Biomolecules
“It is the harmonious and synchronous progress of chemical reactions in body which leads to life”.
A living system grows, sustains and reproduces itself. The most amazing thing about a living system is that it is composed of non-living atoms and molecules. The pursuit of knowledge of what goes on chemically within a living system falls in the domain of biochemistry. Living systems are made up of various complex biomolecules like carbohydrates, proteins, nucleic acids, lipids, etc. Proteins and carbohydrates are essential constituents of our food. These biomolecules interact with each other and constitute the molecular logic of life processes. In addition, some simple molecules like vitamins and mineral salts also play an important role in the functions of organisms. Structures and functions of some of these biomolecules are discussed in this Unit.
14.1 Carbohydrates
Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds. Some common examples are cane sugar, glucose, starch, etc. Most of them have a general formula, Cx(H2O)y, and were considered as hydrates of carbon from where the name carbohydrate was derived. For example, the molecular formula of glucose (C6H12O6) fits into this general formula, C6(H2O)6. But all the compounds which fit into this formula may not be classified as carbohydrates. Acetic acid (CH3COOH) fits into this general formula, C2(H2O)2 but is not a carbohydrate. Similarly, rhamnose, C6H12O5 is a carbohydrate but does not fit in this definition. A large number of their reactions have shown that they contain specific functional groups. Chemically, the carbohydrates may be defined as optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. Some of the carbohydrates,
14.1.1 Classification of Carbohydrates
Carbohydrates are classified on the basis of their behaviour on hydrolysis. They have been broadly divided into following three groups.
(i) Monosaccharides: A carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or ketone is called a monosaccharide. About 20 monosaccharides are known to occur in nature. Some common examples are glucose, fructose, ribose, etc.
(ii) Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides. They are further classified as disaccharides, trisaccharides, tetrasaccharides, etc., depending upon the number of monosaccharides, they provide on hydrolysis. Amongst these the most common are disaccharides. The two monosaccharide units obtained on hydrolysis of a disaccharide may be same or different. For example, sucrose on hydrolysis gives one molecule each of glucose and fructose whereas maltose gives two molecules of glucose only.
(iii) Polysaccharides: Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Some common examples are starch, cellulose, glycogen, gums, etc. Polysaccharides are not sweet in taste, hence they are also called non-sugars.
The carbohydrates may also be classified as either reducing or non- reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.
In disaccharides, if the reducing groups of monosaccharides i.e.,aldehydic or ketonic groups are bonded, these are non-reducing sugarse.g. sucrose. On the other hand, sugars in which these functional groups are free, are called reducing sugars, for example, maltose and lactose.
(i) Monosaccharides: A carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or ketone is called a monosaccharide. About 20 monosaccharides are known to occur in nature. Some common examples are glucose, fructose, ribose, etc.
(ii) Oligosaccharides: Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides. They are further classified as disaccharides, trisaccharides, tetrasaccharides, etc., depending upon the number of monosaccharides, they provide on hydrolysis. Amongst these the most common are disaccharides. The two monosaccharide units obtained on hydrolysis of a disaccharide may be same or different. For example, sucrose on hydrolysis gives one molecule each of glucose and fructose whereas maltose gives two molecules of glucose only.
(iii) Polysaccharides: Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Some common examples are starch, cellulose, glycogen, gums, etc. Polysaccharides are not sweet in taste, hence they are also called non-sugars.
The carbohydrates may also be classified as either reducing or non- reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.
In disaccharides, if the reducing groups of monosaccharides i.e.,aldehydic or ketonic groups are bonded, these are non-reducing sugarse.g. sucrose. On the other hand, sugars in which these functional groups are free, are called reducing sugars, for example, maltose and lactose.
14.1.2 Monosaccharides
Monosaccharides are further classified on the basis of number of carbon atoms and the functional group present in them. If a monosaccharide contains an aldehyde group, it is known as an aldose and if it contains a keto group, it is known as a ketose. Number of carbon atoms constituting the monosaccharide is also introduced in the name as is evident from the examples given in Table 14.1
| CARBON ATOM | GENERAL TERM | ALDEHYDE | KETONE |
|---|---|---|---|
| 3 | Triose | Aldotriose | Ketotriose |
| 4 | Tetrose | Aldotetrose | Ketotetrose |
| 5 | Pentose | Aldopentose | Ketopentose |
| 6 | Hexose | Aldohexose | Ketohexose |
| 7 | Heptose | Aldoheptose | Ketoheptose |
I. Glucose
Glucose occurs freely in nature as well as in the combined form. It is present in sweet fruits and honey. Ripe grapes also contain glucose in large amounts. It is prepared as follows:
Glucose occurs freely in nature as well as in the combined form. It is present in sweet fruits and honey. Ripe grapes also contain glucose in large amounts. It is prepared as follows:
14.1.3 Preparation of Glucose
1. From sucrose (Cane sugar): If sucrose is boiled with dilute HCl or H2SO4 in alcoholic solution, glucose and fructose are obtained in equal amounts.
2. From starch: Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute H2SO4 at 393 K under pressure.
14.1.4 Structure of Glucose
Glucose is an aldohexose and is also known as dextrose. It is the monomer of many of the larger carbohydrates, namely starch, cellulose. It is probably the most abundant organic compound on earth. It was assigned the structure given below on the basis of the following evidences:
1. Its molecular formula was found to be C6H12O6.
2. On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.
1. Its molecular formula was found to be C6H12O6.
2. On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.
3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of a carbonyl group (>C = 0) in glucose.
4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.
5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.
6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.
The exact spatial arrangement of different —OH groups was given by Fischer after studying many other properties. Its configuration is correctly represented as I. So gluconic acid is represented as II and saccharic acid as III.
Glucose is correctly named as D(+)-glucose. ‘D’ before the name of glucose represents the configuration whereas ‘(+)’ represents dextrorotatory nature of the molecule. It may be remembered that ‘D’ and ‘L’ have no relation with the optical activity of the compound. The meaning of D– and L– notations is given as follows.
The letters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer. This refers to their relation with a particular isomer of glyceraldehyde. Glyceraldehyde contains one asymmetric carbon atom and exists in two enantiomeric forms as shown below.
The letters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer. This refers to their relation with a particular isomer of glyceraldehyde. Glyceraldehyde contains one asymmetric carbon atom and exists in two enantiomeric forms as shown below.
All those compounds which can be chemically correlated to (+) isomer of glyceraldehyde are said to have D-configuration whereas those which can be correlated to (–) isomer of glyceraldehyde are said to have L—configuration. For assigning the configuration of monosaccharides, it is the lowest asymmetric carbon atom (as shown below) which is compared. As in (+) glucose, —OH on the lowest asymmetric carbon is on the right side which is comparable to (+) glyceraldehyde, so it is assigned D-configuration. For this comparison, the structure is written in a way that most oxidised carbon is at the top.
14.1.5 Cyclic Structure of Glucose
The structure (I) of glucose explained most of its properties but the following reactions and facts could not be explained by this structure.
1. Despite having the aldehyde group, glucose does not give 2,4-DNP test, Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3.
2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group.
3. Glucose is found to exist in two different crystalline forms which are named as α and β. The α-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the β-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.
1. Despite having the aldehyde group, glucose does not give 2,4-DNP test, Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3.
2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group.
3. Glucose is found to exist in two different crystalline forms which are named as α and β. The α-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the β-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.
This behaviour could not be explained by the open chain structure (I) for glucose. It was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in ring formation. This explains the absence of —CHO group and also existence of glucose in two forms as shown below. These two cyclic forms exist in equilibrium with open chain structure.
The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C1, called anomeric carbon (the aldehyde carbon before cyclisation). Such isomers, i.e., α-form and β-form, are called anomers. The six membered cyclic structure of glucose is called pyranose structure (α– or β–), in analogy with pyran. Pyran is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring. The cyclic structure of glucose is more correctly represented by Haworth structure as given below.
II. Fructose
Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide, sucrose.
Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide, sucrose.
14.1.6 Structure of Fructose
Fructose also has the molecular formula C6H12O6 and on the basis of its reactions it was found to contain a ketonic functional group at carbon number 2 and six carbons in straight chain as in the case of glucose. It belongs to D-series and is a laevorotatory compound. It is appropriately written as D-(–)-fructose. Its open chain structure is as shown.
It also exists in two cyclic forms which are obtained by the addition of ) group. The ring, thus formed is a five membered —OH at C5 to the (ring and is named as furanose with analogy to the compound furan. Furan is a five membered cyclic compound with one oxygen and four carbon atoms.
The cyclic structures of two anomers of fructose are represented by Haworth structures as given.
14.1.7 Disaccharides
You have already read that disaccharides on hydrolysis with dilute acids or enzymes yield two molecules of either the same or different monosaccharides. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.
(i) Sucrose: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose. These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose and C2 of β-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non reducing sugar.
(i) Sucrose: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose. These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose and C2 of β-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non reducing sugar.
These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose and C2 of β-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non reducing sugar.
Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–) and the product is named as invert sugar.
(ii) Maltose: Another disaccharide, maltose is composed of two α-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II). The free aldehyde group can be produced at C1 of second glucose in solution and it shows reducing properties so it is a reducing sugar.
(iii) Lactose: It is more commonly known as milk sugar since this disaccharide is found in milk. It is composed of β-D-galactose and β-D-glucose. The linkage is between C1 of galactose and C4 of glucose. Hence it is also a reducing sugar.
14.1.8 Polysaccharides
Polysaccharides contain a large number of monosaccharide units joined together by glycosidic linkages. These are the most commonly encountered carbohydrates in nature. They mainly act as the food storage or structural materials.
(i) Starch: Starch is the main storage polysaccharide of plants. It is the most important dietary source for human beings. High content of starch is found in cereals, roots, tubers and some vegetables. It is a polymer of α-glucose and consists of two components— Amylose and Amylopectin. Amylose is water soluble component which constitutes about 15-20% of starch. Chemically amylose is a long unbranched chain with 200-1000 α-D-(+)-glucose units held by C1– C4 glycosidic linkage.
(i) Starch: Starch is the main storage polysaccharide of plants. It is the most important dietary source for human beings. High content of starch is found in cereals, roots, tubers and some vegetables. It is a polymer of α-glucose and consists of two components— Amylose and Amylopectin. Amylose is water soluble component which constitutes about 15-20% of starch. Chemically amylose is a long unbranched chain with 200-1000 α-D-(+)-glucose units held by C1– C4 glycosidic linkage.
Amylopectin is insoluble in water and constitutes about 80- 85% of starch. It is a branched chain polymer of α-D-glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage.
(ii) Cellulose: Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chainpolysaccharide composed only of β-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.
(iii) Glycogen: The carbohydrates are stored in animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin and is rather more highly branched. It is present in liver, muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose. Glycogen is also found in yeast and fungi.
14.1.9 Importance of Carbohydrates
Carbohydrates are essential for life in both plants and animals. They form a major portion of our food. Honey has been used for a long time as an instant source of energy by ‘Vaids’ in ayurvedic system of medicine. Carbohydrates are used as storage molecules as starch in plants and glycogen in animals. Cell wall of bacteria and plants is made up of cellulose. We build furniture, etc. from cellulose in the form of wood and clothe ourselves with cellulose in the form of cotton fibre. They provide raw materials for many important industries like textiles, paper, lacquers and breweries.
Two aldopentoses viz. D-ribose and 2-deoxy-D-ribose (Section 14.5.1, Class XII) are present in nucleic acids. Carbohydrates are found in biosystem in combination with many proteins and lipids.
Two aldopentoses viz. D-ribose and 2-deoxy-D-ribose (Section 14.5.1, Class XII) are present in nucleic acids. Carbohydrates are found in biosystem in combination with many proteins and lipids.
Intext Question
14.1 Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
14.2 What are the expected products of hydrolysis of lactose?
14.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
14.1 Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
14.2 What are the expected products of hydrolysis of lactose?
14.3 How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
14.2 proteins
Proteins are the most abundant biomolecules of the living system. Chief sources of proteins are milk, cheese, pulses, peanuts, fish, meat, etc. They occur in every part of the body and form the fundamental basis of structure and functions of life. They are also required for growth and maintenance of body. The word protein is derived from Greek word, “proteios” which means primary or of prime importance. All proteins are polymers of α-amino acids.
14.2.1 Amino Acids
Amino acids contain amino (–NH2) and carboxyl (–COOH) functional groups. Depending upon the relative position of amino group with respect to carboxyl group, the amino acids can beclassified as α, β, γ, δ and so on. Only α-amino acids are obtained on hydrolysis of proteins. They may contain other functional groups also.
All α-amino acids have trivial names, which (R = side chain) usually reflect the property of that compound or its source. Glycine is so named since it has sweet taste (in Greek glykos means sweet) and tyrosine was first obtained from cheese (in Greek, tyros means cheese.) Amino acids are generally represented by a three letter symbol, sometimes one letter symbol is also used. Structures of some commonly occurring amino acids along with their 3-letter and 1-letter symbols are given in Table 14.2.
All α-amino acids have trivial names, which (R = side chain) usually reflect the property of that compound or its source. Glycine is so named since it has sweet taste (in Greek glykos means sweet) and tyrosine was first obtained from cheese (in Greek, tyros means cheese.) Amino acids are generally represented by a three letter symbol, sometimes one letter symbol is also used. Structures of some commonly occurring amino acids along with their 3-letter and 1-letter symbols are given in Table 14.2.
*essential amino acid, a = entire structure
| NAME OF THE AMINO ACIDS | CHARACTERISTICS FEATURE OF SIDE CHAIN, R | THREE LETTER SYMBOL | ONE LETTER CODE |
|---|---|---|---|
| 1. Glycine | H | Gly | G |
| 2. Alanine | -CH3 | Ala | A |
| 3. Valine* | (H3C)2CH- | Val | V |
| 4. Leucine* | (H3C)2CH-CH2- | Leu | L |
| 5. Isoleucine* |  | Ile | I |
| 6. Arginine* |  | Arg | R |
| 7. Lysine* | H2N-(CH2)4- | Lys | K |
| 8. Glutamic acid | HOOC-CH2-CH2- | Glu | E |
| 9. Aspartic acid | HOOC-CH2- | Asp | D |
| 10. Glutamine |  | Gln | Q |
| 11. Asparagine |  | Asn | N |
| 12. Threonine* | H3C-CHOH- | Thr | T |
| 13. Serine | HO-CH2- | Ser | S |
| 14. Cysteine | HS-CH2- | Cys | C |
| 15. Methionine* | H3C-S-CH2-CH2- | Met | M |
| 16. Phenylalanine* | C6H5-CH2- | Phe | F |
| 17. Tyrosine | (p)HO-C6H4-CH2- | Tyr | Y |
| 18. Tryptophan* |  | Trp | W |
| 19. Histidine* |  | His | H |
| 20. Proline |  | Pro | P |
14.2.2 Classification of Amino Acids
Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Equal number of amino and carboxyl groups makes it neutral; more number of amino than carboxyl groups makes it basic and more carboxyl groups as compared to amino groups makes it acidic. The amino acids, which can be synthesised in the body, are known as non- essential amino acids. On the other hand, those which cannot be synthesised in the body and must be obtained through diet, are known as essential amino acids (marked with asterisk in Table 14.2).
Amino acids are usually colourless, crystalline solids. These are water-soluble, high melting solids and behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges.
In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids and bases.
Except glycine, all other naturally occurring α-amino acids are optically active, since the α-carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms. Most naturally occurring amino acids have L-configuration. L-Aminoacids are represented by writing the –NH2group on left hand side.
Except glycine, all other naturally occurring α-amino acids are optically active, since the α-carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms. Most naturally occurring amino acids have L-configuration. L-Aminoacids are represented by writing the –NH2group on left hand side.
14.2.3 Structure of Proteins
You have already read that proteins are the polymers of α-amino acids and they are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between –COOH group and –NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other. This results in the elimination of a water molecule and formation of a peptide bond –CO–NH–. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.
If a third amino acid combines to a dipeptide, the product is called a tripeptide. A tripeptide contains three amino acids linked by two peptide linkages. Similarly when four, five or six amino acids are linked, the respective products are known as tetrapeptide, pentapeptide or hexapeptide, respectively. When the number of such amino acids is more than ten, then the products are called polypeptides. A polypeptide with more than hundred amino acid residues, having molecular mass higher than 10,000u is called a protein. However, the distinction between a polypeptide and a protein is not very sharp. Polypeptides with fewer amino acids are likely to be called proteins if they ordinarily have a well defined conformation of a protein such as insulin which contains 51 amino acids.
Proteins can be classified into two types on the basis of their molecular shape.
Proteins can be classified into two types on the basis of their molecular shape.
(a) Fibrous proteins
When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, then fibre– like structure is formed. Such proteins are generally insoluble in water. Some common examples are keratin (present in hair, wool, silk) and myosin (present in muscles), etc.
(b) Globular proteins
This structure results when the chains of polypeptides coil around to give a spherical shape. These are usually soluble in water. Insulin and albumins are the common examples of globular proteins. Structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quaternary, each level being more complex than the previous one.
(i) Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e., the sequence of amino acids creates a different protein.
(i) Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e., the sequence of amino acids creates a different protein.
(ii) Secondary structure of proteins: The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures viz. α-helix and β-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between
and –NH– groups of the peptide bond.
and –NH– groups of the peptide bond.
α-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with the –NH group of each amino acid residue hydrogen bonded to the
of an adjacent turn of the helix as shown in Fig.14.1.
of an adjacent turn of the helix as shown in Fig.14.1.
In β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as β-pleated sheet.
(iii) Tertiary structure of proteins: The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous and globular. The main forces which stabilise the 2° and 3° structures of proteins are hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction.
(iv) Quaternary structure of proteins: Some of the proteins are composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other is known as quaternary structure.
A diagrammatic representation of all these four structures is given in Figure 14.3 where each coloured ball represents an amino acid.
A diagrammatic representation of all these four structures is given in Figure 14.3 where each coloured ball represents an amino acid.
14.2.4 Denaturation of Proteins
Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation 2° and 3° structures are destroyed but 1o structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.
Intext Questions
14.4 The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
14.5 Where does the water present in the egg go after boiling the egg?
14.4 The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
14.5 Where does the water present in the egg go after boiling the egg?
14.3 Enzymes
Life is possible due to the coordination of various chemical reactions in living organisms. An example is the digestion of food, absorption of appropriate molecules and ultimately production of energy. This process involves a sequence of reactions and all these reactions occur in the body under very mild conditions. This occurs with the help of certain biocatalysts called enzymes. Almost all the enzymes are globular proteins. Enzymes are very specific for a particular reaction and for a particular substrate. They are generally named after the compound or class of compounds upon which they work. For example, the enzyme that catalyses hydrolysis of maltose into glucose is named as maltase.
Sometimes enzymes are also named after the reaction, where they are used. For example, the enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate are named as oxidoreductase enzymes. The ending of the name of an enzyme is -ase.
14.3.1 Mechanism of Enzyme Action
Enzymes are needed only in small quantities for the progress of a reaction. Similar to the action of chemical catalysts, enzymes are said to reduce the magnitude of activation energy. For example, activation energy for acid hydrolysis of sucrose is 6.22 kJ mol–1, while the activation energy is only 2.15 kJ mol–1 when hydrolysed by the enzyme, sucrase. Mechanism for the enzyme action has been discussed in Unit 5.
14.4 Vitamins
It has been observed that certain organic compounds are required in small amounts in our diet but their deficiency causes specific diseases. These compounds are called vitamins. Most of the vitamins cannot be synthesised in our body but plants can synthesise almost all of them, so they are considered as essential food factors. However, the bacteria of the gut can produce some of the vitamins required by us. All the vitamins are generally available in our diet. Different vitamins belong to various chemical classes and it is difficult to define them on the basis of structure. They are generally regarded as 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. Vitamins are designated by alphabets A, B, C, D, etc. Some of them are further named as sub-groups e.g. B1, B2, B6, B12, etc. Excess of vitamins is also harmful and vitamin pills should not be taken without the advice of doctor.
The term “Vitamine” was coined from the word vital + amine since the earlier identified compounds had amino groups. Later work showed that most of them did not contain amino groups, so the letter ‘e’ was dropped and the term vitamin is used these days.
The term “Vitamine” was coined from the word vital + amine since the earlier identified compounds had amino groups. Later work showed that most of them did not contain amino groups, so the letter ‘e’ was dropped and the term vitamin is used these days.
14.4.1 Classification of Vitamins
Vitamins are classified into two groups depending upon their solubility in water or fat.
(i) Fat soluble vitamins: Vitamins which are soluble in fat and oils but insoluble in water are kept in this group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues.
(ii) Water soluble vitamins: B group vitamins and vitamin C are soluble in water so they are grouped together. Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body.
Some important vitamins, their sources and diseases caused by their deficiency are listed in Table 14.3.
(i) Fat soluble vitamins: Vitamins which are soluble in fat and oils but insoluble in water are kept in this group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues.
(ii) Water soluble vitamins: B group vitamins and vitamin C are soluble in water so they are grouped together. Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body.
Some important vitamins, their sources and diseases caused by their deficiency are listed in Table 14.3.
| SL. NO. | NAME OF VITAMINES | SOURCES | DEFICIENCY DISEASES |
|---|---|---|---|
| 1. | Vitamin A | Fish liver oil, carrots, butter and milk | Xerophthalmia (hardening of cornea of eye) Night blindness |
| 2. | Vitamin B1(Thiamine) | Yeast, milk, green vegetables and cereals | Beri beri (loss of appetite, retarded growth) |
| 3. | Vitamin B2(Riboflavin) | Milk, eggwhite, liver, kidney | Cheilosis (fissuring at corners of mouth and lips), digestive disorders and burning sensation of the skin. |
| 4. | Vitamin B6(Pyridoxine) | Yeast, milk, egg yolk, cereals and grams | Convulsions |
| 5. | Vitamin B12 | Meat, fish, egg and curd | Pernicious anaemia (RBC deficient in haemoglobin) |
| 6. | Vitamin C (Ascorbic acid) | Citrus fruits, amla and green leafy vegetables | Scurvy (bleeding gums) |
| 7. | Vitamin D | Exposure to sunlight, fish and egg yolk | Rickets (bone deformities in children) and osteomalacia (soft bones and joint pain in adults) |
| 8. | Vitamin E | Vegetable oils like wheat germ oil, sunflower oil, etc. | Increased fragility of RBCs and muscular weakness |
| 9. | Vitamin K | Green leafy vegetables | Increased blood clotting time |
14.5: Nucleic Acids
Every generation of each and every species resembles its ancestors in many ways. How are these characteristics transmitted from one generation to the next? It has been observed that nucleus of a living cell is responsible for this transmission of inherent characters, also called heredity. The particles in nucleus of the cell, responsible for heredity, are called chromosomes which are made up of proteins and another type of biomolecules called nucleic acids. These are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Since nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides.
James Dewey Watson
Born in Chicago, Illinois, in 1928, Dr Watson received his Ph.D. (1950) from Indiana University in Zoology. He is best known for his discovery of the structure of DNA for which he shared with Francis Crick and Maurice Wilkins the 1962 Nobel prize in Physiology and Medicine. They proposed that DNA molecule takes the shape of a double helix, an elegantly simple structure that resembles a gently twisted ladder. The rails of the ladder are made of alternating units of phosphate and the sugar deoxyribose; the rungs are each composed of a pair of purine/ pyrimidine bases. This research laid the foundation for the emerging field of molecular biology. The complementary pairing of nucleotide bases explains how identical copies of parental DNA pass on to two daughter cells. This research launched a revolution in biology that led to modern recombinant DNA techniques.
Born in Chicago, Illinois, in 1928, Dr Watson received his Ph.D. (1950) from Indiana University in Zoology. He is best known for his discovery of the structure of DNA for which he shared with Francis Crick and Maurice Wilkins the 1962 Nobel prize in Physiology and Medicine. They proposed that DNA molecule takes the shape of a double helix, an elegantly simple structure that resembles a gently twisted ladder. The rails of the ladder are made of alternating units of phosphate and the sugar deoxyribose; the rungs are each composed of a pair of purine/ pyrimidine bases. This research laid the foundation for the emerging field of molecular biology. The complementary pairing of nucleotide bases explains how identical copies of parental DNA pass on to two daughter cells. This research launched a revolution in biology that led to modern recombinant DNA techniques.
14.5.1 Chemical Composition of Nucleic Acids
Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric acid and nitrogen containing heterocyclic compounds (called bases). In DNA molecules, the sugar moiety is β-D-2-deoxyribose whereas in RNA molecule, it is β-D-ribose.
DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also contains four bases, the first three bases are same as in DNA but the fourth one is uracil (U).
14.5.2 Structure of Nucleic Acids
A unit formed by the attachment of a base to 1′ position of sugar is known as nucleoside. In nucleosides, the sugar carbons are numbered as 1′, 2′, 3′, etc. in order to distinguish these from the bases (Fig. 14.5a). When nucleoside is linked to phosphoric acid at 5′-position of sugar moiety, we get a nucleotide (Fig. 14.5).
Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of the pentose sugar. The formation of a typical dinucleotide is shown in Fig. 14.6.
A simplified version of nucleic acid chain is as shown below.
Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure also. James Watson and Francis Crick gave a double strand helix structure for DNA (Fig. 14.7). Two nucleic acid chains are wound about each other and held together
by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.
Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure also. James Watson and Francis Crick gave a double strand helix structure for DNA (Fig. 14.7). Two nucleic acid chains are wound about each other and held together
by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.
In secondary structure of RNA, helices are present which are only single stranded. Sometimes they fold back on themselves to form a double helix structure. RNA molecules are of three types and they perform different functions. They are named as messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).
Har Gobind Khorana
Har Gobind Khorana, was born in 1922. He obtained his M.Sc. degree from Punjab University in Lahore. He worked with Professor Vladimir Prelog, who moulded Khorana’s thought and philosophy towards science, work and effort. After a brief stay in India in 1949, Khorana went back to England and worked with Professor
G.W. Kenner and Professor A.R.Todd. It was at Cambridge, U.K. that he got interested in both proteins and nucleic acids. Dr Khorana shared the Nobel Prize for Medicine and Physiology in 1968 with Marshall Nirenberg and Robert Holley for cracking the genetic code.
G.W. Kenner and Professor A.R.Todd. It was at Cambridge, U.K. that he got interested in both proteins and nucleic acids. Dr Khorana shared the Nobel Prize for Medicine and Physiology in 1968 with Marshall Nirenberg and Robert Holley for cracking the genetic code.
DNA Fingerprinting
It is known that every individual has unique fingerprints. These occur at the tips of the fingers and have been used for identification for a long time but these
can be altered by surgery. A sequence of bases on DNA is also unique for a person and information regarding this is called DNA fingerprinting. It is same for
every cell and cannot be altered by any known treatment. DNA fingerprinting is now used
(i) in forensic laboratories for identification of criminals.
(ii) to determine paternity of an individual.
(iii) to identify the dead bodies in any accident by comparing the DNA’s of parents or children.
(iv) to identify racial groups to rewrite biological evolution.
It is known that every individual has unique fingerprints. These occur at the tips of the fingers and have been used for identification for a long time but these
can be altered by surgery. A sequence of bases on DNA is also unique for a person and information regarding this is called DNA fingerprinting. It is same for
every cell and cannot be altered by any known treatment. DNA fingerprinting is now used
(i) in forensic laboratories for identification of criminals.
(ii) to determine paternity of an individual.
(iii) to identify the dead bodies in any accident by comparing the DNA’s of parents or children.
(iv) to identify racial groups to rewrite biological evolution.
DNA is the chemical basis of heredity and may be regarded as the reserve of genetic information. DNA is exclusively responsible for maintaining the identity of different species of organisms over millions of years. A DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells. Another important function of nucleic acids is the protein synthesis in the cell. Actually, the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.
Intext Questions
14.6 Why cannot vitamin C be stored in our body?
14.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
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?
14.6 Why cannot vitamin C be stored in our body?
14.7 What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
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?
Summary
Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules which provide such units on hydrolysis. They are broadly classified into three groups — monosaccharides, disaccharides and polysaccharides. Glucose, the most important source of energy for mammals, is obtained by the digestion of starch. Monosaccharides are held together by glycosidic linkages to form disaccharides or polysaccharides.
Proteins are the polymers of about twenty different α-amino acids which are linked by peptide bonds. Ten amino acids are called essential amino acids because they cannot be synthesised by our body, hence must be provided through diet. Proteins perform various structural and dynamic functions in the organisms.
Proteins which contain only α-amino acids are called simple proteins. The secondary or tertiary structure of proteins get disturbed on change of pH or temperature and they are not able to perform their functions. This is called denaturation of proteins. Enzymes are biocatalysts which speed up the reactions in biosystems. They are very specific and selective in their action and chemically all enzymes are proteins. Vitamins are accessory food factors required in the diet. They are classified as fat soluble (A, D, E and K) and water soluble (Β group and C). Deficiency of vitamins leads to many diseases.
Nucleic acids are the polymers of nucleotides which in turn consist of a base, a pentose sugar and phosphate moiety. Nucleic acids are responsible for the transfer of characters from parents to offsprings. There are two types of nucleic acids — DNA and RNA. DNA contains a five carbon sugar molecule called 2-deoxyribose whereas RNA contains ribose. Both DNA and RNA contain adenine, guanine and cytosine. The fourth base is thymine in DNA and uracil in RNA. The structure of DNA is a double strand whereas RNA is a single strand molecule. DNA is the chemical basis of heredity and have the coded message for proteins to be synthesised in the cell. There are three types of RNA — mRNA, rRNA and tRNA which actually carry out the protein synthesis in the cell.
Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules which provide such units on hydrolysis. They are broadly classified into three groups — monosaccharides, disaccharides and polysaccharides. Glucose, the most important source of energy for mammals, is obtained by the digestion of starch. Monosaccharides are held together by glycosidic linkages to form disaccharides or polysaccharides.
Proteins are the polymers of about twenty different α-amino acids which are linked by peptide bonds. Ten amino acids are called essential amino acids because they cannot be synthesised by our body, hence must be provided through diet. Proteins perform various structural and dynamic functions in the organisms.
Proteins which contain only α-amino acids are called simple proteins. The secondary or tertiary structure of proteins get disturbed on change of pH or temperature and they are not able to perform their functions. This is called denaturation of proteins. Enzymes are biocatalysts which speed up the reactions in biosystems. They are very specific and selective in their action and chemically all enzymes are proteins. Vitamins are accessory food factors required in the diet. They are classified as fat soluble (A, D, E and K) and water soluble (Β group and C). Deficiency of vitamins leads to many diseases.
Nucleic acids are the polymers of nucleotides which in turn consist of a base, a pentose sugar and phosphate moiety. Nucleic acids are responsible for the transfer of characters from parents to offsprings. There are two types of nucleic acids — DNA and RNA. DNA contains a five carbon sugar molecule called 2-deoxyribose whereas RNA contains ribose. Both DNA and RNA contain adenine, guanine and cytosine. The fourth base is thymine in DNA and uracil in RNA. The structure of DNA is a double strand whereas RNA is a single strand molecule. DNA is the chemical basis of heredity and have the coded message for proteins to be synthesised in the cell. There are three types of RNA — mRNA, rRNA and tRNA which actually carry out the protein synthesis in the cell.
Exercises
14.1 What are monosaccharides?
14.1 What are monosaccharides?
14.2 What are reducing sugars?
14.3 Write two main functions of carbohydrates in plants.
14.4 Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
14.5 What do you understand by the term glycosidic linkage?
14.6 What is glycogen? How is it different from starch?
14.7 What are the hydrolysis products of
(i) sucrose and
(ii) lactose?
(i) sucrose and
(ii) lactose?
14.8 What is the basic structural difference between starch and cellulose?
14.9 What happens when D-glucose is treated with the following reagents?
(i) HI
(ii) Bromine water
(iii) HNO3
(i) HI
(ii) Bromine water
(iii) HNO3
14.10 Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
14.11 What are essential and non-essential amino acids? Give two examples of each type.
14.12 Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
14.13 What are the common types of secondary structure of proteins?
14.14 What type of bonding helps in stabilising the α-helix structure of proteins?
14.15 Differentiate between globular and fibrous proteins.
14.16 How do you explain the amphoteric behaviour of amino acids?
14.17 What are enzymes?
14.18 What is the effect of denaturation on the structure of proteins?
14.19 How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
14.20 Why are vitamin A and vitamin C essential to us? Give their important sources.
14.21 What are nucleic acids? Mention their two important functions.
14.22 What is the difference between a nucleoside and a nucleotide?
14.23 The two strands in DNA are not identical but are complementary. Explain.
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?
TESTS FOR CARBOHYDRATES FATS AND PROTEINS
EXPERIMENT 11.1
Aim
Aim
To study the characteristics of carbohydrates, fats and proteins in pure form and detection of their presence in the given foodstuffs.
I. TEST FOR CARBOHYDRATES, FATS AND PROTEINS IN PURE FORM
A. Tests for Carbohydrates
Theory
Carbohydrates are optically active polyhydroxy aldehydes, polyhydroxy ketones or compounds, which give these units as hydrolysis product. Starch, cellulose
and sugars are the familiar examples of carbohydrates. Carbohydrates are classified on the basis of number of polyhydroxy aldehyde or ketone units obtained from them on hydrolysis. Three broad classes are as follows :
A. Tests for Carbohydrates
Theory
Carbohydrates are optically active polyhydroxy aldehydes, polyhydroxy ketones or compounds, which give these units as hydrolysis product. Starch, cellulose
and sugars are the familiar examples of carbohydrates. Carbohydrates are classified on the basis of number of polyhydroxy aldehyde or ketone units obtained from them on hydrolysis. Three broad classes are as follows :
(i) Monosaccharides : These cannot be hydrolysed further to polyhydroxy aldehydes or ketones.
(ii) Oligosaccharides : These yield 2-10 monosaccharide units on hydrolysis. Common amongst these are disaccharides, which produce two monosaccharide units.
(iii) Polysaccharides : These yield large number of monosaccharide units on hydrolysis.
(ii) Oligosaccharides : These yield 2-10 monosaccharide units on hydrolysis. Common amongst these are disaccharides, which produce two monosaccharide units.
(iii) Polysaccharides : These yield large number of monosaccharide units on hydrolysis.
Monosaccharides are further classified on the basis of number of carbon atoms and functional group present in them. If a monosaccharide contains aldehydic
group it is called aldose. If it contains keto group it is called ketose. Carbohydrates of all classes give Molisch’s test. Carbohydrates, which are sweet in taste, are called sugars. Glucose, fructose (fruit sugar) and sucrose (table sugar) are examples of sugars. Sugars are classified into two major categories: reducing sugars and non-reducing sugars. Reducing property of sugars is detected by the three tests namely Fehling’s test, Benedict’s test and Tollen’s test.
group it is called aldose. If it contains keto group it is called ketose. Carbohydrates of all classes give Molisch’s test. Carbohydrates, which are sweet in taste, are called sugars. Glucose, fructose (fruit sugar) and sucrose (table sugar) are examples of sugars. Sugars are classified into two major categories: reducing sugars and non-reducing sugars. Reducing property of sugars is detected by the three tests namely Fehling’s test, Benedict’s test and Tollen’s test.
Prepare 1% stock solution of glucose, fructose and sucrose in separate beakers and divide each of the solutions into test tubes marked A, B, C and D etc. and
perform the following tests.
perform the following tests.
I. Theory of Molisch’s test
On adding concentrated sulphuric acid to the aqueous solution of carbohydrate containing alcoholic solution of 1-naphthol, a deep violet colour appears at the junction of the two liquids. Concentrated sulphuric acid hydrolyses glycosidic bonds of carbohydrate to give monosaccharides which are dehydrated to an aldehyde known as furfural which undergoes reaction with 1–naphthol to give a unstable condensation product of deep violet colour. This test may be given by some other organic compounds also. The following reaction takes place :
Material Required
Procedure
Add 2-3 drops of alcoholic solution of 1% 1-naphthol in test tube ‘A’ and then pour 2 mL conc. H2SO4 down the sides of the test tube so that it forms a separate layer at the bottom of the test tube. The formation of a purple ring at the interface of the two layers confirms the presence of carbohydrates.
II. Theory of test for reducing sugars
A. Fehling’s test and Benedict test
A. Fehling’s test and Benedict test
Black copper (II) oxide is formed on heating a suspension of copper hydroxide in alkaline solution.
If some reducing agent is present in the reaction medium, then orange-red copper (I) oxide is precipated.
Reducing sugars contain aldehydic group or α-hydroxy ketonic group, therefore in alkaline medium reduce Cu2+ ions. But if the reaction is carried out directly in the presence of an alkali then, copper (II) hydroxide gets precipitated. To overcome this problem, copper (II) ions are complexed with tartarate ions (Fehling’s reagent) or citrate ions (Benedict’s solution). Both the complex ions are soluble in alkaline medium and yield Cu2+ ions in such a low concentration that solubility product of cupric hydroxide is not reached.
Reducing sugars react with Fehling’s reagent according to the following reaction:
The discharge of blue colour due to Cu2+ ions and appearance of orange-red precipitate of Cu2O, indicates the reducing property of sugars.
???????????
Sometimes, the cuprous precipitate comes down as yellow cuprous hydroxide, but on warming this is converted to orangered copper (I) oxide.
In some cases, this reaction may be used as a quantitative analytical process for the determination of reducing sugars in blood and urine etc.
All monosaccharides are reducing sugars. Disaccharides which contain a free hemi-acetal group ( 
) are also mild reducing sugars. Most naturally occurring disaccharides are reducing sugars (sucrose is an exception).
) are also mild reducing sugars. Most naturally occurring disaccharides are reducing sugars (sucrose is an exception).
B. Tollen’s test
Tollen’s reagent is ammoniacal solution of silver nitrate. A reducing sugar, reduces silver ion to metallic silver which gets deposited on the inner surface of the test tube in the form of silver mirror. The reaction occurs as follows:
RCHO + 2 [Ag (NH3)2]+ + 2OH– → 2Ag + RCOONH4 + H2O + 3NH3
Material Required
Procedure
A. Fehling’s test
A. Fehling’s test
Mix 1 mL each of Fehling’s solutions A and B in a test tube and add the mixture to test tube B. Heat the content of the test tube on a water bath. The formation of a orange-red precipitate indicates the presence of reducing sugar.
B. Benedict’s test
Add 1 mL of Benedict’s reagent to test tube C and heat the mixture to boiling in a water bath for 2 minutes. The formation of a orangered precipitate due to the formation of copper (I) oxide indicates the presence of reducing sugar.
C. Tollen’s test
Prepare Tollen’s reagent by adding sodium hydroxide solution dropwise to 1 mL aqueous silver nitrate solution to get the precipitate of silver oxide. Now add ammonium hydroxide solution while shaking the mixture so that initially formed silver oxide precipitate dissolves. Add the reagent to the sugar solution contained in a test tube and warm the reaction mixture on a water bath. Formation of silver mirror on the walls of the test tube shows the presence of reducing sugar.
Caution!
Never heat the test tube on direct flame as it may cause explosion.
Never heat the test tube on direct flame as it may cause explosion.
III. Theory of test to distinguish Monosaccharide from Disaccharide
Barfoed’s test
The reagent is cupric acetate in acetic acid solution. It is weakly acidic and is reduced by only monosaccharides. Prolonged boiling may hydrolyse disaccharides and false positive test may be obtained. Monosaccharides react with this reagent within 5 minutes to give a brick red precipitate of copper (I) oxide. Disaccharides take a longer time to react because aldehyde function is masked in the acetal linkage.
The precipitate of cuprous oxide obtained is less dense and its colour is brick-red instead of orange-red.
Material Required
Procedure
Take 10 drops of 1% sugar solution in a test tube and add 1 mL Barfoed’s reagent. Heat the the content of the test tube in a water bath to boiling for five minutes. The formation of orange-red precipitate indicates positive test for monosaccharides. Disaccharides do not give this test.
Test for Sucrose
Hydrolyse the sucrose for performing the test by adding 5 drops of concentrated HCl to 5 mL of 1% sucrose solution and heating the mixture in a boiling water bath. Cool the mixture and add NaOH solution to obtain neutral or slightly alkaline solution. Perform the tests for reducing sugar and Seliwanoff’s test given
below with the hydrolysed product and record your results.
below with the hydrolysed product and record your results.
IV. Test to distinguish Ketose from Aldose
Seliwanoff’s test
Seliwanoff’s test
Ketoses dehydrate very rapidly under acidic conditions to give furfural, which reacts with resorcinol (1,3-dihydroxy benzene) to give a coloured product.
Ketohexoses give red colour and ketopentoses give blue-green colour. Aldoses take longer time to produce colour because under the same conditions, aldoses form furfural slowly, probably because β-elimination is required before dehydration to furfural. Therefore prolonged heating should be avoided.
Material Required
Procedure
Add 2 mL of Seliwanoff’s reagent to 10 drops of 1% sugar solution taken in a test tube. Heat the test tube in boiling water for two minutes. Ketohexoses give red colour. Ketopentose gives blue-green colour. Aldoses do not give colour within two minutes.
Add 2 mL of Seliwanoff’s reagent to 10 drops of 1% sugar solution taken in a test tube. Heat the test tube in boiling water for two minutes. Ketohexoses give red colour. Ketopentose gives blue-green colour. Aldoses do not give colour within two minutes.
V. Theory of test for Polysaccharides (Starch)
Starch gives blue colour with iodine solution due to the formation of a complex known as starch iodide complex. Starch is present in wheat, rice, maize, potatoes etc.
Material Required
Procedure
Iodine test
Iodine test
Make a suspension of (0.5 g) starch in 5 mL water and pour it in 50 mL boiling water to get an aqueous colloidal solution. To this add a few drops of aqueous iodine solution. The appearance of blue colour indicates the presence of starch.
B. Test for Oils and Fats
Theory
These are the esters of glycerol and long chain fatty acids and are known as triglycerides. Triglycerides which are liquids at room temperature are oils and those that are solids are called fats. Oils are of plant origin and fats are of animal origin. Triglycerides in which three acyl groups are same are called simple triglycerides and a triglyceride in which three acyl groups are different is called mixed triglyceride. Many naturally occurring fatty acids contain two or three double bonds. The fats from which these come are called polyunsaturated fats or oils. while oils are glycerides of unsaturated fatty acids. Fats and oils are insoluble in water.
Theory
These are the esters of glycerol and long chain fatty acids and are known as triglycerides. Triglycerides which are liquids at room temperature are oils and those that are solids are called fats. Oils are of plant origin and fats are of animal origin. Triglycerides in which three acyl groups are same are called simple triglycerides and a triglyceride in which three acyl groups are different is called mixed triglyceride. Many naturally occurring fatty acids contain two or three double bonds. The fats from which these come are called polyunsaturated fats or oils. while oils are glycerides of unsaturated fatty acids. Fats and oils are insoluble in water.
On heating with potassium hydrogen sulphate, oils and fats give characteristic odour of acrolein. This is the test for glycerol present either free or combined as an ester. On heating with potassium hydrogen sulphate, glycerol is dehydrated and acrolein is formed which has a pungent odour. The reaction is as follows :
Material Required
Procedure
Add a few crystals (0.5g) of dry potassium hydrogen sulphate to 3 mL of mustard oil/ghee taken in a test tube and heat the content of the test tube gently. A pungent smell confirms the presence of an oil or a fat.
Add a few crystals (0.5g) of dry potassium hydrogen sulphate to 3 mL of mustard oil/ghee taken in a test tube and heat the content of the test tube gently. A pungent smell confirms the presence of an oil or a fat.
C. Tests for Proteins
Theory
Proteins are complex organic compounds containing nitrogen and are made up of amino acids. Proteins are present in egg albumin, soya beans, pulses, fish, milk etc. Their presence can be confirmed by several tests. Due to the presence of characteristic side chains in them, certain amino acids exhibit typical colour reactions that form the basis for their identification. Proteins also respond to the colour reactions of amino acids, but can be distinguished from amino acids by biuret reaction, and coagulation reaction.
Theory
Proteins are complex organic compounds containing nitrogen and are made up of amino acids. Proteins are present in egg albumin, soya beans, pulses, fish, milk etc. Their presence can be confirmed by several tests. Due to the presence of characteristic side chains in them, certain amino acids exhibit typical colour reactions that form the basis for their identification. Proteins also respond to the colour reactions of amino acids, but can be distinguished from amino acids by biuret reaction, and coagulation reaction.
I. Biuret test for peptide bonds
Alkaline copper sulphate reacts with compounds containing two or more peptide bonds to form complexes of violet colour.
Alkaline copper sulphate reacts with compounds containing two or more peptide bonds to form complexes of violet colour.
The name of the test comes from the name of the compound, biuret, which gives this test. The reaction is not absolutely specific for peptide bond because many compounds containing two carbonyl groups linked through nitrogen or carbon atoms give a positive result.
II. Ninhydrin reaction
Ninhydrin is a powerful oxidizing agent and reacts with proteins to give a blue-violet compound called Rhumann’s purple.
Ninhydrin is a powerful oxidizing agent and reacts with proteins to give a blue-violet compound called Rhumann’s purple.
III. Xanthoproteic reaction
Aromatic groups of either the free aminoacid or protein, undergo nitration on heating with concentrated nitric acid. The salts of these derivatives are orange in colour.
Aromatic groups of either the free aminoacid or protein, undergo nitration on heating with concentrated nitric acid. The salts of these derivatives are orange in colour.
Material Required
Procedure
A. Biuret test
A. Biuret test
Prepare 0.5% (w/V) solution of casein or egg albumin in 0.1 M NaOH solution. Take 2-3 mL of the solution and add about 2 mL of 10% sodium hydroxide solution to it. Add a few drops of copper reagent and warm the mixture for about 5 minutes. Appearance of violet colour due to the formation of a complex species of Cu2+ ions with – CONH – group confirms the presence of protein in the sample.
B. Ninhydrin reaction
Take 2-3 mL of an aqueous solution of egg albumin in a test tube. Add 3-4 drops of ninhydrin solution to it and heat. Appearance of blue colour indicates the presence of protein.
C. Xanthoproteic test
Take 1 mL of an aqueous solution of egg albumin in a test tube and add a few drops of concentrated nitric acid. Heat the reaction mixture for a few minutes on a Bunsen flame. A yellow colour appears. Cool the test tube under running tap and add a few drops of 10M sodium hydroxide solution, an orange colour appears.
II. TEST FOR CARBOHYDRATES, FATS AND PROTEINS IN FOODSTUFFS
(i) Take the samples of milk, wheat flour, rice and gram flour powder of legumes to test for the presence of carbohydrates, fats and proteins.
(ii) Take 0.5 mL sample of milk to carry out each of the tests.
(iii) For wheat flour, rice flour, gram flour and legume powder, add 100 mg of the sample in 10 mL of distilled water and boil the suspensions, to get a colloidal solution. Perform the tests with this collidal solution and record the results in Table 11.1.
(ii) Take 0.5 mL sample of milk to carry out each of the tests.
(iii) For wheat flour, rice flour, gram flour and legume powder, add 100 mg of the sample in 10 mL of distilled water and boil the suspensions, to get a colloidal solution. Perform the tests with this collidal solution and record the results in Table 11.1.
Table 11.1 : Test for carbohydrates, fats and proteins in the different samples of food materials
| Sample | CarbohydratesPresent/Absent | FatsPresent/Absent | ProteinPresent/Absent | Milk | Wheat flour | Rice flour | Gram flour/td> | Legumes |
From the experiments, it will be observed that food materials like wheat flour, gram flour and legumes contain carbohydrates and proteins. The rice flour contains carbohydrates, while milk contains fats and proteins. Similarly, other food materials may be tested for the presence of carbohydrates, fats and proteins.
Precautions
(a) Shake the mixture thoroughly while preparing the extract of gram, wheat and rice flour.
(b) Always use fresh reagents to carry out the tests.
(c) Use only required quantities of reagents.
(a) Shake the mixture thoroughly while preparing the extract of gram, wheat and rice flour.
(b) Always use fresh reagents to carry out the tests.
(c) Use only required quantities of reagents.
Discussion Questions
(i) How will you distinguish between sucrose and glucose?
(ii) Explain why does fructose reduce Fehling’s solution and Tollen’s reagent inspite of the presence of ketonic group?
(iii) What is the role of tartarate and citrate ions in Fehling’s reagent and Benedict’s reagent respectively?
(ii) Explain why does fructose reduce Fehling’s solution and Tollen’s reagent inspite of the presence of ketonic group?
(iii) What is the role of tartarate and citrate ions in Fehling’s reagent and Benedict’s reagent respectively?
I. Multiple Choice Questions (Type-I)
1. Glycogen is a branched chain polymer of α-D-glucose units in which chain is formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of C1-C6 glycosidic linkage. Structure of glycogen is similar to __________.
(i) Amylose
(ii) Amylopectin
(iii) Cellulose
(iv) Glucose
(ii) Amylopectin
(iii) Cellulose
(iv) Glucose
2. Which of the following polymer is stored in the liver of animals?
(i) Amylose
(ii) Cellulose
(iii) Amylopectin
(iv) Glycogen
(ii) Cellulose
(iii) Amylopectin
(iv) Glycogen
3. Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives _________.
(i) 2 molecules of glucose
(ii) 2 molecules of glucose + 1 molecule of fructose
(iii) 1 molecule of glucose + 1 molecule of fructose
(iv) 2 molecules of fructose
(ii) 2 molecules of glucose + 1 molecule of fructose
(iii) 1 molecule of glucose + 1 molecule of fructose
(iv) 2 molecules of fructose
4. Which of the following pairs represents anomers?
5. Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet structure. α-helix structure of protein is stabilised by :
(i) Peptide bonds
(ii) van der Waals forces
(iii) Hydrogen bonds
(iv) Dipole-dipole interactions
(ii) van der Waals forces
(iii) Hydrogen bonds
(iv) Dipole-dipole interactions
6. In disaccharides, if the reducing groups of monosaccharides i.e. aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar?
7. Which of the following acids is a vitamin?
(i) Aspartic acid
(ii) Ascorbic acid
(iii) Adipic acid
(iv) Saccharic acid
(ii) Ascorbic acid
(iii) Adipic acid
(iv) Saccharic acid
8. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present?
(i) 5′ and 3′
(ii) 1′ and 5′
(iii) 5′ and 5′
(iv) 3′ and 3′
(ii) 1′ and 5′
(iii) 5′ and 5′
(iv) 3′ and 3′
9. Nucleic acids are the polymers of ______________.
(i) Nucleosides
(ii) Nucleotides
(iii) Bases
(iv) Sugars
(ii) Nucleotides
(iii) Bases
(iv) Sugars
10. Which of the following statements is not true about glucose?
(i) It is an aldohexose.
(ii) On heating with HI it forms n-hexane.
(iii) It is present in furanose form.
(iv) It does not give 2,4-DNP test.
(ii) On heating with HI it forms n-hexane.
(iii) It is present in furanose form.
(iv) It does not give 2,4-DNP test.
11. Each polypeptide in a protein has aminoacids linked with each other in a specific sequence. This sequence of amino acids is said to be ____________.
(i) primary structure of proteins.
(ii) secondary structure of proteins.
(iii) tertiary structure of proteins.
(iv) quaternary structure of proteins.
(ii) secondary structure of proteins.
(iii) tertiary structure of proteins.
(iv) quaternary structure of proteins.
12. DNA and RNA contain four bases each. Which of the following bases is not present in RNA?
(i) Adenine
(ii) Uracil
(iii) Thymine
(iv) Cytosine
(ii) Uracil
(iii) Thymine
(iv) Cytosine
13. Which of the following B group vitamins can be stored in our body?
(i) Vitamin B1
(ii) Vitamin B2
(iii) Vitamin B6
(iv) Vitamin B12
(ii) Vitamin B2
(iii) Vitamin B6
(iv) Vitamin B12
14. Which of the following bases is not present in DNA?
(i) Adenine
(ii) Thymine
(iii) Cytosine
(iv) Uracil
(ii) Thymine
(iii) Cytosine
(iv) Uracil
15. Three cyclic structures of monosaccharides are given below which of these are anomers.
(i) I and II
(ii) II and III
(iii) I and III
(iv) III is anomer of I and II
(ii) II and III
(iii) I and III
(iv) III is anomer of I and II
16. Which of the following reactions of glucose can be explained only by its cyclic structure?
(i) Glucose forms pentaacetate.
(ii) Glucose reacts with hydroxylamine to form an oxime.
(iii) Pentaacetate of glucose does not react with hydroxylamine.
(iv) Glucose is oxidised by nitric acid to gluconic acid.
(ii) Glucose reacts with hydroxylamine to form an oxime.
(iii) Pentaacetate of glucose does not react with hydroxylamine.
(iv) Glucose is oxidised by nitric acid to gluconic acid.
17. Optical rotations of some compounds along with their structures are given below which of them have D configuration.
(i) I, II, III
(ii) II, III
(iii) I, II
(iv) III
(ii) II, III
(iii) I, II
(iv) III
18. Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.
(i) ‘a’ carbon of glucose and ‘a’ carbon of fructose.
(ii) ‘a’ carbon of glucose and ‘e’ carbon of fructose.
(iii) ‘a’ carbon of glucose and ‘b’ carbon of fructose.
(iv) ‘f ’ carbon of glucose and ‘f ’ carbon of fructose.
(ii) ‘a’ carbon of glucose and ‘e’ carbon of fructose.
(iii) ‘a’ carbon of glucose and ‘b’ carbon of fructose.
(iv) ‘f ’ carbon of glucose and ‘f ’ carbon of fructose.
19. Three structures are given below in which two glucose units are linked. Which of these linkages between glucose units are between C1 and C4 and which linkages are between C1 and C6?
(i) (A) is between C1 and C4, (B) and (C) are between C1 and C6
(ii) (A) and (B) are between C1 and C4, (C) is between C1 and C6
(iii) (A) and (C) are between C1 and C4, (B) is between C1 and C6
(iv) (A) and (C) are between C1 and C6, (B) is between C1 and C4
(ii) (A) and (B) are between C1 and C4, (C) is between C1 and C6
(iii) (A) and (C) are between C1 and C4, (B) is between C1 and C6
(iv) (A) and (C) are between C1 and C6, (B) is between C1 and C4
II. Multiple Choice Questions (Type-II)
Note : In the following questions two or more options may be correct.
20. Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a __________.
(i) monosaccharide
(ii) disaccharide
(iii) reducing sugar
(iv) non-reducing sugar
(ii) disaccharide
(iii) reducing sugar
(iv) non-reducing sugar
21. Proteins can be classified into two types on the basis of their molecular shape i.e., fibrous proteins and globular proteins. Examples of globular proteins are :
(i) Insulin
(ii) Keratin
(iii) Albumin
(iv) Myosin
(ii) Keratin
(iii) Albumin
(iv) Myosin
22. Which of the following carbohydrates are branched polymer of glucose?
(i) Amylose
(ii) Amylopectin
(iii) Cellulose
(iv) Glycogen
(ii) Amylopectin
(iii) Cellulose
(iv) Glycogen
23. Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following are acidic?
(iii) H2N—CH2—CH2—CH2—COOH
24.
Lysine,
is _______________.
Lysine,
is _______________.
(i) α-Amino acid
(ii) Basic amino acid
(iii) Amino acid synthesised in body
(iv) β-Amino acid
(ii) Basic amino acid
(iii) Amino acid synthesised in body
(iv) β-Amino acid
25. Which of the following monosaccharides are present as five membered cyclic structure (furanose structure)?
(i) Ribose
(ii) Glucose
(iii) Fructose
(iv) Galactose
(ii) Glucose
(iii) Fructose
(iv) Galactose
26. In fibrous proteins, polypeptide chains are held together by ___________.
(i) van der Waals forces
(ii) disulphide linkage
(iii) electrostatic forces of attraction
(iv) hydrogen bonds
(ii) disulphide linkage
(iii) electrostatic forces of attraction
(iv) hydrogen bonds
27. Which of the following are purine bases?
(i) Guanine
(ii) Adenine
(iii) Thymine
(iv) Uracil
(ii) Adenine
(iii) Thymine
(iv) Uracil
28. Which of the following terms are correct about enzyme?
(i) Proteins
(ii) Dinucleotides
(iii) Nucleic acids
(iv) Biocatalysts
(ii) Dinucleotides
(iii) Nucleic acids
(iv) Biocatalysts
III. Short Answer Type
29. Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called?
30. How do you explain the presence of all the six carbon atoms in glucose in a straight chain?
31. In nucleoside a base is attached at 1′ position of sugar moiety. Nucleotide is formed by linking of phosphoric acid unit to the sugar unit of nucleoside. At which position of sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide?
32. Name the linkage connecting monosaccharide units in polysaccharides.
33. Under what conditions glucose is converted to gluconic and saccharic acid?
34. Monosaccharides contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will
you place fructose?
35. The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicate the correlation of configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has ‘D’ or ‘L’ configuration.
30. How do you explain the presence of all the six carbon atoms in glucose in a straight chain?
31. In nucleoside a base is attached at 1′ position of sugar moiety. Nucleotide is formed by linking of phosphoric acid unit to the sugar unit of nucleoside. At which position of sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide?
32. Name the linkage connecting monosaccharide units in polysaccharides.
33. Under what conditions glucose is converted to gluconic and saccharic acid?
34. Monosaccharides contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will
you place fructose?
35. The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicate the correlation of configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has ‘D’ or ‘L’ configuration.
36. Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration?
37. Which sugar is called invert sugar? Why is it called so?
38. Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of amino group with respect to carboxyl group. Which type of amino acids form polypetide chain in proteins?
37. Which sugar is called invert sugar? Why is it called so?
38. Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of amino group with respect to carboxyl group. Which type of amino acids form polypetide chain in proteins?
39. α-Helix is a secondary structure of proteins formed by twisting of polypeptide chain into right handed screw like structures. Which type of interactions are responsible for making the α-helix structure stable?
40. Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate.
41. During curdling of milk, what happens to sugar present in it?
42. How do you explain the presence of five —OH groups in glucose molecule?
43. Why does compound (A) given below not form an oxime?
40. Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate.
41. During curdling of milk, what happens to sugar present in it?
42. How do you explain the presence of five —OH groups in glucose molecule?
43. Why does compound (A) given below not form an oxime?
44. Why must vitamin C be supplied regularly in diet?
45. Sucrose is dextrorotatory but the mixture obtained after hydrolysis is laevorotatory. Explain.
46. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.
47. Structures of glycine and alanine are given below. Show the peptide linkage in glycylalanine.
45. Sucrose is dextrorotatory but the mixture obtained after hydrolysis is laevorotatory. Explain.
46. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.
47. Structures of glycine and alanine are given below. Show the peptide linkage in glycylalanine.
48. Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to a physical change like change in temperature or a chemical change like, change in pH, denaturation of protein takes place. Explain the cause.
49. Activation energy for the acid catalysed hydrolysis of sucrose is 6.22 kJ mol–1, while the activation energy is only 2.15 kJ mol–1 when hydrolysis is catalysed by the enzyme sucrase. Explain.
50. How do you explain the presence of an aldehydic group in a glucose molecule?
51. Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
52. What are glycosidic linkages? In which type of biomolecules are they present?
53. Which monosaccharide units are present in starch, cellulose and glucose and which linkages link these units?
54. How do enzymes help a substrate to be attacked by the reagent effectively?
55. Describe the term D- and L- configuration used for amino acids with examples.
56. How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions.
57. Coagulation of egg white on boiling is an example of denaturation of protein. Explain it in terms of structural changes.
50. How do you explain the presence of an aldehydic group in a glucose molecule?
51. Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
52. What are glycosidic linkages? In which type of biomolecules are they present?
53. Which monosaccharide units are present in starch, cellulose and glucose and which linkages link these units?
54. How do enzymes help a substrate to be attacked by the reagent effectively?
55. Describe the term D- and L- configuration used for amino acids with examples.
56. How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions.
57. Coagulation of egg white on boiling is an example of denaturation of protein. Explain it in terms of structural changes.
IV. Matching Type
Note : Match the items of Column I and Column II in the following questions.
More than one option in Column II may match with the items given in Column I.
58. Match the vitamins given in Column I with the deficiency disease they cause given in Column II.
| Column I (Vitamins) | Column II (Diseases) | ||
| (i) | Vitamin A | (a) | Pernicious anaemia |
| (ii) | Vitamin B1 | (b) | Increased blood clotting time |
| (iii) | Vitamin B12 | (c) | Xerophthalmia |
| (iv) | Vitamin C | (d) | Rickets |
| (v) | Vitamin D | (e) | Muscular weakness |
| (vi) | Vitamin E | (f) | Night blindness |
| (vii) | Vitamin K | (g) | Beri Beri |
| (h) | Bleeding gums | ||
| (i) | Osteomalacia |
59. Match the following enzyms given in Column I with the reactions they catalyse given in Column II.
| Column I (Enzymes) | Column II (Reactions) | ||
| (i) | Invertase | (a) | Decomposition of urea into NH3 and CO2 |
| (ii) | Maltase | (b) | Conversion of glucose into ethyl alcohol |
| (iii) | Pepsin | (c) | Hydrolysis of maltose into glucose |
| (iv) | Urease | (d) | Hydrolysis of cane sugar |
| (v) | Zymase | (e) | Hydrolysis of proteins into peptides |
V. Assertion and Reason Type
Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(i) Assertion and reason both are correct statements and reason explains the assertion.
(ii) Both assertion and reason are wrong statements.
(iii) Assertion is correct statement and reason is wrong statement.
(iv) Assertion is wrong statement and reason is correct statement.
(v) Assertion and reason both are correct statements but reason does not explain assertion.
(ii) Both assertion and reason are wrong statements.
(iii) Assertion is correct statement and reason is wrong statement.
(iv) Assertion is wrong statement and reason is correct statement.
(v) Assertion and reason both are correct statements but reason does not explain assertion.
60. Assertion : D (+) – Glucose is dextrorotatory in nature.
Reason : ‘D’ represents its dextrorotatory nature.
Reason : ‘D’ represents its dextrorotatory nature.
61. Assertion : Vitamin D can be stored in our body.
Reason : Vitamin D is fat soluble vitamin.
Reason : Vitamin D is fat soluble vitamin.
62. Assertion : β-glycosidic linkage is present in maltose,
Reason : Maltose is composed of two glucose units in which C–1 of one glucose unit is linked to C–4 of another glucose unit.
63. Assertion : All naturally occurring α-aminoacids except glycine are optically active.
Reason : Most naturally occurring amino acids have L-configuration.
Reason : Most naturally occurring amino acids have L-configuration.
64. Assertion : Deoxyribose, C5H10O4 is not a carbohydrate.
Reason : Carbohydrates are hydrates of carbon so compounds which follow Cx(H2O)yformula are carbohydrates.
Reason : Carbohydrates are hydrates of carbon so compounds which follow Cx(H2O)yformula are carbohydrates.
65. Assertion : Glycine must be taken through diet.
Reason : It is an essential amino acid.
Reason : It is an essential amino acid.
66. Assertion : In presence of enzyme, substrate molecule can be attacked by the reagent effectively.
Reason : Active sites of enzymes hold the substrate molecule in a suitable position.
Reason : Active sites of enzymes hold the substrate molecule in a suitable position.
VI. Long Answer Type
67. Write the reactions of D-glucose which can’t be explained by its open-chain structure. How can cyclic structure of glucose explain these reactions?
68. On the basis of which evidences D-glucose was assigned the following structure?
68. On the basis of which evidences D-glucose was assigned the following structure?
69. Carbohydrates are essential for life in both plants and animals. Name the carbohydrates that are used as storage molecules in plants and animals, also name the carbohydrate which is present in wood or in the fibre of cotton cloth.
70. Explain the terms primary and secondary structure of proteins. What is the difference between α-helix and β-pleated sheet structure of proteins?
71. Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule? Draw a diagram to show pairing of nucleotide bases in double helix of DNA.
70. Explain the terms primary and secondary structure of proteins. What is the difference between α-helix and β-pleated sheet structure of proteins?
71. Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule? Draw a diagram to show pairing of nucleotide bases in double helix of DNA.
ANSWERS
I. Multiple Choice Questions (Type-I)
1. (ii) 2. (iv) 3. (iii)
4. (iii), Hint : Cyclic hemiacetal forms of monosaccharide which differ only in the configuration of the hydroxyl group at C1 are anomers.
5. (iii), Hint : In α-helix, hydrogen bonds are present between –NH group of one amino acid residue to the >C== O group of another aminoacid residue.
6. (ii) 7. (ii) 8. (i) 9. (ii) 10. (iii) 11. (i) 12. (iii) 13. (iv) 14. (iv) 15. (i)
16. (iii) 17. (i) 18. (iii) 19. (iii)
4. (iii), Hint : Cyclic hemiacetal forms of monosaccharide which differ only in the configuration of the hydroxyl group at C1 are anomers.
5. (iii), Hint : In α-helix, hydrogen bonds are present between –NH group of one amino acid residue to the >C== O group of another aminoacid residue.
6. (ii) 7. (ii) 8. (i) 9. (ii) 10. (iii) 11. (i) 12. (iii) 13. (iv) 14. (iv) 15. (i)
16. (iii) 17. (i) 18. (iii) 19. (iii)
II. Multiple Choice Questions (Type-II)
20. (ii), (iv) 21. (i), (iii) 22. (ii), (iv) 23. (ii), (iv)
24. (i), (ii) 25. (i), (iii) 26. (ii), (iv) 27. (i), (ii) 28. (i), (iv)
24. (i), (ii) 25. (i), (iii) 26. (ii), (iv) 27. (i), (ii) 28. (i), (iv)
III. Short Answer Type
29. Lactose, two monosaccharide units are present. Such oligosaccharides are called disaccharides.
30. On prolonged heating with HI, glucose gives n-hexane.
30. On prolonged heating with HI, glucose gives n-hexane.
31. Phosphoric acid is linked at 5′-position of sugar moiety of nucleoside to give a nucleotide.
32. Glycosidic linkage.
33. Glucose is converted to gluconic acid by bromine water and to saccharic acid by conc. HNO3.
34. Fructose is a ketohexose.
35. ‘L’ configuration
36. ‘D’ configuration
37. Sucrose, see page no. 409 of NCERT textbook for the explanation.
33. Glucose is converted to gluconic acid by bromine water and to saccharic acid by conc. HNO3.
34. Fructose is a ketohexose.
35. ‘L’ configuration
36. ‘D’ configuration
37. Sucrose, see page no. 409 of NCERT textbook for the explanation.
38. α-amino acids, 
39. In α-helix,a polypeptide chain is stabilised by the formation of hydrogen bonds between —NH— group of amino acids in one turn with the >C== O groups of amino acids belonging to adjacent turn.
40. Oxidoreductase
41. Lactic acid.
42. Glucose gives pentaacetate derivative on acetylation with acetic anhydride. This confirms the presence of five —OH groups.
43. Glucose pentaacetate (structure A) doesn’t have a free —OH group at C1 and so can’t be converted to the open chain form to give —CHO group and hence doesn’t form the oxime.
40. Oxidoreductase
41. Lactic acid.
42. Glucose gives pentaacetate derivative on acetylation with acetic anhydride. This confirms the presence of five —OH groups.
43. Glucose pentaacetate (structure A) doesn’t have a free —OH group at C1 and so can’t be converted to the open chain form to give —CHO group and hence doesn’t form the oxime.
44. Vitamin C is water soluble therefore it is readily excreted in urine and can’t be stored in our body.
45. On hydrolysis sucrose (dextrorotatory), gives glucose (dextrorotatory, + 52.5°) and fructose (laevorotatory, – 92.4°). Since laevorotation of fructose is more than the dextrorotation of glucose, the mixture is laevorotatory.
45. On hydrolysis sucrose (dextrorotatory), gives glucose (dextrorotatory, + 52.5°) and fructose (laevorotatory, – 92.4°). Since laevorotation of fructose is more than the dextrorotation of glucose, the mixture is laevorotatory.
46. In aqueous solution, the carboxyl group loses a proton and amino group accepts a proton to form a zwitter ion.
47. In glycylalanine, carboxyl group of glycine combines with the amino group of alanine.
48. Due to physical or chemical change, hydrogen bonds in proteins are disturbed, globules unfold and helix gets uncoiled therefore protein loses its biological activity. This is called denaturation of proteins.
49. Enzymes, the biocatalysts, reduce the magnitude of activation energy by providing alternative path. In the hydrolysis of sucrose the enzyme sucrase reduces the activation energy from 6.22 kJ mol–1 to 2.15 kJ mol–1.
50. Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin so it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a six carbon carboxylic acid. This indicates that carbonyl group present in glucose is an aldehydic group.
51. See page no. 420 of NCERT textbook.
52. See page no. 409 of NCERT textbook.
53. In starch and glycogen, glycosidic α-linkage is present and in cellulose, glycosidic β-linkage is present between glucose units.
54. Active site of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively.
55. See the NCERT textbook for Class XII.
56. For answer see page no. 406 of NCERT textbook for Class XII.
57. For answer see page no. 416-417 of NCERT textbook for Class XII.
50. Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin so it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a six carbon carboxylic acid. This indicates that carbonyl group present in glucose is an aldehydic group.
51. See page no. 420 of NCERT textbook.
52. See page no. 409 of NCERT textbook.
53. In starch and glycogen, glycosidic α-linkage is present and in cellulose, glycosidic β-linkage is present between glucose units.
54. Active site of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively.
55. See the NCERT textbook for Class XII.
56. For answer see page no. 406 of NCERT textbook for Class XII.
57. For answer see page no. 416-417 of NCERT textbook for Class XII.
IV. Matching Type
58. (i) → (c), (f) (ii) → (g) (iii) → (a) (iv) → (h) (v) → (d), (i) (vi) → (e), (vii) → (b)
59. (i) → (d) (ii) → (c) (iii) → (e) (iv) → (a) (v) → (b)
59. (i) → (d) (ii) → (c) (iii) → (e) (iv) → (a) (v) → (b)
V. Assertion and Reason Type
60. (iii) 61. (i) 62. (iv) 63. (v) 64. (ii) 65. (ii) 66. (i)
No comments:
Post a Comment