NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules

Here we provide NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules pdf, free NCERT solutions for Class 12 Chemistry Chapter 14 Biomolecules book pdf download. Now you will get step by step solution to each question.

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules

Biomolecules is a very important chapter and requires detailed understanding of the concepts. This topic is often asked in the examination and the students need to be thorough with the concepts.

The students can learn the structure, properties and classification of various biomolecules such as carbohydrates, nucleic acids, etc. These concepts are also taught in further studies. Therefore, the students need to be thorough with the basics.

NCERT INTEXT QUESTIONS

Question 1.
Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
Both glucose (C₆H₁₂O₆) and sucrose (C_l2H₂₂O_n) are organic compounds and are expected to be insoluble in water. But quite surprisingly, they readily dissolve in water. This is due to the presence of a number of OH groups (five in case of glucose and eight in sucrose) which are of polar nature. These are involved in the intermolecular hydrogen bonding with the molecules of H₂O (water). As a result, both of them readily dissolve in water.
Benzene (C₆H₆) and cyclohexane (C₆H₁₂) are hydrocarbons which donot have any polar group. They therefore, donot dissolve in water since there is hardly any scope of hydrogen bonding in their molecules with those of H₂O (water).

Question 2.
What are the expected products of hydrolysis of lactose ?
Answer:
The hydrolysis of lactose (disaccharide) can be done either with dilute HC1 or with enzyme emulsin. D-glucose and D-galactose are the products of hydrolysis. Both of them are monosaccharides with molecular formula C6Hi₂0₆.

Question 3.
How do you explain the absence of aldehydic group in the pentaacetate of D-glucose ?
Answer:
Glucose, as we know is an aldohexose and it is expected to give the characteristic reactions of the aldehydic group e.g., action with NH₂OH, HCN, Tollen’s reagent, Fehling reagent etc. However, the pentaacetyl glucose formed by the acylation of glucose with acetic anhydride does not give these reactions.

This means that the aldehydic group is either absent or is not available in the pentaacetyl glucose for chemical reactions. In fact, the aldehydic group is a part of the hemiacetal structure which the pentaacetyl derivative has. It is therefore, not free or available to take part in these reactions.

Question 4.
The melting points and solubility of amino acids in water are generally higher than those of corresponding haloacids. Explain.
Answer:
The amino acids exist as zwitter ions, H₃N⁺ — CHR-COO-. Due to this dipolar salt like character, they have strong dipole-dipole attractions. Therefore, their melting points are higher than corresponding haloacids which do not have salt like character.
Due to salt like character, amino acids intereact strongly with water. As a result, their solubility in water is higher than corresponding haloacids which do not have salt like character.

Question 5.
Where does water in the egg go after boiling the egg ?
Answer:
Upon boiling the egg, denaturation of globular protein present in it occurs. Water present probably gets either absorbed or adsorbed during denaturation and disappears.

Question 6.
Explain why vitamin C can not be stored in the body.
Answer:
Vitamin C is mainly ascorbic acid which is water soluble. It is readily excreted through urine and cannot be stored in the body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
The products obtained are 2-deoxy-D-ribose, . phosphoric acid and thymine.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained? What does this fact suggest about the structure of RNA?
Answer:
As we know, a molecule of DNA has a double strand structure and the four complementary bases pair each other. Cytosine (C) always pairs up with guanine (G) while thymine (T) is paired up with adinine (A). Because of the presence of the double strand structure, when a molecule of DNA is hydrolysed, in each pair the molar ratio of the bases remains the same. However, this is not seen when RNA is subjected to hydrolysis. This suggests that RNA has not a double strand structure like DNA. It exists as a single strand.

NCERT Exercises

Question 1.
What are monosaccharides?
Answer:
The simplest carbohydrates which cannot be decomposed into smaller products are known as monosaccharides.
the simple carbohydrates which cannot be hydrolvsed into still simpler compounds. About twenty different monosaccharides occur in nature. Monosaccharides are generally crystalline solids, sweet in taste and soluble in water. Their soluhility in water is due to extensive hydrogen bonding between water molecules and the -OH groups present in the molecules.

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduces Fehling’s solution to red precipitate of Cu₂0 or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars. Thus, D – (+) – glucose, D-(-)-fructose, D – (+) – maltose and D – (+) – lactose are reducing sugars.

Question 3.
Write two major functions of carbohydrates in plants. (C.B.S.E. Delhi 2008)
Answer:
(i) Cell walls of plants are made up of cellulose.
(ii) In the form of starch, carbohydrates act as storage molecules in plants.

Question 4.
Classify the following into monosaccharides and disaccharides :
Ribose, 2-deoxyribose,
maltose, galactose,
fructose and lactose.
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose.
Disaccharides: Maltose, lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage is used to link different monosaccharides in disaccharides and polysaccharides through oxygen atom. For example, glycosidic linkage is present in sucrose, lactose, maltose etc. These are all disaccharides.

Question 6.
What is glycogen? How is it different from starch?
Answer:
Glycogen is a condensation polymer of α-D glucose. Starch is not a single compound but is a mixture of two components—a water soluble component called amyldse (15- 20%) and water insoluble component amylopectin (80 – 85%). Amylose is a linear polymer of α – D – glucose. But both glycogen and amylopectin are branched polymers of α – D – glucose; father glycogen is more highly branched than amylopectin as amylopectin chains consists of 20 – 25 glucose units, glycogen chains consist of 10 – 14 glucose units.

Question 7.
What are the hydrolysis products of starch and lactose ?
Answer:
Starch upon hydrolysis gives α – D(+) glucose which is the constituent of both amylose and amylopectin. Lactose upon hydrolysis gives galactose and glucose.Upon hydrolysis, cellulose gives only D(+) glucose. This means that only D(+) glucose units are present in cellulose but unlike starch these are -D(+) glucose molecules and not a – D(+) glucose molecules. The X – ray analysis has shown that there are large linear chains of 3 – D( +) glucose molecules lying side by side in the form of bundles held together by hydrogen bonding in the neighbouring hydroxyl groups. 

Question 8.
What is the basic structural difference between starch and cellulose ?
Answer:
The basic difference in starch and cellulose lies in the nature of the glucose molecules present. Starch consists of two compounds i.e., amylose and amylopectin. In both of them, α-D(+) glucose molecules are present Amylose contains linear chains of these molecules linked in C₁-C₄ manner. In amylopectin, these linear chains are further linked in C₁-C₆ manner. In cellulose, β-D(+) glucose molecules are linked to one another in C₁-C₄ manner. Thus, starch and cellulose have different arrangement of the constituents.

Question 9.
What happens when D-glucose is treated with (i) HI (ii) Bromine water (iii) HN0₃?
Answer:
(i) Reaction with HI

(ii) Reaction with bromine water

(iii) Reaction with HNO₃

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:
(a) D (+) – glucose does not undergo certain characteristic reactions of aldehydes, e.g., glucose does not form NaHSO₃ addition product.
(b)Glucose reacts with NH₂OH to form an oxime but glucose pentaacetate does not. This implies that the aldehydic group is absent in glucose pentaacetate.
(c)D – (+) – glucose exists in two stereoisomeric forms, i.e., α -glucose and β-glucose.
(d)Both α – D – glucose and β – D – glucose undergo mutarotation in aqueous solution. Although the crystalline forms of α-  and β -D (+) – glucose are quite stable in aqueous solution but each form slowly changes into an equilibrium mixture of both.
(e)D (+) – glucose forms two isomeric methyl glucosides. Aldehydes normally react with two moles of methanol per mole of the aldehyde to form an acetal but D (+) – glucose when treated with methanol in presence of dry HCl gas, reacts with only one mole of methanol per mole of glucose to form a mixture of two methyl D – glucosides i. e., methyl – α – D – glucoside (melting point 43 8 K, specific rotation +158°) and methyl – β – D – glucoside (melting point 308 K, specific rotation – 33°).

Question 11.
What are essential and non-essential amino acids?
Answer:
Essential amino acids are those which the body fails to synthesise e.g. valine (val) and leucine (Leu). Non-essential amino acids are the acids which are synthesised by the body e.g. glycine (Gly) and alanine (Ala).

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide Linkage: Proteins are the polymers of a-amino acids which are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and -NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other This results in the elimination of a water molecule and formation of a peptide bond -CO-NH-. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.

(ii) Primary Structure: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e.; the sequence of amine acids creates a different protein.

(iii) Denaturation: Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change 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 denatu ration of protein. During denaturation 2° and 3° structures are destroyed but 1° structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 13.
What are the common types of secondary structures of proteins ?
Answer:
Secondary structure of protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :

• α-helix structure
• β-pleated sheet structure.

Secondary Structure of Proteins:
The long, flexible peptide chains of proteins are folded into the relatively rigid regular conformations called the
secondary structure. It refers to the conformation which the polvpeptide chains assume as a result of hydrogen bonding
between the > C= O and > N-H groups of different peptide bonds.
The type of secondary structure a protein will acquire, in general depends upon the size of the R-group. If the size of the
R-groups are quite large, the protein will acquire ct-helix structure. If on the other hand, the size of the R-groups are relatively
smaller, the protein will acquire a β – flat sheet structure.

(a) α-Helix structure: If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group
of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide
chain coils up into a spiral structure called right handed ct- helix structure. This type of structure is adopted by most of the
fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be
stretched. During this process, the weak hydrogen bonds causing the a – helix are broken. This tends to increase the length of
the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

(b) β—Flat sheet or β—Pleated sheet structure: If R-groups are relatively small, the peptide chains lie side by side in a zig
zag manner with alternate R-groups on the same side situated at fixed distances apart. The two such neighbouring chains are held together by intermolecular hydrogen bonds. A number of such chains can be inter-bonded and this results in the formation of a flat sheet structure These chains may contract or bend a little in order to accommodate moderate sized R-groups. As a result, the sheet bends into parallel folds to form pleated sheet structure known as β – pleated sheet structure. These sheets are then stacked one above the other like the pages of a book to form a three dimensional structure. The protein fibrion in silk fibre has a β – pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its β – sheet structure. For example, silk is non-elastic since stretching leads to pulling the peptide covalent bonds. On the other hand, it can be bent easily like a stack of pages because during this process, the sheets slide over each other.

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilized by intramolecular H-bonding between C = O of one amino acid residue and the N – H of the fourth amino acid residue in the chain. This causes the polypeptide chain to coil up into a spiral structure called right handed α- helix structure.

Question 15.
Differentiate between globular proteins and fibrous proteins. (Jharkhand Board 2014; C.B.S.E. Delhi 2015)
Answer:

Question 16.
How do you explain amphoteric behaviour of amino acids?
Answer:
Amino acids have basic (NH₂) and acidic (COOH) groups. These are, therefore, amphoteric in nature. However, they exhibit these characters in the amino acid molecule itself i.e., NH₂ group (basic) accepts a proton from COOH group (acidic) in the same molecule. Therefore, a-amino acid exists as a dipolar ion.

Question 17.
What are enzymes?
Answer:
We have learnt in the study of carbohydrates that these are body fuels i.e., they provide necessary energy to the body and
keeps it working. Actually, human body is just like a furnace in which chemical reactions take place and are responsible for the
digestion of food, absorption of appropriate molecules and production of energy. The entire process involves a series of reactions that are catalysed by biocatalysts known as enzymes. Thus, enzymes may be detined as:
biological or bio—cataiysts which catalyse the reactions in living beings.
All enzymes are basically globular proteins. Enzymes are very specific for a particular reaction as well as for a particular
substrate. These are generally named after the compounds or the class of substances with which they are linked or work.

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, die globular proteins (soluble in H₂O) are converted into fibrous proteins (insoluble in H₂O) and their biological activity is lost. For example, boiled egg which contains coagulated proteins cannot be hatched.

Question 19.
How are vitamins classified ? Name the vitamin responsible for coagulation of blood? (C.B.S.E. Outside Delhi 2015)
Answer:
Vitamins are broadly classified as water soluble and water insoluble. Apart from these, they have been classified on the basis of composition. Vitamin K is responsible for the blood clotting.

Question 20.
Why are vitamin A and vitamin C essential to us ? Mention their sources.
Answer:
vitamin A: Soluble in water but insoluble in oils and fas. Destroyed by cooking or prolonged exposure to air. it increases resistance of the body towards diseases. Maintains healthy skin and helps in the healing of cuts and abrasions. It is available in

vitamin C:  Soluble in oils and fats but insoluble in water, stable to heat. Promotes growth and improves vision. ¡t also increases resistance to disease. It is available in Citrus fruits (oranges, lemon, grape fruit, lime etc.), amia, cabbage. guava etc.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biologically important polymers which are present in all living cells.

• Nucleic acids play a vital role in the transmission of heredity characteristics.
• Nucleic acids help in the biosynthesis of proteins.

Two important biologicaL functions of nucleic acids are: (i) Replication and (ii) protein synthesis.
These are briefly discussed.
Replication: Replication may be defined as: the process by which a single DNA molecule produces two identical copies of itself.

Replication is an enzyme catalysed process.  The process of replication starts with the partial unwinding of the two strands of the DNA double helix through breaking of the hydrogen bonds between pairs of bases. Each strand then acts as the template (or pattern) for the synthesis of two new strands of DNA in the celllix environment. The specificity of base pairing ensures that each new strand is complementary to its old template strand. As a result, two identical copies of DNA from the original DNA are produced. Each of these two copies are then passed on to the two new cells resulting from cell division. In this way hereditary characters are transmitted from one cell to another.

Question 22.
What is the difference between a nucleoside and nucleotide?
Answer:
A nucleoside contains a pentose sugar and base (purine or pyrimidine) while in the nucleotide, a phosphoric acid component is also present.
Arrangement of constituents in Nucleic Acids. These are, intact, three building blocks in nucleic acid. A combination ol
base and sugar is known as nucleoside. Similarly, base, sugar and phosphates form nucleotides while nucleic acids are
polynucleotides which means that these are the polymers of nucleotides.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The trands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:

Question 25.
What are the different types of RNA found in the cell?
Answer:
Three types of RNA are present in the cell. These are:
messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).

All Chapter NCERT Solutions For Class 12 Chemistry

—————————————————————————–

All Subject NCERT Solutions For Class 12

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share ncertsolutionsfor.com to your friends.