Answer the following in detail.

Question 14 Marks

A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these giving rise to tertiary level structure of protein.

Answer

Tertiary level structure from amino residues.

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Question 24 Marks

(1) Is galactose an aldohexose or a ketohexose?
(2) Which carbon in galactose has different configuration compared to glucose?
(3) Draw Haworth formulae of α-D-galactose and β-D-galactose.
(4) Which disaccharides among sucrose, maltose and lactose is/are expected to give positive Fehling test?
(5) What are the expected products of hydrolysis of lactose?

Answer

Galactose is an aldohexose.
Fourth carbon in galactose has different configuration compared to glucose.
Maltose and lactose are expected to give positive Fehling solution test.
The expected products of hydrolysis of lactose are D – ( +) glucose and D – ( +) galactose.

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Question 34 Marks

Observe the following structural formulae carefully and answer the questions.

(1) How many OH groups are present in glucose, fructose and ribose respectively?
(2) Which other functional groups are present in these three compounds?

Answer

1) Glucose contains five hydroxyl (- OH) groups.
Fructose contains five hydroxyl ( – OH) groups.
Ribose contains four hydroxyl ( – OH) groups.

(2) Glucose contains aldehyde ( – CHO) as other functional group.

Fructose contains ketonic group as other functional group. Ribose contains aldehyde ( – CHO) as other functional group.

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Question 44 Marks

Explain double helix.###State the salient features of the Watson and Crick mode of DNA.

Answer

The Salient features are :

DNA is made of two polynucleotide strands that wind into a right-handed double helix.
The two strands run in opposite directions: one from the Y end to the 3’ end, while the other from the 3’ end to the Y end.
Pcrpcndicular to the axis of the helix, the sugar – phosphate backbone lies on the outside of the helix and the bases lic on the inside.
The hydrogen bonding between the hases of the two DNA strands stabilizes the double helix. This gives rise to a ladder-like structure of DNA double helix.
Adenine always forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosinc. Thus A – T arid C – G arc complementary hase pairs and the Two strands of the double helix arc complementary to each other.

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Question 54 Marks

What is meant by nucleosides?###Write the structure of nucleoside. Give examples.

Answer

A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base.

A nucleoside is formed when 1 -position of a pyrimidine (cytosine, thymine or uracil) or 9 -position of guanine or adenine base is attached to $C$ - I of sugar by $\beta$-linkage.

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Question 64 Marks

Explain chemical composition of nucleic acids.

Answer

Nucleic acids have a polynucleotide structure. Nucleic acids (RNA and DNA) consists of three components :
(1) monosaccharide (sugar)
(2) nitrogen containing base and
(3) phosphate group.
(1) Monosaccharides : Nucleotides of both RNA consist of five membered monosaccharide ring (furanose), called as simply sugar component.

In RNA, the sugar component of nucleotide units is D-ribose and in DNA 2-deoxy-D-ribose. 2 - deoxy means no $- OH$ group at $C _2$ position.
(2) Nitrogen containing base : Total five nitrogen - containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base in designated by a one-letter symbol. Uracil (U) acurs only in RNA while thymine $( T )$ ocurs only in DNA.

(3) Phosphate group : The sugar units are joined to phosphate through $C _3$ and $C _5$ hydroxyl groups.

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Question 74 Marks

Draw a neat labelled diagram for the secondary structure of protein.

Answer

Secondary structure of proteins : The three-dimensional arrangement of Idalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between $N - H$ proton of one amide linkage and $C = O$ oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. $\alpha$-helix and $\beta$-pleated sheet.
$\alpha$-Helix: In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clakwise spiral known as a-helixn. The characteristic features of $\alpha$-helical structure of protein are :

(1) Each turn of the helix has 3.6 amino acids.
(2) $AC = O$ group of one amino acid is hydrogen bonded to $N - H$ group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.
$\beta$-Pleated sheet: In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called $\beta$-pleated sheets. The $\beta$-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of $\beta$-pleated sheet structure are :

The C = O and N – H bonds lie in the planes of the sheet.
Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

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Question 84 Marks

Distinguish between globular and fibrous proteins.

Answer

Globular proteins	Fibrous proteins
(1) The chains of polypeptides of protein coil around to give a spherical shape.	(1) The proteins in which the polypeptide chains lie parallel to form fibre like structure.
(2) Globular proteins are soluble in water.	(2) Fibrous proteins are insoluble in water.
(3) They are sensitive to small changes of temperature and pH.	(3) They are stable to moderate changes of temperature and pH.
(4) They possess biological activity.	(4) They do not possess biological activity.

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Question 94 Marks

Write the classification of amino acids, giving examples.

Answer

The amino acids are of three types : acidic, basic and neutral. The symbol ' $R$ ' in the structure of aamino acids represents side chain and may contain additional functional groups.

(1) Acidic amino acids : If ' $R$ ' contains a carboxyl $(- COOH )$ group the amino acid is acidic amino acid, i.e. If carboxyl groups are more in number than amino groups, then amino acids are acidic in nature.
Examples : Glutamic acid $HOOC - CH _2- CH _2-$ : Aspartic acid $HOO - CH _2-$

(2) Basic amino acids : If ' $R^{\prime}$ contains an amino $\left(1^{\circ}, 2^{\circ}\right.$, or $\left.3^{\circ}\right)$ group, it is called basic amino acid i.e. If amino groups are more in number than carboxyl groups then amino acids are basic in nature.

(3) Neutral amino acids : The other amino acids having neutral or no functional group in 'R' are called neutral amino acids, i.e. The amino acids having equal number of amino and carboxyl groups are neutral amino acids.
Examples : Alanine $CH _3$-; Valine $\left( CH _3\right)_2- CH$

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Question 104 Marks

Explain the structure of starch.

Answer

Starch is found in cereal grains, roots, tubers, potatoes, etc. It is a polymer of $\alpha$-D-glucose and consists of two components, amylose and amylopectin.

Amylose is water soluble component forms blue coloured complex with iodine. It constitutes about $20 \%$ of starch. Amylose contains 200 to $1000 \alpha$-D-glucose units linked together by glycosidic linkage between C-I of one unit and C-4 of another unit. i.e. $\alpha-1,4$ glycosidic linkages.

Amylopectin is insoluble in water and constitutes about $80 \%$ starch which forms blue-violet coloured complex with iodine. It is a branched chain polymer. In amylopectin, $\alpha$-D-glucose molecules are linked together by glycosidic linkage between $C _1$ - of one unit and $C -4$ of another unit to form long chain and branching acurs by glycosidic linkage between C-I and C6 glycosidic linkage.

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Question 114 Marks

Explain the structure of sucrose.

Answer

Sucrose is a hexasaccharide and has molecular formula $C _{12} H _{22} O _{11}$. The structure of shcrose contains glycosidic linkage between $C -1$ of $\alpha$-glucose and $C -2$ of $\beta$-fructose. Since aldehyde and ketone groups of both monosaccharide units are involved in the formation of glycosidic bond, sucrose is a nonreducing sugar.
Sucrose is dextrorotatory, on hydrolysis with dilute acid or an enzyme invertase gives equimolar mixture of dextrorotatory glucose and laevorotatory fructose.

The solution is laevorotatory because laevo rotation of fructose $\left(-92.4^{\circ}\right)$ is more than dextrorotation of glucose $(+52.50)$, hence the sign of rotation is changed from $(+)$ to $(-)$ after hydrolysis, the product is called invert sugar.

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Question 124 Marks

Draw mirror images of glucose and fructose.

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Question 134 Marks

Explain Haworth formula of glycopyranose.

Answer

In the Haworth formula the pyranose ring is considered to be in a perpendicular plane with respect to the plane of paper. The carbons and oxygen in the ring are in the places as they appear in figure. The lower side of the ring is called $\alpha$-side and the upper side is the $\beta$-side. The $\alpha$-anomer has its anomeric hydroxyl $(- OH )$ group (at $C -1$ ) on the $\alpha$-side, whereas the $\beta$-anomer has its anomeric hydroxyl $(- OH )$ group (at $C -1)$ on the $\beta$-side. The groups which appear on right side in the Fischer projection formula appear on $\alpha$-side in the Haworth formula, and the groups which appear on left side in the fischer projection formula appear on a $\beta$-side in the Haworth formula.

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Question 144 Marks

Explain ring structure of glucose.

Answer

Glucose has two cyclic structures (II and III) which are in equilibrium with each other through the open chain structure (I) in aqueous solution.
The ring structure of glucose is formed by reaction between the formyl ( $- CHO$ ) group and the alcoholic $(- OH )$ group at $C -5$. Thus, the ring structure is called a hemiacetal. The two hemiacetal structures (II and III) differ only in the configuration of $C - I$ (Fig.), the additional chiral centre resulting from ring closure. The two ring structures are called $\alpha$ - and $\beta$-anomers of glucose and $C - I$ is called the anomeric carbon. The ring of the cyclic structure of glucose contains five carbons and one oxygen. Thus, it is a six membered ring. It is called pyranose structure, in analogy with the six membered heterayclic compound pyran (IV). Hence glucose is also called glucopyranose.

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Question 154 Marks

Assign D/L configuration to the following monosaccharides.

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Question 164 Marks

Explain D and L configuration in sugars.

Answer

The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms

The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as (-) glyceraldehyde and it corelated as L-configuration i.e., L-glyceraldehyde.
In Fischer projection formula, a monosaccharide is assigned $D$-configuration if the $(- OH )$ hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the $- OH$ group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., $- CHO$ ) is at the top.

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Question 174 Marks

Describe the action of following reagents on glucose :
(1) $HI$
(2) Hydroxyl amine $\left( NH _2 OH \right)$
(3) Hydrogen cyanide
(4) Bromine water
(5) dil. Nitric acid
(6) Acetic anhydride.

Answer

(1) Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain.

(2) Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of CHO group in glucose.

(3) Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin

(4) Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group.

(5) Action of dll. nitric acid: Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group $\left(- CH _2 OH \right)$ in glucose.

(6) Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups.

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Question 184 Marks

Classify the following carbohydrates.
(1) Cellulose,
(2) Maltose,
(3) Raffinose,
(4) Fructose.

Answer

Carbohydrates	Class
(1) Cellulose	Polysaccharide
(2) Maltose	Disaccharide
(3) Raffinose	Trisaccharide
(4) Fructose	Monosaccharide

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Question 194 Marks

Classify the following carbohydrates into Monosaccharide, Disaccharide, Oligosaccharide, Polysaccharide.
(1) Glucose
(2) Starch
(3) Sucrose
(4) Maltose
(5) Galactose
(6) Lactose
(7) Ribose.

Answer

Carbohydrates	Class
(1) Glucose	Monosaccharide
(2) Starch	Polysaccharide
(3) Sucrose	Disaccharide
(4) Maltose	Disaccharide
(5) Galactose	Monosaccharide
(6) Lactose	Disaccharide
(7) Ribose	Monosaccharide

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Question 204 Marks

How are carbohydrates classified?###Classification of carbohydrates with examples.

Answer

Carbohydrates are classified as monosaccharides oligosaccharides and polysaccharides.
(1) Monosaccharides : These carbohydrates cannot be further hydrolysed into smaller units. They are basic units of all carbohydrates, and are called monosaccharides.
Examples : Glucose, fructose, ribose
(2) Oligosaccharides : An oligosaccharide is a carbohydrate (sugar) which on hydrolysis gives two to ten monosaccharide units.
Depending on the number of monosaccharides produced on hydrolysis, oligosaccharides are further classified as :

Oligosaccharide is homogeneous. In this, each molecule of oligosaccharide contains the same number of monosaccharide units joined together in the same order as every other molecule of the same oligosaccharide.
(3) Polysaccharides : These are carbohydrates which on hydrolysis give a large number of monosaccharides.

Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally αcurring polymers of carbohydrates.

Example : Cellulose, starch, glycogen.

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Chemistry Answer the following in detail.

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