. By using this method, we can detect the individual specific variation in human DNA.
3. In this method, the extracted DNA is cut at specific points of varying lengths in the formation of DNA fragments of varying lengths which were analysed by technique called gel electrophoresis. This method separates the fragments based on their size. The gel containing the DNA fragments are then transferred to a nylon sheet using a technique called blotting. Then the fragments will undergo autoradiography in which they were exposed to DNA probes.

4. A piece of X-ray film was then exposed to the fragments, and a dark mark was produced at any point where a radioactive probe had become attached. The resultant pattern of marks could then be compared with other samples.
5. DNA finger printing is based on slight sequence differences between individuals. These methods are providing decisive in court cases world wide.
Common Errors
1. Carbohydrates \& Enzyme names are different.
2. Peptide bond and amide bond are different.
3. Vitamins: Solubility based classification
4. Main source of vitamin
Rectifications
1. Carbohydrates name should end as - ose.
Eg. Diatose, maltose
Enzyme name should end as - ase.
Eg. Diastase, maltase

2.

3. Vitamin A, D, E, K-fat soluble. Vitamin C, B complex are water soluble.
4. Milk:
Vitamin A $\mathrm{A}_1, \mathrm{~B}_2, \mathrm{~B}_6, \mathrm{~B}_7, \mathrm{D}$ (6 Vitamins)
Green Vegetables:
$\mathrm{B}_1, \mathrm{~B}_3, \mathrm{~B}_7, \mathrm{~K}$ (4 vitamins)
Liver:
$
\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3, \mathrm{~B}_7
$

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SaraNexGen