Learning Outcomes

The module will give students the fundamental knowledge about biomolecules including an understanding:-

• to recognize the structures of different classes of carbohydrates, bonding between them, their properties and biological importance.

• to recognize the structure of 20 different amino acids, peptide linkage, polypeptide i.e. protein, classification of proteins based upon their function, structure and significance.

• of enzymes as biological catalysts, their structural features, characteristics and biological role.

• to identify simple units present in nucleic acids, structure of a polynucleotide, comparison of DNA and RNA, their functions.

• of vitamins, their types and functions.

The present module will prepare students for more advanced studies in biochemistry and molecular biology.

Biomolecules

Living beings contain a wide variety of organic macromolecules such as carbohydrates, proteins, enzymes, lipids and nucleic acids. These organic molecules which interact with each other and constitute the molecular logic of life processes, are known as biomolecules.

1. Carbohydrates: Chemically, carbohydrates are defined as optically active polyhydroxy aldehyde or ketone or the compounds which produce such units on hydrolysis. Carbohydrates form a very large group of naturally occurring organic compounds eg. cane sugar, starch, glucose etc. These are also known as saccharides; some of them are sweet to taste and hence are also called sugars.

Importance of Carbohydrates:

• Essential for life in both animals and plants.

• They are stored in plants as starch and in animals as glycogen.

• Cell wall of plants and bacteria, cotton fibre, furniture and many other wood items are made up of cellulose.

• Most carbohydrates (of food) are converted into glucose to release energy. Glycolysis is the metabolic pathway through which glucose is broken down to pyruvate to provide energy. Under aerobic conditions (in presence of $\mathrm{O} _{2}$ ) glucose is completely broken down to $\mathrm{CO} _{2} & \mathrm{H} _{2} \mathrm{O}$ and releases lots of energy in the form of ATP, called cellular respiration. In absence of oxygen, glucose may form alcohol and $\mathrm{CO} _{2}$, called alcoholic fermentation or Lactic acid and $\mathrm{CO} _{2}$, called lactic acid fermentation with release of lesser amount of energy.

1.1 Mono Saccharides: Simplest type that can not be hydrolysed. Water soluble and crystalline.

Different types of monosaccharides

1.1.2 Glucose (Dextrose or grape sugar) : It is an aldose sugar, occurs in nature in free as well as in combined forms. It is prepared by acid hydrolysis of sucrose and starch :

(a) $\underset{\text { sucrose }}{\mathrm{C} _{12} \mathrm{H} _{22} \mathrm{O} _{11}}+\mathrm{H} _{2} \mathrm{O} \xrightarrow[\Delta]{\mathrm{H}^{+}} \underset{\text { glucose }}{\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}}+\underset{\text { fructose }}{\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}}$

(b) $\underset{\text { starch }}{\left(\mathrm{C} _{6} \mathrm{H} _{10} \mathrm{O} _{5}\right) _{n}}+\mathrm{nH} _{2} \mathrm{O} \xrightarrow[\text { 393 K, 2-3 atm }]{\mathrm{H}^{+}} \underset{\text { glucose }}{\mathrm{nC} _{6} \mathrm{H} _{12} \mathrm{O} _{6}}$

Naturally occuring glucose is dextrorotatory and it belongs to ’ $\mathrm{D}$ ’ family

Structure of glucose : Two types

1 Open chain and

2 Cyclic

Observations in favour of open chain structure:

$\rightarrow$ Its molecular formula is $\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}$; proved by elemental analysis & molecular weight determinations.

The exact spatial arrangement of $5-\mathrm{OH}$ groups in glucose was given by Fischer, after studing many other properties of it. The projection formula of glucose:

According to this structure, glucose has 4 chiral carbons $(2,3,4,5)$. Therefore, having $2^{4}=16$ stereo isomers. Glucose is correctly named as $\mathrm{D}(+)$-glucose. ’ $\mathrm{D}$ ’ refers to the configuration whereas ’ $(+)$ ’ represents dextrorotatory nature of the molecule.

The $D$ and $L$ designations of sugars is based upon the configuration of $D$ and $L$ glyceraldehydes. $D$ refers to an arrangement about a chiral carbon that is identical to the $3 \mathrm{D}$ arrangement of atoms in $\mathrm{D}(+)$-glyceraldehyde, in which - $\mathrm{OH}$ group is present on right of the chiral centre in its fischer projection. Similarly, L-refers to the arrangement around a chiral carbon, identical to the L(-)glyceraldehyde. For assigning the configuration, the lowest asymmetric carbon (or carbon farthest to the carbonyl group) in the fischer projection of the compound / sugar, is compared.

Objections against open chain structure :

a) Although glucose has an aldehydic group, it does not form sodium bisulphite adduct with $\mathrm{NaHSO} _{3}$ and hydrazone with 2, 4-dinitrophenyl hydrazine. It does not give reaction with Schiff’s reagent.

b) Its penta acetate does not react with hydroxylamine.

c) The existence of glucose in $\alpha$ and $\beta$ - anomeric forms can not be explained by open chain structure.

Mutarotation : The specific rotation of a freshly prepared $\alpha-D$-glucose gradually decreases from $+112^{\circ}$ to $+52.5^{\circ}$ while that of $\beta-D$-glucose increases from +19.2 to $+52.5^{\circ}$ with time. This phenomenon is known as mutarotation. All reducing sugars, monosaccharides and disaccharides undergo mutarotation in aqueous solution. To understand this phenomenon we need to know the cyclic structure of glucose.

Cyclic structure of glucose: The limitations shown by the open chain structure were overcome by the cyclic structure. It was proposed that $-\mathrm{OH}$ group at $\mathrm{C}-5$ can add to $-\mathrm{CHO}$ group to give a cyclic hemiacetal. This explains the absence of $-\mathrm{CHO}$ group and existence of glucose in $\alpha$ and $\beta$ - anomeric forms

This six membered cyclic structure is known as pyranose ( $\alpha$ or $\beta$ ), due to its analogy with the structure of pyran;

The two cyclic hemiacetal forms of glucose $(\alpha & \beta)$ are called anomers, as these differ only in the configuration of $-\mathrm{OH}$ group at $\mathrm{C}-1$ (anomeric carbon). In the aqueous solution, these two forms are in equilibrium through the open chain form i.e.

$\underset{(36 \%)}{\alpha\text{-D-glucose}} \rightleftharpoons \underset{(<0.5 \%)}{\text{openchainform}} \rightleftharpoons \underset{(64 \%)}{\beta\text{-D-glucose}}$

Due to presence of open chain form, glucose has a free - $\mathrm{CHO}$ group and hence forms an oxime (reaction of $-\mathrm{CHO}$ group).

Cyclic structure of glucose is represented by Haworth projection:

1.1.3 Fructose and its structure : It is a ketohexose and is obtained by hydrolysis of sucrose. Its molecular formula is found to be $\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}$. Its open chain structure is established as :

Naturally occurring fructose is levorotatory and belongs to D-family. Fructose also exits in two cyclic forms like glucose via five membered furanose ring with an analogy to furan ring;

1.2 Disaccharides : These are composed of two similar or different units of monosaccharides, joined together by an oxide linkage known as glycosidic linkage. Upon hydrolysis these form respective monosaccharide units.

eg. Sucrose: $\xrightarrow{\mathrm{H}^{+}} \mathrm{D}-(+)-$ glucose $+\mathrm{D}-(-)-$ fructose

It is nonreducing sugar because the reducing groups on anomeric carbons ( $\mathrm{C}-1$ of glucose $\&$ C-2 of fructose) are involved in glycosidic linkage.

Invert sugar : The hydrolysis product of dextrorotatory sucrose is known as invert sugar. It is equimolar mixture of $D-(+)-$ glucose & D-(-)-fructose. Because dextro rotation of glucose $\left(+52.4^{0}\right)$ is less than laevorotation of fructose $\left(-92.5^{\circ}\right)$ therefore, sign of rotation changes during hydrolysis from dextro (+) to laevo (-).

Maltose $\xrightarrow[\mathrm{H} _{2} \mathrm{O}]{\mathrm{H}^{+}} \alpha-D$-glucose $+\alpha-D$-glucose

Maltose is reducing sugar because reducing part or anomeric carbon ( $\mathrm{C}-1$ ) of one glucose unit is not involved is glycosidic linkage and free to give reduction reactions.

$\underset{\text { (milk sugar) }}{\text { Lactose }} \xrightarrow[\mathrm{H} _{2} \mathrm{O}]{\mathrm{H}^{+}} \beta-(\mathrm{D})$-galactose $+\beta-(\mathrm{D})$-glucose

Lactose is a reducing sugar because the anomeric carbon of the glucose unit ( $\mathrm{C}-1$ ) is free and not involved in glycosidic linkage.

1.3 Polysaccharides : These are long chain polymer of monosaccharides joined together by glycosidic linkages. Polysaccharides on hydrolysis give hundreds to thousands of molecules of monosaccharides.

1.3.1 Cellulose : It is a linear condensation polymer of $\beta-D-g$ lucose and is the chief structural material of all plants. It is also the chief component of cotton, wood, jute, etc. Cellulose is not digestible since human digestive system does not contain the enzyme cellulase which can hydrolyse cellulose into glucose.

1.3.2 Starch : It is the chief storage polysaccharide of plants. It is actually a mixture of two components, i.e., water soluble amylose $(15-20 \%$ ) and water insoluble amylopection $(80-85 \%)$. Amylose is a linear polymer in which $\mathrm{C} _{1}$ of one glucose unit is connected to $\mathrm{C} _{4}$ of the other through $\alpha$-glycosidic linkage. Amylopectin is a highly branched chain polymer. It consists of a large number of short chains, each containing 20-25 glucose units which are joined together through $\alpha$-glycosidic linkages involving $C _{1}$ of one glucose unit with $C _{4}$ of the other. The $\mathrm{C} _{1}$ of terminal glucose unit in each chain is further linked to $\mathrm{C} _{6}$ of some other glucose unit in the next chain through $\mathrm{C} _{1}-\mathrm{C} _{6} \alpha-$ glycosidic linkage.

1.3.3 Glycogen : It is also known as animal starch, having structural similarity to amylopectin. It is present in liver, muscles and brain. When body needs energy, glycogen breaks down to give glucose.

2) Proteins:

Proteins are high molecular mass complex biopolymer of $\alpha$-aminoacids, present in all living cells. These are also known as polypeptides.

2.1 Amino Acids:

These are the organic compounds containing $-\mathrm{NH} _{2}$ and $-\mathrm{COOH}$ functional groups. Hydrolysis of protein gives only $\alpha$-aminoacids ( $-\mathrm{NH} _{2}$ group is present on $\alpha$-carbon) therefore, they are also called building blocks of proteins. $\alpha$-amino acids can exist in two stereoisomeric forms $D &$, but all the naturally occuring $\alpha$-amino acids belong to $L$-series. (glycine is optically inactive)

Zwitter ion structure of aminoacids : In aquous solution aminoacid can exist as a dipolar ion, known as zwitter ion due to transfer of $\mathrm{H}^{+}$from $-\mathrm{COOH}$ group to $-\mathrm{NH} _{2}$ group within the molecule. Zwitter ions are dipolar, amphoteric and have no net charge

Isoelectric point : In acidic medium, $-\mathrm{COO}^{-}$group acts as the base and accepts a proton. As a result, $\alpha$-amino acids exist as cation (I) in the acidic medium and migrate towards cathode under the influence of an electric field.

In alkaline medium, $\mathrm{NH} _{3}$ group acts as the acid and thus loses a proton. As a result, $\alpha$-amino acids exist as anion (III) and thus migrate towards the anode under the influence of an electric field. At some intermediate value of $\mathrm{pH}$, the concentration of the cationic form (I) and anionic form (III) becomes equal and hence there is no net migration of the $\alpha$-amino acid under the influence of an electric field. This $\mathrm{pH}$ is called the Isoelectric point. At this $\mathrm{pH}$, the amino acid exists as dipolar ion or zwitterion (II) and hence has the minimum solubility.

Essential and non-essential amino acids : Out of 20 amino acids required for protein synthesis, human body can synthesize only 10 . These ten amino acids which the body can synthesize are called non-essential or dispensable amino acids. The remaining ten amino acids (valine, leucine, isoleucine, phenylalanine, methionine, tryptophan, threonine, Iysine, arginine and histidine) which the body cannot synthesize are called essential or indispensable amino acids. These must be supplied in the human diet.

Natural Amino Acids

S. No.	Amino acid	Three letter symbol	One letter code	Side chain(R)	Isoelectric Point point (pl)*
	a). Neutral amino acids
1.	Glycine	Gly	G	$-\mathrm{H}$	6.1
2 .	Alanine	Ala	A	$-\mathrm{CH} _{3}$	6.1
3.	Valine $^{*}$	Val	V	$-\mathrm{CH}\left(\mathrm{CH} _{3}\right) _{2}$	6.0
4.	Leucine $^{\star}$	Leu	L	$-\mathrm{CH} _{2} \mathrm{CH}\left(\mathrm{CH} _{3}\right) _{2}$	6.0
5.	Isoleucine*	ile	I		6.0
6.	Phenylalanine*	Phe	F	$\mathrm{CH} _{2} \mathrm{C} _{6} \mathrm{H} _{5}$ or $\mathrm{CH} _{2} \mathrm{Ph}^{\star *}$	5.9
7 .	Methionine*	Met	M	$-\mathrm{CH} _{2} \mathrm{CH} _{2}-\mathrm{S}-\mathrm{CH} _{3}$ $-\mathrm{CH} _{2}$	5.7
8.	Tryptophan*	Trp	W		5.9
9.	Serine	Ser	$S$	$-\mathrm{CH} _{2} \mathrm{OH}$	5.7
10	Cysteine	Cys	C	$-\mathrm{CH} _{2} \mathrm{SH}$	5.1
11.	Glutamine	Gln	$Q$	$-\mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{CONH} _{2}$ $-\mathrm{CHCH} _{3}$	5.7
12.	Threonine*	Thr	T	$\mathrm{OH}$	5.7
13.	Tyrosine	Tyr	Y		5.6
14.	Proline	Pro.	$P$	(It is the complete structure)	6.3
15.	Aspargine	Asn	$\mathrm{N}$	$-\mathrm{CH} _{2} \mathrm{CONH} _{2}$	5.4
	b). Acidic amino acids
16.	Aspartic acid	Asp	$\mathrm{D}$	$-\mathrm{CH} _{2} \mathrm{COOH}$	3.0
17.	Glutamic acid	Glu	$E$	$-\mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{COOH}$	3.2
	c). Basic amino acids
18.	Lysine $^{*}$	Lys	K	$-\mathrm{CH} _{2}\left(\mathrm{CH} _{2}\right) _{3} \mathrm{NH} _{2}$	9.7
19.	Arginine $^{\star}$	Arg	$\mathrm{R}$		10.8
20.	Histidine*	His	$\mathrm{H}$		7.6

${ }^{*}$ Essential amino acids

2.2 Classification of Proteins:

1. on the basis of molecular structure proteins are classified as fibrous and globular proteins :-

a) Fibrous proteins: These consist of linear thread-like molecules which tend to lie side by side to form fibres. The polypeptide chains in them are held together usually at many points by $\mathrm{H}$-bonds. Examples are : Keratin in skin, hair, nails, wool, collagen in tendons, fibroin in silk and myosin in muscles. These are insoluble in water. They are stable to moderate changes in temperature and $\mathrm{pH}$.

b) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule a spheroidal shape. This folding occurs due to the following four types of forces: (i) disulphide bridging (ii) intramolecular $\mathrm{H}$-bonding (iii) van der waals interactions and (iv) dipolar interactions. Unlike fibrous proteins, globular proteins are insoluble in water and are sensitive to small changes in temperature and $\mathrm{pH}$. Examples are enzymes, hormones (insulin, thyroglobulin), antibodies, haemoglobin, fibrinogen, albumin, etc.

2. Classification of proteins on the basis of their biological functions :-

2.3 Structure of Proteins:

Peptide linkage : Proteins are polymers of $\alpha$-aminoacids joined together by peptide linkage / bond. It is an amide bond formed between $-\mathrm{COOH}$ group of one amino acid and $-\mathrm{NH} _{2}$ group of the other aminoacid. The product of the reaction between two similar or different aminoacid is a dipeptide. When three amino acids link together by two peptide bond they form tripeptide and as the number of amino acid increases we get tetra, penta or hexapeptides.

Polypeptide : Proteins are polypeptides as they are formed by linking together a large number of $\alpha$-aminoacid (more than 10) through peptide bond.

Structure and shape of proteins can be studied at four different levels :

a) Primary structure : The sequence in which the various aminoacids are linked to one another in a protein is called its primary structure.

b) Secondary structure : It refers to the shape in which a long polypeptide chain can exist due to $\mathrm{H}$-bonding.

These are of two types:

i) $\alpha$-Helix : If the size of the side groups (alkyl) is large, intramolecular $\mathrm{H}$-bonds are formed between $\mathrm{C}=0$ of one molecule and $\mathrm{NH}$ of the fourth amino acid in the polypeptide chain giving right handed $\alpha$-helix structure to the protein molecule. Examples are : $\alpha$-keratin in hair, nail, wool skin, and myosin in muscles.

ii) $\beta$-Flat sheet or $\beta$-pleated sheet structure : If the size of the side groups (alkyl) is small, intermolecular $\mathrm{H}$-bonds are formed between $\mathrm{C}=0$ of one polypeptide chain with $\mathrm{NH}$ of the other chain giving a $\beta$-flat sheet structure to the protein molecule. If the size of groups $\mathrm{R}$ is moderate, the polypeptide chains contract a little giving $\beta$-pleated sheet structure to protein molecule, i.e., silk protein fibroin.

c) Tertiary structure of a protein refers to its complete three dimensional structure.

d) Quaternary structure : Certain proteins exist as assemblies of two or more polypeptide chains called subunits. Quaternary structure refers to the number of subunits and their spatial arrangement w.r.t. each other in an aggregate protein molecule.

2.4 Denaturation of proteins:

On heating or on treatment with mineral acids (change in temperature & pH) the water soluble globular proteins undergo coagulation or precipitation with loss of biological activity to give water insoluble fibrous proteins. This process is called denaturation and the coagulated proteins thus formed is called the denaturated protein. During denaturation the secondary and tertiary structure of the proteins change while primary structure remains intact. Examples of denaturation are (a) coagulation of albumins present in the white of an egg when the egg is boiled hard (b) formation of cheese from milk on adding lemon juice when the globular milk protein lactalbumin becomes fibrous (c) curding of milk, which is caused by the change in $\mathrm{pH}$, due to formation of lactic acid by lactic acid bacteria present in milk.

Due to the change in $\mathrm{pH}$ and temperature, the hydrogen bonds and other vander walls forces of attraction within protein molecules are disturbed and broken, thereby causing the uncoiling of polypeptide chains and destroying its secondary and tertiary structure and affecting its biological activity.

2.5 Characterization of Proteins:

a) Biuret test: Protein when reacts with biuret reagent $\left(\mathrm{CuSO} _{4}+\mathrm{NaOH}\right)$, a blue-violet colour is obtained.

b) Ninhydrin test : Protein when treated with an ammonical solution of ninhydrin gives colours ranging from deep blue to violet pink or even yellow and red in some cases.

3. Enzymes :

Enzymes are biological catalysts. They are primarily globular proteins and catalyse vital metabolic processes. Simple enzymes are comprised of Proteins only eg. amylase while conjugated enzymes are comprised of both protein and nonprotein part

$\begin{array}{lll}\text { Holoenzyme } && = && \text { Apoenzyme } && + && \text { prosthetic group } / \text { co-factor }\end{array}$

$\begin{array}{lll} \text { (Conjugated enzyme) } & \text { (Protein part) } &&&&& \text { (Non protein part) } \end{array}$

3.1 Characteristics of Enzymes:

(a) Enzymes, being proteneous in nature, exhibit properties of proteins. Like proteins, they are also sensitive towards change in temperature & pH.

(b) They speed up rate of chemical reactions by lowering the activation energy.

(c) Enzymes are highly specific both in binding chiral substrate and in catalysing their reactions.

(d) They don’t get destroyed or get consumed during the reaction. They are regenerated at the end of reaction.

e. Metabolic reactions are reversible and hence the enzymes catalyse reaction in both the directions.

f. Enzyme inhibitors are the substances that reduce activity of an enzyme.

3.2 Some reactions catalysed by enzymes:

4. Vitamins :

Vitamins are required in very small amounts for the life, growth and health of human beings and animals. They can not be synthegized in the body (except vitamin D) and hence have to be supplied in the diet.

4.1 Classification of Vitamins:

a) Water soluble vitamins:

They must be supplied regularly in diet because they are regularly excreted in urine and cannot be stored (expect vitamin $\mathrm{B} _{12}$ ) in our body. Eg. Vitamin B-complex and vitamin C.

b) Fat soluble vitamins:

They are stored in liver and adipose tissue. Eg. Vitamin A, D, E, K.

Hypervitaminoses - Excess intake of vitamin $A$ and $D$ is harmful and may cause hypervitaminoses

Avitaminoses - Multiple deficiencies caused by lack of more than one vitamins.

Some vitamins like C, D, E acts as antioxidants. Vitamin $\mathrm{B} _{12}$ contains cobalt and it can be stored in liver for many months.

5. Nucleic Acids :

Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins. They are biopolymers in which the repeating structural unit of monomeric unit is a nucleotide. That is why nucleic acids are also called polynucleotides.

Each nucleotide consists of three components:

a) A pentose sugar, i.e., D-(-)-ribose or 2’-deoxy-D-(-)-ribose.Both these sugars are found in the furanose form. Ribose is present in RNA and Deoxyribose is present in DNA.

b) A heterocyclic base - Two types

i) Purines: Adenine (A) and Guanine (G);

ii) Pyrimidines: Uracil (U), Thymine ( $\mathrm{T}$ ) and Cytosine (C)

c) Phosphoric acid

Structures of Bases :

Purine bases :

Pyrimidine bases :

Nucleosides contain only two components : a pentose sugar and a nitrogenous base i.e., purine or pyrimidine. Their general structure is : Sugar-Base.

Nucleotides Contain all the three components : a pentose sugar, a nitrogenous base and a phosphoric acid group.

General Structure of a Nucleotide

Difference between DNA and RNA

5.1 Structure of DNA

a) Primary structure

Sequence in which the four nitrogen bases are attached to the sugar phosphate backbone of a nucleotide chain.

b) Secondary structure

DNA consists of two strands of polnucleotides coiled around each other in form of a double helix.

Chargaff rule- Base composition in DNA varied from one species to other but amount of adenine is equal to thymine $(A=T)$ and cytosine is equal to guanine $(C=G)$. Total amount of purines is equal to pyrimidines $(A+G=C+T)$; $A T$ / $C G$ ratio is different for different species.

5.3 Watson and Crick Model of DNA:

DNA consists of two right handed polynucleotide strands. The sugar phosphate units and base units of each strand are pointed into the interior of the helix. A purine base is always paired with a pyrimidine base.

C and $G$ pair through 3 hydrogen bonds.

A and $T$ pair through 2 hydrogen bonds.

The two strands of DNA are complementary and are not idential since the base sequence of one strand automatically fixes that of the other due to the base pairing principle. Distance between two adjacent base pairs is $0.34 \mathrm{~nm}$. Diameter of the helix is $2 \mathrm{~nm}$. Distance between any two successive turns of the helix is $3.4 \mathrm{~nm}$ suggesting that there are ten base pairs in each turn.

5.4 Functions of Nucleic acids

a) Replication: The genetic information of the cell is contained in the sequence of the bases $\mathrm{A}, \mathrm{T}, \mathrm{G}$ and $\mathrm{C}$ in the DNA molecule. When a cell divides, DNA molecules replicate and make exact copies of themselves so that each daughter cell will have DNA identical to that of the parentcell.

b) Protein synthesis: Occurs in two steps, i.e., Transcription and Translation.

Gene : Each segment of a DNA molecule that codes for a specific protein or a polypeptide is called a gene.

Genetic code : The relation between the nucleotide triplets and the amino acids is called the genetic code.

Mutations : A mutation is a chemical change in the sequence of nitrogenous bases along the DNA strands that can lead to the synthesis of proteins with altered amino acid sequence. These changes are caused by radiation, chemical agents or viruses.

Solved Problems :

Question 1- D-glucose reacts with hydroxylamine to form oxime but pentacectate of glucose does not give reactions of-CHO group why?

Show Answer

Answer- This is because cyclic (hemiacetal) structure of D-glucose forms an open chain structure in aqueous medium, which then reacts with $\mathrm{NH} _{2} \mathrm{OH}$ to give an oxime (reaction of $-\mathrm{CHO}$ group) while, pentaacetate of glucose can not form an open chain structure because anomeric carbon is involved in acetylation.

Question 2- Glucose is water soluble but cyclo hexane is insoluble in water. Why?

Question 3- Explain higher melting points and higher solubility of amino acids in water than that of the corresponding halo acids.

Question 4- Although, fructose is a ketonic sugar yet it reduces tollen’s and fehling’s reagent. Why?

Question 5- Human beings can not digest cellulose while cattle and other ruminants can digest cellulose. Why?

Question 6- Why is invert sugar sweeter than sucrose?

Question 7- Give some examples of non sugar compounds, sweeter than sucrose.

Question 8- Write chair conformations of $\alpha$ and $\beta$ (D) glucopyranose.

Show Answer

Answer-

Practice Problems:

Question 1- The human body does not produce:

(a) Enzyme

(b) DNA

(c) Vitamins

(d) Hormones

Question 2- During the process of digestion, the proteins present in food are hydrolysed to aminoacids. The two enzymes involved in the process are:

(a) Invertase and zymase

(b) Amylase & maltase

(c) Diastase and lipase

(d) Pepsin & trypsin

Question 3- Fibrous protein such as silk fibroin consists of chain arranged in

(a) $\alpha$-helix

(b) $\beta$-helix

(c) $\beta$-pleated sheet

(d) none of these

Question 4- In DNA, the complementary bases are :-

(a) Uracil, Adenine; Cytocine, guanine

(b) Adenine, thymine; guanine, cytosine

(c) Adenine, thymine; guanine, uracil

(d) Adenine, guanine; thymine, cytosine

Question 5- The segment of DNA which acts as the instrumental manual for the synthesis of the protein is:

(a) Nucleoside

(b) Nucleotide

(c) Ribose

(d) Gene

Question 6- Haemoglobin is a protein with

(a) Primary structure

(b) Secondary structure

(c) Tertiary structure

(d) Quarternary structure

Question 7- Which one of the following does not exhibit the phenomenon of mutarotation?

(a) (-) Fructose

(b) (+) Sucrose

(c) $(+)$ Lactose

(d) $(+)$ Maltose

Question 8- Which of the following is not a fat soluble vitamin?

(a) Vitamin A

(b) Vitamin B-complex

(c) Vitamin D

(d) Vitamin E

Question 9- Which of the following statements about denaturation are correct?

A- Denaturation causes loss of secondary and tertiary structures of the protein.

B- Denaturation leads to conversion of double strand of DNA into single strand.

C- Denaturation affects primary structure which gets distorted.

(a) $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$

(b) $\mathrm{A}$ and $\mathrm{C}$

(c) $\mathrm{B}$ and $\mathrm{C}$

(d) $\mathrm{A}$ and $\mathrm{B}$

Question 10- The pyrimidine bases present in DNA are

(a) Cytosine and adenine

(b) Cytosine and guanine

(c) Cytosine and thymine

(d) Cytosine and uracil

Question 11- The secondary structure of protein refers to

(a) Sequence of amino acids

(b) Fixed configuration of the polypeptide back bone.

(c) Helical back-bone

(d) Hydrophobic interactions

Question 12- $\alpha-D(+)-$ glucose and $\beta-D-(+)-$ glucose are

(a) Epimers

(b) Anomers

(c) Enantiomers

(d) Conformers

Question 13- Biuret test is not given by

(a) Proteins

(b) Carbohydrates

(c) Polypeptides

(d) Urea

Question 14- The change in optical rotation of freshly prepared solution of glucose is known as :

(a) Specific rotation

(b) Mutarotation

(c) Tautomerism

(d) Racemisation

Question 15- The complementary base sequence for the DNA strand 5’-GCAT-3’, is:

(a) $5^{\prime}-$ GCAT-3'

(b) $\quad 5^{\prime}-\mathrm{ATGC}-3^{\prime}$

(c) $5^{\prime}-$ CGTA-3'

(d) $5^{\prime}-$ TACG-3'

Hint : As per Watson & Crick model of DNA, it has two polypeptide chains running in opposite directions i.e. one chain $3^{\prime}-5^{\prime}$ & other $5^{\prime}-3^{\prime}$ joined to gether by $\mathrm{H}$-bonding between complementary base pairs.

Question 16- Which of the following is a non reducing sugar

(a) Glucose

(b) Sucrose

(c) Maltose

(d) Lactose

Question 17- Number of asymmetric carbon in a–D–(+)–glucose are

(a) 4

(b) 6

(c) 5

(d) 3

Question 18- Presence of carbonyl group in glucose can be shown by its reaction with

(a) $\mathrm{NH} _{2} \mathrm{OH}$

(b) $\mathrm{HCN}$

(c) Tollen’s reagent

(d) All of these

Question 19- Which of the following is the monomer of cellulose

(a) Amylose

(b) Glycogen

(c) $\beta-D$-glucose

(d) Amylopectin

Question 20- Vitamin $B$, is chemically known as

(a) Ascorbic acid

(b) Carotenoids

(c) Thiamine

(d) Pyridoxine

Question 21- The deficiency of vitamin $\mathrm{K}$ causes

(a) Haemorrhage

(b) Increase in blood cloting time

(c) Inflamation of teeth and tongue

(d) both $(a) \& (b)$

Question 22- Which of the following vitamins is oil soluble :

(a) $\mathrm{A}$

(b) $\mathrm{B} _{6}$

(c) $B _{12}$

(d) $\mathrm{C}$

Question 23- Which of the following, if taken excessively, can accumulate in body and cause toxicity

(a) Vitamin C

(b) Vitamin D

(c) Vitamin $B _{2}$

(d) Vitamin K

Question 24- Milk contains vitamins

(a) $A, D & E$

(b) $\mathrm{A}, \mathrm{B} _{12}$ and $\mathrm{D}$

(c) $\mathrm{C}, \mathrm{D}$ and $\mathrm{K}$

(d) $ B, B _{6}$ and $D$

Question 25- Which of the following base is not present in DNA

(a) Thymine

(b) Uracil

(c) Adenine

(d) Guanine

Question 26- Pyrimidine bases present in RNA are

(a) Adenine and guanine

(b) Thymine and Uracil

(c) Uracil and Cytosine

(d) Thymine and Cytosine

Question 27- The enzyme present in human saliva is

(a) Pepsin

(b) Trypsin

(c) $\alpha$-amylase

(d) Zymase

Question 28- Which polysaccharide is stored in the liver of animals

(a) Starch

(b) Cellulose

(c) Amylose

(d) Glycogen

Question 29- Deficiency of which vitamin causes beri-beri and pain in joints

(a) $\mathrm{A}$

(b) $\mathrm{B}$

(c) $\mathrm{C}$

(d) $\mathrm{K}$

Question 30- Which of the following is an essential aminoacid :

(a) Valine

(b) Glycine

(c) Proline

(d) Tyrosine

Question 31- The helical structure of protein is stabilized by

(a) Dipeptide bonds

(b) Hydrogen bonds

(c) Ether bonds

(d) Peptide bonds

Question 32- Which of the following gives positive fehling test

(a) Sucrose

(b) Glucose

(c) Fat

(d) Protein

Question 33- Haemoglobin is

(a) a vitamin

(b) a carborydrate

(c) an enzyme

(d) a globular protein

Question 34- Vitamin $B _{12}$ contains

(a) $\mathrm{Fe}$

(b) Co

(c) $\mathrm{Zn}$

(d) $\mathrm{Ca}$

Question 35- D(+) glucose reacts with hydroxyl amine and yields an oxime. Structure of the oxime would be

Question 36- A peptide bond is

(a) Covalent

(b) Planar

(c) Having partial double bond character

(d) All of the above

Question 37- Fructose reduces tollen’s reagent due to

(a) Asymmetric carbons

(b) Primary alcoholic group

(c) Secondary alcoholic group

(d) Enolisation of fructose followed by conversion to addehyde by base

Question 38- The synthesis of daughter DNA from parent DNA is called:

(a) Translation

(b) Replication

(c) Transcription

(d) Mutation

Question 39- RNA and DNA are chiral molecules. Their chirality is due to

(a) D-sugar component

(b) L-sugar component

(c) chiral bases

(d) chiral phosphate ester units

Question 40- In living systems the function of enzyme is to

(a) Provide energy

(b) Transport oxygen

(c) Catatyse biochemical reactions

(d) Provide immunity

Question 41- The correct statement regarding RNA & DNA, respectively is -

(a) The sugar component in RNA is 2’-deoxyribose and in DNA is Arabinose

(b) The sugar component in RNA is Arabinose and is DNA in 2’-deoxyribose

(c) The sugar component in RNA is Ribose and is DNA in 2’-deoxyribose

(d) The sugar component in RNA is 2’-deoxyribose and in DNA is Ribose

Question 42- Which of the following sets of monosaccharides forms sucrose

(a) $\beta$-D-glucopyranose and $\alpha$-D-fructofuranose

(b) $\alpha$-D-glucopyranose and $\beta$-D-fructopyranose

(c) $\alpha$-D-glucopyranose and $\beta$-D-fructofuranose

(d) $\alpha$-D-galactopyranose and $\alpha$-D-glucopyranose

Question 43- Which of the following has maximum laevorotation

(a) D-glucose

(b) D-fructose

(c) Sucrose

(d) Invert sugar

Question 44- D-glucose and D-mannose are

(a) Anomers

(b) Epimers

(c) Enantiomers

(d) Polysaccharide

Question 45- $\alpha$-maltose contains :

(a) $\alpha-1,2-$ glycosidic linkage

(b) $\beta-1,2$-glycosidic linkage

(c) $\alpha-1,4-$ glycosidic linkage

(d) $\beta-1,4-$ glycosidic linkage

Question 46-. Which of the following hexoses will form the same osazone when treated with excess phenyl hydrazine

(a) D-glucose and D-fructose

(b) D-glucose and D-galactose

(c) D-glucose and D-lactose

(d) D-fructose and D-galactose

Question 47- Which of the following have D-configuration

(a) $1,2,3$

(b) 2,3

(c) 1,2

(d) 3

Question 48- Lysine is least soluble in the $\mathrm{pH}$ range :

(a) $3-4$

(b) $5-6$

(c) $6-7$

(d) $8-9$

Question 49- Which of the following statements is correct

(a) All amino acids are optically active.

(b) All amino acids except glycine are optically active.

(c) All amino acids except lysine are optically active.

(d) All amino acids except glutamic acid are optically active.

Question 50- The amino acids in Haemoglobin (or any protein) uniformly have which of the following configurations:-

(a) $\mathrm{L}$

(b) $\mathrm{S}$

(c) $\mathrm{D}$

(d) $\mathrm{R}$