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Chapter 5 · Biomolecules

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#1 SUB 1M

Question

What is Biochemistry?

Answer

The study of chemistry with in a living system is known as biochemistry.
#2 SUB 1M

Question

Living systems are made up of which Biomolecules?

Answer

Living systems are made up of various complex biomolecules like carbohydrates, proteins, nucleic acids, lipids etc.
#3 SUB 2M

Question

What are monosaccharides?

Answer

Monosaccharides are carbohydrates that can not be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone.
#4 SUB 4M

Question

Describe detailed information on classification of carbohydrates.

Answer

Carbohydrates are classified on the basis of their behaviour on hydrolysis. They have been broadly divided into following 3 groups
(i) Monosaccharide compounds: A carbohydrate that can not 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) Oligosaccharide compounds: 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, one molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose where as maltose gives two molecules of only glucose.
(iii) Polysaccharide compounds: Carbohydrate 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.
#5 SUB 3M

Question

Give primary information about carbohydrate compounds?

Answer

Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds.
Some common examples of carbohydrates 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. For example 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.
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, which are sweet in taste, are also called sugar. The most common sugar, used in our homes is named as sucrose where as the sugar present in milk is known as lactose.
Carbohydrates are also called saccharides.
#6 SUB 2M 🖼 2

Question

Note on preparation of glucose
OR
Explain the preparation of glucose with it's
equations.

Answer

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:
(1) From sucrose (sugar cane): If sucrose is boiled with dilute HCl or H2SO4 in alcoholic solution, glucose and fructose are obtained in equal amounts.
C12H22O11 + H2O C6H12O6 + C6H12O6
Sucrose Glucose + Fructose
(2) From starch: Commercial glucose is obtained by hydrolysis of starch by boiling it with dilute H2SO4 at 393 K under pressure.
(C6H10O5)n + nH2O nC6H12O6
Starch or cellulose Glucose
#7 SUB 2M

Question

Explain the classification of monosaccharide compounds.

Answer

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 ketose.
Number of carbon atoms constituting the monosaccharide is also introduced in the name. e.g. 3 carbon containing monosaccharide is known as triose and 4 carbon containing monosaccharide is known as tetrose.
Different types of monosaccharide compounds are given below in table:

Carbon

atoms

General term

Aldehyde

Ketone

3

Triose

Aldotriose

Ketotriose

4

Tetrose

Aldotetrose

Ketotetrose

5

Pentose

Aldopentose

Ketopentose

6

Hexose

Aldohexose

Ketohexose

7

Heptose

Aldoheptose

Ketoheptose

#8 SUB 3M 🖼 10

Question

Explain by giving an equation how the presence of carbonyl group and five –OH groups in the structure of glucose is determined.
OR
Write the equation reactions of glucose with NH2OH, HCN, and acetic anhydride and state what information is obtained about the structure of glucose through it.
OR
Explain presence of Five –OH group in structure of glucose.

Answer

Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of carbonyl group in glucose.
Acetylation of glucose with acetic anhydride gives glucose penta–acetate 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.
#9 SUB 🖼 3

Question

Draw the stereo structure of glucose, gluconic acid and saccharic acid given by Fischer.

Answer

The exact spatial arrangement of different –OH groups was given by Fischer after studying many other properties.
It's configuration is correctly represented as I. So gluconic acid is represented as II and saccharic acid III.
I II III
#10 SUB 🖼 4

Question

Explain the D and L notation method of spatial arrangement with respect to glucose.

Answer

The letters 'D' or 'L' before the name of any compound indicate the relative configuration of a particular stereoisomer of a compound with respect to configuration of some other compound, configuration of which is known.
In the case of carbohydrates, 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.
(+)–Glyceraldehyde (–)–Glyceraldehyde
(+) Isomer of glyceraldehyde has 'D' configuration it means that the –OH group lies on right hand side in the structure.
All those compounds which can be chemically correlated to D(+) isomer of glyceraldehyde are said to have D-configuration.
Where as those which can be correlated to L (–) isomer of glyceraldehyde are said to have L configuration. In 'L(–)' isomer –OH group is on left hand side.
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 compared to (+) glyceraldehyde. So, (+) glucose is assigned D-configuration.
D(+)–Glyceraldehyde D(+)–Glucose
Other asymmetric carbon atoms of glucose are not considered for this comparison. Also, the structure of glucose and glyceraldehyde is written in a way that most oxidised carbon (in this case -CHO) is at the top.
#11 SUB 3M 🖼 6

Question

Note on: Cyclic structure of glucose.

Answer

Limitations of open chain structure of glucose shows that free –CHO group is absent in 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 C5 is involved in ring formation.
This explains the absence of –CHO group and also existence of glucose in two forms as shown below:
aD(+)–Glucose bD(+)–Glucose
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 isomer i.e. a-form and b-form are called anomers.
The six membered cyclic structure of glucose is called pyranose structure (a- or b-) 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.
#12 SUB 🖼 4

Question

What is fructose? Explain the structure of fructose.

Answer

Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide sucrose.
It is natural monosaccharide found in fruits, honey and vegetables. In it's pure form it is used as a sweetner.
Structure of fructose:
Fructose has the molecular formula C6H12O6.
On the basis of it's 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.
D(–)–Fructose
It also exists in two cyclic forms which are obtained by the addition of –OH at C5 to the group.
The ring, thus formed is a five membered 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.
a–D-Fructofuranose b–D-Fructofuranose
#13 SUB 3M 🖼 1

Question

What do you understand by the term glycosidic linkage?

Answer

The two monosaccharides are joined together by an oxide linkage formed by the loss of water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage. In maltose the glycosidic linkage is as shown below.
#14 SUB 2M

Question

What are reducing sugars?
OR
Explain the classification of carbohydrate compounds on the basis of their reducing nature.

Answer

Carbohydrate compounds are classified as reducing sugars and non reducing sugars.
Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.
#15 SN 3M

Question

Write short note on: Sucrose.
OR
Explain: Sucrose is a non reducing sugar (figure not essential).

Answer

One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D–(+)glucose and D–(–)–fructose.
C12H22O11 + H2O C6H12O6 + C6H12O6
Sucrose D–(+)–Glucose D–(–)–Fructose
These two monosaccharides are held together by a glycosidic linkage between C1 of a–D Glucose and C2 of b–D 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.
#16 SN

Question

Short note on: Maltose

Answer

Maltose is composed of two a-D-glucose.
C12H22O11 + H2O C6H12O6 + C6H12O6
Maltose aDGlucose aD–Glucose
In maltose 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.
#17 SN

Question

Write Short note on: Lactose

Answer

It is more commonly known as milk sugar since this disaccharide is found in milk.
It is composed of b-D-galactose and b-D-Glucose
C12H22O11 + H2O C6H12O6 + C6H12O6
Lactose b-D-galactose b-D-glucose
The linkage is between C1 of galactose and C4 of glucose.
Free aldehyde group may be produced at C–1 of glucose unit, hence it is also a reducing sugar.
#18 SUB 🖼 2

Question

Write detailed note on: Starch

Answer

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 a-glucose and consists of two components amylose and amylopectin.
  • Amylose is water soluble component which constitutes about 15–20% of starch.
  • Chemically amylose is along unbranched chain with 200–1000 a-D-(+)-glucose units held together by C1-C4 glycosidic linkage.
Amylopectin is insoluble in water and constituents about 80–85% of starch.
It is a branched chain polymer of aDglucose units in which chain is formed by C1C4 glycosidic linkage where as branching occurs by C1C6 glycosidic linkage.
#19 SN 🖼 1

Question

Write Short note on: Cellulose.

Answer

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 chain polysaccharide composed only of bDglucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.
#20 SUB 3M 🖼 2

Question

What is the basic structural difference between starch and cellulose?

Answer

Starch consists of two components amylose and amylopectin.
Amylose is a long linear chain of aD(+)glucose units joined by C1C4 glycosidic linkage (alink).
Amylopectin is a branched chain polymer of aDglucose units, in which the chain is formed by glycosidic C1C4 glycosidic linkage and the branching occurs by C1C6 glycosidic linkage.
On the other hand cellulose is a straight chain polysaccharide of bDglucose units joined by C1C4 glycosidic linkage (blink).
#21 SN

Question

Write short note on: Glycogen

Answer

“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 down glycogen into glucose.
#22 SUB 2M

Question

What is glycogen? How is it different from starch?

Answer

The carbohydrates are stored in animal body as glycogen.
It is also known as animal starch because it’s structure is similar to amylopectin and is highly branched.
A major difference between glycogen and starch is their chain length.
The amylopectin chain consists of 20–25 aD glucose units while glycogen has a chain of 10–14 aDglucose units.
Glycogen is more branched than amylopectin.
It is present in liver, muscle and brain.
When the body needs glucose, enzymes break down glycogen to form glucose.
Glycogen is also found in yeast and fungi.
Starch is the most important dietary source for human beings.
High content of starch is found in cereals, roots, tubers and some vegetables.
#23 SUB 2M

Question

Write the two main functions of carbohydrates in plants.

Answer

The two main functions of carbohydrates are as follows:
(i) As a constitutive component of plant cell wall: Cellulose, a polysaccharide is used to build the cell wall.
(ii) As a storage of food: Polysaccharides such as starch as storage molecule.
#24 SUB 3M

Question

Explain the importance of carbohydrate compounds.

Answer

Carbohydrates are essential for life in both plants and animals.
They form a major portion of our food.
Honey has been used for 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 cloth ourselves 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 are present in nucleic acids.
Carbohydrates are found in biosystem in combination with many proteins and lipids.
#25 SUB 2M

Question

Give primary information about the protein compounds.

Answer

Protein 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 the body.
The word protein is derived from greek word "Proteios" which means primary or of prime importance.
All proteins are polymers of aamino acids.
#26 SUB 2M 🖼 3

Question

How do you explain the amphoteric behaviour of amino acids?
OR
Explain the Zwitter ion formation in an amino acid compounds.

Answer

Amino acids are usually colourless, crystalline solids.
These are water-soluble, high melting solids and behaves like salts rather than simple amines or carboxylic acids.
This behaviour is due to the presence of both acidic (Carboxylic acid 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.
#27 SUB 🖼 1

Question

What are amino acid compounds? Explain the nomenclature of amino acid compounds.

Answer

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 be classified as a, b, g, d, and so on.
Only a amino acids are obtained on hydrolysis of proteins.
They may contain other functional groups also.
All a amino acids have trivial names, which usually reflect the property of that compound or it's 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.
e.g. glycine 3-letter symbol is 'Gly' and one letter symbol is "G".
#28 SUB 2M

Question

What are essential and non essential amino acids? Give 2 examples of each type.
OR
Write note on classification of amino acid compounds.

Answer

Amino acid compounds can be classified on the basis of two things:
Relative number of amino and carboxyl group available in their molecule.
According to the need of them in our body.
(i) 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. e.g., Glycine, Alanine, Leucine etc.
More number of carboxyl than amino groups makes it acidic e.g., Glutamic acid, Aspartic acid.
More number of amino than carboxyl groups makes it basic. e.g., Arginine, Lysine, Tryptophan and Histidine.
(ii) Based on need of amino acids can be classified in two types. Essential and non-essential.
The amino acids which can be synthesized in the body are known as non essential amino acids. e.g., Glycine, Alanine, Glutamine etc.
The amino acids which can not be synthesized by the body but must be obtained through diet are known as essential amino acid. e.g., Valine, Leucine, Isoleucine. etc...
#29 SUB 4M ▦ 1

Question

Give the names naturally occurring amino acid compounds along with their 3 letter and 1-letter symbol.

Amino Acids

Side Chain R

Symbol

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

H2N––CH2–CH2

GIn

Q

11. Aspargine

H2N––CH2

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

a = Entire structure

Pro

P

Answer

In the above table * means Essential Amino Acids.
#30 SUB 2M

Question

Explain the property of optical activity of an amino acids.

Answer

Except glycine, all other naturally occurring a-amino acids are optically active since the a-carbon atom is asymmetric.
These exist both in "D" and "L" forms.
Most naturally occurring amino acids have
L configuration.
L amino acids are represented by writing the –NH2 group on left hand side.
#31 SUB 3M 🖼 1

Question

What is a peptide bond or peptide chain? Elaborate on how protein compounds are formed from amino acid compounds.

Answer

Proteins are the polymers of a-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 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.
The third amino acids 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 to 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 10,000 u 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.
#32 SUB 3M 🖼 4

Question

What are the common types of secondary structure of proteins?
OR
Write note on secondary structure of proteins.

Answer

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. a - helix and b - pleated sheet structure.
The 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.
α - Helix structure: a - 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 diagram.
α-Helix structure of proteins
β-Pleated sheet structure: In b-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 k nown as b-pleated sheet.
#33 SUB

Question

Write a note on primary structure of protein compounds.

Answer

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 this sequence of amino acids is said to be the primary structure of that protein.”
Any change in the primary structure i.e., the sequence of amino acids creates a different protein.
#34 SUB 2M 🖼 1

Question

What type of bonding helps in stabilising the α-helix structure of proteins?

Answer

H-bonding helps in stabilising the a-helix structure of proteins.
A polypeptide chains forms all possible hydrogen bonds in a-helix structure.
In this structure, polypeptide chain twists 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.
#35 SUB 2M 🖼 1

Question

Write a note on tertiary structure of protein compound.

Answer

“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 and structures of proteins are hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction.
Tertiary structure
#36 SUB 1M

Question

What is quaternary structure of proteins?

Answer

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.
e.g. haemoglobin
#37 SUB 2M

Question

Differentiate between globular and fibrous proteins.
OR
Explain the classification of protein compound based on its molecular shape.

Answer

Proteins can be classified into two types on the basis of their molecular shape.
(a) Fibrous protein compounds: 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 example are keratin (present in hair, wool, silk) and myosin (present in muscles) etc.
(b) Globular protein compounds: This structures 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.
#38 SUB 2M

Question

What is the effect of denaturation of the structure of proteins?
OR
Explain the denaturation of protein compounds.

Answer

“Protein found in a biological system with a unique 3-D 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 and structures are destroyed but structure remains intact.
The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk is caused due to the formation of lactic acid by the bacteria present in milk.
#39 SUB 1M

Question

What are enzymes?

Answer

Enzymes are proteins that catalyse biological reactions.
They are very specific in nature and catalyse only a particular substrate. Enzymes are usually named after the particular substrate or class of substrate and sometime after the particular reaction.
#40 SUB 2M

Question

Explain the mechanism of enzyme action.

Answer

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, while the activation energy is only 2.15 KJ / mol. when hydrolysed by the enzyme sucrase.
#41 SUB 2M 🖼 1

Question

Explain the nomenclature of Enzymes.

Answer

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.
C12H22O11 + H2O 2C6H12O6
Maltose Glucose
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’.
#42 SUB 3M

Question

What are vitamin compounds? Give primary information about vitamin compounds.

Answer

“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 can not 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.
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.
#43 SUB 2M

Question

How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
OR
Explain the classification of vitamin compounds.

Answer

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.
Vitamin K increases blood clotting time.
#44 SUB 2M

Question

Why B complex vitamine compounds are essential for us? Describe their important sources.

Answer

Deficiency of vitamin B1 (thiamine) leads to the disease like Beriberi (loss of appetite, retarded growth). So, it is essential for us. Important sources are yeast, milk, green vegetables and cereals. etc....
Deficiency of vitamin B2 (Riboflavin) leads to the cheilosis (fissuring at corners of mouth and lips), digestive disorders and burning sensation of the skin, so it is essential for us. Important sources are milk, egg white, liver and kidney.
Deficiency of vitamin B6 (Pyridoxine) leads to convulsions so it is essential for us. Important sources are yeast, milk, egg yolk, cereals and grams.
Deficiency of vitamin B12 leads to pernicious anaemia (RBC deficient in haemoglobin), so it is essential for us. Important sources are meat, fish, egg and curd.
#45 SUB 3M

Question

Why vitamin D, E and K compounds are essential for us? Describe their important sources.

Answer

Deficiency of vitamin D leads to Rickets (bone deformities in children) and osteomalacia (soft bones and joint pain in adults), so it is essential for us. Important sources are exposure to sunlight, fish and egg yolk.
Deficiency of vitamin E leads to increased fragility of RBCs and muscular weakness, so it is essential for us. Important sources are vegetable oils like wheat germ oil, sunflower oil etc.
Deficiency of vitamin K leads to increased blood clotting time so it is essential for us. Important sources are green leafy vegetables.
#46 SUB 3M 🖼 7

Question

Write a note on chemical composition of nucleic acid compounds.

Answer

Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric acid and nitrogen containing hetero cyclic compounds (called bases).
In DNA molecules the sugar moiety is b-D-2 deoxy-ribose where as in RNA molecule it is b-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).
#47 SUB 2M

Question

What are the different types of RNA found in the cell?
OR
Write a note on structure of RNA.

Answer

In secondary structure of RNA, helices are present which are only single stranded.
Sometimes they fold back on themselves.
RNA molecules are of 3 types and they perform different functions.
They are named as messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).
#48 SUB 2M

Question

Give primary information about nucleic acid compounds.

Answer

Every generation of each and every species resembles its ancestors in many ways.
These characteristics transmitted from one generation to the next generation.
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).
Nucleic acids are long chain polymers of nucleotides, so they are also called Polynucleotides.
#49 SUB 🖼 2

Question

Explain the structure of nucleic acid compounds.

Answer

A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside.
In nucleosides the sugar carbons are number as 1', 2', 3' etc. In order to distinguish these from the bases - when nucleoside is linked to phosphoric acid at 5'-position of sugar moiety, we got a nucleotide.
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 figure.
A simplified version of nucleic acid chain is as shown below.
#50 SUB 2M

Question

What is the difference between a nucleoside and a nucleotide?

Answer

A nucleoside is formed by the attachment of a base to 1’ position of sugar.
On the other hand all three basic component of nucleic acids (i.e. pentose sugar, phosphoric acid and base) are present in nucleotide.
#51 SUB 2M 🖼 4

Question

Write a note on: Double Helical structure of DNA.
5'
5'
T
G
C
G
A
T
C
C
A
C
G
T
T
G
C
A
A
G
A
T
C
A
T
3'
C
T
A
T
A
G
A
T
3'

Answer

Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure too.
James Watson and Francis Crick gave a double strand helix structure for DNA. Two nucleic acid chains are wound about each other and held together by hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine where as cytosine forms hydrogen bonds with guanine.
#52 SUB 2M

Question

Describe the biological functions of nucleic acid compounds.

Answer

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.
DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells.
Actually, the proteins are synthesized by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.
#53 SUB 2M

Question

What are hormones?

Answer

Hormones are molecules that act as intracellular messengers. These are produced by endocrine glands in the body and are poured directly in the blood stream which transports them to the site of action.
In terms of chemical nature, some of these are steroids. e.g. estrogens and androgens, some are polypeptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine.
#54 SUB 3M

Question

What are the functions of hormones in body ?

Answer

Hormones have several functions in the body.
They help to maintain the balance of biological activities in the body.
The role of insulin in keeping the blood glucose level with in the narrow limit is a example of this function.
Insulin is released in response to the rapid rise in blood glucose level.
On the other hand hormone glucagon tends to increase the glucose level in the blood.
The two hormones together regulate the glucose level in the blood.
Epinephrine and norepinephrine mediate responses to external stimuli.
Growth hormone and sex hormones play role in growth and development.
Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine.
Abnormally low level of thyroxine leads to hypothyroidism which is characterized by lethargyness and obesity.
Increased level of thyroxine causes hyperthyroidism.
Low level of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This condition is largely being controlled by adding sodium iodide to commercial table salt. (‘‘Iodized’’ salt).
#55 SUB 3M

Question

Give the function of steroid hormones.

Answer

Steroid hormones are produced by adrenal cortex and gonads (testes in males and ovaries in female)
Hormones released by the adrenal cortex play very important role in the function of the body.
For example, glucocorticoids control the carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.
The mineralocorticoid control the level of excretion of water and salt by the kidney. If adrenal cortex does not function properly then one of the results by addison’s disease characterized by hypoglycemia, weakness and increased susceptibility to stress.
The disease is fatal unless it is treated by glucocorticoids and mineralocorticoids.
Hormones released by gonads are responsible for development of secondary sex characters.
Testosterone is the major sex hormone produced in males. It is responsible for development of secondary male characteristics and estradiol is the main female sex hormones. It is responsible for development of secondary female characteristics and participates in the control of menstrual cycle.
Progesterone is responsible for preparing the uterus for implantation of fertilized egg.
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#56 SUB

Question

Glucose or Sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.

Answer

A glucose molecules contains five –OH groups while a sucrose molecule contains eight –OH groups. Thus, glucose and sucrose undergo extensive H–bonding with water. Hence, these are soluble in water. But, cyclohexane and benzene do not contain –OH groups. Hence, they can not undergo H–bonding with water and as a result, they are insoluble in water.
#57 SUB

Question

What are the expected products of hydrolysis of lactose?

Answer

Lactose is composed of bD–galactose + bD–glucose, thus on hydrolysis it gives bD–galactose and bD–glucose.
C12H22O11 + H2O C6H12O6 + C6H12O6
Lactose bD–galactose bD–glucose
#58 SUB 2M

Question

How do you explain the absence of aldehyde group in the penta-acetate of D-glucose?

Answer

D–glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehyde (–CHO) group of carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure to give an oxime. But penta acetate of
D
glucose does not react with NH2OH. This is because penta acetate does not form an open chain structure.
#59 SUB 2M

Question

The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

Answer

Both acidic (Carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids.
In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept, thus giving rise to a dipolar ion known as zwitter ion.
Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water.
But halo acids do not exhibit such dipolar behaviour. For this reason the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids.
#60 SUB 1M

Question

Where does the water present in the egg go after boiling the egg?

Answer

When an egg is boiled, the proteins present inside the egg gets denatured and coagulate. After boiling the egg the water present in it is absorbed by the coagulated protein through H-bonding.
#61 SUB 1M

Question

Why cannot vitamin C be stored in our body?

Answer

Vitamin C cannot be stored in our body because it is water soluble. As a result it is readily excreted in the urine.
#62 SUB 1M

Question

What products would be formed when a nucleotide form DNA containing thymine is hydrolysed?

Answer

When a nucleotide for the DNA containing thymine is hydrolyzed, thymine b–D–2 deoxyribose and phosphoric acid are obtained as products.
#63 SUB 1M

Question

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?

Answer

A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine (A) always pairs with thymine (T) while cytosine (C) always pairs with guanine (G). So on hydrolysis of DNA, the quantity of adenine produced is equal of that of thymine (T) and Adenine (A). Similarly, the quantity of cytosine (C) is equal to that of guanine (G). But when RNA is hydrolysed there is no relationship among the quantities of the different bases obtained. Hence, RNA is single stranded.
Class 12 Chemistry (Part 2) 021
S src: NCERT Textbook Exercise Questions And Answers match 75% type: 25 Q ⤓ Export ZIP
#64 SUB

Question

What are monosaccharides?

Answer

Refer Que. no. 3 Page no. 307
#65 SUB

Question

What are reducing sugars?

Answer

Refer Que. no. 14 Page no. 311
#66 SUB 2M

Question

Write the two main functions of carbohydrates in plants.

Answer

Refer Que. no. 23 Page no. 313
#67 SUB 2M

Question

Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxy ribose; maltose; galactose; fructose and lactose.

Answer

Monosaccharide compounds:
Ribose, 2-deoxyribose, galactose, and fructose.
Disaccharide compounds: Maltose and lactose
#68 SUB

Question

What do you understand by the term glycosidic linkage?

Answer

Refer Que. no. 13 Page no. 310
#69 SUB 2M

Question

What is glycogen? How is it different from starch?

Answer

Refer Que. no. 22 Page no. 313
#70 SUB 2M

Question

What are the hydrolysis products of
(i) sucrose and (ii) lactose?

Answer

(i) On hydrolysis sucrose give one molecule of
a–D glucose and b–D fructose.
C12H22O11 + H2O C6H12O6 + C6H12O6
Sucrose a–D Glucose b–D Fructose
(ii) On hydrolysis lactose gives one molecule of
β-D-galactose and β-D-glucose.
C12H22O11 + H2O C6H12O6 + C6H12O6
Lactose β–D– β–D–
Galactose Glucose
#71 SUB

Question

What is the basic structural difference between starch and cellulose?

Answer

Refer Que. no. 20 Page no. 313
#72 SUB 3M 🖼 1

Question

What happens when D-glucose is treated with the following reagents?
(i) HI (ii) Bromine water (iii) HNO3

Answer

(i) Reactions with HI: When D-glucose is heated with HI for long time, n-Hexane is formed, it shows all 6–carbon atoms are linked in linear chain.
(ii) Reactions with Br2 water: Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with mild oxidising agent like Br2 water. This indicates that the carbonyl group is present as an aldehyde group.
(iii) Reactions with HNO3: On oxidation with HNO3 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.
#73 SUB 3M

Question

Enumerate the reactions of D-glucose which can not be explained by its open chain structure.

Answer

Open chain structure of glucose has free aldehyde (–CHO) group even though following reactions and facts could not be explained by this structure.
(i) Despite having the aldehyde group, glucose does not give Schiff’s test and its does not form the hydrogensulphate addition product with NaHSO3.
(ii) The penta-acetate of glucose does not react with hydroxyl amine indicating the absence of free –CHO group.
(iii) Glucose is found to exist in two different crystalline forms which are named as a and b. The a form of glucose (m.p 419K) is obtained by crystallization from concentrated solution of glucose at 303 K while the b-form (m.p 423K) is obtained by crystallization from hot and saturated aqueous solution at 371 K.
#74 SUB 2M

Question

What are essential and non essential amino acids? Give 2 examples of each type.

Answer

Refer Que. no. 28 Page no. 314
#75 SUB

Question

Define the following as related to proteins.
(i) Peptide linkage (ii) Primary structure
(iii) Denaturation.

Answer

(i) Peptide linkage: The amide formed between –COOH group of one molecule of an amino acid and –NH2 group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.
(ii) Primary structure: The primary structure of protein refers to the specific sequence in which various amino acids are present in it, i.e., the sequence of linkage between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.
(iii) Denaturation: In a biological system, a protein is found to have a unique 3-D structure and a unique biological activity. In such a situation, the protein is called native protein.
However, when the native protein is subjected to physical changes such as change in temperature or chemical change such as change in pH, its H-bonds are disturbed.
The disturbance unfolds the globules and uncoils the helix. As a result the protein loss its biological activity.
This loss of biological activity by the protein is called denaturation.
During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered.
One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.
#76 SUB 3M

Question

What are the common types of secondary structure of proteins?

Answer

Refer Que. no. 32 Page no. 316
#77 SUB 2M

Question

What type of bonding helps in stabilising the a-helix structure of proteins?

Answer

Refer Que. no. 34 Page no. 317
#78 SUB 2M

Question

Differentiate between globular and fibrous proteins.

Answer

Refer Que. no. 37 Page no. 317
#79 SUB 2M

Question

How do you explain the amphoteric behaviour of amino acids?

Answer

Refer Que. no. 26 Page no. 314
#80 SUB 1M

Question

What are enzymes?

Answer

Refer Que. no. 39 Page no. 317
#81 SUB 2M

Question

What is the effect of denaturation of the structure of proteins?

Answer

Refer Que. no. 38 Page no. 317
#82 SUB 2M

Question

How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Answer

Refer Que. no. 43 Page no. 318
#83 SUB 2M

Question

Why are vitamin A and vitamin C essential to us? Give their important sources?
[Topic 10.4] [2 Marks]

Answer

Deficiency of vitamin A leads to xerophthalmia (hardening of cornea of eye) and night blindness so it is essential for us. Important sources are fish liver oil, carrots butter and milk.
Deficiency of vitamin C leads to scurvy (bleeding gums) so it is essential for us - Important sources are citrus fruits, amla and green leafy vegetables.
#84 SUB 2M

Question

What are nucleic acids? Mention their two important function.

Answer

A polymer made up of pentose sugar, Heterocyclic bases and phosphate ion containing nucleotide is known as nucleic acids.
Nucleic acids are responsible for protein synthesis in a cell. It is responsible for the transmission of inherent characters from one generation to the next. This process of transmission is called heredity.
#85 SUB 2M

Question

What is the difference between a nucleoside and a nucleotide?

Answer

Refer Que. no. 50 Page no. 320
#86 SUB 2M

Question

The two strands in DNA are not identical but are complementary. Explain.

Answer

In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bonds with guanine, while adenine forms hydrogen bonds with thymine. As a result the two strands are complementary to each other.
#87 SUB 2M ▦ 2

Question

Write the important structural and functional differences between DNA and RNA.

Answer

Structural differences

DNA

RNA

(i) The sugar moiety in DNA molecules is bD2–deoxyribose

(i) The sugar moiety in RNA molecules is bD–ribose.

(ii) DNA contains cytosine
and thymine as a pyrimidine base where
as guanine and adenine
as a purine base.

(ii) RNA contains cytosine and uracil as a pyrimidine base where as guanine and adenine as a purine base.

(iii) The helical structure of DNA is double stranded.

(iii) The helical structure of RNA is single stranded.

(iv) DNA molecule is too
large. It’s molecular
mass is 6 × 106 – 16 × 106 u.

(iv) RNA molecule is quite smaller. It’s molecular mass is 20000–40000 u.

Functional differences

DNA

RNA

(i) DNA molecules are capable of self-replication.

(i) RNA molecules are not capable of self-replication.

(ii) DNA is responsible for the transmission of genetic characters.

(ii) RNA is responsible for the synthesis of protein compounds.

#88 SUB 2M

Question

What are the different types of RNA found in the cell?

Answer

Refer Que. no. 47 Page no. 319
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#89 SUB 1M 🖼 15

Question

Intermediate End product
compound
What substances are Z and Y respectively?
81. Haworth projection belongs to which of the following?
(a) Maltose, Glucose
(b) Sucrose, Glucose
(c) Lactose, Fructose
(d) Maltose, Fructose

Answer

(A) a-D-(+)-glucose (B) b-D-(+)-galactose
(C) a-D-(+)-galactose (D) b-D-(+)-glucose
#90 MCQ ⚠ needs answer review 1M ▦ 2

Question

Egg yolk 2. Tomato 3. Sunlight
182. Connect vitamins with diseases caused by its deficiency:
183. Nucleic acid is a polymer of whom?
184. In DNA, two hydrogen bonds are found between which bases?
185. If the sequence of bases in one chain of nucleotides in DNA is CAGCTG, then what will be the sequence of bases in the corresponding chain of nucleotides?
186. Which base is a derivative of purine?
187. In DNA the base units are joined to each other by_________.
188. Which of the given pairs is correct for H-bonding in nucleotide?
189. RNA differs from DNA, because RNA has:
190. Which nitrogen base is absent in DNA?
191. Purine base is _________.
192. _________ base is present in RNA.
193. Which base is bicyclic?
194. Which is purine base?
195. Which of the following base is absent in DNA?
196. Which base is join with adenine in DNA?
197. Which nitrogen base is absent in RNA?
198. Which biomolecule is responsible for heredity?
199. Which sugar is present in RNA?
200. What is formed by the hydrolysis of DNA and RNA?
201. Which base is absent in DNA?
202. Which base is absent in RNA?
203. How the structure of DNA suggested?
204. How many bonds are possible between C and G in the structure of DNA?
205. Which pyrimidine base is present in DNA?
206. Which component is present in nucleotide?
207. Which base is present in RNA instead of thymine?
208. Two nucleotides join each other by which linkage?
209. What is the function of DNA in any living organisms?
210. Which statement is incorrect about ribose sugar?
211. Match Column I and Column II correctly.
212. Among the following statements select the correct option using T for true statement and F for false statement :
(i) Chromosomes at the center of a living cell are responsible for heredity.
(ii) Nucleotide chains in DNA are linked to each other by diester bonds.
(iii) A unit formed by base joining to 5th carbon of pentose sugar is called nucleoside.
(iv) RNA has a uracil pyridine base.
213. Which hormone plays an important role in the metabolism of carbohydrates in the body?
214. Which of the following is amine containing hormone?
215. Which of the following is steroid hormone?
216. Which hormone controls the amount of water and salts excreted by the kidneys?
217. Addison's disease is caused by whose mal function?
218. What is the main sex hormone produced in men?
219. Assertion : Proteins lose their biological activity when they undergo denaturation.
Reason : Denaturation involves the disruption of hydrogen bonds and changes in the secondary and tertiary structures of the protein.
220. Assertion : The primary structure of a protein is intact during denaturation.
Reason : Denaturation breaks down the peptide bonds that hold amino acids together in a protein chain.
221. Assertion : Glucose can exist in both α and β-anomeric forms due to the formation of a cyclic structure.
Reason : The anomeric carbon is the carbon that forms the glycosidic bond in carbohydrates.
222. Assertion : The anomeric carbon in glucose is also called the glycosidic carbon.
Reason : The anomeric carbon is involved in the formation of the cyclic structure of glucose.
223. Assertion : Reducing sugars can reduce Tollen’s reagent.
Reason : Reducing sugars have a free aldehyde or ketone group that can donate electrons to reduce Tollen’s reagent.
224. Assertion : The furanose form of glucose is more common in nature than the pyranose form.
Reason : The pyranose form of glucose is a six-membered ring, which is more stable than the furanose form.
225. Assertion : Denaturation of proteins can be reversed under certain conditions.
Reason : Denaturation disrupts hydrogen bonds, but these bonds can reform under specific conditions, allowing the protein to regain its original structure.
226. Assertion : All vitamins are synthesized in the body and do not need to be obtained from the diet.
Reason : Vitamins are essential nutrients that the body requires in small amounts for various metabolic processes.
227. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) Proteins can sometimes regain their original structure and function if denaturation conditions are removed.
(ii) The primary structure of a protein is changed during denaturation.
(iii) Glucose in its cyclic form can exist as either a furanose or pyranose structure.
(iv) All vitamins are water-soluble and cannot be stored in the body.
228. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) The anomeric carbon in a glucose molecule is located at carbon atom C-5.
(ii) Carbohydrates are classified based on the number of sugar units they contain.
(iii) Denaturation of a protein affects only the primary structure.
(iv) α-D-glucose and β-D-glucose are anomers due to the difference in the configuration at C-1.
229. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) Vitamin-C can be synthesized by the human body.
(ii) Enzymes are proteins that act as catalysts in biochemical reactions.
(iii) Pentose sugars are always five-membered ring structures.
(iv) The glycosidic bond forms between the anomeric carbon of one sugar and the hydroxyl group of another.
230. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) The term reducing sugar refers to sugars that can reduce certain metal ions.
(ii) The secondary structure of proteins involves the folding of the polypeptide chain into alpha helices or beta sheets.
(iii) Proteins are made up of α-amino acids, which contain an amine group and a carboxyl group.
(iv) A furanose structure in carbohydrates has a six-membered ring.
231. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) Proteins can have more than one polypeptide chain, forming a quaternary structure.
(ii) In the cyclic form of glucose, the anomeric carbon is C-2.
(iii) Vitamins are not essential for metabolic processes.
(iv) Denaturation of proteins is a reversible process in some cases.
232. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) α-D-glucose and β-D-glucose have the same configuration at the C-1 carbon atom.
(ii) All carbohydrates reduce Fehling’s solution.
(iii) The pyranose form of glucose is a six-membered ring structure.
(iv) The structure of a protein’s primary structure is determined by the sequence of nucleotides in its DNA.
233. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) Proteins are polymers of α-amino acids.
(ii) Vitamin-A is water-soluble and can be stored in the body.
(iii) All pentoses exist in the pyranose form in their free state.
(iv) Glycosidic bonds are formed between the anomeric carbon of one sugar and a hydroxyl group of another.
234. Select the correct option using T (True) or
F (False) symbol for the following statements.
(i) The denaturation of proteins only affects the primary structure.
(ii) The Haworth projection represents the cyclic structure of carbohydrates.
(iii) The β-D-glucose anomer has the hydroxyl group on C-1 pointing up in the cyclic form.
(iv) Furanose structures are six-membered rings.
Class 12 Chemistry (Part 2) 022

Section X

Section Y

(i)

Vitamin A

(A)

Pernicious Anemia

(ii)

Vitamin B1

(B)

Bleeding

(iii)

Vitamin E

(C)

Beriberi

(iv)

Vitamin K

(D)

Impotence

(E)

Night blindness

I

II

(i)

Glycosidic linkage

(P)

Linkage between two nucleotides

(ii)

Peptide bond

(Q)

Insulin

(iii)

Disulphide linkage

(R)

Linkage between two amino acids

(iv)

Phosphodiester linkage

(S)

Linkage
between two monosaccharides

Options

  1. (A) FTTF
  2. (B) TFFT
  3. (C) TTTF
  4. (D) FTFT

Answer

not detected

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#117 CS

Question

Carbohydrates are optically active polyhydroxy aldehydes and ketones. They are also called saccharides. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. Glucose, the most important source of energy for mammals, is obtained by the hydrolysis of starch. Vitamins are necessary food factors required in the diet.
Proteins are the polymers of α-amino acids and perform various structural and dynamic functions in the organisms. Deficiency of vitamins leads to many diseases.
(a) What are reducing sugars?
(b) Why are carbohydrates considered optically active, and what is their structural significance?
(c) What are proteins made of?
(d) Why is vitamin C not stored in our body?

Answer

(a) Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent.
(b) Carbohydrates are optically active due to the presence of chiral carbon atoms, and their polyhydroxy aldehyde or ketone structure enables energy production and other biological functions
(c) Proteins are polymers of α–amino acids.
(d) Vitamin C cannot be stored in our body because it is water-soluble and is excreted in urine.
#118 CS 🖼 1

Question

Pentoses and hexoses undergo intramolecular hemiacetal or hemiketal formation due to the combination of the hydroxyl group (-OH) with the carbonyl group. The resulting structure is either a five- or six-membered ring containing an oxygen atom. In their free state, all pentoses and hexoses exist predominantly in the pyranose form (resembling pyran). However, in the combined state, some of them exist as five-membered cyclic structures called furanose.
The cyclic structure of glucose is represented by the Haworth projection.
α–D–glucose and βDglucose differ in the configuration at the anomeric (C1) carbon atom and are therefore called anomers. The C1 carbon atom is referred to as the anomeric carbon.
The six-membered cyclic structure of glucose is known as the pyranose structure.
(a) What causes pentoses and hexoses to form cyclic structures?
(b) What is the difference between pyranose and furanose structures?
(c) What are anomers, and how do α and β-D-glucose differ?
(d) In the structure of α-D-glucose, which carbon is the anomeric carbon?

Answer

(a) Pentoses and hexoses form cyclic structures due to intramolecular hemiacetal or hemiketal formation involving the hydroxyl and carbonyl groups.
(b) Pyranose structures are six-membered rings resembling pyran, while furanose structures are five-membered rings resembling furan.
(c) Anomers are isomers that differ at the anomeric (C–1) carbon α and β–D–glucose have different configurations at this carbon.
(d) The anomeric carbon is the C–1 carbon, which is bonded to the oxygen atom in the ring and determines the α or β configuration.
#119 CS

Question

When a protein in its native state is exposed to physical factors such as temperature changes or chemical factors like pH alterations, its hydrogen bonds are disrupted. As a result, the protein’s globular structure unfolds, its helix uncoils, and it loses its biological activity. This process is known as denaturation of protein. During denaturation, the secondary and tertiary structures are altered, but the primary structure remains unchanged. Examples of protein denaturation include the coagulation of egg white upon boiling, the curdling of milk, and the formation of cheese when acid is added to milk.
(a) What is protein denaturation?
(b) What causes denaturation of proteins?
(c) Which structures of a protein are affected during denaturation?
(d) Give examples of protein denaturation in daily life.

Answer

(a) Protein denaturation is the loss of biological activity due to disruption of hydrogen bonds, altering secondary and tertiary structures but not the primary structure.
(b) Denaturation is caused by physical factors like temperature changes or chemical factors such as pH changes.
(c) Denaturation affects the secondary and tertiary structures, but the primary structure remains unchanged.
(d) Examples include the coagulation of egg white upon boiling, curdling of milk, and cheese formation when acid is added to milk. (Globular proteins are converted into fibrous proteins and Fibrous proteins are converted into globular protein.
#120 CS 🖼 1

Question

A medical research insitute is studying how carbohydrates, protein and enzymes behave inside human cells. To understand metabolic pathways, Scientist analyze.
Structure of monosaccarides.
Mutarotation
Glycosidic bond formation
Denaturation and struction of protein Enzyme specificity and inhibition.
During experimentation :
1. An aqueous glucose solution rotates plane - polarised light and its rotation changes with time mutarotation.
2. Hydrolysis of sucrose gives glucose and fructose.
3. Heating egg protein leads to denaturation.
4. Enzyme show heighest activity at optimum pH and temperature.
5. A competitive inhibitor reduces enzyme activity by occupying the active site.
Additional data :
Glucose exists mostly in cyclic hemiacetal form.
Proteins contain peptide bonds (–Co–NH–)
Enzymes are hihgly specific to substrates.
(a) Explain why glucose shows mutarotation in aqueous solution.
(b) Why is sucrose a non-reducing sugar even through it contains glucose and fructose ?
(c) How does denaturation affect the biological activity of proteins ?
(d) Explain the difference between competitive and non-competitive. enzyme inhibition.

Answer

(a) In water, glucose interconverts between α - glucose and β - glucose.
This happens via open-chain aldehyle form, causing a change in specific rotation until equilibrium is reached.
(b) In sucrose, the anomeric carbon of glucose
(1) and fructose (2) are both involved in glycosidic bonding.
Thus, neither ring can open no free aldehyde or keto group cannot reduce Tollen’s or fehling’s reagent.
(c) Denaturation breaks, hydrogen bonds, hydrophobic interactions and salt bridges. So, protein losses tertiary / Secondary structure, Active site destroyed and loss of biological function.
(d)
#121 CS

Question

A biotechnology firm studies nucleic acids, vitamins, and amino acids to design new nutritional supplements :
They analyze : Structure of DNA - RNA, Base pairing rules, essential vs non-essential amino acids, vitamin deficiencies and hydrolysis of nucleotides.
Key Observations :
1. DNA contain A, T, G, C, where as RNA contain A, V, G, C
2. Adenine paires with thymine by two H-bonds; guanine pairs with cytosine by 3 H-bonas
3. Vit. C deficiency Seurvy
4. Thiamin deficiency beriberi
5. Hydrolysis of RNA gives a mixture of ribose, phosphate and nitrogenous bases.
Amino acid studid :
Amino acid pl Type
Glycine 6.0 Neutral
Lysin 9.7 Basic
Glutamic Acid 3.2 Acidic
(a) Why is DNA more stable than RNA ? Explain using base pairing and sugar structure.
(b) What products are obtained on complete hydrolysis of RNA
(c) Arrang the amino acid in increasing order of isoelectric points (pl)
(d) Explain why glutamic acid behaves as an acidic amino acid.

Answer

(a) Reason,
(i) DNA contain deoxyribose, lacking 2’ - OH (less reactive)
(ii) RNA has 2’ - OH (prone to hydrolysis)
(iii) G C pairing in DNA (stronger)
(iv) DNA is double stranded
(b) Ribose sugar, Nitrognous bases (A, V, G, C) and phosphoric Acid.
(c) Increasing pl order :
Glutamic acid (3.2) < Glycine (6.0)
(d) Glutamic Acid is acidic, because of...
(i) it contain an extra –COOH group
(ii) Losses H+ easily
(iii) forms stable carboxylate ion.
Carbohydrate: Polyhydroxy aldehyde or ketone, or a compound that yields them on hydrolysis.
Monosaccharide: The simplest carbohydrate unit that cannot be hydrolysed further (e.g., glucose, fructose).
Disaccharide: Carbohydrate made of two monosaccharides linked by a glycosidic bond (e.g., sucrose, lactose).
Polysaccharide: Long-chain polymer of monosaccharide units (e.g., starch, glycogen, cellulose).
Reducing Sugar: Sugar capable of reducing Tollens’/Fehling’s reagent due to free aldehyde or hemiacetal group.
Glycosidic Bond: Covalent linkage between the anomeric carbon of one sugar and the –OH of another.
Anomers: Isomers differing at the anomeric carbon (α- and β-forms) in cyclic sugars.
Amino Acid: Organic molecule containing –NH2
and –COOH groups on the same (α) carbon.
Peptide Bond: –CO–NH– linkage formed between two amino acids during protein formation.
Protein Structure: Primary (sequence), secondary (α-helix/β-sheet), tertiary (3D folding), and quaternary (multi-chain assembly).
Denaturation: Loss of protein structure and function due to heat, pH, or chemical action.
Enzyme: Biological catalyst (mostly proteins) that speeds up biochemical reactions with high specificity.
Cofactor / Coenzyme: Non-protein component required for enzyme activity (metal ion or organic molecule).
Vitamin: Essential micronutrient needed in small amounts; classified as water-soluble (B, C) or fat-soluble (A, D, E, K).
Hormone: Chemical messenger produced by endocrine glands that regulates physiological processes.
Nucleoside: Molecule consisting of a nitrogenous base attached to a pentose sugar (no phosphate).
Nucleotide: Phosphate + sugar + nitrogenous base; basic unit of DNA and RNA.
DNA: Double-stranded nucleic acid carrying genetic information; shows A–T and G–C base pairing.
RNA: Single-stranded nucleic acid involved in protein synthesis; contains uracil instead of thymine.

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