- Class 12 Chapter 14 Biomolecules Notes
- NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules
- CBSE Class 12 Chemistry Important Questions Chapter 14 – Biomolecules
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Class 12 Chapter 14 Biomolecules Notes
The complex organic substances like carbohydrates, proteins etc which combine in a specific manner to produce living systems and maintain it are called biomolecules. The branch of chemistry that deals with the study of chemical reactions that occur in living organisms is called biomolecules.
- They are polyhydroxy-aldehydes or ketones or substances which give these substances on hydrolysis and contain at least one chiral atom.
- They have general formula of C x (H2O)
- Rhamnose, deoxyribose, rhamnohexose do not obey this formula but are carbohydrates.
Types of carbohydrates
- These are simplest carbohydrate which can’t be hydrolyzed further into smaller compounds.
- They are called as aldose or ketose depending upon whether they have aldehyde or ketone group.
- Depending upon the number of carbon atoms present they are called as triose, tetrose etc.
- All monosaccharide’s are sweet smelling crystalline, water soluble and are also capable of diffusing through cell membranes.
For example: Glucose is aldohexose while fructose is a ketohexose. Both of them have 6 carbon atoms. The simplest monosaccharide is a triose (n=3).
Example: Glyceraldehyde and Dihydroxyacetone. They have one or more asymmetric carbon and are optically active.
Their structures are:
All naturally occurying monosaccharides belong to D—series that is OH group at their penultimate C-atom.
Open chain structures:
- D-glucose and D-mannose, differ only in configuration at C-2 and are known as epimers.
- Similarly D-glucose and D-galactose, differ in configuration around C-4 atom and are also known as epimers.
- Thus a pair of diastereomers, differing only in configuration around C-2 or any other chiral carbon except C-1 is called epimers.
Close chain structure
- All the pentose’s and hexoses exist in cyclic hemiacetal structure.
- In free state, they have generally six membered cyclic structure known as pyranose form and in combined state, some of them have 5 membered cyclic structure called as furanose .
Due to cyclic hemiacetal or hemiketal structure all the pentoses and hexoses exist in two stereoisomeric forms
- Alpha form
- Beta form
- Both alpha and beta form are Anomers.
- Their structure is given below :
These carbohydrates on hydrolysis give 2 to 9 molecules of monosaccharides.
They are further of few types:
- Disaccharides (C12H22o11): On hydrolysis, they give 2 molecules of monosaccrides which are held together by Glycosidic linkage
Example: sucrose etc
- Trisaccharides (C18H32o16): On hydrolysis, they form three molecules of monosaccharides.
- Tetra-saccharides: (C24H42o21): Such as stachyose which gives four monosaccrides on hydrolysis.
These are the carbohydrates which on hydrolysis, yield more than nine monosaccharides molecules.
Example: Starch etc
- Glucose exist in two forms : i.e. alpha –D glucose with specific rotation of 112degree and beta D-glucose with specific rotation of +19 degree.
- However, when either of these two forms is dissolved in water and allowed to stand, it gets converted into same equilibrium mixture of both the alpha and beta forms with a small amount of open chain form having specific rotation of 52.7 degree.
- As a result of this, equilibrium the specific rotation of freshly prepared solution of alpha glucose decreases from +112 degree to 52.7 degree while that for beta glucose increases from +19 to 52.7 degrees.
- The phenomenon of change of change in specific rotation of optically active compounds with time to an equilibrium value is known as Mutarotation.
- The alpha D (+) glucose and beta (+) glucose, differ in configuration at C-1 carbon and the compounds differing in configuration at C-1 are called Anomers.
Fructose: It is represented by six membered ring as shown:
Beta –D+(fructose) furanose structure
Fructose is assigned furanose structure.
Lactose: It is made up of molecule and a molecule of galactose. The units are linked together.
Maltose: It has the structure as shown below
- On treatment, with acid or with enzyme maltose gets hydrolysed to 2 molecules.
- That is alpha D-glucose.
- Since one of the glucose units exist in hemiacetal form it is a reducing sugar.
Sucrose: It has a structure shown below
- On hydrolysis, with dilute mineral acid or enzyme invertase sucrose gives glucose and fructose in equal amounts.
- Sucrose and glucose are dextrorotatory while fructose is laevorotatory and has higher value of specific rotation.
- Thus the process is accompanied by inversion of optical activity. The mixture formed is invert sugar. That is sucrose àglucose + fructose
(water and invertase )
Sweetness of sugars
- All the monosaccharide and disaccharides are sweet in taste and hence also known as sugars.
- Sucrose is given sweetness value of 100. The sweetness of other sugars is compared with the value of sucrose.
- The sweetness of fructose -173, invert sugar 130, sucrose 100, glucose 74,glactose 32,maltose 32 and that of lactose is 16.
- All the monosaccharide and disaccharides are reducing agents due to hemiacetal and hemiketal forms which easily change in to aldehydic form in the alkaline medium.
- Although fructose doesn’t contain any aldehydic group yet it gives Tollen’s reagent test and Fehling’s solution test because under the basic conditions of reagent the fructose gets converted into the mixture of glucose and mannose both of which contains aldehydic group.
- This is called Lobry De Bruyn Van Eikensten rearrangement.
- The alpha and beta glucose reacts with one molecule of ethanol to form the corresponding methyl glucosides.
- When glucose is treated with methanol in presence of HCl the hemiacetal form changes to acetal form.
Starch: It serves as a storage polysaccharide in plants .It consist of two components of alpha glucose.
- It is a linear polymer of glucose and is soluble in water.
- Its percentage in starch is about 10-20 %.
- These are linked together by alpha linkage involving C-1 of glucose unit to C-4 of the other.
- It is branched chain polymer of alpha glucose and is insoluble in water.
- There are about 1000 units of glucose.
(c ) Cellulose
- It is found in all plants
- It constitutes 50% of total organic matter in the living beings.
- Cotton is pure cellulose.
- Cellulose is linear polymer of beta D-glucose.
- The chains are arranged to form bundles and linked together by hydrogen bonds between glucose molecules of adjacent organic solvents.
- When it is treated with concentrated sulphuric acid in cold it slowly passes into solution.
- This solution when diluted with water gives starch like substance amyloid which is known as parchment paper.
- On boiling with water it is hydrolyzed into D-glucose.
- Cellulose gives many useful products when treated with different chemicals like rayon, gum, cotton etc.
- Cellulose is directly used in making cloth and paper.
- In glycogen there are about 25 glucose units. Its structure is similar to amylopectin and is a condensation polymer of alpha glucose.
- Glycogen in short term food storage in animals.
Tests for carbohydrates
- For this Molisch test is performed.
- In it Molisch reagent is used which is 10%alcoholic solution of alpha naphthol and it is added to aqueous solution of carbohydrate followed by concentrated sulphuric along the sides of tube.
- As a result a violet ring is formed at the junction of two layers.
Introduction of proteins
- They are high weight polymers.
- They are polyamides that contain C, H, N, O and S.
- Proteins are derived from alpha amino carboxylic acid monomers.
- A simple protein may contain hundred even thousands of amino acid units.
- In living organisms twenty alpha amino acids occur which combine to form different protein molecules.
- A simple amino acid can be represented as R-CH(NH)2COOH (carboxy group and amino group is present in it).
Alpha amino acids
- The acid in which NH2 group is present at carbon atom adjacent to the COOH group are called alpha amino acids.
- Alpha amino acids are the building blocks of proteins the alpha carbon of all amino acids is chiral, hence all amino acids exhibit stereoisomerism that is existence of D and L types of structures.
- All the naturally occurring amino acids, belong to L form category.
- In L amino acid NH2 group lies left to the chiral carbon as shown :
H2N-CH2R (C0OH) L –amino acid
- Due to transfer of proton from carboxy to amino group, alpha amino acid exists as dipolar ion or called as Zwitter ion.
- Due to the zwitter ion structure, alpha amino acids are high meting crystalline solids and moderately soluble in water.
- In acidic medium caboxylate ion group act as a base and accepts a proton. Thus alpha amino acids exist as cations (I) under the influence of electric field.
(zwitter ion )
- In alkaline medium, NH3+groups act as an acid and thus loses a proton.
- Due to this, alpha amino acids exist as anions (III) and thus migrate towards the anode in the electric field.
- However, at some intermediate value of ph, the concentration of cationic form (I) and the anionic form (III) become equal.
- Hence, there is no migration in electric field. This pH is known as isoelectric point.
Types of amino acids
The amino acids are of two types:
- Non essential
Essential amino acids
- The amino acids that can’t be made in our body and must be supplied from outside.
- The lack of these amino acids in diet can cause lot of diseases like kwashiorkor (the disease in which water balance in body is disturbed).
- The essential amino acids are 10 in number out of all.
Non essential amino acids
- They are those that can be synthesized in our body.
- Out of total 20 amino acids the 10 can be synthesized in the body.
List of different types of amino acids:
- The interaction between amino group and carboxyl of amino group give compounds called peptides.
- The amide group –CONH in each such compound is called peptide linkage.
- Depending upon the number of amino acid, residues per molecule they are called as dipeptide, tripeptide etc.
- Peptides of molecular weight up to 10,000 are known as polypeptides and those with higher than 10,000 are called as proteins.
- Each polypeptide chain has a free amino group at one end and a free carboxyl group at other end.
- They are collectively called as end groups.
- The amino acid unit having –NH2group is called N-terminal end and the amino acid unit having free –COOH group is called C-terminal end.
- In this NH-CH-CO is the repeating unit in polypeptide.
Structure of proteins
Proteins have three structures:
- Primary structure
- Secondary structure
- Tertiary structure
Primary structure: The sequence in which amino acids are arranged in protein is called primary structure. The sequence determines the function of a protein.
Secondary structure: The fixed configuration of polypeptide skeleton is known as secondary structure.
- There are two types of secondary structure :
- Alpha helix
- Beta pleated sheet
Alpha helix: If the size of group R is large, intermolecular hydrogen bonds are formed between CO of one amino acid residue and NH of the fourth amino acid residue in polypeptide chain which gives right handed alpha helix structure to the protein molecule.
Example: Alpha keratin in hair etc it is also elastic.
Beta pleated sheet: If the size of group R is small, intermolecular hydrogen bonds are formed between CO of one polypeptide chain with NH of the other chain. Thus the chains are bonded together forming a sheet which can slide over each other to form a three dimensional structure called beta pleated sheet.
Tertiary structure: It implies the three dimensional structure of proteins.
There are two types tertiary structure:
- Fibrous and globular.
- Proteins contain one or more polypeptide chains.
- A protein having one polypeptide chain is known as monomeric while that having more than one polypeptide chains is called oligomeric.
- The constituent peptide chains of an oligomeric protein are called protomers which are held together by weak forces.
- Native state: At normal pH and temperature each protein takes a shape which is energetically most stable.
- In amino acid the shape is specific and is known as native state.
- Globular proteins are tightly folded and give rise to spherical form.
Forces that stabilize protein structures
The forces that are present are as follows:
- Hydrogen bonding
- Anionic bonding
- Hydrophobic bonding
- Covalent bonding
Hydrogen bonding: These forces operate between a partially positive hydrogen and partially negative atom like O or N on the same or on another molecule.
Anionic bonding: A bonding between cation and anion of side chains resulting in side linkage.
Hydrophobic bonding: Some side chains in same amino acid are hydrophobic. In aqueous solutions proteins fold in such a way that these chains get clustered inside the folds .The polar side chains which are hydrophilic lie on the outside or surface of proteins.
Covalent bonding: The bond occurs between S atoms of two residues between two adjacent chains.
- Insulin which contains 51 amino acids is arranged in two polypeptide chains containing 21 and 30 amino acid residues connected by S-S cross links.
Denaturation of proteins
- The globular proteins, which are soluble in water on heating or on treatment of mineral acids or bases undergo coagulation or precipitation to give fibrous proteins which are insoluble in water.
- After coagulation, proteins lose their biological activity this is called denaturation.
- It can be reversible or irreversible.
- Coagulation of lactalbumin to form cheese and coagulation of albumins are examples of denaturation.
Classification of proteins
On the basis of composition, proteins are of following types :
- Simple proteins
- Conjugated proteins
- Derived proteins
- Fibrous proteins
- Globular proteins
Simple proteins: On hydrolysis they give only amino acids.
Example: Globulins and albumin
Conjugated proteins: They contain non protein group attached to the protein part. These non protein groups are called prosthetic groups.
Example: Nucleo-protein contains nucleic acid, phosphor-protein contains phosphoric acid contains phosphoric acid, glycol-proteins contains carbohydrates etc.
Derived proteins: These are the degradation products obtained by the hydrolysis of simple and conjugated proteins.
Example: Peptides, peptones etc
Fibrous proteins: They are long and thread like and tend to lie side by side to form fibers .In some cases, they are held together by hydrogen bonds at many points .these proteins serves as a chief structural material of animal tissues .
Globular proteins: The molecules of these proteins are folded into compact units and form spheroid shapes .Intermolecular forces are weak. These proteins are soluble in water or aqueous solution of acids, bases or salts .Globular proteins make up all enzymes, hormones ,fibrinogen etc.
Role of proteins
- They act as enzymes and transport agents.
- They are structural materials for nails, hair etc.
- Antibodies formed in body are globular proteins.
- They are metabolic regulators like insulin etc.
Hydrolysis of proteins
- Proteins are hydrolyses when boiled with acids or alkalis or when treated with enzymes .the hydrolysis takes place as:
Proteins àproteasesàpeptones àpolypeptides àsimple peptides àamino acids
- Every protein has an isoelectric point at which their ionization is minimum. Proteins have charged groups i.e. NH3 and COO– at the ends of peptide chain.
- They are amphoteric in nature.
- Protein accepts a proton in strong basic solution.
- The pH at which the protein molecule has no net charge is known as isoelectric point.
They are biological catalysts .Chemically all enzymes are globular proteins
Some important enzymes with their function:
- Lactase : convert lactose àglucose + glactose
- Invertase : convert sucrose à glucose and fructose
- Maltase : convert maltose à2glucose
- Emulsion: convert cellulose àn glucose
- Urease : convert urea àcarbon dioxide and water
- And many more.
Some industrial enzymes invertase and zymase are present in yeast .Enzyme diastase is used as converting starch to maltose , a sugar .the names of all enzymes end with the suffix “ase”.
Properties of enzymes
- Enzymes are required only in small amounts.
- They are highly specific.
- Enzymes are efficient catalyst: they speed up reaction.
- They work at optimum pH, at optimum temperature.
- Their mechanism is controlled by various mechanisms and stopped by various organic and inorganic compounds.
- The action of enzymes follows lock and key mechanism .however enzyme action is inhibited by certain organic molecules called inhibitors.
- They are the chemical substances which are needed in small amounts for the growth of human beings.
- They can’t be synthesized in our body therefore need to taken from outsource.
- Their deficiency can cause one or other type of disease.
The following vitamins with their function and deficiency disease are listed below:
The diseases caused by them with their symptoms:
- Nucleic acids are the polymers in which nucleotides are monomers. These are biomolecules present in nuclei of all living cells in the form of nucleoproteins .They are also called as polynucleotides .
They help in the role of transmission of hereditary characters and synthesis of proteins.
Each nucleotide consists of 3 parts:
- A pentose sugar
- A nitrogenous base
- A phosphate group
- The nitrogenous base and a pentose sugar are called as nucleoside.
Nitrogenous bases are of two types: Purines and Pyrimidines
- Purines: adenine and guanine
- Pyrimidines: cytosine , thiamine and uracil
Please note that Purines and Pyrimidines are linked together by hydrogen bonds
- Adenine always bond with thiamine by double bond or vice versa.
- Cytosine always pairs with guanine by triple bond or vice versa.
Types of nucleic acids
- Deoxyribonucleic acid (DNA)
- Ribonucleic acid (RNA)
- It occurs in nucleus of cell. It has double stranded helical structure
- Deoxyribose sugar
- Nitrogenous bases :
- Purines (adenine and guanine ), Pyrimidines (thiamine and cytosine )
- A phosphate group
- It can undergo replication
- It helps in transfer of genetic information from parents to offspring
- It occurs in cytoplasm of cell
It consist of:
- Ribose sugar
- Nitrogenous base
- Purines: adenine and guanine
- Pyrimidines: cytosine and uracil
- A phosphate group
- It has a single strand helical structure
- It doesn’t undergo replication
- It controls synthesis of proteins
The structure of deoxyribose and ribose sugar is given:
Structure of nucleic acids
- Primary structure
- The nucleic acids are formed by the condensation of thousands of molecules of nucleotides.
- On hydrolysis the nucleotides produces phosphoric acid and nucleoside .it means nucleosides on hydrolyses form Purine and Pyrimidines base and sugar moiety.
- A nucleic acid– NH3ànucleotides –aq NH3à nucleosides + phosphoric acid –dilute HCL–àPurines + Pyrimidines + sugar.
- Nucleotides are building blocks of nucleic acids.
- These nucleotides are linked together with one another in a particular sequence, phosphate groups forming bridges between C-5 of the sugar residue of the one nucleoside and C-3 of the sugar residue of the other nucleoside.
- The manner in which the sugar, phosphate and bases are linked with one another in nucleic acids is known as primary structure of nucleic acids.
- Secondary structure:
- Watson and Crick explained the double helix structure of DNA. The nucleotides in each strand are connected by phosphate ester bond and bases of one strand by hydrogen bonds.
- Adenine pairs with thiamine through two double hydrogen bonds whereas cytosine pairs with guanine by triple hydrogen bonds.
- The two strands of DNA are complementary to each other that is if one side there is Purine then on other side at same position Pyrimidine is present. For example if base sequence on strand is ACTCGCCA, then on the other strand the sequence will be complementary that is: TGAGCGGT
- The primary and secondary structure is shown below:
Watson and Crick model of DNA
Functions of nucleic acids
- Replication: The genetic information of cell is contained in the sequence of bases A, T, C and G in DNA molecule .In the division of cell, DNA molecules replicate and makes exact copies of themselves so that each daughter cell will have DNA identical to that of the parent cell.
- Protein synthesis: The specific information coded on DNA has to be translated and expressed in the form of synthesis of specific proteins which performs various functions in the cell. This synthesis is done in two steps:
- Transcription and translation.
- Gene and genetic code: Each segment of DNA molecule that codes for specific protein or a polypeptide is known as The relationship between nucleotides triplets and the amino acids are called the genetic code .This is gene and genetic code.
- Mutation: It is a chemical change in DNA molecule, which leads to the synthesis of proteins with a changed amino acid sequence.
- These changes are caused by radiation, viruses or chemical agents.
- The majority of changes in DNA are replicated by special enzymes in the cell, but if there is failure to repair by the enzymes then it can cause mutation.
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules
NCERT INTEXT QUESTIONS ( Biomolecules )
14.1. Glucose or sucrose are soluble in water but cyclohexane and benzene (simple six membred ring compounds) are insoluble in water Explain.
Ans: The .solubility of a solute in a given solvent follows the rule ‘ Like dissolves like’.Glucose contains five and sucrose contains eight -OH groups. These -OH groups form H-bonds with water. As a result of this extensive intermoleeular H-bonding, glucose and sucrose are soluble in water.On the other hand, benzene and cyclohexane do not contain -OH bonds and hence do not form H-bonds with water. Moreover, they are non-polar molecules and hence do not dissolve in polar water molecules.
14.2. What are the expected products of hydrolysis of lactose?
Ans: Lactose being a disaccharide gives two molecules of monosaccharides Le. one molecule each of D-(+) – glucose and D-(+)-galactbse.
14.3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Ans: The cyclic hemiacetal form of glucose contains an -OH group at C-l which gets hydrolysed in aqueous solution to produce open chain aldehydic form which then reacts with NH2OH -to form corresponding oxime. Thus, glucose contains an aldehydic group. However, when glucose is reacted with acetic anhydride, the -OH group at C-l along with the other -OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate.
Since the penta acetate of1 glucose does not contain a free -OH group at C-l, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not react with NH2OH to form glucose oxime. The reactions are shown as:
14.4. The melting points and solubility in water of a-amino acids are generally higher than those of corresponding haloacids. Explain.
Ans: a-amino acids as we all know, are dipolar in nature (N+H3-CHR-COO– ) and have strong dipolar interactions. As a result, these are high melting solids. These are also involved in intermolecular hydrogen bonding with the molecules of water and are therefore, water soluble. On the contrary, the haloacids RCH(X)COOH are not dipolar like a-amino acids. Moreover, only the carboxyl group of haloacids are involved in hydrogen bonding with the molecules of water and not the halogen atoms. These have therefore, comparatively less melting points and are also soluble in water to smaller extent.
14.5. Where does the water present in the egg go after boiling the egg?
Ans: When egg is boiled, proteins first undergo denaturation and then coagulation and the water present in the egg gets absorbed in coagulated protein, probably through H- bonding
14.6. Why cannot Vitamin C be stored in our body?
Ans: Vitamin C cannot be stored in the body because it is water soluble and is, therefore, easily excreted in urine.
14.7. Which products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Ans: Upon hydrolysis, nucleotide from DNA would form 2-deoxyribose and phosphoric acid along-with thymine.
14.8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Ans: A DNA molecule has two strands in which the four complementary bases pair each other, i.e., cytosine (C) always pair with guanine (G) while thymine (T) always pairs with adenine (A). Thus, when a DNA molecule is hydrolysed, the molar amounts of cytosine is always equal to that of guanine and that of adenine is always equal to thymine.In RNA, there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base pairing principle, i.e. A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.
NCERT EXRECISES ( Biomolecules )
14.1. What are monosaccharides ?
Ans: Monosaccharides are carbohydrates Which cannot be hydrolysed to smaller molecules.Their general formula is (CH2O)n Where n=3-7 These are of two types: Those which contain an aldehyde group (-CHO) are called aldoses and those which contain a keto (C=O) group are called ketoses.
They are further classified as trioses , tetroses ,pentoses , hexoses and heptoses according as they contain 3,4,5,6, and 7 carbon atoms respectively.For example.
14.2. What are reducing sugars?
Ans: Reducing sugars are those which can act as reducing agents. They contain in them a reducing group which may be aldehydic (-CHO) or ketonic (>C=0) group. The characteristic reactions of reducing sugars are with Tollen’s reagent and Fehling solution. Non-reducing sugars donot give these reactions. For example, glucose, fructose, lactose etc. are reducing sugars. Sucrose is regarded as a non-reducing sugar because both glucose and fructose are linked through their aldehydic and ketonic groups by glycosidic linkage. Since these groups are not free, sucrose is a non-reducing sugar.
14.3. Write two main functions of carbohydrates in plants.
Ans: Two major functions of carbohydrates in plants are following
(a)Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls.
(b)Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and act as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.
14.4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Ans: Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose. Disaccharides: Maltose and lactose.
14.5. What do you understand by the term glycosidic linkage?
Ans: The ethereal or oxide linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of disaccharide is called the glycosidic linkage. The glycosidic linkage in maltpse molecule is shown below:
14.6. What is glycogen? How is it different from starch?
Ans: The carbohydrates are stored in animal body as glycogen. It is also called animal starch and its structure is similar toamylopectin which means that it is a branched chain polymer of α-D-glucose units in which the chain is formed by C1 – C4 glycosidic linkage whereas branching occurs by the formation of C1– C6 glycosidic linkage. One main difference between glycogen and amylopectin is the length of the chain. In amylopectin, the chain consists of 20 – 25 α – D – glucose molecules whereas in glycogen, there are 10 -14 molecules of α – D – glucose present. Glycogen is more branched than amylopectin. It is present mainly in liver, muscles and also in brain. Glycogen gets converted into glucose when the body needs it with the help of certain enzymes present in the body. Glycogen has also been found to be present in yeast and fungi.
Starch is a major source of carbohydrates which are very much essential to the human body since they supply energy to the body. It occurs as granules mainly in seeds, fruits, tubers and also in the roots of the plants. The chief commercial sources of starch are wheat, maize, rice, potatoes etc.
14.7. What are the hydrolysis products of (i) sucrose, and (ii) lactose?
Ans: Both sucrose and lactose are disaccharides. Sucrose on hydrolysis gives one molecule each of glucose and fructose but lactose on hydrolysis gives one molecule each of glucose and galactose.
14.8. What is the basic structural difference between starch and cellulose?
Ans: Starch consists of amylose and amylopectin. Amylose is a linear polymer of α-D-glucose while cellulose is a linear polymer of β -D- glucose. In amylose, C -1 of one glucose unit is connected to C – 4 of the other through α-glycosidic linkage. However in cellulose, C – 1 of one glucose unit is connected to C-4 of the other through β – glycosidic linkage. Amylopectin on the other hand has highly branched structure.
14.9. What happens when D-glucose is treated with . the following reagents.
(ii) Bromine water
14.10. Enumerate the reactions of D-glucose which cannot be explained with open chain structure. (C.B.S.E. Delhi 2008, C.B.S.E. Sample Paper 2011)
Open chain structure of D-glucose contains a free aldehydic group (- CHO). However, it does not give the following reactions:
- D(+) glucose does not react with 2, 4 D.N.P.
- D(+) glucose does not react with NaHSO3.
- D(+) glucose does not restore the pink colour to Schiff’s reagent.
- Penia acetyl glucose formed by carrying acetylation with acetic anhydride does not react with hydroxyl amine
(NH2OH) which is the characteristic reaction of all aldehydes.
- D( +) glucose is found to exist in two different crystalline forms which are named as α and β. Both these forms have actually been isolated. For example, α form with m.p. 419 K is obtained by the crystallisation of the saturated solution of glucose prepared at 303 K. Similarly, β-form with m.p. 423 K is isolated by carrying out the crystallisation of the saturated solution of glucose prepared at 371 K. Apart from that the a-form has a specific rotation (α) equal to + 112° while the β- form has specific rotation (α) equal to + 19°.
In the light of the limitations stated above, Tollen stated that an open chain structure for D(+) glucose is probably not practicable. He proposed a cyclic structure which is a hemiacetal structure. In this structure, the aldehydic (CHO) group
is involved in the form of a ring with the -OH group attached to C5 carbon. It is a six membered ring, often called ô-
oxide ring. The ring structure accounts for the two isomeric forms a and shown below.
14.11. What are essential and non-essential amino acids? Give two examples of each type.
Ans: α-Amino acids which are needed for good health and proper growth of human beings but are not synthesized by the human body are called- essential amino acids. For example, valine, leucine, phenylalanine, etc. On the other hand, α-amino acids which are needed for health and growth of human beings and are synthesized by the human body are called non-essential amino acids. For example, glycine, alanine, aspartic acid etc.
14.12. Define the following as related to proteins:
(i) Peptide linkage
(ii) Primary structure
Ans: (i) Peptide bond: Proteins are condensation polymers of α-amino acids in which the same or different α-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between – COOH group of one α-amino acid and -NH-, group of the other α-amino acid by lo;ss of a molecule of water. For example,
(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each . polypeptide chain has a large number of α-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of α-amino acids creates a different protein.
(iii) Denaturation: Each protein in the biological system has a unique three-dimensional structure and has specific biologicalactivity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in pH, etc., hydrogen bonds gets broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity, is called denaturation. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact.
The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.
14.13. What are the common types of secondary structure of proteins?
Secondary structure of protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :
- α-helix structure
- β-pleated sheet structure.
Secondary Structure of Proteins:
The long, flexible peptide chains of proteins are folded into the relatively rigid regular conformations called the
secondary structure. It refers to the conformation which the polvpeptide chains assume as a result of hydrogen bonding
between the > C= O and > N-H groups of different peptide bonds.
The type of secondary structure a protein will acquire, in general depends upon the size of the R-group. If the size of the
R-groups are quite large, the protein will acquire ct-helix structure. If on the other hand, the size of the R-groups are relatively
smaller, the protein will acquire a β – flat sheet structure.
(a) α-Helix structure: If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group
of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide
chain coils up into a spiral structure called right handed ct- helix structure. This type of structure is adopted by most of the
fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be
stretched. During this process, the weak hydrogen bonds causing the a – helix are broken. This tends to increase the length of
the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.
(b) β—Flat sheet or β—Pleated sheet structure: If R-groups are relatively small, the peptide chains lie side by side in a zig
zag manner with alternate R-groups on the same side situated at fixed distances apart. The two such neighbouring chains are held together by intermolecular hydrogen bonds. A number of such chains can be inter-bonded and this results in the formation of a flat sheet structure These chains may contract or bend a little in order to accommodate moderate sized R-groups. As a result, the sheet bends into parallel folds to form pleated sheet structure known as β – pleated sheet structure. These sheets are then stacked one above the other like the pages of a book to form a three dimensional structure. The protein fibrion in silk fibre has a β – pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its β – sheet structure. For example, silk is non-elastic since stretching leads to pulling the peptide covalent bonds. On the other hand, it can be bent easily like a stack of pages because during this process, the sheets slide over each other.
14.14. What types of bonding helps in stabilising the α-helix structure of proteins?
Ans: α-helix structure of proteins is stabilised through hydrogen bonding. (a) α -Helix structure. If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group of one amino acid unit and the > N- H group of the fourth amino acid unit within the same chain. As such the polypeptide chain coils up into a spiral structure called right handed a—helix structure. This type of structure is adopted by most of the fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be stretched. During this process, the weak hydrogen bonds causing the α-helix are broken. This tends to increase the length of the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.
14.15: Differentiate between globular and fibrous proteins.
Ans. (i) Fibrous proteins: These proteins consist of linear thread like molecules which tend to lie side by side (parallel) to form fibres. The polypeptide chains in them are held together usually at many points by hydrogen bonds and some disulphide bonds. As a result,intermolecular forces of attraction are very’ strong and hence fibrous proteins are insoluble in water. Further, these proteins are stable to moderate changes in temperature and pH. Fibrous proteins serve as the chief structural material of animal tissues.For example, keratin in skin, hair, nails and wool, collagen in tendons, fibrosis in silk and myosin in muscles.
(ii) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule an almost spheroidal shape. The folding takes place in such a manner that hydrophobic (non-polar) parts are pushed inwards and hydrophilic (polar) parts are pushed outwards. As a result, water molecules interact strongly with the polar groups and hence globular protein are water soluble. As compared to fibrous proteins, these are very sensitive to small changes of temperature and pH. This class of proteins include all enzymes, many hormones such as insulin from pancreas, thyroglobulin from thyroid gland, etc.
14.16. How do you explain the amphoteric behaviour of amino acids?
Ans: Amino acids contain an acidic (carboxyl group) and basic (amino group) group in the same molecule. In aqueous solution, they neutralize each other. The carboxyl group loses a proton while the amino group accepts it. As a result, a dipolar or zwitter ion is formed.
In zwitter ionjc form, a-amino acid show amphoteric behaviour as they react with both acids and bases.
14.17. What are enzymes?
Ans: Enzymes are biological catalyst. Each biological reaction requires a different enzyme. Thus, as compared to conventional catalyst enzymes are very specific and efficient in their action. Each type of enzyme has its own specific optimum conditions of concentration, pH and temperature at which it works best.
14.18. What is the effect of denaturation on the structure of proteins?
Ans: Denaturation of proteins is done either by change in temperature (upon heating) or by bringing a change in the pH of the medium. As a result, the hydrogen bonding is disturbed and the proteins lose their biological activity i.e., their nature changes. During the denaturation, both the tertiary and secondary structures of proteins are destroyed while the primary structures remain intact.
14.19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Ans: Vitamins are classified into two groups depending upon their solubility in water or fat: (i) Water soluble vitamins: These include vitamin B-complex (B1, B2, B5, i.e., nicotinic acid,B6, B12, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C.
(ii) Fat soluble vitamins: These include vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. Vitamin K is responsible for coagulation of blood.
14.20. Why are vitamin A and vitamin C essential to us? Give their important sources.
Ans: Vitamin A is essential for us because its deficiency causes xerophthalmia (hardening of cornea of eye) and night blindness.
Sources: Fish liver oil, carrots, butter, milk, etc. Vitamin C is essential for us because its deficiency causes scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Sources: Citrous fruits, amla, green leafy vegetables etc.
14.21. What are nucleic acids ? Mention their two important functions.
Ans: Nucleic acids are biomolecules which are found in the nuclei of all living cell in form of nucleoproteins or chromosomes (proteins contains nucleic acids as the prosthetic group).
Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid.(RNA).
The two main functions of nucleic acids are:
(a) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to its unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.
(b) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually the proteins are synthesized by various RNA molecules (r-RNA, m-RNA) and t-RNA) in the cell but the message for the synthesis of a particular protein is coded in DNA.
14.22. What is the difference between a nucleoside and a nucleotide?
Ans: A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. It is formed when 1- position of pyrimidine (cytosine, thiamine or uracil) or 9-position of purine (guanine or adenine) base is attached to C -1 of sugar (ribose or deoxyribose) by a β-linkage. Nucleic acids are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide.
A nucleotide contains all the three basic . components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C5, – OH group of the pentose sugar by phosphoric acid.
14.23. The two strands in DNA are not identical but are complementary. Explain.
Ans: The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,C = G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A = T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and not identical.
14.24. Write the important structural and functional differences between DNA and RNA.
14.25. What are the different types of RNA found in the cell?
Ans: There are three types of RNA:
(a) Ribosomal RNA (r RNA)
(b) Messenger RNA (m RNA)
(c) Transfer RNA (t RNA)
CBSE Class 12 Chemistry Important Questions Chapter 14 – Biomolecules
1 Mark Questions ( Biomolecules )
1.Give some examples of bimolecules
Ans. Examples of biomolecules –
carbohydrates, proteins, Nucleic acids, Lipids, enzymes etc.
2.What are carbohydrates?
Ans. Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis.
3.Give one example of each- Monosaccharide, disaccharide and polysaccharide
Ans. Monosaccharide – Glucose, Fructose etc.
Disaccharide – Sucrose, maltose etc.
Polysaccharide – Cellulose, starch etc
4. Which disaccharides are non – reducing sugars?
Ans. In disaccharides, if the reducing groups of monosaccharides, i.e. aldehydic or ketonic groups are bonded eg. In sucrose, these are non- reducing.
5. Classify the following as monosaccharides disaccharides and polysaccharides
Glucose, Sucrose, maltose, ribose, glycogen, lactose, fructose.
6. What is the meaning of statement- Glucose is an aldohexose.
Ans. Glucose is an aldohexose means that it contains six carbon atoms and aldehyde group.
7. Why are polysaccharides considered non- sugars?
Ans. Polysaccharides are not sweet in taste & hence are called non – sugars.
8. Give two examples of reducing sugars
Ans. Examples of reducing sugars –
Maltose and Lactose.
9. Which sugar is present in milk?
Ans. In milk, lactose is present.
10. Name the reagents used to check the reducing nature of carbohydrates.
Ans. Tollen’s reagent and Fehlings solution can be used to check reducing nature of sugars.
11. Name the different types of RNA molecules found in the cells of organisms
Ans. The different types of RNA molecules are transfer RNA (+ RNA), messenger RNA (m–RNA), ribosomal RNA (r- RNA)
12. What are the three components of nucleic acids?
Ans. The three components of nucleic acid are base, sugar and phosphate group.
13. Name different bases present an (i) DNA (ii) RNA
Ans. Bases present in DNA – Thyamine (T), Adenine (A), Guanine (G), Cytosin (C) and in RNA are uracil (U), Adenine (A), Cytosine (C) and Guanine (G).
14. What is nucleoside?
Ans. The molecules in which one of the organic base is combined with sugar are called nucleosides.
15. What type of bonding occurs between two nucleotides?
Ans. The two nucleotides are joined by phosphodiester linkage.
16. Write the sequence of bases in the complementary strand of the given strand –
A G G C T T A A C C T
Ans. The sequence of bases in the complementary sequence is –
T C C G A A T T G G A
17. Name the various sugars present in RNA & DNA.
Ans. The various sugars present in nucleic acids are Ribose in RNA and deoxyribose in DNA.
18. Write functional differences between RNA & DNA.
Ans. DNA is very important for passing of hereditary information from one generation to other. In RNA protein synthesis takes place.
19. What is the basic unit of proteins?
Ans. The basic unit of all proteins in – amino acids.
20. Give an example of zwitter ion?
21. Write the name of bond between the two – amino acids.
Ans. – amino acids are connected by peptide linkage.
22.What is the information given by primary structure of proteins?
Ans. Primary structure of proteins tells about the sequence in which various amino acids are linked with each other.
23.Name the forces responsible for secondary and tertiary structure.
Ans. The forces which are responsible for tertiary structure of proteins are hydrogen bonds, disulphide linkage, vanderwalls and electrostatic forces of attraction.
24.Which vitamins cannot be stored in our body?
Ans. Water soluble vitamins cannot be stored in our body as they are excreted in urine.
25.Where are fat soluble vitamins like A, D, E, and K stored in our body?
Ans. Fat soluble vitamins are stored in liver and adipose (fat storing) tissues in our body.
26.What are enzymes?
Ans. Enzymes are biocatalyst which are very specific for a particular reaction and a particular substrate. Almost all enzymes are globular proteins.
27.Where does the water present in the egg go after boiling the egg?
Ans.When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding.
28. Why cannot vitamin C be stored in our body?
Ans.Vitamin C cannot be stored in our body because it is water soluble. As a result, it is readily excreted in the urine.
29. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Ans. When a nucleotide from the DNA containing thymine is hydrolyzed, thymine -D-2-deoxyribose and phosphoric acid are obtained as products.
30.What are reducing sugars?
Ans. Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.
31. Write two main functions of carbohydrates in plants.
Ans.Two main functions of carbohydrates in plants are:
(i) Polysaccharides such as starch serve as storage molecules.
(ii) Cellulose, a polysaccharide, is used to build the cell wall.
32. Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose
Ans. Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose
2 Marks Questions ( Biomolecules )
1. How are amino acids classified?
Ans. Amino acids are classified as essential and non – essential amino acids.
The amino acids which can be synthesized in the body are known as non – essential amino acids e.g. Asparetic acid, Glycine etc.
The amino acids which cannot be synthesized in the body and must be obtained through diet are known as essential amino acids. E.g. Histidine, lysine.
3. Differentiate between – helical and – pleated sheet structure.
|– helical structure||– pleated sheet structure|
|1. In this structure, formation of hydrogen Bonding between amide groups within the same chain causes the peptide chains to coil up into a spiral structure like a right handed screw.e.g. – keratin, skin, wool etc.||1. In this structure, the long peptide chains lie side by side in a zig-zag manner to form a flat sheet. Each chain is held to the two neighboring Chains by hydrogen bonds. These sheets can slide upon one another in three dimensional structures.e.g. fibroin present in silk etc.|
7. Give an example of enzyme catalysed reaction.
Ans. Example of enzyme catalysed reaction –
8. What are vitamins? Give two examples.
Ans. Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism. e.g. vitamins A, B, C, D etc.
9. How are vitamins classified?
Ans. Vitamins are classified into two groups depending upon their solubility’s-
(i) Fat soluble vitamins – which are soluble in fats and oils. e.g. vitamins A, D, E & K.
(ii) Water soluble vitamins – which are soluble in water e.g. vitamins B& C.
10. Write the disease caused by deficiency of vitamins A, , , , C, D E and K.
|ACDEK||Xerophthalmia, Night blindnessdigestive disordersConvulsionsPernicious anaemiaScurvyRicketsMuscular weaknessIncreased blood clotting time.|
11. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Ans. A glucose molecule contains five -OH groups while a sucrose molecule contains eight -OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water.
Hence, these are soluble in water.
But cyclohexane and benzene do not contain -OH groups. Hence, they cannot undergo H-bonding with water and as a result, are insoluble in water.
12. What is glycogen? How is it different from starch?
Ans. Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen.
Starch is a carbohydrate consisting of two components – amylose (15 – 20%) and amylopectin (80 – 85%).
However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.
13. What are essential and non-essential amino acids? Give two examples of each type.
Ans. Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food. For example: valine and leucine
Non-essential amino acids are also required by the human body, but they can be synthesised in the body. For example: glycine, and alanine
14. What type of bonding helps in stabilising the -helix structure of proteins?
Ans. The H-bonds formed between the -NH group of each amino acid residue and
the group of the adjacent turns of the -helix help in stabilising the helix.
15. What is the effect of denaturation on the structure of proteins?
Ans. As a result of denaturation, globules get unfolded and helixes get uncoiled. Secondary and
tertiary structures of protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation, secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.
16. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Ans. On the basis of their solubility in water or fat, vitamins are classified into two groups.
(i) Fat-soluble vitamins: Vitamins that are soluble in fat and oils, but not in water, belong to this group. For example: Vitamins A, D, E, and K
(ii) Water-soluble vitamins: Vitamins that are soluble in water belong to this group. For example: B group vitamins (, etc.) and vitamin C
However, biotin or vitamin H is neither soluble in water nor in fat.
Vitamin K is responsible for the coagulation of blood.
17. Why are vitamin A and vitamin C essential to us? Give their important sources.
Ans. The deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and night blindness. The deficiency of vitamin C leads to scurvy (bleeding gums).
The sources of vitamin A are fish liver oil, carrots, butter, and milk. The sources of vitamin C are citrus fruits, amla, and green leafy vegetables.
18. The two strands in DNA are not identical but are complementary. Explain.
Ans.In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other.
19. What are the different types of RNA found in the cell?
Ans. (i) Messenger RNA (m-RNA)
(ii) Ribosomal RNA (r-RNA)
(iii) Transfer RNA (t-RNA)
3 Marks Questions ( Biomolecules )
1.What are the expected products of hydrolysis of lactose?
Ans. Lactose is composed of -D-galactose and -D-glucose. Thus, on hydrolysis, it gives -D-galactose and -D-glucose.
2.The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Ans. Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitter ion.
Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour.
For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids.
3. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Ans. A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine and similarly, the quantity of cytosine is equal to that of guanine.
But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.
4. What are monosaccharides?
Ans. Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone.
Monosaccharides are classified on the bases of number of carbon atoms and the functional group present in them. Monosaccharides containing an aldehyde group are known as aldoses and those containing a keto group are known as ketoses. Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and heptoses according to the number of carbon atomsthey contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is called aldotriose.
5. What do you understand by the term glycosidic linkage?
Ans. Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
For example, in a sucrose molecule, two monosaccharide units, -glucose and -fructose, are joined together by a glycosidic linkage.
6. What are the hydrolysis products of (i)sucrose and (ii)lactose?
Ans. (i) On hydrolysis, sucrose gives one molecule of -D glucose and one molecule of -fructose.
(ii) The hydrolysis of lactose gives -galactose and -glucose.
7. What happens when D-glucose is treated with the following reagents?
(i) HI (ii) Bromine water (iii)
Ans. (i) When D-glucose is heated with HI for a long time, n-hexane is formed.
(ii) When D-glucose is treated with water, D- gluconic acid is produced.
(iii) On being treated with , D-glucose get oxidised to give saccharic acid.
8. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Ans. (1) Aldehydes give 2, 4-DNP test, Schiff’s test, and react with to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
(2) The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free -CHO group is absent from glucose.
(3) Glucose exists in two crystalline forms – and . Theform (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behavior cannot be explained by the open chain structure of glucose.
9. Differentiate between globular and fibrous proteins.
10. How do you explain the amphoteric behavior of amino acids?
Ans.I n aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitter ion.
Therefore, in zwitter ionic form, the amino acid can act both as an acid and as a base.
Thus, amino acids show amphoteric behaviour.
11. What are enzymes?
Ans. Enzymes are proteins that catalyse biological reactions. They are very specific in nature and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after the particular substrate or class of substrate and sometimes after the particular reaction.
For example, the enzyme used to catalyse the hydrolysis of maltose into glucose is named as maltase.
Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named as oxidoreductase enzymes.
The name of an enzyme ends with .
12. What are nucleic acids? Mention their two important functions.
Ans. Nucleic acids are biomolecules found in the nuclei of all living cells, as one of the constituents of chromosomes. There are mainly two types of nucleic acids – deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Nucleic acids are also known as polynucleotides as they are long-chain polymers of nucleotides.
Two main functions of nucleic acids are:
(i) DNA is responsible for the transmission of inherent characters from one generation to the next. This process of transmission is called heredity.
(ii) Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell. Even though the proteins are actually synthesised by the various RNA molecules in a cell, the message for the synthesis of a particular protein is present in DNA.
13. What is the difference between a nucleoside and a nucleotide?
Ans. A nucleoside is formed by the attachment of a base to position of sugar.
Nucleoside = Sugar + Base
On the other hand, all the three basic components of nucleic acids (i.e., pentose sugar, phosphoric acid, and base) are present in a nucleotide.
Nucleotide = Sugar + Base + Phosphoric acid