Biomolecules

Amylopectin is a branched chain polymer of α-D glucose units. Here, the linear chains are formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of C1-C6 glycosidic linkage. Thus, structure of glycogen is similar to amylopectin. But in reality, they are not same. Amylopectin is the main constituent of starch, main storage polysaccharides in plants, whereas glycogen is the polysaccharides that are stored in plants. Thus, option (ii) is correct.

Option (i) is incorrect because amylose is a long unbranched chain polymer of α-D glucose units in which chain is formed by C1—C4 glycosidic linkage.

Option (iii) is incorrect because cellulose is composed of β-D-Glucose.

Option (iv) is incorrect because glycogen and amylopectin are polymers of α-D-glucose units.

Glycogen is the carbohydrates that are stored in the animal body and therefore, they are also called animal starch. They are mainly present in liver, muscles and brain. Thus, option (iv) is correct.

Options (i), (ii) and (iii) are incorrect because they are mainly found in plants

Sucrose (cane sugar) is a disaccharide that means sucrose on hydrolysis produce two monosaccharide molecules.

Sucrose on hydrolysis produces equimolar mixture of D-glucose and D-fructose. These two monosaccharaides are held together by a glycosidic linkage between C1 of α-D-glucose and C2 of β-D-fructose Thus, option (iii) is correct.

Options (i), (ii) and (iv) are incorrect because sucrose on hydrolysis produces 1 molecule of D-glucose and 1 molecule of D-fructose.

Anomers are cyclic monosaccharides that differ in configuration only at the acetal or hemiacetal carbon or at the anomeric carbon atoms.

The cyclic monosaccharides given in option (iii) differ in configuration only at C1 carbon atoms. Thus, option (iii) is correct.

Options (i) and (ii) are incorrect because the pairs of monosaccharides given in these options are not a cyclic monosaccharides.

Option (iv) is incorrect because they are enantiomers, i.e., mirror image of one another.

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 stabilized by Hydrogen bonds. This hydrogen bonds are formed between backbone N-H group of an amino acid with the backbone C=O group of every fourth amino acid in the peptide bond. Thus, option (iii) is correct.

Option (i) is incorrect because peptide bonds are present between the amino acids.

Options (ii) and (iv) are incorrect because α-helix structure of protein is mainly stabilized by Hydrogen bonds and not by van der Waals forces or Dipole-dipole interactions.

Reducing sugars are sugars that are capable of acting as a reducing agent because they have a free aldehyde or a free ketone group.

Non-reducing sugars are sugars that are not capable of acting as a reducing agent because they don’t have a free aldehyde or a free ketone group.

The sugar given in option (ii) is a non-reducing sugar because the reducing group of glucose and fructose are involved in the glycosidic linkage. Thus, this sugar lacks a free aldehyde or a free ketone group. In other words, the reducing groups of monosaccharides are involved in glycosidic bond in this disaccharide. Hence, option (ii) is correct.

Options (i), (iii) and (iv) are incorrect because they are reducing sugars as they have a free aldehyde or a free ketone group.

Ascorbic acid is also called as vitamin C. They are found in fish, egg yolks and are synthesized in our body in presence of the sunlight. Thus, option (ii) is correct.

Option (i) is incorrect because aspartic acid is an amino acid.

Option (iii) is incorrect because adipic acid is a dicarboxylic acid.

Option (iv) is incorrect because saccharic acid is also a dicarboxylic acid.

Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. These linkages are present between 5’ and 3’ carbon atoms of pentose sugars of nucleotides. The 5’ carbon atom of pentose sugar in one nucleotide is attached to the 3’ carbon atom of pentose sugar of the second nucleotide. Thus, option (i) is correct.

Option (ii) is incorrect because 1’ carbon atom of pentose sugar of nucleotides is attached to the nitrogen bases.

Options (iii) and (iv) are incorrect because phosphodiester linkages are present between 5’ and 3’ carbon atoms of pentose sugars of nucleotides.

Nucleic acids are the polymers of nucleotides and they are joined together by a phosphodiester linkage. Nucleotides are the structural units of nucleic acids and are composed of a nitrogenous base, a five carbon sugar and a phosphate group. Thus, option (ii) is correct.

Option (i) is incorrect because nucleosides are composed of a nitrogenous base and a five carbon sugar. They don’t have the phosphate groups. So, they don’t have a phosphodiester linkage.

Options (iii) and (iv) are incorrect because nucleic acids are the polymers of nucleotides and not just nitrogenous bases or pentose sugars.

Glucose is present in pyranose form, i.e., five-membered ring structure. Thus, option (iii) is correct because the given statement is incorrect.

Glucose is a monosaccharide that is composed of six carbon atoms with an aldehyde group. Thus, it is an aldohexose. So, this statement is true and therefore, option (i) is incorrect.

Option (ii) is incorrect because the given statement is true. On heating glucose with HI, it forms n-hexane. This reaction suggests that all the six carbon atoms in glucose are linked in a straight chain.

Option (iv) is incorrect because the given statement is true. The aldehyde group is not free in the cyclic structure. So, it does not give 2, 4-DNP test.

Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be primary structure of proteins. Here, amino acids are joined together by a bond called peptide bond or peptide linkage. Thus, option (i) is correct.

Option (ii) is incorrect because primary structure of proteins is then coiled into secondary structure by the hydrogen bonding in the backbone of the peptides.

Option (iii) is incorrect because tertiary structure of proteins represents the overall folding of the polypeptide chains.

Option (iv) is incorrect because quaternary structure of proteins are composed of two or more polypeptide chains.

DNA and RNA contain four bases each. DNA contains four nitrogenous bases, namely Adenine (A), Guanine (G), Cytosine (C) and Thymine (T).

The nitrogenous bases in RNA are Adenine (A), Guanine (G), Cytosine (C) and Uracil (U). Thymine is not present in RNA. Thus, option (iii) is correct.

Options (i), (ii) and (iv) are incorrect because these nitrogenous bases are there in RNA.

B group vitamins are water-soluble vitamins. They are readily excreted out of the body and therefore, B group vitamins cannot be stored in our body but our body can store Vitamin B₁₂ because they are insoluble in water. Thus, option (iv) is correct.

Options (i), (ii) and (iii) are incorrect because our body cannot store these vitamins.

DNA and RNA contain four bases each. The nitrogenous bases in RNA are Adenine (A), Guanine (G), Cytosine (C) and Uracil (U).

DNA contains four nitrogenous bases, namely Adenine (A), Guanine (G), Cytosine (C) and Thymine (T). Uracil is not present in DNA. Thus, option (iv) is correct.

Options(i), (ii) and (iii) are incorrect because these nitrogenous bases are there in DNA.

Anomers are cyclic monosaccharides that differ in configuration only at the acetal or hemiacetal carbon or at the anomeric carbon atoms.

The cyclic monosaccharides given in I and II differ in configuration only at C1 atom. Therefore, they are anomers. Thus, option (i) is correct.

Options (ii) and (iii) are incorrect because the numbers of carbon atoms in both the cyclic compounds in each of the options are different. They differ from each other at C1 position Thus, they are not anomers.

Option (iv) is incorrect because III is not an anomer of I and II. They differ in the number of carbon atoms and they differ from each other at C1 position.

The pentaacetate of glucose does not react with hydroxylamine. This indicates that the aldehyde in glucose is not free and is involved in the formation of cyclic structure. Hence, this reaction of glucose can only be explained by its cyclic structure. Thus, option (iii) is correct.

Option (i) is incorrect because the capability of glucose to form pentaacetate tells us about the 5 hydroxyl group of the glucose.

Option (ii) is incorrect because this reaction confirms the presence of a carbonyl group in glucose.

Option (iv) is incorrect because this reaction indicates the presence of a primary alcohol in the glucose.

All the three monosaccharides have D configuration because the hydroxyl group on the lowest asymmetric carbon is on the right hand side. Thus, option (i) is correct.

Options (ii), (iii) and (iv) are incorrect because all the three monosaccharides have D configuration.

Anomers are cyclic monosaccharides that differ in configuration only at the acetal or hemiacetal carbon or at the anomeric carbon atoms.

Here, the anmeric carbon atoms are: ‘a’ carbon of glucose and ‘b’ carbon of fructose.

This disaccharide on hydrolysis yields α-glucose and β-fructose.

Thus, option (iii) is correct.

Options (i), (ii) and (iv) are incorrect because ‘a’ carbon of glucose and ‘b’ carbon of fructose are the anomeric carbon atoms.

The linkages between C1 and C4 of glucose are represented in the picture as ‘A’ and ‘C’ and ‘B’ is the linkage between C1 and C6 of the glucose units. Also, upon looking at the structures it is clear that option (iii) is correct.

Option (ii) is incorrect because ‘B’ is the linkage between C1 and C6 of the glucose units.

Option (i) is incorrect because ‘C’ is the linkage between C1 and C4 of the glucose units.

Option (iv) is incorrect because ‘A’ is the linkage between C1 and C4 of the glucose units.

Sucrose yields two monosaccharides on hydrolysis, glucose and fructose. So Sucrose is a disaccharide. So option (ii) is correct.

Monosaccharide contains only one sugar ring. Sucrose contains two monosaccharide glucose and fructose. Therefore, option (i) is wrong.

A reducing sugar has free ketone or aldehyde groups and therefore contain a hemiacetal instead of an acetal. Sucrose doesn’t have a hemiacetal group. So it behaves as a non-reducing sugar. Therefore option (iv) is correct.

Reducing sugar contains hemiacetal instead of acetal group. So option (iii) is wrong.

(Sucrose)

Two monosaccharides α-D-glucose and ß-D-fructose are joined with each other to form disaccharide sucrose.

In globular proteins, polypeptide chains are coiled with each other to give spherical shape and they are soluble in water. Insulin and Albumin have a spherical shape. So options (i) and (iii) are correct.

But Myosin has long and narrow shape and they are water-insoluble, which is the characteristic of a fibrous protein. Therefore, option (iv) is wrong.

Amylose is an unbranched linear polymer having 200-1000 α-d-glucose units of glucose joined by C1-C4 glycosidic linkage. So option (i) is wrong.

Cellulose is comprised of ß-D-glucose units as monomer joined through glycosidic linkage to form a straight chain. So option (iii) is wrong.

Amylopectin is a branched polymer of α-d-glucose where branched-chain is formed by C1-C6 glycosidic linkage. So (ii) is correct.

Glycogen is more highly branched polysaccharide of glucose similar to Amylopectin. Thus, option (iv) is also correct.

Amino acids in which the number of –COOH and –NH2 groups are equal are called neutral amino acids. In (i) and (iii) each contains an equal number of -COOH and –NH2 group. So both (i) and (iii) are incorrect.

Acidic amino acids are those in which the number of –COOH groups are more than –NH2 groups.

In both amino acids (ii) and (iv), number of –COOH groups and –NH2 group is 2 and 1 respectively. So they are acidic amino acids.Therefore options (ii) and (iv) are correct.

α-C atom

(Lysine)

In Lysine, the amino group –NH2 is attached with an alpha Carbon atom to –COOH group. So this is an alpha(α) amino acid. So (i) is correct and (iv) is wrong.

Lysine contains a greater number of –NH2 groups than –COOH groups. So it is a basic amino acid. Therefore,(ii) is a basic amino acid.

Lysine is not synthesized in our body, it has to be supplied from outside food sources. So it is an essential amino acid. Therefore (iii) is also not correct.

Both monosaccharides Ribose and Fructose have a cyclic structure. They both contain a five-membered ring which is polyhydroxy carbonyl compounds.

So both option (i) and (iii) are correct.

Glucose and galactose both are monosaccharides containing the six-membered ring. So (ii) and (iv) are not correct.

In fibrous proteins, polypeptide chains are held together by disulphide and hydrogen bond in a parallel manner, due to which fibrous like structure is obtained. So options (ii) and (iv) are correct.

Vander Waals forces and electrostatic forces of attraction are not present in the fibrous protein. Therefore, (i) and (iii) are wrong.

Purine is a heterocyclic aromatic organic compound, consists of a pyrimidine ring fused to an imidazole ring.

Therefore, (i) and (ii) are the correct options.

Pyrimidine is also a heterocyclic aromatic compound. This contains two nitrogen atoms similar to benzene. Uracil and Thymine have the following structures.

Thus, (ii) and (iv) are incorrect options.

Enzymes are mainly globular proteins containing 100-1000 different amino acids. They act as biocatalysts in a living organism and highly specific. They catalyse various chemical reactions in our body. For example, enzyme Trypsin breaks proteins into amino acids. So (i) and (iv) are correct.

Nucleic acids are long-chain polymers of nucleotides. Nucleotides consist of the sugar ring, base and phosphodiester linkage. So their structures are not similar to proteins.

Therefore, options (ii) and (iii) are wrong.

The sugar present in milk is lactose. Lactose contains two monosaccharides, glucose and galactose.

Oligosaccharides containing two monosaccharide units are called disaccharides.

When glucose is heated for a prolonged time with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.

A nucleoside is formed when a nitrogenous base is attached to a 1’ position of a five carbon sugar. Phosphoric acid is linked to the 5’ carbon of the sugar in a nucleoside molecule to give a nucleotide molecule.

The monosaccharide units in polysaccharides are linked by glycosidic bonds. A glycosidic linkage is when an oxide linkage is formed between two monosaccharide units with the loss of a water molecule.

Glucose is converted to gluconic acid which is a six carbon carboxylic acid, on treatment with a mild oxidizing agent like Br2 water.

Glucose is converted to saccharic acid, which is a dicarboxylic acid, on treatment with nitric acid.

When monosaccharides contain an aldehyde group, they are known as aldoses. When containing a ketone group, they are known as ketoses. Monosaccharides are also classified depending on the number of carbon atoms the molecule contains. The molecular formula of fructose is C6H12O6. It also contains a ketone group, hence it is placed in the class of ketohexoses. The structure of fructose is given below.

The –OH group is linked on the left side of C5 carbon atom. Hence, the given compound has ‘L’ configuration.

Aldopentoses like ribose and 2-deoxyribose are the sugar moieties in nucleic acids like RNA and DNA respectively. The structures of these sugars are given below. The configuration of both the aldopentoses is D-configuration. Ribose is named β-D-ribose and 2-deoxyribose is β-D-2-deoxyribose.

Sucrose is also known as invert sugar.

It is derived from sugarcane and sugar beet naturally. Aqueous solution of sucrose is dextrorotatory and rotates plane polarized light entering the solution 66.5° to the right. When sucrose is hydrolysed with dilute acids or invertase enzyme, it gives two products in equimolar concentration, dextrorotatory D-(+)-glucose and laevorotatory D-(-)-fructose. The laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the resulting equimolar mixture of the products is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–) and thus the product is named as invert sugar.

α-amino acids, alpha-amino acids, where the amino acid is linked to the α-carbon in the molecule are the type of amino acids which form a polypeptide chain. α-amino acids are obtained on hydrolysis of proteins. β-, - , δ- are non-proteinogenic amino acids, and are classified according to the carbon group to which an amino group is attached to. They are called non-proteinogenic as they do not form peptide chains, as the different positions of the –NH2 and the –COOH group do not allow the formation of a fixed polypeptide structure.

A stable α-Helix is formed as a right handed screw helix structure as the –NH group of each amino acid residue hydrogen is bonded to the –C=O of an adjacent turn of the helix.

Structure of stable α-Helix. Image from Wikipedia using Wikipedia Commons License.

The name given to the class of enzymes which catalyse redox reactions are known as enzyme oxidoreductases. Examples of oxidoreductases include Alcohol Dehydrogenase, which helps in reducing alcohol levels in human body when alcohol is ingested.

During curdling of milk, which is caused due to bacteria, sugar present in milk, lactose, is converted to lactic acid.

Molecular structure of lactose and lactic acid. Both images obtained from Wikipedia under Creative Commons license.

When glucose is treated with acetic anhydride (CH3CO)2O, in the presence of ZnCl2, it undergoes acetylation to form glucose pentaacetate which confirms the presence of five –OH groups.

The given compound is glucose pentaacetate. Glucose contains a free –C=O group, and formation of oxime from glucose confirms presence of free carbonyl group. The given compound does not have a free carbonyl group and thus does not form an oxime on reaction with hydroxylamine.

Vitamin C is a water soluble vitamin, and hence the excess is excreted regularly from the body. Since it cannot be stored in the body, vitamin C must be supplied regularly in diet.

Aqueous solution of sucrose is dextrorotatory and rotates plane polarized light entering the solution 66.5° to the right. When sucrose is hydrolysed with dilute acids or invertase enzyme, it gives two products in equimolar concentration, dextrorotatory D-(+)-glucose and laevorotatory D-(-)-fructose. The laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the resulting equimolar mixture of the products is laevorotatory with resulting rotation of -39.9°. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–). Thus the hydrolysed mixture is laevorotatory.

An amino acid contains an –NH2 group as well as –COOH. In aqueous solution of the amino acid, the –COOH group loses a proton [H]+ and the –NH2 gains a proton to form a zwitter ion which is a salt.

The hydroxyl group of glycine is linked to amine group of alanine by peptide (-CONH) linkage to form glycylalanine.

The amino acid residues of proteins are joined by hydrogen bonds and various other intermolecular forces. On physical or chemical change, the hydrogen bonds are disturbed and native protein unfolds. This process is called denaturation where secondary and tertiary structures are destroyed but primary structure remains intact.

Enzymes are biocatalysts, which in small quantities provide an alternate path to reduce the activation energy of the reaction. Using enzyme sucrase, the hydrolysis of sucrose is much faster than conventional acidic hydrolysis.

The presence of a carbonyl group in a glucose molecule can be confirmed by treating glucose with hydroxylamine to give a monoxime. It can also be treated with hydrogen cyanide to give cyanohydrin. To deduce that the carbonyl group is an aldehyde, glucose can be treated with bromine water, which undergoes mild oxidation to give the carboxylic acid gluconic acid, which confirms the presence of an aldehyde group.

Nucleosides are linked to phosphoric acid at 5’ carbon of the sugar moiety to form a nucleotide. To form a dinucleotide, two nucleotides are linked by phosphodiester linkage between the 5’ and 3’ carbon atom of adjacent pentose sugars. Phosphoric acid is involved in the formation of this linkage.

When two monosaccharides are linked to each other with a bond oxide linkage formed by the loss of a water molecule, the bond is a glycosidic linkage.

They are present in disaccharides, oligosaccharides and polysaccharides.

In starch, α-glucose monosaccharides are present, while in cellulose, β-D-glucose monosaccharides are present. In starch and glycogen, α-glycosidic linkages link the units, while in cellulose, β-glycosidic bonds link the units.

Starch Cellulose Glucose

α-glucose monosaccharides are present β-D-glucose monosaccharides are present. Glucose itself is a monosaccharide.

α-glycosidic linkages link the units connecting the C1 of one unit and C4 of the next unit. β-glycosidic bonds link the units connecting the C1 of one unit and C4 of the next unit.

The active sites of enzymes hold the substrate molecule in a position suitable for the reagent to attack and form product.

Every amino acid except glycine occurs in D- and L- forms. Most naturally occurring amino acids are present in L- forms. D/L configuration is determined by analogy with the structure of glyceraldehyde. One example of an amino acid having D and L isomers is the chiral amino acid alanine, which has two optical isomers, and they are labeled according to which isomer of glyceraldehyde they come from.

Image from Libretexts Chemistry under Creative Commons License.

When glucose is treated with acetic anhydride (CH3CO)2O in the presence of pyridine or few drops of conc. H2SO4, it undergoes oxidation to give glucose penta acetate, indicating the presence of five –OH groups, in which one group is a primary 1° –OH group and four groups are secondary 2° –OH groups. When glucose is oxidized with concentrated HNO3, it gets oxidized to form saccharic acid, a dicarboxylic acid. This reaction proves that primary –OH group gets oxidized to –COOH but only in the presence of strong acids, the secondary –OH group gets oxidized.

The contents of an egg are of two types, the egg yolk (yellow) and the albumin (white). Both of these components are globular proteins in their native form. When egg whites are boiled, they are denatured, resulting into coagulated white protein insoluble in water. In denaturation of protein, the secondary and tertiary structures of egg albumin are converted to primary structure, destroying all the hydrogen bonds that held the structure. The globular protein gets converted into insoluble fibrous protein and the biological activity is lost.

(i) Vitamin A – (c) Xerophthalmia , (f) Night blindness

Vitamin A plays an important role in our vision. Deficiency of vitamin A leads to various disorders like xerophthalmia which is a medical condition where eyes fail to produce tears and it can develop into night blindness which can damage the cornea.

(ii) Vitamin B₁ – (g) Beri Beri

Beri Beri is a deficiency disease which is caused due to the deficiency of vitamin B1. It is a condition which features problems related to the peripheral nerves and it can even lead to paralysis.

(iii) Vitamin B₁₂ – (a) Pernicious anaemia

Pernicious anaemia is a deficiency disease caused due to the deficiency of vitamin B₁₂. It is a condition where the body cannot produce enough red blood cells due to the absence of vitamin B₁₂. As a result of which, due to the lack of oxygen, we feel tired and get fatigued easily.

(iv) Vitamin C – (h) Bleeding gums

Bleeding gums are caused due to the absence of vitamin C in our diet. The condition is known as scurvy. Along with swollen and bleeding gums, it can also lead to loss of teeth.

(v) Vitamin D – (i) Osteomalacia , (d) Rickets

The deficiency of vitamin D leads to muscle weakness and bone pain. The softening of bones, due to the deficiency of calcium and phosphorus in our body leads to a condition known as osteomalacia or rickets.

(vi) Vitamin E – (e) Muscular weakness

Deficiency of vitamin E leads to problems related to nervous system along with disorientation and vision problems with muscular weakness.

(vii) Vitamin K – (b) Increased blood clotting time

Vitamin K plays a very important role in helping the blood clot during any injury and prevents excess blood loss and its deficiency leads to increase in the time required for blood to clot.

(i) Invertase – (d) Hydrolysis of cane sugar

Invertase is an enzyme which catalyzes the hydrolysis of cane sugar (sucrose) into fructose and glucose.

(ii) Maltase – (c) Hydrolysis of maltose into glucose

As we get to know from the name, maltase is an enzyme which catalyzes the hydrolysis of maltose to glucose.

(iii) Pepsin – (e) Hydrolysis of proteins into peptides

Pepsin is one of the main digestive enzymes and it helps in digestion of the proteins in food. It breaks down proteins into smaller peptides.

(iv) Urease – (a) Decomposition of urea into NH₃ and CO₂.

Urease is an enzyme which catalyzes the hydrolysis of urea into ammonia and carbon dioxide. It is found in numerous bacteria, fungi, plant, etc.

(v) Zymase – (b) Conversion of glucose into ethyl alcohol

Zymase is an enzyme which catalyzes the fermentation of glucose into ethyl alcohol and carbon dioxide. This process occurs in yeast.

The D-L system corresponds to the configuration of the molecule and the spatial arrangement of its atoms around the chirality center. The D-L configuration is not to be confused with the d- l- naming which means dextrorotatory and levorotatory respectively. (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane polarized light clockwise (+) or counterclockwise (-). So, D(+) – glucose is dextrorotatory in nature but ‘D’ does not represent its dextrorotatory nature.

Vitamins which can dissolve in fats and oils are known as fat soluble vitamins. They are absorbed along with the fat and can be stored in our body. Eg. Vitamin A,D,E,K.

We find α-glycosidic linkage present in maltose.

In maltose, two glucose units are joined by a glycosidic linkage between the α-anomeric form of C-1 on one glucose unit and the hydroxyl oxygen atom on C-4 of the adjacent glucose unit. Such a linkage is called α-1,4-glycosidic bond.

All naturally occurring α-amino acids except glycine are optically active. This is because glycine does not have all four different substituent which means that the carbon atom is achiral.

Glycine

A carbohydrate is a biomolecule consisting of carbon, oxygen and hydrogen atoms. Deoxyribose is therefore a 5 carbon carbohydrate molecule that forms the phosphate backbone of DNA molecule. Carbohydrates, which are polyhydroxy aldehyde or polyhydroxy ketone, on hydrolysis, yields these simple molecules.

Glycine is a non-essential amino acid. Due to which, it is not required to take them in our diet because they are produced in our body. Whereas, essential amino acids are those which our human body cannot produce so they need to come from food. Eg. Histidine, leucine, lysine, etc.

Enzymes hold the active sites of the substrate molecule in such a way so that the reagent can attack the substrate site effectively. Stereospecificity is one of the main properties of enzyme catalyzed reactions.

The open chain structure if glucose could explain most of its properties but failed to explain the following:-

• In spite of the presence of an aldehyde group, glucose does not restore the pink colour of Schiff’s reagent, does not give 2,4-DNP test and also it does not form any addition product with sodium hydrogen sulphite and ammonia.

• Glucose pentaacetate does not react with hydroxyl amine thus it indicates the absence of free –CHO group.

• An aqueous solution of glucose shows mutorotation, which means that the specific rotation decreases from +110° to +52.5° for α-glucose and increases from +19.7° to +52.5° for β-glucose.

• Glucose exists in two crystalline forms alpha(α) and beta(β). The α-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K.

The above facts of glucose can be explained in terms of cyclic structure of glucose which is formed through intramolecular hemiacetal formation which leads to cyclization.

Glucose

The reactions of glucose suggest that its molecule contains one primary (-CH₂OH) and four secondary (-C-HOH) hydroxyl groups. The evidences that support the following structure of glucose are:-

• Molecular formula of glucose is C₆H₁₂O₆.

• Reduction - Its reduction to sorbitol in the presence of Ni or sodium amalgam as catalyst.

• Reaction with HI

• Acetylation – A pentaacetyl derivative of glucose is obtained when glucose reacts with acetic anhydride in presence of sulphuric acid. As we know that the presence of two or more hydroxyl groups present on same carbon atom makes the molecule unstable, but since glucose is stable, we conclude that the 5-OH are present on different carbon atoms.

• Oxidation - Glucose on oxidation with mild oxidizing agents like Tollen’s reagent or Fehling’s solution give gluconic acid

On the basis of all these above evidences, the structure of glucose is supported.

Carbohydrates are essential for life in both plants and animals. They are used as storage cells in plants and animals are :-

• In plants, we find mainly starch, cellulose, sucrose, etc.

• Glycogen, also known as animal starch is found inside the bodies of animals. It is present in liver, muscles, brain and when body needs glucose, enzymes break down glycogen to glucose.

• In wood, or in fibre of cotton cloth, we find cellulose.

Primary structure of proteins refer to the sequence in which the amino acids are arranged in them. Proteins have specific sequence of amino acids which are essential for a particular biological activity. The change in even one amino acid can alter the biological activity.

Secondary structures of protein refer to the arrangement of the polypeptide chain giving rise to a particular shape, which arises due to hydrogen bonding.

The two common secondary structures of proteins are α-helix and β-pleated sheet structure.

An α helix is a right-handed helix that is held together by hydrogen bonding. In this structure the hydrogen bond is formed between the N-H on one amino acid and the C=O on another amino acid 4 residues away.

The β sheet is formed when beta strands are linked together by hydrogen bonds, forming a pleated sheet of amino acid residues. Again, the hydrogen bonds are between the N-H group of one amino acid and the C=O group of another.

Structures of α-helix and β-pleated sheet.

The complete hydrolysis of DNA molecule yields a pentose sugar, phosphoric acid and nitrogen containing heterocyclic biological compound called bases.

Ribose

Deoxyribose

Phosphoric acid

The four nitrogenous bases produced are – Adenine(A), Guanine(G), Cytosine(C), Thymine(T).

Adenine

Guanine

Cytosine

Thymine

Adenine-Thymine pairing

Guanine-Cytosine pairing

Pairing of nucleotides in a DNA molecule

Double helical structure of DNA